problem_id int64 1 978 | question stringlengths 86 2.11k | source stringlengths 19 76 | solution stringlengths 94 14.7k | asymptote_code stringlengths 44 17.8k | solution_image_url stringclasses 16
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895 | Frieda the frog begins a sequence of hops on a $3 \times 3$ grid of squares, moving one square on each hop and choosing at random the direction of each hop-up, down, left, or right. She does not hop diagonally. When the direction of a hop would take Frieda off the grid, she "wraps around" and jumps to the opposite edge. For example if Frieda begins in the center square and makes two hops "up", the first hop would place her in the top row middle square, and the second hop would cause Frieda to jump to the opposite edge, landing in the bottom row middle square. Suppose Frieda starts from the center square, makes at most four hops at random, and stops hopping if she lands on a corner square. What is the probability that she reaches a corner square on one of the four hops?
$\textbf{(A)} ~\frac{9}{16}\qquad\textbf{(B)} ~\frac{5}{8}\qquad\textbf{(C)} ~\frac{3}{4}\qquad\textbf{(D)} ~\frac{25}{32}\qquad\textbf{(E)} ~\frac{13}{16}$ | 2021 AMC 12A Problem 23 | Imagine an infinite grid of $2$ by $2$ squares such that there is a $2$ by $2$ square centered at $(3x, 3y)$ for all ordered pairs of integers $(x, y).$
It is easy to see that the problem is equivalent to Frieda moving left, right, up, or down on this infinite grid starting at $(0, 0)$ (minus the teleportations). Since counting the complement set is easier, we'll count the number of $4$-step paths such that Frieda never reaches a corner point.
In other words, since the reachable corner points are $(\pm 1, \pm 1), (\pm 1, \pm 2), (\pm 2, \pm 1),$ and $(\pm 2, \pm 2),$ Frieda can only travel along the collection of points included in $S$, where $S$ is all points on $x=0$ and $y=0$ such that $|y|<4$ and $|x|<4$, respectively, plus all points on the big square with side length $6$ centered at $(0, 0).$ We then can proceed with casework:
Case $1$: Frieda never reaches $(0, \pm 3)$ nor $(\pm 3, 0).$
When Frieda only moves horizontally or vertically for her four moves, she can do so in $2^4 - 4 = 12$ ways for each case . Thus, $12 \cdot 2$ total paths for the subcase of staying in one direction. (For instance, all length $4$ combinations of $F$ and $B$ except $FFFF$, $BBBB$, $FFFB$, and $BBBF$ for the horizontal direction.)
There is another subcase where she changes directions during her path. There are four symmetric cases for this subcase depending on which of the four quadrants Frieda hugs. For the first quadrant, the possible paths are $FBUD$, $FBUU$, $UDFB$, and $UDFF.$ Thus, a total of $4 \cdot 4 = 16$ ways for this subcase.
Total for Case $1$: $24 + 16 = 40$
Case $2$: Frieda reaches $(0, \pm 3)$ or $(\pm 3, 0)$.
Once Frieda reaches one of the points listed above (by using three moves), she has four choices for her last move. Thus, a total of $4 \cdot 4 = 16$ paths for this case.
Our total number of paths never reaching coroners is thus $16+40=56,$ making for an answer of \[\frac{4^4-56}{4^4} = \boxed{\textbf{(D)} ~\frac{25}{32}}.\]
-fidgetboss_4000 | dot((0,0)); draw((-1,-1)--(-1,1)--(1,1)--(1,-1)--cycle); draw((-4,-1)--(-4,1)--(-2,1)--(-2,-1)--cycle); draw((2,-1)--(2,1)--(4,1)--(4,-1)--cycle); draw((-1,2)--(-1,4)--(1,4)--(1,2)--cycle); draw((-1,-4)--(-1,-2)--(1,-2)--(1,-4)--cycle); draw((-4,2)--(-2,2)--(-2,4)--(-4,4)--cycle); draw((4,-2)--(2,-2)--(2,-4)--(4,-4)--cycle); draw((4,2)--(2,2)--(2,4)--(4,4)--cycle); draw((-4,-2)--(-2,-2)--(-2,-4)--(-4,-4)--cycle); draw((-3,-3)--(3,-3)--(3,3)--(-3,3)--cycle); draw((-4,0)--(4,0)); draw((0,4)--(0,-4)); | [] |
895 | Frieda the frog begins a sequence of hops on a $3 \times 3$ grid of squares, moving one square on each hop and choosing at random the direction of each hop-up, down, left, or right. She does not hop diagonally. When the direction of a hop would take Frieda off the grid, she "wraps around" and jumps to the opposite edge. For example if Frieda begins in the center square and makes two hops "up", the first hop would place her in the top row middle square, and the second hop would cause Frieda to jump to the opposite edge, landing in the bottom row middle square. Suppose Frieda starts from the center square, makes at most four hops at random, and stops hopping if she lands on a corner square. What is the probability that she reaches a corner square on one of the four hops?
$\textbf{(A)} ~\frac{9}{16}\qquad\textbf{(B)} ~\frac{5}{8}\qquad\textbf{(C)} ~\frac{3}{4}\qquad\textbf{(D)} ~\frac{25}{32}\qquad\textbf{(E)} ~\frac{13}{16}$ | 2021 AMC 12A Problem 23 | Let us go through the frog’s jumps step by step to find the probability that it will land in a corner. Denote the frog’s position as $F$:
First notice that since the frog only moves up, down, left or right, it’s original hop will not matter by symmetry, because all first hops will bring the frog at the edge next to two corners. WLOG we let the frog hop up:
Notice that in its second hop, there are two ways for it to move to a corner (left and right) out of four, and thus the probability of Frieda reaching a corner on the second hop is $\frac{1}{2}$.
There are two other cases for Frieda’s second hop: it could either wrap around to another edge square or move back to the middle. We analyze each of these cases individually, noting that each has probability $\frac{1}{4}$:
In this case, there are again $2$ choices to go to a corner on the third hop, go back to the previous square, or go to the middle. Here there is a $\frac{1}{2}$ probability that Frieda will reach a corner on the third hop. If Frieda hops back with a $\frac{1}{4}$ probability, she will again have a $\frac{1}{2}$ chance of reaching a corner on her fourth hop, but after four hops she stops hopping, so there are no more chances for her to reach a corner if she hops back. If she hops to the middle, then her fourth hop cannot reach a corner and that case yields $0$ probability of reaching a corner.
To sum up this case, the total probability of her reaching a corner is $\frac{1}{4}\left(\frac{1}{2}+\frac{1}{4}\cdot \frac{1}{2}\right)$.
Finally suppose Frieda goes back to her initial position after $2$ steps. Then her third step again doesn’t matter because whatever she does, she will be next to $2$ corners, so the probability of her reaching a corner on her fourth hop in this case is$\frac{1}{4}\cdot\frac{1}{2}$.
Summing up all the cases, we get
\[\frac{1}{2}+\frac{1}{4}\left(\frac{1}{2}+\frac{1}{4}\cdot\frac{1}{2}\right)+\frac{1}{4}\cdot\frac{1}{2} = \frac{1}{2}+\frac{1}{4}\left(\frac{1}{2}+\frac{1}{8}\right)+\frac{1}{8} = \frac{5}{4}\cdot\left(\frac{1}{2}+\frac{1}{8}\right) = \frac{5}{4} \cdot \frac{5}{8} = \boxed{\textbf{(D)} ~\frac{25}{32}}.\]
~KingRavi | // Block 1
label("$F$",(0,0)); draw((-1,-1)--(-1,1)--(1,1)--(1,-1)--cycle); draw((1,-1)--(1,1)--(3,1)--(3,-1)--cycle); draw((-1,1)--(-1,3)--(1,3)--(1,1)--cycle); draw((-3,-1)--(-3,1)--(-1,1)--(-1,-1)--cycle); draw((-1,-3)--(-1,-1)--(1,-1)--(1,-3)--cycle); draw((-3,-3)--(3,-3)--(3,3)--(-3,3)--cycle);
// Block 2
label("$F$",(0,2)); draw((-1,-1)--(-1,1)--(1,1)--(1,-1)--cycle); draw((1,-1)--(1,1)--(3,1)--(3,-1)--cycle); draw((-1,1)--(-1,3)--(1,3)--(1,1)--cycle); draw((-3,-1)--(-3,1)--(-1,1)--(-1,-1)--cycle); draw((-1,-3)--(-1,-1)--(1,-1)--(1,-3)--cycle); draw((-3,-3)--(3,-3)--(3,3)--(-3,3)--cycle);
// Block 3
label("$F$",(0,-2)); draw((-1,-1)--(-1,1)--(1,1)--(1,-1)--cycle); draw((1,-1)--(1,1)--(3,1)--(3,-1)--cycle); draw((-1,1)--(-1,3)--(1,3)--(1,1)--cycle); draw((-3,-1)--(-3,1)--(-1,1)--(-1,-1)--cycle); draw((-1,-3)--(-1,-1)--(1,-1)--(1,-3)--cycle); draw((-3,-3)--(3,-3)--(3,3)--(-3,3)--cycle); | [] |
896 | Semicircle $\Gamma$ has diameter $\overline{AB}$ of length $14$. Circle $\Omega$ lies tangent to $\overline{AB}$ at a point $P$ and intersects $\Gamma$ at points $Q$ and $R$. If $QR=3\sqrt3$ and $\angle QPR=60^\circ$, then the area of $\triangle PQR$ equals $\tfrac{a\sqrt{b}}{c}$, where $a$ and $c$ are relatively prime positive integers, and $b$ is a positive integer not divisible by the square of any prime. What is $a+b+c$?
$\textbf{(A) }110 \qquad \textbf{(B) }114 \qquad \textbf{(C) }118 \qquad \textbf{(D) }122\qquad \textbf{(E) }126$ | 2021 AMC 12A Problem 24 | Let $O=\Gamma$ be the center of the semicircle and $X=\Omega$ be the center of the circle.
Applying the Extended Law of Sines to $\triangle PQR,$ we find the radius of $\odot X:$ \[XP=\frac{QR}{2\cdot\sin \angle QPR}=\frac{3\sqrt3}{2\cdot\frac{\sqrt3}{2}}=3.\]
Alternatively, by the Inscribed Angle Theorem, $\triangle QRX$ is a $30^\circ\text{-}30^\circ\text{-}120^\circ$ triangle with base $QR=3\sqrt3.$ Dividing $\triangle QRX$ into two congruent $30^\circ\text{-}60^\circ\text{-}90^\circ$ triangles, we get that the radius of $\odot X$ is $XQ=XR=3$ by the side-length ratios.
Let $M$ be the midpoint of $\overline{QR}.$ By the Perpendicular Chord Bisector Converse, we have $\overline{OM}\perp\overline{QR}$ and $\overline{XM}\perp\overline{QR}.$ Together, points $O, X,$ and $M$ must be collinear.
By the SAS Congruence, we have $\triangle QXM\cong\triangle RXM,$ both of which are $30^\circ\text{-}60^\circ\text{-}90^\circ$ triangles. By the side-length ratios, we obtain $RM=\frac{3\sqrt3}{2}, RX=3,$ and $XM=\frac{3}{2}.$ By the Pythagorean Theorem on right $\triangle ORM,$ we get $OM=\frac{13}{2}$ and $OX=OM-XM=5.$ By the Pythagorean Theorem on right $\triangle OXP,$ we get $OP=4.$
Let $C$ be the foot of the perpendicular from $P$ to $\overline{QR},$ and $D$ be the foot of the perpendicular from $X$ to $\overline{PC},$ as shown below:
Clearly, quadrilateral $XDCM$ is a rectangle. Since $\angle XPD=\angle OXP$ by alternate interior angles, we have $\triangle XPD\sim\triangle OXP$ by the AA Similarity, with the ratio of similitude $\frac{XP}{OX}=\frac 35.$ Therefore, we get $PD=\frac 95$ and $PC=PD+DC=PD+XM=\frac 95 + \frac 32 = \frac{33}{10}.$
The area of $\triangle PQR$ is \[\frac12\cdot QR\cdot PC=\frac12\cdot3\sqrt3\cdot\frac{33}{10}=\frac{99\sqrt3}{20},\] from which the answer is $99+3+20=\boxed{\textbf{(D) } 122}.$
~MRENTHUSIASM | // Block 1
/* Made by MRENTHUSIASM */
size(300);
pair O, X, A, B, P, Q, R, M, C, D;
O = (0,0);
X = (4,3);
A = (-7,0);
B = (7,0);
P = (4,0);
Q = intersectionpoints(Circle(O,7),Circle(X,3))[0];
R = intersectionpoints(Circle(O,7),Circle(X,3))[1];
M = midpoint(Q--R);
C = foot(P,Q,R);
D = foot(X,P,C);
fill(P--Q--R--cycle,yellow);
dot("$O$",O,S);
dot("$X$",X,N);
dot("$A$",A,SW);
dot("$B$",B,SE);
dot("$P$",P,S);
dot("$Q$",Q,E);
dot("$R$",R,N);
dot("$M$",M,dir(M));
dot("$C$",C,NE);
dot("$D$",D,SE);
markscalefactor=0.0375;
draw(rightanglemark(O,M,R),red);
draw(rightanglemark(P,C,M),red);
draw(rightanglemark(P,D,X),red);
draw(rightanglemark(O,P,X),red);
draw(P--Q--R--cycle);
draw(arc(O, 7, 0, 180)^^A--B^^Circle(X,3));
draw(O--M^^X--P);
draw(P--C^^X--D,dashed);
// Block 2
/* Made by MRENTHUSIASM */ size(300); pair O, X, A, B, P, Q, R, M, C, D; O = (0,0); X = (4,3); A = (-7,0); B = (7,0); P = (4,0); Q = intersectionpoints(Circle(O,7),Circle(X,3))[0]; R = intersectionpoints(Circle(O,7),Circle(X,3))[1]; M = midpoint(Q--R); C = foot(P,Q,R); D = foot(X,P,C); fill(P--Q--R--cycle,yellow); dot("$O$",O,S); dot("$X$",X,N); dot("$A$",A,SW); dot("$B$",B,SE); dot("$P$",P,S); dot("$Q$",Q,E); dot("$R$",R,N); dot("$M$",M,dir(M)); dot("$C$",C,NE); dot("$D$",D,SE); markscalefactor=0.0375; draw(rightanglemark(O,M,R),red); draw(rightanglemark(P,C,M),red); draw(rightanglemark(P,D,X),red); draw(rightanglemark(O,P,X),red); draw(P--Q--R--cycle); draw(arc(O, 7, 0, 180)^^A--B^^Circle(X,3)); draw(O--M^^X--P); draw(P--C^^X--D,dashed); | [] |
896 | Semicircle $\Gamma$ has diameter $\overline{AB}$ of length $14$. Circle $\Omega$ lies tangent to $\overline{AB}$ at a point $P$ and intersects $\Gamma$ at points $Q$ and $R$. If $QR=3\sqrt3$ and $\angle QPR=60^\circ$, then the area of $\triangle PQR$ equals $\tfrac{a\sqrt{b}}{c}$, where $a$ and $c$ are relatively prime positive integers, and $b$ is a positive integer not divisible by the square of any prime. What is $a+b+c$?
$\textbf{(A) }110 \qquad \textbf{(B) }114 \qquad \textbf{(C) }118 \qquad \textbf{(D) }122\qquad \textbf{(E) }126$ | 2021 AMC 12A Problem 24 | Suppose we label the points as shown in the diagram above, where $C$ is the center of the semicircle and $O$ is the center of the circle tangent to $\overline{AB}$. Since $\angle QPR = 60^{\circ}$, we have $\angle QOR = 2\cdot 60^{\circ}=120^{\circ}$ and $\triangle QOR$ is a $30-30-120$ triangle, which can be split into two $30-60-90$ triangles by the altitude from $O$. Since $QR=3\sqrt{3},$ we know $OQ=OR=\tfrac{3\sqrt{3}}{\sqrt{3}}=3$ by $30-60-90$ triangles. The area of this part of $\triangle PQR$ is $\frac{1}{2}bh=\tfrac{3\sqrt{3}}{2}\cdot\tfrac{3}{2}=\tfrac{9\sqrt{3}}{4}$. We would like to add this value to the sum of the areas of the other two parts of $\triangle PQR$.
To find the areas of the other two parts of $\triangle PQR$ using the $\sin$ area formula, we need the sides and included angles. Here we know the sides but what we don't know are the angles. So it seems like we will have to use an angle from another triangle and combine them with the angles we already know to find these angles easily. We know that $\angle QOR = 120^{\circ}$ and triangles $\triangle COQ$ and $\triangle COR$ are congruent as they share a side, $CQ=CR,$ and $OQ=OR$. Therefore $\angle COQ = \angle COR = 120^{\circ}$. Suppose $CO=x$. Then $3^{2}+x^{2}-6x\cos{120^{\circ}}=7^{2}$, and since $\cos{120^{\circ}}=-\tfrac{1}{2}$, this simplifies to $x^{2}+3x=7^{2}-3^{2}\rightarrow x^{2}+3x-40=0$. This factors nicely as $(x-5)(x+8)=0$, so $x=5$ as $x$ can't be $-8$. Since $CO=5, OP=3$ and $\angle OPC=90^{\circ}$, we now know that $\triangle OPC$ is a $3-4-5$ right triangle. This may be useful info for later as we might use an angle in this triangle to find the areas of the other two parts of $\triangle PQR$.
Let $\angle POC = \alpha$. Then $\sin\alpha = \tfrac{4}{5}, \cos\alpha = \tfrac{3}{5}, \angle QOP = 120+\alpha,$ and $\angle POR = 120-\alpha$. The sum of the areas of $\triangle QOP$ and $\triangle POR$ is $3\cdot 3\cdot\tfrac{1}{2}\cdot\left[\sin(120-\alpha)+\sin(120+\alpha)\right]=\tfrac{9}{2}\left[\sin(120-\alpha)+\sin(120+\alpha)\right],$ which we will add to $\tfrac{9\sqrt{3}}{4}$ to get the area of $\triangle PQR$. Observe that \[\sin(120-\alpha) = \sin 120\cos\alpha-\sin\alpha\cos 120 = \tfrac{\sqrt{3}}{2}\cdot\tfrac{3}{5}-\tfrac{4}{5}\cdot\tfrac{-1}{2}=\tfrac{3\sqrt{3}}{10}+\tfrac{4}{10}=\tfrac{3\sqrt{3}+4}{10}\]and similarly $\sin(120+\alpha)=\tfrac{3\sqrt{3}-4}{10}$. Adding these two gives $\tfrac{3\sqrt{3}}{5}$ and multiplying that by $\tfrac{9}{2}$ gets us $\tfrac{27\sqrt{3}}{10},$ which we add to $\tfrac{9\sqrt{3}}{4}$ to get $\tfrac{54\sqrt{3}+45\sqrt{3}}{20}=\tfrac{99\sqrt{3}}{20}$. The answer is $99+3+20=102+20=\boxed{\textbf{(D)} ~122}.$
~sugar_rush | // Block 1
size(150);
draw(circle((7,0),7));
pair A = (0, 0);
pair B = (14, 0);
draw(A--B);
draw(circle((11,3),3));
label("$C$", (7, 0), S);
label("$O$", (11, 3), E);
label("$P$", (11, 0), S);
pair C = (7, 0);
pair O = (11, 3);
pair P = (11, 0);
pair Q = intersectionpoints(circle(C, 7), circle(O, 3))[1];
pair R = intersectionpoints(circle(C, 7), circle(O, 3))[0];
draw(C--O);
draw(C--Q);
draw(C--R);
draw(Q--R);
draw(O--P);
draw(O--Q);
draw(O--R);
draw(P--Q);
draw(P--R);
label("$Q$", Q, N);
label("$R$", R, E);
// Block 2
size(150); draw(circle((7,0),7)); pair A = (0, 0); pair B = (14, 0); draw(A--B); draw(circle((11,3),3)); label("$C$", (7, 0), S); label("$O$", (11, 3), E); label("$P$", (11, 0), S); pair C = (7, 0); pair O = (11, 3); pair P = (11, 0); pair Q = intersectionpoints(circle(C, 7), circle(O, 3))[1]; pair R = intersectionpoints(circle(C, 7), circle(O, 3))[0]; draw(C--O); draw(C--Q); draw(C--R); draw(Q--R); draw(O--P); draw(O--Q); draw(O--R); draw(P--Q); draw(P--R); label("$Q$", Q, N); label("$R$", R, E); | [] |
896 | Semicircle $\Gamma$ has diameter $\overline{AB}$ of length $14$. Circle $\Omega$ lies tangent to $\overline{AB}$ at a point $P$ and intersects $\Gamma$ at points $Q$ and $R$. If $QR=3\sqrt3$ and $\angle QPR=60^\circ$, then the area of $\triangle PQR$ equals $\tfrac{a\sqrt{b}}{c}$, where $a$ and $c$ are relatively prime positive integers, and $b$ is a positive integer not divisible by the square of any prime. What is $a+b+c$?
$\textbf{(A) }110 \qquad \textbf{(B) }114 \qquad \textbf{(C) }118 \qquad \textbf{(D) }122\qquad \textbf{(E) }126$ | 2021 AMC 12A Problem 24 | Define points as shown above, where $N=\overleftrightarrow{PA}\cap\overleftrightarrow{QR}$. The area of $\triangle PQR$ is simply\[\dfrac{1}{2}PX\cdot QR=\dfrac{3\sqrt{3}}{2}PX;\]
it remains to compute the value of $PX$. Note that $PX$ is simply a weighted average of $BT$ and $CS;$ it is $\dfrac{CP}{BP}$ times closer to $BT$ than it is to $CS$. Observe that\[CS=\sqrt{CQ^{2}-\left(\dfrac{1}{2}QR\right)^{2}}=\sqrt{7^{2}-\left(\dfrac{3\sqrt{3}}{2}\right)^{2}}=6.5\]
since the radius of $\Gamma$ is $7$ as its diameter is $14$. Note also by the Extended Law of Sines the radius of $\Omega$ is $\dfrac{3\sqrt{3}}{2\sin 60^{\circ}}=3,$ so $OS=3\cos 60^{\circ}=1.5$. Since $C, O,$ and $S$ are collinear by symmetry we have $CO=CS-OS=5,$ so $CP=\sqrt{5^{2}-3^{2}}=4$ and $BP=7-4=3$. Therefore, $\triangle OPC$ is a $3\text{-}4\text{-}5$ right triangle; $\triangle OPC\sim\triangle NSC$ since $\angle OPC=\angle CSN=90^{\circ}$ and $\angle OCP=\angle NCS=\sin^{-1}\left(\dfrac{3}{5}\right)$. Therefore $\dfrac{CN}{CS}=\cfrac{CO}{CP}=\dfrac{5}{4}$ so $CN=\dfrac{5}{4}CS=\dfrac{65}{8}$. Since $\triangle BTN\sim\triangle CSN,$ we have $\dfrac{BT}{BN}=\dfrac{CS}{CN}=\dfrac{4}{5}$. Therefore \[BT=\dfrac{4}{5}BN=\dfrac{4}{5}\left(CN-7\right)=\dfrac{4}{5}\cdot\dfrac{9}{8}=\dfrac{36}{40}=0.9;\]
so $PX$ is $\dfrac{4}{3}$ times as close to $0.9$ as to $6.5;$ we can compute $PX=\dfrac{4}{7}BT+\dfrac{3}{7}CS=\dfrac{4}{7}\cdot0.9+\dfrac{3}{7}\cdot6.5=3.3$. The area of $\triangle PQR$ is\[\dfrac{3\sqrt{3}}{2}\cdot 3.3=\dfrac{99\sqrt{3}}{20}\]and $99+3+20=\boxed{\textbf{(D)} ~122}$.
~sugar_rush | size(300); pair C = (7, 0); draw(arc(C, 7, 0, 180)); pair A = (0, 0), B = (14, 0); draw(A--B); draw(circle((11,3),3)); label("$A$", A, SSE); label("$B$", B, SSW); label("$C$", (A+B)/2, S); label("$O$", (11, 3), E); label("$P$", (11, 0), S); pair O = (11, 3), P = (11, 0), Q = intersectionpoints(circle(C, 7), circle(O, 3))[1], R = intersectionpoints(circle(C, 7), circle(O, 3))[0], S = (Q+R)/2, N = (121/8, 0), T = (8/11)*N + (3/11)*R, X = (4/7)*T + (3/7)*S; draw(C--O, blue); draw(O--S, red); draw(C--Q); draw(C--R); draw(Q--N--B); draw(O--P); draw(O--Q); draw(O--R); draw(P--Q--R--cycle); draw(B--T); draw(P--X); label("$Q$", Q, NNE); label("$R$", R, E); label("$S$", S, ENE); label("$N$", N, SSE); label("$T$", T, ENE); label("$X$", X, NE); draw(rightanglemark(P, X, Q)); draw(rightanglemark(B, T, R)); draw(rightanglemark(C, S, Q)); | [] |
897 | In a plane, four circles with radii $1,3,5,$ and $7$ are tangent to line $\ell$ at the same point $A,$ but they may be on either side of $\ell$. Region $S$ consists of all the points that lie inside exactly one of the four circles. What is the maximum possible area of region $S$?
$\textbf{(A) }24\pi \qquad \textbf{(B) }32\pi \qquad \textbf{(C) }64\pi \qquad \textbf{(D) }65\pi \qquad \textbf{(E) }84\pi$ | 2021 AMC 10B Problem 7 | Suppose that line $\ell$ is horizontal, and each circle lies either north or south to $\ell.$ We construct the circles one by one:
Without the loss of generality, we draw the circle with radius $7$ north to $\ell.$
To maximize the area of region $S,$ we draw the circle with radius $5$ south to $\ell.$
Now, we need to subtract the circle with radius $3$ at least. The optimal situation is that the circle with radius $3$ encompasses the circle with radius $1,$ in which we do not need to subtract more. That is, the two smallest circles are on the same side of $\ell,$ but can be on either side.
The diagram below shows one possible configuration of the four circles:
Together, the answer is $\pi\cdot7^2+\pi\cdot5^2-\pi\cdot3^2=\boxed{\textbf{(D) }65\pi}.$
~samrocksnature ~MRENTHUSIASM | // Block 1
/* diagram made by samrocksnature, edited by MRENTHUSIASM */
pair A=(10,0);
pair B=(-10,0);
draw(A--B);
filldraw(circle((0,7),7),yellow);
filldraw(circle((0,-5),5),yellow);
filldraw(circle((0,-3),3),white);
filldraw(circle((0,-1),1),white);
dot((0,0));
label("$A$",(0,0),(0,1.5));
label("$\ell$",(10,0),(1.5,0));
// Block 2
/* diagram made by samrocksnature, edited by MRENTHUSIASM */ pair A=(10,0); pair B=(-10,0); draw(A--B); filldraw(circle((0,7),7),yellow); filldraw(circle((0,-5),5),yellow); filldraw(circle((0,-3),3),white); filldraw(circle((0,-1),1),white); dot((0,0)); label("$A$",(0,0),(0,1.5)); label("$\ell$",(10,0),(1.5,0)); | [] |
898 | Mr. Zhou places all the integers from $1$ to $225$ into a $15$ by $15$ grid. He places $1$ in the middle square (eighth row and eighth column) and places other numbers one by one clockwise, as shown in part in the diagram below. What is the sum of the greatest number and the least number that appear in the second row from the top?
$\textbf{(A)} ~367 \qquad\textbf{(B)} ~368 \qquad\textbf{(C)} ~369 \qquad\textbf{(D)} ~379 \qquad\textbf{(E)} ~380$ | 2021 AMC 10B Problem 8 | In the diagram below, the red arrows indicate the progression of numbers. In the second row from the top, the greatest number and the least number are $D$ and $E,$ respectively. Note that the numbers in the yellow cells are consecutive odd perfect squares, as we can prove by induction.
By observations, we proceed as follows:
\begin{alignat*}{6} A=15^2=225, \ B=13^2=169 \quad &\implies \quad &C &= \hspace{1mm}&&A-14\hspace{1mm} &= 211& \\ \quad &\implies \quad &D &= &&C-1 &= 210& \\ \quad &\implies \quad &E &= &&B-12 &= 157&. \end{alignat*}
Therefore, the answer is $D+E=\boxed{\textbf{(A)} ~367}.$
~MRENTHUSIASM | // Block 1
/* Made by MRENTHUSIASM */
size(11.5cm);
for (real i=7.5; i<=14.5; ++i)
{
fill((i+0.5,i+0.5)--(i-0.5,i+0.5)--(i-0.5,i-0.5)--(i+0.5,i-0.5)--cycle,yellow);
}
fill((2,14)--(1,14)--(1,13)--(2,13)--cycle,green);
fill((1,14)--(0,14)--(0,13)--(1,13)--cycle,green);
label("$A$",(14.5,14.5));
label("$B$",(13.5,13.5));
label("$C$",(0.5,14.5));
label("$E$",(1.5,13.5));
label("$D$",(0.5,13.5));
add(grid(15,15,linewidth(1.25)));
draw((1.5,13.5)--(14.5,13.5)--(14.5,0.5)--(0.5,0.5)--(0.5,14.5)--(14.5,14.5),red+linewidth(1.125),EndArrow);
// Block 2
/* Made by MRENTHUSIASM */ size(11.5cm); for (real i=7.5; i<=14.5; ++i) { fill((i+0.5,i+0.5)--(i-0.5,i+0.5)--(i-0.5,i-0.5)--(i+0.5,i-0.5)--cycle,yellow); } fill((2,14)--(1,14)--(1,13)--(2,13)--cycle,green); fill((1,14)--(0,14)--(0,13)--(1,13)--cycle,green); label("$A$",(14.5,14.5)); label("$B$",(13.5,13.5)); label("$C$",(0.5,14.5)); label("$E$",(1.5,13.5)); label("$D$",(0.5,13.5)); add(grid(15,15,linewidth(1.25))); draw((1.5,13.5)--(14.5,13.5)--(14.5,0.5)--(0.5,0.5)--(0.5,14.5)--(14.5,14.5),red+linewidth(1.125),EndArrow); | [] |
898 | Mr. Zhou places all the integers from $1$ to $225$ into a $15$ by $15$ grid. He places $1$ in the middle square (eighth row and eighth column) and places other numbers one by one clockwise, as shown in part in the diagram below. What is the sum of the greatest number and the least number that appear in the second row from the top?
$\textbf{(A)} ~367 \qquad\textbf{(B)} ~368 \qquad\textbf{(C)} ~369 \qquad\textbf{(D)} ~379 \qquad\textbf{(E)} ~380$ | 2021 AMC 10B Problem 8 | In the diagram below, the red arrows indicate the progression of numbers. In the second row from the top, the greatest number and the least number are $C$ and $G,$ respectively.
By observations, we proceed as follows:
\begin{alignat*}{6} A=15^2=225 \quad &\implies \quad &B &= \hspace{1mm}&&A-14\hspace{1mm} &= 211& \\ \quad &\implies \quad &C &= &&B-1 &= 210& \\ \quad &\implies \quad &D &= &&C-13 &= 197& \\ \quad &\implies \quad &E &= &&D-14 &= 183& \\ \quad &\implies \quad &F &= &&E-13 &= 170& \\ \quad &\implies \quad &G &= &&F-13 &= 157&. \end{alignat*}
Therefore, the answer is $C+G=\boxed{\textbf{(A)} ~367}.$
~MRENTHUSIASM ~Dynosol | // Block 1
/* Made by MRENTHUSIASM */
size(11.5cm);
fill((2,14)--(1,14)--(1,13)--(2,13)--cycle,green);
fill((1,14)--(0,14)--(0,13)--(1,13)--cycle,green);
label("$A$",(14.5,14.5));
label("$B$",(0.5,14.5));
label("$C$",(0.5,13.5));
label("$D$",(0.5,0.5));
label("$E$",(14.5,0.5));
label("$F$",(14.5,13.5));
label("$G$",(1.5,13.5));
add(grid(15,15,linewidth(1.25)));
draw((1.5,13.5)--(14.5,13.5)--(14.5,0.5)--(0.5,0.5)--(0.5,14.5)--(14.5,14.5),red+linewidth(1.125),EndArrow);
// Block 2
/* Made by MRENTHUSIASM */ size(11.5cm); fill((2,14)--(1,14)--(1,13)--(2,13)--cycle,green); fill((1,14)--(0,14)--(0,13)--(1,13)--cycle,green); label("$A$",(14.5,14.5)); label("$B$",(0.5,14.5)); label("$C$",(0.5,13.5)); label("$D$",(0.5,0.5)); label("$E$",(14.5,0.5)); label("$F$",(14.5,13.5)); label("$G$",(1.5,13.5)); add(grid(15,15,linewidth(1.25))); draw((1.5,13.5)--(14.5,13.5)--(14.5,0.5)--(0.5,0.5)--(0.5,14.5)--(14.5,14.5),red+linewidth(1.125),EndArrow); | [] |
898 | Mr. Zhou places all the integers from $1$ to $225$ into a $15$ by $15$ grid. He places $1$ in the middle square (eighth row and eighth column) and places other numbers one by one clockwise, as shown in part in the diagram below. What is the sum of the greatest number and the least number that appear in the second row from the top?
$\textbf{(A)} ~367 \qquad\textbf{(B)} ~368 \qquad\textbf{(C)} ~369 \qquad\textbf{(D)} ~379 \qquad\textbf{(E)} ~380$ | 2021 AMC 10B Problem 8 | From the full diagram below, the answer is $210+157=\boxed{\textbf{(A)} ~367}.$
This solution is extremely time-consuming and error-prone. Please do not try it in a real competition unless you have no other options.
~MRENTHUSIASM ~Taco12 | // Block 1
/* Made by MRENTHUSIASM */
size(11.5cm);
fill((2,14)--(1,14)--(1,13)--(2,13)--cycle,green);
fill((1,14)--(0,14)--(0,13)--(1,13)--cycle,green);
add(grid(15,15,linewidth(1.25)));
int adj = 1;
int curDown = 2;
int curLeft = 4;
int curUp = 6;
int curRight = 8;
label("$1$",(7.5,7.5));
for (int len = 3; len<=15; len+=2)
{
for (int i=1; i<=len-1; ++i)
{
label("$"+string(curDown)+"$",(7.5+adj,7.5+adj-i));
label("$"+string(curLeft)+"$",(7.5+adj-i,7.5-adj));
label("$"+string(curUp)+"$",(7.5-adj,7.5-adj+i));
label("$"+string(curRight)+"$",(7.5-adj+i,7.5+adj));
++curDown;
++curLeft;
++curUp;
++curRight;
}
++adj;
curDown = len^2 + 1;
curLeft = len^2 + len + 2;
curUp = len^2 + 2*len + 3;
curRight = len^2 + 3*len + 4;
}
// Block 2
/* Made by MRENTHUSIASM */ size(11.5cm); fill((2,14)--(1,14)--(1,13)--(2,13)--cycle,green); fill((1,14)--(0,14)--(0,13)--(1,13)--cycle,green); add(grid(15,15,linewidth(1.25))); int adj = 1; int curDown = 2; int curLeft = 4; int curUp = 6; int curRight = 8; label("$1$",(7.5,7.5)); for (int len = 3; len<=15; len+=2) { for (int i=1; i<=len-1; ++i) { label("$"+string(curDown)+"$",(7.5+adj,7.5+adj-i)); label("$"+string(curLeft)+"$",(7.5+adj-i,7.5-adj)); label("$"+string(curUp)+"$",(7.5-adj,7.5-adj+i)); label("$"+string(curRight)+"$",(7.5-adj+i,7.5+adj)); ++curDown; ++curLeft; ++curUp; ++curRight; } ++adj; curDown = len^2 + 1; curLeft = len^2 + len + 2; curUp = len^2 + 2*len + 3; curRight = len^2 + 3*len + 4; } | [] |
899 | Three equally spaced parallel lines intersect a circle, creating three chords of lengths $38,38,$ and $34$. What is the distance between two adjacent parallel lines?
$\textbf{(A) }5\frac12 \qquad \textbf{(B) }6 \qquad \textbf{(C) }6\frac12 \qquad \textbf{(D) }7 \qquad \textbf{(E) }7\frac12$ | 2021 AMC 10B Problem 14 | Since two parallel chords have the same length ($38$), they must be equidistant from the center of the circle. Let the perpendicular distance of each chord from the center of the circle be $d$. Thus, the distance from the center of the circle to the chord of length $34$ is
\[2d + d = 3d\]
and the distance between each of the chords is just $2d$. Let the radius of the circle be $r$. Drawing radii to the points where the lines intersect the circle, we create two different right triangles:
- One with base $\frac{38}{2}= 19$, height $d$, and hypotenuse $r$ ($\triangle RAO$ on the diagram)
- Another with base $\frac{34}{2} = 17$, height $3d$, and hypotenuse $r$ ($\triangle LBO$ on the diagram)
By the Pythagorean theorem, we can create the following system of equations:
\[19^2 + d^2 = r^2\]
\[17^2 + (2d + d)^2 = r^2\]
Solving, we find $d = 3$, so $2d = \boxed{\textbf{(B)}\ 6}$.
~Joeya (Solution)
~Jamess2022 (burntTacos) (Diagram)
~lpieleanu (Minor Edits) | size(8cm); pair O = (0, 0), A = (0, 3), B = (0, 9), R = (19, 3), L = (17, 9); draw(O--A--B); draw(O--R); draw(O--L); label("$A$", A, NE); label("$B$", B, N); label("$R$", R, NE); label("$L$", L, NE); label("$O$", O, S); label("$d$", O--A, W); label("$2d$", A--B, W); label("$r$", O--R, S); label("$r$", O--L, NW); dot(O); dot(A); dot(B); dot(R); dot(L); draw(circle((0, 0), sqrt(370))); draw(-R -- (R.x, -R.y)); draw((-R.x, R.y) -- R); draw((-L.x, L.y) -- L); | [] |
899 | Three equally spaced parallel lines intersect a circle, creating three chords of lengths $38,38,$ and $34$. What is the distance between two adjacent parallel lines?
$\textbf{(A) }5\frac12 \qquad \textbf{(B) }6 \qquad \textbf{(C) }6\frac12 \qquad \textbf{(D) }7 \qquad \textbf{(E) }7\frac12$ | 2021 AMC 10B Problem 14 | If $d$ is the requested distance, and $r$ is the radius of the circle, Stewart's Theorem applied to $\triangle OCD$ with cevian $\overleftrightarrow{OP}$ gives \[19\cdot 38\cdot 19 + \tfrac{1}{2}d\cdot 38\cdot\tfrac{1}{2}d=19r^{2}+19r^{2}.\] This simplifies to $13718+\tfrac{19}{2}d^{2}=38r^{2}$. Similarly, another round of Stewart's Theorem applied to $\triangle OEF$ with cevian $\overleftrightarrow{OQ}$ gives \[17\cdot 34\cdot 17 + \tfrac{3}{2}d\cdot 34\cdot\tfrac{3}{2}d=17r^{2}+17r^{2}.\] This simplifies to $9826+\tfrac{153}{2}d^{2}=34r^{2}$. Dividing the top equation by $38$ and the bottom equation by $34$ results in the system of equations
\begin{align*} 361+\tfrac{1}{4}d^{2} &= r^{2} \\ 289+\tfrac{9}{4}d^{2} &= r^{2} \\ \end{align*}
By transitive, $361+\tfrac{1}{4}d^{2}=289+\tfrac{9}{4}d^{2}$. Therefore $(\tfrac{9}{4}-\tfrac{1}{4})d^{2}=361-289\rightarrow 2d^{2}=72\rightarrow d^{2}=36\rightarrow d=\boxed{\textbf{(B)} ~6}.$
~Punxsutawney Phil | real r=sqrt(370); draw(circle((0, 0), r)); pair A = (-19, 3); pair B = (19, 3); draw(A--B); pair C = (-19, -3); pair D = (19, -3); draw(C--D); pair E = (-17, -9); pair F = (17, -9); draw(E--F); pair O = (0, 0); pair P = (0, -3); pair Q = (0, -9); draw(O--Q); draw(O--C); draw(O--D); draw(O--E); draw(O--F); label("$O$", O, N); label("$C$", C, SW); label("$D$", D, SE); label("$E$", E, SW); label("$F$", F, SE); label("$P$", P, SW); label("$Q$", Q, S); | [] |
900 | The figure is constructed from $11$ line segments, each of which has length $2$. The area of pentagon $ABCDE$ can be written as $\sqrt{m} + \sqrt{n}$, where $m$ and $n$ are positive integers. What is $m + n ?$
$\textbf{(A)} ~20 \qquad\textbf{(B)} ~21 \qquad\textbf{(C)} ~22 \qquad\textbf{(D)} ~23 \qquad\textbf{(E)} ~24$ | 2021 AMC 10B Problem 20 | Draw diagonals $AC$ and $AD$ to split the pentagon into three parts. We can compute the area for each triangle and sum them up at the end. For triangles $ABC$ and $ADE$, they each have area $2\cdot\frac{1}{2}\cdot\frac{4\sqrt{3}}{4}=\sqrt{3}$. For triangle $ACD$, we can see that $AC=AD=2\sqrt{3}$ and $CD=2$. Using Pythagorean Theorem, the altitude for this triangle is $\sqrt{11}$, so the area is $\sqrt{11}$. Adding each part up, we get $2\sqrt{3}+\sqrt{11}=\sqrt{12}+\sqrt{11} \implies \boxed{\textbf{(D)} ~23}$.
Note: Another easier way to find the areas would be to use the formula $A=\frac12ab\sin C$ - erringbubble | /* Made by samrocksnature, adapted by Tucker, then adjusted by samrocksnature again, then adjusted by erics118 xD*/ pair A=(-2.4638,4.10658); pair B=(-4,2.6567453480756127); pair C=(-3.47132,0.6335248637894945); pair D=(-1.464483379039766,0.6335248637894945); pair E=(-0.956630463955801,2.6567453480756127); pair F=(-1.85,2); pair G=(-3.1,2); draw(A--G--A--F, lightgray); draw(B--F--C, lightgray); draw(E--G--D, lightgray); dot(F^^G, lightgray); draw(A--B--C--D--E--A); draw(A--C--A--D); label("A",A,N); label("B",B,W); label("C",C,S); label("D",D,S); label("E",E,dir(0)); dot(A^^B^^C^^D^^E); | [] |
901 | Let $S$ be the set of lattice points in the coordinate plane, both of whose coordinates are integers between $1$ and $30,$ inclusive. Exactly $300$ points in $S$ lie on or below a line with equation $y=mx.$ The possible values of $m$ lie in an interval of length $\frac ab,$ where $a$ and $b$ are relatively prime positive integers. What is $a+b?$
$\textbf{(A)} ~31 \qquad \textbf{(B)} ~47 \qquad \textbf{(C)} ~62\qquad \textbf{(D)} ~72 \qquad \textbf{(E)} ~85$ | 2021 AMC 10B Problem 25 | Based on area ratios between a square of side length $30$ and a triangle with base $30$, we estimate that the slope $m$ of the line we want is approximately $\frac{2}{3}$. Following this estimate, we see if there are approximately $30 \cdot 30 - 300 = 600$ lattice points above the line $y=\frac{2}{3}x$.
Counting the number of lattice points with $x=1$ above the line, the number of lattice points with $x=2$ above the line, and so on, we find that the total number of lattice points above the line is $30+29+28+28+27+26+26 \ldots+ 10$, with the even integers repeating every third term. We see that the average of the $30$ terms is $20$, which means that exactly $20 \cdot 30 = 600$ lattice points above the line as desired. This gives a lower bound because any decrease in the slope of the line would cause points that were already on the line to shift to being above it.
To find the upper bound, notice that each lattice point less than $1$ unit above the line is either $\frac{1}{3}$ or $\frac{2}{3}$ above. Since the slope through a point is the y-coordinate divided by the x-coordinate, a shift in the slope will increase the y-value of the higher x-coordinates more than the y-value of the lower x-coordinates. So, we turn our attention to $x=28, 29, 30$ for which the line $y=\frac{2}{3}x$ intersects at $y= \frac{56}{3}, \frac{58}{3}, 20$. The point $(30,20)$ is already counted, and we can clearly see that if we slowly increase the slope of the line, we will first hit the point $(28,19)$ since $(28, \frac{56}{3})$ is the closest to it. The equation of the line which goes through both the origin and $(28,19)$ is $y=\frac{19}{28}x$. This gives an upper bound of $m=\frac{19}{28}$.
Taking the upper bound of $m$ and subtracting the lower bound yields $\frac{19}{28}-\frac{2}{3}=\frac{1}{84}$. The answer is therefore $1+84=$ $\boxed{\textbf{(E)} ~85}$.
~theAJL ~Minor edits by Eric_Zang
Diagram | /* Created by Brendanb4321 */ import graph; size(16cm); defaultpen(fontsize(9pt)); xaxis(0,30,Ticks(1.0)); yaxis(0,25,Ticks(1.0)); draw((0,0)--(30,20)); draw((0,0)--(30,30/28*19), dotted); for (int i = 1; i<=30; ++i) { for (int j = 1; j<=2/3*i+1; ++j) { dot((i,j)); } } dot((28,19), red); label("$m=2/3$", (32,20)); label("$m=19/28$", (32.3,20.8)); | [] |
901 | Let $S$ be the set of lattice points in the coordinate plane, both of whose coordinates are integers between $1$ and $30,$ inclusive. Exactly $300$ points in $S$ lie on or below a line with equation $y=mx.$ The possible values of $m$ lie in an interval of length $\frac ab,$ where $a$ and $b$ are relatively prime positive integers. What is $a+b?$
$\textbf{(A)} ~31 \qquad \textbf{(B)} ~47 \qquad \textbf{(C)} ~62\qquad \textbf{(D)} ~72 \qquad \textbf{(E)} ~85$ | 2021 AMC 10B Problem 25 | It's easier to calculate the number of lattice points inside a rectangle with vertices $(0,0)$, $(p,0)$, $(p,q)$, $(0,q)$. Those lattice points are divided by the diagonal $y = \frac{p}{q} \cdot x$ into $2$ halves. In this problem the number of lattice points on or below the diagonal and $x \ge 1$ is
$\frac{1}{2} [(p+1)(q+1) - d] + d - (p+1)$, $d$ is the number of lattice points on the diagonal, $d \ge 2$
$(p+1)(q+1)$ is the total number of lattice points inside the rectangle. Subtract the number of lattice points on the diagonal, divided by 2 is the number of lattice points below the diagonal, add the number of lattice points on the diagonal, and subtract the lattice points on the $x$ axis, then we get the total number of lattice points on or below the diagonal and $x \ge 1$.
There are $900$ lattice points in total. $300$ is $\frac{1}{3}$ of $900$. The $x$ coordinate of the top-right vertex of the rectangle is $30$, $\frac{1}{2} \cdot 30 \cdot 20 = 300$. I guess the $y$ coordinate of the top-right vertex of the rectangle is $20$. Now I am going to verify that. The slope of the diagonal is $\frac{20}{30} = \frac{2}{3}$, there are $11$ lattice points on the diagonal. Substitute $(p,q)=(30, 20)$, $d=11$ to the above formula:
$\frac{1}{2} [(30+1)(10+1) - 11] + 11 - (30+1) = 300$
Because there are $11$ lattice points on line $y = \frac{2}{3}x$, if $m < \frac{2}{3}$, then the number of lattice points on or below the line is less than $300$. So $m = \frac{2}{3}$ is the lower bound.
Now I am going to calculate the upper bound. From $\frac{b}{a} < \frac{b+1}{a+1}$,
$\frac{2}{3} = \frac{18}{27} < \frac{19}{28}$
$\frac{2}{3} = \frac{20}{30} < \frac{21}{31} < \frac{19}{28}$
If $m = \frac{21}{31}$, I will calculate by using the rectangle with blue vertex $(p,q) = (31, 21)$, then subtract lattice points on line $x = 31$, which is $21$. There are 2 lattice points on the diagonal, $d=2$.
$\frac{1}{2} [(31+1)(21+1) - 2] + 2 - (31+1) - 21 = 300$, same as that of $m = \frac{2}{3}$
If $m = \frac{19}{28}$, I will calculate by using the rectangle with red vertex $(p,q) = (28, 19)$, then add lattice points on line $x = 29$ and $x = 30$, which is $19 + 20 = 39$. There are 2 lattice points on the diagonal, $d=2$.
$\frac{1}{2} [(28+1)(19+1) - 2] + 2 - (28+1) + 39 = 301$, $1$ more than that of $m = \frac{2}{3}$
When $m$ increases, more lattice points falls below the line $y = mx$. Any value larger than $\frac{19}{28}$ has more than $301$ lattice points on or below $y = \frac{19}{28} x$. So the upper bound is $\frac{19}{28}$.
$\frac{19}{28}-\frac{2}{3}=\frac{1}{84}$, $\boxed{\textbf{(E)} ~85}$.
~isabelchen | /* Created by Brendanb4321, modified by isabelchen */ import graph; size(18cm); defaultpen(fontsize(9pt)); xaxis(0,31,Ticks(1.0)); yaxis(0,22,Ticks(1.0)); draw((0,0)--(30,20)); draw((0,0)--(30,30*19/28), dotted); draw((0,0)--(31,31*21/31), dotted); for (int i = 1; i<=31; ++i) { for (int j = 1; j<=2/3*i+1; ++j) { dot((i,j)); } } dot((28,19), red); dot((31,21), blue); label("$m=2/3$", (33,20)); label("$m=21/31$", (33,21)); label("$m=19/28$", (33,22)); | [] |
901 | Let $S$ be the set of lattice points in the coordinate plane, both of whose coordinates are integers between $1$ and $30,$ inclusive. Exactly $300$ points in $S$ lie on or below a line with equation $y=mx.$ The possible values of $m$ lie in an interval of length $\frac ab,$ where $a$ and $b$ are relatively prime positive integers. What is $a+b?$
$\textbf{(A)} ~31 \qquad \textbf{(B)} ~47 \qquad \textbf{(C)} ~62\qquad \textbf{(D)} ~72 \qquad \textbf{(E)} ~85$ | 2021 AMC 10B Problem 25 | The lower bound of $m$ is $\frac23 = \frac{20}{30}$. Inside the rectangle with vertices $(0,0)$, $(30,0)$, $(30,20)$, $(0, 20)$ and diagonal $y = \frac23 x$, there are $(30-1)(20-1) = 551$ lattice points inside, not including the edges. There are $9$ lattice points on diagonal $y = \frac23 x$ inside the rectangle, $551 + 9 = 560$. Half of the $560$ lattice points are below diagonal $y = \frac23 x$, $560 \cdot \frac12 = 280$. There are $20$ lattice points on edge $x = 30$, $280 + 20 = 300$. Once $m < \frac23$, the $9$ lattice points on diagonal $y = \frac23 x$ will be above the new diagonal, making the number of lattice points on and below the diagonal less than $300$.
Now we are going to calculate the upper bound by the following formula:
The number of lattice points inside rectangle $(0,0)$, $(a,0)$, $(a, b)$, $(0, b)$ and below diagonal $y = \frac{b}{a}x$ is $\frac{(a-1)(b-1)}{2}$, where $a$ and $b$ are relatively prime.
There are $(a-1)(b-1)$ lattice points inside the rectangle. Because $a$ and $b$ are relatively prime, the only lattice points on the diagonal are $(0,0)$ and $(a,b)$. By symmetry, half of the lattice points are below the diagonal.
$\frac{2}{3} = \frac{20}{30} < \frac{21}{31} < \frac{19}{28}$
When $a = 31$, $b = 21$, $\frac{(31-1)(21-1)}{2} = 300$.
When $a = 28$, $b = 19$, $\frac{(28-1)(19-1)}{2} = 243$. Below the line $y = \frac{19}{28} x$, there are $19$ lattice points on line $x = 28$, $19$ lattice points on line $x = 29$, $20$ lattice points on line $x = 30$, $243 + 19 + 19 + 20 = 301$.
More lattice points fall below the line $y = mx$ as $m$ increases. There are more than $301$ lattice points on and below the line for any $m$ greater than $\frac{19}{28}$. Therefore, the upper bound is $\frac{19}{28}$.
$\frac{19}{28}-\frac{2}{3}=\frac{1}{84}$, so $1+84=\boxed{\textbf{(E)} ~85}$.
~isabelchen
Remark
$\lfloor \frac{b}{a} k \rfloor$ is the number of lattice points on line $x = k$, below line $y = \frac{b}{a} x$ and above the $x$ axis, where $k$ is an integer and $0<k<a$. Therefore, $\sum_{k=1}^{a-1} \lfloor \frac{b}{a} k \rfloor$ is the number of lattice points inside the rectangle $(0,0)$, $(a,0)$, $(a, b)$, $(0, b)$, below diagonal $y = \frac{b}{a} x$. If $a$ and $b$ are relatively prime, $\sum_{k=1}^{a-1} \lfloor \frac{b}{a} k \rfloor = \frac{(a-1)(b-1)}{2}$, as explained in solution 5. This problem is about finding the upper and lower bound of $\frac{b}{a}$, given $\sum_{k=1}^{30} \lfloor \frac{b}{a} k \rfloor = 300$. The same problem can have geometric representation as stated in the original problem, or algebraic representation as stated here.
~isabelchen | /* Created by isabelchen */ import graph; size(8cm); defaultpen(fontsize(9pt)); xaxis(0,8); yaxis(0,6); draw((0,0)--(7,5)); draw((7,0)--(7,5)); draw((0,5)--(7,5)); dot((0,0)); dot((1,0)); dot((2,0)); dot((3,0)); dot((4,0)); dot((5,0)); dot((6,0)); dot((7,0)); dot((8,0)); dot((0,1)); dot((1,1)); dot((2,1)); dot((3,1)); dot((4,1)); dot((5,1)); dot((6,1)); dot((7,1)); dot((8,1)); dot((0,2)); dot((1,2)); dot((2,2)); dot((3,2)); dot((4,2)); dot((5,2)); dot((6,2)); dot((7,2)); dot((8,2)); dot((0,3)); dot((1,3)); dot((2,3)); dot((3,3)); dot((4,3)); dot((5,3)); dot((6,3)); dot((7,3)); dot((8,3)); dot((0,4)); dot((1,4)); dot((2,4)); dot((3,4)); dot((4,4)); dot((5,4)); dot((6,4)); dot((7,4)); dot((8,4)); dot((0,5)); dot((1,5)); dot((2,5)); dot((3,5)); dot((4,5)); dot((5,5)); dot((6,5)); dot((7,5)); dot((8,5)); dot((0,6)); dot((1,6)); dot((2,6)); dot((3,6)); dot((4,6)); dot((5,6)); dot((6,6)); dot((7,6)); dot((8,6)); label("$(0,0)$", (0,0), SW); label("$(a, b)$", (7,5), NE); dot((7,0)); label("$a$", (7,0), S); dot((0,5)); label("$b$", (0,5), W); | [] |
902 | Three equally spaced parallel lines intersect a circle, creating three chords of lengths $38,38,$ and $34$. What is the distance between two adjacent parallel lines?
$\textbf{(A) }5\frac12 \qquad \textbf{(B) }6 \qquad \textbf{(C) }6\frac12 \qquad \textbf{(D) }7 \qquad \textbf{(E) }7\frac12$ | 2021 AMC 12B Problem 8 | Since two parallel chords have the same length ($38$), they must be equidistant from the center of the circle. Let the perpendicular distance of each chord from the center of the circle be $d$. Thus, the distance from the center of the circle to the chord of length $34$ is
\[2d + d = 3d\]
and the distance between each of the chords is just $2d$. Let the radius of the circle be $r$. Drawing radii to the points where the lines intersect the circle, we create two different right triangles:
- One with base $\frac{38}{2}= 19$, height $d$, and hypotenuse $r$ ($\triangle RAO$ on the diagram)
- Another with base $\frac{34}{2} = 17$, height $3d$, and hypotenuse $r$ ($\triangle LBO$ on the diagram)
By the Pythagorean theorem, we can create the following system of equations:
\[19^2 + d^2 = r^2\]
\[17^2 + (2d + d)^2 = r^2\]
Solving, we find $d = 3$, so $2d = \boxed{\textbf{(B)}\ 6}$.
~Joeya (Solution)
~Jamess2022 (burntTacos) (Diagram)
~lpieleanu (Minor Edits) | // Block 1
size(8cm);
pair O = (0, 0), A = (0, 3), B = (0, 9), R = (19, 3), L = (17, 9);
draw(O--A--B);
draw(O--R);
draw(O--L);
label("$A$", A, NE);
label("$B$", B, N);
label("$R$", R, NE);
label("$L$", L, NE);
label("$O$", O, S);
label("$d$", O--A, W);
label("$2d$", A--B, W);
label("$r$", O--R, S);
label("$r$", O--L, NW);
dot(O);
dot(A);
dot(B);
dot(R);
dot(L);
draw(circle((0, 0), sqrt(370)));
draw(-R -- (R.x, -R.y));
draw((-R.x, R.y) -- R);
draw((-L.x, L.y) -- L);
// Block 2
size(8cm); pair O = (0, 0), A = (0, 3), B = (0, 9), R = (19, 3), L = (17, 9); draw(O--A--B); draw(O--R); draw(O--L); label("$A$", A, NE); label("$B$", B, N); label("$R$", R, NE); label("$L$", L, NE); label("$O$", O, S); label("$d$", O--A, W); label("$2d$", A--B, W); label("$r$", O--R, S); label("$r$", O--L, NW); dot(O); dot(A); dot(B); dot(R); dot(L); draw(circle((0, 0), sqrt(370))); draw(-R -- (R.x, -R.y)); draw((-R.x, R.y) -- R); draw((-L.x, L.y) -- L); | [] |
902 | Three equally spaced parallel lines intersect a circle, creating three chords of lengths $38,38,$ and $34$. What is the distance between two adjacent parallel lines?
$\textbf{(A) }5\frac12 \qquad \textbf{(B) }6 \qquad \textbf{(C) }6\frac12 \qquad \textbf{(D) }7 \qquad \textbf{(E) }7\frac12$ | 2021 AMC 12B Problem 8 | If $d$ is the requested distance, and $r$ is the radius of the circle, Stewart's Theorem applied to $\triangle OCD$ with cevian $\overleftrightarrow{OP}$ gives \[19\cdot 38\cdot 19 + \tfrac{1}{2}d\cdot 38\cdot\tfrac{1}{2}d=19r^{2}+19r^{2}.\] This simplifies to $13718+\tfrac{19}{2}d^{2}=38r^{2}$. Similarly, another round of Stewart's Theorem applied to $\triangle OEF$ with cevian $\overleftrightarrow{OQ}$ gives \[17\cdot 34\cdot 17 + \tfrac{3}{2}d\cdot 34\cdot\tfrac{3}{2}d=17r^{2}+17r^{2}.\] This simplifies to $9826+\tfrac{153}{2}d^{2}=34r^{2}$. Dividing the top equation by $38$ and the bottom equation by $34$ results in the system of equations
\begin{align*} 361+\tfrac{1}{4}d^{2} &= r^{2} \\ 289+\tfrac{9}{4}d^{2} &= r^{2} \\ \end{align*}
By transitive, $361+\tfrac{1}{4}d^{2}=289+\tfrac{9}{4}d^{2}$. Therefore $(\tfrac{9}{4}-\tfrac{1}{4})d^{2}=361-289\rightarrow 2d^{2}=72\rightarrow d^{2}=36\rightarrow d=\boxed{\textbf{(B)} ~6}.$
~Punxsutawney Phil | // Block 1
real r=sqrt(370);
draw(circle((0, 0), r));
pair A = (-19, 3);
pair B = (19, 3);
draw(A--B);
pair C = (-19, -3);
pair D = (19, -3);
draw(C--D);
pair E = (-17, -9);
pair F = (17, -9);
draw(E--F);
pair O = (0, 0);
pair P = (0, -3);
pair Q = (0, -9);
draw(O--Q);
draw(O--C);
draw(O--D);
draw(O--E);
draw(O--F);
label("$O$", O, N);
label("$C$", C, SW);
label("$D$", D, SE);
label("$E$", E, SW);
label("$F$", F, SE);
label("$P$", P, SW);
label("$Q$", Q, S);
// Block 2
real r=sqrt(370); draw(circle((0, 0), r)); pair A = (-19, 3); pair B = (19, 3); draw(A--B); pair C = (-19, -3); pair D = (19, -3); draw(C--D); pair E = (-17, -9); pair F = (17, -9); draw(E--F); pair O = (0, 0); pair P = (0, -3); pair Q = (0, -9); draw(O--Q); draw(O--C); draw(O--D); draw(O--E); draw(O--F); label("$O$", O, N); label("$C$", C, SW); label("$D$", D, SE); label("$E$", E, SW); label("$F$", F, SE); label("$P$", P, SW); label("$Q$", Q, S); | [] |
903 | How many values of $\theta$ in the interval $0<\theta\le 2\pi$ satisfy \[1-3\sin\theta+5\cos3\theta = 0?\]
$\textbf{(A) }2 \qquad \textbf{(B) }4 \qquad \textbf{(C) }5\qquad \textbf{(D) }6 \qquad \textbf{(E) }8$ | 2021 AMC 12B Problem 13 | We rearrange to get \[5\cos3\theta = 3\sin\theta-1.\]
We can graph two functions in this case: $y=5\cos{3x}$ and $y=3\sin{x} -1$.
Using transformation of functions, we know that $5\cos{3x}$ is just a cosine function with amplitude $5$ and period $\frac{2\pi}{3}$. Similarly, $3\sin{x} -1$ is just a sine function with amplitude $3$ and shifted $1$ unit downward:
So, we have $\boxed{\textbf{(D) }6}$ solutions.
~Jamess2022 (burntTacos) | // Block 1
import graph;
size(400,200,IgnoreAspect);
real Sin(real t) {return 3*sin(t) - 1;}
real Cos(real t) {return 5*cos(3*t);}
draw(graph(Sin,0, 2pi),red,"$3\sin{x} -1 $");
draw(graph(Cos,0, 2pi),blue,"$5\cos{3x}$");
xaxis("$x$",BottomTop,LeftTicks);
yaxis("$y$",LeftRight,RightTicks(trailingzero));
add(legend(),point(E),20E,UnFill);
// Block 2
import graph; size(400,200,IgnoreAspect); real Sin(real t) {return 3*sin(t) - 1;} real Cos(real t) {return 5*cos(3*t);} draw(graph(Sin,0, 2pi),red,"$3\sin{x} -1 $"); draw(graph(Cos,0, 2pi),blue,"$5\cos{3x}$"); xaxis("$x$",BottomTop,LeftTicks); yaxis("$y$",LeftRight,RightTicks(trailingzero)); add(legend(),point(E),20E,UnFill); | [] |
904 | Let $ABCD$ be a rectangle and let $\overline{DM}$ be a segment perpendicular to the plane of $ABCD$. Suppose that $\overline{DM}$ has integer length, and the lengths of $\overline{MA},\overline{MC},$ and $\overline{MB}$ are consecutive odd positive integers (in this order). What is the volume of pyramid $MABCD?$
$\textbf{(A) }24\sqrt5 \qquad \textbf{(B) }60 \qquad \textbf{(C) }28\sqrt5\qquad \textbf{(D) }66 \qquad \textbf{(E) }8\sqrt{70}$ | 2021 AMC 12B Problem 14 | Let $MA=a$ and $MD=d.$ It follows that $MC=a+2$ and $MB=a+4.$
As shown below, note that $\triangle MAD$ and $\triangle MBC$ are both right triangles.
By the Pythagorean Theorem, we have
\begin{alignat*}{6} AD^2 &= MA^2 - MD^2 &&= a^2 - d^2, \\ BC^2 &= MB^2 - MC^2 &&= (a+4)^2 - (a+2)^2. \end{alignat*}
Since $AD=BC$ in rectangle $ABCD,$ we equate the expressions for $AD^2$ and $BC^2,$ then rearrange and factor:
\begin{align*} a^2 - d^2 &= (a+4)^2 - (a+2)^2 \\ a^2 - d^2 &= 4a + 12 \\ a^2 - 4a - d^2 &= 12 \\ (a-2)^2 - d^2 &= 16 \\ (a+d-2)(a-d-2) &= 16. \end{align*}
As $a+d-2$ and $a-d-2$ have the same parity, we get $a+d-2=8$ and $a-d-2=2,$ from which $(a,d)=(7,3).$
Applying the Pythagorean Theorem to right $\triangle MAD$ and right $\triangle MCD,$ we obtain $AD=2\sqrt{10}$ and $CD=6\sqrt2,$ respectively.
Let the brackets denote areas. Together, the volume of pyramid $MABCD$ is \[\frac13\cdot [ABCD]\cdot MD = \frac13\cdot (AD\cdot CD)\cdot MD = \boxed{\textbf{(A) }24\sqrt5}.\]
~Lopkiloinm ~MRENTHUSIASM | // Block 1
size(300);
import graph3;
import solids;
currentprojection=orthographic((0.5,-0.25,-0.5));
triple A, B, C, D, M;
A = (-2sqrt(10),0,0);
B = (-2sqrt(10),-6sqrt(2),0);
C = (0,-6sqrt(2),0);
D = (0,0,0);
M = (0,0,3);
draw(surface(M--A--D--cycle),yellow);
draw(surface(M--B--C--cycle),yellow);
draw(D--A--B--C^^A--M^^B--M^^C--M^^D--M);
draw(C--D,dashed);
dot(A^^B^^C^^D^^M,linewidth(4.5));
label("$A$",A,2*dir(A-B));
label("$B$",B,2*dir(B-A));
label("$C$",C,2*dir(C-D));
label("$D$",D,2*dir(D-C));
label("$M$",M,2*dir((1,1,0)));
label("$a$",midpoint(M--A),2*dir((-1,-1,0)),red);
label("$d$",midpoint(M--D),2*dir((1,1,0)),red);
label("$a+2$",midpoint(M--C),2*dir((1,-1,0)),red);
label("$a+4$",midpoint(M--B),2*dir((-1,1,0)),red);
// Block 2
size(300); import graph3; import solids; currentprojection=orthographic((0.5,-0.25,-0.5)); triple A, B, C, D, M; A = (-2sqrt(10),0,0); B = (-2sqrt(10),-6sqrt(2),0); C = (0,-6sqrt(2),0); D = (0,0,0); M = (0,0,3); draw(surface(M--A--D--cycle),yellow); draw(surface(M--B--C--cycle),yellow); draw(D--A--B--C^^A--M^^B--M^^C--M^^D--M); draw(C--D,dashed); dot(A^^B^^C^^D^^M,linewidth(4.5)); label("$A$",A,2*dir(A-B)); label("$B$",B,2*dir(B-A)); label("$C$",C,2*dir(C-D)); label("$D$",D,2*dir(D-C)); label("$M$",M,2*dir((1,1,0))); label("$a$",midpoint(M--A),2*dir((-1,-1,0)),red); label("$d$",midpoint(M--D),2*dir((1,1,0)),red); label("$a+2$",midpoint(M--C),2*dir((1,-1,0)),red); label("$a+4$",midpoint(M--B),2*dir((-1,1,0)),red); | [] |
904 | Let $ABCD$ be a rectangle and let $\overline{DM}$ be a segment perpendicular to the plane of $ABCD$. Suppose that $\overline{DM}$ has integer length, and the lengths of $\overline{MA},\overline{MC},$ and $\overline{MB}$ are consecutive odd positive integers (in this order). What is the volume of pyramid $MABCD?$
$\textbf{(A) }24\sqrt5 \qquad \textbf{(B) }60 \qquad \textbf{(C) }28\sqrt5\qquad \textbf{(D) }66 \qquad \textbf{(E) }8\sqrt{70}$ | 2021 AMC 12B Problem 14 | First and foremost, the diagram must be drawn.
With the diagram, an impulse to utilize Pythagorean theorem is created. Thereby, the shaded right triangles may be used.
The following equations could be obtained from Pythagorean Theorem.
\begin{align*} I^2+a^2&=x^2 \\ I^2+b^2&=(x+2)^2 \\ I^2+a^2+b^2&=(x+4)^2 \\ \end{align*}
Because $x$ and $I$ are integers, writing an equation in terms of only $x$ and $I$ may be beneficial for trial and error.
\begin{align*} x^2+(x+2)^2&=(x+4)^2+I^2 \\ 2x^2+4x+4&=x^2+8x+16+I^2 \\ x^2-4x-12&=I^2 \\ (x-6)(x+2)&=I^2 \end{align*}
$x=7$ seems valid since $I=3$ when $x=7$. Therefore, when $x=7$, $a=2\sqrt{10}$ and $b=6\sqrt{2}$. We could double check that $40+72+9=11^2$. Thereby,
\[\frac{a\cdot b\cdot I}{3}=\frac{2\sqrt{10}\cdot6\sqrt{2}\cdot3}{3}=\boxed{\textbf{(A) }24\sqrt5}.\]
~MaPhyCom | // Block 1
import graph;
size(250);
pair A = (0,0);
pair B = (4.5,0);
pair C = (6,1.1);
pair D = (1.5,1.1);
pair M = (1.5,3.45);
draw(A--B);
draw(C--B);
draw(A--D, dashed);
draw(C--D, dashed);
draw(M--A);
draw(M--B);
draw(M--C);
draw(M--D, dashed);
draw((1.5,1.3)--(1.7,1.3)--(1.7,1.1));
dot(A,linewidth(4.5));
dot(B,linewidth(4.5));
dot(C,linewidth(4.5));
dot(D,linewidth(4.5));
dot(M,linewidth(4.5));
label("$A$", A, W);
label("$B$", B, SE);
label("$C$", C, E);
label("$D$", D, NW);
label("$M$", M, N);
label("$x$", midpoint(M--A), NW, red);
label("$x+4$", midpoint(M--B), SE, red);
label("$x+2$", midpoint(M--C), NE, red);
label("$a$", midpoint(A--D), W, red);
label("$b$", midpoint(A--B), S, red);
label("$I$", midpoint(M--D), E, red);
// Block 2
import graph;
size(250);
pair A = (0,0);
pair B = (4.5,0);
pair C = (6,1.1);
pair D = (1.5,1.1);
pair M = (1.5,3.45);
draw(A--B);
draw(C--B);
draw(A--D, dashed);
draw(C--D, dashed);
draw(D--B, dashed);
draw(M--A);
draw(M--B);
draw(M--C);
draw(M--D, dashed);
draw(M--A--D--cycle, blue);
draw(M--B--D--cycle, green);
draw(M--C--D--cycle, red);
draw((1.5,1.3)--(1.7,1.3)--(1.7,1.1));
dot(A,linewidth(4.5));
dot(B,linewidth(4.5));
dot(C,linewidth(4.5));
dot(D,linewidth(4.5));
dot(M,linewidth(4.5));
label("$A$", A, W);
label("$B$", B, SE);
label("$C$", C, E);
label("$D$", D, NW);
label("$M$", M, N);
label("$x$", midpoint(M--A), NW, red);
label("$x+4$", midpoint(M--B), SE, red);
label("$x+2$", midpoint(M--C), NE, red);
label("$a$", midpoint(A--D), W, red);
label("$b$", midpoint(A--B), S, red);
label("$I$", midpoint(M--D), E, red);
// Block 3
import graph; size(250); pair A = (0,0); pair B = (4.5,0); pair C = (6,1.1); pair D = (1.5,1.1); pair M = (1.5,3.45); draw(A--B); draw(C--B); draw(A--D, dashed); draw(C--D, dashed); draw(M--A); draw(M--B); draw(M--C); draw(M--D, dashed); draw((1.5,1.3)--(1.7,1.3)--(1.7,1.1)); dot(A,linewidth(4.5)); dot(B,linewidth(4.5)); dot(C,linewidth(4.5)); dot(D,linewidth(4.5)); dot(M,linewidth(4.5)); label("$A$", A, W); label("$B$", B, SE); label("$C$", C, E); label("$D$", D, NW); label("$M$", M, N); label("$x$", midpoint(M--A), NW, red); label("$x+4$", midpoint(M--B), SE, red); label("$x+2$", midpoint(M--C), NE, red); label("$a$", midpoint(A--D), W, red); label("$b$", midpoint(A--B), S, red); label("$I$", midpoint(M--D), E, red);
// Block 4
import graph; size(250); pair A = (0,0); pair B = (4.5,0); pair C = (6,1.1); pair D = (1.5,1.1); pair M = (1.5,3.45); draw(A--B); draw(C--B); draw(A--D, dashed); draw(C--D, dashed); draw(D--B, dashed); draw(M--A); draw(M--B); draw(M--C); draw(M--D, dashed); draw(M--A--D--cycle, blue); draw(M--B--D--cycle, green); draw(M--C--D--cycle, red); draw((1.5,1.3)--(1.7,1.3)--(1.7,1.1)); dot(A,linewidth(4.5)); dot(B,linewidth(4.5)); dot(C,linewidth(4.5)); dot(D,linewidth(4.5)); dot(M,linewidth(4.5)); label("$A$", A, W); label("$B$", B, SE); label("$C$", C, E); label("$D$", D, NW); label("$M$", M, N); label("$x$", midpoint(M--A), NW, red); label("$x+4$", midpoint(M--B), SE, red); label("$x+2$", midpoint(M--C), NE, red); label("$a$", midpoint(A--D), W, red); label("$b$", midpoint(A--B), S, red); label("$I$", midpoint(M--D), E, red); | [] |
905 | The figure is constructed from $11$ line segments, each of which has length $2$. The area of pentagon $ABCDE$ can be written as $\sqrt{m} + \sqrt{n}$, where $m$ and $n$ are positive integers. What is $m + n ?$
$\textbf{(A)} ~20 \qquad\textbf{(B)} ~21 \qquad\textbf{(C)} ~22 \qquad\textbf{(D)} ~23 \qquad\textbf{(E)} ~24$ | 2021 AMC 12B Problem 15 | Draw diagonals $AC$ and $AD$ to split the pentagon into three parts. We can compute the area for each triangle and sum them up at the end. For triangles $ABC$ and $ADE$, they each have area $2\cdot\frac{1}{2}\cdot\frac{4\sqrt{3}}{4}=\sqrt{3}$. For triangle $ACD$, we can see that $AC=AD=2\sqrt{3}$ and $CD=2$. Using Pythagorean Theorem, the altitude for this triangle is $\sqrt{11}$, so the area is $\sqrt{11}$. Adding each part up, we get $2\sqrt{3}+\sqrt{11}=\sqrt{12}+\sqrt{11} \implies \boxed{\textbf{(D)} ~23}$.
Note: Another easier way to find the areas would be to use the formula $A=\frac12ab\sin C$ - erringbubble | // Block 1
/* Made by samrocksnature, adapted by Tucker, then adjusted by samrocksnature again, then adjusted by erics118 xD*/
pair A=(-2.4638,4.10658);
pair B=(-4,2.6567453480756127);
pair C=(-3.47132,0.6335248637894945);
pair D=(-1.464483379039766,0.6335248637894945);
pair E=(-0.956630463955801,2.6567453480756127);
pair F=(-1.85,2);
pair G=(-3.1,2);
draw(A--G--A--F, lightgray);
draw(B--F--C, lightgray);
draw(E--G--D, lightgray);
dot(F^^G, lightgray);
draw(A--B--C--D--E--A);
draw(A--C--A--D);
label("A",A,N);
label("B",B,W);
label("C",C,S);
label("D",D,S);
label("E",E,dir(0));
dot(A^^B^^C^^D^^E);
// Block 2
/* Made by samrocksnature, adapted by Tucker, then adjusted by samrocksnature again, then adjusted by erics118 xD*/ pair A=(-2.4638,4.10658); pair B=(-4,2.6567453480756127); pair C=(-3.47132,0.6335248637894945); pair D=(-1.464483379039766,0.6335248637894945); pair E=(-0.956630463955801,2.6567453480756127); pair F=(-1.85,2); pair G=(-3.1,2); draw(A--G--A--F, lightgray); draw(B--F--C, lightgray); draw(E--G--D, lightgray); dot(F^^G, lightgray); draw(A--B--C--D--E--A); draw(A--C--A--D); label("A",A,N); label("B",B,W); label("C",C,S); label("D",D,S); label("E",E,dir(0)); dot(A^^B^^C^^D^^E); | [] |
906 | Let $S$ be the sum of all positive real numbers $x$ for which\[x^{2^{\sqrt2}}=\sqrt2^{2^x}.\]Which of the following statements is true?
$\textbf{(A) }S<\sqrt2 \qquad \textbf{(B) }S=\sqrt2 \qquad \textbf{(C) }\sqrt2<S<2\qquad \textbf{(D) }2\le S<6 \qquad \textbf{(E) }S\ge 6$ | 2021 AMC 12B Problem 21 | We rewrite the right side without using square roots, then take the base-$2$ logarithm for both sides:
\begin{align*} x^{2^{\sqrt2}}&=\left(2^\frac12\right)^{2^x} \\ x^{2^{\sqrt2}}&=2^{\frac12\cdot2^x} \\ x^{2^{\sqrt2}}&=2^{2^{x-1}} \\ \log_2{\left(x^{2^{\sqrt2}}\right)}&=\log_2{\left(2^{2^{x-1}}\right)} \\ 2^{\sqrt2}\log_2{x}&=2^{x-1}. \hspace{20mm} (*) \end{align*}
By observations, $x=\sqrt2$ is one solution. Graphing $f(x)=2^{\sqrt2}\log_2{x}$ and $g(x)=2^{x-1},$ we conclude that $(*)$ has two solutions, with the smaller solution $x=\sqrt2.$ We construct the following table of values:
\[\begin{array}{c|c|c|c} & & & \\ [-2ex] \boldsymbol{x} & \boldsymbol{f(x)=2^{\sqrt2}\log_2{x}} & \boldsymbol{g(x)=2^{x-1}} & \textbf{Comparison} \\ [1ex] \hline & & & \\ [-1ex] 1 & 0 & 1 & \\ [1ex] \sqrt2 & 2^{\sqrt2-1} & 2^{\sqrt2-1} & f\left(\sqrt2\right)=g\left(\sqrt2\right) \\ [1ex] 2 & 2^{\sqrt2} & 2 & f(2)>g(2) \\ [1ex] 3 & 2^{\sqrt2}\log_2{3} & 2^2 & \\ [1ex] 4 & 2^{\sqrt2+1} & 2^3 & f(4)<g(4) \\ [1ex] \end{array}\]
Let $x=t$ be the larger solution. Since exponential functions outgrow logarithmic functions, we have $f(x)<g(x)$ for all $x>t.$ By the Intermediate Value Theorem, we get $t\in(2,4),$ from which \[S=\sqrt2+t\in\left(\sqrt2+2,\sqrt2+4\right).\]
Finally, approximating with $\sqrt2\approx1.414$ results in $\boxed{\textbf{(D) }2\le S<6}.$
The graphs of $y=f(x)$ and $y=g(x)$ are shown below:
~MRENTHUSIASM | // Block 1
/* Made by MRENTHUSIASM */
size(1200,200);
int xMin = 0;
int xMax = 5;
int yMin = 0;
int yMax = 5;
//Draws the horizontal gridlines
void horizontalLines()
{
for (int i = yMin+1; i < yMax; ++i)
{
draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4));
}
}
//Draws the vertical gridlines
void verticalLines()
{
for (int i = xMin+1; i < xMax; ++i)
{
draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4));
}
}
//Draws the horizontal ticks
void horizontalTicks()
{
for (int i = yMin+1; i < yMax; ++i)
{
draw((-1/8,i)--(1/8,i), black+linewidth(1));
}
}
//Draws the vertical ticks
void verticalTicks()
{
for (int i = xMin+1; i < xMax; ++i)
{
draw((i,-1/8)--(i,1/8), black+linewidth(1));
}
}
horizontalLines();
verticalLines();
horizontalTicks();
verticalTicks();
draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5));
draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5));
label("$x$",(xMax,0),(2,0));
label("$y$",(0,yMax),(0,2));
real f(real x) {return 2^sqrt(2)*log(x)/log(2);};
real g(real x) {return 2^(x-1);};
draw(graph(f,1,3.65),red,"$y=2^{\sqrt2}\log_2{x}$");
draw(graph(g,0,3.32),blue,"$y=2^{x-1}$");
pair A, B;
A = intersectionpoint(graph(f,1,2),graph(g,1,2));
B = intersectionpoint(graph(f,2,4),graph(g,2,4));
dot(A,linewidth(4.5));
dot(B,linewidth(4.5));
label("$0$",(0,0),2.5*SW);
label("$\sqrt2$",(A.x,0),2.25*S);
label("$t$",(B.x,0),3*S);
label("$4$",(4,0),3*S);
label("$4$",(0,4),3*W);
draw(A--(A.x,0),dashed);
draw(B--(B.x,0),dashed);
add(legend(),point(E),40E,UnFill);
// Block 2
/* Made by MRENTHUSIASM */ size(1200,200); int xMin = 0; int xMax = 5; int yMin = 0; int yMax = 5; //Draws the horizontal gridlines void horizontalLines() { for (int i = yMin+1; i < yMax; ++i) { draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4)); } } //Draws the vertical gridlines void verticalLines() { for (int i = xMin+1; i < xMax; ++i) { draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4)); } } //Draws the horizontal ticks void horizontalTicks() { for (int i = yMin+1; i < yMax; ++i) { draw((-1/8,i)--(1/8,i), black+linewidth(1)); } } //Draws the vertical ticks void verticalTicks() { for (int i = xMin+1; i < xMax; ++i) { draw((i,-1/8)--(i,1/8), black+linewidth(1)); } } horizontalLines(); verticalLines(); horizontalTicks(); verticalTicks(); draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("$x$",(xMax,0),(2,0)); label("$y$",(0,yMax),(0,2)); real f(real x) {return 2^sqrt(2)*log(x)/log(2);}; real g(real x) {return 2^(x-1);}; draw(graph(f,1,3.65),red,"$y=2^{\sqrt2}\log_2{x}$"); draw(graph(g,0,3.32),blue,"$y=2^{x-1}$"); pair A, B; A = intersectionpoint(graph(f,1,2),graph(g,1,2)); B = intersectionpoint(graph(f,2,4),graph(g,2,4)); dot(A,linewidth(4.5)); dot(B,linewidth(4.5)); label("$0$",(0,0),2.5*SW); label("$\sqrt2$",(A.x,0),2.25*S); label("$t$",(B.x,0),3*S); label("$4$",(4,0),3*S); label("$4$",(0,4),3*W); draw(A--(A.x,0),dashed); draw(B--(B.x,0),dashed); add(legend(),point(E),40E,UnFill); | [] |
907 | Let $ABCD$ be a parallelogram with area $15$. Points $P$ and $Q$ are the projections of $A$ and $C,$ respectively, onto the line $BD;$ and points $R$ and $S$ are the projections of $B$ and $D,$ respectively, onto the line $AC.$ See the figure, which also shows the relative locations of these points.
Suppose $PQ=6$ and $RS=8,$ and let $d$ denote the length of $\overline{BD},$ the longer diagonal of $ABCD.$ Then $d^2$ can be written in the form $m+n\sqrt p,$ where $m,n,$ and $p$ are positive integers and $p$ is not divisible by the square of any prime. What is $m+n+p?$
$\textbf{(A) }81 \qquad \textbf{(B) }89 \qquad \textbf{(C) }97\qquad \textbf{(D) }105 \qquad \textbf{(E) }113$ | 2021 AMC 12B Problem 24 | First and foremost, the condition given in the problem could be represented. Moreover, let $QB=x$ and $CQ=h$ for convenience.
Using the property of a parallelogram, it could be inferred that $\frac{h(3+x)}{2}=\frac{15}{4}$. Moreover, Pythagorean Theorem manifests that $\left(\frac{4}{3}h\right)^2+4^2=(3+x)^2$.
\begin{align*} 3+x&=\frac{15}{2h} \\ \left(\frac{4}{3}h\right)^2+4^2&=\left(\frac{15}{2h}\right)^2 \\ \frac{16h^2}{9}+16&=\frac{225}{4h^2} \\ 64h^4+576h^2-2025&=0 \\ h^2&\Rightarrow\frac{9\sqrt{41}-36}{8} \end{align*}
\begin{align*} \left(\frac{4}{3}h\right)^2+4^2&=(3+x)^2 \\ \frac{16}{9}\cdot\frac{9\sqrt{41}-36}{8}+4^2&=(3+x)^2 \\ 2\sqrt{41}-8+16&=(3+x)^2 \\ (3+x)^2&=8+2\sqrt{41} \\ d^2=4(3+x)^2&=32+8\sqrt{41} \\ \end{align*}
Thus, $32+8+41=\boxed{\textbf{(A) }81}$.
~MaPhyCom | // Block 1
import olympiad;
size(10cm);
defaultpen(linewidth(1));
pair A = (-2.0833333333, -1.3020833333);
pair B = (3.35, 0.0);
pair C = (2.0833333333, 1.3020833333);
pair D = (-3.35, 0.0);
pair P = (-2.0833333333, 0.0);
pair Q = (2.0833333333, 0.0);
pair R = (2.4109375, 1.5052083333);
pair S = (-2.4109375, -1.5052083333);
pair O = (0,0);
draw(A--B--C--D--cycle);
draw(A--C);
draw(D--O);
draw(A--P, dashed);
draw(D--S, dashed);
draw(R--S);
draw(O--C--Q--cycle, red+1.2);
draw(O--B--R--cycle, blue+0.8);
draw(rightanglemark(A,P,Q,8));
draw(rightanglemark(C,Q,D,8));
draw(rightanglemark(D,S,R,8));
draw(rightanglemark(B,R,S,8));
label("$A$", A, S);
label("$B$", B, E);
label("$C$", C, N);
label("$D$", D, W);
label("$P$", P, NE);
label("$Q$", Q, S);
label("$R$", R, N);
label("$S$", S, S);
label("$3$", midpoint(P--O), S);
label("$3$", midpoint(O--Q), S);
label("$x$", midpoint(Q--B), S);
label("$h$", midpoint(C--Q), E);
label("$\frac{4}{3}h$", midpoint(R--B), E);
label("$8$", midpoint(R--S), N+0.4*dir(90));
// Block 2
import olympiad; size(10cm); defaultpen(linewidth(1)); pair A = (-2.0833333333, -1.3020833333); pair B = (3.35, 0.0); pair C = (2.0833333333, 1.3020833333); pair D = (-3.35, 0.0); pair P = (-2.0833333333, 0.0); pair Q = (2.0833333333, 0.0); pair R = (2.4109375, 1.5052083333); pair S = (-2.4109375, -1.5052083333); pair O = (0,0); draw(A--B--C--D--cycle); draw(A--C); draw(D--O); draw(A--P, dashed); draw(D--S, dashed); draw(R--S); draw(O--C--Q--cycle, red+1.2); draw(O--B--R--cycle, blue+0.8); draw(rightanglemark(A,P,Q,8)); draw(rightanglemark(C,Q,D,8)); draw(rightanglemark(D,S,R,8)); draw(rightanglemark(B,R,S,8)); label("$A$", A, S); label("$B$", B, E); label("$C$", C, N); label("$D$", D, W); label("$P$", P, NE); label("$Q$", Q, S); label("$R$", R, N); label("$S$", S, S); label("$3$", midpoint(P--O), S); label("$3$", midpoint(O--Q), S); label("$x$", midpoint(Q--B), S); label("$h$", midpoint(C--Q), E); label("$\frac{4}{3}h$", midpoint(R--B), E); label("$8$", midpoint(R--S), N+0.4*dir(90)); | [] |
908 | Let $S$ be the set of lattice points in the coordinate plane, both of whose coordinates are integers between $1$ and $30,$ inclusive. Exactly $300$ points in $S$ lie on or below a line with equation $y=mx.$ The possible values of $m$ lie in an interval of length $\frac ab,$ where $a$ and $b$ are relatively prime positive integers. What is $a+b?$
$\textbf{(A)} ~31 \qquad \textbf{(B)} ~47 \qquad \textbf{(C)} ~62\qquad \textbf{(D)} ~72 \qquad \textbf{(E)} ~85$ | 2021 AMC 12B Problem 25 | Based on area ratios between a square of side length $30$ and a triangle with base $30$, we estimate that the slope $m$ of the line we want is approximately $\frac{2}{3}$. Following this estimate, we see if there are approximately $30 \cdot 30 - 300 = 600$ lattice points above the line $y=\frac{2}{3}x$.
Counting the number of lattice points with $x=1$ above the line, the number of lattice points with $x=2$ above the line, and so on, we find that the total number of lattice points above the line is $30+29+28+28+27+26+26 \ldots+ 10$, with the even integers repeating every third term. We see that the average of the $30$ terms is $20$, which means that exactly $20 \cdot 30 = 600$ lattice points above the line as desired. This gives a lower bound because any decrease in the slope of the line would cause points that were already on the line to shift to being above it.
To find the upper bound, notice that each lattice point less than $1$ unit above the line is either $\frac{1}{3}$ or $\frac{2}{3}$ above. Since the slope through a point is the y-coordinate divided by the x-coordinate, a shift in the slope will increase the y-value of the higher x-coordinates more than the y-value of the lower x-coordinates. So, we turn our attention to $x=28, 29, 30$ for which the line $y=\frac{2}{3}x$ intersects at $y= \frac{56}{3}, \frac{58}{3}, 20$. The point $(30,20)$ is already counted, and we can clearly see that if we slowly increase the slope of the line, we will first hit the point $(28,19)$ since $(28, \frac{56}{3})$ is the closest to it. The equation of the line which goes through both the origin and $(28,19)$ is $y=\frac{19}{28}x$. This gives an upper bound of $m=\frac{19}{28}$.
Taking the upper bound of $m$ and subtracting the lower bound yields $\frac{19}{28}-\frac{2}{3}=\frac{1}{84}$. The answer is therefore $1+84=$ $\boxed{\textbf{(E)} ~85}$.
~theAJL ~Minor edits by Eric_Zang
Diagram | /* Created by Brendanb4321 */ import graph; size(16cm); defaultpen(fontsize(9pt)); xaxis(0,30,Ticks(1.0)); yaxis(0,25,Ticks(1.0)); draw((0,0)--(30,20)); draw((0,0)--(30,30/28*19), dotted); for (int i = 1; i<=30; ++i) { for (int j = 1; j<=2/3*i+1; ++j) { dot((i,j)); } } dot((28,19), red); label("$m=2/3$", (32,20)); label("$m=19/28$", (32.3,20.8)); | [] |
908 | Let $S$ be the set of lattice points in the coordinate plane, both of whose coordinates are integers between $1$ and $30,$ inclusive. Exactly $300$ points in $S$ lie on or below a line with equation $y=mx.$ The possible values of $m$ lie in an interval of length $\frac ab,$ where $a$ and $b$ are relatively prime positive integers. What is $a+b?$
$\textbf{(A)} ~31 \qquad \textbf{(B)} ~47 \qquad \textbf{(C)} ~62\qquad \textbf{(D)} ~72 \qquad \textbf{(E)} ~85$ | 2021 AMC 12B Problem 25 | It's easier to calculate the number of lattice points inside a rectangle with vertices $(0,0)$, $(p,0)$, $(p,q)$, $(0,q)$. Those lattice points are divided by the diagonal $y = \frac{p}{q} \cdot x$ into $2$ halves. In this problem the number of lattice points on or below the diagonal and $x \ge 1$ is
$\frac{1}{2} [(p+1)(q+1) - d] + d - (p+1)$, $d$ is the number of lattice points on the diagonal, $d \ge 2$
$(p+1)(q+1)$ is the total number of lattice points inside the rectangle. Subtract the number of lattice points on the diagonal, divided by 2 is the number of lattice points below the diagonal, add the number of lattice points on the diagonal, and subtract the lattice points on the $x$ axis, then we get the total number of lattice points on or below the diagonal and $x \ge 1$.
There are $900$ lattice points in total. $300$ is $\frac{1}{3}$ of $900$. The $x$ coordinate of the top-right vertex of the rectangle is $30$, $\frac{1}{2} \cdot 30 \cdot 20 = 300$. I guess the $y$ coordinate of the top-right vertex of the rectangle is $20$. Now I am going to verify that. The slope of the diagonal is $\frac{20}{30} = \frac{2}{3}$, there are $11$ lattice points on the diagonal. Substitute $(p,q)=(30, 20)$, $d=11$ to the above formula:
$\frac{1}{2} [(30+1)(10+1) - 11] + 11 - (30+1) = 300$
Because there are $11$ lattice points on line $y = \frac{2}{3}x$, if $m < \frac{2}{3}$, then the number of lattice points on or below the line is less than $300$. So $m = \frac{2}{3}$ is the lower bound.
Now I am going to calculate the upper bound. From $\frac{b}{a} < \frac{b+1}{a+1}$,
$\frac{2}{3} = \frac{18}{27} < \frac{19}{28}$
$\frac{2}{3} = \frac{20}{30} < \frac{21}{31} < \frac{19}{28}$
If $m = \frac{21}{31}$, I will calculate by using the rectangle with blue vertex $(p,q) = (31, 21)$, then subtract lattice points on line $x = 31$, which is $21$. There are 2 lattice points on the diagonal, $d=2$.
$\frac{1}{2} [(31+1)(21+1) - 2] + 2 - (31+1) - 21 = 300$, same as that of $m = \frac{2}{3}$
If $m = \frac{19}{28}$, I will calculate by using the rectangle with red vertex $(p,q) = (28, 19)$, then add lattice points on line $x = 29$ and $x = 30$, which is $19 + 20 = 39$. There are 2 lattice points on the diagonal, $d=2$.
$\frac{1}{2} [(28+1)(19+1) - 2] + 2 - (28+1) + 39 = 301$, $1$ more than that of $m = \frac{2}{3}$
When $m$ increases, more lattice points falls below the line $y = mx$. Any value larger than $\frac{19}{28}$ has more than $301$ lattice points on or below $y = \frac{19}{28} x$. So the upper bound is $\frac{19}{28}$.
$\frac{19}{28}-\frac{2}{3}=\frac{1}{84}$, $\boxed{\textbf{(E)} ~85}$.
~isabelchen | // Block 1
/* Created by Brendanb4321, modified by isabelchen */
import graph;
size(18cm);
defaultpen(fontsize(9pt));
xaxis(0,31,Ticks(1.0));
yaxis(0,22,Ticks(1.0));
draw((0,0)--(30,20));
draw((0,0)--(30,30*19/28), dotted);
draw((0,0)--(31,31*21/31), dotted);
for (int i = 1; i<=31; ++i) {
for (int j = 1; j<=2/3*i+1; ++j) {
dot((i,j));
}
}
dot((28,19), red);
dot((31,21), blue);
label("$m=2/3$", (33,20));
label("$m=21/31$", (33,21));
label("$m=19/28$", (33,22));
// Block 2
/* Created by Brendanb4321, modified by isabelchen */ import graph; size(18cm); defaultpen(fontsize(9pt)); xaxis(0,31,Ticks(1.0)); yaxis(0,22,Ticks(1.0)); draw((0,0)--(30,20)); draw((0,0)--(30,30*19/28), dotted); draw((0,0)--(31,31*21/31), dotted); for (int i = 1; i<=31; ++i) { for (int j = 1; j<=2/3*i+1; ++j) { dot((i,j)); } } dot((28,19), red); dot((31,21), blue); label("$m=2/3$", (33,20)); label("$m=21/31$", (33,21)); label("$m=19/28$", (33,22)); | [] |
908 | Let $S$ be the set of lattice points in the coordinate plane, both of whose coordinates are integers between $1$ and $30,$ inclusive. Exactly $300$ points in $S$ lie on or below a line with equation $y=mx.$ The possible values of $m$ lie in an interval of length $\frac ab,$ where $a$ and $b$ are relatively prime positive integers. What is $a+b?$
$\textbf{(A)} ~31 \qquad \textbf{(B)} ~47 \qquad \textbf{(C)} ~62\qquad \textbf{(D)} ~72 \qquad \textbf{(E)} ~85$ | 2021 AMC 12B Problem 25 | The lower bound of $m$ is $\frac23 = \frac{20}{30}$. Inside the rectangle with vertices $(0,0)$, $(30,0)$, $(30,20)$, $(0, 20)$ and diagonal $y = \frac23 x$, there are $(30-1)(20-1) = 551$ lattice points inside, not including the edges. There are $9$ lattice points on diagonal $y = \frac23 x$ inside the rectangle, $551 + 9 = 560$. Half of the $560$ lattice points are below diagonal $y = \frac23 x$, $560 \cdot \frac12 = 280$. There are $20$ lattice points on edge $x = 30$, $280 + 20 = 300$. Once $m < \frac23$, the $9$ lattice points on diagonal $y = \frac23 x$ will be above the new diagonal, making the number of lattice points on and below the diagonal less than $300$.
Now we are going to calculate the upper bound by the following formula:
The number of lattice points inside rectangle $(0,0)$, $(a,0)$, $(a, b)$, $(0, b)$ and below diagonal $y = \frac{b}{a}x$ is $\frac{(a-1)(b-1)}{2}$, where $a$ and $b$ are relatively prime.
There are $(a-1)(b-1)$ lattice points inside the rectangle. Because $a$ and $b$ are relatively prime, the only lattice points on the diagonal are $(0,0)$ and $(a,b)$. By symmetry, half of the lattice points are below the diagonal.
$\frac{2}{3} = \frac{20}{30} < \frac{21}{31} < \frac{19}{28}$
When $a = 31$, $b = 21$, $\frac{(31-1)(21-1)}{2} = 300$.
When $a = 28$, $b = 19$, $\frac{(28-1)(19-1)}{2} = 243$. Below the line $y = \frac{19}{28} x$, there are $19$ lattice points on line $x = 28$, $19$ lattice points on line $x = 29$, $20$ lattice points on line $x = 30$, $243 + 19 + 19 + 20 = 301$.
More lattice points fall below the line $y = mx$ as $m$ increases. There are more than $301$ lattice points on and below the line for any $m$ greater than $\frac{19}{28}$. Therefore, the upper bound is $\frac{19}{28}$.
$\frac{19}{28}-\frac{2}{3}=\frac{1}{84}$, so $1+84=\boxed{\textbf{(E)} ~85}$.
~isabelchen
Remark
$\lfloor \frac{b}{a} k \rfloor$ is the number of lattice points on line $x = k$, below line $y = \frac{b}{a} x$ and above the $x$ axis, where $k$ is an integer and $0<k<a$. Therefore, $\sum_{k=1}^{a-1} \lfloor \frac{b}{a} k \rfloor$ is the number of lattice points inside the rectangle $(0,0)$, $(a,0)$, $(a, b)$, $(0, b)$, below diagonal $y = \frac{b}{a} x$. If $a$ and $b$ are relatively prime, $\sum_{k=1}^{a-1} \lfloor \frac{b}{a} k \rfloor = \frac{(a-1)(b-1)}{2}$, as explained in solution 5. This problem is about finding the upper and lower bound of $\frac{b}{a}$, given $\sum_{k=1}^{30} \lfloor \frac{b}{a} k \rfloor = 300$. The same problem can have geometric representation as stated in the original problem, or algebraic representation as stated here.
~isabelchen | // Block 1
/* Created by isabelchen */
import graph;
size(8cm);
defaultpen(fontsize(9pt));
xaxis(0,8);
yaxis(0,6);
draw((0,0)--(7,5));
draw((7,0)--(7,5));
draw((0,5)--(7,5));
dot((0,0)); dot((1,0)); dot((2,0)); dot((3,0)); dot((4,0)); dot((5,0)); dot((6,0)); dot((7,0)); dot((8,0)); dot((0,1)); dot((1,1)); dot((2,1)); dot((3,1)); dot((4,1)); dot((5,1)); dot((6,1)); dot((7,1)); dot((8,1)); dot((0,2)); dot((1,2)); dot((2,2)); dot((3,2)); dot((4,2)); dot((5,2)); dot((6,2)); dot((7,2)); dot((8,2)); dot((0,3)); dot((1,3)); dot((2,3)); dot((3,3)); dot((4,3)); dot((5,3)); dot((6,3)); dot((7,3)); dot((8,3)); dot((0,4)); dot((1,4)); dot((2,4)); dot((3,4)); dot((4,4)); dot((5,4)); dot((6,4)); dot((7,4)); dot((8,4)); dot((0,5)); dot((1,5)); dot((2,5)); dot((3,5)); dot((4,5)); dot((5,5)); dot((6,5)); dot((7,5)); dot((8,5)); dot((0,6)); dot((1,6)); dot((2,6)); dot((3,6)); dot((4,6)); dot((5,6)); dot((6,6)); dot((7,6)); dot((8,6));
label("$(0,0)$", (0,0), SW);
label("$(a, b)$", (7,5), NE);
dot((7,0));
label("$a$", (7,0), S);
dot((0,5));
label("$b$", (0,5), W);
// Block 2
/* Created by isabelchen */ import graph; size(8cm); defaultpen(fontsize(9pt)); xaxis(0,8); yaxis(0,6); draw((0,0)--(7,5)); draw((7,0)--(7,5)); draw((0,5)--(7,5)); dot((0,0)); dot((1,0)); dot((2,0)); dot((3,0)); dot((4,0)); dot((5,0)); dot((6,0)); dot((7,0)); dot((8,0)); dot((0,1)); dot((1,1)); dot((2,1)); dot((3,1)); dot((4,1)); dot((5,1)); dot((6,1)); dot((7,1)); dot((8,1)); dot((0,2)); dot((1,2)); dot((2,2)); dot((3,2)); dot((4,2)); dot((5,2)); dot((6,2)); dot((7,2)); dot((8,2)); dot((0,3)); dot((1,3)); dot((2,3)); dot((3,3)); dot((4,3)); dot((5,3)); dot((6,3)); dot((7,3)); dot((8,3)); dot((0,4)); dot((1,4)); dot((2,4)); dot((3,4)); dot((4,4)); dot((5,4)); dot((6,4)); dot((7,4)); dot((8,4)); dot((0,5)); dot((1,5)); dot((2,5)); dot((3,5)); dot((4,5)); dot((5,5)); dot((6,5)); dot((7,5)); dot((8,5)); dot((0,6)); dot((1,6)); dot((2,6)); dot((3,6)); dot((4,6)); dot((5,6)); dot((6,6)); dot((7,6)); dot((8,6)); label("$(0,0)$", (0,0), SW); label("$(a, b)$", (7,5), NE); dot((7,0)); label("$a$", (7,0), S); dot((0,5)); label("$b$", (0,5), W); | [] |
909 | Daniel finds a rectangular index card and measures its diagonal to be $8$ centimeters.
Daniel then cuts out equal squares of side $1$ cm at two opposite corners of the index card and measures the distance between the two closest vertices of these squares to be $4\sqrt{2}$ centimeters, as shown below. What is the area of the original index card?
$\textbf{(A) } 14 \qquad \textbf{(B) } 10\sqrt{2} \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 12\sqrt{2} \qquad \textbf{(E) } 18$ | 2022 AMC 10A Problem 10 | Label the bottom left corner of the larger rectangle (without the square cut out) as $A$ and the top right as $D$. $w$ is the width of the rectangle and $\ell$ is the length. Now we have vertices $E, F, G, H$ as vertices of the irregular octagon created by cutting out the squares. Let $I, J$ be the two closest vertices formed by the squares.
The distance between the two closest vertices of the squares is thus $IJ=\left(4\sqrt{2}\right).$
Substituting, we get:
\[(IJ)^2 = (w-2)^2 + (\ell-2)^2 = \left(4\sqrt{2}\right)^2 = 32 \implies w^2+\ell^2-4w-4\ell = 24.\]
Using the fact that the diagonal of the rectangle is $8,$ we get:
\[w^2+\ell^2 = 64.\]
Subtracting the first equation from the second equation, we get: \[4w+4\ell=40 \implies w+\ell = 10.\]
Squaring yields: \[w^2 + 2w\ell + \ell^2 = 100.\]
Subtracting the second equation from this, we get: $2w\ell = 36,$ and thus the area of the original rectangle is $w\ell = \boxed{\textbf{(E) } 18}.$
~USAMO333
Edits and Diagram by ~KingRavi and ~MRENTHUSIASM
Minor edit by yanes04 | // Block 1
/* Edited by MRENTHUSIASM */
size(250);
real x, y;
x = 6;
y = 3;
draw((0,0)--(x,0));
draw((0,0)--(0,y));
draw((0,y)--(x,y));
draw((x,0)--(x,y));
draw((0.5,0)--(0.5,0.5)--(0,0.5));
draw((x-0.5,y)--(x-0.5,y-0.5)--(x,y-0.5));
draw((0.5,0.5)--(x-0.5,y-0.5),dashed,Arrows());
draw((x,0)--(0,y),dashed,Arrows());
label("$1$",(x-0.5,y-0.25),W);
label("$1$",(x-0.25,y-0.5),S);
label("$8$",midpoint((0.5,y-0.5)--(x/2,y/2)),(0,2.5));
label("$4\sqrt{2}$",midpoint((0.5,0.5)--(x/2,y/2)),S);
label("$A$",(0,0),SW);
label("$E$",(0,0.5),W);
label("$F$",(0.5,0),S);
label("$I$",(0.5,0.5),N);
label("$D$",(x,y),NE);
label("$G$",(x-0.5,y),N);
label("$H$",(x,y-0.5),E);
label("$J$",(x-0.5,y-0.5),S);
Label L1 = Label("$w$", align=(0,0), position=MidPoint, filltype=Fill(3,0,white));
Label L2 = Label("$\ell$", align=(0,0), position=MidPoint, filltype=Fill(0,3,white));
draw((0,-1)--(x,-1), L=L1, arrow=Arrows(),bar=Bars(15));
draw((x+1,0)--(x+1,y), L=L2, arrow=Arrows(),bar=Bars(15));
// Block 2
/* Edited by MRENTHUSIASM */ size(250); real x, y; x = 6; y = 3; draw((0,0)--(x,0)); draw((0,0)--(0,y)); draw((0,y)--(x,y)); draw((x,0)--(x,y)); draw((0.5,0)--(0.5,0.5)--(0,0.5)); draw((x-0.5,y)--(x-0.5,y-0.5)--(x,y-0.5)); draw((0.5,0.5)--(x-0.5,y-0.5),dashed,Arrows()); draw((x,0)--(0,y),dashed,Arrows()); label("$1$",(x-0.5,y-0.25),W); label("$1$",(x-0.25,y-0.5),S); label("$8$",midpoint((0.5,y-0.5)--(x/2,y/2)),(0,2.5)); label("$4\sqrt{2}$",midpoint((0.5,0.5)--(x/2,y/2)),S); label("$A$",(0,0),SW); label("$E$",(0,0.5),W); label("$F$",(0.5,0),S); label("$I$",(0.5,0.5),N); label("$D$",(x,y),NE); label("$G$",(x-0.5,y),N); label("$H$",(x,y-0.5),E); label("$J$",(x-0.5,y-0.5),S); Label L1 = Label("$w$", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); Label L2 = Label("$\ell$", align=(0,0), position=MidPoint, filltype=Fill(0,3,white)); draw((0,-1)--(x,-1), L=L1, arrow=Arrows(),bar=Bars(15)); draw((x+1,0)--(x+1,y), L=L2, arrow=Arrows(),bar=Bars(15)); | [] |
910 | Let $\triangle ABC$ be a scalene triangle. Point $P$ lies on $\overline{BC}$ so that $\overline{AP}$ bisects $\angle BAC.$ The line through $B$ perpendicular to $\overline{AP}$ intersects the line through $A$ parallel to $\overline{BC}$ at point $D.$ Suppose $BP=2$ and $PC=3.$ What is $AD?$
$\textbf{(A) } 8 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 10 \qquad \textbf{(D) } 11 \qquad \textbf{(E) } 12$ | 2022 AMC 10A Problem 13 | Since there is only one possible value of $AD$, we assume $\angle{B}=90^{\circ}$. By the angle bisector theorem, $\frac{AB}{AC}=\frac{2}{3}$, so $AB=2\sqrt{5}$ and $AC=3\sqrt{5}$. Now observe that $\angle{BAD}=90^{\circ}$. Let the intersection of $BD$ and $AP$ be $X$. Then $\angle{ABD}=90^{\circ}-\angle{BAX}=\angle{APB}$. Consequently, \[\bigtriangleup DAB \sim \bigtriangleup ABP\] and therefore $\frac{DA}{AB} = \frac{AB}{BP}$, so $AD=\boxed{\textbf{(C) }10}$, and we're done!
~Bxiao31415 | // Block 1
size(300);
pair A, B, C, P, XX, D;
B = (0,0);
P = (2,0);
C = (5,0);
A=(0,4.47214);
D = A + (10,0);
draw(A--B--C--cycle, linewidth(1));
dot("$A$", A, N);
dot("$B$", B, SW);
dot("$C$", C, E);
dot("$P$", P, S);
dot("$D$", D, E);
markscalefactor = 0.1;
draw(anglemark(B,A,P));
markscalefactor = 0.12;
draw(anglemark(P,A,C));
draw(P--A--D--B, linewidth(1));
XX = intersectionpoints(A--P,B--D)[0];
dot("$X$", XX, dir(150));
markscalefactor = 0.03;
draw(rightanglemark(A,B,C));
draw(rightanglemark(D,XX,P));
// Block 2
size(300); pair A, B, C, P, XX, D; B = (0,0); P = (2,0); C = (5,0); A=(0,4.47214); D = A + (10,0); draw(A--B--C--cycle, linewidth(1)); dot("$A$", A, N); dot("$B$", B, SW); dot("$C$", C, E); dot("$P$", P, S); dot("$D$", D, E); markscalefactor = 0.1; draw(anglemark(B,A,P)); markscalefactor = 0.12; draw(anglemark(P,A,C)); draw(P--A--D--B, linewidth(1)); XX = intersectionpoints(A--P,B--D)[0]; dot("$X$", XX, dir(150)); markscalefactor = 0.03; draw(rightanglemark(A,B,C)); draw(rightanglemark(D,XX,P)); | [] |
911 | A bowl is formed by attaching four regular hexagons of side $1$ to a square of side $1$. The edges of the adjacent hexagons coincide, as shown in the figure. What is the area of the octagon obtained by joining the top eight vertices of the four hexagons, situated on the rim of the bowl?
$\textbf{(A) }6\qquad\textbf{(B) }7\qquad\textbf{(C) }5+2\sqrt{2}\qquad\textbf{(D) }8\qquad\textbf{(E) }9$ | 2022 AMC 10A Problem 21 | We extend line segments $\ell,m,$ and $n$ to their point of concurrency, as shown below:
We claim that lines $\ell,m,$ and $n$ are concurrent: In the lateral faces of the bowl, we know that lines $\ell$ and $m$ must intersect, and lines $\ell$ and $n$ must intersect. In addition, line $\ell$ intersects the top plane of the bowl at exactly one point. Since lines $m$ and $n$ are both in the top plane of the bowl, we conclude that lines $\ell,m,$ and $n$ are concurrent.
In the lateral faces of the bowl, the dashed red line segments create equilateral triangles. So, the dashed red line segments all have length $1.$ In the top plane of the bowl, we know that $\overleftrightarrow{m}\perp\overleftrightarrow{n}.$ So, the dashed red line segments create an isosceles triangle with leg-length $1.$
Note that octagon has four pairs of parallel sides, and the successive side-lengths are $1,\sqrt2,1,\sqrt2,1,\sqrt2,1,\sqrt2,$ as shown below:
The area of the octagon is \[3^2-4\cdot\left(\frac12\cdot1^2\right)=\boxed{\textbf{(B) }7}.\]
~MRENTHUSIASM | // Block 1
/* Made by AoPS; edited by MRENTHUSIASM */
import three;
size(225);
currentprojection=
orthographic(camera=(-5.52541796301147,-2.61548797564715,1.6545450372312),
up=(0.00247902062334861,0.000877141782387748,0.00966536329192992),
target=(0,0,0),
zoom=0.570588560870951);
currentpen = black+1.5bp;
triple A = O;
triple M = (X+Y)/2;
triple B = (-1/2,-1/2,1/sqrt(2));
triple C = (-1,0,sqrt(2));
triple D = (0,-1,sqrt(2));
transform3 rho = rotate(90,M,M+Z);
//arrays of vertices for the lower level (the square), the middle level,
//and the interleaves vertices of the upper level (the octagon)
triple[] lVs = {A};
triple[] mVs = {B};
triple[] uVsl = {C};
triple[] uVsr = {D};
for(int i = 0; i < 3; ++i){
lVs.push(rho*lVs[i]);
mVs.push(rho*mVs[i]);
uVsl.push(rho*uVsl[i]);
uVsr.push(rho*uVsr[i]);
}
lVs.cyclic = true;
uVsl.cyclic = true;
draw(surface(uVsl[0]--uVsr[0]--uVsl[1]--uVsr[1]--uVsl[2]--uVsr[2]--uVsl[3]--uVsr[3]--cycle),yellow);
for(int i : new int[] {0,1,2,3}){
draw(uVsl[i]--uVsr[i]);
draw(uVsr[i]--uVsl[i+1]);
}
draw(lVs[0]--lVs[1]^^lVs[0]--lVs[3]);
for(int i : new int[] {0,1,3}){
draw(lVs[0]--lVs[i]);
draw(lVs[i]--mVs[i]);
draw(mVs[i]--uVsl[i]);
}
for(int i : new int[] {0,3}){
draw(mVs[i]--uVsr[i]);
}
for(int i : new int[] {1,3}) draw(lVs[2]--lVs[i],dashed);
draw(lVs[2]--mVs[2],dashed);
draw(mVs[2]--uVsl[2]^^mVs[2]--uVsr[2],dashed);
draw(mVs[1]--uVsr[1],dashed);
draw(uVsl[1]--uVsr[0],red+2bp);
draw(uVsr[3]--uVsl[4],red+2bp);
draw(lVs[0]--mVs[0],red+2bp);
draw(uVsr[0]--uVsr[0]+uVsr[0]-uVsl[1],red+dashed+2bp);
draw(uVsl[4]--uVsl[4]+uVsl[4]-uVsr[3],red+dashed+2bp);
draw(mVs[0]--mVs[0]+mVs[0]-lVs[0],red+dashed+2bp);
label("$\ell$",midpoint(lVs[0]--mVs[0]),(1,2,0),red);
label("$m$",midpoint(uVsr[3]--uVsl[4]),(1,0,-2),red);
label("$n$",midpoint(uVsl[1]--uVsr[0]),(-1,0,-2),red);
// Block 2
/* Made by AoPS; edited by MRENTHUSIASM */
size(225);
real r = 1/3;
draw((0,r)--(0,0)--(r,0)^^(1-r,0)--(1,0)--(1,r)^^(1,1-r)--(1,1)--(1-r,1)^^(r,1)--(0,1)--(0,1-r),red+2bp+dashed);
fill((r,0)--(1-r,0)--(1,r)--(1,1-r)--(1-r,1)--(r,1)--(0,1-r)--(0,r)--cycle,yellow);
draw((r,0)--(1-r,0)--(1,r)--(1,1-r)--(1-r,1)--(r,1)--(0,1-r)--(0,r)--cycle,black+2bp);
label("$1$",(0.5,0),S);
label("$1$",(1,0.5),E);
label("$1$",(0.5,1),N);
label("$1$",(0,0.5),W);
label("$\sqrt2$",(1-r/2,r/2),NW);
label("$\sqrt2$",(1-r/2,1-r/2),SW);
label("$\sqrt2$",(r/2,1-r/2),SE);
label("$\sqrt2$",(r/2,r/2),NE);
label("$1$",(1-r/2,0),S);
label("$1$",(1,r/2),E);
label("$1$",(1,1-r/2),E);
label("$1$",(1-r/2,1),N);
label("$1$",(r/2,1),N);
label("$1$",(0,1-r/2),W);
label("$1$",(0,r/2),W);
label("$1$",(r/2,0),S);
// Block 3
/* Made by AoPS; edited by MRENTHUSIASM */ import three; size(225); currentprojection= orthographic(camera=(-5.52541796301147,-2.61548797564715,1.6545450372312), up=(0.00247902062334861,0.000877141782387748,0.00966536329192992), target=(0,0,0), zoom=0.570588560870951); currentpen = black+1.5bp; triple A = O; triple M = (X+Y)/2; triple B = (-1/2,-1/2,1/sqrt(2)); triple C = (-1,0,sqrt(2)); triple D = (0,-1,sqrt(2)); transform3 rho = rotate(90,M,M+Z); //arrays of vertices for the lower level (the square), the middle level, //and the interleaves vertices of the upper level (the octagon) triple[] lVs = {A}; triple[] mVs = {B}; triple[] uVsl = {C}; triple[] uVsr = {D}; for(int i = 0; i < 3; ++i){ lVs.push(rho*lVs[i]); mVs.push(rho*mVs[i]); uVsl.push(rho*uVsl[i]); uVsr.push(rho*uVsr[i]); } lVs.cyclic = true; uVsl.cyclic = true; draw(surface(uVsl[0]--uVsr[0]--uVsl[1]--uVsr[1]--uVsl[2]--uVsr[2]--uVsl[3]--uVsr[3]--cycle),yellow); for(int i : new int[] {0,1,2,3}){ draw(uVsl[i]--uVsr[i]); draw(uVsr[i]--uVsl[i+1]); } draw(lVs[0]--lVs[1]^^lVs[0]--lVs[3]); for(int i : new int[] {0,1,3}){ draw(lVs[0]--lVs[i]); draw(lVs[i]--mVs[i]); draw(mVs[i]--uVsl[i]); } for(int i : new int[] {0,3}){ draw(mVs[i]--uVsr[i]); } for(int i : new int[] {1,3}) draw(lVs[2]--lVs[i],dashed); draw(lVs[2]--mVs[2],dashed); draw(mVs[2]--uVsl[2]^^mVs[2]--uVsr[2],dashed); draw(mVs[1]--uVsr[1],dashed); draw(uVsl[1]--uVsr[0],red+2bp); draw(uVsr[3]--uVsl[4],red+2bp); draw(lVs[0]--mVs[0],red+2bp); draw(uVsr[0]--uVsr[0]+uVsr[0]-uVsl[1],red+dashed+2bp); draw(uVsl[4]--uVsl[4]+uVsl[4]-uVsr[3],red+dashed+2bp); draw(mVs[0]--mVs[0]+mVs[0]-lVs[0],red+dashed+2bp); label("$\ell$",midpoint(lVs[0]--mVs[0]),(1,2,0),red); label("$m$",midpoint(uVsr[3]--uVsl[4]),(1,0,-2),red); label("$n$",midpoint(uVsl[1]--uVsr[0]),(-1,0,-2),red);
// Block 4
/* Made by AoPS; edited by MRENTHUSIASM */ size(225); real r = 1/3; draw((0,r)--(0,0)--(r,0)^^(1-r,0)--(1,0)--(1,r)^^(1,1-r)--(1,1)--(1-r,1)^^(r,1)--(0,1)--(0,1-r),red+2bp+dashed); fill((r,0)--(1-r,0)--(1,r)--(1,1-r)--(1-r,1)--(r,1)--(0,1-r)--(0,r)--cycle,yellow); draw((r,0)--(1-r,0)--(1,r)--(1,1-r)--(1-r,1)--(r,1)--(0,1-r)--(0,r)--cycle,black+2bp); label("$1$",(0.5,0),S); label("$1$",(1,0.5),E); label("$1$",(0.5,1),N); label("$1$",(0,0.5),W); label("$\sqrt2$",(1-r/2,r/2),NW); label("$\sqrt2$",(1-r/2,1-r/2),SW); label("$\sqrt2$",(r/2,1-r/2),SE); label("$\sqrt2$",(r/2,r/2),NE); label("$1$",(1-r/2,0),S); label("$1$",(1,r/2),E); label("$1$",(1,1-r/2),E); label("$1$",(1-r/2,1),N); label("$1$",(r/2,1),N); label("$1$",(0,1-r/2),W); label("$1$",(0,r/2),W); label("$1$",(r/2,0),S); | [] |
911 | A bowl is formed by attaching four regular hexagons of side $1$ to a square of side $1$. The edges of the adjacent hexagons coincide, as shown in the figure. What is the area of the octagon obtained by joining the top eight vertices of the four hexagons, situated on the rim of the bowl?
$\textbf{(A) }6\qquad\textbf{(B) }7\qquad\textbf{(C) }5+2\sqrt{2}\qquad\textbf{(D) }8\qquad\textbf{(E) }9$ | 2022 AMC 10A Problem 21 | You could also think of this problem as folding the four triangles connecting the hexagon faces to cover parts of of the octagon:
So the octagon is made up of five unit squares and four $45-45-90$ triangles with legs of length $1.$ So the area is $\boxed{7}.$
~grogg007 | // Block 1
size(180);
import graph;
pair A = (0, 0);
pair B = A + dir(0)*1;
pair C = B + dir(-45)*sqrt(2);
pair D = C + dir(-90)*1;
pair E = D + dir(-135)*sqrt(2);
pair F = E + dir(180)*1;
pair G = F + dir(135)*sqrt(2);
pair H = G + dir(90)*1;
pair I1 = intersectionpoint(B--E, C--H);
pair I2 = intersectionpoint(B--E, D--G);
pair I3 = intersectionpoint(D--G, A--F);
pair I4 = intersectionpoint(A--F, C--H);
fill(B--C--I1--cycle, rgb(1,0.8,0.8));
fill(D--E--I2--cycle, rgb(1,0.8,0.8));
fill(F--G--I3--cycle, rgb(1,0.8,0.8));
fill(A--H--I4--cycle, rgb(1,0.8,0.8));
draw(A--B--C--D--E--F--G--H--cycle, linewidth(1));
draw(A--F, dashed+red);
draw(B--E, dashed+blue);
draw(C--H, dashed+green);
draw(D--G, dashed+purple);
label("$A$", A, NW);
label("$B$", B, NE);
label("$C$", C, SE);
label("$D$", D, S);
label("$E$", E, SW);
label("$F$", F, W);
label("$G$", G, NW);
label("$H$", H, N);
label("$1$", midpoint(A--B), N);
label("$\sqrt{2}$", midpoint(B--C), dir(-22));
label("$1$", midpoint(C--D), E);
label("$\sqrt{2}$", midpoint(D--E), dir(-135));
label("$1$", midpoint(E--F), S);
label("$\sqrt{2}$", midpoint(F--G), dir(135));
label("$1$", midpoint(G--H), W);
label("$\sqrt{2}$", midpoint(H--A), dir(45));
// Block 2
size(180); import graph; pair A = (0, 0); pair B = A + dir(0)*1; pair C = B + dir(-45)*sqrt(2); pair D = C + dir(-90)*1; pair E = D + dir(-135)*sqrt(2); pair F = E + dir(180)*1; pair G = F + dir(135)*sqrt(2); pair H = G + dir(90)*1; pair I1 = intersectionpoint(B--E, C--H); pair I2 = intersectionpoint(B--E, D--G); pair I3 = intersectionpoint(D--G, A--F); pair I4 = intersectionpoint(A--F, C--H); fill(B--C--I1--cycle, rgb(1,0.8,0.8)); fill(D--E--I2--cycle, rgb(1,0.8,0.8)); fill(F--G--I3--cycle, rgb(1,0.8,0.8)); fill(A--H--I4--cycle, rgb(1,0.8,0.8)); draw(A--B--C--D--E--F--G--H--cycle, linewidth(1)); draw(A--F, dashed+red); draw(B--E, dashed+blue); draw(C--H, dashed+green); draw(D--G, dashed+purple); label("$A$", A, NW); label("$B$", B, NE); label("$C$", C, SE); label("$D$", D, S); label("$E$", E, SW); label("$F$", F, W); label("$G$", G, NW); label("$H$", H, N); label("$1$", midpoint(A--B), N); label("$\sqrt{2}$", midpoint(B--C), dir(-22)); label("$1$", midpoint(C--D), E); label("$\sqrt{2}$", midpoint(D--E), dir(-135)); label("$1$", midpoint(E--F), S); label("$\sqrt{2}$", midpoint(F--G), dir(135)); label("$1$", midpoint(G--H), W); label("$\sqrt{2}$", midpoint(H--A), dir(45)); | [] |
911 | A bowl is formed by attaching four regular hexagons of side $1$ to a square of side $1$. The edges of the adjacent hexagons coincide, as shown in the figure. What is the area of the octagon obtained by joining the top eight vertices of the four hexagons, situated on the rim of the bowl?
$\textbf{(A) }6\qquad\textbf{(B) }7\qquad\textbf{(C) }5+2\sqrt{2}\qquad\textbf{(D) }8\qquad\textbf{(E) }9$ | 2022 AMC 10A Problem 21 | Notice that $AD \parallel PQ$, $BC \parallel PQ$, so $AD \parallel BC$. Obviously, $AD = BC$, so $ABCD$ is a parallelogram, we get \[CD = AB = \sqrt{1^2+1^2} = \sqrt{2}.\]
Note that octagon has four pairs of parallel sides, and the successive side-lengths are $1,\sqrt2,1,\sqrt2,1,\sqrt2,1,\sqrt2,$ as shown below:
The area of the octagon is $\boxed{\textbf{(B) }7}.$
~reda_mandymath | // Block 1
import three;
size(225);
currentprojection=
orthographic(camera=(-5.52541796301147,-2.61548797564715,1.6545450372312),
up=(0.00247902062334861,0.000877141782387748,0.00966536329192992),
target=(0,0,0),
zoom=0.570588560870951);
currentpen = black+1.5bp;
triple A = O;
triple M = (X+Y)/2;
triple B = (-1/2,-1/2,1/sqrt(2));
triple C = (-1,0,sqrt(2));
triple D = (0,-1,sqrt(2));
transform3 rho = rotate(90,M,M+Z);
//arrays of vertices for the lower level (the square), the middle level,
//and the interleaves vertices of the upper level (the octagon)
triple[] lVs = {A};
triple[] mVs = {B};
triple[] uVsl = {C};
triple[] uVsr = {D};
for(int i = 0; i < 3; ++i){
lVs.push(rho*lVs[i]);
mVs.push(rho*mVs[i]);
uVsl.push(rho*uVsl[i]);
uVsr.push(rho*uVsr[i]);
}
lVs.cyclic = true;
uVsl.cyclic = true;
for(int i : new int[] {0,1,2,3}){
draw(uVsl[i]--uVsr[i]);
draw(uVsr[i]--uVsl[i+1]);
}
draw(lVs[0]--lVs[1]^^lVs[0]--lVs[3]);
for(int i : new int[] {0,1,3}){
draw(lVs[0]--lVs[i]);
draw(lVs[i]--mVs[i]);
draw(mVs[i]--uVsl[i]);
}
for(int i : new int[] {0,3}){
draw(mVs[i]--uVsr[i]);
}
for(int i : new int[] {1,3}) draw(lVs[2]--lVs[i],dashed);
draw(lVs[2]--mVs[2],dashed);
draw(mVs[2]--uVsl[2]^^mVs[2]--uVsr[2],dashed);
draw(mVs[1]--uVsr[1],dashed);
//Comment two lines below to remove red edges
draw(lVs[1]--lVs[3],red+2bp+dashed);
draw(uVsl[2]--uVsr[2],red+2bp);
draw(uVsl[2]--lVs[1],red+2bp+dashed);
draw(uVsr[2]--lVs[3],red+2bp+dashed);
label("$A$", lVs[3], SW);
label("$B$", lVs[1], SE);
label("$C$", uVsl[2], NE);
label("$D$", uVsr[2], NW);
label("$P$", lVs[2], NE);
label("$Q$", mVs[2], N);
// Block 2
size(225);
real r = 1/3;
draw((r,0)--(1-r,0)--(1,r)--(1,1-r)--(1-r,1)--(r,1)--(0,1-r)--(0,r)--cycle,black+2bp);
draw((r,0)--(r,1)^^(1-r,0)--(1-r,1)^^(0,r)--(1,r)^^(0,1-r)--(1,1-r), dashed);
label("$1$",(0.5,0),S);
label("$1$",(1,0.5),E);
label("$1$",(0.5,1),N);
label("$1$",(0,0.5),W);
label("$\sqrt2$",(1-r/2,r/2),SE);
label("$\sqrt2$",(1-r/2,1-r/2),NE);
label("$\sqrt2$",(r/2,1-r/2),NW);
label("$\sqrt2$",(r/2,r/2),SW);
label("$1$",(r/2,1/2));
label("$1$",(1-r/2,1/2));
label("$1$",(1/2,1/2));
label("$1$",(1/2,r/2));
label("$1$",(1/2,1-r/2));
real rr=r*2/3;
label("$\frac{1}{2}$",(rr,rr));
label("$\frac{1}{2}$",(1-rr,1-rr));
label("$\frac{1}{2}$",(rr,1-rr));
label("$\frac{1}{2}$",(1-rr,rr));
label("$C$",(r,1),N);
label("$D$",(0,1-r),W);
// Block 3
import three; size(225); currentprojection= orthographic(camera=(-5.52541796301147,-2.61548797564715,1.6545450372312), up=(0.00247902062334861,0.000877141782387748,0.00966536329192992), target=(0,0,0), zoom=0.570588560870951); currentpen = black+1.5bp; triple A = O; triple M = (X+Y)/2; triple B = (-1/2,-1/2,1/sqrt(2)); triple C = (-1,0,sqrt(2)); triple D = (0,-1,sqrt(2)); transform3 rho = rotate(90,M,M+Z); //arrays of vertices for the lower level (the square), the middle level, //and the interleaves vertices of the upper level (the octagon) triple[] lVs = {A}; triple[] mVs = {B}; triple[] uVsl = {C}; triple[] uVsr = {D}; for(int i = 0; i < 3; ++i){ lVs.push(rho*lVs[i]); mVs.push(rho*mVs[i]); uVsl.push(rho*uVsl[i]); uVsr.push(rho*uVsr[i]); } lVs.cyclic = true; uVsl.cyclic = true; for(int i : new int[] {0,1,2,3}){ draw(uVsl[i]--uVsr[i]); draw(uVsr[i]--uVsl[i+1]); } draw(lVs[0]--lVs[1]^^lVs[0]--lVs[3]); for(int i : new int[] {0,1,3}){ draw(lVs[0]--lVs[i]); draw(lVs[i]--mVs[i]); draw(mVs[i]--uVsl[i]); } for(int i : new int[] {0,3}){ draw(mVs[i]--uVsr[i]); } for(int i : new int[] {1,3}) draw(lVs[2]--lVs[i],dashed); draw(lVs[2]--mVs[2],dashed); draw(mVs[2]--uVsl[2]^^mVs[2]--uVsr[2],dashed); draw(mVs[1]--uVsr[1],dashed); //Comment two lines below to remove red edges draw(lVs[1]--lVs[3],red+2bp+dashed); draw(uVsl[2]--uVsr[2],red+2bp); draw(uVsl[2]--lVs[1],red+2bp+dashed); draw(uVsr[2]--lVs[3],red+2bp+dashed); label("$A$", lVs[3], SW); label("$B$", lVs[1], SE); label("$C$", uVsl[2], NE); label("$D$", uVsr[2], NW); label("$P$", lVs[2], NE); label("$Q$", mVs[2], N);
// Block 4
size(225); real r = 1/3; draw((r,0)--(1-r,0)--(1,r)--(1,1-r)--(1-r,1)--(r,1)--(0,1-r)--(0,r)--cycle,black+2bp); draw((r,0)--(r,1)^^(1-r,0)--(1-r,1)^^(0,r)--(1,r)^^(0,1-r)--(1,1-r), dashed); label("$1$",(0.5,0),S); label("$1$",(1,0.5),E); label("$1$",(0.5,1),N); label("$1$",(0,0.5),W); label("$\sqrt2$",(1-r/2,r/2),SE); label("$\sqrt2$",(1-r/2,1-r/2),NE); label("$\sqrt2$",(r/2,1-r/2),NW); label("$\sqrt2$",(r/2,r/2),SW); label("$1$",(r/2,1/2)); label("$1$",(1-r/2,1/2)); label("$1$",(1/2,1/2)); label("$1$",(1/2,r/2)); label("$1$",(1/2,1-r/2)); real rr=r*2/3; label("$\frac{1}{2}$",(rr,rr)); label("$\frac{1}{2}$",(1-rr,1-rr)); label("$\frac{1}{2}$",(rr,1-rr)); label("$\frac{1}{2}$",(1-rr,rr)); label("$C$",(r,1),N); label("$D$",(0,1-r),W); | [] |
912 | Suppose that $13$ cards numbered $1, 2, 3, \ldots, 13$ are arranged in a row. The task is to pick them up in numerically increasing order, working repeatedly from left to right. In the example below, cards $1, 2, 3$ are picked up on the first pass, $4$ and $5$ on the second pass, $6$ on the third pass, $7, 8, 9, 10$ on the fourth pass, and $11, 12, 13$ on the fifth pass. For how many of the $13!$ possible orderings of the cards will the $13$ cards be picked up in exactly two passes?
$\textbf{(A) } 4082 \qquad \textbf{(B) } 4095 \qquad \textbf{(C) } 4096 \qquad \textbf{(D) } 8178 \qquad \textbf{(E) } 8191$ | 2022 AMC 10A Problem 22 | For $1\leq k\leq 12,$ suppose that cards $1, 2, \ldots, k$ are picked up on the first pass. It follows that cards $k+1,k+2,\ldots,13$ are picked up on the second pass.
Once we pick the spots for the cards on the first pass, there is only one way to arrange all $\boldsymbol{13}$ cards.
For each value of $k,$ there are $\binom{13}{k}-1$ ways to pick the $k$ spots for the cards on the first pass: We exclude the arrangement
in which the cards are arranged such that the first pass consists of all $13$ cards.
Therefore, the answer is \[\sum_{k=1}^{12}\left[\binom{13}{k}-1\right] = \left[\sum_{k=1}^{12}\binom{13}{k}\right]-12 = \left[\sum_{k=0}^{13}\binom{13}{k}\right]-14 = 2^{13} - 14 = \boxed{\textbf{(D) } 8178}.\]
~MRENTHUSIASM | // Block 1
size(11cm);
draw((0,0)--(2,0)--(2,3)--(0,3)--cycle);
label("1", (1,1.5));
draw((3,0)--(5,0)--(5,3)--(3,3)--cycle);
label("2", (4,1.5));
draw((6,0)--(8,0)--(8,3)--(6,3)--cycle);
label("3", (7,1.5));
draw((9,0)--(11,0)--(11,3)--(9,3)--cycle);
label("4", (10,1.5));
draw((12,0)--(14,0)--(14,3)--(12,3)--cycle);
label("5", (13,1.5));
draw((15,0)--(17,0)--(17,3)--(15,3)--cycle);
label("6", (16,1.5));
draw((18,0)--(20,0)--(20,3)--(18,3)--cycle);
label("7", (19,1.5));
draw((21,0)--(23,0)--(23,3)--(21,3)--cycle);
label("8", (22,1.5));
draw((24,0)--(26,0)--(26,3)--(24,3)--cycle);
label("9", (25,1.5));
draw((27,0)--(29,0)--(29,3)--(27,3)--cycle);
label("10", (28,1.5));
draw((30,0)--(32,0)--(32,3)--(30,3)--cycle);
label("11", (31,1.5));
draw((33,0)--(35,0)--(35,3)--(33,3)--cycle);
label("12", (34,1.5));
draw((36,0)--(38,0)--(38,3)--(36,3)--cycle);
label("13", (37,1.5));
// Block 2
size(11cm); draw((0,0)--(2,0)--(2,3)--(0,3)--cycle); label("1", (1,1.5)); draw((3,0)--(5,0)--(5,3)--(3,3)--cycle); label("2", (4,1.5)); draw((6,0)--(8,0)--(8,3)--(6,3)--cycle); label("3", (7,1.5)); draw((9,0)--(11,0)--(11,3)--(9,3)--cycle); label("4", (10,1.5)); draw((12,0)--(14,0)--(14,3)--(12,3)--cycle); label("5", (13,1.5)); draw((15,0)--(17,0)--(17,3)--(15,3)--cycle); label("6", (16,1.5)); draw((18,0)--(20,0)--(20,3)--(18,3)--cycle); label("7", (19,1.5)); draw((21,0)--(23,0)--(23,3)--(21,3)--cycle); label("8", (22,1.5)); draw((24,0)--(26,0)--(26,3)--(24,3)--cycle); label("9", (25,1.5)); draw((27,0)--(29,0)--(29,3)--(27,3)--cycle); label("10", (28,1.5)); draw((30,0)--(32,0)--(32,3)--(30,3)--cycle); label("11", (31,1.5)); draw((33,0)--(35,0)--(35,3)--(33,3)--cycle); label("12", (34,1.5)); draw((36,0)--(38,0)--(38,3)--(36,3)--cycle); label("13", (37,1.5)); | [] |
913 | Isosceles trapezoid $ABCD$ has parallel sides $\overline{AD}$ and $\overline{BC},$ with $BC < AD$ and $AB = CD.$ There is a point $P$ in the plane such that $PA=1, PB=2, PC=3,$ and $PD=4.$ What is $\tfrac{BC}{AD}?$
$\textbf{(A) }\frac{1}{4}\qquad\textbf{(B) }\frac{1}{3}\qquad\textbf{(C) }\frac{1}{2}\qquad\textbf{(D) }\frac{2}{3}\qquad\textbf{(E) }\frac{3}{4}$ | 2022 AMC 10A Problem 23 | Consider the reflection $P^{\prime}$ of $P$ over the perpendicular bisector of $\overline{BC}$, creating two new isosceles trapezoids $DAPP^{\prime}$ and $CBPP^{\prime}$. Under this reflection, $P^{\prime}A=PD=4$, $P^{\prime}D=PA=1$, $P^{\prime}C=PB=2$, and $P^{\prime}B=PC=3$.
Since $DAPP'$ and $CBPP'$ are isosceles trapezoids, they are cyclic. Using Ptolemy's theorem on $DAPP'$, we get that $(PP')(AD) + (PA)(P'D) = (AP')(PD)$, so
\[PP' \cdot AD + 1 \cdot 1 = 4 \cdot 4.\] Then, using Ptolemy's theorem again on $CBPP'$, we get that $(BC)(PP') + (BP)(CP') = (BP')(CP)$, so
\[PP' \cdot BC + 2 \cdot 2 = 3 \cdot 3.\] Thus, $PP^{\prime}\cdot AD=15$ and $PP^{\prime}\cdot BC=5$; dividing these two equations and taking the reciprocal yields $\frac{BC}{AD}=\boxed{\textbf{(B) }\frac{1}{3}}$.
(diagram by cinnamon_e) | // Block 1
size(300);
pair A = (0,0);
pair B = (1, 2);
pair C = (2,2);
pair D = (3,0);
label("$A$", A, SW);
label("$B$", B, NW);
label("$C$", C, NE);
label("$D$", D, SE);
draw(A--B--C--D--cycle, blue);
pair P = (0.8, 0.6);
dot("$P$", P, NW);
draw(P--A, magenta);
draw(P--B, magenta);
draw(P--C);
draw(P--D);
label("$1$", P--A, NW);
label("$2$", P--B, E);
label("$3$", P--C, NW);
label("$4$", P--D, S);
pair P1 = (2.2, 0.6);
dot("$P'$", P1, NE);
draw(P1--D, magenta);
draw(P1--C, magenta);
draw(P1--A);
draw(P1--B);
label("$1$", P1--D, NE);
label("$2$", P1--C, E);
label("$3$", P1--B, NE);
label("$4$", P1--A, SE);
draw(P--P1, dashed+magenta);
// Block 2
size(300); pair A = (0,0); pair B = (1, 2); pair C = (2,2); pair D = (3,0); label("$A$", A, SW); label("$B$", B, NW); label("$C$", C, NE); label("$D$", D, SE); draw(A--B--C--D--cycle, blue); pair P = (0.8, 0.6); dot("$P$", P, NW); draw(P--A, magenta); draw(P--B, magenta); draw(P--C); draw(P--D); label("$1$", P--A, NW); label("$2$", P--B, E); label("$3$", P--C, NW); label("$4$", P--D, S); pair P1 = (2.2, 0.6); dot("$P'$", P1, NE); draw(P1--D, magenta); draw(P1--C, magenta); draw(P1--A); draw(P1--B); label("$1$", P1--D, NE); label("$2$", P1--C, E); label("$3$", P1--B, NE); label("$4$", P1--A, SE); draw(P--P1, dashed+magenta); | [] |
913 | Isosceles trapezoid $ABCD$ has parallel sides $\overline{AD}$ and $\overline{BC},$ with $BC < AD$ and $AB = CD.$ There is a point $P$ in the plane such that $PA=1, PB=2, PC=3,$ and $PD=4.$ What is $\tfrac{BC}{AD}?$
$\textbf{(A) }\frac{1}{4}\qquad\textbf{(B) }\frac{1}{3}\qquad\textbf{(C) }\frac{1}{2}\qquad\textbf{(D) }\frac{2}{3}\qquad\textbf{(E) }\frac{3}{4}$ | 2022 AMC 10A Problem 23 | Extend $AB$ and $CD$ to a point $Q$ as shown, and let $PQ = s$. Then let $BQ=CQ = a$ and $AB=DC = b$. Notice that $\frac{BC}{AD} = \frac{QC}{QD} = \frac{a}{a+b}$ by similar triangles.
By Stewart's theorem on $APQ$ and $DPQ$, we have \begin{align*} ab(a+b) + 9(a+b) &= 16a + s^2b \\ ab(a+b) + 4(a+b) &= a + s^2b \\ \end{align*}
Subtracting, $5(a+b) = 15a$, and so $\frac{BC}{AD} = \frac{a}{a+b} = \frac{5}{15} = \boxed{\textbf{(B) }\frac{1}{3}}$.
~kred9 (minor edit by gwang2008) | // Block 1
size(7.5cm);
draw((0,0)--(4.2,0));
draw((0,0)--(1.4,2)--(2.8,2)--(4.2,0));
draw((1.4,2)--(2.1,3)--(2.8,2));
draw((0,0)--(1,0.5)--(1.4,2)--(1,0.5)--(2.8,2)--(1,0.5)--(4.2,0));
label("$A$",(0,0),SW);
label("$B$",(1.4,2),NW);
label("$C$",(2.8,2),NE);
label("$D$",(4.2,0),SE);
label("$P$",(1,0.5),NW);
label("$Q$",(2.1,3),N);
draw((2.1,3)--(1,0.5));
// Block 2
size(7.5cm); draw((0,0)--(4.2,0)); draw((0,0)--(1.4,2)--(2.8,2)--(4.2,0)); draw((1.4,2)--(2.1,3)--(2.8,2)); draw((0,0)--(1,0.5)--(1.4,2)--(1,0.5)--(2.8,2)--(1,0.5)--(4.2,0)); label("$A$",(0,0),SW); label("$B$",(1.4,2),NW); label("$C$",(2.8,2),NE); label("$D$",(4.2,0),SE); label("$P$",(1,0.5),NW); label("$Q$",(2.1,3),N); draw((2.1,3)--(1,0.5)); | [] |
914 | Five rectangles, $A$, $B$, $C$, $D$, and $E$, are arranged in a square as shown below. These rectangles have dimensions $1\times6$, $2\times 4$, $5\times6$, $2\times7$, and $2\times3$, respectively. (The figure is not drawn to scale.) Which of the five rectangles is the shaded one in the middle?
$\textbf{(A) }A\qquad\textbf{(B) }B \qquad\textbf{(C) }C \qquad\textbf{(D) }D\qquad\textbf{(E) }E$ | 2022 AMC 12A Problem 3 | Note that rectangle $D$ must be on the edge. Without loss of generality, let the top-left rectangle be $D,$ as shown below:
It is clear that $x=1,$ so we can determine Rectangle $A.$
Continuing with a similar process, we can determine Rectangles $C,E,$ and $B,$ in this order. The answer is $\boxed{\textbf{(B) }B}$ as shown below.
~MRENTHUSIASM | // Block 1
size(175);
currentpen = black+1.25bp;
fill((3,2.5)--(3,4.5)--(5.3,4.5)--(5.3,2.5)--cycle,gray);
draw((0,0)--(7,0)--(7,7)--(0,7)--(0,0));
draw((3,0)--(3,4.5));
draw((0,4.5)--(5.3,4.5));
draw((5.3,7)--(5.3,2.5));
draw((7,2.5)--(3,2.5));
label("$7$",midpoint((0,7)--(5.3,7)),N,blue);
label("$x$",midpoint((5.3,7)--(7,7)),N,blue);
label("$2$",midpoint((0,4.5)--(0,7)),W,blue);
label("$x+5$",midpoint((0,0)--(0,4.5)),W,blue);
label("$D$",midpoint((0,7)--(5.3,4.5)),red);
// Block 2
size(175);
currentpen = black+1.25bp;
fill((3,2.5)--(3,4.5)--(5.3,4.5)--(5.3,2.5)--cycle,gray);
draw((0,0)--(7,0)--(7,7)--(0,7)--(0,0));
draw((3,0)--(3,4.5));
draw((0,4.5)--(5.3,4.5));
draw((5.3,7)--(5.3,2.5));
draw((7,2.5)--(3,2.5));
label("$7$",midpoint((0,7)--(5.3,7)),N,blue);
label("$1$",midpoint((5.3,7)--(7,7)),N,blue);
label("$2$",midpoint((0,4.5)--(0,7)),W,blue);
label("$6$",midpoint((0,0)--(0,4.5)),W,blue);
label("$6$",midpoint((7,7)--(7,2.5)),E,blue);
label("$2$",midpoint((7,2.5)--(7,0)),E,blue);
label("$2$",midpoint((5.3,4.5)--(5.3,7)),E,blue);
label("$4$",midpoint((5.3,4.5)--(5.3,2.5)),E,blue);
label("$1$",midpoint((5.3,2.5)--(7,2.5)),S,blue);
label("$5$",midpoint((0,4.5)--(3,4.5)),N,blue);
label("$2$",midpoint((3,4.5)--(5.3,4.5)),N,blue);
label("$5$",midpoint((0,0)--(3,0)),S,blue);
label("$3$",midpoint((3,0)--(7,0)),S,blue);
label("$2$",midpoint((3,0)--(3,2.5)),W,blue);
label("$4$",midpoint((3,2.5)--(3,4.5)),W,blue);
label("$2$",midpoint((3,2.5)--(5.3,2.5)),S,blue);
label("$D$",midpoint((0,7)--(5.3,4.5)),red);
label("$A$",midpoint((5.3,7)--(7,2.5)),red);
label("$C$",midpoint((0,4.5)--(3,0)),red);
label("$E$",midpoint((3,2.5)--(7,0)),red);
label("$B$",midpoint((3,4.5)--(5.3,2.5)),red);
// Block 3
size(175); currentpen = black+1.25bp; fill((3,2.5)--(3,4.5)--(5.3,4.5)--(5.3,2.5)--cycle,gray); draw((0,0)--(7,0)--(7,7)--(0,7)--(0,0)); draw((3,0)--(3,4.5)); draw((0,4.5)--(5.3,4.5)); draw((5.3,7)--(5.3,2.5)); draw((7,2.5)--(3,2.5)); label("$7$",midpoint((0,7)--(5.3,7)),N,blue); label("$x$",midpoint((5.3,7)--(7,7)),N,blue); label("$2$",midpoint((0,4.5)--(0,7)),W,blue); label("$x+5$",midpoint((0,0)--(0,4.5)),W,blue); label("$D$",midpoint((0,7)--(5.3,4.5)),red);
// Block 4
size(175); currentpen = black+1.25bp; fill((3,2.5)--(3,4.5)--(5.3,4.5)--(5.3,2.5)--cycle,gray); draw((0,0)--(7,0)--(7,7)--(0,7)--(0,0)); draw((3,0)--(3,4.5)); draw((0,4.5)--(5.3,4.5)); draw((5.3,7)--(5.3,2.5)); draw((7,2.5)--(3,2.5)); label("$7$",midpoint((0,7)--(5.3,7)),N,blue); label("$1$",midpoint((5.3,7)--(7,7)),N,blue); label("$2$",midpoint((0,4.5)--(0,7)),W,blue); label("$6$",midpoint((0,0)--(0,4.5)),W,blue); label("$6$",midpoint((7,7)--(7,2.5)),E,blue); label("$2$",midpoint((7,2.5)--(7,0)),E,blue); label("$2$",midpoint((5.3,4.5)--(5.3,7)),E,blue); label("$4$",midpoint((5.3,4.5)--(5.3,2.5)),E,blue); label("$1$",midpoint((5.3,2.5)--(7,2.5)),S,blue); label("$5$",midpoint((0,4.5)--(3,4.5)),N,blue); label("$2$",midpoint((3,4.5)--(5.3,4.5)),N,blue); label("$5$",midpoint((0,0)--(3,0)),S,blue); label("$3$",midpoint((3,0)--(7,0)),S,blue); label("$2$",midpoint((3,0)--(3,2.5)),W,blue); label("$4$",midpoint((3,2.5)--(3,4.5)),W,blue); label("$2$",midpoint((3,2.5)--(5.3,2.5)),S,blue); label("$D$",midpoint((0,7)--(5.3,4.5)),red); label("$A$",midpoint((5.3,7)--(7,2.5)),red); label("$C$",midpoint((0,4.5)--(3,0)),red); label("$E$",midpoint((3,2.5)--(7,0)),red); label("$B$",midpoint((3,4.5)--(5.3,2.5)),red); | [] |
914 | Five rectangles, $A$, $B$, $C$, $D$, and $E$, are arranged in a square as shown below. These rectangles have dimensions $1\times6$, $2\times 4$, $5\times6$, $2\times7$, and $2\times3$, respectively. (The figure is not drawn to scale.) Which of the five rectangles is the shaded one in the middle?
$\textbf{(A) }A\qquad\textbf{(B) }B \qquad\textbf{(C) }C \qquad\textbf{(D) }D\qquad\textbf{(E) }E$ | 2022 AMC 12A Problem 3 | Let's label some points:
By finding the dimensions of the middle rectangle, we need to find the dimensions of the other four rectangles. We have a rule: \[AB + BC = CD + DE = EF + FG = GH + AH.\]
Let's make a list of all dimensions of the rectangles from the diagram. We have to fill in the dimensions from up above, but still apply to the rule:
\begin{align*} AB&\times AH \\ CD&\times BC \\ EF&\times DE \\ GH&\times FG \end{align*}
By applying the rule, we get $AB=2, BC=6, CD=5, DE=3, EF=2, FG=6, GH=1$, and $AH=7$.
By substitution, we get this list
\begin{align*} 2&\times 7 \\ 5&\times 6 \\ 2&\times 3 \\ 1&\times 6 \\ \end{align*}
(This also tells us that the diagram is not drawn to scale.)
Notice how the only dimension not used in the list was $2\times 4$ and that corresponds with B so the answer is, $\boxed{\textbf{(B) }B}.$
~ghfhgvghj10 & Education, the study of everything. | // Block 1
size(175);
currentpen = black+1.25bp;
fill((3,2.5)--(3,4.5)--(5.3,4.5)--(5.3,2.5)--cycle,gray);
draw((0,0)--(7,0)--(7,7)--(0,7)--(0,0));
draw((3,0)--(3,4.5));
draw((0,4.5)--(5.3,4.5));
draw((5.3,7)--(5.3,2.5));
draw((7,2.5)--(3,2.5));
label("$A$",(0,0),SW);
label("$B$",(3,0),S);
label("$C$",(7,0),SE);
label("$D$",(7,2.5),(1,0));
label("$E$",(7,7),NE);
label("$F$",(5.5,7),N);
label("$G$",(0,7),NW);
label("$H$",(0,4.5),W);
// Block 2
size(175); currentpen = black+1.25bp; fill((3,2.5)--(3,4.5)--(5.3,4.5)--(5.3,2.5)--cycle,gray); draw((0,0)--(7,0)--(7,7)--(0,7)--(0,0)); draw((3,0)--(3,4.5)); draw((0,4.5)--(5.3,4.5)); draw((5.3,7)--(5.3,2.5)); draw((7,2.5)--(3,2.5)); label("$A$",(0,0),SW); label("$B$",(3,0),S); label("$C$",(7,0),SE); label("$D$",(7,2.5),(1,0)); label("$E$",(7,7),NE); label("$F$",(5.5,7),N); label("$G$",(0,7),NW); label("$H$",(0,4.5),W); | [] |
915 | The taxicab distance between points $(x_1, y_1)$ and $(x_2, y_2)$ in the coordinate plane is given by \[|x_1 - x_2| + |y_1 - y_2|.\]
For how many points $P$ with integer coordinates is the taxicab distance between $P$ and the origin less than or equal to $20$?
$\textbf{(A)} \, 441 \qquad\textbf{(B)} \, 761 \qquad\textbf{(C)} \, 841 \qquad\textbf{(D)} \, 921 \qquad\textbf{(E)} \, 924$ | 2022 AMC 12A Problem 5 | This solution refers to the Diagram section.
The problem can be visualized as depicted on the right split equally into four "triangular" parts excluding the origin. The "triangular" parts are identical the ones that would be used in a visual proof of the formula for triangular numbers. Becuase of this the number of points in each part is equal to $\frac{n(n+1)}{2}$ where $n$ is the length of a "leg" of the "triangle" which is $20$ for this problem. Substituting and computing, we get $210.$ Multiplying by $4$ and adding $1$ to account for all parts and the origin, we get $210\cdot4 + 1 = \boxed{\textbf{(C)} \, 841}.$
~Apersoma | // Block 1
/* Made by MRENTHUSIASM */
size(350);
for (int y = 20; y >= 1; --y) {
for (int x = 0; x <= 20-y; ++x) {
dot((x,y),green+linewidth(4));
}
}
for (int y = 0; y >= -20; --y) {
for (int x = 1; x <= y+20; ++x) {
dot((x,y),blue+linewidth(4));
}
}
for (int y = 20; y >= 0; --y) {
for (int x = y-20; x <= -1; ++x) {
dot((x,y),purple+linewidth(4));
}
}
for (int y = -1; y >= -20; --y) {
for (int x = -y-20; x <= 0; ++x) {
dot((x,y),red+linewidth(4));
}
}
dot(origin,black+linewidth(4));
// Block 2
/* Made by MRENTHUSIASM */ size(350); for (int y = 20; y >= 1; --y) { for (int x = 0; x <= 20-y; ++x) { dot((x,y),green+linewidth(4)); } } for (int y = 0; y >= -20; --y) { for (int x = 1; x <= y+20; ++x) { dot((x,y),blue+linewidth(4)); } } for (int y = 20; y >= 0; --y) { for (int x = y-20; x <= -1; ++x) { dot((x,y),purple+linewidth(4)); } } for (int y = -1; y >= -20; --y) { for (int x = -y-20; x <= 0; ++x) { dot((x,y),red+linewidth(4)); } } dot(origin,black+linewidth(4)); | [] |
915 | The taxicab distance between points $(x_1, y_1)$ and $(x_2, y_2)$ in the coordinate plane is given by \[|x_1 - x_2| + |y_1 - y_2|.\]
For how many points $P$ with integer coordinates is the taxicab distance between $P$ and the origin less than or equal to $20$?
$\textbf{(A)} \, 441 \qquad\textbf{(B)} \, 761 \qquad\textbf{(C)} \, 841 \qquad\textbf{(D)} \, 921 \qquad\textbf{(E)} \, 924$ | 2022 AMC 12A Problem 5 | This solution refers to the Diagram section.
As shown below, the taxicab distance between each red point and the origin is even, and the taxicab distance between each blue point and the origin is odd.
Note that the red array consists of $21^2=441$ points, and the blue array consists of $20^2=400$ points.
Together, the answer is $441+400=\boxed{\textbf{(C)} \, 841}.$
~MRENTHUSIASM | // Block 1
/* Made by MRENTHUSIASM */
size(350);
for (int y = 20; y >= 0; --y) {
for (int x = y-20; x <= 20-y; x+=2) {
dot((x,y),red+linewidth(4));
}
}
for (int y = -1; y >= -20; --y) {
for (int x = -y-20; x <= y+20; x+=2) {
dot((x,y),red+linewidth(4));
}
}
for (int y = 19; y >= 0; --y) {
for (int x = y-19; x <= 19-y; x+=2) {
dot((x,y),blue+linewidth(4));
}
}
for (int y = -1; y >= -19; --y) {
for (int x = -y-19; x <= y+19; x+=2) {
dot((x,y),blue+linewidth(4));
}
}
draw((20,0)--(0,20)--(-20,0)--(0,-20)--cycle,red+linewidth(1.25));
draw((19,0)--(0,19)--(-19,0)--(0,-19)--cycle,blue+linewidth(1.25));
// Block 2
/* Made by MRENTHUSIASM */ size(350); for (int y = 20; y >= 0; --y) { for (int x = y-20; x <= 20-y; x+=2) { dot((x,y),red+linewidth(4)); } } for (int y = -1; y >= -20; --y) { for (int x = -y-20; x <= y+20; x+=2) { dot((x,y),red+linewidth(4)); } } for (int y = 19; y >= 0; --y) { for (int x = y-19; x <= 19-y; x+=2) { dot((x,y),blue+linewidth(4)); } } for (int y = -1; y >= -19; --y) { for (int x = -y-19; x <= y+19; x+=2) { dot((x,y),blue+linewidth(4)); } } draw((20,0)--(0,20)--(-20,0)--(0,-20)--cycle,red+linewidth(1.25)); draw((19,0)--(0,19)--(-19,0)--(0,-19)--cycle,blue+linewidth(1.25)); | [] |
916 | Let $\mathcal{R}$ be the region in the complex plane consisting of all complex numbers $z$ that can be written as the sum of complex numbers $z_1$ and $z_2$, where $z_1$ lies on the segment with endpoints $3$ and $4i$, and $z_2$ has magnitude at most $1$. What integer is closest to the area of $\mathcal{R}$?
$\textbf{(A) } 13 \qquad \textbf{(B) } 14 \qquad \textbf{(C) } 15 \qquad \textbf{(D) } 16 \qquad \textbf{(E) } 17$ | 2022 AMC 12A Problem 13 | If $z$ is a complex number and $z = a + bi$, then the magnitude (length) of $z$ is $\sqrt{a^2 + b^2}$. Therefore, $z_1$ has a magnitude of 5. If $z_2$ has a magnitude of at most one, that means for each point on the segment given by $z_1$, the bounds of the region $\mathcal{R}$ could be at most 1 away. Alone the line, excluding the endpoints, a rectangle with a width of 2 and a length of 5, the magnitude, would be formed. At the endpoints, two semicircles will be formed with a radius of 1 for a total area of $\pi \approx 3$.
Therefore, the total area is $5(2) + \pi \approx 10 + 3 = \boxed{\textbf{(A) } 13}$.
~juicefruit | // Block 1
size(250);
import TrigMacros;
rr_cartesian_axes(-2,6,-2,6,complexplane=true, usegrid = true);
Label f;
f.p=fontsize(6);
xaxis(-1,5,Ticks(f, 1.0));
yaxis(-1,5,Ticks(f, 1.0));
dot((3,0));
dot((0,4));
draw((0,4)--(3,0), blue);
draw((0.8, 4.6)..(-.6,4.8)..(-.8, 3.4),red);
draw((-.8, 3.4)--(2.2, -0.6), red);
draw((2.2, -0.6)..(3.6,-0.8)..(3.8,0.6), red);
draw((0.8, 4.6)--(3.8,0.6),red);
draw((0.8, 4.6)--(-.8, 3.4),red+dashed);
draw((2.2, -0.6)--(3.8,0.6),red+ dashed);
draw((3,0)--(3,-1),Arrow);
label("1",(3,0)--(3,-1),E);
draw((0,4)--(-.6,4.8),Arrow);
label("1",(0,4)--(-.6,4.8),SW);
draw((1.5,2)--(2.3,2.6),Arrow);
label("1",(1.5,2)--(2.3,2.6),SE);
// Block 2
size(250); import TrigMacros; rr_cartesian_axes(-2,6,-2,6,complexplane=true, usegrid = true); Label f; f.p=fontsize(6); xaxis(-1,5,Ticks(f, 1.0)); yaxis(-1,5,Ticks(f, 1.0)); dot((3,0)); dot((0,4)); draw((0,4)--(3,0), blue); draw((0.8, 4.6)..(-.6,4.8)..(-.8, 3.4),red); draw((-.8, 3.4)--(2.2, -0.6), red); draw((2.2, -0.6)..(3.6,-0.8)..(3.8,0.6), red); draw((0.8, 4.6)--(3.8,0.6),red); draw((0.8, 4.6)--(-.8, 3.4),red+dashed); draw((2.2, -0.6)--(3.8,0.6),red+ dashed); draw((3,0)--(3,-1),Arrow); label("1",(3,0)--(3,-1),E); draw((0,4)--(-.6,4.8),Arrow); label("1",(0,4)--(-.6,4.8),SW); draw((1.5,2)--(2.3,2.6),Arrow); label("1",(1.5,2)--(2.3,2.6),SE); | [] |
917 | Suppose that $13$ cards numbered $1, 2, 3, \ldots, 13$ are arranged in a row. The task is to pick them up in numerically increasing order, working repeatedly from left to right. In the example below, cards $1, 2, 3$ are picked up on the first pass, $4$ and $5$ on the second pass, $6$ on the third pass, $7, 8, 9, 10$ on the fourth pass, and $11, 12, 13$ on the fifth pass. For how many of the $13!$ possible orderings of the cards will the $13$ cards be picked up in exactly two passes?
$\textbf{(A) } 4082 \qquad \textbf{(B) } 4095 \qquad \textbf{(C) } 4096 \qquad \textbf{(D) } 8178 \qquad \textbf{(E) } 8191$ | 2022 AMC 12A Problem 19 | For $1\leq k\leq 12,$ suppose that cards $1, 2, \ldots, k$ are picked up on the first pass. It follows that cards $k+1,k+2,\ldots,13$ are picked up on the second pass.
Once we pick the spots for the cards on the first pass, there is only one way to arrange all $\boldsymbol{13}$ cards.
For each value of $k,$ there are $\binom{13}{k}-1$ ways to pick the $k$ spots for the cards on the first pass: We exclude the arrangement
in which the cards are arranged such that the first pass consists of all $13$ cards.
Therefore, the answer is \[\sum_{k=1}^{12}\left[\binom{13}{k}-1\right] = \left[\sum_{k=1}^{12}\binom{13}{k}\right]-12 = \left[\sum_{k=0}^{13}\binom{13}{k}\right]-14 = 2^{13} - 14 = \boxed{\textbf{(D) } 8178}.\]
~MRENTHUSIASM | size(11cm); draw((0,0)--(2,0)--(2,3)--(0,3)--cycle); label("1", (1,1.5)); draw((3,0)--(5,0)--(5,3)--(3,3)--cycle); label("2", (4,1.5)); draw((6,0)--(8,0)--(8,3)--(6,3)--cycle); label("3", (7,1.5)); draw((9,0)--(11,0)--(11,3)--(9,3)--cycle); label("4", (10,1.5)); draw((12,0)--(14,0)--(14,3)--(12,3)--cycle); label("5", (13,1.5)); draw((15,0)--(17,0)--(17,3)--(15,3)--cycle); label("6", (16,1.5)); draw((18,0)--(20,0)--(20,3)--(18,3)--cycle); label("7", (19,1.5)); draw((21,0)--(23,0)--(23,3)--(21,3)--cycle); label("8", (22,1.5)); draw((24,0)--(26,0)--(26,3)--(24,3)--cycle); label("9", (25,1.5)); draw((27,0)--(29,0)--(29,3)--(27,3)--cycle); label("10", (28,1.5)); draw((30,0)--(32,0)--(32,3)--(30,3)--cycle); label("11", (31,1.5)); draw((33,0)--(35,0)--(35,3)--(33,3)--cycle); label("12", (34,1.5)); draw((36,0)--(38,0)--(38,3)--(36,3)--cycle); label("13", (37,1.5)); | [] |
918 | Isosceles trapezoid $ABCD$ has parallel sides $\overline{AD}$ and $\overline{BC},$ with $BC < AD$ and $AB = CD.$ There is a point $P$ in the plane such that $PA=1, PB=2, PC=3,$ and $PD=4.$ What is $\tfrac{BC}{AD}?$
$\textbf{(A) }\frac{1}{4}\qquad\textbf{(B) }\frac{1}{3}\qquad\textbf{(C) }\frac{1}{2}\qquad\textbf{(D) }\frac{2}{3}\qquad\textbf{(E) }\frac{3}{4}$ | 2022 AMC 12A Problem 20 | Consider the reflection $P^{\prime}$ of $P$ over the perpendicular bisector of $\overline{BC}$, creating two new isosceles trapezoids $DAPP^{\prime}$ and $CBPP^{\prime}$. Under this reflection, $P^{\prime}A=PD=4$, $P^{\prime}D=PA=1$, $P^{\prime}C=PB=2$, and $P^{\prime}B=PC=3$.
Since $DAPP'$ and $CBPP'$ are isosceles trapezoids, they are cyclic. Using Ptolemy's theorem on $DAPP'$, we get that $(PP')(AD) + (PA)(P'D) = (AP')(PD)$, so
\[PP' \cdot AD + 1 \cdot 1 = 4 \cdot 4.\] Then, using Ptolemy's theorem again on $CBPP'$, we get that $(BC)(PP') + (BP)(CP') = (BP')(CP)$, so
\[PP' \cdot BC + 2 \cdot 2 = 3 \cdot 3.\] Thus, $PP^{\prime}\cdot AD=15$ and $PP^{\prime}\cdot BC=5$; dividing these two equations and taking the reciprocal yields $\frac{BC}{AD}=\boxed{\textbf{(B) }\frac{1}{3}}$.
(diagram by cinnamon_e) | size(300); pair A = (0,0); pair B = (1, 2); pair C = (2,2); pair D = (3,0); label("$A$", A, SW); label("$B$", B, NW); label("$C$", C, NE); label("$D$", D, SE); draw(A--B--C--D--cycle, blue); pair P = (0.8, 0.6); dot("$P$", P, NW); draw(P--A, magenta); draw(P--B, magenta); draw(P--C); draw(P--D); label("$1$", P--A, NW); label("$2$", P--B, E); label("$3$", P--C, NW); label("$4$", P--D, S); pair P1 = (2.2, 0.6); dot("$P'$", P1, NE); draw(P1--D, magenta); draw(P1--C, magenta); draw(P1--A); draw(P1--B); label("$1$", P1--D, NE); label("$2$", P1--C, E); label("$3$", P1--B, NE); label("$4$", P1--A, SE); draw(P--P1, dashed+magenta); | [] |
918 | Isosceles trapezoid $ABCD$ has parallel sides $\overline{AD}$ and $\overline{BC},$ with $BC < AD$ and $AB = CD.$ There is a point $P$ in the plane such that $PA=1, PB=2, PC=3,$ and $PD=4.$ What is $\tfrac{BC}{AD}?$
$\textbf{(A) }\frac{1}{4}\qquad\textbf{(B) }\frac{1}{3}\qquad\textbf{(C) }\frac{1}{2}\qquad\textbf{(D) }\frac{2}{3}\qquad\textbf{(E) }\frac{3}{4}$ | 2022 AMC 12A Problem 20 | Extend $AB$ and $CD$ to a point $Q$ as shown, and let $PQ = s$. Then let $BQ=CQ = a$ and $AB=DC = b$. Notice that $\frac{BC}{AD} = \frac{QC}{QD} = \frac{a}{a+b}$ by similar triangles.
By Stewart's theorem on $APQ$ and $DPQ$, we have \begin{align*} ab(a+b) + 9(a+b) &= 16a + s^2b \\ ab(a+b) + 4(a+b) &= a + s^2b \\ \end{align*}
Subtracting, $5(a+b) = 15a$, and so $\frac{BC}{AD} = \frac{a}{a+b} = \frac{5}{15} = \boxed{\textbf{(B) }\frac{1}{3}}$.
~kred9 (minor edit by gwang2008) | size(7.5cm); draw((0,0)--(4.2,0)); draw((0,0)--(1.4,2)--(2.8,2)--(4.2,0)); draw((1.4,2)--(2.1,3)--(2.8,2)); draw((0,0)--(1,0.5)--(1.4,2)--(1,0.5)--(2.8,2)--(1,0.5)--(4.2,0)); label("$A$",(0,0),SW); label("$B$",(1.4,2),NW); label("$C$",(2.8,2),NE); label("$D$",(4.2,0),SE); label("$P$",(1,0.5),NW); label("$Q$",(2.1,3),N); draw((2.1,3)--(1,0.5)); | [] |
919 | Let $c$ be a real number, and let $z_1$ and $z_2$ be the two complex numbers satisfying the equation
$z^2 - cz + 10 = 0$. Points $z_1$, $z_2$, $\frac{1}{z_1}$, and $\frac{1}{z_2}$ are the vertices of (convex) quadrilateral $\mathcal{Q}$ in the complex plane. When the area of $\mathcal{Q}$ obtains its maximum possible value, $c$ is closest to which of the following?
$\textbf{(A) }4.5 \qquad\textbf{(B) }5 \qquad\textbf{(C) }5.5 \qquad\textbf{(D) }6\qquad\textbf{(E) }6.5$ | 2022 AMC 12A Problem 22 | Note that $z=\dfrac c2\pm\dfrac{\sqrt{c^2-40}}2$, so let $z_1=\dfrac c2+\dfrac{\sqrt{c^2-40}}2$ and $z_2=\dfrac c2-\dfrac{\sqrt{c^2-40}}2$. Taking a look at the answer choices, they range between $c=4.5$ to $c=6.5$, and in that range, $c^2$ is always less than $40$. Thus, $c^2-40<0$ for our possible answer choices; we can then rewrite $z_1$ and $z_2$ as $\dfrac c2+\dfrac{\sqrt{40-c^2}}2i$ and $\dfrac c2-\dfrac{\sqrt{40-c^2}}2i$, respectively, with real coefficients.
Let us compute $\dfrac1z$:
\[\dfrac1z=\dfrac1{\frac c2\pm\frac{\sqrt{c^2-40}}2}=\dfrac{\frac c2\mp\frac{\sqrt{c^2-40}}2}{\left(\frac c2\right)^2-\left(\frac{\sqrt{c^2-40}}2\right)^2}=\dfrac{\frac c2\mp\frac{\sqrt{c^2-40}}2}{\frac{c^2}4-\frac{c^2-40}4}=\dfrac{2c\mp2\sqrt{c^2-40}}{40}=\dfrac{c\mp\sqrt{c^2-40}}{20}.\]
Then, $\dfrac1{z_1}=\dfrac{c-\sqrt{c^2-40}}{20}=\dfrac c{20}-\dfrac{\sqrt{40-c^2}}{20}i$ while $\dfrac1{z_2}=\dfrac{c+\sqrt{c^2-40}}{20}=\dfrac c{20}+\dfrac{\sqrt{40-c^2}}{20}i$.
In the complex plane, we can draw a rough sketch of $z_1,z_2,\dfrac1{z_1},\dfrac1{z_2}$:
Note that, here, our quadrilateral is an isoceles trapezoid. The shorter base length is $\left(\dfrac{\sqrt{40-c^2}}{20}-\left(-\dfrac{\sqrt{40-c^2}}{20}\right)\right)=\dfrac1{10}\sqrt{40-c^2}$.
The longer base length is $\left(\dfrac{\sqrt{40-c^2}}2-\left(-\dfrac{\sqrt{40-c^2}}2\right)\right)=\sqrt{40-c^2}$.
The average of the two bases is $\dfrac{11}{20}\sqrt{40-c^2}$.
The height of our trapezoid (which is horizontal parallel to the $x$-axis in our diagram above) is simply $\dfrac c2-\dfrac c{20}=\dfrac9{20}c$.
Since the area of a trapezoid is the product of the average of its bases and its height, we conclude that our trapezoid's area is $\dfrac{99}{400}c\sqrt{40-c^2}$, which is a function of $c$. Thus, let $A(c)=\dfrac{99}{400}c\sqrt{40-c^2}$.
Taking the derivative (FINALLY we get to the Calculus part haha, this definitely wasn't clickbait :3 trust me) with respect to $c$, we find that $\dfrac{dA}{dc}=\dfrac{99}{400}\left(\sqrt{40-c^2}+c\left(\dfrac{-2c}{2\sqrt{40-c^2}}\right)\right)=\dfrac{99}{400}\left(\sqrt{40-c^2}-\dfrac{c^2}{\sqrt{40-c^2}}\right)$.
To find an extremum, we set the derivative equal to zero:
\begin{align*}
\dfrac{dA}{dc}&=\dfrac{99}{400}\left(\sqrt{40-c^2}-\dfrac{c^2}{\sqrt{40-c^2}}\right) \\
0&=\dfrac{99}{400}\left(\sqrt{40-c^2}-\dfrac{c^2}{\sqrt{40-c^2}}\right) \\
\sqrt{40-c^2}-\dfrac{c^2}{\sqrt{40-c^2}}&=0 \\
\sqrt{40-c^2}&=\dfrac{c^2}{\sqrt{40-c^2}} \\
\left(\sqrt{40-c^2}\right)^2&=c^2 \\
c^2&=40-c^2 \\
2c^2&=40 \\
c^2&=20 \\
c&=\sqrt{20} \\
&\approx4.47 \\
\end{align*}
Clearly, this is very close to $\boxed{\textbf{(A)}~4.5}$, so we are done. QED.
~Technodoggo
(Minor Edits by dolphindesigner) | // Block 1
import graph;
unitsize(0.5cm);
/*xaxis(Ticks, xmin=-1,xmax=8);
yaxis(Ticks, ymin=-11,ymax=11);*/
draw((-1,0)--(8,0));
draw((0,-11)--(0,11));
dot((7,10));dot((7,-10));dot((0.7,1));dot((0.7,-1));
draw((7,10)--(7,-10)--(0.7,-1)--(0.7,1)--cycle);
label("$z_1$", (7,10), E);
label("$z_2$", (7,-10), E);
label("$\frac1{z_2}$", (0.7,1), N);
label("$\frac1{z_1}$", (0.7,-1), S);
// Block 2
import graph; unitsize(0.5cm); /*xaxis(Ticks, xmin=-1,xmax=8); yaxis(Ticks, ymin=-11,ymax=11);*/ draw((-1,0)--(8,0)); draw((0,-11)--(0,11)); dot((7,10));dot((7,-10));dot((0.7,1));dot((0.7,-1)); draw((7,10)--(7,-10)--(0.7,-1)--(0.7,1)--cycle); label("$z_1$", (7,10), E); label("$z_2$", (7,-10), E); label("$\frac1{z_2}$", (0.7,1), N); label("$\frac1{z_1}$", (0.7,-1), S); | [] |
920 | The diagram below shows a rectangle with side lengths $4$ and $8$ and a square with side length $5$. Three vertices of the square lie on three different sides of the rectangle, as shown. What is the area of the region inside both the square and the rectangle?
$\textbf{(A) }15\dfrac{1}{8} \qquad \textbf{(B) }15\dfrac{3}{8} \qquad \textbf{(C) }15\dfrac{1}{2} \qquad \textbf{(D) }15\dfrac{5}{8} \qquad \textbf{(E) }15\dfrac{7}{8}$ | 2022 AMC 10B Problem 16 | Let us label the points on the diagram.
By doing some angle chasing using the fact that $\angle ACE$ and $\angle CEG$ are right angles, we find that $\angle BAC = \angle DCE = \angle FEG$. Similarly, $\angle ACB = \angle CED = \angle EGF$. Therefore, $\triangle ABC \sim \triangle CDE \sim \triangle EFG$.
As we are given a rectangle and a square, $AB = 4$ and $AC = 5$. Therefore, $\triangle ABC$ is a $3$-$4$-$5$ right triangle and $BC = 3$.
$CE$ is also $5$. So, using the similar triangles, $CD = 4$ and $DE = 3$.
$EF = DF - DE = 4 - 3 = 1$. Using the similar triangles again, $EF$ is $\frac14$ of the corresponding $AB$. So,
\begin{align*} [\triangle EFG] &= \left(\frac14\right)^2 \cdot [\triangle ABC] \\ &= \frac{1}{16} \cdot 6 \\ &= \frac38. \end{align*}
Finally, we have
\begin{align*} [ACEG] &= [ABDF] - [\triangle ABC] - [\triangle CDE] - [\triangle EFG] \\ &= 7 \cdot 4 - \frac12 \cdot 3 \cdot 4 - \frac12 \cdot 3 \cdot 4 - \frac38 \\ &= 28 - 6 - 6 - \frac38 \\ &= \boxed{\textbf{(D) }15\dfrac{5}{8}}. \end{align*}
~Connor132435 | // Block 1
import olympiad;
size(200);
defaultpen(linewidth(1) + fontsize(10));
pair A = (0,0), B = (1,0), C = (4,0), D = (8,0), K = (0,4), F = (1,4), G = (7.25, 4), H = (8, 4), I = (8,3), J = (5, 7);
fill(F--G--I--C--F--cycle, grey);
markscalefactor=0.05;
draw(A--D--H--K--A^^B--F^^F--C--I--J--F^^rightanglemark(F,J,I)^^rightanglemark(F,B,C)^^anglemark(D,C,I)^^anglemark(B,F,C)^^anglemark(H,I,G));
draw(anglemark(F,C,B)^^anglemark(C,I,D)^^anglemark(I,G,H));
markscalefactor=0.041;
draw(anglemark(F,C,B)^^anglemark(C,I,D)^^anglemark(I,G,H));
label("8",(4,-.5),S);
label("5",(3, 5.5),NW);
label("4",(8.25, 2), E);
label("A", F, NW);
label("B", B, S);
label("C", C, S);
label("D", D, SE);
label("E", I, E);
label("F", H, NE);
label("G", G, NE);
label("4", (1,2), E);
label("5", (2.5,2), SW);
label("3", (2.5,0), S);
label("4", (6,0), S);
label("5", (6,1.5), SE);
label("3", (8, 1.5), E);
label("1", (8, 3.5), E);
// Block 2
import olympiad; size(200); defaultpen(linewidth(1) + fontsize(10)); pair A = (0,0), B = (1,0), C = (4,0), D = (8,0), K = (0,4), F = (1,4), G = (7.25, 4), H = (8, 4), I = (8,3), J = (5, 7); fill(F--G--I--C--F--cycle, grey); markscalefactor=0.05; draw(A--D--H--K--A^^B--F^^F--C--I--J--F^^rightanglemark(F,J,I)^^rightanglemark(F,B,C)^^anglemark(D,C,I)^^anglemark(B,F,C)^^anglemark(H,I,G)); draw(anglemark(F,C,B)^^anglemark(C,I,D)^^anglemark(I,G,H)); markscalefactor=0.041; draw(anglemark(F,C,B)^^anglemark(C,I,D)^^anglemark(I,G,H)); label("8",(4,-.5),S); label("5",(3, 5.5),NW); label("4",(8.25, 2), E); label("A", F, NW); label("B", B, S); label("C", C, S); label("D", D, SE); label("E", I, E); label("F", H, NE); label("G", G, NE); label("4", (1,2), E); label("5", (2.5,2), SW); label("3", (2.5,0), S); label("4", (6,0), S); label("5", (6,1.5), SE); label("3", (8, 1.5), E); label("1", (8, 3.5), E); | [] |
920 | The diagram below shows a rectangle with side lengths $4$ and $8$ and a square with side length $5$. Three vertices of the square lie on three different sides of the rectangle, as shown. What is the area of the region inside both the square and the rectangle?
$\textbf{(A) }15\dfrac{1}{8} \qquad \textbf{(B) }15\dfrac{3}{8} \qquad \textbf{(C) }15\dfrac{1}{2} \qquad \textbf{(D) }15\dfrac{5}{8} \qquad \textbf{(E) }15\dfrac{7}{8}$ | 2022 AMC 10B Problem 16 | Same diagram as Solution 1, but added point $H$, which is $(4,7)$. I also renamed all the points to form coordinates using $B$ as the origin.
In order to find the area, point $G$'s coordinates must be found. Notice how $EH$ and $AG$ intercept at point $G$. This means that we need to find the equations for $EH$ and $AG$ and make a system of linear equations.
Using the slope formula $m=\frac{y_{2} - y_{1}}{x_{2} - x_{1}}$, we get the slope for $EH$, which means $m=\frac{3-7}{7-4} = -\frac{4}{3}$
Then, by using point slope form. $y-y_{1}=m(x-x_{1})$. We can say that the equation for $EH$ is $y-7=-\frac{4}{3}(x-4)$ or in this case, $y=-\frac{4}{3}x+12 \frac{1}{3}$.
And it is easy to figure out that the equation for $AG$ is $y=4$.
The best way to solve the system of linear equations is to substitute the $y$ for the $4$ in equation $EH$.
$4=-\frac{4}{3}x+12 \frac{1}{3}$, so $x=6\frac{1}{4}$ and $y=4$ This would mean $G\left(6\frac{1}{4},4\right)$.
Since we have our $G$ coordinate, we can continue with Solution 3, with the area of the trapezoid $\left(\frac{EG+AC}{2}\right)(CE)$, where $EG=\frac{5}{4}$ (using distance formula for $E$ to $G$), $AC=5$, and $CE=5$.
By substitution, we get $\left(\frac{\frac{5}{4}+5}{2}\right)(5)=$$\boxed{\textbf{(D) }15\dfrac{5}{8}}$.
~ghfhgvghj10 (+ minor edits ~TaeKim) | // Block 1
import olympiad;
size(200);
defaultpen(linewidth(1) + fontsize(10));
pair A = (0,0), B = (1,0), C = (4,0), D = (8,0), K = (0,4), F = (1,4), G = (7.25, 4), H = (8, 4), I = (8,3), J = (5, 7);
fill(F--G--I--C--F--cycle, grey);
markscalefactor=0.05;
draw(A--D--H--K--A^^B--F^^F--C--I--J--F^^rightanglemark(F,J,I)^^rightanglemark(F,B,C)^^anglemark(D,C,I)^^anglemark(B,F,C)^^anglemark(H,I,G));
draw(anglemark(F,C,B)^^anglemark(C,I,D)^^anglemark(I,G,H));
markscalefactor=0.041;
draw(anglemark(F,C,B)^^anglemark(C,I,D)^^anglemark(I,G,H));
label("8",(4,-.5),S);
label("5",(3, 5.5),NW);
label("4",(8.25, 2), E);
label("A(0,4)", F, NW);
label("B(0,0)", B, S);
label("C(3,0)", C, S);
label("D(7,0)", D, SE);
label("E(7,3)", I, E);
label("F(7,4)", H, NE);
label("G", G, NE);
label("4", (1,2), E);
label("5", (2.5,2), SW);
label("3", (2.5,0), S);
label("4", (6,0), S);
label("5", (6,1.5), SE);
label("3", (8, 1.5), E);
label("1", (8, 3.5), E);
label("H(4,7)", (4.65, 7.25), E);
// Block 2
import olympiad; size(200); defaultpen(linewidth(1) + fontsize(10)); pair A = (0,0), B = (1,0), C = (4,0), D = (8,0), K = (0,4), F = (1,4), G = (7.25, 4), H = (8, 4), I = (8,3), J = (5, 7); fill(F--G--I--C--F--cycle, grey); markscalefactor=0.05; draw(A--D--H--K--A^^B--F^^F--C--I--J--F^^rightanglemark(F,J,I)^^rightanglemark(F,B,C)^^anglemark(D,C,I)^^anglemark(B,F,C)^^anglemark(H,I,G)); draw(anglemark(F,C,B)^^anglemark(C,I,D)^^anglemark(I,G,H)); markscalefactor=0.041; draw(anglemark(F,C,B)^^anglemark(C,I,D)^^anglemark(I,G,H)); label("8",(4,-.5),S); label("5",(3, 5.5),NW); label("4",(8.25, 2), E); label("A(0,4)", F, NW); label("B(0,0)", B, S); label("C(3,0)", C, S); label("D(7,0)", D, SE); label("E(7,3)", I, E); label("F(7,4)", H, NE); label("G", G, NE); label("4", (1,2), E); label("5", (2.5,2), SW); label("3", (2.5,0), S); label("4", (6,0), S); label("5", (6,1.5), SE); label("3", (8, 1.5), E); label("1", (8, 3.5), E); label("H(4,7)", (4.65, 7.25), E); | [] |
921 | Each square in a $5 \times 5$ grid is either filled or empty, and has up to eight adjacent neighboring squares, where neighboring squares share either a side or a corner. The grid is transformed by the following rules:
Any filled square with two or three filled neighbors remains filled.
Any empty square with exactly three filled neighbors becomes a filled square.
All other squares remain empty or become empty.
A sample transformation is shown in the figure below.
Suppose the $5 \times 5$ grid has a border of empty squares surrounding a $3 \times 3$ subgrid. How many initial configurations will lead to a transformed grid consisting of a single filled square in the center after a single transformation? (Rotations and reflections of the same configuration are considered different.)
$\textbf{(A)}\ 14 \qquad\textbf{(B)}\ 18 \qquad\textbf{(C)}\ 22 \qquad\textbf{(D)}\ 26 \qquad\textbf{(E)}\ 30$ | 2022 AMC 10B Problem 19 | There are two cases for the initial configuration:
The center square is filled.
Exactly two of the eight adjacent neighboring squares of the center are filled. Clearly, the only possibility is that the squares along one diagonal are filled, as shown below:
In this case, there are $2$ possible initial configurations. All rotations and reflections are considered.
The center square is empty.
Exactly three of the eight adjacent neighboring squares of the center are filled. The possibilities are shown below:
In this case, there are $4+4+4+8=20$ possible initial configurations. All rotations and reflections are considered.
Together, the answer is $2+20=\boxed{\textbf{(C)}\ 22}.$
~mathboy100 ~MRENTHUSIASM | // Block 1
import geometry;
unitsize(0.6cm);
void ds(pair x) {
filldraw(x -- (1,0) + x -- (1,1) + x -- (0,1)+x -- cycle,mediumgray,invisible);
}
ds((1,3));
ds((2,2));
ds((3,1));
for (int i = 0; i <= 5; ++i) {
draw((0,i)--(5,i));
draw((i,0)--(i,5));
}
label("$2$ Configurations", (2.5,-1));
// Block 2
import geometry;
unitsize(0.6cm);
void ds(pair x) {
filldraw(x -- (1,0) + x -- (1,1) + x -- (0,1)+x -- cycle,mediumgray,invisible);
}
ds((1,3));
ds((3,3));
ds((1,1));
for (int i = 0; i <= 5; ++i) {
draw((0,i)--(5,i));
draw((i,0)--(i,5));
}
ds((10,3));
ds((12,3));
ds((11,1));
for (int i = 0; i <= 5; ++i) {
draw((9,i)--(14,i));
draw((i+9,0)--(i+9,5));
}
ds((19,3));
ds((20,1));
ds((21,2));
for (int i = 0; i <= 5; ++i) {
draw((18,i)--(23,i));
draw((i+18,0)--(i+18,5));
}
ds((28,3));
ds((29,1));
ds((30,1));
for (int i = 0; i <= 5; ++i) {
draw((27,i)--(32,i));
draw((i+27,0)--(i+27,5));
}
label("$4$ Configurations", (2.5,-1));
label("$4$ Configurations", (11.5,-1));
label("$4$ Configurations", (20.5,-1));
label("$8$ Configurations", (29.5,-1));
// Block 3
import geometry; unitsize(0.6cm); void ds(pair x) { filldraw(x -- (1,0) + x -- (1,1) + x -- (0,1)+x -- cycle,mediumgray,invisible); } ds((1,3)); ds((2,2)); ds((3,1)); for (int i = 0; i <= 5; ++i) { draw((0,i)--(5,i)); draw((i,0)--(i,5)); } label("$2$ Configurations", (2.5,-1));
// Block 4
import geometry; unitsize(0.6cm); void ds(pair x) { filldraw(x -- (1,0) + x -- (1,1) + x -- (0,1)+x -- cycle,mediumgray,invisible); } ds((1,3)); ds((3,3)); ds((1,1)); for (int i = 0; i <= 5; ++i) { draw((0,i)--(5,i)); draw((i,0)--(i,5)); } ds((10,3)); ds((12,3)); ds((11,1)); for (int i = 0; i <= 5; ++i) { draw((9,i)--(14,i)); draw((i+9,0)--(i+9,5)); } ds((19,3)); ds((20,1)); ds((21,2)); for (int i = 0; i <= 5; ++i) { draw((18,i)--(23,i)); draw((i+18,0)--(i+18,5)); } ds((28,3)); ds((29,1)); ds((30,1)); for (int i = 0; i <= 5; ++i) { draw((27,i)--(32,i)); draw((i+27,0)--(i+27,5)); } label("$4$ Configurations", (2.5,-1)); label("$4$ Configurations", (11.5,-1)); label("$4$ Configurations", (20.5,-1)); label("$8$ Configurations", (29.5,-1)); | [] |
922 | Let $ABCD$ be a rhombus with $\angle ADC = 46^\circ$. Let $E$ be the midpoint of $\overline{CD}$, and let $F$ be the point
on $\overline{BE}$ such that $\overline{AF}$ is perpendicular to $\overline{BE}$. What is the degree measure of $\angle BFC$?
$\textbf{(A)}\ 110 \qquad\textbf{(B)}\ 111 \qquad\textbf{(C)}\ 112 \qquad\textbf{(D)}\ 113 \qquad\textbf{(E)}\ 114$ | 2022 AMC 10B Problem 20 | This solution refers to the Diagram section.
We extend $AD$ and $BE$ to point $G$, as shown below:
We know that $AB=AD=2$ and $CE=DE=1$.
By AA Similarity, $\triangle ABG \sim \triangle DEG$ with a ratio of $2:1$. This implies that $2AD=AG$ and $AD \cong DG$, so $AG=2AD=2\cdot2=4$. That is, $D$ is the midpoint of $AG$.
Note that as $\angle{AFG}$ has an angle of 90 deg and $AG=2DG$, we can redraw our previous diagram, but construct a circle with radius $AD$ or $2$ centered at $D$ and by extending $CD$ to point $H$, which is on the circle, as shown below:
Notice how $F$ and $C$ are on the circle and that $\angle CFE$ intercepts with $\overset{\Large\frown} {CG}$.
Let's call $\angle CFE = \theta$.
Note that $\angle CDG$ also intercepts $\overset{\Large\frown} {CG}$, So $\angle CDG = 2\angle CFE$.
Let $\angle CDG = 2\theta$. Notice how $\angle CDG$ and $\angle ADC$ are supplementary to each other. We conclude that \begin{align*} 2\theta &= 180-\angle ADC \\ 2\theta &= 180-46 \\ 2\theta &= 134 \\ \theta &= 67. \end{align*}
Since $\angle BFC=180-\theta$, we have $\angle BFC=180-67=\boxed{\textbf{(D)} \ 113}$.
~ghfhgvghj10 (If I make any minor mistakes, feel free to make minor fixes and edits).
~mathboy282
Note: You can also find that CFE is half of CDG via circle theorems. We know CDG = 180 - 46 = 134, therefore making CFE 67 and BFC 113.
~meikh_neiht | // Block 1
/*
Made by ghfhgvghj10
Edited by MRENTHUSIASM
*/
size(300);
pair A, B, C, D, E, F, G;
D = origin;
A = 6*dir(46);
C = (6,0);
B = C + (A-D);
E = midpoint(C--D);
F = foot(A,B,E);
G = 6*dir(226);
dot("$A$",A,1.5*NW,linewidth(5));
dot("$B$",B,1.5*NE,linewidth(5));
dot("$C$",C,1.5*SE,linewidth(5));
dot("$D$",D,1.5*NW,linewidth(5));
dot("$E$",E,1.5*S,linewidth(5));
dot("$F$",F,1.5*dir(-20),linewidth(5));
dot("$G$",G,1.5*SW,linewidth(5));
markscalefactor=0.04;
draw(rightanglemark(A,F,B),red);
draw(A--B--C--D--cycle^^A--F--C^^B--E^^D--G^^E--G);
label("$46^{\circ}$",D,3*dir(26),red+fontsize(10));
// Block 2
/*
Made by ghfhgvghj10
Edited by MRENTHUSIASM
*/
size(300);
pair A, B, C, D, E, F, G;
D = origin;
A = 6*dir(46);
C = (6,0);
B = C + (A-D);
E = midpoint(C--D);
F = foot(A,B,E);
G = 6*dir(226);
dot("$A$",A,1.5*NE,linewidth(5));
dot("$B$",B,1.5*NE,linewidth(5));
dot("$C$",C,1.5*SE,linewidth(5));
dot("$D$",D,1.5*NW,linewidth(5));
dot("$E$",E,1.5*S,linewidth(5));
dot("$F$",F,1.5*dir(-20),linewidth(5));
dot("$G$",G,1.5*SW,linewidth(5));
markscalefactor=0.04;
draw(rightanglemark(A,F,B),red);
draw(A--B--C--D--cycle^^A--F--C^^B--E^^D--G^^E--G);
label("$46^{\circ}$",D,3*dir(26),red+fontsize(10));
draw(Circle(D,6),dashed);
// Block 3
/* Made by ghfhgvghj10 Edited by MRENTHUSIASM */ size(300); pair A, B, C, D, E, F, G; D = origin; A = 6*dir(46); C = (6,0); B = C + (A-D); E = midpoint(C--D); F = foot(A,B,E); G = 6*dir(226); dot("$A$",A,1.5*NW,linewidth(5)); dot("$B$",B,1.5*NE,linewidth(5)); dot("$C$",C,1.5*SE,linewidth(5)); dot("$D$",D,1.5*NW,linewidth(5)); dot("$E$",E,1.5*S,linewidth(5)); dot("$F$",F,1.5*dir(-20),linewidth(5)); dot("$G$",G,1.5*SW,linewidth(5)); markscalefactor=0.04; draw(rightanglemark(A,F,B),red); draw(A--B--C--D--cycle^^A--F--C^^B--E^^D--G^^E--G); label("$46^{\circ}$",D,3*dir(26),red+fontsize(10));
// Block 4
/* Made by ghfhgvghj10 Edited by MRENTHUSIASM */ size(300); pair A, B, C, D, E, F, G; D = origin; A = 6*dir(46); C = (6,0); B = C + (A-D); E = midpoint(C--D); F = foot(A,B,E); G = 6*dir(226); dot("$A$",A,1.5*NE,linewidth(5)); dot("$B$",B,1.5*NE,linewidth(5)); dot("$C$",C,1.5*SE,linewidth(5)); dot("$D$",D,1.5*NW,linewidth(5)); dot("$E$",E,1.5*S,linewidth(5)); dot("$F$",F,1.5*dir(-20),linewidth(5)); dot("$G$",G,1.5*SW,linewidth(5)); markscalefactor=0.04; draw(rightanglemark(A,F,B),red); draw(A--B--C--D--cycle^^A--F--C^^B--E^^D--G^^E--G); label("$46^{\circ}$",D,3*dir(26),red+fontsize(10)); draw(Circle(D,6),dashed); | [] |
923 | Let $S$ be the set of circles in the coordinate plane that are tangent to each of the three circles with equations $x^{2}+y^{2}=4$, $x^{2}+y^{2}=64$, and $(x-5)^{2}+y^{2}=3$. What is the sum of the areas of all circles in $S$?
$\textbf{(A)}~48\pi\qquad\textbf{(B)}~68\pi\qquad\textbf{(C)}~96\pi\qquad\textbf{(D)}~102\pi\qquad\textbf{(E)}~136\pi\qquad$ | 2022 AMC 10B Problem 22 | The circles match up as follows: Case $1$ is brown, Case $2$ is blue, Case $3$ is green, and Case 4 is red.
Let $x^2 + y^2 = 64$ be circle $O$, $x^2 + y^2 = 4$ be circle $P$, and $(x-5)^2 + y^2 = 3$ be circle $Q$.
All the circles in S are internally tangent to circle $O$.
There are four cases with two circles belonging to each:
$*$ $P$ and $Q$ are internally tangent to $S$.
$*$ $P$ and $Q$ are externally tangent to $S$.
$*$ $P$ is externally and Circle $Q$ is internally tangent to $S$.
$*$ $P$ is internally and Circle $Q$ is externally tangent to $S$.
Consider Cases $1$ and $4$ together. Since circles $O$ and $P$ have the same center, the line connecting the center of $S$ and the center of $O$ will pass through the tangency point of both $S$ and $O$ and the tangency point of $S$ and $P$. This line will be the diameter of $S$ and have length $r_P + r_O = 10$. Therefore the radius of $S$ in these cases is $5$.
Consider Cases $2$ and $3$ together. Similarly to Cases $1$ and $4$, the line connecting the center of $S$ to the center of $O$ will pass through the tangency points. This time, however, the diameter of $S$ will have length $r_P-r_O=6$. Therefore, the radius of $S$ in these cases is $3$.
The set of circles $S$ consists of $8$ circles - $4$ of which have radius $5$ and $4$ of which have radius $3$.
The total area of all circles in $S$ is $4(5^2\pi + 3^2\pi) = 136\pi \Rightarrow \boxed{\textbf{(E)}}$.
-naman12 | // Block 1
import geometry;
unitsize(0.5cm);
void dc(pair x, pen p) {
pair y = intersectionpoints(circle((0,0),8),(0,0)--1000*x)[0];
draw(circle(x, abs(x-y)),p+linewidth(2));
}
pair O1 = (0,0),O2=(5,0),P1=intersectionpoints(circle(O1,5),circle(O2,3+sqrt(3)))[0],P2=intersectionpoints(circle(O1,3),circle(O2,5+sqrt(3)))[0],P3=intersectionpoints(circle(O1,5),circle(O2,3-sqrt(3)))[0],P4=intersectionpoints(circle(O1,3),circle(O2,5-sqrt(3)))[0];
draw(circle(O1,2));
draw(circle(O1,8));
draw(circle(O2,sqrt(3)));
dc(P1,blue);
dc(P2,red);
dc(P3,mediumgreen);
dc(P4,brown);
// Block 2
import geometry; unitsize(0.5cm); void dc(pair x, pen p) { pair y = intersectionpoints(circle((0,0),8),(0,0)--1000*x)[0]; draw(circle(x, abs(x-y)),p+linewidth(2)); } pair O1 = (0,0),O2=(5,0),P1=intersectionpoints(circle(O1,5),circle(O2,3+sqrt(3)))[0],P2=intersectionpoints(circle(O1,3),circle(O2,5+sqrt(3)))[0],P3=intersectionpoints(circle(O1,5),circle(O2,3-sqrt(3)))[0],P4=intersectionpoints(circle(O1,3),circle(O2,5-sqrt(3)))[0]; draw(circle(O1,2)); draw(circle(O1,8)); draw(circle(O2,sqrt(3))); dc(P1,blue); dc(P2,red); dc(P3,mediumgreen); dc(P4,brown); | [] |
924 | Ant Amelia starts on the number line at $0$ and crawls in the following manner. For $n=1,2,3,$ Amelia chooses a time duration $t_n$ and an increment $x_n$ independently and uniformly at random from the interval $(0,1).$ During the $n$th step of the process, Amelia moves $x_n$ units in the positive direction, using up $t_n$ minutes. If the total elapsed time has exceeded $1$ minute during the $n$th step, she stops at the end of that step; otherwise, she continues with the next step, taking at most $3$ steps in all. What is the probability that Amelia’s position when she stops will be greater than $1$?
$\textbf{(A) }\frac{1}{3} \qquad \textbf{(B) }\frac{1}{2} \qquad \textbf{(C) }\frac{2}{3} \qquad \textbf{(D) }\frac{3}{4} \qquad \textbf{(E) }\frac{5}{6}$ | 2022 AMC 10B Problem 23 | Let $x$ and $y$ be random variables that are independently and uniformly distributed in the interval $(0,1).$ Note that \[P(x+y\leq 1)=\frac{\frac12\cdot1^2}{1^2}=\frac12,\] as shown below:
Let $x,y,$ and $z$ be random variables that are independently and uniformly distributed in the interval $(0,1).$ Note that \[P(x+y+z\leq 1)=\frac{\frac13\cdot\left(\frac12\cdot1^2\right)\cdot1}{1^3}=\frac16,\] as shown below:
We have two cases:
Amelia takes exactly $2$ steps.
We need $x_1+x_2>1$ and $t_1+t_2>1.$ So, the probability is \[P(x_1+x_2>1)\cdot P(t_1+t_2>1)=\left(1-\frac12\right)\cdot\left(1-\frac12\right)=\frac14.\]
Amelia takes exactly $3$ steps.
We need $x_1+x_2+x_3>1$ and $t_1+t_2\leq1.$ So, the probability is \[P(x_1+x_2+x_3>1)\cdot P(t_1+t_2\leq1)=\left(1-\frac16\right)\cdot\frac12=\frac{5}{12}.\]
Together, the answer is $\frac14 + \frac{5}{12} = \boxed{\textbf{(C) }\frac{2}{3}}.$
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
~MRENTHUSIASM | // Block 1
/* Made by MRENTHUSIASM */
size(200);
real xMin = -0.25;
real xMax = 1.25;
real yMin = -0.25;
real yMax = 1.25;
//Draws the horizontal ticks
void horizontalTicks()
{
for (real i = 1; i < yMax; ++i)
{
draw((-1/32,i)--(1/32,i), black+linewidth(1));
}
}
//Draws the vertical ticks
void verticalTicks()
{
for (real i = 1; i < xMax; ++i)
{
draw((i,-1/32)--(i,1/32), black+linewidth(1));
}
}
horizontalTicks();
verticalTicks();
label("$0$",(0,0),2*SW);
label("$1$",(1,0),2*S);
label("$1$",(0,1),2*W);
fill((0,0)--(1,0)--(0,1)--cycle,yellow);
draw((0,1)--(1,1)^^(1,0)--(1,1),dashed);
draw((0,1)--(1,0));
draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(8));
draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(8));
label("$x$",(xMax,0),(2,0));
label("$y$",(0,yMax),(0,2));
// Block 2
/* Made by MRENTHUSIASM */
size(200);
import graph3;
import solids;
currentprojection=orthographic((0.3,0.1,0.1));
draw(surface((1,0,0)--(0,1,0)--(0,0,1)--cycle),yellow);
draw(surface((1,0,0)--(0,1,0)--(0,0,0)--cycle),yellow);
draw(surface((1,0,0)--(0,0,1)--(0,0,0)--cycle),yellow);
draw(surface((0,1,0)--(0,0,1)--(0,0,0)--cycle),yellow);
draw((0,0,1)--(0,1,1)--(1,1,1)--(1,0,1)--cycle,dashed);
draw((0,1,0)--(1,1,0)--(1,0,0),dashed);
draw((0,1,1)--(0,1,0)^^(1,1,1)--(1,1,0)^^(1,0,1)--(1,0,0),dashed);
draw((-0.5,0,0)--(1.5,0,0),linewidth(1.25),EndArrow3(10));
draw((0,-0.5,0)--(0,1.5,0),linewidth(1.25),EndArrow3(10));
draw((0,0,-0.5)--(0,0,1.5),linewidth(1.25),EndArrow3(10));
draw((-0.1,0,1)--(0.1,0,1),linewidth(1));
draw((0,1,-0.1)--(0,1,0.1),linewidth(1));
draw((1,-0.1,0)--(1,0.1,0),linewidth(1));
label("$x$",(1.5,0,0),4*dir((1.5,0,0)));
label("$y$",(0,1.5,0),2*dir((0,1.5,0)));
label("$z$",(0,0,1.5),2*dir((0,0,1.5)));
label("$0$",(0,0,0),2*dir((0,0.5,-0.5)));
label("$1$",(1,0,0),4*dir((0,-1,0)));
label("$1$",(0,1,0),4*dir((0,0,-1)));
label("$1$",(0,0,1),5*dir((-1,0,0)));
draw((1,0,0)--(0,1,0)--(0,0,1)--cycle);
// Block 3
/* Made by MRENTHUSIASM */ size(200); real xMin = -0.25; real xMax = 1.25; real yMin = -0.25; real yMax = 1.25; //Draws the horizontal ticks void horizontalTicks() { for (real i = 1; i < yMax; ++i) { draw((-1/32,i)--(1/32,i), black+linewidth(1)); } } //Draws the vertical ticks void verticalTicks() { for (real i = 1; i < xMax; ++i) { draw((i,-1/32)--(i,1/32), black+linewidth(1)); } } horizontalTicks(); verticalTicks(); label("$0$",(0,0),2*SW); label("$1$",(1,0),2*S); label("$1$",(0,1),2*W); fill((0,0)--(1,0)--(0,1)--cycle,yellow); draw((0,1)--(1,1)^^(1,0)--(1,1),dashed); draw((0,1)--(1,0)); draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(8)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(8)); label("$x$",(xMax,0),(2,0)); label("$y$",(0,yMax),(0,2));
// Block 4
/* Made by MRENTHUSIASM */ size(200); import graph3; import solids; currentprojection=orthographic((0.3,0.1,0.1)); draw(surface((1,0,0)--(0,1,0)--(0,0,1)--cycle),yellow); draw(surface((1,0,0)--(0,1,0)--(0,0,0)--cycle),yellow); draw(surface((1,0,0)--(0,0,1)--(0,0,0)--cycle),yellow); draw(surface((0,1,0)--(0,0,1)--(0,0,0)--cycle),yellow); draw((0,0,1)--(0,1,1)--(1,1,1)--(1,0,1)--cycle,dashed); draw((0,1,0)--(1,1,0)--(1,0,0),dashed); draw((0,1,1)--(0,1,0)^^(1,1,1)--(1,1,0)^^(1,0,1)--(1,0,0),dashed); draw((-0.5,0,0)--(1.5,0,0),linewidth(1.25),EndArrow3(10)); draw((0,-0.5,0)--(0,1.5,0),linewidth(1.25),EndArrow3(10)); draw((0,0,-0.5)--(0,0,1.5),linewidth(1.25),EndArrow3(10)); draw((-0.1,0,1)--(0.1,0,1),linewidth(1)); draw((0,1,-0.1)--(0,1,0.1),linewidth(1)); draw((1,-0.1,0)--(1,0.1,0),linewidth(1)); label("$x$",(1.5,0,0),4*dir((1.5,0,0))); label("$y$",(0,1.5,0),2*dir((0,1.5,0))); label("$z$",(0,0,1.5),2*dir((0,0,1.5))); label("$0$",(0,0,0),2*dir((0,0.5,-0.5))); label("$1$",(1,0,0),4*dir((0,-1,0))); label("$1$",(0,1,0),4*dir((0,0,-1))); label("$1$",(0,0,1),5*dir((-1,0,0))); draw((1,0,0)--(0,1,0)--(0,0,1)--cycle); | [] |
925 | The diagram below shows a rectangle with side lengths $4$ and $8$ and a square with side length $5$. Three vertices of the square lie on three different sides of the rectangle, as shown. What is the area of the region inside both the square and the rectangle?
$\textbf{(A) }15\dfrac{1}{8} \qquad \textbf{(B) }15\dfrac{3}{8} \qquad \textbf{(C) }15\dfrac{1}{2} \qquad \textbf{(D) }15\dfrac{5}{8} \qquad \textbf{(E) }15\dfrac{7}{8}$ | 2022 AMC 12B Problem 13 | Let us label the points on the diagram.
By doing some angle chasing using the fact that $\angle ACE$ and $\angle CEG$ are right angles, we find that $\angle BAC = \angle DCE = \angle FEG$. Similarly, $\angle ACB = \angle CED = \angle EGF$. Therefore, $\triangle ABC \sim \triangle CDE \sim \triangle EFG$.
As we are given a rectangle and a square, $AB = 4$ and $AC = 5$. Therefore, $\triangle ABC$ is a $3$-$4$-$5$ right triangle and $BC = 3$.
$CE$ is also $5$. So, using the similar triangles, $CD = 4$ and $DE = 3$.
$EF = DF - DE = 4 - 3 = 1$. Using the similar triangles again, $EF$ is $\frac14$ of the corresponding $AB$. So,
\begin{align*} [\triangle EFG] &= \left(\frac14\right)^2 \cdot [\triangle ABC] \\ &= \frac{1}{16} \cdot 6 \\ &= \frac38. \end{align*}
Finally, we have
\begin{align*} [ACEG] &= [ABDF] - [\triangle ABC] - [\triangle CDE] - [\triangle EFG] \\ &= 7 \cdot 4 - \frac12 \cdot 3 \cdot 4 - \frac12 \cdot 3 \cdot 4 - \frac38 \\ &= 28 - 6 - 6 - \frac38 \\ &= \boxed{\textbf{(D) }15\dfrac{5}{8}}. \end{align*}
~Connor132435 | import olympiad; size(200); defaultpen(linewidth(1) + fontsize(10)); pair A = (0,0), B = (1,0), C = (4,0), D = (8,0), K = (0,4), F = (1,4), G = (7.25, 4), H = (8, 4), I = (8,3), J = (5, 7); fill(F--G--I--C--F--cycle, grey); markscalefactor=0.05; draw(A--D--H--K--A^^B--F^^F--C--I--J--F^^rightanglemark(F,J,I)^^rightanglemark(F,B,C)^^anglemark(D,C,I)^^anglemark(B,F,C)^^anglemark(H,I,G)); draw(anglemark(F,C,B)^^anglemark(C,I,D)^^anglemark(I,G,H)); markscalefactor=0.041; draw(anglemark(F,C,B)^^anglemark(C,I,D)^^anglemark(I,G,H)); label("8",(4,-.5),S); label("5",(3, 5.5),NW); label("4",(8.25, 2), E); label("A", F, NW); label("B", B, S); label("C", C, S); label("D", D, SE); label("E", I, E); label("F", H, NE); label("G", G, NE); label("4", (1,2), E); label("5", (2.5,2), SW); label("3", (2.5,0), S); label("4", (6,0), S); label("5", (6,1.5), SE); label("3", (8, 1.5), E); label("1", (8, 3.5), E); | [] |
925 | The diagram below shows a rectangle with side lengths $4$ and $8$ and a square with side length $5$. Three vertices of the square lie on three different sides of the rectangle, as shown. What is the area of the region inside both the square and the rectangle?
$\textbf{(A) }15\dfrac{1}{8} \qquad \textbf{(B) }15\dfrac{3}{8} \qquad \textbf{(C) }15\dfrac{1}{2} \qquad \textbf{(D) }15\dfrac{5}{8} \qquad \textbf{(E) }15\dfrac{7}{8}$ | 2022 AMC 12B Problem 13 | Same diagram as Solution 1, but added point $H$, which is $(4,7)$. I also renamed all the points to form coordinates using $B$ as the origin.
In order to find the area, point $G$'s coordinates must be found. Notice how $EH$ and $AG$ intercept at point $G$. This means that we need to find the equations for $EH$ and $AG$ and make a system of linear equations.
Using the slope formula $m=\frac{y_{2} - y_{1}}{x_{2} - x_{1}}$, we get the slope for $EH$, which means $m=\frac{3-7}{7-4} = -\frac{4}{3}$
Then, by using point slope form. $y-y_{1}=m(x-x_{1})$. We can say that the equation for $EH$ is $y-7=-\frac{4}{3}(x-4)$ or in this case, $y=-\frac{4}{3}x+12 \frac{1}{3}$.
And it is easy to figure out that the equation for $AG$ is $y=4$.
The best way to solve the system of linear equations is to substitute the $y$ for the $4$ in equation $EH$.
$4=-\frac{4}{3}x+12 \frac{1}{3}$, so $x=6\frac{1}{4}$ and $y=4$ This would mean $G\left(6\frac{1}{4},4\right)$.
Since we have our $G$ coordinate, we can continue with Solution 3, with the area of the trapezoid $\left(\frac{EG+AC}{2}\right)(CE)$, where $EG=\frac{5}{4}$ (using distance formula for $E$ to $G$), $AC=5$, and $CE=5$.
By substitution, we get $\left(\frac{\frac{5}{4}+5}{2}\right)(5)=$$\boxed{\textbf{(D) }15\dfrac{5}{8}}$.
~ghfhgvghj10 (+ minor edits ~TaeKim) | import olympiad; size(200); defaultpen(linewidth(1) + fontsize(10)); pair A = (0,0), B = (1,0), C = (4,0), D = (8,0), K = (0,4), F = (1,4), G = (7.25, 4), H = (8, 4), I = (8,3), J = (5, 7); fill(F--G--I--C--F--cycle, grey); markscalefactor=0.05; draw(A--D--H--K--A^^B--F^^F--C--I--J--F^^rightanglemark(F,J,I)^^rightanglemark(F,B,C)^^anglemark(D,C,I)^^anglemark(B,F,C)^^anglemark(H,I,G)); draw(anglemark(F,C,B)^^anglemark(C,I,D)^^anglemark(I,G,H)); markscalefactor=0.041; draw(anglemark(F,C,B)^^anglemark(C,I,D)^^anglemark(I,G,H)); label("8",(4,-.5),S); label("5",(3, 5.5),NW); label("4",(8.25, 2), E); label("A(0,4)", F, NW); label("B(0,0)", B, S); label("C(3,0)", C, S); label("D(7,0)", D, SE); label("E(7,3)", I, E); label("F(7,4)", H, NE); label("G", G, NE); label("4", (1,2), E); label("5", (2.5,2), SW); label("3", (2.5,0), S); label("4", (6,0), S); label("5", (6,1.5), SE); label("3", (8, 1.5), E); label("1", (8, 3.5), E); label("H(4,7)", (4.65, 7.25), E); | [] |
926 | The graph of $y=x^2+2x-15$ intersects the $x$-axis at points $A$ and $C$ and the $y$-axis at point $B$. What is $\tan(\angle ABC)$?
$\textbf{(A)}\ \frac{1}{7} \qquad \textbf{(B)}\ \frac{1}{4} \qquad \textbf{(C)}\ \frac{3}{7} \qquad \textbf{(D)}\ \frac{1}{2} \qquad \textbf{(E)}\ \frac{4}{7} \qquad$ | 2022 AMC 12B Problem 14 | From $x^2 + 2x - 15 = (x-3)(x+5)$, we may assume, without loss of generality, that $x$-intercepts of the given parabola are $A(-5,0)$ and $C(3,0)$. And, point $B$ has coordinates $(0,-15)$. Consider complex numbers $z = 3 + i$ and $w = 5 + i$ whose arguments are $\theta \coloneqq \angle OBA$ and $\phi \coloneqq \angle OBC$, respectively. Notice that $\angle ABC = \theta + \phi$ is the argument of the product $zw$ which is \[zw = (3+i)(5+i) = 14 + 8i.\]
Hence \[\tan \angle ABC = \frac{\operatorname{Im}(zw)}{\operatorname{Re}(zw)} = \frac{8}{14} = \boxed{\textbf{(E)}\ \frac{4}{7}}.\]
~VensL. | // Block 1
/* Made by MRENTHUSIASM */
size(300);
real xMin = -15;
real xMax = 15;
real yMin = -17;
real yMax = 17;
draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5));
draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5));
label("$x$",(xMax,0),(2,0));
label("$y$",(0,yMax),(0,2));
real f(real x) { return x^2+2*x-15; }
draw(graph(f,-6.75,4.75),red);
pair A, B, C, O;
A = (-5,0);
B = (0,-15);
C = (3,0);
O = origin;
markscalefactor=0.1;
draw(rightanglemark(B,O,C));
draw(A--B--C);
dot("$A$",A,1.5SW,linewidth(4.5));
dot("$B$",B,1.5SE,linewidth(4.5));
dot("$C$",C,1.5SE,linewidth(4.5));
dot("$O$",O,1.5SW,linewidth(4.5));
label("$y=x^2+2x-15$",(12,9),red);
label("$5$",(-2.5,0),1.5N);
label("$3$",(1.5,0),1.5N);
label("$15$",(0,-6),W);
label("$\theta$",(0,-15),9*dir(100));
label("$\phi$",(0,-15),9*dir(84));
// Block 2
/* Made by MRENTHUSIASM */ size(300); real xMin = -15; real xMax = 15; real yMin = -17; real yMax = 17; draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("$x$",(xMax,0),(2,0)); label("$y$",(0,yMax),(0,2)); real f(real x) { return x^2+2*x-15; } draw(graph(f,-6.75,4.75),red); pair A, B, C, O; A = (-5,0); B = (0,-15); C = (3,0); O = origin; markscalefactor=0.1; draw(rightanglemark(B,O,C)); draw(A--B--C); dot("$A$",A,1.5SW,linewidth(4.5)); dot("$B$",B,1.5SE,linewidth(4.5)); dot("$C$",C,1.5SE,linewidth(4.5)); dot("$O$",O,1.5SW,linewidth(4.5)); label("$y=x^2+2x-15$",(12,9),red); label("$5$",(-2.5,0),1.5N); label("$3$",(1.5,0),1.5N); label("$15$",(0,-6),W); label("$\theta$",(0,-15),9*dir(100)); label("$\phi$",(0,-15),9*dir(84)); | [] |
927 | Each square in a $5 \times 5$ grid is either filled or empty, and has up to eight adjacent neighboring squares, where neighboring squares share either a side or a corner. The grid is transformed by the following rules:
Any filled square with two or three filled neighbors remains filled.
Any empty square with exactly three filled neighbors becomes a filled square.
All other squares remain empty or become empty.
A sample transformation is shown in the figure below.
Suppose the $5 \times 5$ grid has a border of empty squares surrounding a $3 \times 3$ subgrid. How many initial configurations will lead to a transformed grid consisting of a single filled square in the center after a single transformation? (Rotations and reflections of the same configuration are considered different.)
$\textbf{(A)}\ 14 \qquad\textbf{(B)}\ 18 \qquad\textbf{(C)}\ 22 \qquad\textbf{(D)}\ 26 \qquad\textbf{(E)}\ 30$ | 2022 AMC 12B Problem 18 | There are two cases for the initial configuration:
The center square is filled.
Exactly two of the eight adjacent neighboring squares of the center are filled. Clearly, the only possibility is that the squares along one diagonal are filled, as shown below:
In this case, there are $2$ possible initial configurations. All rotations and reflections are considered.
The center square is empty.
Exactly three of the eight adjacent neighboring squares of the center are filled. The possibilities are shown below:
In this case, there are $4+4+4+8=20$ possible initial configurations. All rotations and reflections are considered.
Together, the answer is $2+20=\boxed{\textbf{(C)}\ 22}.$
~mathboy100 ~MRENTHUSIASM | // Block 1
import geometry; unitsize(0.6cm); void ds(pair x) { filldraw(x -- (1,0) + x -- (1,1) + x -- (0,1)+x -- cycle,mediumgray,invisible); } ds((1,3)); ds((2,2)); ds((3,1)); for (int i = 0; i <= 5; ++i) { draw((0,i)--(5,i)); draw((i,0)--(i,5)); } label("$2$ Configurations", (2.5,-1));
// Block 2
import geometry; unitsize(0.6cm); void ds(pair x) { filldraw(x -- (1,0) + x -- (1,1) + x -- (0,1)+x -- cycle,mediumgray,invisible); } ds((1,3)); ds((3,3)); ds((1,1)); for (int i = 0; i <= 5; ++i) { draw((0,i)--(5,i)); draw((i,0)--(i,5)); } ds((10,3)); ds((12,3)); ds((11,1)); for (int i = 0; i <= 5; ++i) { draw((9,i)--(14,i)); draw((i+9,0)--(i+9,5)); } ds((19,3)); ds((20,1)); ds((21,2)); for (int i = 0; i <= 5; ++i) { draw((18,i)--(23,i)); draw((i+18,0)--(i+18,5)); } ds((28,3)); ds((29,1)); ds((30,1)); for (int i = 0; i <= 5; ++i) { draw((27,i)--(32,i)); draw((i+27,0)--(i+27,5)); } label("$4$ Configurations", (2.5,-1)); label("$4$ Configurations", (11.5,-1)); label("$4$ Configurations", (20.5,-1)); label("$8$ Configurations", (29.5,-1)); | [] |
928 | Let $S$ be the set of circles in the coordinate plane that are tangent to each of the three circles with equations $x^{2}+y^{2}=4$, $x^{2}+y^{2}=64$, and $(x-5)^{2}+y^{2}=3$. What is the sum of the areas of all circles in $S$?
$\textbf{(A)}~48\pi\qquad\textbf{(B)}~68\pi\qquad\textbf{(C)}~96\pi\qquad\textbf{(D)}~102\pi\qquad\textbf{(E)}~136\pi\qquad$ | 2022 AMC 12B Problem 21 | The circles match up as follows: Case $1$ is brown, Case $2$ is blue, Case $3$ is green, and Case 4 is red.
Let $x^2 + y^2 = 64$ be circle $O$, $x^2 + y^2 = 4$ be circle $P$, and $(x-5)^2 + y^2 = 3$ be circle $Q$.
All the circles in S are internally tangent to circle $O$.
There are four cases with two circles belonging to each:
$*$ $P$ and $Q$ are internally tangent to $S$.
$*$ $P$ and $Q$ are externally tangent to $S$.
$*$ $P$ is externally and Circle $Q$ is internally tangent to $S$.
$*$ $P$ is internally and Circle $Q$ is externally tangent to $S$.
Consider Cases $1$ and $4$ together. Since circles $O$ and $P$ have the same center, the line connecting the center of $S$ and the center of $O$ will pass through the tangency point of both $S$ and $O$ and the tangency point of $S$ and $P$. This line will be the diameter of $S$ and have length $r_P + r_O = 10$. Therefore the radius of $S$ in these cases is $5$.
Consider Cases $2$ and $3$ together. Similarly to Cases $1$ and $4$, the line connecting the center of $S$ to the center of $O$ will pass through the tangency points. This time, however, the diameter of $S$ will have length $r_P-r_O=6$. Therefore, the radius of $S$ in these cases is $3$.
The set of circles $S$ consists of $8$ circles - $4$ of which have radius $5$ and $4$ of which have radius $3$.
The total area of all circles in $S$ is $4(5^2\pi + 3^2\pi) = 136\pi \Rightarrow \boxed{\textbf{(E)}}$.
-naman12 | import geometry; unitsize(0.5cm); void dc(pair x, pen p) { pair y = intersectionpoints(circle((0,0),8),(0,0)--1000*x)[0]; draw(circle(x, abs(x-y)),p+linewidth(2)); } pair O1 = (0,0),O2=(5,0),P1=intersectionpoints(circle(O1,5),circle(O2,3+sqrt(3)))[0],P2=intersectionpoints(circle(O1,3),circle(O2,5+sqrt(3)))[0],P3=intersectionpoints(circle(O1,5),circle(O2,3-sqrt(3)))[0],P4=intersectionpoints(circle(O1,3),circle(O2,5-sqrt(3)))[0]; draw(circle(O1,2)); draw(circle(O1,8)); draw(circle(O2,sqrt(3))); dc(P1,blue); dc(P2,red); dc(P3,mediumgreen); dc(P4,brown); | [] |
929 | Ant Amelia starts on the number line at $0$ and crawls in the following manner. For $n=1,2,3,$ Amelia chooses a time duration $t_n$ and an increment $x_n$ independently and uniformly at random from the interval $(0,1).$ During the $n$th step of the process, Amelia moves $x_n$ units in the positive direction, using up $t_n$ minutes. If the total elapsed time has exceeded $1$ minute during the $n$th step, she stops at the end of that step; otherwise, she continues with the next step, taking at most $3$ steps in all. What is the probability that Amelia’s position when she stops will be greater than $1$?
$\textbf{(A) }\frac{1}{3} \qquad \textbf{(B) }\frac{1}{2} \qquad \textbf{(C) }\frac{2}{3} \qquad \textbf{(D) }\frac{3}{4} \qquad \textbf{(E) }\frac{5}{6}$ | 2022 AMC 12B Problem 22 | Let $x$ and $y$ be random variables that are independently and uniformly distributed in the interval $(0,1).$ Note that \[P(x+y\leq 1)=\frac{\frac12\cdot1^2}{1^2}=\frac12,\] as shown below:
Let $x,y,$ and $z$ be random variables that are independently and uniformly distributed in the interval $(0,1).$ Note that \[P(x+y+z\leq 1)=\frac{\frac13\cdot\left(\frac12\cdot1^2\right)\cdot1}{1^3}=\frac16,\] as shown below:
We have two cases:
Amelia takes exactly $2$ steps.
We need $x_1+x_2>1$ and $t_1+t_2>1.$ So, the probability is \[P(x_1+x_2>1)\cdot P(t_1+t_2>1)=\left(1-\frac12\right)\cdot\left(1-\frac12\right)=\frac14.\]
Amelia takes exactly $3$ steps.
We need $x_1+x_2+x_3>1$ and $t_1+t_2\leq1.$ So, the probability is \[P(x_1+x_2+x_3>1)\cdot P(t_1+t_2\leq1)=\left(1-\frac16\right)\cdot\frac12=\frac{5}{12}.\]
Together, the answer is $\frac14 + \frac{5}{12} = \boxed{\textbf{(C) }\frac{2}{3}}.$
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
~MRENTHUSIASM | // Block 1
/* Made by MRENTHUSIASM */ size(200); real xMin = -0.25; real xMax = 1.25; real yMin = -0.25; real yMax = 1.25; //Draws the horizontal ticks void horizontalTicks() { for (real i = 1; i < yMax; ++i) { draw((-1/32,i)--(1/32,i), black+linewidth(1)); } } //Draws the vertical ticks void verticalTicks() { for (real i = 1; i < xMax; ++i) { draw((i,-1/32)--(i,1/32), black+linewidth(1)); } } horizontalTicks(); verticalTicks(); label("$0$",(0,0),2*SW); label("$1$",(1,0),2*S); label("$1$",(0,1),2*W); fill((0,0)--(1,0)--(0,1)--cycle,yellow); draw((0,1)--(1,1)^^(1,0)--(1,1),dashed); draw((0,1)--(1,0)); draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(8)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(8)); label("$x$",(xMax,0),(2,0)); label("$y$",(0,yMax),(0,2));
// Block 2
/* Made by MRENTHUSIASM */ size(200); import graph3; import solids; currentprojection=orthographic((0.3,0.1,0.1)); draw(surface((1,0,0)--(0,1,0)--(0,0,1)--cycle),yellow); draw(surface((1,0,0)--(0,1,0)--(0,0,0)--cycle),yellow); draw(surface((1,0,0)--(0,0,1)--(0,0,0)--cycle),yellow); draw(surface((0,1,0)--(0,0,1)--(0,0,0)--cycle),yellow); draw((0,0,1)--(0,1,1)--(1,1,1)--(1,0,1)--cycle,dashed); draw((0,1,0)--(1,1,0)--(1,0,0),dashed); draw((0,1,1)--(0,1,0)^^(1,1,1)--(1,1,0)^^(1,0,1)--(1,0,0),dashed); draw((-0.5,0,0)--(1.5,0,0),linewidth(1.25),EndArrow3(10)); draw((0,-0.5,0)--(0,1.5,0),linewidth(1.25),EndArrow3(10)); draw((0,0,-0.5)--(0,0,1.5),linewidth(1.25),EndArrow3(10)); draw((-0.1,0,1)--(0.1,0,1),linewidth(1)); draw((0,1,-0.1)--(0,1,0.1),linewidth(1)); draw((1,-0.1,0)--(1,0.1,0),linewidth(1)); label("$x$",(1.5,0,0),4*dir((1.5,0,0))); label("$y$",(0,1.5,0),2*dir((0,1.5,0))); label("$z$",(0,0,1.5),2*dir((0,0,1.5))); label("$0$",(0,0,0),2*dir((0,0.5,-0.5))); label("$1$",(1,0,0),4*dir((0,-1,0))); label("$1$",(0,1,0),4*dir((0,0,-1))); label("$1$",(0,0,1),5*dir((-1,0,0))); draw((1,0,0)--(0,1,0)--(0,0,1)--cycle); | [] |
930 | Four regular hexagons surround a square with side length 1, each one sharing an edge with the square,
as shown in the figure below. The area of the resulting 12-sided outer nonconvex polygon can be
written as $m \sqrt{n} + p$, where $m$, $n$, and $p$ are integers and $n$ is not divisible by the square of any prime.
What is $m+n+p$?
$\textbf{(A) } -12 \qquad \textbf{(B) }-4 \qquad \textbf{(C) } 4 \qquad \textbf{(D) }24 \qquad \textbf{(E) }32$ | 2022 AMC 12B Problem 25 | Refer to the diagram above.
Let the origin be at the center of the square, $A$ be the intersection of the top and right hexagons, $B$ be the intersection of the top and left hexagons, and $M$ and $N$ be the top points in the diagram.
By symmetry, $A$ lies on the line $y = x$. The equation of line $AN$ is $y = -x\sqrt{3} + \frac{3}{2}\sqrt{3} - \frac{1}{2}$ (due to it being one of the sides of the top hexagon). Thus, we can solve for the coordinates of $A$ by finding the intersection of the two lines:
\[x = -x\sqrt{3} + \frac{3\sqrt{3} - 1}{2}\]
\[x(\sqrt{3} + 1) = \frac{3\sqrt{3} - 1}{2}\]
\[x = \frac{3\sqrt{3}-1}{2} \cdot \frac{1}{\sqrt{3} + 1}\]
\[= \frac{3\sqrt{3}-1}{2(\sqrt{3} + 1)} \cdot \frac{\sqrt{3} - 1}{\sqrt{3} - 1}\]
\[= \frac{10 - 4\sqrt{3}}{4}\]
\[= \frac{5}{2} - \sqrt{3}\]
\[\therefore A = \left(\frac{5}{2} - \sqrt{3}, \frac{5}{2} - \sqrt{3}\right).\]
This means that we can find the length $AB$, which is equal to $2(\frac{5}{2} - \sqrt{3}) = 5 - 2\sqrt{3}$. We will next find the area of trapezoid $ABMN$. The lengths of the bases are $1$ and $5 - 2\sqrt{3}$, and the height is equal to the $y$-coordinate of $M$ minus the $y$-coordinate of $A$. The height of the hexagon is $\sqrt{3}$ and the bottom of the hexagon lies on the line $y = \frac{1}{2}$. Thus, the $y$-coordinate of $M$ is $\sqrt{3} - \frac{1}{2}$, and the height is $2\sqrt{3} - 3$. We can now find the area of the trapezoid:
\[[ABMN] = (2\sqrt{3} - 3)\left(\frac{1 + 5 - 2\sqrt{3}}{2}\right)\]
\[= (2\sqrt{3} - 3)(3 - \sqrt{3})\]
\[= 6\sqrt{3} + 3\sqrt{3} - 9 - 6\]
\[= 9\sqrt{3} - 15.\]
The total area of the figure is the area of a square with side length $AB$ plus four times the area of this trapezoid:
\[\textrm{Area} = (5 - 2\sqrt{3})^2 + 4(9\sqrt{3} - 15)\]
\[= 37 - 20\sqrt{3} + 36\sqrt{3} - 60\]
\[= 16\sqrt{3} - 23.\]
Our answer is $16 + 3 - 23 = \boxed{\textbf{(B) }-4}$.
~mathboy100 | // Block 1
import geometry;
unitsize(3cm);
draw((0,0) -- (1,0) -- (1,1) -- (0,1) -- cycle);
draw(shift((1/2,1-sqrt(3)/2))*polygon(6));
draw(shift((1/2,sqrt(3)/2))*polygon(6));
draw(shift((sqrt(3)/2,1/2))*rotate(90)*polygon(6));
draw(shift((1-sqrt(3)/2,1/2))*rotate(90)*polygon(6));
draw((0,1-sqrt(3))--(1,1-sqrt(3))--(3-sqrt(3),sqrt(3)-2)--(sqrt(3),0)--(sqrt(3),1)--(3-sqrt(3),3-sqrt(3))--(1,sqrt(3))--(0,sqrt(3))--(sqrt(3)-2,3-sqrt(3))--(1-sqrt(3),1)--(1-sqrt(3),0)--(sqrt(3)-2,sqrt(3)-2)--cycle,linewidth(1.5));
draw((3-sqrt(3),3-sqrt(3)) -- (3-sqrt(3),sqrt(3)-2) -- (sqrt(3)-2,sqrt(3)-2) -- (sqrt(3)-2,3-sqrt(3)) -- cycle,linewidth(1.5));
label("$O (0, 0)$",(0.5,0.5),S);
dot((0.5,0.5));
label("$A$", (3-sqrt(3), 3-sqrt(3)), NE);
label("$B$", (sqrt(3) - 2, 3-sqrt(3)), NW);
label("$M$", (0, sqrt(3)), NW);
label("$N$", (1, sqrt(3)), NE);
// Block 2
import geometry; unitsize(3cm); draw((0,0) -- (1,0) -- (1,1) -- (0,1) -- cycle); draw(shift((1/2,1-sqrt(3)/2))*polygon(6)); draw(shift((1/2,sqrt(3)/2))*polygon(6)); draw(shift((sqrt(3)/2,1/2))*rotate(90)*polygon(6)); draw(shift((1-sqrt(3)/2,1/2))*rotate(90)*polygon(6)); draw((0,1-sqrt(3))--(1,1-sqrt(3))--(3-sqrt(3),sqrt(3)-2)--(sqrt(3),0)--(sqrt(3),1)--(3-sqrt(3),3-sqrt(3))--(1,sqrt(3))--(0,sqrt(3))--(sqrt(3)-2,3-sqrt(3))--(1-sqrt(3),1)--(1-sqrt(3),0)--(sqrt(3)-2,sqrt(3)-2)--cycle,linewidth(1.5)); draw((3-sqrt(3),3-sqrt(3)) -- (3-sqrt(3),sqrt(3)-2) -- (sqrt(3)-2,sqrt(3)-2) -- (sqrt(3)-2,3-sqrt(3)) -- cycle,linewidth(1.5)); label("$O (0, 0)$",(0.5,0.5),S); dot((0.5,0.5)); label("$A$", (3-sqrt(3), 3-sqrt(3)), NE); label("$B$", (sqrt(3) - 2, 3-sqrt(3)), NW); label("$M$", (0, sqrt(3)), NW); label("$N$", (1, sqrt(3)), NE); | [] |
930 | Four regular hexagons surround a square with side length 1, each one sharing an edge with the square,
as shown in the figure below. The area of the resulting 12-sided outer nonconvex polygon can be
written as $m \sqrt{n} + p$, where $m$, $n$, and $p$ are integers and $n$ is not divisible by the square of any prime.
What is $m+n+p$?
$\textbf{(A) } -12 \qquad \textbf{(B) }-4 \qquad \textbf{(C) } 4 \qquad \textbf{(D) }24 \qquad \textbf{(E) }32$ | 2022 AMC 12B Problem 25 | Begin by dividing the figure as shown above. Clearly, the entire figure has 8-fold symmetry. Therefore, we can calculate the area of $ODBA$ and multiply it by 8. We split $[ODBA]$ into $[ODBC]+[ABC]$.
Knowing the side length of the hexagon is $1$, we can use 30-60-90 triangles within the hexagon to find the total distance between opposite edges is $2\cdot \frac{\sqrt{3}}{2}=\sqrt{3}.$ Thus, $OD=\sqrt{3}-\frac{1}{2}$ and $BC=\sqrt{3}-1.$ Recognizing $DB=\frac{1}{2}$ and $ODBC$ is a trapezoid, \[8\cdot[ODBC]=2\left(\sqrt{3}-\frac{1}{2}+\sqrt{3}-1\right)=4\sqrt{3}-3.\]
Next, we aim to find $[ABC]$. By angle chasing, we find $\angle A=105^\circ,$ $\angle B=30^\circ,$ and $\angle C = 45^\circ.$ We can use the law of sines to find $AB$:
\[\frac{\sin(105^\circ)}{\sqrt{3}-1}=\frac{\sin(45^\circ)}{AB}\implies AB=\frac{\sqrt{6}-\sqrt{2}}{2\sin(105^\circ)}.\]
We may not know what $\sin(105^\circ)$ is by memory, but we can cleverly calculate it using a common trig identity:
\begin{align*} \sin(105^\circ)&=\sin(60^\circ+45^\circ),\\ &=\sin(60^\circ)\cos(45^\circ)+\cos(60^\circ)\sin(45^\circ),\\ &=\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}{2}+\frac{1}{2}\cdot\frac{\sqrt{2}}{2},\\ &=\frac{\sqrt{6}+\sqrt{2}}{4}. \end{align*}
With some simplification, we'll find $AB=4-2\sqrt{3}$. Now, we can easily calculate $8\cdot [ABC]$ as \[8\cdot\frac{1}{2}\cdot(\sqrt{3}-1)(4-2\sqrt{3})\sin(30^\circ)=12\sqrt{3}-20.\]
Thus, the area of the dodecagon is $8\cdot [ODBA] = 8\cdot [ODBC] + 8 \cdot [ABC] =4\sqrt{3}-3+12\sqrt{3}-20=16\sqrt{3}-23.$
Finally, we find \[16+3-23=\boxed{\textbf{(B)}\ -4}.\]
~Jackson La Vallee | // Block 1
import geometry;
unitsize(3cm);
draw((0,0) -- (1,0) -- (1,1) -- (0,1) -- cycle);
draw(shift((1/2,1-sqrt(3)/2))*polygon(6));
draw(shift((1/2,sqrt(3)/2))*polygon(6));
draw(shift((sqrt(3)/2,1/2))*rotate(90)*polygon(6));
draw(shift((1-sqrt(3)/2,1/2))*rotate(90)*polygon(6));
draw((0,1-sqrt(3))--(1,1-sqrt(3))--(3-sqrt(3),sqrt(3)-2)--(sqrt(3),0)--(sqrt(3),1)--(3-sqrt(3),3-sqrt(3))--(1,sqrt(3))--(0,sqrt(3))--(sqrt(3)-2,3-sqrt(3))--(1-sqrt(3),1)--(1-sqrt(3),0)--(sqrt(3)-2,sqrt(3)-2)--cycle,linewidth(2));
draw((0.5,sqrt(3))--(0.5,1-sqrt(3)),linewidth(2));
draw((1-sqrt(3),0.5)--(sqrt(3),0.5),linewidth(2));
draw((-2+sqrt(3),-2+sqrt(3))--(3-sqrt(3),3-sqrt(3)),linewidth(2));
draw((1, sqrt(3))--(1,1),linewidth(2));
label("$O$",(0.5,0.5),SE);
dot((0.5,0.5));
label("$A$", (3-sqrt(3), 3-sqrt(3)), NE);
label("$B$", (1, sqrt(3)), NE);
label("$C$", (1,1), E);
label("$D$", (1/2, sqrt(3)), N);
// Block 2
import geometry; unitsize(3cm); draw((0,0) -- (1,0) -- (1,1) -- (0,1) -- cycle); draw(shift((1/2,1-sqrt(3)/2))*polygon(6)); draw(shift((1/2,sqrt(3)/2))*polygon(6)); draw(shift((sqrt(3)/2,1/2))*rotate(90)*polygon(6)); draw(shift((1-sqrt(3)/2,1/2))*rotate(90)*polygon(6)); draw((0,1-sqrt(3))--(1,1-sqrt(3))--(3-sqrt(3),sqrt(3)-2)--(sqrt(3),0)--(sqrt(3),1)--(3-sqrt(3),3-sqrt(3))--(1,sqrt(3))--(0,sqrt(3))--(sqrt(3)-2,3-sqrt(3))--(1-sqrt(3),1)--(1-sqrt(3),0)--(sqrt(3)-2,sqrt(3)-2)--cycle,linewidth(2)); draw((0.5,sqrt(3))--(0.5,1-sqrt(3)),linewidth(2)); draw((1-sqrt(3),0.5)--(sqrt(3),0.5),linewidth(2)); draw((-2+sqrt(3),-2+sqrt(3))--(3-sqrt(3),3-sqrt(3)),linewidth(2)); draw((1, sqrt(3))--(1,1),linewidth(2)); label("$O$",(0.5,0.5),SE); dot((0.5,0.5)); label("$A$", (3-sqrt(3), 3-sqrt(3)), NE); label("$B$", (1, sqrt(3)), NE); label("$C$", (1,1), E); label("$D$", (1/2, sqrt(3)), N); | [] |
930 | Four regular hexagons surround a square with side length 1, each one sharing an edge with the square,
as shown in the figure below. The area of the resulting 12-sided outer nonconvex polygon can be
written as $m \sqrt{n} + p$, where $m$, $n$, and $p$ are integers and $n$ is not divisible by the square of any prime.
What is $m+n+p$?
$\textbf{(A) } -12 \qquad \textbf{(B) }-4 \qquad \textbf{(C) } 4 \qquad \textbf{(D) }24 \qquad \textbf{(E) }32$ | 2022 AMC 12B Problem 25 | Divide this polygon into a 'cross' and $8$ triangles like $\triangle ABC$. $BD = 2 \cdot \frac{ \sqrt{3} }{2} = \sqrt{3}$, $BE = 2BD - CD = 2\sqrt{3} - 1$, $EF = 1$, $[BEFG] = 1 \cdot (2\sqrt{3} - 1) = 2\sqrt{3} - 1$, $[\text{cross}] = 2\cdot [BEFG] - 1 = 2(2\sqrt{3} - 1)- 1 = 4\sqrt{3} - 3$
[Image: images/amc/2022_AMC_12B_Problem_25_0.png]
$\angle ABC = 120^{\circ} - 90^{\circ} = 30^{\circ}$, $\angle ACB = \frac{90^{\circ}}{2} = 45^{\circ}$, $\angle BAC = 180^{\circ} - 30^{\circ} - 45^{\circ} = 105^{\circ}$, $BC = BD - CD = \sqrt{3} - 1$. Draw a perpendicular line $AG$ to side $BC$.
$BG + CG = BC$, $AG \cdot \sqrt{3} + AG = \sqrt{3} - 1$, $AG = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = \frac{(\sqrt{3} - 1)^2}{2} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}$, $[ABC] = \frac12 \cdot AG \cdot BC = \frac12 (2 - \sqrt{3})(\sqrt{3} - 1)$
$[\text{12-sided outer polygon}] = [\text{cross}] + 8 \cdot [ABC] = 4\sqrt{3} - 3 + 8 \cdot \frac12 (2 - \sqrt{3})(\sqrt{3} - 1) = 4\sqrt{3} - 3 + 4 (3\sqrt{3} - 5) = 16\sqrt{3}-23$
\[16+3-23=\boxed{\textbf{(B)}\ -4}\]
~isabelchen | // Block 1
import geometry;
unitsize(3cm);
draw((0,0) -- (1,0) -- (1,1) -- (0,1) -- cycle);
draw(shift((1/2,1-sqrt(3)/2))*polygon(6));
draw(shift((1/2,sqrt(3)/2))*polygon(6));
draw(shift((sqrt(3)/2,1/2))*rotate(90)*polygon(6));
draw(shift((1-sqrt(3)/2,1/2))*rotate(90)*polygon(6));
draw((0,1-sqrt(3))--(1,1-sqrt(3))--(3-sqrt(3),sqrt(3)-2)--(sqrt(3),0)--(sqrt(3),1)--(3-sqrt(3),3-sqrt(3))--(1,sqrt(3))--(0,sqrt(3))--(sqrt(3)-2,3-sqrt(3))--(1-sqrt(3),1)--(1-sqrt(3),0)--(sqrt(3)-2,sqrt(3)-2)--cycle,linewidth(2));
draw((1,1)--(3-sqrt(3),3-sqrt(3)),linewidth(2));
draw((0,1)--(sqrt(3)-2,3-sqrt(3)),linewidth(2));
draw((0,0)--(sqrt(3)-2,sqrt(3)-2),linewidth(2));
draw((1,0)--(3-sqrt(3),sqrt(3)-2),linewidth(2));
draw((0, sqrt(3))--(0, -sqrt(3)+1),linewidth(2));
draw((1, sqrt(3))--(1, -sqrt(3)+1),linewidth(2));
draw((1-sqrt(3),0)--(sqrt(3),0),linewidth(2));
draw((1-sqrt(3),1)--(sqrt(3),1),linewidth(2));
label("$A$", (3-sqrt(3), 3-sqrt(3)), NE);
label("$B$", (1, sqrt(3)), NE);
label("$C$", (1,1), SW);
label("$D$", (1,0), NW);
label("$E$", (1, -sqrt(3)+1), SE);
label("$F$", (0, -sqrt(3)+1), SW);
label("$G$", (0, sqrt(3)), NW);
// Block 2
import geometry; unitsize(3cm); draw((0,0) -- (1,0) -- (1,1) -- (0,1) -- cycle); draw(shift((1/2,1-sqrt(3)/2))*polygon(6)); draw(shift((1/2,sqrt(3)/2))*polygon(6)); draw(shift((sqrt(3)/2,1/2))*rotate(90)*polygon(6)); draw(shift((1-sqrt(3)/2,1/2))*rotate(90)*polygon(6)); draw((0,1-sqrt(3))--(1,1-sqrt(3))--(3-sqrt(3),sqrt(3)-2)--(sqrt(3),0)--(sqrt(3),1)--(3-sqrt(3),3-sqrt(3))--(1,sqrt(3))--(0,sqrt(3))--(sqrt(3)-2,3-sqrt(3))--(1-sqrt(3),1)--(1-sqrt(3),0)--(sqrt(3)-2,sqrt(3)-2)--cycle,linewidth(2)); draw((1,1)--(3-sqrt(3),3-sqrt(3)),linewidth(2)); draw((0,1)--(sqrt(3)-2,3-sqrt(3)),linewidth(2)); draw((0,0)--(sqrt(3)-2,sqrt(3)-2),linewidth(2)); draw((1,0)--(3-sqrt(3),sqrt(3)-2),linewidth(2)); draw((0, sqrt(3))--(0, -sqrt(3)+1),linewidth(2)); draw((1, sqrt(3))--(1, -sqrt(3)+1),linewidth(2)); draw((1-sqrt(3),0)--(sqrt(3),0),linewidth(2)); draw((1-sqrt(3),1)--(sqrt(3),1),linewidth(2)); label("$A$", (3-sqrt(3), 3-sqrt(3)), NE); label("$B$", (1, sqrt(3)), NE); label("$C$", (1,1), SW); label("$D$", (1,0), NW); label("$E$", (1, -sqrt(3)+1), SE); label("$F$", (0, -sqrt(3)+1), SW); label("$G$", (0, sqrt(3)), NW); | ["https://artofproblemsolving.com/wiki/images/thumb/3/39/2022AMC12BP25.png/150px-2022AMC12BP25.png"] |
930 | Four regular hexagons surround a square with side length 1, each one sharing an edge with the square,
as shown in the figure below. The area of the resulting 12-sided outer nonconvex polygon can be
written as $m \sqrt{n} + p$, where $m$, $n$, and $p$ are integers and $n$ is not divisible by the square of any prime.
What is $m+n+p$?
$\textbf{(A) } -12 \qquad \textbf{(B) }-4 \qquad \textbf{(C) } 4 \qquad \textbf{(D) }24 \qquad \textbf{(E) }32$ | 2022 AMC 12B Problem 25 | We calculate the area as the area of the red octagon minus the four purple congruent triangles:
We first find the important angles in the figure. We note that 2 adjacent hexagons are rotated $90^\circ$ with respect to the other, so the angles between any sides is $150^\circ$. In particular, as the purple triangles are isosceles, they have angles $150^\circ,15^\circ$, and $15^\circ$, and the octagon is equiangular (all its angles are $135^\circ$). Thus, we can draw a square around the octagon, and we note that the ``cut out" triangles are all isosceles right triangles.
Now, we calculate the side length of the square. Note that the hexagon has a height of $\sqrt 3$, so the length of a side of the square is $2\sqrt 3-1$. In particular, the horizontal/vertical sides of the octagon have length $1$, so the legs of the isosceles triangles are
\[\frac{2\sqrt3-1-1}2=\sqrt3-1\]Thus, the area of the octagon is
\[(2\sqrt3-1)^2-4\cdot\frac 12(\sqrt3-1)^2=5\]Now, we calculate the area of one of the four isosceles triangles. The base of the triangle is $(\sqrt 3-1)\sqrt 2$, so the area is
\[\frac 14\mathrm{base length}^2\cdot\tan(\mathrm{base angle})=\frac 14((\sqrt3-1)\sqrt2)^2\cdot\tan15^\circ=\frac 14(8-4\sqrt3)(2-\sqrt3)=7-4\sqrt 3\]Thus, the area of the dodecagon is
\[5-4(7-4\sqrt3)=16\sqrt3-23\]Thus the answer is $16+3-23=-4$, or $\boxed{\textbf{(B)}}$.
~cr. naman12 | // Block 1
import geometry;
unitsize(3cm);
draw((1-sqrt(3),1-sqrt(3))--(1-sqrt(3),sqrt(3))--(sqrt(3),sqrt(3))--(sqrt(3),1-sqrt(3))--cycle,dashed);
filldraw((0,1-sqrt(3))--(1,1-sqrt(3))--(sqrt(3),0)--(sqrt(3),1)--(1,sqrt(3))--(0,sqrt(3))--(1-sqrt(3),1)--(1-sqrt(3),0)--cycle,red*0.2+white,red);
filldraw((1,1-sqrt(3))--(3-sqrt(3),sqrt(3)-2)--(sqrt(3),0)--cycle,purple*0.2+white,blue);
filldraw((sqrt(3),1)--(3-sqrt(3),3-sqrt(3))--(1,sqrt(3))--cycle,purple*0.2+white,blue);
filldraw((0,sqrt(3))--(sqrt(3)-2,3-sqrt(3))--(1-sqrt(3),1)--cycle,purple*0.2+white,blue);
filldraw((0,1-sqrt(3))--(1-sqrt(3),0)--(sqrt(3)-2,sqrt(3)-2)--cycle,purple*0.2+white,blue);
draw((0,0) -- (1,0) -- (1,1) -- (0,1) -- cycle);
draw(shift((1/2,1-sqrt(3)/2))*polygon(6));
draw(shift((1/2,sqrt(3)/2))*polygon(6));
draw(shift((sqrt(3)/2,1/2))*rotate(90)*polygon(6));
draw(shift((1-sqrt(3)/2,1/2))*rotate(90)*polygon(6));
// Block 2
import geometry; unitsize(3cm); draw((1-sqrt(3),1-sqrt(3))--(1-sqrt(3),sqrt(3))--(sqrt(3),sqrt(3))--(sqrt(3),1-sqrt(3))--cycle,dashed); filldraw((0,1-sqrt(3))--(1,1-sqrt(3))--(sqrt(3),0)--(sqrt(3),1)--(1,sqrt(3))--(0,sqrt(3))--(1-sqrt(3),1)--(1-sqrt(3),0)--cycle,red*0.2+white,red); filldraw((1,1-sqrt(3))--(3-sqrt(3),sqrt(3)-2)--(sqrt(3),0)--cycle,purple*0.2+white,blue); filldraw((sqrt(3),1)--(3-sqrt(3),3-sqrt(3))--(1,sqrt(3))--cycle,purple*0.2+white,blue); filldraw((0,sqrt(3))--(sqrt(3)-2,3-sqrt(3))--(1-sqrt(3),1)--cycle,purple*0.2+white,blue); filldraw((0,1-sqrt(3))--(1-sqrt(3),0)--(sqrt(3)-2,sqrt(3)-2)--cycle,purple*0.2+white,blue); draw((0,0) -- (1,0) -- (1,1) -- (0,1) -- cycle); draw(shift((1/2,1-sqrt(3)/2))*polygon(6)); draw(shift((1/2,sqrt(3)/2))*polygon(6)); draw(shift((sqrt(3)/2,1/2))*rotate(90)*polygon(6)); draw(shift((1-sqrt(3)/2,1/2))*rotate(90)*polygon(6)); | [] |
930 | Four regular hexagons surround a square with side length 1, each one sharing an edge with the square,
as shown in the figure below. The area of the resulting 12-sided outer nonconvex polygon can be
written as $m \sqrt{n} + p$, where $m$, $n$, and $p$ are integers and $n$ is not divisible by the square of any prime.
What is $m+n+p$?
$\textbf{(A) } -12 \qquad \textbf{(B) }-4 \qquad \textbf{(C) } 4 \qquad \textbf{(D) }24 \qquad \textbf{(E) }32$ | 2022 AMC 12B Problem 25 | [Image: images/amc/2022_AMC_12B_Problem_25_0.jpeg]
Note that each of the green sections is a rectangle, so its interior angles are all $90^{\circ}.$ Since $\angle{ABC}=120^{\circ}$, every one of the orange sections is a $30-60-90$ right triangle.
Define $x$ to be the distance from the corner of the square with side length $1$ to the corner of the larger blue square. Due to the sides of the two squares being parallel to each other, the large blue triangle is a $45-45-90$ right triangle. By $AAA$ similarity, the smaller blue triangles are also $45-45-90,$ and have side lengths of $\frac{x\sqrt{2}}{2}, \frac{x\sqrt{2}}{2},$ and $x$. By $30-60-90$ triangle relations, the largest altitude of the orange triangle is $\frac{x\sqrt{6}}{2}.$
Now, we can find the height of the hexagon to obtain an equation in terms of $x$. Consider a hexagon with side length $1$, where point $P$ is the foot of the perpendicular dropped from $B$, bisecting $\angle{ABC}$:
Note that triangles $\triangle{APB}$ and $\triangle{CPB}$ are congruent $30-60-90$ triangles, by SAS congruence. Since the side length of this hexagon is $1$, the length of $AP$ is $\frac{1}{2}\cdot \sqrt{3} =\frac{\sqrt{3}}{2}$, by $30-60-90$ triangle relations. The height of the hexagon is twice this value, or $\sqrt{3}.$
The height is also equal to the sum of the values along the long blue line, in the first diagram. Therefore, $1+\frac{x\sqrt{2}}{2}+\frac{x\sqrt{6}}{2}=\sqrt{3}.$ Solving and rationalizing, $x=2\sqrt{2}-\sqrt{6}.$
The area of the dodecagon is equal to the sum of the areas of the four rectangles, eight orange triangles, and purple square. In terms of $x$, this is
\[8\cdot \frac{1}{2}\cdot \frac{x\sqrt{2}}{2}\cdot \frac{x\sqrt{6}}{{2}}+4\cdot 1\cdot\frac{x\sqrt{6}}{2}+(x\sqrt{2}+1)^2.\]
Plugging in $x=2\sqrt{2}-\sqrt{6}$, the area of the dodecagon is $16\sqrt{3}-23$. Therefore, the answer is $16+3-23=$ $\boxed{\textbf{(B) }-4}.$
-Benedict T (countmath1) | // Block 1
import geometry;
import fontsize;
defaultpen(fontsize(9pt));
unitsize(3cm);
draw(shift((1/2,1-sqrt(3)/2))*polygon(6));
draw((1,1-sqrt(3))--(1, 1)--cycle,linewidth(1));
draw((3/2,1-sqrt(3)/2)--(1, 1-sqrt(3)/2)--cycle,linewidth(1));
draw((1, 1-sqrt(3)/2)--(1.1, 1-sqrt(3)/2)--(1.1, 1-sqrt(3)/2 +0.1)--(1, 1-sqrt(3)/2+0.1)--cycle,linewidth(1));
draw((1, 1-sqrt(3)/2)--(1.1, 1-sqrt(3)/2)--(1.1, 1-sqrt(3)/2 -0.1)--(1, 1-sqrt(3)/2-0.1)--cycle,linewidth(1));
label("$A$", (1, 1), N);
label("$P$", (1, 1-sqrt(3)/2), W);
label("$C$", (1, 1-sqrt(3)), S);
label("$B$", (3/2, 1-sqrt(3)/2), E);
label("$60^{\circ}$", (3/2-0.025, 1-sqrt(3)/2), NW);
label("$60^{\circ}$", (3/2-0.025, 1-sqrt(3)/2), SW);
// Block 2
import geometry; import fontsize; defaultpen(fontsize(9pt)); unitsize(3cm); draw(shift((1/2,1-sqrt(3)/2))*polygon(6)); draw((1,1-sqrt(3))--(1, 1)--cycle,linewidth(1)); draw((3/2,1-sqrt(3)/2)--(1, 1-sqrt(3)/2)--cycle,linewidth(1)); draw((1, 1-sqrt(3)/2)--(1.1, 1-sqrt(3)/2)--(1.1, 1-sqrt(3)/2 +0.1)--(1, 1-sqrt(3)/2+0.1)--cycle,linewidth(1)); draw((1, 1-sqrt(3)/2)--(1.1, 1-sqrt(3)/2)--(1.1, 1-sqrt(3)/2 -0.1)--(1, 1-sqrt(3)/2-0.1)--cycle,linewidth(1)); label("$A$", (1, 1), N); label("$P$", (1, 1-sqrt(3)/2), W); label("$C$", (1, 1-sqrt(3)), S); label("$B$", (3/2, 1-sqrt(3)/2), E); label("$60^{\circ}$", (3/2-0.025, 1-sqrt(3)/2), NW); label("$60^{\circ}$", (3/2-0.025, 1-sqrt(3)/2), SW); | ["https://artofproblemsolving.com/wiki/images/thumb/f/fb/2022AMC12B.jpeg/509px-2022AMC12B.jpeg"] |
930 | Four regular hexagons surround a square with side length 1, each one sharing an edge with the square,
as shown in the figure below. The area of the resulting 12-sided outer nonconvex polygon can be
written as $m \sqrt{n} + p$, where $m$, $n$, and $p$ are integers and $n$ is not divisible by the square of any prime.
What is $m+n+p$?
$\textbf{(A) } -12 \qquad \textbf{(B) }-4 \qquad \textbf{(C) } 4 \qquad \textbf{(D) }24 \qquad \textbf{(E) }32$ | 2022 AMC 12B Problem 25 | Obviously, each of the hexagon have area of $\frac{3\sqrt{3}}{2}$, which gives a total of $6\sqrt{3}$ (without accounting for the overcounted overlapped areas). Now we calculate the area of the intersection of two hexagons. There are two cases :
Case 1: The hexagons are on the opposite sides
In this case, we see that the two hexagons' intersection is the middle square plus two triangles on its side. The square have area $1$. Each of the triangles are $120-30-30$, and hence its short side is $\frac{\sqrt{3}}{3}$ because its longer side is $1$. Therefore, each of the small triangles have area of $\frac{\sqrt{3}}{12}$, so each of the intersections have $1+\frac{\sqrt{3}}{6}$ area. Since there are $2$ pairs of hexagons on opposite side of each other, that gives a total area of $2+\frac{\sqrt{3}}{3}$.
Case 2: The hexagons are adjacent
In this case, their intersection is a convex hexagon. We cut it in half and we're left with a quadrilateral with two sides with length one (because it shares these sides with one of the hexagons, and for this reason the angle between the two sides is $120$ degrees) and one short side and one long side. Here, we divide the area even further. We draw a line that divide the quadrilateral into a triangle with side lengths $1$, $1$, and $\sqrt{3}$(because it is $30-30-120$) and another triangle with unknown side lengths. Simple angle chasing yields the triangle with unknown side lengths is a $15-75-90$ triangle. If you remembered your sines and cosines, you know that $\sin{15}=\frac{\sqrt{6}-\sqrt{2}}{4}$ or you can calculate it by using the fact that $\sin{15}=\sin{45-30}$. Also, $\sin{75}=\frac{\sqrt{6}+\sqrt{2}}{4}$. Now, because the $15-75-90$ triangle have one side with length $\sqrt{3}$(it shares the side with the other triangle) and it corresponds to the 75 degrees angle, we have
\[\frac{\sqrt{3}}{\frac{\sqrt{6}+\sqrt{2}}{4}}=\frac{x}{\frac{\sqrt{6}-\sqrt{2}}{4}}\]
\[\implies x=\frac{\sqrt{3}{(\sqrt{6}-\sqrt{2})}}{\sqrt{6}+\sqrt{2}}=2\sqrt{3}-3\]
where $x$ denote the shorter leg of the $15-75-90$ triangle. Hence, that triangle have area
\[\frac{\sqrt{3}}{2} x=\frac{\sqrt{3}}{2}(2\sqrt{3}-3)=\frac{6-3\sqrt{3}}{2}\].
The isosceles $120-30-30$ triangle obviously have area $\frac{\sqrt{3}}{4}$, and that gives a total of $\frac{12-5\sqrt{3}}{4}$. We multiply that by $2$ because $\frac{12-5\sqrt{3}}{4}$ is only half of the intersection of the hexagons. We multiply the result by $4$ because there are $4$ pairs of adjacent hexagons. That gives $24-10\sqrt{3}$, which is it for Case 2.
Now that we got the total area of the intersection of two hexagons, we find the area of the intersection of three hexagons. Note that the intersection of any three hexagons is just the middle square plus one small triangle on the side. Again, the middle square have area $1$ and the triangle have area $\frac{\sqrt{3}}{12}$, so the total is $1+\frac{\sqrt{3}}{12}$. We multiply that by $4\choose3$ to get $4+\frac{\sqrt{3}}{3}$.
Then, we calculate the area of the intersection of all four hexagons, which is just the middle square with area $1$.
Finally, we apply PIE. The total area is
\[6\sqrt{3}-[(2+\frac{\sqrt{3}}{3})+(24-10\sqrt{3})]+(4+\frac{\sqrt{3}}{3})-1=16\sqrt{3}-23\],
so our final answer is $16+3-23=\boxed{(B)-4}$. $\blacksquare$ ~Ddk001 | // Block 1
pair[] A,B;
path h1, h2,h3,h4;
h1 = shift((sqrt(3)-1)/2*dir(90))*polygon(6);
h2 = shift((sqrt(3)-1)/2*dir(0))*rotate(90)*polygon(6);
h3 = shift((sqrt(3)-1)/2*dir(-90))*polygon(6);
h4 = shift((sqrt(3)-1)/2*dir(180))*rotate(90)*polygon(6);
B.cyclic=true;
B.push((0,0));
int indexSquare[] = sequence(4);
int indexPolygon[] = sequence(12);
for(int i : indexSquare){
A.push(1/sqrt(2)*dir(45+90*i));
}
draw(A[0]--A[1]--A[2]--A[3]--cycle);
fill(h1,green);
fill(h2,red);
fill(buildcycle(h2,h1),yellow);
draw(h1);
draw(h2);
draw(h3);
draw(h4);
for(int i : sequence(4))
{
for(int j : sequence(3))
{
B.push(B[3*i+j]+((sqrt(3)-1.5)*(j-1)*(j)+4-2*sqrt(3))*dir(90*i+60+30*j*(1.5*(j-1)-1)));
}
}
for(int i=0; i <12; ++i)
{
draw(shift(0.5,0.5-sqrt(3))*(B[i])--shift(0.5,0.5-sqrt(3))*(B[i + 1]),linewidth(2));
}
// Block 2
pair[] A,B; path h1, h2,h3,h4; h1 = shift((sqrt(3)-1)/2*dir(90))*polygon(6); h2 = shift((sqrt(3)-1)/2*dir(0))*rotate(90)*polygon(6); h3 = shift((sqrt(3)-1)/2*dir(-90))*polygon(6); h4 = shift((sqrt(3)-1)/2*dir(180))*rotate(90)*polygon(6); B.cyclic=true; B.push((0,0)); int indexSquare[] = sequence(4); int indexPolygon[] = sequence(12); for(int i : indexSquare){ A.push(1/sqrt(2)*dir(45+90*i)); } draw(A[0]--A[1]--A[2]--A[3]--cycle); fill(h1,green); fill(h2,red); fill(buildcycle(h2,h1),yellow); draw(h1); draw(h2); draw(h3); draw(h4); for(int i : sequence(4)) { for(int j : sequence(3)) { B.push(B[3*i+j]+((sqrt(3)-1.5)*(j-1)*(j)+4-2*sqrt(3))*dir(90*i+60+30*j*(1.5*(j-1)-1))); } } for(int i=0; i <12; ++i) { draw(shift(0.5,0.5-sqrt(3))*(B[i])--shift(0.5,0.5-sqrt(3))*(B[i + 1]),linewidth(2)); } | [] |
931 | A quadrilateral has all integer sides lengths, a perimeter of $26$, and one side of length $4$. What is the greatest possible length of one side of this quadrilateral?
$\textbf{(A) }9\qquad\textbf{(B) }10\qquad\textbf{(C) }11\qquad\textbf{(D) }12\qquad\textbf{(E) }13$ | 2023 AMC 10A Problem 4 | This is an AMC 10 problem 4, so there is no need for any complex formulas. The largest singular side length from a quadrilateral comes from a trapezoid. So we can set the $2$ sides of the trapezoid equal to $4$. Next we can split the trapezoid into $5$ triangles, where each base length of the triangle equals $4$. So the top side equals $8$, and the bottom side length equals $4+4+4$ $=$ $\boxed {\textbf{(D) 12}}$
~ kabbybear | // Block 1
size(120);
draw((0,0)--(2,6),red);
draw((2,6)--(5,6),red);
draw((5,6)--(7,0),red);
draw((0,0)--(7,0),red);
draw((2,6)--(2.67,0),red);
draw((2.67,0)--(3.5,6),red);
draw((3.5,6)--(4.67,0),red);
draw((4.67,0)--(5,6),red);
// Block 2
size(120); draw((0,0)--(2,6),red); draw((2,6)--(5,6),red); draw((5,6)--(7,0),red); draw((0,0)--(7,0),red); draw((2,6)--(2.67,0),red); draw((2.67,0)--(3.5,6),red); draw((3.5,6)--(4.67,0),red); draw((4.67,0)--(5,6),red); | [] |
932 | Let $ABCD$ be a rectangle with $AB = 30$ and $BC = 28$. Point $P$ and $Q$ lie on $\overline{BC}$ and $\overline{CD}$ respectively so that all sides of $\triangle{ABP}, \triangle{PCQ},$ and $\triangle{QDA}$ have integer lengths. What is the perimeter of $\triangle{APQ}$?
$\textbf{(A) } 84 \qquad \textbf{(B) } 86 \qquad \textbf{(C) } 88 \qquad \textbf{(D) } 90 \qquad \textbf{(E) } 92$ | 2023 AMC 10A Problem 17 | We know that all side lengths are integers, so we can test Pythagorean triples for all triangles.
First, we focus on $\triangle{ABP}$. The length of $AB$ is $30$, and the possible (small enough) Pythagorean triples $\triangle{ABP}$ can be are $(3, 4, 5), (5, 12, 13), (8, 15, 17),$ where the length of the longer leg is a factor of $30$. Testing these, we get that only $(8, 15, 17)$ is a valid solution. Thus, we know that $BP = 16$ and $AP = 34$.
Next, we move on to $\triangle{QDA}$. The length of $AD$ is $28$, and the small enough triples are $(3, 4, 5)$ and $(7, 24, 25)$. Testing again, we get that $(3, 4, 5)$ is our triple. We get the value of $DQ = 21$, and $AQ = 35$.
We know that $CQ = CD - DQ$ which is $9$, and $CP = BC - BP$ which is $12$. $\triangle{CPQ}$ is therefore a right triangle with side length ratios ${3, 4, 5}$, and the hypotenuse is equal to $15$.
$\triangle{APQ}$ has side lengths $34, 35,$ and $15,$ so the perimeter is equal to $34 + 35 + 15 = \boxed{\textbf{(A) } 84}.$
~Gabe Horn ~ItsMeNoobieboy ~oinava | // Block 1
/* ~ItsMeNoobieboy */
size(200);
pair A, B, C, D, P, Q;
A = (0,28/30);
B = (1,28/30);
C = (1,0);
D = (0,0);
P = (1,12/30);
Q = (21/30,0);
draw(A--B--C--D--cycle);
draw(A--P--Q--cycle);
dot("$A$",A,NW,linewidth(4));
dot("$B$",B,NE,linewidth(4));
dot("$C$",C,SE,linewidth(4));
dot("$D$",D,SW,linewidth(4));
dot("$P$",P,E,linewidth(4));
dot("$Q$",Q,S,linewidth(4));
label("$30$",midpoint(A--B),N);
label("$16$",midpoint(B--P),E);
label("$34$",midpoint(A--P),NE, red);
label("$28$",midpoint(A--D),W);
label("$21$",midpoint(D--Q),S);
label("$35$",midpoint(A--Q),SW, red);
label("$9$",midpoint(Q--C),S);
label("$12$",midpoint(C--P),E);
label("$15$",midpoint(Q--P),SE, red);
// Block 2
/* ~ItsMeNoobieboy */ size(200); pair A, B, C, D, P, Q; A = (0,28/30); B = (1,28/30); C = (1,0); D = (0,0); P = (1,12/30); Q = (21/30,0); draw(A--B--C--D--cycle); draw(A--P--Q--cycle); dot("$A$",A,NW,linewidth(4)); dot("$B$",B,NE,linewidth(4)); dot("$C$",C,SE,linewidth(4)); dot("$D$",D,SW,linewidth(4)); dot("$P$",P,E,linewidth(4)); dot("$Q$",Q,S,linewidth(4)); label("$30$",midpoint(A--B),N); label("$16$",midpoint(B--P),E); label("$34$",midpoint(A--P),NE, red); label("$28$",midpoint(A--D),W); label("$21$",midpoint(D--Q),S); label("$35$",midpoint(A--Q),SW, red); label("$9$",midpoint(Q--C),S); label("$12$",midpoint(C--P),E); label("$15$",midpoint(Q--P),SE, red); | [] |
933 | The line segment formed by $A(1, 2)$ and $B(3, 3)$ is rotated to the line segment formed by $A'(3, 1)$ and $B'(4, 3)$ about the point $P(r, s)$. What is $|r-s|$?
$\textbf{(A) } \frac{1}{4} \qquad \textbf{(B) } \frac{1}{2} \qquad \textbf{(C) } \frac{3}{4} \qquad \textbf{(D) } \frac{2}{3} \qquad \textbf{(E) } 1$ | 2023 AMC 10A Problem 19 | To find the center of rotation, we find the intersection point of the perpendicular bisectors of $\overline{AA^\prime}$ and $\overline{BB^\prime}$. (this is because $\overline{PA}$ is equal to $\overline{PA^\prime}$ and same with $B$ and $B^\prime$
We can find that the equation of the line $\overline{AA^\prime}$ is $y = -\frac{1}{2}x + \frac{5}{2}$, and that the equation of the line $\overline{BB^\prime}$ is $y = 3$.
When we solve for the perpendicular bisector of $y = -\frac{1}{2}x + \frac{5}{2}$, we determine that it has a slope of 2, and it runs through $(2, 1.5)$. Plugging in $1.5 = 2(2)-n$, we get than $n = \frac{5}{2}$. Therefore our perpendicular bisector is $y=2x-\frac{5}{2}$. Next, we solve for the perpendicular of $y = 3$. We know that it has an undefined slope, and it runs through $(3.5, 3)$. We can determine that our second perpendicular bisector is $x = 3.5$.
Setting the equations equal to each other, we get $y = \frac{9}{2}$. Therefore, $|r - s| = |3.5 - 4.5| = \boxed{\textbf{(E) } 1}$.
~aydenlee & wuwang2002
~minor explanations by pungent_muskrat | // Block 1
pair A=(1,2);
pair B=(3,3);
pair A1=(3,1);
pair B1=(4,3);
dot("A",A,NW);
dot("B",B,S);
dot("A'",A1,S);
dot("B'",B1,E);
draw(A--A1);
draw(B--B1);
draw((3.5,0)--(3.5,6),BeginArrow(5),EndArrow(5));
draw((1,-0.5)--(4,5.5),BeginArrow(5),EndArrow(5));
pair P=(3.5,4.5);
dot("P",P,NW);
// Block 2
pair A=(1,2); pair B=(3,3); pair A1=(3,1); pair B1=(4,3); dot("A",A,NW); dot("B",B,S); dot("A'",A1,S); dot("B'",B1,E); draw(A--A1); draw(B--B1); draw((3.5,0)--(3.5,6),BeginArrow(5),EndArrow(5)); draw((1,-0.5)--(4,5.5),BeginArrow(5),EndArrow(5)); pair P=(3.5,4.5); dot("P",P,NW); | [] |
934 | Each square in a $3\times3$ grid of squares is colored red, white, blue, or green so that every $2\times2$ square contains one square of each color. One such coloring is shown on the right below. How many different colorings are possible?
$\textbf{(A) }24\qquad\textbf{(B) }48\qquad\textbf{(C) }60\qquad\textbf{(D) }72\qquad\textbf{(E) }96$ | 2023 AMC 10A Problem 20 | We can split this problem into $2$ cases as shown in solution 1. We can swap a set of equal colors for another set of equal colors to create a new square.
Square 1:
The first square can be rotated to create another square so we have to multiply the number of arrangements by $2$. We have $4! = 24$ arrangements without rotating and $24\cdot 2 = 48$ arrangements in total for the first square.
Square 2:
There are $4! = 24$ ways to arrange the colors.
In total, we have $48 + 24 = \boxed{\textbf{(D) }72}$ arrangements.
~South (LaTeX and Solution)
Edit by: Mismatchedcubing/Andrew_Lu | // Block 1
unitsize(0.5cm, 0.5cm);
draw((0,0)--(9,0)--(9,9)--(0,9)--cycle);
draw((0,3)--(9,3));
draw((0,6)--(9,6));
draw((3,0)--(3,9));
draw((6,0)--(6,9));
label("R", (1.5,1.5));
label("B", (4.5,1.5));
label("R", (7.5,1.5));
label("G", (1.5,4.5));
label("W", (4.5,4.5));
label("G", (7.5,4.5));
label("B", (1.5,7.5));
label("R", (4.5,7.5));
label("B", (7.5,7.5));
draw((18,0)--(27,0)--(27,9)--(18,9)--cycle);
draw((18,3)--(27,3));
draw((18,6)--(27,6));
draw((21,0)--(21,9));
draw((24,0)--(24,9));
label("R", (19.5,1.5));
label("B", (22.5,1.5));
label("R", (25.5,1.5));
label("G", (19.5,4.5));
label("W", (22.5,4.5));
label("G", (25.5,4.5));
label("R", (19.5,7.5));
label("B", (22.5,7.5));
label("R", (25.5,7.5));
// Block 2
unitsize(0.5cm, 0.5cm); draw((0,0)--(9,0)--(9,9)--(0,9)--cycle); draw((0,3)--(9,3)); draw((0,6)--(9,6)); draw((3,0)--(3,9)); draw((6,0)--(6,9)); label("R", (1.5,1.5)); label("B", (4.5,1.5)); label("R", (7.5,1.5)); label("G", (1.5,4.5)); label("W", (4.5,4.5)); label("G", (7.5,4.5)); label("B", (1.5,7.5)); label("R", (4.5,7.5)); label("B", (7.5,7.5)); draw((18,0)--(27,0)--(27,9)--(18,9)--cycle); draw((18,3)--(27,3)); draw((18,6)--(27,6)); draw((21,0)--(21,9)); draw((24,0)--(24,9)); label("R", (19.5,1.5)); label("B", (22.5,1.5)); label("R", (25.5,1.5)); label("G", (19.5,4.5)); label("W", (22.5,4.5)); label("G", (25.5,4.5)); label("R", (19.5,7.5)); label("B", (22.5,7.5)); label("R", (25.5,7.5)); | [] |
934 | Each square in a $3\times3$ grid of squares is colored red, white, blue, or green so that every $2\times2$ square contains one square of each color. One such coloring is shown on the right below. How many different colorings are possible?
$\textbf{(A) }24\qquad\textbf{(B) }48\qquad\textbf{(C) }60\qquad\textbf{(D) }72\qquad\textbf{(E) }96$ | 2023 AMC 10A Problem 20 | Note that we could fill the 3 by 3 square with numbers of choices, rather than letters or color names. The top-left corner receives a 3 because there are 3 total choices to choose from: R, G and B. The squares bordering them has values of 2 and 1, regardless of order. 2 indicates that the small square can have any color excluding the one existing color, 1 indicates the remaining color of the 2 by 2 square. Finally, the middle square receives 3, since the first 2 by 2 square has RGB already, and it does not matter what color it has. Moving onto the next 2 by 2 square, we see that there are already 2 decided colors, so the other squares must be 2 and 1, again, regardless of the order. The same applys to the third 2 by 2 square, and finally the last square has the value of one, as it is the only square left. Multiplying the numbers, $2\times2\times2\times3\time3\times3$ = $\boxed{\textbf{(D) }72}$ (Similar to Solution 1)
-MEZE_RUN
why are there only 3 choices for the top right corner? you havent placed any colors yet so there should be 4 choices which would make your answer incorrect | // Block 1
unitsize(0.5cm, 0.5cm);
draw((0,0)--(9,0)--(9,9)--(0,9)--cycle);
draw((0,3)--(9,3));
draw((0,6)--(9,6));
draw((3,0)--(3,9));
draw((6,0)--(6,9));
label("2", (1.5,1.5));
label("1", (4.5,1.5));
label("1", (7.5,1.5));
label("2", (1.5,4.5));
label("3", (4.5,4.5));
label("1", (7.5,4.5));
label("3", (1.5,7.5));
label("1", (4.5,7.5));
label("2", (7.5,7.5));
// Block 2
unitsize(0.5cm, 0.5cm); draw((0,0)--(9,0)--(9,9)--(0,9)--cycle); draw((0,3)--(9,3)); draw((0,6)--(9,6)); draw((3,0)--(3,9)); draw((6,0)--(6,9)); label("2", (1.5,1.5)); label("1", (4.5,1.5)); label("1", (7.5,1.5)); label("2", (1.5,4.5)); label("3", (4.5,4.5)); label("1", (7.5,4.5)); label("3", (1.5,7.5)); label("1", (4.5,7.5)); label("2", (7.5,7.5)); | [] |
934 | Each square in a $3\times3$ grid of squares is colored red, white, blue, or green so that every $2\times2$ square contains one square of each color. One such coloring is shown on the right below. How many different colorings are possible?
$\textbf{(A) }24\qquad\textbf{(B) }48\qquad\textbf{(C) }60\qquad\textbf{(D) }72\qquad\textbf{(E) }96$ | 2023 AMC 10A Problem 20 | Let's name the cells $A,B,C,D,E,F,G,H,I$ from the top left to the bottom right.
Case 1. Cell $B$ and cell $H$ have the same color. The middle one cell $E$ has $4$ choices, cell $B$ has $3$ choices, then cell $E$ has $2$ choices and cell $F$ has $2$ choices, this gives $4\cdot 3\cdot 2\cdot 2=48$ ways.
Case 2. Cell $B$ and cell $H$ have different colors. The middle one cell $E$ has $4$ choices, cell $B$ has $3$ choices, cell $H$ has $2$ choices, then cell $D$ and $F$ each can only have one choice (different from $B,E,H).$ This gives
$4\cdot 3\cdot 2=24$ ways.
The answer is thus $48+24=72,$ corresponding to answer choice $\boxed{D}$ | // Block 1
unitsize(0.5cm, 0.5cm);
draw((0,0)--(9,0)--(9,9)--(0,9)--cycle);
draw((0,3)--(9,3));
draw((0,6)--(9,6));
draw((3,0)--(3,9));
draw((6,0)--(6,9));
label("G", (1.5,1.5));
label("H", (4.5,1.5));
label("I", (7.5,1.5));
label("D", (1.5,4.5));
label("E", (4.5,4.5));
label("F", (7.5,4.5));
label("A", (1.5,7.5));
label("B", (4.5,7.5));
label("C", (7.5,7.5));
// Block 2
unitsize(0.5cm, 0.5cm); draw((0,0)--(9,0)--(9,9)--(0,9)--cycle); draw((0,3)--(9,3)); draw((0,6)--(9,6)); draw((3,0)--(3,9)); draw((6,0)--(6,9)); label("G", (1.5,1.5)); label("H", (4.5,1.5)); label("I", (7.5,1.5)); label("D", (1.5,4.5)); label("E", (4.5,4.5)); label("F", (7.5,4.5)); label("A", (1.5,7.5)); label("B", (4.5,7.5)); label("C", (7.5,7.5)); | [] |
935 | Circle $C_1$ and $C_2$ each have radius $1$, and the distance between their centers is $\frac{1}{2}$. Circle $C_3$ is the largest circle internally tangent to both $C_1$ and $C_2$. Circle $C_4$ is internally tangent to both $C_1$ and $C_2$ and externally tangent to $C_3$. What is the radius of $C_4$?
$\textbf{(A) } \frac{1}{14} \qquad \textbf{(B) } \frac{1}{12} \qquad \textbf{(C) } \frac{1}{10} \qquad \textbf{(D) } \frac{3}{28} \qquad \textbf{(E) } \frac{1}{9}$ | 2023 AMC 10A Problem 22 | Let $O$ be the center of the midpoint of the line segment connecting both the centers, say $A$ and $B$.
Let the point of tangency with the inscribed circle and the right larger circles be $T$.
Then $OT = BO + BT = BO + AT - \frac{1}{2} = \frac{1}{4} + 1 - \frac{1}{2} = \frac{3}{4}.$
Since $C_4$ is internally tangent to $C_1$, center of $C_4$, $C_1$ and their tangent point must be on the same line.
Now, if we connect centers of $C_4$, $C_3$ and $C_1$/$C_2$, we get a right angled triangle.
Let the radius of $C_4$ equal $r$. With the pythagorean theorem on our triangle, we have
\[\left(r+\frac{3}{4}\right)^2+\left(\frac{1}{4}\right)^2=(1-r)^2\]
Solving this equation gives us
\[r = \boxed{\textbf{(D) } \frac{3}{28}}\]
~lptoggled
~ShawnX (Diagram)
~ap246 (Minor Changes) | import olympiad; size(10cm); draw(circle((0,0),0.75), gray(0.7)); draw(circle((-0.25,0),1), gray(0.7)); draw(circle((0.25,0),1), gray(0.7)); draw(circle((0,6/7),3/28), gray(0.7)); pair A = (0,0), B = (-0.25,0), C = (0.25,0), D = (0,6/7), EE = (-0.95710678118, 0.70710678118), F = (0.95710678118, -0.70710678118), G = (0,0), T=(0.75,0); dot(D); dot(G); draw(B--EE, dashed+gray(0.7)); draw(C--F, dashed+gray(0.7)); dot(C, gray(0.9)); draw(B--C, gray(0.7)); draw(B--A); draw(A--D); draw(B--D); draw(B--T); label("$\frac{1}{4}$", (-0.125, -0.125)); label("$r + \frac{3}{4}$", (0.2, 3/7)); label("$1 - r$", (-0.29, 3/7)); label("$O$",A,S); label("$A$",B,S); dot("$B$",C,S); dot("$T$",T,E); label("$1$", (-.85, 0.70)); label("$1$", (.85, -.7)); markscalefactor=0.05; | [] |
936 | Six regular hexagonal blocks of side length 1 unit are arranged inside a regular hexagonal frame. Each block lies along an inside edge of the frame and is aligned with two other blocks, as shown in the figure below. The distance from any corner of the frame to the nearest vertex of a block is $\frac{3}{7}$ unit. What is the area of the region inside the frame not occupied by the blocks?
$\textbf{(A)}~\frac{13 \sqrt{3}}{3}\qquad\textbf{(B)}~\frac{216 \sqrt{3}}{49}\qquad\textbf{(C)}~\frac{9 \sqrt{3}}{2} \qquad\textbf{(D)}~ \frac{14 \sqrt{3}}{3}\qquad\textbf{(E)}~\frac{243 \sqrt{3}}{49}$ | 2023 AMC 10A Problem 24 | Examining the red isosceles trapezoid with $1$ and $\dfrac{3}{7}$ as two bases, we know that the side lengths are $\dfrac{4}{7}$ from $30-60-90$ triangle.
We can conclude that the big hexagon has side length 3.
Thus the target area is: area of the big hexagon - 6 * area of the small hexagon.
$\dfrac{3\sqrt{3}}{2}(3^2-6\cdot1^2) = \dfrac{3\sqrt{3}}{2}(3) = \boxed{\textbf{(C)}~\frac{9 \sqrt{3}}{2}}$
~Technodoggo
Solution 1.1
We can extend the line of the parallelogram to the end until it touches the next hexagon and it will make a small equilateral triangle and a longer parallelogram. We can prove that one side of the tiny equilateral triangle is 4/7 by playing around with angles and the parallelogram because it is parallel, we can then use the whole side of the hexagon which is one and subtract 3/7 which is one side of the equilateral triangle which is 4/7. That means the whole side of the big hexagon length is 3 and we can continue with solution 1.
-BrandonChen
Solution 1.3
An alternative diagram, perhaps slightly more intuitive, can be drawn by extending segments to form two equilateral triangles (shown in blue) with side length 1.
The red segment is given to have length \(3/7\). Furthermore, we can find yet another equilateral triangle within the bottom blue triangle that has side length \(1-3/7 = 4/7\), shown in dashed magenta. Adding all four segments (red, blue, blue, magenta), we find the side length of the larger hexagon to be \(3/7+1+1+4/7 = 3\). Then we can continue with Solution 1.
~abghim
(diagram borrowed mostly from ~Technodoggo's Solution 1) | // Block 1
unitsize(1cm);
pair A, B, C, D, E, F, W,X,Y,Z;
real bigSide = 3;
real smallSide = 1;
real angle = 60; // Each external angle for the hexagon
real offset = 3/7; // Offset for the smaller hexagons
// Function to draw a hexagon given a starting point and side length
void drawHexagon(pair start, real side) {
pair current = start;
for (int i = 0; i < 6; ++i) {
pair next = current + side * dir(angle * i);
draw(current--next);
current = next;
}
draw(current--start); // Close the hexagon
}
// Define the first vertex of the big hexagon
A = (0,0);
// Calculate the other vertices of the big hexagon
B = A + bigSide * dir(0);
C = B + bigSide * dir(angle);
D = C + bigSide * dir(2*angle);
E = D + bigSide * dir(3*angle);
F = E + bigSide * dir(4*angle);
// Draw the big hexagon
drawHexagon(A, bigSide);
// Function to calculate the center of a side given two vertices
pair sideCenter(pair start, pair end) {
return (start + end)/2;
}
// Draw the smaller hexagons
drawHexagon(A + offset * dir(0), smallSide);
drawHexagon(B - smallSide * dir(0)+offset*dir(60), smallSide);
drawHexagon(C - smallSide * dir(0)-dir(60)+dir(120)*3/7, smallSide);
drawHexagon(D - 2*smallSide*dir(120)-(2+3/7)*smallSide, smallSide);
drawHexagon(E - 2*smallSide*dir(60)+smallSide-3/7*dir(60), smallSide);
drawHexagon(F + smallSide*dir(-60)+(3/7)*dir(-60), smallSide);
// Optionally, label the vertices of the big hexagon
label("$A$", A, SW);
label("$B$", B, SE);
label("$C$", C, E);
label("$D$", D, NE);
label("$E$", E, NW);
label("$F$", F, W);
void drawTrap(pair W, real side, pen p) {
X = W+(3/7)*side*dir(0);
Y = X+(4/7)*side*dir(60);
Z = Y - side*dir(0);
draw(W--X, p);
draw(X--Y,p);
draw(Y--Z,p);
draw(Z--W,p);
}
W = A+smallSide*dir(120);
drawTrap(W,1, red+2);
pair W2,W3,W4,W5;
W2 = A+3*dir(-90);
W3 = W2+dir(90)*4*sqrt(3)/7;
W4 = W3+dir(0)*6/7;
W5 = W2+dir(0)*6/7;
drawTrap(W2,2,blue+1);
draw(W2--W3,blue+0.5);
draw(W4--W5,blue+0.5);
label("2/7",W3,NW);
label("3/7",W3,NE);
W4 = W3+6/7*dir(0);
label("2/7",W4,NE);
label("4/7",W2+dir(160)*0.5,W);
draw(A -1.5*dir(45)-- F -1.5*dir(45), green+0.5);
pair J,K,L,M,N;
J = ((A/10+9*F/10))-0.25*dir(45);
L = ((A+F)/2)-0.25*dir(45);
K = ((J+L)/2)-0.25*dir(45);
M = ((L+A)/2)-0.25*dir(45);
N = ((A+F)/2)-1.6*dir(45);
label("3/7",J,SW);
label("4/7",L,SW);
label("1",K,SW);
label("1",M,SW);
label("3",N,SW);
// Block 2
unitsize(1cm); pair A, B, C, D, E, F, W,X,Y,Z; real bigSide = 3; real smallSide = 1; real angle = 60; // Each external angle for the hexagon real offset = 3/7; // Offset for the smaller hexagons // Function to draw a hexagon given a starting point and side length void drawHexagon(pair start, real side) { pair current = start; for (int i = 0; i < 6; ++i) { pair next = current + side * dir(angle * i); draw(current--next); current = next; } draw(current--start); // Close the hexagon } // Define the first vertex of the big hexagon A = (0,0); // Calculate the other vertices of the big hexagon B = A + bigSide * dir(0); C = B + bigSide * dir(angle); D = C + bigSide * dir(2*angle); E = D + bigSide * dir(3*angle); F = E + bigSide * dir(4*angle); // Draw the big hexagon drawHexagon(A, bigSide); // Function to calculate the center of a side given two vertices pair sideCenter(pair start, pair end) { return (start + end)/2; } // Draw the smaller hexagons drawHexagon(A + offset * dir(0), smallSide); drawHexagon(B - smallSide * dir(0)+offset*dir(60), smallSide); drawHexagon(C - smallSide * dir(0)-dir(60)+dir(120)*3/7, smallSide); drawHexagon(D - 2*smallSide*dir(120)-(2+3/7)*smallSide, smallSide); drawHexagon(E - 2*smallSide*dir(60)+smallSide-3/7*dir(60), smallSide); drawHexagon(F + smallSide*dir(-60)+(3/7)*dir(-60), smallSide); // Optionally, label the vertices of the big hexagon label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, E); label("$D$", D, NE); label("$E$", E, NW); label("$F$", F, W); void drawTrap(pair W, real side, pen p) { X = W+(3/7)*side*dir(0); Y = X+(4/7)*side*dir(60); Z = Y - side*dir(0); draw(W--X, p); draw(X--Y,p); draw(Y--Z,p); draw(Z--W,p); } W = A+smallSide*dir(120); drawTrap(W,1, red+2); pair W2,W3,W4,W5; W2 = A+3*dir(-90); W3 = W2+dir(90)*4*sqrt(3)/7; W4 = W3+dir(0)*6/7; W5 = W2+dir(0)*6/7; drawTrap(W2,2,blue+1); draw(W2--W3,blue+0.5); draw(W4--W5,blue+0.5); label("2/7",W3,NW); label("3/7",W3,NE); W4 = W3+6/7*dir(0); label("2/7",W4,NE); label("4/7",W2+dir(160)*0.5,W); draw(A -1.5*dir(45)-- F -1.5*dir(45), green+0.5); pair J,K,L,M,N; J = ((A/10+9*F/10))-0.25*dir(45); L = ((A+F)/2)-0.25*dir(45); K = ((J+L)/2)-0.25*dir(45); M = ((L+A)/2)-0.25*dir(45); N = ((A+F)/2)-1.6*dir(45); label("3/7",J,SW); label("4/7",L,SW); label("1",K,SW); label("1",M,SW); label("3",N,SW);
// Block 3
unitsize(1cm); pair A, B, C, D, E, F, W,X,Y,Z; real bigSide = 3; real smallSide = 1; real angle = 60; // Each external angle for the hexagon real offset = 3/7; // Offset for the smaller hexagons // Function to draw a hexagon given a starting point and side length void drawHexagon(pair start, real side) { pair current = start; for (int i = 0; i < 6; ++i) { pair next = current + side * dir(angle * i); draw(current--next); current = next; } draw(current--start); // Close the hexagon } // Define the first vertex of the big hexagon A = (0,0); // Calculate the other vertices of the big hexagon B = A + bigSide * dir(0); C = B + bigSide * dir(angle); D = C + bigSide * dir(2*angle); E = D + bigSide * dir(3*angle); F = E + bigSide * dir(4*angle); // Draw the big hexagon drawHexagon(A, bigSide); // Function to calculate the center of a side given two vertices pair sideCenter(pair start, pair end) { return (start + end)/2; } // Draw the smaller hexagons drawHexagon(A + offset * dir(0), smallSide); drawHexagon(B - smallSide * dir(0)+offset*dir(60), smallSide); drawHexagon(C - smallSide * dir(0)-dir(60)+dir(120)*3/7, smallSide); drawHexagon(D - 2*smallSide*dir(120)-(2+3/7)*smallSide, smallSide); drawHexagon(E - 2*smallSide*dir(60)+smallSide-3/7*dir(60), smallSide); drawHexagon(F + smallSide*dir(-60)+(3/7)*dir(-60), smallSide); // Optionally, label the vertices of the big hexagon label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, E); label("$D$", D, NE); label("$E$", E, NW); label("$F$", F, W); void drawTriangle(pair W, real side, pen p) { X = W+(3/7)*side*dir(0); Y = A+(3/7)*side*dir(0); draw(W--X, p); draw(W--A, p); draw(Y--X, p); draw(A--Y, p); } W = A+11/7*dir(120); drawTriangle(W,1, purple+2); draw(A -1.5*dir(45)-- F -1.5*dir(45), green+0.5); pair J,K,L,M,N; J = ((A/10+9*F/10))-0.25*dir(45); L = ((A+F)/2)-0.25*dir(45); K = ((J+L)/2)-0.25*dir(45); M = ((L+A)/2)-0.25*dir(45); N = ((A+F)/2)-1.6*dir(45); label("3/7",J,SW); label("4/7",L,SW); label("1",K,SW); label("1",M,SW); label("3",N,SW);
// Block 4
// Asymptote code borrowed from Technodoggo's solution unitsize(1cm); pair A, B, C, D, E, F, W,X,Y,Z; real bigSide = 3; real smallSide = 1; real angle = 60; // Each external angle for the hexagon real offset = 3/7; // Offset for the smaller hexagons // Function to draw a hexagon given a starting point and side length void drawHexagon(pair start, real side) { pair current = start; for (int i = 0; i < 6; ++i) { pair next = current + side * dir(angle * i); draw(current--next); current = next; } draw(current--start); // Close the hexagon } // Define the first vertex of the big hexagon A = (0,0); // Calculate the other vertices of the big hexagon B = A + bigSide * dir(0); C = B + bigSide * dir(angle); D = C + bigSide * dir(2*angle); E = D + bigSide * dir(3*angle); F = E + bigSide * dir(4*angle); // Draw the big hexagon drawHexagon(A, bigSide); // Function to calculate the center of a side given two vertices pair sideCenter(pair start, pair end) { return (start + end)/2; } // Draw the smaller hexagons drawHexagon(A + offset * dir(0), smallSide); drawHexagon(B - smallSide * dir(0)+offset*dir(60), smallSide); drawHexagon(C - smallSide * dir(0)-dir(60)+dir(120)*3/7, smallSide); drawHexagon(D - 2*smallSide*dir(120)-(2+3/7)*smallSide, smallSide); drawHexagon(E - 2*smallSide*dir(60)+smallSide-3/7*dir(60), smallSide); drawHexagon(F + smallSide*dir(-60)+(3/7)*dir(-60), smallSide); // Optionally, label the vertices of the big hexagon label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, E); label("$D$", D, NE); label("$E$", E, NW); label("$F$", F, W); pair J,K,L,M,N; J = ((A/10+9*F/10))-0.25*dir(45); L = ((A+F)/2)-0.25*dir(45); K = ((J+L)/2)-0.25*dir(45); M = ((L+A)/2)-0.25*dir(45); // Extending lines to form two triangles X = A+dir(180)*0.57+dir(60)*(1+4/7); Y = A+dir(180)*0.57; draw(X -- Y, blue+1.5); draw(Y -- A+dir(0)*(3/7), blue+1.5); draw(A+dir(0)*(3/7) -- A+dir(0)*(3/7)+dir(120)*1, blue+1.5); draw(A+dir(120)*(4/7) -- A+dir(120)*(4/7)+dir(120), blue+1.5); draw(A+dir(120)*(4/7)+dir(120) -- X, blue+1.5); draw(A+dir(120)*(4/7)+dir(120) -- A+dir(120)*(4/7)+dir(120)+dir(120), blue+1.5); draw(A+dir(120)*(4/7)+dir(120)+dir(120) -- F, red+1.5); draw(A -- A+dir(120)*(4/7), dashed+magenta+1.5+linetype("0 1.4")); draw(A -- A+dir(180)*(4/7), dashed+magenta+1.5+linetype("0 1.4")); draw(A+dir(180)*(4/7) -- A+dir(120)*(4/7), dashed+magenta+1.5+linetype("0 1.4")); | [] |
936 | Six regular hexagonal blocks of side length 1 unit are arranged inside a regular hexagonal frame. Each block lies along an inside edge of the frame and is aligned with two other blocks, as shown in the figure below. The distance from any corner of the frame to the nearest vertex of a block is $\frac{3}{7}$ unit. What is the area of the region inside the frame not occupied by the blocks?
$\textbf{(A)}~\frac{13 \sqrt{3}}{3}\qquad\textbf{(B)}~\frac{216 \sqrt{3}}{49}\qquad\textbf{(C)}~\frac{9 \sqrt{3}}{2} \qquad\textbf{(D)}~ \frac{14 \sqrt{3}}{3}\qquad\textbf{(E)}~\frac{243 \sqrt{3}}{49}$ | 2023 AMC 10A Problem 24 | Note that one can "slide' the small hexagons along their respective edges, and either by sliding them to the center or to the corners, and thus getting that the side length of the larger hexagon is 3. The rest proceeds the same as solution 1.
Solution 2.1 (Clarification)
Notice that when sliding the smaller hexagon along the edge, we see that the contact edge withe the smaller hexagon "in front" of it is $60^{\circ}$, thus meaning the hexagon "in front" is pushed at a speed $\sin{60^{\circ}}$ times the actual speed of the hexagon. We can preform a similar analysis on the hexagon that is being pushed and get that the speed at which that hexagon is moving is $\frac{1}{\sin{60^{\circ}}}$ times the speed it pushed by. As we can see, the two factors cancel out and by the same argument, every small hexagon can move at the same speed while mantaining an edge of contact with the two adjacent hexagons.
Diagram by ihatemath123
Note
The number $\frac{3}{7}$ is irrelevant to solve the problem.
In fact, if the smaller hexagons have side length $x$, the side length of the large hexagon will always be $3x$. | // Block 1
unitsize(1cm); draw(scale(3)*polygon(6)); filldraw(shift(dir(0)*2+dir(120)*0.4)*polygon(6), pink); filldraw(shift(dir(60)*2+dir(180)*0.4)*polygon(6), palered); filldraw(shift(dir(120)*2+dir(240)*0.4)*polygon(6), paleyellow); filldraw(shift(dir(180)*2+dir(300)*0.4)*polygon(6), palegreen); filldraw(shift(dir(240)*2+dir(360)*0.4)*polygon(6), palecyan); filldraw(shift(dir(300)*2+dir(420)*0.4)*polygon(6), paleblue); draw(dir(120)*2+dir(240)*0.4--dir(120)*2,red+linewidth(1.5),EndArrow(size=12)); draw(dir(180)*2+dir(300)*0.4--dir(180)*2,red+linewidth(1.5),EndArrow(size=12)); draw(dir(240)*2+dir(0)*0.4--dir(240)*2,red+linewidth(1.5),EndArrow(size=12)); draw(dir(300)*2+dir(60)*0.4--dir(300)*2,red+linewidth(1.5),EndArrow(size=12)); draw(dir(0)*2+dir(120)*0.4--dir(360)*2,red+linewidth(1.5),EndArrow(size=12)); draw(dir(60)*2+dir(180)*0.4--dir(60)*2,red+linewidth(1.5),EndArrow(size=12));
// Block 2
unitsize(1cm); draw(scale(3)*polygon(6)); filldraw(shift(dir(0)*2+dir(120)*0)*polygon(6), pink); filldraw(shift(dir(60)*2+dir(180)*0)*polygon(6), palered); filldraw(shift(dir(120)*2+dir(240)*0)*polygon(6), paleyellow); filldraw(shift(dir(180)*2+dir(300)*0)*polygon(6), palegreen); filldraw(shift(dir(240)*2+dir(360)*0)*polygon(6), palecyan); filldraw(shift(dir(300)*2+dir(420)*0)*polygon(6), paleblue); | [] |
936 | Six regular hexagonal blocks of side length 1 unit are arranged inside a regular hexagonal frame. Each block lies along an inside edge of the frame and is aligned with two other blocks, as shown in the figure below. The distance from any corner of the frame to the nearest vertex of a block is $\frac{3}{7}$ unit. What is the area of the region inside the frame not occupied by the blocks?
$\textbf{(A)}~\frac{13 \sqrt{3}}{3}\qquad\textbf{(B)}~\frac{216 \sqrt{3}}{49}\qquad\textbf{(C)}~\frac{9 \sqrt{3}}{2} \qquad\textbf{(D)}~ \frac{14 \sqrt{3}}{3}\qquad\textbf{(E)}~\frac{243 \sqrt{3}}{49}$ | 2023 AMC 10A Problem 24 | $(3^2-6\times 1^2)\times\frac{3\sqrt{3}}{2}=(9-6\times 1)\times\frac{3\sqrt{3}}{2}=(9-6)\times\frac{3\sqrt{3}}{2}=3\times\frac{3\sqrt{3}}{2}=\boxed{\textbf{(C) } \frac{9 \sqrt{3}}{2}}$
~~By afly
~(very minor change by Marshall_Huang) | // Block 1
unitsize(1cm);
unitsize(5cm);
draw(scale(3)*polygon(6));
filldraw(shift(dir(0)*2+dir(120)*3/7)*polygon(6), lightgray);
filldraw(shift(dir(60)*2+dir(180)*3/7)*polygon(6), lightgray);
filldraw(shift(dir(120)*2+dir(240)*3/7)*polygon(6), lightgray);
filldraw(shift(dir(180)*2+dir(300)*3/7)*polygon(6), lightgray);
filldraw(shift(dir(240)*2+dir(0)*3/7)*polygon(6), lightgray);
filldraw(shift(dir(300)*2+dir(60)*3/7)*polygon(6), yellow);
pair A = (0,0) + 3 * dir(300);
pair B = A + 3/7 *dir(60);
pair C = B + 1 * dir(180);
pair D = C + 3/7 * dir(240);
pair E = C + 4/7 * dir(120);
pair F = E + 3/7 * dir(240);
pair G = F + 4/7 * dir(240);
pen p = red+linewidth(6);
draw(A--B--C--D--cycle,p);
draw(C--E--F--D--cycle,p);
draw(F--G--D--cycle,p);
label("1",(B + C)/2,dir(90));
label("1",(A + D)/2,dir(270));
label("3/7",(A + B)/2,dir(330));
label("3/7",(C + D)/2,dir(330));
label("4/7",(C+E)/2,dir(30));
label("3/7",(E+F)/2,dir(150));
label("4/7",(F+D)/2,dir(30));
label("4/7",(F+G)/2,dir(150));
label("4/7",(G+D)/2,dir(270));
// Block 2
unitsize(1cm); unitsize(5cm); draw(scale(3)*polygon(6)); filldraw(shift(dir(0)*2+dir(120)*3/7)*polygon(6), lightgray); filldraw(shift(dir(60)*2+dir(180)*3/7)*polygon(6), lightgray); filldraw(shift(dir(120)*2+dir(240)*3/7)*polygon(6), lightgray); filldraw(shift(dir(180)*2+dir(300)*3/7)*polygon(6), lightgray); filldraw(shift(dir(240)*2+dir(0)*3/7)*polygon(6), lightgray); filldraw(shift(dir(300)*2+dir(60)*3/7)*polygon(6), yellow); pair A = (0,0) + 3 * dir(300); pair B = A + 3/7 *dir(60); pair C = B + 1 * dir(180); pair D = C + 3/7 * dir(240); pair E = C + 4/7 * dir(120); pair F = E + 3/7 * dir(240); pair G = F + 4/7 * dir(240); pen p = red+linewidth(6); draw(A--B--C--D--cycle,p); draw(C--E--F--D--cycle,p); draw(F--G--D--cycle,p); label("1",(B + C)/2,dir(90)); label("1",(A + D)/2,dir(270)); label("3/7",(A + B)/2,dir(330)); label("3/7",(C + D)/2,dir(330)); label("4/7",(C+E)/2,dir(30)); label("3/7",(E+F)/2,dir(150)); label("4/7",(F+D)/2,dir(30)); label("4/7",(F+G)/2,dir(150)); label("4/7",(G+D)/2,dir(270)); | [] |
937 | Usain is walking for exercise by zigzagging across a $100$-meter by $30$-meter rectangular field, beginning at point $A$ and ending on the segment $\overline{BC}$. He wants to increase the distance walked by zigzagging as shown in the figure below $(APQRS)$. What angle $\theta = \angle PAB=\angle QPC=\angle RQB=\cdots$ will produce a length that is $120$ meters? (This figure is not drawn to scale. Do not assume that he zigzag path has exactly four segments as shown; there could be more or fewer.)
$\textbf{(A)}~\arccos\frac{5}{6}\qquad\textbf{(B)}~\arccos\frac{4}{5}\qquad\textbf{(C)}~\arccos\frac{3}{10}\qquad\textbf{(D)}~\arcsin\frac{4}{5}\qquad\textbf{(E)}~\arcsin\frac{5}{6}$ | 2023 AMC 12A Problem 15 | By "unfolding" $APQRS$ into a straight line, we get a right angled triangle $ABS'$.
It follows that
\begin{align*} \cos(\theta)&=\frac{100}{120} \\ \theta&=\boxed{\textbf{(A) } \arccos\left(\frac{5}{6}\right)}. \end{align*}
~lptoggled | // Block 1
import olympiad;
draw((-50,15)--(50,15));
draw((50,15)--(50,-15));
draw((50,-15)--(-50,-15));
draw((-50,-15)--(-50,15));
draw((-50,-15)--(-22.5,15));
draw((-22.5,15)--(5,-15));
draw((5,-15)--(32.5,15));
draw((32.5,15)--(50,-4.090909090909));
label("$\theta$", (-41.5,-10.5));
label("$\theta$", (-13,10.5));
label("$\theta$", (15.5,-10.5));
label("$\theta$", (43,10.5));
dot((-50,15));
dot((-50,-15));
dot((50,15));
dot((50,-15));
dot((50,-4.09090909090909));
label("$D$",(-58,15));
label("$A$",(-58,-15));
label("$C$",(58,15));
label("$B$",(58,-15));
label("$S$",(58,-4.0909090909));
dot((-22.5,15));
dot((5,-15));
dot((32.5,15));
dot((5,45));
dot((32.5,75));
dot((50,94.09090909090909));
draw((-22.5,15)--(50,94.09090909090909));
draw((50,-4.09090909090909)--(50,94.09090909090909));
label("$P$",(-22.5,23));
label("$Q$",(5,-23));
label("$R$",(32.5,23));
label("$Q'$",(5,35));
label("$R'$",(32.5,85));
label("$S'$",(58,94.09090909090909));
// Block 2
import olympiad; draw((-50,15)--(50,15)); draw((50,15)--(50,-15)); draw((50,-15)--(-50,-15)); draw((-50,-15)--(-50,15)); draw((-50,-15)--(-22.5,15)); draw((-22.5,15)--(5,-15)); draw((5,-15)--(32.5,15)); draw((32.5,15)--(50,-4.090909090909)); label("$\theta$", (-41.5,-10.5)); label("$\theta$", (-13,10.5)); label("$\theta$", (15.5,-10.5)); label("$\theta$", (43,10.5)); dot((-50,15)); dot((-50,-15)); dot((50,15)); dot((50,-15)); dot((50,-4.09090909090909)); label("$D$",(-58,15)); label("$A$",(-58,-15)); label("$C$",(58,15)); label("$B$",(58,-15)); label("$S$",(58,-4.0909090909)); dot((-22.5,15)); dot((5,-15)); dot((32.5,15)); dot((5,45)); dot((32.5,75)); dot((50,94.09090909090909)); draw((-22.5,15)--(50,94.09090909090909)); draw((50,-4.09090909090909)--(50,94.09090909090909)); label("$P$",(-22.5,23)); label("$Q$",(5,-23)); label("$R$",(32.5,23)); label("$Q'$",(5,35)); label("$R'$",(32.5,85)); label("$S'$",(58,94.09090909090909)); | [] |
938 | Consider the set of complex numbers $z$ satisfying $|1+z+z^{2}|=4$. The maximum value of the imaginary part of $z$ can be written in the form $\tfrac{\sqrt{m}}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$?
$\textbf{(A)}~20\qquad\textbf{(B)}~21\qquad\textbf{(C)}~22\qquad\textbf{(D)}~23\qquad\textbf{(E)}~24$ | 2023 AMC 12A Problem 16 | We are given that $1+z+z^2=c$ where $c$ is some complex number with magnitude $4$. Rearranging the quadratic to standard form and applying the quadratic formula, we have
\[z=\frac{-1\pm \sqrt{1^2-4(1)(1-c)}}{2}=\frac{-1\pm\sqrt{4c-3}}{2}.\]
The imaginary part of $z$ is maximized when $c=-4$. (Why? See note below.)
Thus $z=-\tfrac 12 \pm i \tfrac{\sqrt{19}}{2}$, and so the answer is $\boxed{\textbf{(B)}~21}$.
~cantalon, centslordm
Note: You can cheese/fakesolve $c$ by assuming $c$ is real. But there is a good reason for it, within the parameters of this problem:
$c$ lies on a circle of radius 4, centered at the origin. Thus, ${4c-3}$ is also a circle, centered at $-3$. Luckily, the center of this adjusted circle is a negative real number, so the most negative point on that circle ($w = -19$) has the largest magnitude (and so its square root also has the largest magnitude of any square root), and its positive square root is the most purely imaginary square root ("vertical"). [We ignore the alternative negative square root, since the problem asks for the most positive value.]
Thus, $\text{Im}(\sqrt w) = r \sin(\theta)$ has both the largest $r$ and largest $\sin(\theta)$ of any square root of $4c-3$, and so the largest product $r \sin(\theta)$. (See note2 for more details)
[This argument and overall solution would not work if $-3$ were instead a positive or non-real number, which would happen if the original problem had $|C+Bz+Az^2|=4$ with $B^2-4AC > 0$.]
~oinava
Note 2:
The circle in the graph is all the possible values of $4c - 3$.
Let ${4c - 3} = re ^ {i\theta}$.
${4c - 3} = re ^ {i\theta} \implies \sqrt{4c - 3} = \sqrt{re ^ {i\theta}} = \sqrt{r} e ^ {i\frac{\theta}{2}}$.
Therefore, $Im(\sqrt{4c - 3}) = \sqrt{r}\sin(\frac{\theta}{2})$.
At $w(-19, 0)$ as in the graph, $\theta = \pi$ and $r = 19$.
So $\sqrt{r}$ and $\sin(\frac{\theta}{2})$ are all at their maximum, which means that $Im(\sqrt{4c - 3})$ is at its maximum.
~JiYang | // Block 1
size(250);
import TrigMacros;
rr_cartesian_axes(-22,19,-19,19,complexplane=true, usegrid = false);
Label f;
f.p=fontsize(4);
dot((-3, 0), red);
label("$(-3, 0)$", (-3, 0), NW);
dot((0,0));
draw(circle((-3, 0), 16), red);
dot((-19, 0), blue);
label("$w(-19, 0)$", (-19, 0), NW);
// Block 2
size(250); import TrigMacros; rr_cartesian_axes(-22,19,-19,19,complexplane=true, usegrid = false); Label f; f.p=fontsize(4); dot((-3, 0), red); label("$(-3, 0)$", (-3, 0), NW); dot((0,0)); draw(circle((-3, 0), 16), red); dot((-19, 0), blue); label("$w(-19, 0)$", (-19, 0), NW); | [] |
938 | Consider the set of complex numbers $z$ satisfying $|1+z+z^{2}|=4$. The maximum value of the imaginary part of $z$ can be written in the form $\tfrac{\sqrt{m}}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$?
$\textbf{(A)}~20\qquad\textbf{(B)}~21\qquad\textbf{(C)}~22\qquad\textbf{(D)}~23\qquad\textbf{(E)}~24$ | 2023 AMC 12A Problem 16 | We can write the given condition as \[\left|\left(z+\frac{1}{2}\right)^2 + \frac{3}{4}\right| = 4.\]
Letting $u = \left(z+\frac{1}{2}\right)^2$, the equation $\left|u + \frac{3}{4}\right| = 4$ equates to the circle centered at $-\frac{3}{4}$ with radius $4$ in the complex plane, call it $\omega$. Thus the locus of $\left(z+\frac{1}{2}\right)^2$ is $\omega$. Let $v = z+\frac{1}{2}$, and since the $+\frac{1}{2}$ does not change $z$'s imaginary part, we now need to find $v$ with the largest imaginary part such that $v^2$ lies on $\omega$.
Note that the point on $\omega$ with largest magnitude is $19/4$ and has argument $\pi$, call it $\phi$ (The leftmost point on $\omega$). The value $v'$ with positive imaginary part such that $(v')^2 = \phi$ has an argument of $\frac{\pi}{2}$ and a magnitude of $\frac{\sqrt{19}}{2}$.
Since across all values of $v$ the imaginary part is given by $r\sin{\theta}$ and $v'$ has the largest possible $r$ and the largest possible value of $\sin{\theta},$ it must have the largest imaginary part.
This can non-rigorously be seen by sketching the oval which is the locus of $v$.
This gives $19 + 2 \implies \boxed{\textbf{(B)}~21}$
~AtharvNaphade | // Block 1
size(250);
import TrigMacros;
rr_cartesian_axes(-6,5,-5,5,complexplane=true, usegrid = true);
Label f;
f.p=fontsize(6);
xaxis(-6,5,Ticks(f, 1.0));
yaxis(-5,5,Ticks(f, 1.0));
dot((0,0));
draw(circle((-3/4, 0), 4), red + dashed);
dot((-19/4, 0), blue);
label("$\phi$", (-19/4, 0), NW);
dot((0, 2.18), blue);
label("$v'$", (0, 2.18), NE);
draw(ellipse((0,0),1.8,2.18), green);
// Block 2
size(250); import TrigMacros; rr_cartesian_axes(-6,5,-5,5,complexplane=true, usegrid = true); Label f; f.p=fontsize(6); xaxis(-6,5,Ticks(f, 1.0)); yaxis(-5,5,Ticks(f, 1.0)); dot((0,0)); draw(circle((-3/4, 0), 4), red + dashed); dot((-19/4, 0), blue); label("$\phi$", (-19/4, 0), NW); dot((0, 2.18), blue); label("$v'$", (0, 2.18), NE); draw(ellipse((0,0),1.8,2.18), green); | [] |
939 | Circle $C_1$ and $C_2$ each have radius $1$, and the distance between their centers is $\frac{1}{2}$. Circle $C_3$ is the largest circle internally tangent to both $C_1$ and $C_2$. Circle $C_4$ is internally tangent to both $C_1$ and $C_2$ and externally tangent to $C_3$. What is the radius of $C_4$?
$\textbf{(A) } \frac{1}{14} \qquad \textbf{(B) } \frac{1}{12} \qquad \textbf{(C) } \frac{1}{10} \qquad \textbf{(D) } \frac{3}{28} \qquad \textbf{(E) } \frac{1}{9}$ | 2023 AMC 12A Problem 18 | Let $O$ be the center of the midpoint of the line segment connecting both the centers, say $A$ and $B$.
Let the point of tangency with the inscribed circle and the right larger circles be $T$.
Then $OT = BO + BT = BO + AT - \frac{1}{2} = \frac{1}{4} + 1 - \frac{1}{2} = \frac{3}{4}.$
Since $C_4$ is internally tangent to $C_1$, center of $C_4$, $C_1$ and their tangent point must be on the same line.
Now, if we connect centers of $C_4$, $C_3$ and $C_1$/$C_2$, we get a right angled triangle.
Let the radius of $C_4$ equal $r$. With the pythagorean theorem on our triangle, we have
\[\left(r+\frac{3}{4}\right)^2+\left(\frac{1}{4}\right)^2=(1-r)^2\]
Solving this equation gives us
\[r = \boxed{\textbf{(D) } \frac{3}{28}}\]
~lptoggled
~ShawnX (Diagram)
~ap246 (Minor Changes) | // Block 1
import olympiad;
size(10cm);
draw(circle((0,0),0.75), gray(0.7));
draw(circle((-0.25,0),1), gray(0.7));
draw(circle((0.25,0),1), gray(0.7));
draw(circle((0,6/7),3/28), gray(0.7));
pair A = (0,0), B = (-0.25,0), C = (0.25,0), D = (0,6/7), EE = (-0.95710678118, 0.70710678118), F = (0.95710678118, -0.70710678118), G = (0,0), T=(0.75,0);
dot(D);
dot(G);
draw(B--EE, dashed+gray(0.7));
draw(C--F, dashed+gray(0.7));
dot(C, gray(0.9));
draw(B--C, gray(0.7));
draw(B--A);
draw(A--D);
draw(B--D);
draw(B--T);
label("$\frac{1}{4}$", (-0.125, -0.125));
label("$r + \frac{3}{4}$", (0.2, 3/7));
label("$1 - r$", (-0.29, 3/7));
label("$O$",A,S);
label("$A$",B,S);
dot("$B$",C,S);
dot("$T$",T,E);
label("$1$", (-.85, 0.70));
label("$1$", (.85, -.7));
markscalefactor=0.05;
// Block 2
import olympiad; size(10cm); draw(circle((0,0),0.75), gray(0.7)); draw(circle((-0.25,0),1), gray(0.7)); draw(circle((0.25,0),1), gray(0.7)); draw(circle((0,6/7),3/28), gray(0.7)); pair A = (0,0), B = (-0.25,0), C = (0.25,0), D = (0,6/7), EE = (-0.95710678118, 0.70710678118), F = (0.95710678118, -0.70710678118), G = (0,0), T=(0.75,0); dot(D); dot(G); draw(B--EE, dashed+gray(0.7)); draw(C--F, dashed+gray(0.7)); dot(C, gray(0.9)); draw(B--C, gray(0.7)); draw(B--A); draw(A--D); draw(B--D); draw(B--T); label("$\frac{1}{4}$", (-0.125, -0.125)); label("$r + \frac{3}{4}$", (0.2, 3/7)); label("$1 - r$", (-0.29, 3/7)); label("$O$",A,S); label("$A$",B,S); dot("$B$",C,S); dot("$T$",T,E); label("$1$", (-.85, 0.70)); label("$1$", (.85, -.7)); markscalefactor=0.05; | [] |
940 | Rows 1, 2, 3, 4, and 5 of a triangular array of integers are shown below.
Each row after the first row is formed by placing a 1 at each end of the row, and each interior entry is 1 greater than the sum of the two numbers diagonally above it in the previous row. What is the units digits of the sum of the 2023 numbers in the 2023rd row?
$\textbf{(A) } 1 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 5 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 9$ | 2023 AMC 12A Problem 20 | Consider Pascal's triangle as the starting point. In the Pascal's triangle depicted below, the sum of the numbers in the $n$th row is $2^{(n-1)}$. For the 2023rd row in the Pascal's triangle, the sum of numbers is $2^{2022}$.
For the triangular array of integers in the problem, 1 is added to each interior entry, which propagates to two numbers diagonally below it in the next row, making the sum of numbers in the next row to increase by 2. And the addition of 1 continues to propagate to the next row, which makes the sum of numbers in the row below the next to increase by 4.
Examine the following triangular array of 1's, which indicates the 1's being added to each position of corresponding numbers in the Pascal's triangle.
For the 3rd row, 1 is added to the original number in the same position in the Pascal's triangle. And the addition of 1 in the 3rd row makes the sum of numbers in the 4th row to increase by 2, and makes the sum of numbers in the 5th row to increase by 4, and so forth. Therefore, the addition of a 1 in the 3rd row makes the sum of numbers in the 2023rd row to increase by $2^{2023-3}=2^{2020}$. And similarly, the addition of each 1 in the 4th row makes the number of numbers in the 2023rd row to increase by $2^{2023-4}=2^{2019}$. The 1's being added between the 3rd and 2022nd rows impact on the sum of numbers in the 2023rd row to increase by $2^{2020} + 2 \cdot 2^{2019} + 3 \cdot 2^{2018} + \dots + 2020 \cdot 2^1 = \sum_{n=1}^{2020}(n \cdot 2^{(2021-n)})$.
Therefore, the sum of numbers in the 2023rd row is the aggregation of the sum of numbers in the 2023rd row in the Pascal's triangle, the impact of addition of 1's between the 3rd row and the 2022nd row, and the addition of 1's on 2021 interior entries in the 2023rd row, which is $2^{2022} + \sum_{n=1}^{2020}(n \cdot 2^{(2021-n)}) + 2021$.
Because $2^{2022} = 2^{2020+2} = 16^{505} \cdot 2^2 = 16^{505} \cdot 4$ and $16^k$ will always ends with 6 as the unit digit, the unit digit of $2^{2022}$ is 4.
For $\sum_{n=1}^{2020}(n \cdot 2^{(2021-n)})$, we can solve the sum of geometric sequence to be ${2^{2022} - 4 - 2020 \cdot 2}$, which has 0 for the unit digit.
Therefore, for the sum of numbers in the 2023rd row, which is $2^{2022} + \sum_{n=1}^{2020}(n \cdot 2^{(2021-n)}) + 2021$, its unit digit is $4 + 0 + 1 = \boxed{\textbf{(C) } 5}$.
~sqroot | // Block 1
size(4.5cm);
label("$1$", (0,0));
label("$1$", (-0.5,-2/3));
label("$1$", (0.5,-2/3));
label("$1$", (-1,-4/3));
label("$2$", (0,-4/3));
label("$1$", (1,-4/3));
label("$1$", (-1.5,-2));
label("$3$", (-0.5,-2));
label("$3$", (0.5,-2));
label("$1$", (1.5,-2));
label("$1$", (-2,-8/3));
label("$4$", (-1,-8/3));
label("$6$", (0,-8/3));
label("$4$", (1,-8/3));
label("$1$", (2,-8/3));
// Block 2
size(4.5cm);
label("$0$", (0,0));
label("$0$", (-0.5,-2/3));
label("$0$", (0.5,-2/3));
label("$0$", (-1,-4/3));
label("$1$", (0,-4/3));
label("$0$", (1,-4/3));
label("$0$", (-1.5,-2));
label("$1$", (-0.5,-2));
label("$1$", (0.5,-2));
label("$0$", (1.5,-2));
label("$0$", (-2,-8/3));
label("$1$", (-1,-8/3));
label("$1$", (0,-8/3));
label("$1$", (1,-8/3));
label("$0$", (2,-8/3));
// Block 3
size(4.5cm); label("$1$", (0,0)); label("$1$", (-0.5,-2/3)); label("$1$", (0.5,-2/3)); label("$1$", (-1,-4/3)); label("$2$", (0,-4/3)); label("$1$", (1,-4/3)); label("$1$", (-1.5,-2)); label("$3$", (-0.5,-2)); label("$3$", (0.5,-2)); label("$1$", (1.5,-2)); label("$1$", (-2,-8/3)); label("$4$", (-1,-8/3)); label("$6$", (0,-8/3)); label("$4$", (1,-8/3)); label("$1$", (2,-8/3));
// Block 4
size(4.5cm); label("$0$", (0,0)); label("$0$", (-0.5,-2/3)); label("$0$", (0.5,-2/3)); label("$0$", (-1,-4/3)); label("$1$", (0,-4/3)); label("$0$", (1,-4/3)); label("$0$", (-1.5,-2)); label("$1$", (-0.5,-2)); label("$1$", (0.5,-2)); label("$0$", (1.5,-2)); label("$0$", (-2,-8/3)); label("$1$", (-1,-8/3)); label("$1$", (0,-8/3)); label("$1$", (1,-8/3)); label("$0$", (2,-8/3)); | [] |
940 | Rows 1, 2, 3, 4, and 5 of a triangular array of integers are shown below.
Each row after the first row is formed by placing a 1 at each end of the row, and each interior entry is 1 greater than the sum of the two numbers diagonally above it in the previous row. What is the units digits of the sum of the 2023 numbers in the 2023rd row?
$\textbf{(A) } 1 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 5 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 9$ | 2023 AMC 12A Problem 20 | The triangle provided by the question may be rewritten with Pascal's Triangle being the starting point.
The numbers in Pascal's Triangle could be written with combinations. Moreover, the following equation is true.
\[\sum_{k=0}^{n}{}_nC_k=2^n\]
Therefore, the sum of every number in $n^{th}$ row of a Pascal's Triangle is $2^{n}$.
Another triangle with the red colored number could be written to evaluate the sum of the red numbers in each row.
The sum of numbers in each line could be investigated.
\[\begin{array}{c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c@{\quad}c} 1 & & 4 & & 11 & & 26 & & 53 & \dots \\ & 3 & & 7 & & 15 & & 27 & \dots & \\ & & 4 & & 8 & & 12 & \dots & & \\ & & & 4 & & 4 & \dots & & & \\ \end{array}\]
Using Newton's Little Formula, it is evident that the following equation is true.
\[a_n=1\cdot{}_{n-1}C_0+3\cdot{}_{n-1}C_1+4\cdot{}_{n-1}C_2+4\cdot{}_{n-1}C_3\]
With all the tools, the last digit, or the remainder when divided by 10, of the sum of the numbers in the 2023rd row may be computed. Let $x$ be the last digit. Moreover, $a_{2020}$ is equivalent to the sum of red numbers in 2023rd row of the original triangle.
\begin{align*}
2^{2020}+1\cdot{}_{2020}C_0+3\cdot{}_{2020}C_1+4\cdot{}_{2020}C_2+4\cdot{}_{2020}C_3 &\equiv x\pmod{10} \\
&\equiv 2^{2022}+1+3\cdot2020+2\cdot2020\cdot2019 \\
&\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }+2\cdot2020\cdot673\cdot2018 \pmod{10} \\
&=2^{2022}+1 \pmod{10} \\
&=4+1 \pmod{10} \\
&=5 \pmod{10}
\end{align*}
Answer: $\boxed{(C) 5}$
~MaPhyCom
~ALGEBRAIC_ALGORITHMIC (latex changes) | // Block 1
usepackage("color");
texpreamble("\usepackage{color}");
size(6cm); // Reduced from 8cm
// Reduced spacing
real dx = 1.2;
real dy = 0.6;
// Level 0
label(Label("$1$", align=Center), (0, 0));
// Level 1
label(Label("$1$", align=Center), (-dx/2, -dy));
label(Label("$1$", align=Center), ( dx/2, -dy));
// Level 2
label(Label("$1$", align=Center), (-dx, -2*dy));
label(Label("$2+\textcolor{red}{1}$", align=Center), (0, -2*dy));
label(Label("$1$", align=Center), (dx, -2*dy));
// Level 3
label(Label("$1$", align=Center), (-1.5*dx, -3*dy));
label(Label("$3+\textcolor{red}{2}$", align=Center), (-0.5*dx, -3*dy));
label(Label("$3+\textcolor{red}{2}$", align=Center), (0.5*dx, -3*dy));
label(Label("$1$", align=Center), (1.5*dx, -3*dy));
// Level 4
label(Label("$1$", align=Center), (-2*dx, -4*dy));
label(Label("$4+\textcolor{red}{3}$", align=Center), (-dx, -4*dy));
label(Label("$4+\textcolor{red}{5}$", align=Center), (0, -4*dy));
label(Label("$4+\textcolor{red}{3}$", align=Center), (dx, -4*dy));
label(Label("$1$", align=Center), (2*dx, -4*dy));
label(Label("$\vdots$", align=Center), (0, -5*dy));
// Block 2
size(2.5cm);
label("$\textcolor{red}{1}$", (0, 0));
label("$\textcolor{red}{2}$", (0.1/2, -0.08));
label("$\textcolor{red}{2}$", (-0.1/2, -0.08));
label("$\textcolor{red}{3}$", (-0.1, -2*0.08));
label("$\textcolor{red}{5}$", ( 0, -2*0.08));
label("$\textcolor{red}{3}$", ( 0.1, -2*0.08));
label("$\vdots$", (0, -3*0.08));
// Block 3
usepackage("color"); texpreamble("\usepackage{color}"); size(6cm); // Reduced from 8cm // Reduced spacing real dx = 1.2; real dy = 0.6; // Level 0 label(Label("$1$", align=Center), (0, 0)); // Level 1 label(Label("$1$", align=Center), (-dx/2, -dy)); label(Label("$1$", align=Center), ( dx/2, -dy)); // Level 2 label(Label("$1$", align=Center), (-dx, -2*dy)); label(Label("$2+\textcolor{red}{1}$", align=Center), (0, -2*dy)); label(Label("$1$", align=Center), (dx, -2*dy)); // Level 3 label(Label("$1$", align=Center), (-1.5*dx, -3*dy)); label(Label("$3+\textcolor{red}{2}$", align=Center), (-0.5*dx, -3*dy)); label(Label("$3+\textcolor{red}{2}$", align=Center), (0.5*dx, -3*dy)); label(Label("$1$", align=Center), (1.5*dx, -3*dy)); // Level 4 label(Label("$1$", align=Center), (-2*dx, -4*dy)); label(Label("$4+\textcolor{red}{3}$", align=Center), (-dx, -4*dy)); label(Label("$4+\textcolor{red}{5}$", align=Center), (0, -4*dy)); label(Label("$4+\textcolor{red}{3}$", align=Center), (dx, -4*dy)); label(Label("$1$", align=Center), (2*dx, -4*dy)); label(Label("$\vdots$", align=Center), (0, -5*dy));
// Block 4
size(2.5cm); label("$\textcolor{red}{1}$", (0, 0)); label("$\textcolor{red}{2}$", (0.1/2, -0.08)); label("$\textcolor{red}{2}$", (-0.1/2, -0.08)); label("$\textcolor{red}{3}$", (-0.1, -2*0.08)); label("$\textcolor{red}{5}$", ( 0, -2*0.08)); label("$\textcolor{red}{3}$", ( 0.1, -2*0.08)); label("$\vdots$", (0, -3*0.08)); | [] |
941 | You are playing a game. A $2 \times 1$ rectangle covers two adjacent squares (oriented either horizontally or vertically) of a $3 \times 3$ grid of squares, but you are not told which two squares are covered. Your goal is to find at least one square that is covered by the rectangle. A "turn" consists of you guessing a square, after which you are told whether that square is covered by the hidden rectangle. What is the minimum number of turns you need to ensure that at least one of your guessed squares is covered by the rectangle?
$\textbf{(A)}~3\qquad\textbf{(B)}~5\qquad\textbf{(C)}~4\qquad\textbf{(D)}~8\qquad\textbf{(E)}~6$ | 2023 AMC 10B Problem 10 | Notice that the $3\times3$ square grid has a total of $12$ possible $2\times1$ rectangles.
Suppose you choose the middle square for one of your turns. The middle square is covered by $4$ rectangles, and each of the remaining $8$ squares is covered by a maximum of $2$ uncounted rectangles. This means that the number of turns is at least $1+\frac{12-4}{2}=1+4=5$.
Now suppose you don't choose the middle square. The squares on the middle of the sides are covered by at most 3 uncounted rectangles, and the squares on the corners are covered by at most 2 uncounted rectangles. In this case, we see that the least number of turns needed to account for all 12 rectangles is $12\div 3=4.$ To prove that choosing only side squares indeed does cover all 12 rectangles, we need to show that the 3 rectangles per square that cover each side square do not overlap. Drawing the rectangles that cover one square, we see they form a $T$ shape and they do not cover any other side square. Hence, our answer is $4.$ | draw((0,0)--(0.5,0)--(0.5,0.5)--(0,0.5)--(0,0)); draw((0,1)--(0.5,1)--(0.5,1.5)--(0,1.5)--(0,1)); draw((0.5,0.5)--(1,0.5)--(1,1)--(0.5,1)--(0.5,0.5)); draw((1,0)--(1.5,0)--(1.5,0.5)--(1,0.5)--(1,0)); draw((1,1)--(1.5,1)--(1.5,1.5)--(1,1.5)--(1,1)); draw((0,0.5)--(0.5,0.5)--(0.5,1)--(0,1)--(0,0.5)); draw((0.5,0)--(1,0)--(1,0.5)--(0.5,0.5)--(0.5,0)); draw((0.5,1)--(1,1)--(1,1.5)--(0.5,1.5)--(0.5,1)); draw((1,0.5)--(1.5,0.5)--(1.5,1)--(1,1)--(1,0.5)); filldraw((0,0.5)--(0.5,0.5)--(0.5,1)--(0,1)--(0,0.5)--cycle, red, black+linewidth(1)); filldraw((0,0)--(0.5,0)--(0.5,0.5)--(0,0.5)--(0,0)--cycle, red, black+linewidth(1)); filldraw((0,1)--(0.5,1)--(0.5,1.5)--(0,1.5)--(0,1)--cycle, red, black+linewidth(1)); filldraw((0.5,0.5)--(1,0.5)--(1,1)--(0.5,1)--(0.5,0.5)--cycle, red, black+linewidth(1)); | [] |
941 | You are playing a game. A $2 \times 1$ rectangle covers two adjacent squares (oriented either horizontally or vertically) of a $3 \times 3$ grid of squares, but you are not told which two squares are covered. Your goal is to find at least one square that is covered by the rectangle. A "turn" consists of you guessing a square, after which you are told whether that square is covered by the hidden rectangle. What is the minimum number of turns you need to ensure that at least one of your guessed squares is covered by the rectangle?
$\textbf{(A)}~3\qquad\textbf{(B)}~5\qquad\textbf{(C)}~4\qquad\textbf{(D)}~8\qquad\textbf{(E)}~6$ | 2023 AMC 10B Problem 10 | The $3 \times 3$ grid can be colored like a checkerboard with alternating black and white squares.
Let the top left square be white, and the rest of the squares alternate colors, as shown below:
Each $2 \times 1$ rectangle always covers $1$ white square and $1$ black square.
You can ensure that at least one of your guessed squares is covered by the rectangle by choosing either each of the white squares ($5$ turns) or each of the black squares ($4$ turns).
Since it is ideal to be the most efficient with our turns, by choosing all the black squares, we guarantee that one of the $\boxed{\textbf{(C)}~4}$ squares are of the $2 \times 1$ rectangle.
~ CherryBerry
(Minor edits by NSAoPS) | /* Diagram by MRENTHUSIASM */ import olympiad; unitsize(25); fill((1,2)--(1,3)--(2,3)--(2,2)--cycle, mediumgray); fill((1,0)--(1,1)--(2,1)--(2,0)--cycle, mediumgray); fill((2,1)--(2,2)--(3,2)--(3,1)--cycle, mediumgray); fill((0,1)--(0,2)--(1,2)--(1,1)--cycle, mediumgray); for (int i = 0; i < 4; ++i) { draw((i,0)--(i,3)); draw((0,i)--(3,i)); } | [] |
942 | A rectangular box $\mathcal{P}$ has distinct edge lengths $a$, $b$, and $c$. The sum of the lengths of all $12$ edges of $\mathcal{P}$ is $13$, the areas of all $6$ faces of $\mathcal{P}$ is $\frac{11}{2}$, and the volume of $\mathcal{P}$ is $\frac{1}{2}$. What is the length of the longest interior diagonal connecting two vertices of $\mathcal{P}$?
$\textbf{(A)}~2\qquad\textbf{(B)}~\frac{3}{8}\qquad\textbf{(C)}~\frac{9}{8}\qquad\textbf{(D)}~\frac{9}{4}\qquad\textbf{(E)}~\frac{3}{2}$ | 2023 AMC 10B Problem 17 | We can create three equations using the given information.
\[4a+4b+4c = 13\]
\[2ab+2ac+2bc=\frac{11}{2}\]
\[abc=\frac{1}{2}\]
We also know that we want $\sqrt{a^2 + b^2 + c^2}$ because that is the length that can be found from using the Pythagorean Theorem. We cleverly notice that $a^2 + b^2 + c^2 = (a+b+c)^2 - 2(ab+ac+bc)$. We know that $a+b+c = \frac{13}{4}$ and $2(ab+ac+bc)=\dfrac{11}2$, so $a^2 + b^2 + c^2 = \left(\frac{13}{4}\right)^2 - \frac{11}{2} = \frac{169-88}{16} = \frac{81}{16}$. So our answer is $\sqrt{\frac{81}{16}} = \boxed{\textbf{(D)}~\tfrac94}$.
~lprado
~minor edits and add-ons by Technodoggo, lucaswujc, andliu766, BcMath, and MW2014
Note:This solution does not necessarily use the fact that $abc=\frac{1}{2}$, unlike the second solution. | import geometry; pair A = (-3, 4); pair B = (-3, 5); pair C = (-1, 4); pair D = (-1, 5); pair AA = (0, 0); pair BB = (0, 1); pair CC = (2, 0); pair DD = (2, 1); draw(D--AA,dashed); draw(A--B); draw(A--C); draw(B--D); draw(C--D); draw(A--AA); draw(B--BB); draw(C--CC); draw(D--DD); // Dotted vertices dot(A); dot(B); dot(C); dot(D); dot(AA); dot(BB); dot(CC); dot(DD); draw(AA--BB); draw(AA--CC); draw(BB--DD); draw(CC--DD); label("a",midpoint(D--DD),E); label("b",midpoint(CC--DD),E); label("c",midpoint(AA--CC),S); | [] |
943 | Sonya the frog chooses a point uniformly at random lying within the square
$[0, 6]$ $\times$ $[0, 6]$ in the coordinate plane and hops to that point. She then randomly
chooses a distance uniformly at random from $[0, 1]$ and a direction uniformly at
random from {north, south, east, west}. All of her choices are independent. She now
hops the distance in the chosen direction. What is the probability that she lands
outside the square?
$\textbf{(A) } \frac{1}{6} \qquad \textbf{(B) } \frac{1}{12} \qquad \textbf{(C) } \frac{1}{4} \qquad \textbf{(D) } \frac{1}{10} \qquad \textbf{(E) } \frac{1}{9}$ | 2023 AMC 10B Problem 19 | WLOG, we assume Sonya jumps $0.5$ units every time, since that is her expected value.
If Sonya is within $0.5$ blocks of an edge, she can jump off the board. Let us examine the region that is at most $0.5$ blocks from exactly one edge.
If Sonya starts in this region, she has a $\dfrac14$ chance of landing outside (there's exactly one direction she can hop to get out). The total area of this region is $4\cdot0.5\cdot5=10.$ For this region, Sonya has a $\dfrac14$ chance, so we multiply $10$ by $\dfrac14$ to get $2.5.$
If Sonya is in one of the corner squares, she can go two directions to get out, so she has a $\dfrac24=\dfrac12$ chance to get out. The total area is $0.5\cdot0.5\cdot4=1$, so this region yields $\dfrac12\cdot1=\dfrac12.$
Adding the two, we get $3$. This is out of $36$ square units of area, so our answer is thus $\boxed{\textbf{(B) }\tfrac{1}{12}}.$
~Technodoggo
Note: When Sonya is within 0 to 1 units away from the border in a given direction, the probability that she will jump outside the border can be thought of as a function of distance from the border. It is easy to see that if Sonya is \(x\) units away from the border, a jump distance in the interval \([x, 1]\) will move her outside. Thus the probability is a function \(P(x) = 1-x\). The probability in an entire region 0 to 1 units away from the border is an integral of the function, \(\int^1_0 (1-x) dx = [x-\frac{1}{2}x^2]^1_0 = 1/2\). This is why the probability is \(1/2\).
~abghim | // Block 1
import graph;
Label f;
xaxis(0,6,Ticks(f, 6.0, 0.5));
yaxis(0,6,Ticks(f, 6.0, 0.5));
draw((0,0)--(6,0)--(6,6)--(0,6)--cycle);
filldraw((0,0.5)--(0.5,0.5)--(0.5,5.5)--(0,5.5)--cycle,gray);
filldraw((0.5,0)--(0.5,0.5)--(5.5,0.5)--(5.5,0)--cycle,gray);
filldraw((6,0.5)--(5.5,0.5)--(5.5,5.5)--(6,5.5)--cycle,gray);
filldraw((0.5,6)--(0.5,5.5)--(5.5,5.5)--(5.5,6)--cycle,gray);
// Block 2
import graph; Label f; xaxis(0,6,Ticks(f, 6.0, 0.5)); yaxis(0,6,Ticks(f, 6.0, 0.5)); draw((0,0)--(6,0)--(6,6)--(0,6)--cycle); filldraw((0,0.5)--(0.5,0.5)--(0.5,5.5)--(0,5.5)--cycle,gray); filldraw((0.5,0)--(0.5,0.5)--(5.5,0.5)--(5.5,0)--cycle,gray); filldraw((6,0.5)--(5.5,0.5)--(5.5,5.5)--(6,5.5)--cycle,gray); filldraw((0.5,6)--(0.5,5.5)--(5.5,5.5)--(5.5,6)--cycle,gray); | [] |
943 | Sonya the frog chooses a point uniformly at random lying within the square
$[0, 6]$ $\times$ $[0, 6]$ in the coordinate plane and hops to that point. She then randomly
chooses a distance uniformly at random from $[0, 1]$ and a direction uniformly at
random from {north, south, east, west}. All of her choices are independent. She now
hops the distance in the chosen direction. What is the probability that she lands
outside the square?
$\textbf{(A) } \frac{1}{6} \qquad \textbf{(B) } \frac{1}{12} \qquad \textbf{(C) } \frac{1}{4} \qquad \textbf{(D) } \frac{1}{10} \qquad \textbf{(E) } \frac{1}{9}$ | 2023 AMC 10B Problem 19 | We know that Sonya can only jump out if she is within 1 unit of the border. We can calculate the probability that Sonya can jump out. $\newline$
$\newline$
The total area of the colored regions is $20$, so the probability that Sonya lands in a colored region is $\frac{20}{36} = \frac{5}{9}$. We can calculate the probability that Sonya gets out of each type of region. $\newline$
Case 1: Sonya chooses a blue region.
The probability that Sonya chooses a blue region is $\frac{16}{20} = \frac{4}{5}$. One direction can let her out, so the probability that she chooses the right one is $\frac{1}{4}$. Finally, the probability that Sonya chooses a distance to get her out is $\frac{1}{2}$. So, the probability that she chooses a blue region and gets out is $\frac{5}{9} \cdot \frac{4}{5} \cdot \frac{1}{4} \cdot \frac{1}{2} = \frac{1}{18}$. $\newline$
Case 2: Sonya choose a red region.
The probability that Sonya chooses a red region is $\frac{4}{20} = \frac{1}{5}$. Two directions can let her out, so the probability that she chooses one of them is $\frac{1}{2}$. Finally, the probability that Sonya chooses a distance to get her out is $\frac{1}{2}$. So, the probability that she chooses a red region and gets out is $\frac{5}{9} \cdot \frac{1}{5} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{36}$. $\newline$
So, the probability that Sonya gets out is $\frac{1}{18} + \frac{1}{36} = \frac{3}{36} =$ $\boxed{\textbf{(B) }\tfrac{1}{12}}.$
~MathCactus0_0 | // Block 1
import graph;
draw((0,0)--(0,6)--(6,6)--(6,0)--cycle);
filldraw((1,0)--(1,1)--(5,1)--(5,0)--cycle,blue);
filldraw((0,1)--(1,1)--(1,5)--(0,5)--cycle,blue);
filldraw((1,5)--(1,6)--(5,6)--(5,5)--cycle,blue);
filldraw((5,1)--(6,1)--(6,5)--(5,5)--cycle,blue);
filldraw((0,0)--(0,1)--(1,1)--(1,0)--cycle,red);
filldraw((0,5)--(1,5)--(1,6)--(0,6)--cycle,red);
filldraw((5,0)--(5,1)--(6,1)--(6,0)--cycle,red);
filldraw((5,5)--(6,5)--(6,6)--(5,6)--cycle,red);
// Block 2
import graph; draw((0,0)--(0,6)--(6,6)--(6,0)--cycle); filldraw((1,0)--(1,1)--(5,1)--(5,0)--cycle,blue); filldraw((0,1)--(1,1)--(1,5)--(0,5)--cycle,blue); filldraw((1,5)--(1,6)--(5,6)--(5,5)--cycle,blue); filldraw((5,1)--(6,1)--(6,5)--(5,5)--cycle,blue); filldraw((0,0)--(0,1)--(1,1)--(1,0)--cycle,red); filldraw((0,5)--(1,5)--(1,6)--(0,6)--cycle,red); filldraw((5,0)--(5,1)--(6,1)--(6,0)--cycle,red); filldraw((5,5)--(6,5)--(6,6)--(5,6)--cycle,red); | [] |
944 | What is the perimeter of the boundary of the region consisting of all points which can be expressed as $(2u-3w, v+4w)$ with $0\le u\le1$, $0\le v\le1,$ and $0\le w\le1$?
$\textbf{(A) } 10\sqrt{3} \qquad \textbf{(B) } 13 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 16$ | 2023 AMC 10B Problem 24 | Notice that we are given a parametric form of the region, and $w$ is used in both $x$ and $y$. We first fix $u$ and $v$ to $0$, and graph $(-3w,4w)$ from $0\le w\le1$. When $w$ is $0$, we have the point $(0,0)$, and when $w$ is $1$, we have the point $(-3,4)$. We see that since this is a directly proportional function, we can just connect the dots like this:
Now, when we vary $2u$ from $0$ to $2$, this line is translated to the right $2$ units:
We know that any points in the region between the line (or rather segment) and its translation satisfy $w$ and $u$, so we shade in the region:
We can also shift this quadrilateral one unit up, because of $v$. Thus, this is our figure:
The length of the boundary is simply $1+2+5+1+2+5$ ($5$ can be obtained by Pythagorean theorem since we have side lengths $3$ and $4$.). This equals $\boxed{\textbf{(E) }16.}$
~Technodoggo ~ESAOPS | // Block 1
import geometry;
pair A = (-3, 4);
pair B = (-3, 5);
pair C = (-1, 4);
pair D = (-1, 5);
pair AA = (0, 0);
pair BB = (0, 1);
pair CC = (2, 0);
pair DD = (2, 1);
//draw(A--B--D--C--cycle);
draw(A--B);
label("1",midpoint(A--B),W);
label("2",midpoint(D--B),N);
draw(A--C,dashed);
draw(B--D);
draw(C--D, dashed);
draw(A--AA);
label("5",midpoint(A--AA),W);
draw(B--BB,dashed);
draw(C--CC,dashed);
draw(D--DD);
label("5",midpoint(D--DD),E);
label("1",midpoint(CC--DD),E);
label("2",midpoint(AA--CC),S);
// Dotted vertices
dot(A); dot(B); dot(C); dot(D);
dot(AA); dot(BB); dot(CC); dot(DD);
draw(AA--BB,dashed);
draw(AA--CC);
draw(BB--DD,dashed);
draw(CC--DD);
label("(0,0)",AA,W);
label("(-3,4)",A,SW);
label("(-1,5)",D,E);
label("(2,1)",DD,NE);
// Block 2
import graph;
Label f;
size(5cm);
unitsize(0.7cm);
xaxis(-5,5,Ticks(f, 5.0, 1.0));
yaxis(-5,5,Ticks(f, 5.0, 1.0));
draw((0,0)--(-3,4));
// Block 3
import graph;
Label f;
unitsize(0.7cm);
size(5cm);
xaxis(-5,5,Ticks(f, 5.0, 1.0));
yaxis(-5,5,Ticks(f, 5.0, 1.0));
draw((0,0)--(-3,4));
draw((2,0)--(-1,4));
// Block 4
import graph;
Label f;
unitsize(0.7cm);
size(5cm);
xaxis(-5,5,Ticks(f, 5.0, 1.0));
yaxis(-5,5,Ticks(f, 5.0, 1.0));
draw((0,0)--(-3,4));
draw((2,0)--(-1,4));
filldraw((0,0)--(-3,4)--(-1,4)--(2,0)--cycle, gray);
// Block 5
import graph;
Label f;
unitsize(0.7cm);
size(5cm);
xaxis(-5,5,Ticks(f, 5.0, 1.0));
yaxis(-5,5,Ticks(f, 5.0, 1.0));
draw((0,0)--(-3,4));
draw((2,0)--(-1,4));
filldraw((0,0)--(-3,4)--(-1,4)--(2,0)--cycle, gray);
filldraw((0,1)--(-3,5)--(-1,5)--(2,1)--cycle, gray);
draw((0,0)--(0,1),black+dashed);
draw((2,0)--(2,1),black+dashed);
draw((-3,4)--(-3,5),black+dashed);
// Block 6
import graph;
Label f;
unitsize(0.7cm);
size(5cm);
xaxis(-5,5,Ticks(f, 5.0, 1.0));
yaxis(-5,5,Ticks(f, 5.0, 1.0));
draw((0,0)--(-3,4));
draw((1,0)--(-2,4));
filldraw((0,0)--(2,0)--(2,1)--(-1,5)--(-3,5)--(-3,4)--cycle, gray);
// Block 7
import geometry; pair A = (-3, 4); pair B = (-3, 5); pair C = (-1, 4); pair D = (-1, 5); pair AA = (0, 0); pair BB = (0, 1); pair CC = (2, 0); pair DD = (2, 1); //draw(A--B--D--C--cycle); draw(A--B); label("1",midpoint(A--B),W); label("2",midpoint(D--B),N); draw(A--C,dashed); draw(B--D); draw(C--D, dashed); draw(A--AA); label("5",midpoint(A--AA),W); draw(B--BB,dashed); draw(C--CC,dashed); draw(D--DD); label("5",midpoint(D--DD),E); label("1",midpoint(CC--DD),E); label("2",midpoint(AA--CC),S); // Dotted vertices dot(A); dot(B); dot(C); dot(D); dot(AA); dot(BB); dot(CC); dot(DD); draw(AA--BB,dashed); draw(AA--CC); draw(BB--DD,dashed); draw(CC--DD); label("(0,0)",AA,W); label("(-3,4)",A,SW); label("(-1,5)",D,E); label("(2,1)",DD,NE);
// Block 8
import graph; Label f; size(5cm); unitsize(0.7cm); xaxis(-5,5,Ticks(f, 5.0, 1.0)); yaxis(-5,5,Ticks(f, 5.0, 1.0)); draw((0,0)--(-3,4));
// Block 9
import graph; Label f; unitsize(0.7cm); size(5cm); xaxis(-5,5,Ticks(f, 5.0, 1.0)); yaxis(-5,5,Ticks(f, 5.0, 1.0)); draw((0,0)--(-3,4)); draw((2,0)--(-1,4));
// Block 10
import graph; Label f; unitsize(0.7cm); size(5cm); xaxis(-5,5,Ticks(f, 5.0, 1.0)); yaxis(-5,5,Ticks(f, 5.0, 1.0)); draw((0,0)--(-3,4)); draw((2,0)--(-1,4)); filldraw((0,0)--(-3,4)--(-1,4)--(2,0)--cycle, gray);
// Block 11
import graph; Label f; unitsize(0.7cm); size(5cm); xaxis(-5,5,Ticks(f, 5.0, 1.0)); yaxis(-5,5,Ticks(f, 5.0, 1.0)); draw((0,0)--(-3,4)); draw((2,0)--(-1,4)); filldraw((0,0)--(-3,4)--(-1,4)--(2,0)--cycle, gray); filldraw((0,1)--(-3,5)--(-1,5)--(2,1)--cycle, gray); draw((0,0)--(0,1),black+dashed); draw((2,0)--(2,1),black+dashed); draw((-3,4)--(-3,5),black+dashed);
// Block 12
import graph; Label f; unitsize(0.7cm); size(5cm); xaxis(-5,5,Ticks(f, 5.0, 1.0)); yaxis(-5,5,Ticks(f, 5.0, 1.0)); draw((0,0)--(-3,4)); draw((1,0)--(-2,4)); filldraw((0,0)--(2,0)--(2,1)--(-1,5)--(-3,5)--(-3,4)--cycle, gray); | [] |
944 | What is the perimeter of the boundary of the region consisting of all points which can be expressed as $(2u-3w, v+4w)$ with $0\le u\le1$, $0\le v\le1,$ and $0\le w\le1$?
$\textbf{(A) } 10\sqrt{3} \qquad \textbf{(B) } 13 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 16$ | 2023 AMC 10B Problem 24 | We can find the "boundary points" and work with our intuition to solve the problem. We set each of $u, v, w$ equal to $0, 1$ for a total of $8$ combinations in $u, v, w$. We now test each one.
Case 1: $u = 0, v = 0, w = 0 \implies (0, 0)$
Case 2: $u = 0, v = 0, w = 1 \implies (-3, 4)$
Case 3: $u = 0, v = 1, w = 0 \implies (0, 1)$
Case 4: $u = 0, v = 1, w = 1 \implies (-3, 5)$
Case 5: $u = 1, v = 0, w = 0 \implies (2, 0)$
Case 6: $u = 1, v = 0, w = 1 \implies (-1, 4)$
Case 7: $u = 1, v = 1, w = 0 \implies (2, 1)$
Case 8: $u = 1, v = 1, w = 1 \implies (-1, 5)$
When graphed on a coordinate plane, the points appear as follows.
Notice how there are two distinct rectangles visible in the figure. This leads us to believe that the region tracks the motion of this region as it travels in space. To understand why this is true, we can imagine a fixed $w$ (as it is present in both the $x$ and $y$ coordinates). Then if we hold one of $u$ or $v$ fixed and let the other vary, we get a straight line parallel to the $x$ or $y$ axis respectively. If we let the other vary, we get the other type of straight line. Together, they form a rectangular region. In addition, $w$ serves as a diagonal translation, so if we now let $w$ vary, it traces out the motion of the rectangle. Keeping this in mind, we connect the dots.
Each of the diagonal sides have length $5$ by the distance formula on $(0,0)$ and $(-3,4)$ (the other diagonal side is congruent), so our total perimeter is $2 + 1 + 5 + 2 + 1 + 5 = \boxed{\textbf{(E)}~ 16}$.
~ cxsmi | // Block 1
import graph;
import geometry;
Label f;
size(5cm);
unitsize(0.7cm);
xaxis(-5,5,Ticks(f, 5.0, 1.0));
yaxis(-5,5,Ticks(f, 5.0, 1.0));
pair A = (0, 0);
dot (A);
pair B = (-3, 4);
dot (B);
pair C = (0, 1);
dot (C);
pair D = (-3, 5);
dot (D);
pair E = (2, 0);
dot (E);
pair F = (-1, 4);
dot (F);
pair G = (2, 1);
dot (G);
pair H = (-1, 5);
dot (H);
// Block 2
import graph;
import geometry;
Label f;
size(5cm);
unitsize(0.7cm);
xaxis(-5,5,Ticks(f, 5.0, 1.0));
yaxis(-5,5,Ticks(f, 5.0, 1.0));
draw((0,0)--(-3,4)--(-3,5)--(-1,5)--(2,1)--(2,0)--cycle, gray);
// Block 3
import graph; import geometry; Label f; size(5cm); unitsize(0.7cm); xaxis(-5,5,Ticks(f, 5.0, 1.0)); yaxis(-5,5,Ticks(f, 5.0, 1.0)); pair A = (0, 0); dot (A); pair B = (-3, 4); dot (B); pair C = (0, 1); dot (C); pair D = (-3, 5); dot (D); pair E = (2, 0); dot (E); pair F = (-1, 4); dot (F); pair G = (2, 1); dot (G); pair H = (-1, 5); dot (H);
// Block 4
import graph; import geometry; Label f; size(5cm); unitsize(0.7cm); xaxis(-5,5,Ticks(f, 5.0, 1.0)); yaxis(-5,5,Ticks(f, 5.0, 1.0)); draw((0,0)--(-3,4)--(-3,5)--(-1,5)--(2,1)--(2,0)--cycle, gray); | [] |
945 | A regular pentagon with area $\sqrt{5}+1$ is printed on paper and cut out. The five vertices of the pentagon are folded into the center of the pentagon, creating a smaller pentagon. What is the area of the new pentagon?
$\textbf{(A)}~4-\sqrt{5}\qquad\textbf{(B)}~\sqrt{5}-1\qquad\textbf{(C)}~8-3\sqrt{5}\qquad\textbf{(D)}~\frac{\sqrt{5}+1}{2}\qquad\textbf{(E)}~\frac{2+\sqrt{5}}{3}$ | 2023 AMC 10B Problem 25 | We can find the area of the red pentagon by taking the area of the total pentagon and subtracting the area outside the red pentagon.
The area outside the red pentagon is the sum of the larger isosceles triangles, but this double counts the overlapping regions of the small isosceles triangles, so we have to subtract those out.
We have $[FGHIJ] = [ABCDE]-(5 \cdot[DKT]-5 \cdot [PFK])$
Lets focus on finding the area of each individual triangle:
Notice that we have no information about the side length, so instead we let the side length be $s$. Now we can drop an altitude from $O$ to the base of the triangle, and we know this altitude must split the base of the pentagon in half, so we can create a right triangle. Furthermore, draw a line from $O$ to $D$. This must bisect angle $D$ which is $108$ degrees, so we create $36-54-90$ triangles. Specifically, we know $\angle ODK = 54^{\circ}$, $\angle DOU = 36^{\circ}$, and $\angle DTK = 36^{\circ}$ because $\triangle DTK$ is isosceles and we know the vertex angle is $108^{\circ}$. We encode this information in the diagram below:
Since $\triangle DKT$ is isosceles, the area of $\triangle DVT$ is half the area of $\triangle DKT$. Similarly, the area of $\triangle UFK$ is half that of $\triangle PFK$. Thus:
\[[FGHIJ] = [ABCDE]-(5 \cdot[DKT]-5 \cdot [PFK]) \implies [FGHIJ] = [ABCDE]-(10 \cdot [DVT]-10 \cdot [UFK])\]
We also know that since we dropped an altitude from $O$ to $U$, the area of $\triangle ODU$ must be half of a fifth of the total area of the pentagon. Therefore we can rewrite the above equation as \[[FGHIJ]=10 \cdot ([ODU]-[DVT]+[UFK])\]
Now notice that $\triangle ODU ~ \triangle TDV ~ \triangle KFU$ by AA similarity. Therefore, if we can write the areas of the latter two triangles as a ratio of the first triangle, we can express the whole equation in terms of $[ODU]$, and by extension $[ABCDE]$, which we know. To find these ratios, we can find the side length ratios and square them because the triangles are similar.
We already know $DU = \frac{s}{2}$, so let's try to find it's analogous side for $\triangle TDV$ and $\triangle KFU$. These sides are $DV$ and $FU$, respectively.
First, $\frac{s}{2} = OD \cdot cos(54^{\circ})$, so $OD = \frac{s}{2} \cdot sec(54^{\circ})$. Then notice that $DV = \frac{OD}{2}$ because we have to fold $D$ to hit $O$, so the folding crease has to be exactly halfway between $O$ and $D$. Therefore, \[DV = \frac{s}{4} \cdot sec(54^{\circ}) \implies \frac{DU}{DV} = \frac{\frac{s}{2}}{\frac{s}{4} \cdot sec(54^{\circ})} = 2 \cdot cos(54^{\circ})\]
Now the ratio between the area of two similar triangles is the square of the ratio of their analogous side lengths. Thus
\[\frac{[ODU]}{[TDV]} = 4 \cdot cos^2(54^{\circ}) \implies \frac{10 \cdot [ODU]}{10 \cdot [TDV]} = 4 \cdot cos^2(54^{\circ})\]
\[\implies \frac{[ABCDE]}{10 \cdot [TDV]} = 4 \cdot cos^2(54^{\circ}) \implies \frac{\sqrt{5}+1}{10 \cdot [TDV]} = 4 \cdot cos^2(54^{\circ})\]
\[\implies 10 \cdot [TDV] = \frac{\sqrt{5}+1}{4 \cdot cos^2(54^{\circ})} \implies 10 \cdot [TDV] = \frac{\sqrt{5}+1}{4} \cdot sec^2(54^{\circ})\]
Now let's move on and calculate the ratio of the other side length. Calculating $FU$ is slightly tricker.
First, we find $TD$: $TD \cdot cos(54^{\circ}) = DV = \frac{s}{4} sec^54^{\circ}) \implies TD = \frac{s}{4} \cdot sec^2(54^{\circ})$. Now since $\triangle DTK$ is isosceles, $TD = DK$ and $UK = DK-DU = TD-DU = \frac{s}{4} \cdot sec^2(54^{\circ})-\frac{s}{2} = \frac{s}{4} \cdot (sec^2(54^{\circ})-2) = \frac{s}{4} \cdot (tan^2(54^{\circ})-1)$.
Now $FU = UK \cdot tan(36^{\circ}) \implies FU \frac{s}{4} \cdot (tan^2(54^{\circ})-1) \cdot tan(36^{\circ})$. Now note that $tan(x) \cdot tan(90-x) = 1$ because opposite over adjacent cancel each other out in a right triangle. Thus, $FU = \frac{s}{4} \cdot (tan(54^{\circ})-tan(36^{\circ}))$
Now, \[\frac{DU}{FU} = \frac{\frac{s}{2}}{\frac{s}{4} \cdot (tan(54^{\circ})-tan(36^{\circ}))} = \frac{2}{tan(54^{\circ})-tan(36^{\circ})}\]
\[\implies \frac{[DU]}{[FU]} = \frac{4}{(tan(54^{\circ})-tan(36^{\circ}))^2} \implies 10 \cdot [FU] = \frac{1+\sqrt{5}}{4} \cdot (tan(54^{\circ})-tan(36^{\circ}))^2\]
Now we go back to our first equation and plug in our values:
\[[FGHIJ] = [ABCDE]-(10 \cdot [DVT]-10 \cdot [UFK]) \implies [FGHIJ] = 1+\sqrt{5}-\frac{\sqrt{5}+1}{4} \cdot sec^2(54^{\circ})+\frac{1+\sqrt{5}}{4} \cdot (tan(54^{\circ})-tan(36^{\circ}))^2\]
\[\implies [FGHIJ] = \frac{1+\sqrt{5}}{4} \cdot (4-sec^2(54^{\circ})+(tan(54^{\circ})-tan(36^{\circ}))^2)\]
Note $(tan(x)-tan(90-x))^2 = tan^2(x)-2 \cdot tan(x) \cdot tan(90-x)+tan^2(90-x) = tan^2(x)-2+tan^2(90-x)$.
Also note that $tan^2(x)+1 = sec^2(x)$.
Thus \[[FGHIJ] = \frac{1+\sqrt{5}}{4} \cdot (2-sec^2(54^{\circ})+tan^2(54^{\circ})+tan^2(36^{\circ}))\]
\[\implies [FGHIJ] = \frac{1+\sqrt{5}}{4} \cdot (2+(-1)+tan^2(36^{\circ})) \implies [FGHIJ] = \frac{1+\sqrt{5}}{4} \cdot (1+tan^2(36^{\circ}))\].
Now all that remains is to find $tan^2(36^{\circ})$. We can use the tan addition formula to find the general form of $tan(5x)$ or remember question 25 from this year's AMC 12A. We have that \[tan(5x) = \frac{5tan(x)-10tan^3(x)+tan^5(x)}{1-10tan^2(x)+5tan^4(x)}\].
Plug in $x=36$. Then we have \[tan(180^{\circ}) = \frac{5tan(36^{\circ})-10tan^3(36^{\circ})+tan^5(36^{\circ})}{1-10tan^2(36^{\circ})+5tan^4(36^{\circ})}\]
Now let $y = tan(36^{\circ})$. We have the equation \[\frac{5y-10y^3+y^5}{1-10y^2+5y^4} = 0 \implies 5y-10y^3+y^5 = 0\]
\[\implies 5-10y^2+y^4 = 0 \implies 5-10z+z^2 = 0\]
Where we let $z = y^2$. Using the quadratic formula, we have \[z = \frac{10 \pm \sqrt{80}}{2} = 5 \pm 2\sqrt{5}\]
Now since $y = tan(36^{\circ})$, $z = tan^2(36^{\circ})$, which is what we were looking for. Notice that $tan(0^{\circ}) = 0$ and $tan(45^{\circ}) = 1$, so $tan(36^{\circ})$ is between $0$ and $1$, and so is it's square. Thus $z = 5 - 2\sqrt{5}$, not the other root.
Finally:
\[[FGHIJ] = \frac{1+\sqrt{5}}{4} \cdot (1+tan^2(36^{\circ})) = \frac{1+\sqrt{5}}{4} \cdot (1+5-2\sqrt{5})\]
\[\implies [FGHIJ] = \frac{1+\sqrt{5}}{4} \cdot (6-2\sqrt{5}) = \frac{4\sqrt{5}-4}{4} = \sqrt{5}-1\]
Therefore, \[[FGHIJ] = \sqrt{5}-1 = \boxed{B}\]
~KingRavi | // Block 1
unitsize(5cm); // Define the vertices of the pentagons pair A, B, C, D, E; pair F, G, H, I, J; // Calculate the vertices of the larger pentagon A = dir(90); B = dir(90 - 72); C = dir(90 - 2*72); D = dir(90 - 3*72); E = dir(90 - 4*72); // Draw the larger pentagon draw(A--B--C--D--E--cycle); pair O = (A+B+C+D+E)/5; pair AA,OO; real gap = 0.02; AA = A+(0,0); OO = O+(0,0); label("$O$", O, S); pair OOO, OAO; OOO = O+(gap,0); OAO = (O+A)/2 + (gap,0); dot(O); label("$A$", (0,1), E); label("$B$", B, S); label("$C$", C, S); label("$D$", D, S); label("$E$", E, W); real scaleFactor = 1/1.618; // Adjust this value as needed // Rotate the smaller pentagon by 180 degrees F = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 + 180); G = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 72 + 180); H = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 2*72 + 180); I = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 3*72 + 180); J = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 4*72 + 180); pair K, L, M, N, O, P, Q, R, S, T, U, V; real newScaleFactor = 0.8507; K = newScaleFactor*dir(270+18); L = newScaleFactor*dir(270+72+18); M = newScaleFactor*dir(270+72+72+18); N = newScaleFactor*dir(270+72+72+72+18); O = newScaleFactor*dir(270+72+72+72+72+18); P = newScaleFactor*dir(270-18); Q = newScaleFactor*dir(270+72-18); R = newScaleFactor*dir(270+72+72-18); S = newScaleFactor*dir(270+72+72+72-18); T = newScaleFactor*dir(270+72+72+72+72-18); label("$K$", K, S); label("$L$", L, S); label("$M$", M, S); label("$N$", N, S); label("$O$", O, W); label("$P$", P, S); label("$Q$", Q, E); label("$R$", R, S); label("$S$", S, S); label("$T$", T, W); draw(K--T, dashed); draw(S--O, dashed); draw(P--L, dashed); draw(Q--M, dashed); draw(R--N, dashed); label("$F$", F, S); label("$G$", G, S); label("$H$", H, S); label("$I$", I, S); label("$J$", J, S); // Draw the smaller pentagon draw(F--G--H--I--J--cycle,red);
// Block 2
unitsize(5cm); // Define the vertices of the pentagons pair A, B, C, D, E; pair F, G, H, I, J; // Calculate the vertices of the larger pentagon A = dir(90); B = dir(90 - 72); C = dir(90 - 2*72); D = dir(90 - 3*72); E = dir(90 - 4*72); pair O = (A+B+C+D+E)/5; pair AA,OO; real gap = 0.02; AA = A+(0,0); OO = O+(0,0); label("$O$", O, S); pair OOO, OAO; OOO = O+(gap,0); OAO = (O+A)/2 + (gap,0); dot(O); label("$D$", D, S); real scaleFactor = 1/1.618; // Adjust this value as needed // Rotate the smaller pentagon by 180 degrees F = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 + 180); G = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 72 + 180); H = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 2*72 + 180); I = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 3*72 + 180); J = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 4*72 + 180); label("$F$", F, N); pair K, L, M, N, O, P, Q, R, S, T, U, V; real newScaleFactor = 0.8507; K = newScaleFactor*dir(270+18); T = newScaleFactor*dir(270+72+72+72+72-18); label("$K$", K, E); label("$T$", T, W); draw(K--T); draw(K--D); draw(D--T);
// Block 3
unitsize(5cm); // Define the vertices of the pentagons pair A, B, C, D, E; pair F, G, H, I, J; // Calculate the vertices of the larger pentagon A = dir(90); B = dir(90 - 72); C = dir(90 - 2*72); D = dir(90 - 3*72); E = dir(90 - 4*72); pair O = (A+B+C+D+E)/5; pair AA,OO; real gap = 0.02; AA = A+(0,0); OO = O+(0,0); label("$O$", O, E); pair OOO, OAO; OOO = O+(gap,0); OAO = (O+A)/2 + (gap,0); dot(O); label("$D$", D, S); real scaleFactor = 1/1.618; // Adjust this value as needed // Rotate the smaller pentagon by 180 degrees F = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 + 180); G = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 72 + 180); H = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 2*72 + 180); I = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 3*72 + 180); J = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 4*72 + 180); label("$F$", (0.1,-1/1.618), E); pair K, L, M, N, O, P, Q, R, S, T, U, V; real newScaleFactor = 0.8507; K = newScaleFactor*dir(270+18); T = newScaleFactor*dir(270+72+72+72+72-18); label("$K$", K, S); label("$T$", T, W); draw(K--T); draw(K--D); draw(D--T); pair U; U=(0,-0.809); label("$U$",(0,-0.9), S); draw(O--U); draw(O--D); pair V; V = midpoint(O--D); label("$V$", V+(0,0.05), N); markscalefactor = 0.005; draw(rightanglemark(D,U,O)); draw(rightanglemark(F,V,O)); draw(rightanglemark(K,U,F)); draw(anglemark(U,D,O)); label("$54^{\circ}$", D+(0.05,0),NE); draw(anglemark(D,O,U)); label("$36^{\circ}$", O-(0,0.2),SW); draw(anglemark(D,T,F)); label("$36^{\circ}$", T+(0.1,-0.17),SE); label("$\frac{s}{2}$", D+(0.3,-0.1), S); | [] |
945 | A regular pentagon with area $\sqrt{5}+1$ is printed on paper and cut out. The five vertices of the pentagon are folded into the center of the pentagon, creating a smaller pentagon. What is the area of the new pentagon?
$\textbf{(A)}~4-\sqrt{5}\qquad\textbf{(B)}~\sqrt{5}-1\qquad\textbf{(C)}~8-3\sqrt{5}\qquad\textbf{(D)}~\frac{\sqrt{5}+1}{2}\qquad\textbf{(E)}~\frac{2+\sqrt{5}}{3}$ | 2023 AMC 10B Problem 25 | Let $r_b$ and $r_s$ be the circumradius of the big and small pentagon, respectively. Let $a_s$ be the apothem of the smaller pentagon and $A_s$ and $A_b$ be the areas of the smaller and larger pentagon, respectively.
From the diagram:
\begin{align*} \cos{36^\circ} &= \dfrac{a_s}{r_s} = \dfrac{\phi}{2} = \dfrac{\sqrt{5}+1}{4}\\ a_s &= \dfrac{r_b}{2}\\ A_s &= \left(\dfrac{r_s}{r_b}\right)^2A_b\\ &=\left(\dfrac{a_s}{\cos{36^\circ} r_b}\right)^2 \left(1+\sqrt{5}\right)\\ &=\left(\dfrac{a_s}{\dfrac{\phi}{2} r_b}\right)^2 \left(1+\sqrt{5}\right)\\ &=\left(\dfrac{1}{2 \dfrac{\phi}{2}}\right)^2 \left(1+\sqrt{5}\right)\\ &=\left(\dfrac{2}{\sqrt{5}+1}\right)^2 \left(1+\sqrt{5}\right)\\ &=\dfrac{4}{\sqrt{5}+1} \\ &=\dfrac{4\left(\sqrt{5}-1\right)}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)} \\ &=\sqrt{5}-1 \end{align*}
\[\boxed{\textbf{(B) }\sqrt{5}-1}\]
~Technodoggo | unitsize(5cm); // Define the vertices of the pentagons pair A, B, C, D, E; pair F, G, H, I, J; // Calculate the vertices of the larger pentagon A = dir(90); B = dir(90 - 72); C = dir(90 - 2*72); D = dir(90 - 3*72); E = dir(90 - 4*72); // Draw the larger pentagon draw(A--B--C--D--E--cycle); pair O = (A+B+C+D+E)/5; pair AA,OO; real gap = 0.02; AA = A+(0,0); OO = O+(0,0); draw(AA--OO, blue); pair OOO, OAO; OOO = O+(gap,0); OAO = (O+A)/2 + (gap,0); draw(OOO--OAO,green); dot(O); dot((O+A)/2); label("$r_b$", (O+A)*.7, E,blue); label("$a_s$", (O+A)*.2 +(0+0.18,0.05), E,green); label("$r_s$", O+(-0.175,0.2), E,pink); label("$A$", (0,0), E); real scaleFactor = 1/1.618; // Adjust this value as needed // Rotate the smaller pentagon by 180 degrees F = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 + 180); G = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 72 + 180); H = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 2*72 + 180); I = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 3*72 + 180); J = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 4*72 + 180); // Draw the smaller pentagon draw(F--G--H--I--J--cycle,red); draw(arc(O,(H+I)*.5*.6,H*.6)); label("$36^\circ$",O+(+0.05,0.15),NW); draw(O--H,pink); | [] |
946 | You are playing a game. A $2 \times 1$ rectangle covers two adjacent squares (oriented either horizontally or vertically) of a $3 \times 3$ grid of squares, but you are not told which two squares are covered. Your goal is to find at least one square that is covered by the rectangle. A "turn" consists of you guessing a square, after which you are told whether that square is covered by the hidden rectangle. What is the minimum number of turns you need to ensure that at least one of your guessed squares is covered by the rectangle?
$\textbf{(A)}~3\qquad\textbf{(B)}~5\qquad\textbf{(C)}~4\qquad\textbf{(D)}~8\qquad\textbf{(E)}~6$ | 2023 AMC 12B Problem 5 | Notice that the $3\times3$ square grid has a total of $12$ possible $2\times1$ rectangles.
Suppose you choose the middle square for one of your turns. The middle square is covered by $4$ rectangles, and each of the remaining $8$ squares is covered by a maximum of $2$ uncounted rectangles. This means that the number of turns is at least $1+\frac{12-4}{2}=1+4=5$.
Now suppose you don't choose the middle square. The squares on the middle of the sides are covered by at most 3 uncounted rectangles, and the squares on the corners are covered by at most 2 uncounted rectangles. In this case, we see that the least number of turns needed to account for all 12 rectangles is $12\div 3=4.$ To prove that choosing only side squares indeed does cover all 12 rectangles, we need to show that the 3 rectangles per square that cover each side square do not overlap. Drawing the rectangles that cover one square, we see they form a $T$ shape and they do not cover any other side square. Hence, our answer is $4.$ | // Block 1
draw((0,0)--(0.5,0)--(0.5,0.5)--(0,0.5)--(0,0));
draw((0,1)--(0.5,1)--(0.5,1.5)--(0,1.5)--(0,1));
draw((0.5,0.5)--(1,0.5)--(1,1)--(0.5,1)--(0.5,0.5));
draw((1,0)--(1.5,0)--(1.5,0.5)--(1,0.5)--(1,0));
draw((1,1)--(1.5,1)--(1.5,1.5)--(1,1.5)--(1,1));
draw((0,0.5)--(0.5,0.5)--(0.5,1)--(0,1)--(0,0.5));
draw((0.5,0)--(1,0)--(1,0.5)--(0.5,0.5)--(0.5,0));
draw((0.5,1)--(1,1)--(1,1.5)--(0.5,1.5)--(0.5,1));
draw((1,0.5)--(1.5,0.5)--(1.5,1)--(1,1)--(1,0.5));
filldraw((0,0.5)--(0.5,0.5)--(0.5,1)--(0,1)--(0,0.5)--cycle, red, black+linewidth(1));
filldraw((0,0)--(0.5,0)--(0.5,0.5)--(0,0.5)--(0,0)--cycle, red, black+linewidth(1));
filldraw((0,1)--(0.5,1)--(0.5,1.5)--(0,1.5)--(0,1)--cycle, red, black+linewidth(1));
filldraw((0.5,0.5)--(1,0.5)--(1,1)--(0.5,1)--(0.5,0.5)--cycle, red, black+linewidth(1));
// Block 2
draw((0,0)--(0.5,0)--(0.5,0.5)--(0,0.5)--(0,0)); draw((0,1)--(0.5,1)--(0.5,1.5)--(0,1.5)--(0,1)); draw((0.5,0.5)--(1,0.5)--(1,1)--(0.5,1)--(0.5,0.5)); draw((1,0)--(1.5,0)--(1.5,0.5)--(1,0.5)--(1,0)); draw((1,1)--(1.5,1)--(1.5,1.5)--(1,1.5)--(1,1)); draw((0,0.5)--(0.5,0.5)--(0.5,1)--(0,1)--(0,0.5)); draw((0.5,0)--(1,0)--(1,0.5)--(0.5,0.5)--(0.5,0)); draw((0.5,1)--(1,1)--(1,1.5)--(0.5,1.5)--(0.5,1)); draw((1,0.5)--(1.5,0.5)--(1.5,1)--(1,1)--(1,0.5)); filldraw((0,0.5)--(0.5,0.5)--(0.5,1)--(0,1)--(0,0.5)--cycle, red, black+linewidth(1)); filldraw((0,0)--(0.5,0)--(0.5,0.5)--(0,0.5)--(0,0)--cycle, red, black+linewidth(1)); filldraw((0,1)--(0.5,1)--(0.5,1.5)--(0,1.5)--(0,1)--cycle, red, black+linewidth(1)); filldraw((0.5,0.5)--(1,0.5)--(1,1)--(0.5,1)--(0.5,0.5)--cycle, red, black+linewidth(1)); | [] |
946 | You are playing a game. A $2 \times 1$ rectangle covers two adjacent squares (oriented either horizontally or vertically) of a $3 \times 3$ grid of squares, but you are not told which two squares are covered. Your goal is to find at least one square that is covered by the rectangle. A "turn" consists of you guessing a square, after which you are told whether that square is covered by the hidden rectangle. What is the minimum number of turns you need to ensure that at least one of your guessed squares is covered by the rectangle?
$\textbf{(A)}~3\qquad\textbf{(B)}~5\qquad\textbf{(C)}~4\qquad\textbf{(D)}~8\qquad\textbf{(E)}~6$ | 2023 AMC 12B Problem 5 | The $3 \times 3$ grid can be colored like a checkerboard with alternating black and white squares.
Let the top left square be white, and the rest of the squares alternate colors, as shown below:
Each $2 \times 1$ rectangle always covers $1$ white square and $1$ black square.
You can ensure that at least one of your guessed squares is covered by the rectangle by choosing either each of the white squares ($5$ turns) or each of the black squares ($4$ turns).
Since it is ideal to be the most efficient with our turns, by choosing all the black squares, we guarantee that one of the $\boxed{\textbf{(C)}~4}$ squares are of the $2 \times 1$ rectangle.
~ CherryBerry
(Minor edits by NSAoPS) | // Block 1
/* Diagram by MRENTHUSIASM */
import olympiad;
unitsize(25);
fill((1,2)--(1,3)--(2,3)--(2,2)--cycle, mediumgray);
fill((1,0)--(1,1)--(2,1)--(2,0)--cycle, mediumgray);
fill((2,1)--(2,2)--(3,2)--(3,1)--cycle, mediumgray);
fill((0,1)--(0,2)--(1,2)--(1,1)--cycle, mediumgray);
for (int i = 0; i < 4; ++i) {
draw((i,0)--(i,3));
draw((0,i)--(3,i));
}
// Block 2
/* Diagram by MRENTHUSIASM */ import olympiad; unitsize(25); fill((1,2)--(1,3)--(2,3)--(2,2)--cycle, mediumgray); fill((1,0)--(1,1)--(2,1)--(2,0)--cycle, mediumgray); fill((2,1)--(2,2)--(3,2)--(3,1)--cycle, mediumgray); fill((0,1)--(0,2)--(1,2)--(1,1)--cycle, mediumgray); for (int i = 0; i < 4; ++i) { draw((i,0)--(i,3)); draw((0,i)--(3,i)); } | [] |
947 | What is the maximum area of an isosceles trapezoid that has legs of length $1$ and one base twice as long as the other?
$\textbf{(A) }\frac 54 \qquad \textbf{(B) } \frac 87 \qquad \textbf{(C)} \frac{5\sqrt2}4 \qquad \textbf{(D) } \frac 32 \qquad \textbf{(E) } \frac{3\sqrt3}4$ | 2023 AMC 12B Problem 11 | Let the length of the shorter base of the trapezoid be $2x$ and the height of the trapezoid be $y$.
Each leg has length $1$ if and only if $x^2+y^2=1$, where $x$ and $y$ are positive real numbers. The general solution to this equation is \[(x,y)=(\cos t,\sin t)\] for any number $0<t<\frac{\pi}{2}$ so that $x$ and $y$ are positive. The area to maximize is \[\frac{1}{2}(2x+4x)y=3xy\] Hence, we maximize $3\sin t\cos t$ for $0<t<\frac{\pi}{2}$.
\begin{align*} 3xy &= 3\sin t\cos t \\ &= \frac{3}{2}(2\sin t\cos t) \\ &= \frac{3}{2}\sin(2t) \end{align*}
The maximum of $\sin(2t)$ is $1$, thus the maximum of $3xy$ is $\boxed{\text{(D) }\frac{3}{2}}$ which occurs at $t=\frac{\pi}{4}$, satisfying the inequality $0<t<\frac{\pi}{2}$.
~Robabob1 | unitsize(100); pair A=(-1, 0), B=(1, 0), C=(0.5, 0.5), D=(-0.5, 0.5); draw(A--B--C--D--cycle, black); label("$2x$",(0,0.58),(0,0)); label("$2x$",(0,-0.08),(0,0)); label("$x$",(-0.75,-0.08),(0,0)); label("$x$",(0.75,-0.08),(0,0)); draw(D--(-0.5,0),black); draw(C--(0.5,0),black); label("$y$",(0.58,0.25)); label("$y$",(-0.42,0.25)); | [] |
948 | A rectangular box $\mathcal{P}$ has distinct edge lengths $a$, $b$, and $c$. The sum of the lengths of all $12$ edges of $\mathcal{P}$ is $13$, the areas of all $6$ faces of $\mathcal{P}$ is $\frac{11}{2}$, and the volume of $\mathcal{P}$ is $\frac{1}{2}$. What is the length of the longest interior diagonal connecting two vertices of $\mathcal{P}$?
$\textbf{(A)}~2\qquad\textbf{(B)}~\frac{3}{8}\qquad\textbf{(C)}~\frac{9}{8}\qquad\textbf{(D)}~\frac{9}{4}\qquad\textbf{(E)}~\frac{3}{2}$ | 2023 AMC 12B Problem 13 | We can create three equations using the given information.
\[4a+4b+4c = 13\]
\[2ab+2ac+2bc=\frac{11}{2}\]
\[abc=\frac{1}{2}\]
We also know that we want $\sqrt{a^2 + b^2 + c^2}$ because that is the length that can be found from using the Pythagorean Theorem. We cleverly notice that $a^2 + b^2 + c^2 = (a+b+c)^2 - 2(ab+ac+bc)$. We know that $a+b+c = \frac{13}{4}$ and $2(ab+ac+bc)=\dfrac{11}2$, so $a^2 + b^2 + c^2 = \left(\frac{13}{4}\right)^2 - \frac{11}{2} = \frac{169-88}{16} = \frac{81}{16}$. So our answer is $\sqrt{\frac{81}{16}} = \boxed{\textbf{(D)}~\tfrac94}$.
~lprado
~minor edits and add-ons by Technodoggo, lucaswujc, andliu766, BcMath, and MW2014
Note:This solution does not necessarily use the fact that $abc=\frac{1}{2}$, unlike the second solution. | // Block 1
import geometry;
pair A = (-3, 4);
pair B = (-3, 5);
pair C = (-1, 4);
pair D = (-1, 5);
pair AA = (0, 0);
pair BB = (0, 1);
pair CC = (2, 0);
pair DD = (2, 1);
draw(D--AA,dashed);
draw(A--B);
draw(A--C);
draw(B--D);
draw(C--D);
draw(A--AA);
draw(B--BB);
draw(C--CC);
draw(D--DD);
// Dotted vertices
dot(A); dot(B); dot(C); dot(D);
dot(AA); dot(BB); dot(CC); dot(DD);
draw(AA--BB);
draw(AA--CC);
draw(BB--DD);
draw(CC--DD);
label("a",midpoint(D--DD),E);
label("b",midpoint(CC--DD),E);
label("c",midpoint(AA--CC),S);
// Block 2
import geometry; pair A = (-3, 4); pair B = (-3, 5); pair C = (-1, 4); pair D = (-1, 5); pair AA = (0, 0); pair BB = (0, 1); pair CC = (2, 0); pair DD = (2, 1); draw(D--AA,dashed); draw(A--B); draw(A--C); draw(B--D); draw(C--D); draw(A--AA); draw(B--BB); draw(C--CC); draw(D--DD); // Dotted vertices dot(A); dot(B); dot(C); dot(D); dot(AA); dot(BB); dot(CC); dot(DD); draw(AA--BB); draw(AA--CC); draw(BB--DD); draw(CC--DD); label("a",midpoint(D--DD),E); label("b",midpoint(CC--DD),E); label("c",midpoint(AA--CC),S); | [] |
949 | A regular pentagon with area $\sqrt{5}+1$ is printed on paper and cut out. The five vertices of the pentagon are folded into the center of the pentagon, creating a smaller pentagon. What is the area of the new pentagon?
$\textbf{(A)}~4-\sqrt{5}\qquad\textbf{(B)}~\sqrt{5}-1\qquad\textbf{(C)}~8-3\sqrt{5}\qquad\textbf{(D)}~\frac{\sqrt{5}+1}{2}\qquad\textbf{(E)}~\frac{2+\sqrt{5}}{3}$ | 2023 AMC 12B Problem 25 | We can find the area of the red pentagon by taking the area of the total pentagon and subtracting the area outside the red pentagon.
The area outside the red pentagon is the sum of the larger isosceles triangles, but this double counts the overlapping regions of the small isosceles triangles, so we have to subtract those out.
We have $[FGHIJ] = [ABCDE]-(5 \cdot[DKT]-5 \cdot [PFK])$
Lets focus on finding the area of each individual triangle:
Notice that we have no information about the side length, so instead we let the side length be $s$. Now we can drop an altitude from $O$ to the base of the triangle, and we know this altitude must split the base of the pentagon in half, so we can create a right triangle. Furthermore, draw a line from $O$ to $D$. This must bisect angle $D$ which is $108$ degrees, so we create $36-54-90$ triangles. Specifically, we know $\angle ODK = 54^{\circ}$, $\angle DOU = 36^{\circ}$, and $\angle DTK = 36^{\circ}$ because $\triangle DTK$ is isosceles and we know the vertex angle is $108^{\circ}$. We encode this information in the diagram below:
Since $\triangle DKT$ is isosceles, the area of $\triangle DVT$ is half the area of $\triangle DKT$. Similarly, the area of $\triangle UFK$ is half that of $\triangle PFK$. Thus:
\[[FGHIJ] = [ABCDE]-(5 \cdot[DKT]-5 \cdot [PFK]) \implies [FGHIJ] = [ABCDE]-(10 \cdot [DVT]-10 \cdot [UFK])\]
We also know that since we dropped an altitude from $O$ to $U$, the area of $\triangle ODU$ must be half of a fifth of the total area of the pentagon. Therefore we can rewrite the above equation as \[[FGHIJ]=10 \cdot ([ODU]-[DVT]+[UFK])\]
Now notice that $\triangle ODU ~ \triangle TDV ~ \triangle KFU$ by AA similarity. Therefore, if we can write the areas of the latter two triangles as a ratio of the first triangle, we can express the whole equation in terms of $[ODU]$, and by extension $[ABCDE]$, which we know. To find these ratios, we can find the side length ratios and square them because the triangles are similar.
We already know $DU = \frac{s}{2}$, so let's try to find it's analogous side for $\triangle TDV$ and $\triangle KFU$. These sides are $DV$ and $FU$, respectively.
First, $\frac{s}{2} = OD \cdot cos(54^{\circ})$, so $OD = \frac{s}{2} \cdot sec(54^{\circ})$. Then notice that $DV = \frac{OD}{2}$ because we have to fold $D$ to hit $O$, so the folding crease has to be exactly halfway between $O$ and $D$. Therefore, \[DV = \frac{s}{4} \cdot sec(54^{\circ}) \implies \frac{DU}{DV} = \frac{\frac{s}{2}}{\frac{s}{4} \cdot sec(54^{\circ})} = 2 \cdot cos(54^{\circ})\]
Now the ratio between the area of two similar triangles is the square of the ratio of their analogous side lengths. Thus
\[\frac{[ODU]}{[TDV]} = 4 \cdot cos^2(54^{\circ}) \implies \frac{10 \cdot [ODU]}{10 \cdot [TDV]} = 4 \cdot cos^2(54^{\circ})\]
\[\implies \frac{[ABCDE]}{10 \cdot [TDV]} = 4 \cdot cos^2(54^{\circ}) \implies \frac{\sqrt{5}+1}{10 \cdot [TDV]} = 4 \cdot cos^2(54^{\circ})\]
\[\implies 10 \cdot [TDV] = \frac{\sqrt{5}+1}{4 \cdot cos^2(54^{\circ})} \implies 10 \cdot [TDV] = \frac{\sqrt{5}+1}{4} \cdot sec^2(54^{\circ})\]
Now let's move on and calculate the ratio of the other side length. Calculating $FU$ is slightly tricker.
First, we find $TD$: $TD \cdot cos(54^{\circ}) = DV = \frac{s}{4} sec^54^{\circ}) \implies TD = \frac{s}{4} \cdot sec^2(54^{\circ})$. Now since $\triangle DTK$ is isosceles, $TD = DK$ and $UK = DK-DU = TD-DU = \frac{s}{4} \cdot sec^2(54^{\circ})-\frac{s}{2} = \frac{s}{4} \cdot (sec^2(54^{\circ})-2) = \frac{s}{4} \cdot (tan^2(54^{\circ})-1)$.
Now $FU = UK \cdot tan(36^{\circ}) \implies FU \frac{s}{4} \cdot (tan^2(54^{\circ})-1) \cdot tan(36^{\circ})$. Now note that $tan(x) \cdot tan(90-x) = 1$ because opposite over adjacent cancel each other out in a right triangle. Thus, $FU = \frac{s}{4} \cdot (tan(54^{\circ})-tan(36^{\circ}))$
Now, \[\frac{DU}{FU} = \frac{\frac{s}{2}}{\frac{s}{4} \cdot (tan(54^{\circ})-tan(36^{\circ}))} = \frac{2}{tan(54^{\circ})-tan(36^{\circ})}\]
\[\implies \frac{[DU]}{[FU]} = \frac{4}{(tan(54^{\circ})-tan(36^{\circ}))^2} \implies 10 \cdot [FU] = \frac{1+\sqrt{5}}{4} \cdot (tan(54^{\circ})-tan(36^{\circ}))^2\]
Now we go back to our first equation and plug in our values:
\[[FGHIJ] = [ABCDE]-(10 \cdot [DVT]-10 \cdot [UFK]) \implies [FGHIJ] = 1+\sqrt{5}-\frac{\sqrt{5}+1}{4} \cdot sec^2(54^{\circ})+\frac{1+\sqrt{5}}{4} \cdot (tan(54^{\circ})-tan(36^{\circ}))^2\]
\[\implies [FGHIJ] = \frac{1+\sqrt{5}}{4} \cdot (4-sec^2(54^{\circ})+(tan(54^{\circ})-tan(36^{\circ}))^2)\]
Note $(tan(x)-tan(90-x))^2 = tan^2(x)-2 \cdot tan(x) \cdot tan(90-x)+tan^2(90-x) = tan^2(x)-2+tan^2(90-x)$.
Also note that $tan^2(x)+1 = sec^2(x)$.
Thus \[[FGHIJ] = \frac{1+\sqrt{5}}{4} \cdot (2-sec^2(54^{\circ})+tan^2(54^{\circ})+tan^2(36^{\circ}))\]
\[\implies [FGHIJ] = \frac{1+\sqrt{5}}{4} \cdot (2+(-1)+tan^2(36^{\circ})) \implies [FGHIJ] = \frac{1+\sqrt{5}}{4} \cdot (1+tan^2(36^{\circ}))\].
Now all that remains is to find $tan^2(36^{\circ})$. We can use the tan addition formula to find the general form of $tan(5x)$ or remember question 25 from this year's AMC 12A. We have that \[tan(5x) = \frac{5tan(x)-10tan^3(x)+tan^5(x)}{1-10tan^2(x)+5tan^4(x)}\].
Plug in $x=36$. Then we have \[tan(180^{\circ}) = \frac{5tan(36^{\circ})-10tan^3(36^{\circ})+tan^5(36^{\circ})}{1-10tan^2(36^{\circ})+5tan^4(36^{\circ})}\]
Now let $y = tan(36^{\circ})$. We have the equation \[\frac{5y-10y^3+y^5}{1-10y^2+5y^4} = 0 \implies 5y-10y^3+y^5 = 0\]
\[\implies 5-10y^2+y^4 = 0 \implies 5-10z+z^2 = 0\]
Where we let $z = y^2$. Using the quadratic formula, we have \[z = \frac{10 \pm \sqrt{80}}{2} = 5 \pm 2\sqrt{5}\]
Now since $y = tan(36^{\circ})$, $z = tan^2(36^{\circ})$, which is what we were looking for. Notice that $tan(0^{\circ}) = 0$ and $tan(45^{\circ}) = 1$, so $tan(36^{\circ})$ is between $0$ and $1$, and so is it's square. Thus $z = 5 - 2\sqrt{5}$, not the other root.
Finally:
\[[FGHIJ] = \frac{1+\sqrt{5}}{4} \cdot (1+tan^2(36^{\circ})) = \frac{1+\sqrt{5}}{4} \cdot (1+5-2\sqrt{5})\]
\[\implies [FGHIJ] = \frac{1+\sqrt{5}}{4} \cdot (6-2\sqrt{5}) = \frac{4\sqrt{5}-4}{4} = \sqrt{5}-1\]
Therefore, \[[FGHIJ] = \sqrt{5}-1 = \boxed{B}\]
~KingRavi | // Block 1
unitsize(5cm);
// Define the vertices of the pentagons
pair A, B, C, D, E;
pair F, G, H, I, J;
// Calculate the vertices of the larger pentagon
A = dir(90);
B = dir(90 - 72);
C = dir(90 - 2*72);
D = dir(90 - 3*72);
E = dir(90 - 4*72);
// Draw the larger pentagon
draw(A--B--C--D--E--cycle);
pair O = (A+B+C+D+E)/5;
pair AA,OO;
real gap = 0.02;
AA = A+(0,0);
OO = O+(0,0);
label("$O$", O, S);
pair OOO, OAO;
OOO = O+(gap,0);
OAO = (O+A)/2 + (gap,0);
dot(O);
label("$A$", (0,1), E);
label("$B$", B, S);
label("$C$", C, S);
label("$D$", D, S);
label("$E$", E, W);
real scaleFactor = 1/1.618; // Adjust this value as needed
// Rotate the smaller pentagon by 180 degrees
F = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 + 180);
G = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 72 + 180);
H = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 2*72 + 180);
I = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 3*72 + 180);
J = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 4*72 + 180);
pair K, L, M, N, O, P, Q, R, S, T, U, V;
real newScaleFactor = 0.8507;
K = newScaleFactor*dir(270+18);
L = newScaleFactor*dir(270+72+18);
M = newScaleFactor*dir(270+72+72+18);
N = newScaleFactor*dir(270+72+72+72+18);
O = newScaleFactor*dir(270+72+72+72+72+18);
P = newScaleFactor*dir(270-18);
Q = newScaleFactor*dir(270+72-18);
R = newScaleFactor*dir(270+72+72-18);
S = newScaleFactor*dir(270+72+72+72-18);
T = newScaleFactor*dir(270+72+72+72+72-18);
label("$K$", K, S);
label("$L$", L, S);
label("$M$", M, S);
label("$N$", N, S);
label("$O$", O, W);
label("$P$", P, S);
label("$Q$", Q, E);
label("$R$", R, S);
label("$S$", S, S);
label("$T$", T, W);
draw(K--T, dashed);
draw(S--O, dashed);
draw(P--L, dashed);
draw(Q--M, dashed);
draw(R--N, dashed);
label("$F$", F, S);
label("$G$", G, S);
label("$H$", H, S);
label("$I$", I, S);
label("$J$", J, S);
// Draw the smaller pentagon
draw(F--G--H--I--J--cycle,red);
// Block 2
unitsize(5cm);
// Define the vertices of the pentagons
pair A, B, C, D, E;
pair F, G, H, I, J;
// Calculate the vertices of the larger pentagon
A = dir(90);
B = dir(90 - 72);
C = dir(90 - 2*72);
D = dir(90 - 3*72);
E = dir(90 - 4*72);
pair O = (A+B+C+D+E)/5;
pair AA,OO;
real gap = 0.02;
AA = A+(0,0);
OO = O+(0,0);
label("$O$", O, S);
pair OOO, OAO;
OOO = O+(gap,0);
OAO = (O+A)/2 + (gap,0);
dot(O);
label("$D$", D, S);
real scaleFactor = 1/1.618; // Adjust this value as needed
// Rotate the smaller pentagon by 180 degrees
F = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 + 180);
G = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 72 + 180);
H = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 2*72 + 180);
I = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 3*72 + 180);
J = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 4*72 + 180);
label("$F$", F, N);
pair K, L, M, N, O, P, Q, R, S, T, U, V;
real newScaleFactor = 0.8507;
K = newScaleFactor*dir(270+18);
T = newScaleFactor*dir(270+72+72+72+72-18);
label("$K$", K, E);
label("$T$", T, W);
draw(K--T);
draw(K--D);
draw(D--T);
// Block 3
unitsize(5cm);
// Define the vertices of the pentagons
pair A, B, C, D, E;
pair F, G, H, I, J;
// Calculate the vertices of the larger pentagon
A = dir(90);
B = dir(90 - 72);
C = dir(90 - 2*72);
D = dir(90 - 3*72);
E = dir(90 - 4*72);
pair O = (A+B+C+D+E)/5;
pair AA,OO;
real gap = 0.02;
AA = A+(0,0);
OO = O+(0,0);
label("$O$", O, E);
pair OOO, OAO;
OOO = O+(gap,0);
OAO = (O+A)/2 + (gap,0);
dot(O);
label("$D$", D, S);
real scaleFactor = 1/1.618; // Adjust this value as needed
// Rotate the smaller pentagon by 180 degrees
F = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 + 180);
G = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 72 + 180);
H = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 2*72 + 180);
I = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 3*72 + 180);
J = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 4*72 + 180);
label("$F$", (0.1,-1/1.618), E);
pair K, L, M, N, O, P, Q, R, S, T, U, V;
real newScaleFactor = 0.8507;
K = newScaleFactor*dir(270+18);
T = newScaleFactor*dir(270+72+72+72+72-18);
label("$K$", K, S);
label("$T$", T, W);
draw(K--T);
draw(K--D);
draw(D--T);
pair U;
U=(0,-0.809);
label("$U$",(0,-0.9), S);
draw(O--U);
draw(O--D);
pair V;
V = midpoint(O--D);
label("$V$", V+(0,0.05), N);
markscalefactor = 0.005;
draw(rightanglemark(D,U,O));
draw(rightanglemark(F,V,O));
draw(rightanglemark(K,U,F));
draw(anglemark(U,D,O));
label("$54^{\circ}$", D+(0.05,0),NE);
draw(anglemark(D,O,U));
label("$36^{\circ}$", O-(0,0.2),SW);
draw(anglemark(D,T,F));
label("$36^{\circ}$", T+(0.1,-0.17),SE);
label("$\frac{s}{2}$", D+(0.3,-0.1), S);
// Block 4
unitsize(5cm); // Define the vertices of the pentagons pair A, B, C, D, E; pair F, G, H, I, J; // Calculate the vertices of the larger pentagon A = dir(90); B = dir(90 - 72); C = dir(90 - 2*72); D = dir(90 - 3*72); E = dir(90 - 4*72); // Draw the larger pentagon draw(A--B--C--D--E--cycle); pair O = (A+B+C+D+E)/5; pair AA,OO; real gap = 0.02; AA = A+(0,0); OO = O+(0,0); label("$O$", O, S); pair OOO, OAO; OOO = O+(gap,0); OAO = (O+A)/2 + (gap,0); dot(O); label("$A$", (0,1), E); label("$B$", B, S); label("$C$", C, S); label("$D$", D, S); label("$E$", E, W); real scaleFactor = 1/1.618; // Adjust this value as needed // Rotate the smaller pentagon by 180 degrees F = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 + 180); G = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 72 + 180); H = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 2*72 + 180); I = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 3*72 + 180); J = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 4*72 + 180); pair K, L, M, N, O, P, Q, R, S, T, U, V; real newScaleFactor = 0.8507; K = newScaleFactor*dir(270+18); L = newScaleFactor*dir(270+72+18); M = newScaleFactor*dir(270+72+72+18); N = newScaleFactor*dir(270+72+72+72+18); O = newScaleFactor*dir(270+72+72+72+72+18); P = newScaleFactor*dir(270-18); Q = newScaleFactor*dir(270+72-18); R = newScaleFactor*dir(270+72+72-18); S = newScaleFactor*dir(270+72+72+72-18); T = newScaleFactor*dir(270+72+72+72+72-18); label("$K$", K, S); label("$L$", L, S); label("$M$", M, S); label("$N$", N, S); label("$O$", O, W); label("$P$", P, S); label("$Q$", Q, E); label("$R$", R, S); label("$S$", S, S); label("$T$", T, W); draw(K--T, dashed); draw(S--O, dashed); draw(P--L, dashed); draw(Q--M, dashed); draw(R--N, dashed); label("$F$", F, S); label("$G$", G, S); label("$H$", H, S); label("$I$", I, S); label("$J$", J, S); // Draw the smaller pentagon draw(F--G--H--I--J--cycle,red);
// Block 5
unitsize(5cm); // Define the vertices of the pentagons pair A, B, C, D, E; pair F, G, H, I, J; // Calculate the vertices of the larger pentagon A = dir(90); B = dir(90 - 72); C = dir(90 - 2*72); D = dir(90 - 3*72); E = dir(90 - 4*72); pair O = (A+B+C+D+E)/5; pair AA,OO; real gap = 0.02; AA = A+(0,0); OO = O+(0,0); label("$O$", O, S); pair OOO, OAO; OOO = O+(gap,0); OAO = (O+A)/2 + (gap,0); dot(O); label("$D$", D, S); real scaleFactor = 1/1.618; // Adjust this value as needed // Rotate the smaller pentagon by 180 degrees F = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 + 180); G = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 72 + 180); H = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 2*72 + 180); I = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 3*72 + 180); J = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 4*72 + 180); label("$F$", F, N); pair K, L, M, N, O, P, Q, R, S, T, U, V; real newScaleFactor = 0.8507; K = newScaleFactor*dir(270+18); T = newScaleFactor*dir(270+72+72+72+72-18); label("$K$", K, E); label("$T$", T, W); draw(K--T); draw(K--D); draw(D--T);
// Block 6
unitsize(5cm); // Define the vertices of the pentagons pair A, B, C, D, E; pair F, G, H, I, J; // Calculate the vertices of the larger pentagon A = dir(90); B = dir(90 - 72); C = dir(90 - 2*72); D = dir(90 - 3*72); E = dir(90 - 4*72); pair O = (A+B+C+D+E)/5; pair AA,OO; real gap = 0.02; AA = A+(0,0); OO = O+(0,0); label("$O$", O, E); pair OOO, OAO; OOO = O+(gap,0); OAO = (O+A)/2 + (gap,0); dot(O); label("$D$", D, S); real scaleFactor = 1/1.618; // Adjust this value as needed // Rotate the smaller pentagon by 180 degrees F = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 + 180); G = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 72 + 180); H = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 2*72 + 180); I = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 3*72 + 180); J = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 4*72 + 180); label("$F$", (0.1,-1/1.618), E); pair K, L, M, N, O, P, Q, R, S, T, U, V; real newScaleFactor = 0.8507; K = newScaleFactor*dir(270+18); T = newScaleFactor*dir(270+72+72+72+72-18); label("$K$", K, S); label("$T$", T, W); draw(K--T); draw(K--D); draw(D--T); pair U; U=(0,-0.809); label("$U$",(0,-0.9), S); draw(O--U); draw(O--D); pair V; V = midpoint(O--D); label("$V$", V+(0,0.05), N); markscalefactor = 0.005; draw(rightanglemark(D,U,O)); draw(rightanglemark(F,V,O)); draw(rightanglemark(K,U,F)); draw(anglemark(U,D,O)); label("$54^{\circ}$", D+(0.05,0),NE); draw(anglemark(D,O,U)); label("$36^{\circ}$", O-(0,0.2),SW); draw(anglemark(D,T,F)); label("$36^{\circ}$", T+(0.1,-0.17),SE); label("$\frac{s}{2}$", D+(0.3,-0.1), S); | [] |
949 | A regular pentagon with area $\sqrt{5}+1$ is printed on paper and cut out. The five vertices of the pentagon are folded into the center of the pentagon, creating a smaller pentagon. What is the area of the new pentagon?
$\textbf{(A)}~4-\sqrt{5}\qquad\textbf{(B)}~\sqrt{5}-1\qquad\textbf{(C)}~8-3\sqrt{5}\qquad\textbf{(D)}~\frac{\sqrt{5}+1}{2}\qquad\textbf{(E)}~\frac{2+\sqrt{5}}{3}$ | 2023 AMC 12B Problem 25 | Let $r_b$ and $r_s$ be the circumradius of the big and small pentagon, respectively. Let $a_s$ be the apothem of the smaller pentagon and $A_s$ and $A_b$ be the areas of the smaller and larger pentagon, respectively.
From the diagram:
\begin{align*} \cos{36^\circ} &= \dfrac{a_s}{r_s} = \dfrac{\phi}{2} = \dfrac{\sqrt{5}+1}{4}\\ a_s &= \dfrac{r_b}{2}\\ A_s &= \left(\dfrac{r_s}{r_b}\right)^2A_b\\ &=\left(\dfrac{a_s}{\cos{36^\circ} r_b}\right)^2 \left(1+\sqrt{5}\right)\\ &=\left(\dfrac{a_s}{\dfrac{\phi}{2} r_b}\right)^2 \left(1+\sqrt{5}\right)\\ &=\left(\dfrac{1}{2 \dfrac{\phi}{2}}\right)^2 \left(1+\sqrt{5}\right)\\ &=\left(\dfrac{2}{\sqrt{5}+1}\right)^2 \left(1+\sqrt{5}\right)\\ &=\dfrac{4}{\sqrt{5}+1} \\ &=\dfrac{4\left(\sqrt{5}-1\right)}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)} \\ &=\sqrt{5}-1 \end{align*}
\[\boxed{\textbf{(B) }\sqrt{5}-1}\]
~Technodoggo | // Block 1
unitsize(5cm);
// Define the vertices of the pentagons
pair A, B, C, D, E;
pair F, G, H, I, J;
// Calculate the vertices of the larger pentagon
A = dir(90);
B = dir(90 - 72);
C = dir(90 - 2*72);
D = dir(90 - 3*72);
E = dir(90 - 4*72);
// Draw the larger pentagon
draw(A--B--C--D--E--cycle);
pair O = (A+B+C+D+E)/5;
pair AA,OO;
real gap = 0.02;
AA = A+(0,0);
OO = O+(0,0);
draw(AA--OO, blue);
pair OOO, OAO;
OOO = O+(gap,0);
OAO = (O+A)/2 + (gap,0);
draw(OOO--OAO,green);
dot(O);
dot((O+A)/2);
label("$r_b$", (O+A)*.7, E,blue);
label("$a_s$", (O+A)*.2 +(0+0.18,0.05), E,green);
label("$r_s$", O+(-0.175,0.2), E,pink);
label("$A$", (0,0), E);
real scaleFactor = 1/1.618; // Adjust this value as needed
// Rotate the smaller pentagon by 180 degrees
F = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 + 180);
G = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 72 + 180);
H = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 2*72 + 180);
I = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 3*72 + 180);
J = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 4*72 + 180);
// Draw the smaller pentagon
draw(F--G--H--I--J--cycle,red);
draw(arc(O,(H+I)*.5*.6,H*.6));
label("$36^\circ$",O+(+0.05,0.15),NW);
draw(O--H,pink);
// Block 2
unitsize(5cm); // Define the vertices of the pentagons pair A, B, C, D, E; pair F, G, H, I, J; // Calculate the vertices of the larger pentagon A = dir(90); B = dir(90 - 72); C = dir(90 - 2*72); D = dir(90 - 3*72); E = dir(90 - 4*72); // Draw the larger pentagon draw(A--B--C--D--E--cycle); pair O = (A+B+C+D+E)/5; pair AA,OO; real gap = 0.02; AA = A+(0,0); OO = O+(0,0); draw(AA--OO, blue); pair OOO, OAO; OOO = O+(gap,0); OAO = (O+A)/2 + (gap,0); draw(OOO--OAO,green); dot(O); dot((O+A)/2); label("$r_b$", (O+A)*.7, E,blue); label("$a_s$", (O+A)*.2 +(0+0.18,0.05), E,green); label("$r_s$", O+(-0.175,0.2), E,pink); label("$A$", (0,0), E); real scaleFactor = 1/1.618; // Adjust this value as needed // Rotate the smaller pentagon by 180 degrees F = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 + 180); G = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 72 + 180); H = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 2*72 + 180); I = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 3*72 + 180); J = (1 - scaleFactor) * (0,0) + scaleFactor * dir(90 - 4*72 + 180); // Draw the smaller pentagon draw(F--G--H--I--J--cycle,red); draw(arc(O,(H+I)*.5*.6,H*.6)); label("$36^\circ$",O+(+0.05,0.15),NW); draw(O--H,pink); | [] |
950 | One side of an equilateral triangle of height $24$ lies on line $\ell$. A circle of radius $12$ is tangent to line $\ l$ and is externally tangent to the triangle. The area of the region exterior to the triangle and the circle and bounded by the triangle, the circle, and line $\ell$ can be written as $a \sqrt{b} - c \pi$, where $a$, $b$, and $c$ are positive integers and $b$ is not divisible by the square of any prime. What is $a + b + c$?
$\textbf{(A)}~72\qquad\textbf{(B)}~73\qquad\textbf{(C)}~74\qquad\textbf{(D)}~75\qquad\textbf{(E)}~76$ | 2024 AMC 10A Problem 14 | So, we're here to over-complicate things. Let's start with the circle. Recognize that since the triangle is equilateral, all angles measure $60^\circ$ Observe that the two tangent lines form a $120^\circ$ angle due to the Angle Sum Theorem. Therefore, the area within the circle is the area within a $60^\circ$ arc.
Let's complicate finding the area of the circle.
Now, instead of using the equation of a circle we set \[x^2 + y^2 = 144\]
then we solve for $y$: \[y = \pm\sqrt{144 - x^2}\] then we want the bottom of this circle so we do the negative root $y$ = $-\sqrt{144 - x^2}$ Now, we define this as a function $f(x)$
\[f(x) = -\sqrt{144 - x^2}\] but hold up! waaait a minute! We can make the circle centered at $(12,0)$ because we're just like that. So, to translate it 12 right we must do $f(x - 12)$ so \[f(x - 12) = -\sqrt{144 - (x - 12)^2} = -\sqrt{-x^2 + 24x}\] and because there is a $60 ^\circ$ arc we want $\frac{1}{3}$ of this semicircle so we compute \[\frac{1}{3} \cdot \int_{0}^{24} -\sqrt{-x^2 + 24x} \ \ dx\] then we need to undo our steps and negate the integral(this is done when it's converted to polar) \[\frac{1}{3} \cdot \int_{0}^{24} -\sqrt{-x^2 + 24x} \ \ dx \rightarrow \frac{1}{3} \cdot \int_{0}^{24} -\sqrt{144-(x-12)^2}\ \ dx \rightarrow \int_0^{\pi/3} \frac{1}{2} \cdot 12^2 \ d\theta = \int_0^{\pi/3} 72 \ d\theta\]
Recall that
\[A= \frac{1}{2} \cdot \int_{\theta_1}^{\theta_2} [f(\theta)]^2 \ \ d\theta\]
Therefore,
\[\int_0^{\pi/3} 72 \ d\theta = 72 \cdot \frac{\pi}{3} = 24\pi\]
Now, we must find the area enclosed by $2$ radii extending to the two points of tangency and their respective tangent lines. So, the two radii are $12$, and they form a $60^\circ$ angle. Therefore, you have an equilateral triangle of side length $12$, and its height is $6\sqrt{3}$ using $30^\circ \ 60^\circ \ 90^\circ$
Now, you can just see that the area is $36\sqrt{3}$, but we won't be doing that here. We can define an absolute value function $g(x)$ such that $g(x)$ has zeros at $-6$ and $6$, and a y-intercept at $(0, 6\sqrt{3})$. This would create our triangle above the $x$-axis. We can see that its slope is $\sqrt{3}$ for $x<0$ and $-\sqrt{3}$ for $x>0$. Then, we can see that it's translated $6\sqrt{3}$ up from its parent function $h(x)=|x|$. Therefore, the function is $g(x) = -|x\sqrt{3}|+6\sqrt{3}$. Now, we integrate over $x\in[-6, 6]$
\[\int_{-6}^{6} -|x\sqrt{3}|+6\sqrt{3} \ \ dx\]
Now, we split this into a piecewise function over two integrals
\[\int_{-6}^{0} x\sqrt{3}+6\sqrt{3} \ \ dx \ + \ \int_{0}^{6} -x\sqrt{3}+6\sqrt{3} \ \ dx\]
Let's apply the power rule to each term. We get
\[\int_{-6}^{0} x\sqrt{3}+6\sqrt{3} \ \ dx= \left[\frac{x^2\sqrt3}{2}+6x\sqrt{3} \right]_{-6}^0 = \left(\frac{0^2\sqrt3}{2}+6\cdot0\sqrt{3}\right) - \left(\frac{(-6)^2\sqrt3}{2}+6\cdot-6\sqrt{3}\right) =\]
\[0 - (18\sqrt{3} - 36\sqrt3) = 0 - (-18\sqrt{3}) = 18\sqrt{3}\]
and, we could say that they are mirror images of each other, but that would make this faster (it's called inefficient for a reason)
\[\int_{0}^{6} -x\sqrt{3}+6\sqrt{3} \ \ dx = \left[\frac{-x^2\sqrt3}{2}+6x\sqrt{3} \right]_{0}^6 = \left(\frac{-6^2\sqrt3}{2}+6\cdot6\sqrt{3}\right) - \left(\frac{(0)^2\sqrt3}{2}+6\cdot0\sqrt{3}\right) =\]
\[(-18\sqrt{3}+36\sqrt{3}) - 0 = 18\sqrt{3}\]
Therefore, by the substitution property equality,
\[\int_{-6}^{0} x\sqrt{3}+6\sqrt{3} \ \ dx \ + \ \int_{0}^{6} -x\sqrt{3}+6\sqrt{3} \ \ dx = 18\sqrt{3}+18\sqrt{3}=36\sqrt{3}\]
Then, there is another triangle we must define above the $x$-axis. It is enclosed by the equilateral triangle of height $24$, line $\ell$, and the triangle we just found the area of. This triangle has base $12$ and height of $\frac{6}{\sqrt{3}}$ using $30^\circ \ 60^\circ \ 90^\circ$ triangle ratios, which rationalizes to $2\sqrt{3}$. So, you can tell the area is $12\sqrt{3}$, but that would make our solution faster.
Since half the base is $6$, we can tell the zeros are $x= -6$ and $x=6$, and the $y$-intercept is $2\sqrt{3}$
We can tell the slope is $\frac{\sqrt{3}}{3}$ for $x<0$ and $-\frac{\sqrt{3}}{3}$ for $x>0$
Then, we define another function $f(x)= -|\frac{1}{3} \cdot x\sqrt{3}| + 2\sqrt{3}$
Now, we integrate over $x\in[-6, 6]$
So, we must compute \[\int_{-6}^{6} -|\frac{1}{3} \cdot x\sqrt{3}| + 2\sqrt{3} \ dx\]
We split the function into a piecewise over two integrals
\[\int_{-6}^{0} \ \frac{1}{3} \cdot x\sqrt{3} + 2\sqrt{3} \ dx + \int_{0}^{6} -\frac{1}{3} \cdot x\sqrt{3} + 2\sqrt{3} \ \ dx\]
Now, we apply the reverse power rule to each term, we get
\[\int_{-6}^{0} \ \frac{1}{3} \cdot x\sqrt{3} + 2\sqrt{3} \ \ dx = \left[ \frac{\sqrt{3}}{{3}} \cdot \frac{x^2}{2}+2x\sqrt{3} \right]_{-6}^{0} =\left[ \frac{\sqrt{3}}{{6}} \cdot{x^2}+2x\sqrt{3} \right]_{-6}^{0} =\]
\[\left(\frac{\sqrt{3}}{{6}} \cdot{0^2}+2\cdot0\sqrt{3}\right)-\left(\frac{\sqrt{3}}{{6}} \cdot{6^2}+2\cdot-6\sqrt{3}\right)= 0-\left(\frac{\sqrt{3}}{{6}} \cdot{6^2}+2\cdot-6\sqrt{3}\right) =\]
\[0- (6\sqrt{3}-12\sqrt{3})=6\sqrt{3}\]
Then
\[\int_{0}^{6} -\frac{1}{3} \cdot x\sqrt{3} + 2\sqrt{3} \ \ dx = \left[ -\frac{\sqrt{3}}{{3}} \cdot \frac{x^2}{2}+2x\sqrt{3} \right]_{0}^{6} = \left[ -\frac{\sqrt{3}}{{6}} \cdot{x^2}+2x\sqrt{3} \right]_{0}^{6} =\]
\[\left(-\frac{\sqrt{3}}{{6}} \cdot{6^2}+2\cdot6\sqrt{3}\right) - \left(-\frac{\sqrt{3}}{{6}} \cdot{0^2}+2\cdot0\sqrt{3} \right)=(-6\sqrt{3} + 12\sqrt{3})-0= 6\sqrt{3}\]
Now, by the substitution property of equality,
\[\int_{-6}^{0} \frac{1}{3} \cdot x\sqrt{3} + 2\sqrt{3} \ \ dx + \int_{0}^{6} -\frac{1}{3} \cdot x\sqrt{3} + 2\sqrt{3} \ \ dx = 6\sqrt{3}+6\sqrt{3}=12\sqrt{3}\]
Now, the total area enclosed by the two circle radii of length $12$, line $\ell$, and the isosceles triangle of height $24$ is simply
\[36\sqrt{3}+12\sqrt{3}=48\sqrt{3}\]
Now, we must subtract the circle's sector area of $24\pi$ to get the expression
\[48\sqrt{3}-24\pi\]
Which is in the form
\[a\sqrt{b}-c\pi\]
That the problem asks for. Hence,
\[a=48\]
\[b=3\]
\[c=24\]
The problem asks for $a+b+c$, thus
\[a+b+c=48+3+24=75 \implies \boxed{\textbf{(D)}~75}\]
-shockfront99
PS: I've never taken calculus, so I could be pretty wrong | // Block 1
size(200);
pair A = (-48,0);
pair B = (-24,0);
pair C = (-36,24);
pair O = (-16.5,12); // center of the circle
draw(A--B--C--cycle, linewidth(1));
draw(circle(O,12), linewidth(1));
draw((-48,0) -- (10,0));
draw((-36,24) -- (-36, 0));
dot(O);
draw(O -- (-16.5,0));
pair dir = rotate(205)*(1,0);
draw(O -- O + 12*dir);
label("$24$", midpoint((-36,24) -- (-36, 0)), E);
label("$\ell$", (10,0), E);
// Block 2
size(200); pair A = (-48,0); pair B = (-24,0); pair C = (-36,24); pair O = (-16.5,12); // center of the circle draw(A--B--C--cycle, linewidth(1)); draw(circle(O,12), linewidth(1)); draw((-48,0) -- (10,0)); draw((-36,24) -- (-36, 0)); dot(O); draw(O -- (-16.5,0)); pair dir = rotate(205)*(1,0); draw(O -- O + 12*dir); label("$24$", midpoint((-36,24) -- (-36, 0)), E); label("$\ell$", (10,0), E); | [] |
951 | Let $\alpha$ be the radian measure of the smallest angle in a $3{-}4{-}5$ right triangle. Let $\beta$ be the radian measure of the smallest angle in a $7{-}24{-}25$ right triangle. In terms of $\alpha$, what is $\beta$?
$\textbf{(A) }\frac{\alpha}{3}\qquad \textbf{(B) }\alpha - \frac{\pi}{8}\qquad \textbf{(C) }\frac{\pi}{2} - 2\alpha \qquad \textbf{(D) }\frac{\alpha}{2}\qquad \textbf{(E) }\pi - 4\alpha\qquad$ | 2024 AMC 12A Problem 10 | We can scale the $3$-$4$-$5$ triangle up by a factor of $6$ to make its side lengths $18,24,$ and $30,$ then glue its side of length $24$ to the corresponding side in the $7$-$24$-$25$ triangle:
Angles $\angle DAB$ and $\angle BDA$ are complementary in $\triangle ABD,$ so $\angle DAB = \frac{\pi}{2} - \alpha.$ We also have $AC = 18 + 7 = 25 = CD,$ so $\triangle ACD$ is isosceles. That means that its base angles $\angle CDA$ and $\angle CAD$ are congruent, so $\alpha + \beta = \frac{\pi}{2} - \alpha,$ and hence $\beta = \boxed{\textbf{(C) }\frac{\pi}{2}-2\alpha}.$
~MartianTom | // Block 1
pair A = (0,0);
pair B = (18,0);
pair C = (25,0);
pair D = (18,24);
draw(A--C--D--cycle);
draw(B--D);
draw(rightanglemark(C,B,D,50));
label("A", A, SW);
label("B", B, S);
label("C", C, SE);
label("D", D, N);
label("18", A--B, S);
label("7", B--C, S);
label("25", C--D, NE);
label("30", D--A, NW);
label("24", B--D, W);
label("$\alpha$", D, 6*dir(250));
label("$\beta$", D, 9*dir(278));
label("$\frac{\pi}{2} - \alpha$", A, 4*dir(25));
// Block 2
pair A = (0,0); pair B = (18,0); pair C = (25,0); pair D = (18,24); draw(A--C--D--cycle); draw(B--D); draw(rightanglemark(C,B,D,50)); label("A", A, SW); label("B", B, S); label("C", C, SE); label("D", D, N); label("18", A--B, S); label("7", B--C, S); label("25", C--D, NE); label("30", D--A, NW); label("24", B--D, W); label("$\alpha$", D, 6*dir(250)); label("$\beta$", D, 9*dir(278)); label("$\frac{\pi}{2} - \alpha$", A, 4*dir(25)); | [] |
952 | The graph of $y=e^{x+1}+e^{-x}-2$ has an axis of symmetry. What is the reflection of the point $(-1,\tfrac{1}{2})$ over this axis?
$\textbf{(A) }\left(-1,-\frac{3}{2}\right)\qquad\textbf{(B) }(-1,0)\qquad\textbf{(C) }\left(-1,\frac{1}{2}\right)\qquad\textbf{(D) }\left(0,\frac{1}{2}\right)\qquad\textbf{(E) }\left(3,\frac{1}{2}\right)$ | 2024 AMC 12A Problem 13 | Consider the graphs of $y=e^{x+1}-1$ and $y=e^{-x}-1$. A rough sketch will show that they intercept somewhere between -1 and 0 and the axis of symmetry is vertical. Thus, $\boxed{\textbf{(D) }\left(0,\dfrac12\right)}$ is the only possible answer.
Note: You can more rigorously think about the solution by noting that since the derivative of the power that e is raised to in one equation is equal to the derivative of the power that e is raised to multiplied by $-1$ and both equations are subtracted by 1, then the sum of both equations will be the same from one side of the interception to the other. Setting both equations equal to each other, it is trivial to see $x=-1/2$, giving us the axis of symmetry.
~woeIsMe
(graph by Technodoggo) | // Block 1
unitsize(2cm);
real e = 2.71828;
real f1(real x) {return e^(x+1)-1;}
real f2(real x) {return e^(0-x)-1;}
draw(graph(f1,-1.5,0.5));
draw(graph(f2,-1.5,0.5));
xaxis(-2,1,Ticks());
yaxis(f2(0.5),f1(0.5),Ticks());
draw((-0.5,f2(0.5))--(-0.5,f1(0.5)),red+dashed);
/*graph by Technodoggo, 9 November 2024*/
// Block 2
unitsize(2cm); real e = 2.71828; real f1(real x) {return e^(x+1)-1;} real f2(real x) {return e^(0-x)-1;} draw(graph(f1,-1.5,0.5)); draw(graph(f2,-1.5,0.5)); xaxis(-2,1,Ticks()); yaxis(f2(0.5),f1(0.5),Ticks()); draw((-0.5,f2(0.5))--(-0.5,f1(0.5)),red+dashed); /*graph by Technodoggo, 9 November 2024*/ | [] |
953 | On top of a rectangular card with sides of length $1$ and $2+\sqrt{3}$, an identical card is placed so that two of their diagonals line up, as shown ($\overline{AC}$, in this case).
Continue the process, adding a third card to the second, and so on, lining up successive diagonals after rotating clockwise. In total, how many cards must be used until a vertex of a new card lands exactly on the vertex labeled $B$ in the figure?
$\textbf{(A) }6\qquad\textbf{(B) }8\qquad\textbf{(C) }10\qquad\textbf{(D) }12\qquad\textbf{(E) }\text{No new vertex will land on }B.$ | 2024 AMC 12A Problem 18 | By the Pythagorean Theorem, $AC = \sqrt{(2 + \sqrt{3})^2 + 1} = \sqrt{8 + 4\sqrt{3}}$. So we have
\[\arcsin \angle BCA = \frac{1}{\sqrt{8 + 4\sqrt{3}}} = \frac{1}{\sqrt{(\sqrt{6} + \sqrt{2})^2}} = \frac{\sqrt{6} - \sqrt{2}}{4}.\]
We note that \[\sin 15^\circ = \cos 75^\circ = \cos(30^\circ + 45^\circ) = \cos30^\circ \cos45^\circ - \sin45^\circ \sin30^\circ = \frac{\sqrt{6} - \sqrt{2}}{4}.\]
Therefore, $\angle BCA = 15^\circ.$
Call the rotated rectangle given in the problem's diagram $AB'CD.$ Since $\angle ACB' = \angle BCA = 15^\circ,$ we have $\angle BCB' = 30^\circ.$ So the angle formed between $B$ and $B'$ is $30^\circ \cdot 2 = 60^\circ$.
In general, when the rectangle is rotated, the angle formed between the left vertex of the original rectangle and the right vertex of the new rectangle is $60^\circ.$ So we can inscribe the rectangles in a circle with $2 \cdot 360/60 = 12$ equally spaced points, and the rotated rectangles will have vertices on these points.
We get the equation $P = 2(R + 1),$ where $R$ is the number of rectangles used and $P$ is the number of points used on the circle for those rectangles (for example, 1 rectangle uses $4$ points, $2$ rectangles use $6$, etc). The vertex will land on $B$ after we have used all $12$ points on the circle plus the additional $2$ points we get from creating a new rotated rectangle, so $P = 12 + 2 = 14.$ Then $R = \boxed{\textbf{(A) }6}$.
~grogg007 | // Block 1
size(150);
defaultpen(fontsize(10pt));
pair O = (0,0);
pair[] pts;
for (int i = 0; i < 12; ++i) {
pts[i] = dir(30 * i);
}
fill(pts[9]--pts[10]--pts[3]--pts[4]--cycle, paleblue);
fill(pts[8]--pts[9]--pts[2]--pts[3]--cycle, palegreen);
fill(pts[7]--pts[8]--pts[1]--pts[2]--cycle, palered);
draw(Circle(O,1));
draw(O--pts[4], blue+1.5);
draw(O--pts[2], blue+1.5);
for (int i = 0; i < 12; ++i) {
dot(pts[i], black + 3);
}
dot(O, black + 3);
label("$O$", O, SW);
label("$B$", pts[4], dir(30 * 4));
label("$A$", pts[3], dir(30 * 3));
label("$B'$", pts[2], dir(30 * 2));
label("$C$", pts[9], dir(30 * 9));
label("$D$", pts[8], dir(30 * 8));
real r = 0.2;
draw(arc(O, r, 120, 60), red+1.5);
label("$60^\circ$", O + r*dir(90) + (0,0.05), red);
// Block 2
size(150); defaultpen(fontsize(10pt)); pair O = (0,0); pair[] pts; for (int i = 0; i < 12; ++i) { pts[i] = dir(30 * i); } fill(pts[9]--pts[10]--pts[3]--pts[4]--cycle, paleblue); fill(pts[8]--pts[9]--pts[2]--pts[3]--cycle, palegreen); fill(pts[7]--pts[8]--pts[1]--pts[2]--cycle, palered); draw(Circle(O,1)); draw(O--pts[4], blue+1.5); draw(O--pts[2], blue+1.5); for (int i = 0; i < 12; ++i) { dot(pts[i], black + 3); } dot(O, black + 3); label("$O$", O, SW); label("$B$", pts[4], dir(30 * 4)); label("$A$", pts[3], dir(30 * 3)); label("$B'$", pts[2], dir(30 * 2)); label("$C$", pts[9], dir(30 * 9)); label("$D$", pts[8], dir(30 * 8)); real r = 0.2; draw(arc(O, r, 120, 60), red+1.5); label("$60^\circ$", O + r*dir(90) + (0,0.05), red); | [] |
954 | Cyclic quadrilateral $ABCD$ has lengths $BC=CD=3$ and $DA=5$ with $\angle CDA=120^\circ$. What is the length of the shorter diagonal of $ABCD$?
$\textbf{(A) }\frac{31}7 \qquad \textbf{(B) }\frac{33}7 \qquad \textbf{(C) }5 \qquad \textbf{(D) }\frac{39}7 \qquad \textbf{(E) }\frac{41}7 \qquad$ | 2024 AMC 12A Problem 19 | ~diagram by erics118
First, $\angle CBA=60 ^\circ$ by properties of cyclic quadrilaterals.
Let $AC=u$. Apply the Law of Cosines on $\triangle ACD$:
\[u^2=3^2+5^2-2(3)(5)\cos120^\circ\]
\[u=7\]
Let $AB=v$. Apply the Law of Cosines on $\triangle ABC$:
\[7^2=3^2+v^2-2(3)(v)\cos60^\circ\]
\[v=\frac{3\pm13}{2}\]
\[v=8\]
By Ptolemy’s Theorem,
\[AB \cdot CD+AD \cdot BC=AC \cdot BD\]
\[8 \cdot 3+5 \cdot 3=7BD\]
\[BD=\frac{39}{7}\]
Since $\frac{39}{7}<7$,
The answer is $\boxed{\textbf{(D) }\frac{39}{7}}$.
~lptoggled,eevee9406, meh494 | // Block 1
import geometry;
size(200);
pair A = (-1.66, 0.33);
pair B = (-9.61277, 1.19799);
pair C = (-7.83974, 3.61798);
pair D = (-4.88713, 4.14911);
draw(circumcircle(A, B, C));
draw(A--C);
draw(A--D);
draw(C--D);
draw(B--C);
draw(A--B);
label("$A$", A, E);
label("$B$", B, W);
label("$C$", C, NW);
label("$D$", D, N);
label("$7$", midpoint(A--C), SW);
label("$5$", midpoint(A--D), NE);
label("$3$", midpoint(C--D)+ dir(135)*0.3, N);
label("$3$", midpoint(B--C)+dir(180)*0.3, NW);
label("$8$", midpoint(A--B), S);
markangle(Label("$60^\circ$", Relative(0.5)), A, B, C, radius=10);
markangle(Label("$120^\circ$", Relative(0.5)), C, D, A, radius=10);
// Block 2
import geometry; size(200); pair A = (-1.66, 0.33); pair B = (-9.61277, 1.19799); pair C = (-7.83974, 3.61798); pair D = (-4.88713, 4.14911); draw(circumcircle(A, B, C)); draw(A--C); draw(A--D); draw(C--D); draw(B--C); draw(A--B); label("$A$", A, E); label("$B$", B, W); label("$C$", C, NW); label("$D$", D, N); label("$7$", midpoint(A--C), SW); label("$5$", midpoint(A--D), NE); label("$3$", midpoint(C--D)+ dir(135)*0.3, N); label("$3$", midpoint(B--C)+dir(180)*0.3, NW); label("$8$", midpoint(A--B), S); markangle(Label("$60^\circ$", Relative(0.5)), A, B, C, radius=10); markangle(Label("$120^\circ$", Relative(0.5)), C, D, A, radius=10); | [] |
955 | Points $P$ and $Q$ are chosen uniformly and independently at random on sides $\overline {AB}$ and $\overline{AC},$ respectively, of equilateral triangle $\triangle ABC.$ Which of the following intervals contains the probability that the area of $\triangle APQ$ is less than half the area of $\triangle ABC?$
$\textbf{(A) } \left[\frac 38, \frac 12\right] \qquad \textbf{(B) } \left(\frac 12, \frac 23\right] \qquad \textbf{(C) } \left(\frac 23, \frac 34\right] \qquad \textbf{(D) } \left(\frac 34, \frac 78\right] \qquad \textbf{(E) } \left(\frac 78, 1\right]$ | 2024 AMC 12A Problem 20 | Let $\overline{AP}=x$ and $\overline{AQ}=y$. Applying the sine formula for a triangle's area, we see that
\[[\Delta APQ]=\dfrac12\cdot x\cdot y\sin(\angle PAQ)=\dfrac{xy}2\sin(60^\circ)=\dfrac{\sqrt3}4xy.\]
Without loss of generality, we let $AB=BC=CA=1$, and thus $[\Delta ABC]=\dfrac{\sqrt3}4$; we therefore require $\dfrac{\sqrt3}4xy\le\dfrac12\cdot\dfrac{\sqrt3}4\implies xy\le\dfrac12$ for $0\le x,y\le1$. (Note: You can skip most of this by using triangle area ratios ~A_MatheMagician.)
A quick rough sketch of $y=\dfrac1{2x}$ on the square given by $x,y\in[0,1]$ reveals that the curve intersects the boundaries at $(0.5,1)$ and $(1,0.5)$, and it is actually quite (very) obvious that the area bounded by the inequality $xy\le0.5$ and the aforementioned unit square is more than $\dfrac34$ but less than $\dfrac78$ (cf. the diagram below). Thus, our answer is $\boxed{\textbf{(D) }\left(\dfrac34,\dfrac78\right]}$.
~Technodoggo | // Block 1
/*Asymptote visual by Technodoggo, 7 November 2024*/
unitsize(8cm);
draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);
label("$0$",(-0.05,-0.05));
label("$1$",(1,-0.05),S);
label("$1$",(-0.05,1),W);
draw((-0.05,0)--(1,0)--(1,-0.05));
draw((0,-0.05)--(0,1)--(-0.05,1));
real f(real x) {return 1/(2*x);}
path c = graph(f, 0.5,1)--(1,0)--(0,0)--(0,1)--cycle;
filldraw(c,blue+white);
draw((0.5,1)--(0.5,0.5)--(1,0.5),white+dashed+1.1);
draw((0.5,1)--(1,0.5),red+dashed+1.1);
// Block 2
/*Asymptote visual by Technodoggo, 7 November 2024*/ unitsize(8cm); draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); label("$0$",(-0.05,-0.05)); label("$1$",(1,-0.05),S); label("$1$",(-0.05,1),W); draw((-0.05,0)--(1,0)--(1,-0.05)); draw((0,-0.05)--(0,1)--(-0.05,1)); real f(real x) {return 1/(2*x);} path c = graph(f, 0.5,1)--(1,0)--(0,0)--(0,1)--cycle; filldraw(c,blue+white); draw((0.5,1)--(0.5,0.5)--(1,0.5),white+dashed+1.1); draw((0.5,1)--(1,0.5),red+dashed+1.1); | [] |
956 | Two straight pipes (circular cylinders), with radii $1$ and $\frac{1}{4}$, lie parallel and in contact on a flat floor. The figure below shows a head-on view. What is the sum of the possible radii of a third parallel pipe lying on the same floor and in contact with both?
$\textbf{(A)}~\frac{1}{9} \qquad\textbf{(B)}~1 \qquad\textbf{(C)}~\frac{10}{9} \qquad\textbf{(D)}~\frac{11}{9} \qquad\textbf{(E)}~\frac{19}{9}$ | 2024 AMC 10B Problem 21 | Notice that the sum of radii of two circles tangent to each other will equal to the distance from center to center. Set the center of the big circle be at $(0,1).$ Since both circles are tangent to a line (in this case, $y=0$), the y-coordinates of the centers are just its radius.
Hence, the center of the smaller circle is at $\left(x_2, \frac14\right)$. From the the sum of radii and distance formula, we have:
\[1+\frac14 = \sqrt{x_2^2 + \left(\frac34\right)^2} \Rightarrow x_2 = 1.\]
So, the coordinates of the center of the smaller circle are $(1, \frac14).$ Now, let $(x_3,r_3)$ be the coordinates of the new circle. Then, from the fact that sum of radii of this circle and the circle with radius $1$ is equal to the distance from the two centers, you have:
\[\sqrt{(x_3-0)^2+(r_3-1)^2} = 1+r_3.\]
Similarly, from the fact that the sum of radii of this circle and the circle with radius $\frac14$, you have:
\[\sqrt{(x_3-1)^2+\left(r_3-\frac14\right)^2}= \frac14 + r_3.\]
Squaring the first equation, you have:
\[x_3^2+r_3^2-2r_3+1=1+2r_3+r_3^2 \Rightarrow 4r_3=x_3^2 \Rightarrow x_3=2\sqrt{r_3}.\]
Squaring the second equation, you have:
\[x_3^2-2x_3+1+r_3^2-\frac{r_3}{2}+\frac{1}{16}=\frac{1}{16}+\frac{r_3}{2}+r_3^2 \Rightarrow x_3^2-2x_3+1=r_3\]
Plugging in from the first equation we have
\[r_3-1=x_3^2-2x_3=4r_3-4\sqrt{r_3} \Rightarrow 3r_3-4\sqrt{r_3}+1=0 \Rightarrow (3\sqrt{r_3}-1)(\sqrt{r_3}-1)=0 \Rightarrow r_3=1, \frac19.\]
Summing these two yields $\boxed{\frac{10}{9}}.$
~mathboy282
Diagram
~Elephant200 | // By Elephant200 size(8cm); draw(circle((0,1),1), linewidth(1.2)); draw((-1,0)--(3,0), linewidth(1.2)); draw(circle((1,1/4),1/4), linewidth(1.2)); draw(circle((2/3,1/9),1/9), red+linewidth(1.2)); draw(circle((2,1),1), red+linewidth(1.2)); | [] |
956 | Two straight pipes (circular cylinders), with radii $1$ and $\frac{1}{4}$, lie parallel and in contact on a flat floor. The figure below shows a head-on view. What is the sum of the possible radii of a third parallel pipe lying on the same floor and in contact with both?
$\textbf{(A)}~\frac{1}{9} \qquad\textbf{(B)}~1 \qquad\textbf{(C)}~\frac{10}{9} \qquad\textbf{(D)}~\frac{11}{9} \qquad\textbf{(E)}~\frac{19}{9}$ | 2024 AMC 10B Problem 21 | We can immediately see that we can put a circle with radius $1$ outside of the two circles already drawn that are tangent to the circles. If you brought a ruler with you, you can confirm this. We now find the radius of the circle between the circles of radius $1$ and $\frac{1}{4}$.. By drawing out the diagram, a geometry hack is to draw lines connecting the centers of the circles. Using the fact that tangent lines to the radius are perpendicular, we can construct a trapezoid with heights of lengths $1$ and $\frac{1}{4}$, a slant base of length $1 + \frac{1}{4} = \frac{5}{4}$. To find the base of the trapezoid, we can construct a right triangle with the slant base as the hypotenuse and a leg of length $1 - \frac{1}{4} = \frac{3}{4}$. Therefore, we see we have a $3-4-5$ right triangle and the length of the base of the trapezoid is conveniently just $1$. Let the radius of the circle between the two given circles $r$. By drawing the center of this circle and connecting the radius to the radii of the other two circles, we can construct two more trapezoids similar to the initially big one we constructed: one with perpendicular heights of lengths $1$ and $r$ and a slant base of $1 + r$, and another with perpendicular heights of lengths $\frac{1}{4}$ and $r$ and a slant base of length $r + \frac{1}{4}$. To find the bases of these trapezoids,(which we know sums to $1$ as we found earlier), we can use a similar trick by constructing right triangles. We can construct two right triangles: one with hypotenuse $1 + r$ and a leg of length $1 - r$, and another with hypotenuse $r + \frac{1}{4}$ and a leg of length $\frac{1}{4} - r$. We can use the Pythagorean Theorem to find the other legs of these right triangles and we can set the sum to equal $1$. Conveniently, the leg lengths turns out of be $2\sqrt{r}$ and $\sqrt{r}$, respectively, and therefore the sum will just be $3\sqrt{r}$. Since this is equal to $1$, we can solve $3\sqrt{r} = 1$ to get the radius to be equal to $\frac{1}{9}$. Now, we add $1$ and $\frac{1}{9}$ to get $\frac{10}{9}$ and our answer is just $\boxed{\textbf{(C) }\frac{10}{9}}$.
credit to Elephant200 for the diagram
~ilikemath247365 | // Block 1
// By Elephant200
size(8cm);
draw(circle((0,1),1), linewidth(1.2));
draw((-1,0)--(3,0), linewidth(1.2));
draw(circle((1,1/4),1/4), linewidth(1.2));
draw(circle((2/3,1/9),1/9), red+linewidth(1.2));
draw(circle((2,1),1), red+linewidth(1.2));
// Block 2
// By Elephant200 size(8cm); draw(circle((0,1),1), linewidth(1.2)); draw((-1,0)--(3,0), linewidth(1.2)); draw(circle((1,1/4),1/4), linewidth(1.2)); draw(circle((2/3,1/9),1/9), red+linewidth(1.2)); draw(circle((2,1),1), red+linewidth(1.2)); | [] |
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