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
values |
|---|---|---|---|---|---|
727 | The closed curve in the figure is made up of 9 congruent circular arcs each of length $\frac{2\pi}{3}$, where each of the centers of the corresponding circles is among the vertices of a regular hexagon of side 2. What is the area enclosed by the curve?
$\textbf{(A)}\ 2\pi+6\qquad\textbf{(B)}\ 2\pi+4\sqrt{3}\qquad\textbf{(C)}\ 3\pi+4\qquad\textbf{(D)}\ 2\pi+3\sqrt{3}+2\qquad\textbf{(E)}\ \pi+6\sqrt{3}$ | 2012 AMC 12A Problem 14 | We can draw the hexagon between the centers of the circles, and compute its area. The hexagon is made of $6$ equilateral triangles each with length $2$, so the area is:
\[\frac{\sqrt{3}}{4} \cdot 2^2 \cdot 6=6 \sqrt{3}.\]
Then, we add the areas of the three sectors outside the hexagon:
\[\frac 23 \pi \cdot 3=2\pi.\]
We now subtract the areas of the three sectors inside the hexagon but outside the figure (which is $\pi$) to get the area enclosed in the curved figure:
\[6 \sqrt{3}+2\pi-\pi=\pi+6\sqrt{3}.\]
Hence, our answer is $\boxed{\textbf{(E)}\ \pi+6\sqrt{3}},$ and we are done.
\[\]
(Minor edits, made by dbnl.) | unitsize(2cm); defaultpen(fontsize(6pt)); dotfactor=4; label("$\circ$",(0,1)); label("$\circ$",(0.865,0.5)); label("$\circ$",(-0.865,0.5)); label("$\circ$",(0.865,-0.5)); label("$\circ$",(-0.865,-0.5)); label("$\circ$",(0,-1)); dot((0,1.5)); dot((-0.4325,0.75)); dot((0.4325,0.75)); dot((-0.4325,-0.75)); dot((0.4325,-0.75)); dot((-0.865,0)); dot((0.865,0)); dot((-1.2975,-0.75)); dot((1.2975,-0.75)); draw(Arc((0,1),0.5,210,-30)); draw(Arc((0.865,0.5),0.5,150,270)); draw(Arc((0.865,-0.5),0.5,90,-150)); draw(Arc((0.865,-0.5),0.5,90,-150)); draw(Arc((0,-1),0.5,30,150)); draw(Arc((-0.865,-0.5),0.5,330,90)); draw(Arc((-0.865,0.5),0.5,-90,30)); draw((0,1)--(0.865, 0.5)--(0.865,-0.5)--(0,-1)--(-0.865,-0.5)--(-0.865,0.5)--(0,1)); | [] |
727 | The closed curve in the figure is made up of 9 congruent circular arcs each of length $\frac{2\pi}{3}$, where each of the centers of the corresponding circles is among the vertices of a regular hexagon of side 2. What is the area enclosed by the curve?
$\textbf{(A)}\ 2\pi+6\qquad\textbf{(B)}\ 2\pi+4\sqrt{3}\qquad\textbf{(C)}\ 3\pi+4\qquad\textbf{(D)}\ 2\pi+3\sqrt{3}+2\qquad\textbf{(E)}\ \pi+6\sqrt{3}$ | 2012 AMC 12A Problem 14 | As you can see, this diagram looks like a fidget spinner ;). Fidget spinners aside, we need to add stuff to our diagram to make it look easier. In the directions, they were talking about the centers of each arc create a hexagon, so let's add that to our diagram.
The side length of the hexagon is 2 and if we plug it in to the area of a regular hexagon formula $\frac{3\sqrt 3}{2}s^2$ we get $6\sqrt 3$.
Note that the interior angles of a regular hexagon is 120 because of the formula $\frac{180(n-2)}{n}$ where n is the number of sides. Knowing that, each sector is $\frac{1}{3}$ of a circle. This would mean the three sectors inside the hexagon altogether equal a full circle. Knowing that the radius is 1, we can use the area of a circle $\pi r^2$ and subtract it to $6\sqrt 3$. Thus we get the total area of $6\sqrt 3 - \pi$.
Notice that we have three sectors exterior to the hexagon. Realize that the central angles of a circle always sum up to 360. Since we know one of the central angles is equal to 120, then we subtract 360 to 120 which gives us 240. Knowing that, each sector is $\frac{2}{3}$ of a circle and since there is 3 of them, $\frac{2}{3}*3=2$ circles. To find the area of those circles, we have to use $\pi r^2$ again, but since there are 2 circles, then it is $2\pi r^2$, which gives us $2\pi$.
Now we have enough information to find the total area,
$(6\sqrt 3 -\pi+2\pi)=\textbf{(E)}\ \pi+6\sqrt{3}$
~ghfhgvghj10 | defaultpen(fontsize(6pt)); dotfactor=4; label("$\circ$",(0,1)); label("$\circ$",(0.865,0.5)); label("$\circ$",(-0.865,0.5)); label("$\circ$",(0.865,-0.5)); label("$\circ$",(-0.865,-0.5)); label("$\circ$",(0,-1)); dot((0,1.5)); dot((-0.432,0.75)); dot((0.4325,0.75)); dot((-0.4325,-0.75)); dot((0.4325,-0.75)); dot((-0.865,0)); dot((0.865,0)); dot((-1.2975,-0.75)); dot((1.2975,-0.75)); draw(Arc((0,1),0.5,210,-30)); draw(Arc((0.865,0.5),0.5,150,270)); draw(Arc((0.865,-0.5),0.5,90,-150)); draw(Arc((0.865,-0.5),0.5,90,-150)); draw(Arc((0,-1),0.5,30,150)); draw(Arc((-0.865,-0.5),0.5,330,90)); draw(Arc((-0.865,0.5),0.5,-90,30)); draw((0,1)--(0.865, 0.5)--(0.865,-0.5)--(0,-1)--(-0.865,-0.5)--(-0.865,0.5)--(0,1)); | [] |
728 | Distinct planes $p_1,p_2,....,p_k$ intersect the interior of a cube $Q$. Let $S$ be the union of the faces of $Q$ and let $P =\bigcup_{j=1}^{k}p_{j}$. The intersection of $P$ and $S$ consists of the union of all segments joining the midpoints of every pair of edges belonging to the same face of $Q$. What is the difference between the maximum and minimum possible values of $k$?
$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 23\qquad\textbf{(E)}\ 24$ | 2012 AMC 12A Problem 22 | We need two different kinds of planes that only intersect $Q$ at the mentioned segments (we call them traces in this solution). These will be all the possible $p_j$'s.
First, there are two kinds of segments joining the midpoints of every pair of edges belonging to the same face of $Q$: long traces are those connecting the midpoint of opposite sides of the same face of $Q$, and short traces are those connecting the midpoint of adjacent sides of the same face of $Q$.
Suppose $p_j$ contains a short trace $t_1$ of a face of $Q$. Then it must also contain some trace $t_2$ of an adjacent face of $Q$, where $t_2$ share a common endpoint with $t_1$. So, there are three possibilities for $t_2$, each of which determines a plane $p_j$ containing both $t_1$ and $t_2$.
Case 1: $t_2$ makes an acute angle with $t_1$. In this case, $p_j \cap Q$ is an equilateral triangle made by three short traces. There are $8$ of them, corresponding to the $8$ vertices.
Case 2: $t_2$ is a long trace. $p \cap Q$ is a rectangle. Each pair of parallel faces of $Q$ contributes $4$ of these rectangles so there are $12$ such rectangles.
Case 3: $t_2$ is the short trace other than the one described in case 1, i.e. $t_2$ makes an obtuse angle with $t_1$. It is possible to prove that $p \cap Q$ is a regular hexagon (See note #1 for a proof) and there are $4$ of them.
Case 4: $p_j$ contains no short traces. This can only make $p_j \cap Q$ be a square enclosed by long traces. There are $3$ such squares.
In total, there are $8+12+4+3=27$ possible planes in $P$. So the maximum of $k$ is $27$.
On the other hand, the most economic way to generate these long and short traces is to take all the planes in case 3 and case 4. Overall, they intersect at each trace exactly once (there is a quick way to prove this. See note #2 below.) and also covered all the $6\times 4 + 4\times 3 = 36$ traces. So the minimum of $k$ is $7$. The answer to this problem is then $27-7=20$ ... $\framebox{C}$.
Note 1: Indeed, let $t_1=AB$ where $B=t_1\cap t_2$, and $C$ be the other endpoint of $t_2$ that is not $B$. Draw a line through $C$ parallel to $t_1$. This line passes through the center $O$ of the cube and therefore we see that the reflection of $A,B,C$, denoted by $A', B', C'$, respectively, lie on the same plane containing $A,B,C$. Thus $p_j \cap Q$ is the regular hexagon $ABCA'B'C'$. To count the number of these hexagons, just notice that each short trace uniquely determine a hexagon (by drawing the plane through this trace and the center), and that each face has $4$ short traces. Therefore, there are $4$ such hexagons.
Note 2: The quick way to prove the fact that none of the planes described in case 3 and case 4 share the same trace is as follows: each of these plane contains the center and therefore the intersection of each pair of them is a line through the center, which obviously does not contain any traces. | // Block 1
pair A=(0,0), B=(2,0), C=(2,2), D=(0,2), E=(-1,3), F=(1,3), G=(1,1), H=(-1,1), I=(1,0), J=(2,1), K=(1,2), L=(0,1), M=(-0.5,0.5), N=(-1,2), O=(-0.5,2.5), P=(0,3), Q=(1.5,2.5), R=(1,2), S=(1.5,0.5), T=(0,1);
draw(A--B--C--D--E--F);
draw(H--A);
draw(A--D);
draw(H--E);
draw(F--C);
draw(H--G--F);
draw(G--B);
label("\(A\)",A,SW);
label("\(B\)",B,SE);
label("\(C\)",C,NE);
label("\(D\)",D,SW);
label("\(E\)",E,NW);
label("\(F\)",F,NE);
label("\(G\)",G,NE);
label("\(H\)",H,SW);
// Block 2
pair A=(0,0), B=(2,0), C=(2,2), D=(0,2), E=(-1,3), F=(1,3), G=(1,1), H=(-1,1), I=(1,0), J=(2,1), K=(1,2), L=(0,1), M=(-0.5,0.5), N=(-1,2), O=(-0.5,2.5), P=(0,3), Q=(1.5,2.5), R=(1,2), S=(1.5,0.5), T=(0,1);
draw(A--B--C--D--E--F);
draw(H--A);
draw(A--D);
draw(H--E);
draw(F--C);
draw(I--J--K--L--I);
draw(I--K);
draw(J--L);
draw(K--Q--P--O--K);
draw(K--P);
draw(Q--O);
draw(O--L--M--N--O);
draw(O--M);
draw(L--N);
label("\(A\)",A,SW);
label("\(B\)",B,SE);
label("\(C\)",C,NE);
label("\(D\)",D,SW);
label("\(E\)",E,NW);
label("\(F\)",F,NE);
label("\(H\)",H,SW);
// Block 3
pair A=(0,0), B=(2,0), C=(2,2), D=(0,2), E=(-1,3), F=(1,3), G=(1,1), H=(-1,1), I=(1,0), J=(2,1), K=(1,2), L=(0,1), M=(-0.5,0.5), N=(-1,2), O=(-0.5,2.5), P=(0,3), Q=(1.5,2.5), R=(1,2), S=(1.5,0.5), T=(0,1); draw(A--B--C--D--E--F); draw(H--A); draw(A--D); draw(H--E); draw(F--C); draw(H--G--F); draw(G--B); label("\(A\)",A,SW); label("\(B\)",B,SE); label("\(C\)",C,NE); label("\(D\)",D,SW); label("\(E\)",E,NW); label("\(F\)",F,NE); label("\(G\)",G,NE); label("\(H\)",H,SW);
// Block 4
pair A=(0,0), B=(2,0), C=(2,2), D=(0,2), E=(-1,3), F=(1,3), G=(1,1), H=(-1,1), I=(1,0), J=(2,1), K=(1,2), L=(0,1), M=(-0.5,0.5), N=(-1,2), O=(-0.5,2.5), P=(0,3), Q=(1.5,2.5), R=(1,2), S=(1.5,0.5), T=(0,1); draw(A--B--C--D--E--F); draw(H--A); draw(A--D); draw(H--E); draw(F--C); draw(I--J--K--L--I); draw(I--K); draw(J--L); draw(K--Q--P--O--K); draw(K--P); draw(Q--O); draw(O--L--M--N--O); draw(O--M); draw(L--N); label("\(A\)",A,SW); label("\(B\)",B,SE); label("\(C\)",C,NE); label("\(D\)",D,SW); label("\(E\)",E,NW); label("\(F\)",F,NE); label("\(H\)",H,SW); | [] |
729 | Let $S$ be the square one of whose diagonals has endpoints $(1/10,7/10)$ and $(-1/10,-7/10)$. A point $v=(x,y)$ is chosen uniformly at random over all pairs of real numbers $x$ and $y$ such that $0 \le x \le 2012$ and $0\le y\le 2012$. Let $T(v)$ be a translated copy of $S$ centered at $v$. What is the probability that the square region determined by $T(v)$ contains exactly two points with integer coefficients in its interior?
$\textbf{(A)}\ \frac{1}{8}\qquad\textbf{(B) }\frac{7}{50}\qquad\textbf{(C) }\frac{4}{25}\qquad\textbf{(D) }\frac{1}{4}\qquad\textbf{(E) }\frac{8}{25}$ | 2012 AMC 12A Problem 23 | The unit square's diagonal has a length of $\sqrt{0.2^2 + 1.4^2} = \sqrt{2}$. Because $S$ square is not parallel to the axis, the two points must be adjacent.
Now consider the unit square $U$ with vertices $(0,0), (1,0), (1,1)$ and $(0,1)$. Let us first consider only two vertices, $(0,0)$ and $(1,0)$. We want to find the area of the region within $U$ that the point $v=(x,y)$ will create the translation of $S$, $T(v)$ such that it covers both $(0,0)$ and $(1,0)$. By symmetry, there will be three equal regions that cover the other pairs of adjacent vertices.
For $T(v)$ to contain the point $(0,0)$, $v$ must be inside square $S$. Similarly, for $T(v)$ to contain the point $(1,0)$, $v$ must be inside a translated square $S$ with center at $(1,0)$, which we will call $S'$. Therefore, the area we seek is Area$(U \cap S \cap S')$.
To calculate the area, we notice that Area$(U \cap S \cap S') = \frac{1}{2} \cdot$ Area$(S \cap S')$ by symmetry. Let $S_1 = (0.1, 0.7), S_2 = (0.7, -0.1), S'_1 = (1.1, 0.7), S'_2 = (0.3, 0.1)$. Let $M = (0.7, 0.4)$ be the midpoint of $S'_1S'_2$, and $N = (0.7, 0.7)$ along the line $S_1S'_1$. Let $I$ be the intersection of $S$ and $S'$ within $U$, and $J$ be the intersection of $S$ and $S'$ outside $U$. Therefore, the area we seek is $\frac{1}{2} \cdot$ Area$(S \cap S') = \frac{1}{2} [IS'_2JS_2]$. Because $S_2, M, N$ all have $x$ coordinate $0.7$, they are collinear. Noting that the side length of $S$ and $S'$ is $1$ (as shown above), we also see that $S_2M = MS'_1 = 0.5$, so $\triangle{S'_1NM} \cong \triangle{S_2IM}$. If follows that $IS_2 = NS'_1 = 1.1 - 0.7 = 0.4$ and $IS'_2 = MS'_2 - MI = MS'_2 - MN = 0.5 - 0.3 = 0.2$. Therefore, the area is $\frac{1}{2} \cdot$ Area$(S \cap S') = \frac{1}{2} [IS'_2JS_2] = \frac{1}{2} \cdot 0.2 \cdot 0.4 = 0.04$.
Because there are three other regions in the unit square $U$ that we need to count, the total area of $v$ within $U$ such that $T(v)$ contains two adjacent lattice points is $0.04 \cdot 4 = 0.16$.
By periodicity, this probability is the same for all $0 \le x \le 2012$ and $0 \le y \le 2012$. Therefore, the answer is $0.16 = \boxed{\frac{4}{25} \textbf{(C)} }$ | // Block 1
pair A=(0.1,0.7), C=(-0.1,-0.7), B=(-0.7,0.1), D=(0.7,-0.1), X=(1,0), W=(-1,0), Y=(0,1), Z=(0,-1);
draw (A--B--C--D--A);
draw(A--C);
draw(B--D);
draw(W--X);
draw(Y--Z);
label("\((0.1,0.7)\)",A,NE);
label("\((-0.1,-0.7)\)",C,SW);
label("\(x\)",X,NW);
label("\(y\)",Y,NE);
// Block 2
pair A=(0.1,0.7), C=(-0.1,-0.7), B=(-0.7,0.1), D=(0.7,-0.1), X=(1,0), W=(-1,0), Y=(0,1), Z=(0,-1); draw (A--B--C--D--A); draw(A--C); draw(B--D); draw(W--X); draw(Y--Z); label("\((0.1,0.7)\)",A,NE); label("\((-0.1,-0.7)\)",C,SW); label("\(x\)",X,NW); label("\(y\)",Y,NE); | [] |
730 | Point $B$ is due east of point $A$. Point $C$ is due north of point $B$. The distance between points $A$ and $C$ is $10\sqrt 2$, and $\angle BAC = 45^\circ$. Point $D$ is $20$ meters due north of point $C$. The distance $AD$ is between which two integers?
$\textbf{(A)}\ 30\ \text{and}\ 31 \qquad\textbf{(B)}\ 31\ \text{and}\ 32 \qquad\textbf{(C)}\ 32\ \text{and}\ 33 \qquad\textbf{(D)}\ 33\ \text{and}\ 34 \qquad\textbf{(E)}\ 34\ \text{and}\ 35$ | 2012 AMC 10B Problem 12 | If point B is due east of point A and point C is due north of point B, $\angle CBA$ is a right angle. And if $\angle BAC = 45^\circ$, $\triangle CBA$ is a 45-45-90 triangle. Thus, the lengths of sides $CB$, $BA$, and $AC$ are in the ratio $1:1:\sqrt 2$, and $CB$ is $10 \sqrt 2 \div \sqrt 2 = 10$.
$\triangle DBA$ is clearly a right triangle with $C$ on the side $DB$. $DC$ is 20, so $DB = DC + CB = 20 + 10 = 30$.
By the Pythagorean Theorem, $DA = \sqrt {DB^2 + BA^2} = \sqrt {30^2 + 10 ^2} = \sqrt {1000}$.
$31^2 = 961$, and $32^2 = 1024$. Thus, $\sqrt {1000}$ must be between $31$ and $32$. The answer is $\boxed {B}$. | // Block 1
unitsize(4);
pair A=(0,0);
label ("A",(0,0),W);
pair B=(10,0);
label ("B",(10,0),E);
pair C=(10,10);
label ("C",(10,10),E);
pair D=(10,30);
label ("D",(10,30),E);
dot(A);
dot(B);
dot(C);
dot(D);
draw(A--B);
draw(A--C);
draw(A--D);
draw(C--D);
draw(B--C);
// Block 2
unitsize(4); pair A=(0,0); label ("A",(0,0),W); pair B=(10,0); label ("B",(10,0),E); pair C=(10,10); label ("C",(10,10),E); pair D=(10,30); label ("D",(10,30),E); dot(A); dot(B); dot(C); dot(D); draw(A--B); draw(A--C); draw(A--D); draw(C--D); draw(B--C); | [] |
731 | Two equilateral triangles are contained in square whose side length is $2\sqrt 3$. The bases of these triangles are the opposite side of the square, and their intersection is a rhombus. What is the area of the rhombus?
$\text{(A) } \frac{3}{2} \qquad \text{(B) } \sqrt 3 \qquad \text{(C) } 2\sqrt 2 - 1 \qquad \text{(D) } 8\sqrt 3 - 12 \qquad \text{(E)} \frac{4\sqrt 3}{3}$ | 2012 AMC 10B Problem 14 | Observe that the rhombus is made up of two congruent equilateral triangles with side length equal to GF. Since AE has length $\sqrt{3}$ and triangle AEF is a 30-60-90 triangle, it follows that EF has length 1. By symmetry, HG also has length 1. Thus GF has length $2\sqrt{3} - 2$. The formula for the area of an equilateral triangle of length $s$ is $\frac{\sqrt{3}}{4}s^2$. It follows that the area of the rhombus is:
$2\times\frac{\sqrt{3}}{4}(2\sqrt{3}-2)^2 = \boxed{\mathbf{(D)} 8\sqrt{3} - 12}.$ | // Block 1
size(8cm);
pair A, B, C, D, E, F, G, H, BF, AF;
A = (0,0);
B = (1,0);
C = (1,1);
D = (0,1);
E = (1/2,0);
H = (1/2,1);
G = (1/2,1/2^(1/2));
F = (1/2,1-(1/2^(1/2)));
AF = (3^(1/2)/2,1/2);
BF = (1-3^(1/2)/2,1/2);
draw(A--B--C--D--A--AF--D);
draw(C--BF--B);
draw(H--E,linetype("8 8"));
label("$A$",A,SW);
label("$B$",B,SE);
label("$C$",C,NE);
label("$D$",D,NW);
label("$E$",E,S);
label("$H$",H,N);
label("$G$",G+1/20,E);
label("$F$",F-1/20,W);
// Block 2
size(8cm); pair A, B, C, D, E, F, G, H, BF, AF; A = (0,0); B = (1,0); C = (1,1); D = (0,1); E = (1/2,0); H = (1/2,1); G = (1/2,1/2^(1/2)); F = (1/2,1-(1/2^(1/2))); AF = (3^(1/2)/2,1/2); BF = (1-3^(1/2)/2,1/2); draw(A--B--C--D--A--AF--D); draw(C--BF--B); draw(H--E,linetype("8 8")); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,NE); label("$D$",D,NW); label("$E$",E,S); label("$H$",H,N); label("$G$",G+1/20,E); label("$F$",F-1/20,W); | [] |
732 | In rectangle $ABCD$, $AB=6$, $AD=30$, and $G$ is the midpoint of $\overline{AD}$. Segment $AB$ is extended 2 units beyond $B$ to point $E$, and $F$ is the intersection of $\overline{ED}$ and $\overline{BC}$. What is the area of quadrilateral $BFDG$?
$\textbf{(A)}\ \frac{133}{2}\qquad\textbf{(B)}\ 67\qquad\textbf{(C)}\ \frac{135}{2}\qquad\textbf{(D)}\ 68\qquad\textbf{(E)}\ \frac{137}{2}$ | 2012 AMC 10B Problem 19 | Note that the area of $BFDG$ equals the area of $ABCD-\triangle AGB-\triangle DCF$.
Since $AG=\frac{AD}{2}=15,$ $\triangle AGB=\frac{15\times 6}{2}=45$. Now, $\triangle AED\sim \triangle BEF$, so $\frac{AE}{BE}=4=\frac{AD}{BF}=\frac{30}{BF}\implies BF=7.5$ and $FC=22.5,$ so $\triangle DCF=\frac{22.5\times6}{2}=\frac{135}{2}.$
Therefore, \begin{align*}BFDG&=ABCD-\triangle AGB-\triangle DCF \\ &=180-45-\frac{135}{2} \\ &=\boxed{\frac{135}{2}}\end{align*}
hence our answer is $\fbox{C}$ | // Block 1
unitsize(10);
pair B=(0,0);
pair A=(0,6);
pair C=(30,0);
pair D=(30,6);
pair G=(15,6);
pair E=(0,-2);
pair F=(15/2,0);
dot(A);
dot(B);
dot(C);
dot(D);
dot(G);
dot(E);
dot(F);
label("A",(0,6),NW);
label("B",(0,0),W);
label("C",(30,0),E);
label("D",(30,6),NE);
label("G",(15,6),N);
label("E",(0,-2),SW);
label("F",(15/2,0),N);
label("15",(A--G),N);
label("15",(G--D),N);
label("6",(A--B),W);
label("2",(B--E),W);
label("6",(D--C),E);
draw(A--B);
draw(B--C);
draw(C--D);
draw(D--A);
draw(E--D);
draw(B--E);
draw(B--G);
// Block 2
unitsize(10); pair B=(0,0); pair A=(0,6); pair C=(30,0); pair D=(30,6); pair G=(15,6); pair E=(0,-2); pair F=(15/2,0); dot(A); dot(B); dot(C); dot(D); dot(G); dot(E); dot(F); label("A",(0,6),NW); label("B",(0,0),W); label("C",(30,0),E); label("D",(30,6),NE); label("G",(15,6),N); label("E",(0,-2),SW); label("F",(15/2,0),N); label("15",(A--G),N); label("15",(G--D),N); label("6",(A--B),W); label("2",(B--E),W); label("6",(D--C),E); draw(A--B); draw(B--C); draw(C--D); draw(D--A); draw(E--D); draw(B--E); draw(B--G); | [] |
733 | Four distinct points are arranged on a plane so that the segments connecting them have lengths $a$, $a$, $a$, $a$, $2a$, and $b$. What is the ratio of $b$ to $a$?
$\textbf{(A)}\ \sqrt{3}\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \sqrt{5}\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ \pi$ | 2012 AMC 10B Problem 21 | When you see that there are lengths a and 2a, one could think of 30-60-90 triangles. Since all of the other's lengths are a, you could think that $b=\sqrt{3}a$.
Drawing the points out, it is possible to have a diagram where $b=\sqrt{3}a$. It turns out that $a,$ $2a,$ and $b$ could be the lengths of a 30-60-90 triangle, and the other 3 $a\text{'s}$ can be the lengths of an equilateral triangle formed from connecting the dots.
So, $b=\sqrt{3}a$, so $b:a= \boxed{\textbf{(A)} \: \sqrt{3}}$ | // Block 1
draw((0, 0)--(1/2, sqrt(3)/2)--(1, 0)--cycle);
draw((1/2, sqrt(3)/2)--(2, 0)--(1,0));
label("$a$", (0, 0)--(1, 0), S);
label("$a$", (1, 0)--(2, 0), S);
label("$a$", (0, 0)--(1/2, sqrt(3)/2), NW);
label("$a$", (1, 0)--(1/2, sqrt(3)/2), NE);
label("$b=\sqrt{3}a$", (1/2, sqrt(3)/2)--(2, 0), NE);
// Block 2
draw((0, 0)--(1/2, sqrt(3)/2)--(1, 0)--cycle); draw((1/2, sqrt(3)/2)--(2, 0)--(1,0)); label("$a$", (0, 0)--(1, 0), S); label("$a$", (1, 0)--(2, 0), S); label("$a$", (0, 0)--(1/2, sqrt(3)/2), NW); label("$a$", (1, 0)--(1/2, sqrt(3)/2), NE); label("$b=\sqrt{3}a$", (1/2, sqrt(3)/2)--(2, 0), NE); | [] |
734 | A solid tetrahedron is sliced off a solid wooden unit cube by a plane passing through two nonadjacent vertices on one face and one vertex on the opposite face not adjacent to either of the first two vertices. The tetrahedron is discarded and the remaining portion of the cube is placed on a table with the cut surface face down. What is the height of this object?
$\textbf{(A)}\ \frac{\sqrt{3}}{3} \qquad\textbf{(B)}\ \frac{2 \sqrt{2}}{3}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ \frac{2 \sqrt{3}}{3}\qquad\textbf{(E)}\ \sqrt{2}$ | 2012 AMC 10B Problem 23 | We can approach this problem similar to the solution above, by attacking it on the Cartesian plane. As mentioned before, it has been proved that the space diagonal intersects the centroid of triangle $BGF$ (I have defined the tetrahedron as cutting through points $B$, $G$, $F$, and $E$). We can label $E$ as $(0, 0, 0)$, $F$ as $(1, 0, 0)$, $B$ as $(0, 1, 0)$, and $G$ as $(0, 0, 1)$. Therefore, the centroid of triangle $BGF$ would be the average of these three points, or ($\frac{1}{3}$, $\frac{1}{3}$, $\frac{1}{3}$). Since $C$ is defined as $(1, 1, 1)$, and the intersection point passes through
($\frac{1}{3}$, $\frac{1}{3}$, $\frac{1}{3}$) (or one-third of the space diagonal), we can conclude that the altitude is $\boxed{\textbf{(D)}}$, or $\frac{2}{3}$ of $\sqrt3$. | // Block 1
import three; draw((1,1,1)--(1,0,1)--(1,0,0)--(0,0,0)--(0,0,1)--(0,1,1)--(1,1,1)--(1,1,0)--(0,1,0)--(0,1,1)); draw((0,0,1)--(1,0,1)); draw((1,0,0)--(1,1,0)); draw((0,0,0)--(0,1,0));
label("A",(0,0,0),S);
label("B",(1,0,0),W);
label("C",(0,0,1),N);
label("D",(1,0,1),NW);
label("E",(1,1,0),S);
label("F",(0,1,0),E);
label("G",(1,1,1),SE);
label("H",(0,1,1),NE);
// Block 2
import three; draw((1,1,1)--(1,0,1)--(1,0,0)--(0,0,0)--(0,0,1)--(0,1,1)--(1,1,1)--(1,1,0)--(0,1,0)--(0,1,1)); draw((0,0,1)--(1,0,1)); draw((1,0,0)--(1,1,0)); draw((0,0,0)--(0,1,0)); label("A",(0,0,0),S); label("B",(1,0,0),W); label("C",(0,0,1),N); label("D",(1,0,1),NW); label("E",(1,1,0),S); label("F",(0,1,0),E); label("G",(1,1,1),SE); label("H",(0,1,1),NE); | [] |
735 | A bug travels from A to B along the segments in the hexagonal lattice pictured below. The segments marked with an arrow can be traveled only in the direction of the arrow, and the bug never travels the same segment more than once. How many different paths are there?
$\textbf{(A)}\ 2112\qquad\textbf{(B)}\ 2304\qquad\textbf{(C)}\ 2368\qquad\textbf{(D)}\ 2384\qquad\textbf{(E)}\ 2400$ | 2012 AMC 10B Problem 25 | There is $1$ way to get to any of the red arrows. From the first (top) red arrow, there are $2$ ways to get to each of the first and the second (top 2) blue arrows; from the second (bottom) red arrow, there are $3$ ways to get to each of the first and the second blue arrows. So there are in total $5$ ways to get to each of the blue arrows.
From each of the first and second blue arrows, there are respectively $4$ ways to get to each of the first and the second green arrows; from each of the third and the fourth blue arrows, there are respectively $8$ ways to get to each of the first and the second green arrows. Therefore there are in total $5 \cdot (4+4+8+8) = 120$ ways to get to each of the green arrows.
Finally, from each of the first and second green arrows, there are respectively $2$ ways to get to the first orange arrow; from each of the third and the fourth green arrows, there are $3$ ways to get to the first orange arrow. Therefore there are $120 \cdot (2+2+3+3) = 1200$ ways to get to each of the orange arrows, hence $2400$ ways to get to the point $B$. $\boxed{\textbf{(E)}\ 2400}$ | // Block 1
size(10cm);
draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)--(10.0,6.928203230275509)--(12.0,6.928203230275509));
draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)--(10.0,6.928203230275509)--(12.0,6.928203230275509));
draw((3.0,-1.7320508075688772)--(4.0,0.0)--(6.0,0.0)--(7.0,1.7320508075688772)--(9.0,1.7320508075688772)--(10.0,3.4641016151377544)--(12.0,3.464101615137755)--(13.0,5.196152422706632)--(15.0,5.196152422706632));
draw((6.0,-3.4641016151377544)--(7.0,-1.7320508075688772)--(9.0,-1.7320508075688772)--(10.0,0.0)--(12.0,0.0)--(13.0,1.7320508075688772)--(15.0,1.7320508075688776)--(16.0,3.464101615137755)--(18.0,3.4641016151377544));
draw((9.0,-5.196152422706632)--(10.0,-3.464101615137755)--(12.0,-3.464101615137755)--(13.0,-1.7320508075688776)--(15.0,-1.7320508075688776)--(16.0,0)--(18.0,0.0)--(19.0,1.7320508075688772)--(21.0,1.7320508075688767));
draw((12.0,-6.928203230275509)--(13.0,-5.196152422706632)--(15.0,-5.196152422706632)--(16.0,-3.464101615137755)--(18.0,-3.4641016151377544)--(19.0,-1.7320508075688772)--(21.0,-1.7320508075688767)--(22.0,0));
draw((0.0,-0.0)--(1.0,-1.7320508075688772)--(3.0,-1.7320508075688772)--(4.0,-3.4641016151377544)--(6.0,-3.4641016151377544)--(7.0,-5.196152422706632)--(9.0,-5.196152422706632)--(10.0,-6.928203230275509)--(12.0,-6.928203230275509));
draw((3.0,1.7320508075688772)--(4.0,-0.0)--(6.0,-0.0)--(7.0,-1.7320508075688772)--(9.0,-1.7320508075688772)--(10.0,-3.4641016151377544)--(12.0,-3.464101615137755)--(13.0,-5.196152422706632)--(15.0,-5.196152422706632));
draw((6.0,3.4641016151377544)--(7.0,1.7320508075688772)--(9.0,1.7320508075688772)--(10.0,-0.0)--(12.0,-0.0)--(13.0,-1.7320508075688772)--(15.0,-1.7320508075688776)--(16.0,-3.464101615137755)--(18.0,-3.4641016151377544));
draw((9.0,5.1961524)--(10.0,3.464101)--(12.0,3.46410)--(13.0,1.73205)--(15.0,1.732050)--(16.0,0)--(18.0,-0.0)--(19.0,-1.7320)--(21.0,-1.73205080));
draw((12.0,6.928203)--(13.0,5.1961524)--(15.0,5.1961524)--(16.0,3.464101615)--(18.0,3.4641016)--(19.0,1.7320508)--(21.0,1.732050)--(22.0,0));
dot((0,0));
dot((22,0));
label("$A$",(0,0),WNW);
label("$B$",(22,0),E);
filldraw((2.0,1.7320508075688772)--(1.6,1.2320508075688772)--(1.75,1.7320508075688772)--(1.6,2.232050807568877)--cycle,red);
filldraw((5.0,3.4641016151377544)--(4.6,2.9641016151377544)--(4.75,3.4641016151377544)--(4.6,3.9641016151377544)--cycle,black);
filldraw((8.0,5.196152422706632)--(7.6,4.696152422706632)--(7.75,5.196152422706632)--(7.6,5.696152422706632)--cycle,blue);
filldraw((11.0,6.928203230275509)--(10.6,6.428203230275509)--(10.75,6.928203230275509)--(10.6,7.428203230275509)--cycle,black);
filldraw((4.6,0.0)--(5.0,-0.5)--(4.85,0.0)--(5.0,0.5)--cycle,white);
filldraw((8.0,1.732050)--(7.6,1.2320)--(7.75,1.73205)--(7.6,2.2320)--cycle,blue);
filldraw((11.0,3.4641016)--(10.6,2.9641016)--(10.75,3.46410161)--(10.6,3.964101)--cycle,black);
filldraw((14.0,5.196152422706632)--(13.6,4.696152422706632)--(13.75,5.196152422706632)--(13.6,5.696152422706632)--cycle,green);
filldraw((8.0,-1.732050)--(7.6,-2.232050)--(7.75,-1.7320508)--(7.6,-1.2320)--cycle,blue);
filldraw((10.6,0.0)--(11,-0.5)--(10.85,0.0)--(11,0.5)--cycle,white);
filldraw((14.0,1.7320508075688772)--(13.6,1.2320508075688772)--(13.75,1.7320508075688772)--(13.6,2.232050807568877)--cycle,green);
filldraw((17.0,3.464101615137755)--(16.6,2.964101615137755)--(16.75,3.464101615137755)--(16.6,3.964101615137755)--cycle,black);
filldraw((11.0,-3.464101615137755)--(10.6,-3.964101615137755)--(10.75,-3.464101615137755)--(10.6,-2.964101615137755)--cycle,black);
filldraw((14.0,-1.7320508075688776)--(13.6,-2.2320508075688776)--(13.75,-1.7320508075688776)--(13.6,-1.2320508075688776)--cycle,green);
filldraw((16.6,0)--(17,-0.5)--(16.85,0)--(17,0.5)--cycle,white);
filldraw((20.0,1.7320508075688772)--(19.6,1.2320508075688772)--(19.75,1.7320508075688772)--(19.6,2.232050807568877)--cycle,orange);
filldraw((14.0,-5.196152422706632)--(13.6,-5.696152422706632)--(13.75,-5.196152422706632)--(13.6,-4.696152422706632)--cycle,green);
filldraw((17.0,-3.464101615137755)--(16.6,-3.964101615137755)--(16.75,-3.464101615137755)--(16.6,-2.964101615137755)--cycle,black);
filldraw((20.0,-1.7320508075688772)--(19.6,-2.232050807568877)--(19.75,-1.7320508075688772)--(19.6,-1.2320508075688772)--cycle,orange);
filldraw((2.0,-1.7320508075688772)--(1.6,-1.2320508075688772)--(1.75,-1.7320508075688772)--(1.6,-2.232050807568877)--cycle,red);
filldraw((5.0,-3.4641016)--(4.6,-2.964101)--(4.75,-3.4641)--(4.6,-3.9641016)--cycle,black);
filldraw((8.0,-5.1961524)--(7.6,-4.6961524)--(7.75,-5.19615242)--(7.6,-5.696152422)--cycle,blue);
filldraw((11.0,-6.9282032)--(10.6,-6.4282032)--(10.75,-6.928203)--(10.6,-7.428203)--cycle,black);
// Block 2
size(10cm); draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)--(10.0,6.928203230275509)--(12.0,6.928203230275509)); draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)--(10.0,6.928203230275509)--(12.0,6.928203230275509)); draw((3.0,-1.7320508075688772)--(4.0,0.0)--(6.0,0.0)--(7.0,1.7320508075688772)--(9.0,1.7320508075688772)--(10.0,3.4641016151377544)--(12.0,3.464101615137755)--(13.0,5.196152422706632)--(15.0,5.196152422706632)); draw((6.0,-3.4641016151377544)--(7.0,-1.7320508075688772)--(9.0,-1.7320508075688772)--(10.0,0.0)--(12.0,0.0)--(13.0,1.7320508075688772)--(15.0,1.7320508075688776)--(16.0,3.464101615137755)--(18.0,3.4641016151377544)); draw((9.0,-5.196152422706632)--(10.0,-3.464101615137755)--(12.0,-3.464101615137755)--(13.0,-1.7320508075688776)--(15.0,-1.7320508075688776)--(16.0,0)--(18.0,0.0)--(19.0,1.7320508075688772)--(21.0,1.7320508075688767)); draw((12.0,-6.928203230275509)--(13.0,-5.196152422706632)--(15.0,-5.196152422706632)--(16.0,-3.464101615137755)--(18.0,-3.4641016151377544)--(19.0,-1.7320508075688772)--(21.0,-1.7320508075688767)--(22.0,0)); draw((0.0,-0.0)--(1.0,-1.7320508075688772)--(3.0,-1.7320508075688772)--(4.0,-3.4641016151377544)--(6.0,-3.4641016151377544)--(7.0,-5.196152422706632)--(9.0,-5.196152422706632)--(10.0,-6.928203230275509)--(12.0,-6.928203230275509)); draw((3.0,1.7320508075688772)--(4.0,-0.0)--(6.0,-0.0)--(7.0,-1.7320508075688772)--(9.0,-1.7320508075688772)--(10.0,-3.4641016151377544)--(12.0,-3.464101615137755)--(13.0,-5.196152422706632)--(15.0,-5.196152422706632)); draw((6.0,3.4641016151377544)--(7.0,1.7320508075688772)--(9.0,1.7320508075688772)--(10.0,-0.0)--(12.0,-0.0)--(13.0,-1.7320508075688772)--(15.0,-1.7320508075688776)--(16.0,-3.464101615137755)--(18.0,-3.4641016151377544)); draw((9.0,5.1961524)--(10.0,3.464101)--(12.0,3.46410)--(13.0,1.73205)--(15.0,1.732050)--(16.0,0)--(18.0,-0.0)--(19.0,-1.7320)--(21.0,-1.73205080)); draw((12.0,6.928203)--(13.0,5.1961524)--(15.0,5.1961524)--(16.0,3.464101615)--(18.0,3.4641016)--(19.0,1.7320508)--(21.0,1.732050)--(22.0,0)); dot((0,0)); dot((22,0)); label("$A$",(0,0),WNW); label("$B$",(22,0),E); filldraw((2.0,1.7320508075688772)--(1.6,1.2320508075688772)--(1.75,1.7320508075688772)--(1.6,2.232050807568877)--cycle,red); filldraw((5.0,3.4641016151377544)--(4.6,2.9641016151377544)--(4.75,3.4641016151377544)--(4.6,3.9641016151377544)--cycle,black); filldraw((8.0,5.196152422706632)--(7.6,4.696152422706632)--(7.75,5.196152422706632)--(7.6,5.696152422706632)--cycle,blue); filldraw((11.0,6.928203230275509)--(10.6,6.428203230275509)--(10.75,6.928203230275509)--(10.6,7.428203230275509)--cycle,black); filldraw((4.6,0.0)--(5.0,-0.5)--(4.85,0.0)--(5.0,0.5)--cycle,white); filldraw((8.0,1.732050)--(7.6,1.2320)--(7.75,1.73205)--(7.6,2.2320)--cycle,blue); filldraw((11.0,3.4641016)--(10.6,2.9641016)--(10.75,3.46410161)--(10.6,3.964101)--cycle,black); filldraw((14.0,5.196152422706632)--(13.6,4.696152422706632)--(13.75,5.196152422706632)--(13.6,5.696152422706632)--cycle,green); filldraw((8.0,-1.732050)--(7.6,-2.232050)--(7.75,-1.7320508)--(7.6,-1.2320)--cycle,blue); filldraw((10.6,0.0)--(11,-0.5)--(10.85,0.0)--(11,0.5)--cycle,white); filldraw((14.0,1.7320508075688772)--(13.6,1.2320508075688772)--(13.75,1.7320508075688772)--(13.6,2.232050807568877)--cycle,green); filldraw((17.0,3.464101615137755)--(16.6,2.964101615137755)--(16.75,3.464101615137755)--(16.6,3.964101615137755)--cycle,black); filldraw((11.0,-3.464101615137755)--(10.6,-3.964101615137755)--(10.75,-3.464101615137755)--(10.6,-2.964101615137755)--cycle,black); filldraw((14.0,-1.7320508075688776)--(13.6,-2.2320508075688776)--(13.75,-1.7320508075688776)--(13.6,-1.2320508075688776)--cycle,green); filldraw((16.6,0)--(17,-0.5)--(16.85,0)--(17,0.5)--cycle,white); filldraw((20.0,1.7320508075688772)--(19.6,1.2320508075688772)--(19.75,1.7320508075688772)--(19.6,2.232050807568877)--cycle,orange); filldraw((14.0,-5.196152422706632)--(13.6,-5.696152422706632)--(13.75,-5.196152422706632)--(13.6,-4.696152422706632)--cycle,green); filldraw((17.0,-3.464101615137755)--(16.6,-3.964101615137755)--(16.75,-3.464101615137755)--(16.6,-2.964101615137755)--cycle,black); filldraw((20.0,-1.7320508075688772)--(19.6,-2.232050807568877)--(19.75,-1.7320508075688772)--(19.6,-1.2320508075688772)--cycle,orange); filldraw((2.0,-1.7320508075688772)--(1.6,-1.2320508075688772)--(1.75,-1.7320508075688772)--(1.6,-2.232050807568877)--cycle,red); filldraw((5.0,-3.4641016)--(4.6,-2.964101)--(4.75,-3.4641)--(4.6,-3.9641016)--cycle,black); filldraw((8.0,-5.1961524)--(7.6,-4.6961524)--(7.75,-5.19615242)--(7.6,-5.696152422)--cycle,blue); filldraw((11.0,-6.9282032)--(10.6,-6.4282032)--(10.75,-6.928203)--(10.6,-7.428203)--cycle,black); | [] |
735 | A bug travels from A to B along the segments in the hexagonal lattice pictured below. The segments marked with an arrow can be traveled only in the direction of the arrow, and the bug never travels the same segment more than once. How many different paths are there?
$\textbf{(A)}\ 2112\qquad\textbf{(B)}\ 2304\qquad\textbf{(C)}\ 2368\qquad\textbf{(D)}\ 2384\qquad\textbf{(E)}\ 2400$ | 2012 AMC 10B Problem 25 | For every blue arrow, there are $2\cdot 2=4$ ways to reach it without using the reverse arrow since the bug can choose any of $2$ red arrows to pass through and $2$ black arrows to pass through. If the bug passes through the white arrow, the red arrow that the bug travels through must be the closest to the first black arrow. Otherwise, the bug will have to travel through both red segments, which is impossible because now there is no path to take after the bug emerges from the reverse arrow. Similarly, with the blue segments, the second black arrow taken must be the one that is closest to the blue arrow that was taken. Also, it is trivial that the two black arrows taken must be different. Therefore, if the reverse arrow is taken, the blue arrow taken determines the entire path and there is $1$ path for every arrow. Since the bug cannot return once it takes a blue arrow, the answer must be divisible by $5$. $\boxed{\textbf{(E)}\ 2400}$ is the only answer that is. | // Block 1
size(6cm);
draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632));
draw((0.0,0.0)--(1.0,-1.7320508075688772)--(3.0,-1.7320508075688772)--(4.0,-3.4641016151377544)--(6.0,-3.4641016151377544)--(7.0,-5.196152422706632)--(9.0,-5.196152422706632));
draw((6.0,-3.4641016151377544)--(7.0,-1.7320508075688772)--(9.0,-1.7320508075688772));
draw((4.0, 0.0)--(6.0,0.0));
draw((6.0,3.4641016151377544)--(7.0,1.7320508075688772)--(9.0,1.7320508075688772));
draw((3.0,1.7320508075688772)--(4.0,-0.0)--(3.0,-1.7320508075688772), red);
draw((7.0,1.7320508075688772)--(6.0,-0.0)--(7.0,-1.7320508075688772), blue);
dot((0,0));
label("$A$",(0,0),WNW);
filldraw((2.0,1.7320508075688772)--(1.6,1.2320508075688772)--(1.75,1.7320508075688772)--(1.6,2.232050807568877)--cycle,red);
filldraw((5.0,3.4641016151377544)--(4.6,2.9641016151377544)--(4.75,3.4641016151377544)--(4.6,3.9641016151377544)--cycle,black);
filldraw((8.0,5.196152422706632)--(7.6,4.696152422706632)--(7.75,5.196152422706632)--(7.6,5.696152422706632)--cycle,blue);
filldraw((4.6,0.0)--(5.0,-0.5)--(4.85,0.0)--(5.0,0.5)--cycle,white);
filldraw((8.0,1.732050)--(7.6,1.2320)--(7.75,1.73205)--(7.6,2.2320)--cycle,blue);
filldraw((8.0,-1.732050)--(7.6,-2.232050)--(7.75,-1.7320508)--(7.6,-1.2320)--cycle,blue);
filldraw((2.0,-1.7320508075688772)--(1.6,-1.2320508075688772)--(1.75,-1.7320508075688772)--(1.6,-2.232050807568877)--cycle,red);
filldraw((5.0,-3.4641016)--(4.6,-2.964101)--(4.75,-3.4641)--(4.6,-3.9641016)--cycle,black);
filldraw((8.0,-5.1961524)--(7.6,-4.6961524)--(7.75,-5.19615242)--(7.6,-5.696152422)--cycle,blue);
// Block 2
size(6cm); draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)); draw((0.0,0.0)--(1.0,-1.7320508075688772)--(3.0,-1.7320508075688772)--(4.0,-3.4641016151377544)--(6.0,-3.4641016151377544)--(7.0,-5.196152422706632)--(9.0,-5.196152422706632)); draw((6.0,-3.4641016151377544)--(7.0,-1.7320508075688772)--(9.0,-1.7320508075688772)); draw((4.0, 0.0)--(6.0,0.0)); draw((6.0,3.4641016151377544)--(7.0,1.7320508075688772)--(9.0,1.7320508075688772)); draw((3.0,1.7320508075688772)--(4.0,-0.0)--(3.0,-1.7320508075688772), red); draw((7.0,1.7320508075688772)--(6.0,-0.0)--(7.0,-1.7320508075688772), blue); dot((0,0)); label("$A$",(0,0),WNW); filldraw((2.0,1.7320508075688772)--(1.6,1.2320508075688772)--(1.75,1.7320508075688772)--(1.6,2.232050807568877)--cycle,red); filldraw((5.0,3.4641016151377544)--(4.6,2.9641016151377544)--(4.75,3.4641016151377544)--(4.6,3.9641016151377544)--cycle,black); filldraw((8.0,5.196152422706632)--(7.6,4.696152422706632)--(7.75,5.196152422706632)--(7.6,5.696152422706632)--cycle,blue); filldraw((4.6,0.0)--(5.0,-0.5)--(4.85,0.0)--(5.0,0.5)--cycle,white); filldraw((8.0,1.732050)--(7.6,1.2320)--(7.75,1.73205)--(7.6,2.2320)--cycle,blue); filldraw((8.0,-1.732050)--(7.6,-2.232050)--(7.75,-1.7320508)--(7.6,-1.2320)--cycle,blue); filldraw((2.0,-1.7320508075688772)--(1.6,-1.2320508075688772)--(1.75,-1.7320508075688772)--(1.6,-2.232050807568877)--cycle,red); filldraw((5.0,-3.4641016)--(4.6,-2.964101)--(4.75,-3.4641)--(4.6,-3.9641016)--cycle,black); filldraw((8.0,-5.1961524)--(7.6,-4.6961524)--(7.75,-5.19615242)--(7.6,-5.696152422)--cycle,blue); | [] |
736 | Square $PQRS$ lies in the first quadrant. Points $(3,0), (5,0), (7,0),$ and $(13,0)$ lie on lines $SP, RQ, PQ$, and $SR$, respectively. What is the sum of the coordinates of the center of the square $PQRS$?
$\textbf{(A)}\ 6\qquad\textbf{(B) }\frac{31}5\qquad\textbf{(C) }\frac{32}5\qquad\textbf{(D) }\frac{33}5\qquad\textbf{(E) }\frac{34}5$ | 2012 AMC 12B Problem 17 | Construct the midpoints $E=(4,0)$ and $F=(10,0)$ and triangle $\triangle EMF$ as in the diagram, where $M$ is the center of square $PQRS$. Also construct points $G$ and $H$ as in the diagram so that $BG\parallel PQ$ and $CH\parallel QR$.
Observe that $\triangle AGB\sim\triangle CHD$ while $PQRS$ being a square implies that $GB=CH$. Furthermore, $CD=6=3\cdot AB$, so $\triangle CHD$ is 3 times bigger than $\triangle AGB$. Therefore, $HD=3\cdot GB=3\cdot HC$. In other words, the longer leg is 3 times the shorter leg in any triangle similar to $\triangle AGB$.
Let $K$ be the foot of the perpendicular from $M$ to $EF$, and let $x=EK$. Triangles $\triangle EKM$ and $\triangle MKF$, being similar to $\triangle AGB$, also have legs in a 1:3 ratio, therefore, $MK=3x$ and $KF=9x$, so $10x=EF=6$. It follows that $EK=0.6$ and $MK=1.8$, so the coordinates of $M$ are $(4+0.6,1.8)=(4.6,1.8)$ and so our answer is $4.6+1.8 = 6.4 =$ $\boxed{\mathbf{(C)}\ 32/5}$. | // Block 1
size(14cm);
pair A=(3,0),B=(5,0),C=(7,0),D=(13,0),EE=(4,0),F=(10,0),P=(3.4,1.2),Q=(5.2,0.6),R=(5.8,2.4),SS=(4,3),M=(4.6,1.8),G=(3.2,0.6),H=(7.6,1.8);
dot(A^^B^^C^^D^^EE^^F^^P^^Q^^R^^SS^^M^^G^^H);
draw(A--SS--D--cycle);
draw(P--Q--R^^B--Q--C);
draw(EE--M--F^^G--B^^C--H,dotted);
label("A",A,SW);
label("B",B,S);
label("C",C,S);
label("D",D,SE);
label("E",EE,S);
label("F",F,S);
label("P",P,W);
label("Q",Q,NW);
label("R",R,NE);
label("S",SS,N);
label("M",M,S);
label("G",G,W);
label("H",H,NE);
// Block 2
size(14cm); pair A=(3,0),B=(5,0),C=(7,0),D=(13,0),EE=(4,0),F=(10,0),P=(3.4,1.2),Q=(5.2,0.6),R=(5.8,2.4),SS=(4,3),M=(4.6,1.8),G=(3.2,0.6),H=(7.6,1.8); dot(A^^B^^C^^D^^EE^^F^^P^^Q^^R^^SS^^M^^G^^H); draw(A--SS--D--cycle); draw(P--Q--R^^B--Q--C); draw(EE--M--F^^G--B^^C--H,dotted); label("A",A,SW); label("B",B,S); label("C",C,S); label("D",D,SE); label("E",EE,S); label("F",F,S); label("P",P,W); label("Q",Q,NW); label("R",R,NE); label("S",SS,N); label("M",M,S); label("G",G,W); label("H",H,NE); | [] |
736 | Square $PQRS$ lies in the first quadrant. Points $(3,0), (5,0), (7,0),$ and $(13,0)$ lie on lines $SP, RQ, PQ$, and $SR$, respectively. What is the sum of the coordinates of the center of the square $PQRS$?
$\textbf{(A)}\ 6\qquad\textbf{(B) }\frac{31}5\qquad\textbf{(C) }\frac{32}5\qquad\textbf{(D) }\frac{33}5\qquad\textbf{(E) }\frac{34}5$ | 2012 AMC 12B Problem 17 | Let the four points be labeled $P_1$, $P_2$, $P_3$, and $P_4$, respectively. Let the lines that go through each point be labeled $L_1$, $L_2$, $L_3$, and $L_4$, respectively. Since $L_1$ and $L_2$ go through $SP$ and $RQ$, respectively, and $SP$ and $RQ$ are opposite sides of the square, we can say that $L_1$ and $L_2$ are parallel with slope $m$. Similarly, $L_3$ and $L_4$ have slope $-\frac{1}{m}$. Also, note that since square $PQRS$ lies in the first quadrant, $L_1$ and $L_2$ must have a positive slope. Using the point-slope form, we can now find the equations of all four lines: $L_1: y = m(x-3)$, $L_2: y = m(x-5)$, $L_3: y = -\frac{1}{m}(x-7)$, $L_4: y = -\frac{1}{m}(x-13)$.
Since $PQRS$ is a square, it follows that $\Delta x$ between points $P$ and $Q$ is equal to $\Delta y$ between points $Q$ and $R$. Our approach will be to find $\Delta x$ and $\Delta y$ in terms of $m$ and equate the two to solve for $m$. $L_1$ and $L_3$ intersect at point $P$. Setting the equations for $L_1$ and $L_3$ equal to each other and solving for $x$, we find that they intersect at $x = \frac{3m^2 + 7}{m^2 + 1}$. $L_2$ and $L_3$ intersect at point $Q$. Intersecting the two equations, the $x$-coordinate of point $Q$ is found to be $x = \frac{5m^2 + 7}{m^2 + 1}$. Subtracting the two, we get $\Delta x = \frac{2m^2}{m^2 + 1}$. Substituting the $x$-coordinate for point $Q$ found above into the equation for $L_2$, we find that the $y$-coordinate of point $Q$ is $y = \frac{2m}{m^2+1}$. $L_2$ and $L_4$ intersect at point $R$. Intersecting the two equations, the $y$-coordinate of point $R$ is found to be $y = \frac{8m}{m^2 + 1}$. Subtracting the two, we get $\Delta y = \frac{6m}{m^2 + 1}$. Equating $\Delta x$ and $\Delta y$, we get $2m^2 = 6m$ which gives us $m = 3$. Finally, note that the line which goes though the midpoint of $P_1$ and $P_2$ with slope $3$ and the line which goes through the midpoint of $P_3$ and $P_4$ with slope $-\frac{1}{3}$ must intersect at at the center of the square. The equation of the line going through $(4,0)$ is given by $y = 3(x-4)$ and the equation of the line going through $(10,0)$ is $y = -\frac{1}{3}(x-10)$. Equating the two, we find that they intersect at $(4.6, 1.8)$. Adding the $x$ and $y$-coordinates, we get $6.4 = 32/5$. Thus, answer choice $\boxed{\textbf{(C)}}$ is correct. | size(7cm); pair A=(0,0),B=(1,1.5),D=B*dir(-90),C=B+D-A; draw((-4,-2)--(8,-2), Arrows); draw(A--B--C--D--cycle); pair AB = extension(A,B,(0,-2),(1,-2)); pair BC = extension(B,C,(0,-2),(1,-2)); pair CD = extension(C,D,(0,-2),(1,-2)); pair DA = extension(D,A,(0,-2),(1,-2)); draw(A--AB--B--BC--C--CD--D--DA--A, dotted); dot(AB^^BC^^CD^^DA); | [] |
736 | Square $PQRS$ lies in the first quadrant. Points $(3,0), (5,0), (7,0),$ and $(13,0)$ lie on lines $SP, RQ, PQ$, and $SR$, respectively. What is the sum of the coordinates of the center of the square $PQRS$?
$\textbf{(A)}\ 6\qquad\textbf{(B) }\frac{31}5\qquad\textbf{(C) }\frac{32}5\qquad\textbf{(D) }\frac{33}5\qquad\textbf{(E) }\frac{34}5$ | 2012 AMC 12B Problem 17 | $SP: y = mx - 3m$, $RQ: y = mx-5m$, $PQ: y = -\frac{1}{m}x + \frac{7}{m}$, $SR: y = -\frac{1}{m}x + \frac{13}{m}$
Let $SP = RP = PQ = SR = a$, $\angle GAB = \angle HCD = \theta$, and the slope of $SP$ be $m$.
When the slope of $SP$ is $m$, the slope of $SR$ is $-\frac{1}{m}$, $\tan \theta = m$, $\cot \theta = -\frac{1}{m}$
$\sin \theta = \frac{GB}{AB} = \frac{a}{2}$, $\cos \theta = \frac{HC}{CD} = \frac{a}{6}$
As $\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{a}{2}}{\frac{a}{6}} = 3$, $m = 3$
$SP: y = 3x - 9$, $RQ: y = 3x-15$, $PQ: y = -\frac{1}{3}x + \frac{7}{3}$, $SR: y = -\frac{1}{3}x + \frac{13}{3}$
$3x - 9 = -\frac{1}{3}x + \frac{13}{3}$, $x = 4$, $y = 3$, $S = (4, 3)$
$3x-15 = -\frac{1}{3}x + \frac{7}{3}$, $x = 5.2$, $y = 0.6$, $Q = (5.2, 0.6)$
$M = (4.6, 1.8)$, $4.6 + 1.8 = \boxed{\mathbf{(C)}\ 32/5}$
~isabelchen | // Block 1
size(14cm);
pair A=(3,0),B=(5,0),C=(7,0),D=(13,0),EE=(4,0),F=(10,0),P=(3.4,1.2),Q=(5.2,0.6),R=(5.8,2.4),SS=(4,3),M=(4.6,1.8),G=(3.2,0.6),H=(7.6,1.8);
dot(A^^B^^C^^D^^EE^^F^^P^^Q^^R^^SS^^M^^G^^H);
draw(A--SS--D--cycle);
draw(P--Q--R^^B--Q--C);
draw(EE--M--F^^G--B^^C--H,dotted);
label("A",A,SW);
label("B",B,S);
label("C",C,S);
label("D",D,SE);
label("E",EE,S);
label("F",F,S);
label("P",P,W);
label("Q",Q,NW);
label("R",R,NE);
label("S",SS,N);
label("M",M,S);
label("G",G,W);
label("H",H,NE);
// Block 2
size(14cm); pair A=(3,0),B=(5,0),C=(7,0),D=(13,0),EE=(4,0),F=(10,0),P=(3.4,1.2),Q=(5.2,0.6),R=(5.8,2.4),SS=(4,3),M=(4.6,1.8),G=(3.2,0.6),H=(7.6,1.8); dot(A^^B^^C^^D^^EE^^F^^P^^Q^^R^^SS^^M^^G^^H); draw(A--SS--D--cycle); draw(P--Q--R^^B--Q--C); draw(EE--M--F^^G--B^^C--H,dotted); label("A",A,SW); label("B",B,S); label("C",C,S); label("D",D,SE); label("E",EE,S); label("F",F,S); label("P",P,W); label("Q",Q,NW); label("R",R,NE); label("S",SS,N); label("M",M,S); label("G",G,W); label("H",H,NE); | [] |
736 | Square $PQRS$ lies in the first quadrant. Points $(3,0), (5,0), (7,0),$ and $(13,0)$ lie on lines $SP, RQ, PQ$, and $SR$, respectively. What is the sum of the coordinates of the center of the square $PQRS$?
$\textbf{(A)}\ 6\qquad\textbf{(B) }\frac{31}5\qquad\textbf{(C) }\frac{32}5\qquad\textbf{(D) }\frac{33}5\qquad\textbf{(E) }\frac{34}5$ | 2012 AMC 12B Problem 17 | Let $PQ=x$ and $\angle PCA=\theta.$ Draw the line $BE$ such that $E$ is on $AP$ and $BE\parallel PQ.$ Also, Draw the line $CF$ such that $F$ is on $DR$ and $CF\parallel RQ.$ Then $EB=FC=x$ and $\angle EBA=\angle FDC=\theta.$ Also, note that $AB=2$ and $CD=6.$ Hence:
\[\cos(\theta)=\frac{x}{2},\sin(\theta)=\frac{x}{6}.\]
Thus $\cos(\theta)=3\sin(\theta).$ Since $\theta<\frac{\pi}{2},$ $\sin(\theta)=\sqrt{1-\cos^2(\theta)},$ so $\cos(\theta)=3\sqrt{1-\cos^2(\theta)}.$ Hence, $\cos(\theta)=\frac{3}{\sqrt{10}}$ and $x=\frac{6}{\sqrt{10}}.$ Draw the perpendicular lines $EF_1\perp AD,PF_2\perp AD,RF_3\perp AD.$ Note that:
\[F_1B=BE\cdot\cos(\theta)=\frac{6}{\sqrt{10}}\cdot\frac{3}{\sqrt{10}}=\frac{9}{5}.\]
Hence:
\[EF_1=\sqrt{EB^2-EF_1^2}=\sqrt{\frac{18}{5}-\frac{81}{25}}=\frac{3}{5}.\]
Note that $\triangle EF_1B\sim\triangle PF_2C,$ so:
\[\frac{PF_2}{EF_1}=\frac{F_2C}{F_1B}=\frac{AC}{AB}=2.\]
Hence:
\[PF_2=\frac{6}{5}, F_2C=\frac{18}{5}.\]
So $P$ has coordinates:
\[\left(7-\frac{18}{5},\frac{6}{5}\right)=\left(\frac{17}{5},\frac{6}{5}\right).\]
Also note that $\triangle EF_1B\sim\triangle RF_3D,$ so:
\[\frac{RF_3}{EF_1}=\frac{F_3D}{F_1B}=\frac{BD}{AB}=4.\]
Hence:
\[RF_3=\frac{12}{5}, F_3D=\frac{36}{5}.\]
So $R$ has coordinates:
\[\left(13-\frac{36}{5},\frac{12}{5}\right)=\left(\frac{29}{5},\frac{12}{5}\right).\]
Hence, the center of square $PQRS,$ which is also the midpoint of $PR,$ has coordinates:
\[\left(\frac{\frac{17}{5}+\frac{29}{5}}{2},\frac{\frac{6}{5}+\frac{12}{5}}{2}\right)=\left(\frac{23}{5},\frac{9}{5}\right).\]
We thus see that the answer is:
\[\frac{23}{5}+\frac{9}{5}=\boxed{\text{(C)}\frac{32}{5}}.\] | // Block 1
unitsize(1 cm);
pair P,Q,R,S,A,B,C,D,E,F,F1,F2,F3,M;
P = (3.4,1.2);
Q = (5.2,0.6);
R = (5.8,2.4);
S = (4,3);
A = (3,0);
B = (5,0);
C = (7,0);
D = (13,0);
E = (3.2,0.6);
F = (7.6,1.8);
F1 = (3.2,0);
F2 = (3.4,0);
F3 = (5.8,0);
M = (4.6,1.8);
dot(P);
dot(Q);
dot(R);
dot(S);
dot(A);
dot(B);
dot(C);
dot(D);
dot(E);
dot(F);
dot(F1);
dot(F2);
dot(F3);
dot(M);
draw(P--Q--R--S--cycle);
draw(P--A);
draw(Q--B);
draw(R--D);
draw(Q--C);
draw(A--D);
draw(B--E, dashed);
draw(C--F, dashed);
draw(E--F1, dashed);
draw(P--F2, dashed);
draw(R--F3, dashed);
draw(P--R, dashed);
label("$A$",A,W);
label("$B$",B,SW);
label("$C$",C,SW);
label("$D$",D,SW);
label("$P$",P,W);
label("$Q$",Q,ENE);
label("$R$",R,N);
label("$S$",S,N);
label("$E$",E,NW);
label("$F$",F,NE);
label("$F_1$",F1,SSW);
label("$F_2$",F2,SE);
label("$F_3$",F3,SW);
label("$M$",M,NW);
// Block 2
unitsize(1 cm); pair P,Q,R,S,A,B,C,D,E,F,F1,F2,F3,M; P = (3.4,1.2); Q = (5.2,0.6); R = (5.8,2.4); S = (4,3); A = (3,0); B = (5,0); C = (7,0); D = (13,0); E = (3.2,0.6); F = (7.6,1.8); F1 = (3.2,0); F2 = (3.4,0); F3 = (5.8,0); M = (4.6,1.8); dot(P); dot(Q); dot(R); dot(S); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); dot(F1); dot(F2); dot(F3); dot(M); draw(P--Q--R--S--cycle); draw(P--A); draw(Q--B); draw(R--D); draw(Q--C); draw(A--D); draw(B--E, dashed); draw(C--F, dashed); draw(E--F1, dashed); draw(P--F2, dashed); draw(R--F3, dashed); draw(P--R, dashed); label("$A$",A,W); label("$B$",B,SW); label("$C$",C,SW); label("$D$",D,SW); label("$P$",P,W); label("$Q$",Q,ENE); label("$R$",R,N); label("$S$",S,N); label("$E$",E,NW); label("$F$",F,NE); label("$F_1$",F1,SSW); label("$F_2$",F2,SE); label("$F_3$",F3,SW); label("$M$",M,NW); | [] |
737 | Square $AXYZ$ is inscribed in equiangular hexagon $ABCDEF$ with $X$ on $\overline{BC}$, $Y$ on $\overline{DE}$, and $Z$ on $\overline{EF}$. Suppose that $AB=40$, and $EF=41(\sqrt{3}-1)$. What is the side-length of the square?
$\textbf{(A)}\ 29\sqrt{3} \qquad\textbf{(B)}\ \frac{21}{2}\sqrt{2}+\frac{41}{2}\sqrt{3} \qquad\textbf{(C)}\ \ 20\sqrt{3}+16 \qquad\textbf{(D)}\ 20\sqrt{2}+13 \sqrt{3} \qquad\textbf{(E)}\ 21\sqrt{6}$
(diagram by djmathman) | 2012 AMC 12B Problem 21 | We can, $\textsc{wlog}$, assume $Y$ coincides with $D$ and $CD\parallel AF$ as before. In which case, we will have $BC=EF=41(\sqrt{3}-1)$. So we have square $AXDZ$ inscribed in equiangular hexagon $ABCDEF$ with $X$ on $\overline{BC}$ and $Z$ on $\overline{EF}$.
Let $\angle BXA = \theta$; then $\angle BAX=60^\circ -\theta$. Let $BX=u$. In $\triangle ABX$ we have
\begin{align} \frac{2s}{\sqrt{3}}=\frac{u}{\sin(60^\circ-\theta)}=\frac {40}{\sin\theta} \end{align}
We also have $\angle CXD=90^\circ - \theta$ and $\angle CDX = \theta-30^\circ$. Let $CX=v$. In $\triangle CDX$ we have
\begin{align}\tag{2} \frac{2s}{\sqrt{3}}=\frac{v}{\sin(\theta-30^\circ)}=\frac {CD}{\cos\theta} \end{align}
Now $BC=u+v=41(\sqrt{3}-1)$. From $(1)$ and $(2)$ we get\begin{align*} 41(\sqrt{3}-1) &= \frac{2s}{\sqrt{3}}\left(\sin(60^\circ-\theta)+\sin(\theta-30^\circ)\right) \\ &= \frac{2s}{\sqrt{3}} \cdot \frac{\sqrt{3}-1}2\cdot (\sin\theta + \cos\theta) \end{align*}
From $(1)$ we get $s\sin\theta = 20\sqrt{3}$ and therefore $s\cos\theta = \sqrt{s^2-3\cdot 20^2}$. Thus
\[41(\sqrt{3}-1) = \frac{\sqrt{3}-1}{\sqrt{3}}(20\sqrt{3}+\sqrt{s^2-3\cdot 20^2})\]which simplifies to\[3\cdot 21^2 = s^2-3\cdot 20^2.\]Since $(20, 21, 29)$ is a Pythagorean triple, we get $s=29\sqrt{3}$, i.e. $\framebox{A}$. | size(200); defaultpen(fontsize(10)+linewidth(1)); pair A=origin,B=(2.5,0),C=B+2.5*dir(60), D=C+1.75*dir(120),E=D-(3.19,0),F=E-1.8*dir(60); pair X=waypoint(B--C,0.345),Z=rotate(90,A)*X,Y=rotate(90,Z)*A; pair Cp=extension(B,C,Y,Y+dir(-60)); draw(A--B--Cp--Y--E--F--cycle); draw(A--X--Y--Z--cycle,linewidth(0.9)+linetype("2 2")); dot("$A$",A,W,linewidth(4)); dot("$B$",B,dir(0),linewidth(4)); dot("$C$",Cp,dir(0),linewidth(4)); dot("$E$",E,dir(100),linewidth(4)); dot("$F$",F,W,linewidth(4)); dot("$X$",X,dir(0),linewidth(4)); dot("$D$",Y,N,linewidth(4)); dot("$Z$",Z,W,linewidth(4)); label("$u$", B--X, SE);label("$v$", X--Cp, SE); label("$40$", A--B, S); label("$s$", A--X, NW); label("$s$", Y--X, SW); | [] |
738 | A bug travels from A to B along the segments in the hexagonal lattice pictured below. The segments marked with an arrow can be traveled only in the direction of the arrow, and the bug never travels the same segment more than once. How many different paths are there?
$\textbf{(A)}\ 2112\qquad\textbf{(B)}\ 2304\qquad\textbf{(C)}\ 2368\qquad\textbf{(D)}\ 2384\qquad\textbf{(E)}\ 2400$ | 2012 AMC 12B Problem 22 | There is $1$ way to get to any of the red arrows. From the first (top) red arrow, there are $2$ ways to get to each of the first and the second (top 2) blue arrows; from the second (bottom) red arrow, there are $3$ ways to get to each of the first and the second blue arrows. So there are in total $5$ ways to get to each of the blue arrows.
From each of the first and second blue arrows, there are respectively $4$ ways to get to each of the first and the second green arrows; from each of the third and the fourth blue arrows, there are respectively $8$ ways to get to each of the first and the second green arrows. Therefore there are in total $5 \cdot (4+4+8+8) = 120$ ways to get to each of the green arrows.
Finally, from each of the first and second green arrows, there are respectively $2$ ways to get to the first orange arrow; from each of the third and the fourth green arrows, there are $3$ ways to get to the first orange arrow. Therefore there are $120 \cdot (2+2+3+3) = 1200$ ways to get to each of the orange arrows, hence $2400$ ways to get to the point $B$. $\boxed{\textbf{(E)}\ 2400}$ | size(10cm); draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)--(10.0,6.928203230275509)--(12.0,6.928203230275509)); draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)--(10.0,6.928203230275509)--(12.0,6.928203230275509)); draw((3.0,-1.7320508075688772)--(4.0,0.0)--(6.0,0.0)--(7.0,1.7320508075688772)--(9.0,1.7320508075688772)--(10.0,3.4641016151377544)--(12.0,3.464101615137755)--(13.0,5.196152422706632)--(15.0,5.196152422706632)); draw((6.0,-3.4641016151377544)--(7.0,-1.7320508075688772)--(9.0,-1.7320508075688772)--(10.0,0.0)--(12.0,0.0)--(13.0,1.7320508075688772)--(15.0,1.7320508075688776)--(16.0,3.464101615137755)--(18.0,3.4641016151377544)); draw((9.0,-5.196152422706632)--(10.0,-3.464101615137755)--(12.0,-3.464101615137755)--(13.0,-1.7320508075688776)--(15.0,-1.7320508075688776)--(16.0,0)--(18.0,0.0)--(19.0,1.7320508075688772)--(21.0,1.7320508075688767)); draw((12.0,-6.928203230275509)--(13.0,-5.196152422706632)--(15.0,-5.196152422706632)--(16.0,-3.464101615137755)--(18.0,-3.4641016151377544)--(19.0,-1.7320508075688772)--(21.0,-1.7320508075688767)--(22.0,0)); draw((0.0,-0.0)--(1.0,-1.7320508075688772)--(3.0,-1.7320508075688772)--(4.0,-3.4641016151377544)--(6.0,-3.4641016151377544)--(7.0,-5.196152422706632)--(9.0,-5.196152422706632)--(10.0,-6.928203230275509)--(12.0,-6.928203230275509)); draw((3.0,1.7320508075688772)--(4.0,-0.0)--(6.0,-0.0)--(7.0,-1.7320508075688772)--(9.0,-1.7320508075688772)--(10.0,-3.4641016151377544)--(12.0,-3.464101615137755)--(13.0,-5.196152422706632)--(15.0,-5.196152422706632)); draw((6.0,3.4641016151377544)--(7.0,1.7320508075688772)--(9.0,1.7320508075688772)--(10.0,-0.0)--(12.0,-0.0)--(13.0,-1.7320508075688772)--(15.0,-1.7320508075688776)--(16.0,-3.464101615137755)--(18.0,-3.4641016151377544)); draw((9.0,5.1961524)--(10.0,3.464101)--(12.0,3.46410)--(13.0,1.73205)--(15.0,1.732050)--(16.0,0)--(18.0,-0.0)--(19.0,-1.7320)--(21.0,-1.73205080)); draw((12.0,6.928203)--(13.0,5.1961524)--(15.0,5.1961524)--(16.0,3.464101615)--(18.0,3.4641016)--(19.0,1.7320508)--(21.0,1.732050)--(22.0,0)); dot((0,0)); dot((22,0)); label("$A$",(0,0),WNW); label("$B$",(22,0),E); filldraw((2.0,1.7320508075688772)--(1.6,1.2320508075688772)--(1.75,1.7320508075688772)--(1.6,2.232050807568877)--cycle,red); filldraw((5.0,3.4641016151377544)--(4.6,2.9641016151377544)--(4.75,3.4641016151377544)--(4.6,3.9641016151377544)--cycle,black); filldraw((8.0,5.196152422706632)--(7.6,4.696152422706632)--(7.75,5.196152422706632)--(7.6,5.696152422706632)--cycle,blue); filldraw((11.0,6.928203230275509)--(10.6,6.428203230275509)--(10.75,6.928203230275509)--(10.6,7.428203230275509)--cycle,black); filldraw((4.6,0.0)--(5.0,-0.5)--(4.85,0.0)--(5.0,0.5)--cycle,white); filldraw((8.0,1.732050)--(7.6,1.2320)--(7.75,1.73205)--(7.6,2.2320)--cycle,blue); filldraw((11.0,3.4641016)--(10.6,2.9641016)--(10.75,3.46410161)--(10.6,3.964101)--cycle,black); filldraw((14.0,5.196152422706632)--(13.6,4.696152422706632)--(13.75,5.196152422706632)--(13.6,5.696152422706632)--cycle,green); filldraw((8.0,-1.732050)--(7.6,-2.232050)--(7.75,-1.7320508)--(7.6,-1.2320)--cycle,blue); filldraw((10.6,0.0)--(11,-0.5)--(10.85,0.0)--(11,0.5)--cycle,white); filldraw((14.0,1.7320508075688772)--(13.6,1.2320508075688772)--(13.75,1.7320508075688772)--(13.6,2.232050807568877)--cycle,green); filldraw((17.0,3.464101615137755)--(16.6,2.964101615137755)--(16.75,3.464101615137755)--(16.6,3.964101615137755)--cycle,black); filldraw((11.0,-3.464101615137755)--(10.6,-3.964101615137755)--(10.75,-3.464101615137755)--(10.6,-2.964101615137755)--cycle,black); filldraw((14.0,-1.7320508075688776)--(13.6,-2.2320508075688776)--(13.75,-1.7320508075688776)--(13.6,-1.2320508075688776)--cycle,green); filldraw((16.6,0)--(17,-0.5)--(16.85,0)--(17,0.5)--cycle,white); filldraw((20.0,1.7320508075688772)--(19.6,1.2320508075688772)--(19.75,1.7320508075688772)--(19.6,2.232050807568877)--cycle,orange); filldraw((14.0,-5.196152422706632)--(13.6,-5.696152422706632)--(13.75,-5.196152422706632)--(13.6,-4.696152422706632)--cycle,green); filldraw((17.0,-3.464101615137755)--(16.6,-3.964101615137755)--(16.75,-3.464101615137755)--(16.6,-2.964101615137755)--cycle,black); filldraw((20.0,-1.7320508075688772)--(19.6,-2.232050807568877)--(19.75,-1.7320508075688772)--(19.6,-1.2320508075688772)--cycle,orange); filldraw((2.0,-1.7320508075688772)--(1.6,-1.2320508075688772)--(1.75,-1.7320508075688772)--(1.6,-2.232050807568877)--cycle,red); filldraw((5.0,-3.4641016)--(4.6,-2.964101)--(4.75,-3.4641)--(4.6,-3.9641016)--cycle,black); filldraw((8.0,-5.1961524)--(7.6,-4.6961524)--(7.75,-5.19615242)--(7.6,-5.696152422)--cycle,blue); filldraw((11.0,-6.9282032)--(10.6,-6.4282032)--(10.75,-6.928203)--(10.6,-7.428203)--cycle,black); | [] |
738 | A bug travels from A to B along the segments in the hexagonal lattice pictured below. The segments marked with an arrow can be traveled only in the direction of the arrow, and the bug never travels the same segment more than once. How many different paths are there?
$\textbf{(A)}\ 2112\qquad\textbf{(B)}\ 2304\qquad\textbf{(C)}\ 2368\qquad\textbf{(D)}\ 2384\qquad\textbf{(E)}\ 2400$ | 2012 AMC 12B Problem 22 | For every blue arrow, there are $2\cdot 2=4$ ways to reach it without using the reverse arrow since the bug can choose any of $2$ red arrows to pass through and $2$ black arrows to pass through. If the bug passes through the white arrow, the red arrow that the bug travels through must be the closest to the first black arrow. Otherwise, the bug will have to travel through both red segments, which is impossible because now there is no path to take after the bug emerges from the reverse arrow. Similarly, with the blue segments, the second black arrow taken must be the one that is closest to the blue arrow that was taken. Also, it is trivial that the two black arrows taken must be different. Therefore, if the reverse arrow is taken, the blue arrow taken determines the entire path and there is $1$ path for every arrow. Since the bug cannot return once it takes a blue arrow, the answer must be divisible by $5$. $\boxed{\textbf{(E)}\ 2400}$ is the only answer that is. | size(6cm); draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)); draw((0.0,0.0)--(1.0,-1.7320508075688772)--(3.0,-1.7320508075688772)--(4.0,-3.4641016151377544)--(6.0,-3.4641016151377544)--(7.0,-5.196152422706632)--(9.0,-5.196152422706632)); draw((6.0,-3.4641016151377544)--(7.0,-1.7320508075688772)--(9.0,-1.7320508075688772)); draw((4.0, 0.0)--(6.0,0.0)); draw((6.0,3.4641016151377544)--(7.0,1.7320508075688772)--(9.0,1.7320508075688772)); draw((3.0,1.7320508075688772)--(4.0,-0.0)--(3.0,-1.7320508075688772), red); draw((7.0,1.7320508075688772)--(6.0,-0.0)--(7.0,-1.7320508075688772), blue); dot((0,0)); label("$A$",(0,0),WNW); filldraw((2.0,1.7320508075688772)--(1.6,1.2320508075688772)--(1.75,1.7320508075688772)--(1.6,2.232050807568877)--cycle,red); filldraw((5.0,3.4641016151377544)--(4.6,2.9641016151377544)--(4.75,3.4641016151377544)--(4.6,3.9641016151377544)--cycle,black); filldraw((8.0,5.196152422706632)--(7.6,4.696152422706632)--(7.75,5.196152422706632)--(7.6,5.696152422706632)--cycle,blue); filldraw((4.6,0.0)--(5.0,-0.5)--(4.85,0.0)--(5.0,0.5)--cycle,white); filldraw((8.0,1.732050)--(7.6,1.2320)--(7.75,1.73205)--(7.6,2.2320)--cycle,blue); filldraw((8.0,-1.732050)--(7.6,-2.232050)--(7.75,-1.7320508)--(7.6,-1.2320)--cycle,blue); filldraw((2.0,-1.7320508075688772)--(1.6,-1.2320508075688772)--(1.75,-1.7320508075688772)--(1.6,-2.232050807568877)--cycle,red); filldraw((5.0,-3.4641016)--(4.6,-2.964101)--(4.75,-3.4641)--(4.6,-3.9641016)--cycle,black); filldraw((8.0,-5.1961524)--(7.6,-4.6961524)--(7.75,-5.19615242)--(7.6,-5.696152422)--cycle,blue); | [] |
739 | Let $S=\{(x,y) : x\in \{0,1,2,3,4\}, y\in \{0,1,2,3,4,5\},\text{ and } (x,y)\ne (0,0)\}$.
Let $T$ be the set of all right triangles whose vertices are in $S$. For every right triangle $t=\triangle{ABC}$ with vertices $A$, $B$, and $C$ in counter-clockwise order and right angle at $A$, let $f(t)=\tan(\angle{CBA})$. What is \[\prod_{t\in T} f(t)?\]
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{625}{144}\qquad\textbf{(C)}\ \frac{125}{24}\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ \frac{625}{24}$ | 2012 AMC 12B Problem 25 | This is just another way for the reasoning of solution 1. Picture the question to be a grid of unit squares instead of a coordinate system. Note that the restriction, (This is just to make visualization easier.) Define a "cell" to be a rectangle in the set of $S.$ For example, a cell can be (labeled in a red):
Note that choosing any three points and labeling them according to the problem will result in a product of one. For example, with the cell we just labeled, the four triangles we can create are:
If we define the longer side to be $x$ and the shorter side to be $y,$ then the product will be $\frac{x}{y} \cdot \frac{y}{x} \cdot \frac{x}{y} \cdot \frac{y}{x}=1,$ and we are done.
Otherwise, the three points are not contained in a "cell." This will result in the solution 1 path as described before. Our three points must take the form $(0,5), (*,0), (*,*),$ where $*$ is a number defined by the boundaries of $S.$ Thus, by the three cases, our answer is $\boxed{\textbf{(B)} \ \frac{625}{144}}.$
~wesserwes7254 | // Block 1
unitsize(0.5 cm);
draw((0,1)--(0,5),black);
draw((1,0)--(1,5),black);
draw((2,0)--(2,5),black);
draw((3,0)--(3,5),black);
draw((4,0)--(4,5),black);
draw((0,1)--(5,1),black);
draw((5,0)--(5,5),black);
draw((1,0)--(5,0),black);
draw((0,2)--(5,2),black);
draw((0,3)--(5,3),black);
draw((0,4)--(5,4),black);
draw((0,5)--(5,5),black);
draw((0,1)--(5,1),red);
draw((0,1)--(0,5),red);
draw((0,5)--(5,5),red);
draw((5,5)--(5,1),red);
// Block 2
unitsize(0.5 cm);
draw((0,1)--(5,1),red);
draw((0,1)--(0,5),red);
draw((0,5)--(5,5),red);
draw((5,5)--(5,1),red);
draw((0,1)--(5,1),black);
draw((0,1)--(0,5),black);
draw((0,5)--(5,1),black);
pair A, B, C;
A = (0,1);
B = (0,5);
C = (5,1);
// Block 3
unitsize(0.5 cm);
draw((0,1)--(5,1),red);
draw((0,1)--(0,5),red);
draw((0,5)--(5,5),red);
draw((5,5)--(5,1),red);
draw((0,1)--(5,1),black);
draw((5,1)--(5,5),black);
draw((5,5)--(0,1),black);
// Block 4
unitsize(0.5 cm);
draw((0,1)--(5,1),red);
draw((0,1)--(0,5),red);
draw((0,5)--(5,5),red);
draw((5,5)--(5,1),red);
draw((0,5)--(5,5),black);
draw((5,1)--(5,5),black);
draw((0,5)--(5,1),black);
// Block 5
unitsize(0.5 cm);
draw((0,1)--(5,1),red);
draw((0,1)--(0,5),red);
draw((0,5)--(5,5),red);
draw((5,5)--(5,1),red);
draw((0,5)--(5,5),black);
draw((0,1)--(0,5),black);
draw((0,1)--(5,5),black);
// Block 6
unitsize(0.5 cm); draw((0,1)--(0,5),black); draw((1,0)--(1,5),black); draw((2,0)--(2,5),black); draw((3,0)--(3,5),black); draw((4,0)--(4,5),black); draw((0,1)--(5,1),black); draw((5,0)--(5,5),black); draw((1,0)--(5,0),black); draw((0,2)--(5,2),black); draw((0,3)--(5,3),black); draw((0,4)--(5,4),black); draw((0,5)--(5,5),black); draw((0,1)--(5,1),red); draw((0,1)--(0,5),red); draw((0,5)--(5,5),red); draw((5,5)--(5,1),red);
// Block 7
unitsize(0.5 cm); draw((0,1)--(5,1),red); draw((0,1)--(0,5),red); draw((0,5)--(5,5),red); draw((5,5)--(5,1),red); draw((0,1)--(5,1),black); draw((0,1)--(0,5),black); draw((0,5)--(5,1),black); pair A, B, C; A = (0,1); B = (0,5); C = (5,1);
// Block 8
unitsize(0.5 cm); draw((0,1)--(5,1),red); draw((0,1)--(0,5),red); draw((0,5)--(5,5),red); draw((5,5)--(5,1),red); draw((0,1)--(5,1),black); draw((5,1)--(5,5),black); draw((5,5)--(0,1),black);
// Block 9
unitsize(0.5 cm); draw((0,1)--(5,1),red); draw((0,1)--(0,5),red); draw((0,5)--(5,5),red); draw((5,5)--(5,1),red); draw((0,5)--(5,5),black); draw((5,1)--(5,5),black); draw((0,5)--(5,1),black);
// Block 10
unitsize(0.5 cm); draw((0,1)--(5,1),red); draw((0,1)--(0,5),red); draw((0,5)--(5,5),red); draw((5,5)--(5,1),red); draw((0,5)--(5,5),black); draw((0,1)--(0,5),black); draw((0,1)--(5,5),black); | [] |
740 | A triangle with vertices $(6, 5)$, $(8, -3)$, and $(9, 1)$ is reflected about the line $x=8$ to create a second triangle. What is the area of the union of the two triangles?
$\textbf{(A)}\ 9 \qquad\textbf{(B)}\ \frac{28}{3} \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ \frac{31}{3} \qquad\textbf{(E)}\ \frac{32}{3}$ | 2013 AMC 10A Problem 16 | Let $A$ be at $(6, 5)$, B be at $(8, -3)$, and $C$ be at $(9, 1)$. Reflecting over the line $x=8$, we see that $A' = D = (10,5)$, $B' = B$ (as the x-coordinate of B is 8), and $C' = E = (7, 1)$. Line $AB$ can be represented as $y=-4x+29$, so we see that $E$ is on line $AB$.
We see that if we connect $A$ to $D$, we get a line of length $4$ (between $(6, 5)$ and $(10,5)$). The area of $\triangle ABD$ is equal to $\frac{bh}{2} = \frac{4(8)}{2} = 16$.
Now, let the point of intersection between $AC$ and $DE$ be $F$. If we can just find the area of $\triangle ADF$ and subtract it from $16$, we are done.
We realize that because the diagram is symmetric over $x = 8$, the intersection of lines $AC$ and $DE$ should intersect at an x-coordinate of $8$. We know that the slope of $DE$ is $\frac{5-1}{10-7} = \frac{4}{3}$. Thus, we can represent the line going through $E$ and $D$ as $y - 1=\frac{4}{3}(x - 7)$. Plugging in $x = 8$, we find that the y-coordinate of F is $\frac{7}{3}$. Thus, the height of $\triangle ADF$ is $5 - \frac{7}{3} = \frac{8}{3}$. Using the formula for the area of a triangle, the area of $\triangle ADF$ is $\frac{16}{3}$.
To get our final answer, we must subtract this from $16$. $[ABD] - [ADF] = 16 - \frac{16}{3} = \boxed{\textbf{(E) }\frac{32}{3}}$ | // Block 1
pair A = (6, 5), B = (8, -3), C = (9, 1), D = (10, 5), E = (7, 1), F = (8, 7/3);
draw(A--B--C--cycle^^D--E--B--cycle);
dot(A^^B^^C^^D^^E^^F);
label("$A$",A,NW);
label("$B$",B,S);
label("$C$",C,SE);
label("$D$",D,NE);
label("$E$",E,W);
label("$F$", F, N);
// Block 2
pair A = (6, 5), B = (8, -3), C = (9, 1), D = (10, 5), E = (7, 1), F = (8, 7/3); draw(A--B--C--cycle^^D--E--B--cycle); dot(A^^B^^C^^D^^E^^F); label("$A$",A,NW); label("$B$",B,S); label("$C$",C,SE); label("$D$",D,NE); label("$E$",E,W); label("$F$", F, N); | [] |
740 | A triangle with vertices $(6, 5)$, $(8, -3)$, and $(9, 1)$ is reflected about the line $x=8$ to create a second triangle. What is the area of the union of the two triangles?
$\textbf{(A)}\ 9 \qquad\textbf{(B)}\ \frac{28}{3} \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ \frac{31}{3} \qquad\textbf{(E)}\ \frac{32}{3}$ | 2013 AMC 10A Problem 16 | Since we have to find the area on a coordinate plane, we can use the Shoelace Theorem to find the area of the intersection. When you reflect it, it makes a quadrilateral.
Since it is reflected around $x=8$, the point $(8,-3)$ remains the same on both. The top right corners are $(6,5)$, and its reflection $(10,5)$. Now to find the 4th point, point F, we can use the equation of the line DE($\frac{4}{3}x - \frac{25}{3}$, and substitute $x=8$, to get $\frac{7}{3}$. Now we can use the theorem: $\frac{1}{2}(((6*-3)+(10*5)+(8*5)+(\frac{7}{3}*8))-((8*5)+(6*5)+(10*\frac{7}{3}) + (8* -3))) = \boxed{\textbf{(E) }\frac{32}{3}}$
~idk12345678 | // Block 1
pair A = (6, 5), B = (8, -3), C = (9, 1), D = (10, 5), E = (7, 1), F = (8, 7/3);
draw(A--B--C--cycle^^D--E--B--cycle);
dot(A^^B^^C^^D^^E^^F);
label("$A$",A,NW);
label("$B$",B,S);
label("$C$",C,SE);
label("$D$",D,NE);
label("$E$",E,W);
label("$F$", F, N);
// Block 2
pair A = (6, 5), B = (8, -3), C = (9, 1), D = (10, 5), E = (7, 1), F = (8, 7/3); draw(A--B--C--cycle^^D--E--B--cycle); dot(A^^B^^C^^D^^E^^F); label("$A$",A,NW); label("$B$",B,S); label("$C$",C,SE); label("$D$",D,NE); label("$E$",E,W); label("$F$", F, N); | [] |
741 | Let points $A = (0, 0)$, $B = (1, 2)$, $C=(3, 3)$, and $D = (4, 0)$. Quadrilateral $ABCD$ is cut into equal area pieces by a line passing through $A$. This line intersects $\overline{CD}$ at point $\left(\frac{p}{q}, \frac{r}{s}\right)$, where these fractions are in lowest terms. What is $p+q+r+s$?
$\textbf{(A)}\ 54\qquad\textbf{(B)}\ 58\qquad\textbf{(C)}\ 62\qquad\textbf{(D)}\ 70\qquad\textbf{(E)}\ 75$ | 2013 AMC 10A Problem 18 | First, we shall find the area of quadrilateral $ABCD$. This can be done in any of three ways:
Pick's Theorem: $[ABCD] = I + \dfrac{B}{2} - 1 = 5 + \dfrac{7}{2} - 1 = \dfrac{15}{2}.$
Splitting: Drop perpendiculars from $B$ and $C$ to the x-axis to divide the quadrilateral into triangles and trapezoids, and so the area is $1 + 5 + \dfrac{3}{2} = \dfrac{15}{2}.$
Shoelace Theorem: The area is half of $|1 \cdot 3 - 2 \cdot 3 - 3 \cdot 4| = 15$, or $\dfrac{15}{2}$.
$[ABCD] = \frac{15}{2}$. Therefore, each equal piece that the line separates $ABCD$ into must have an area of $\frac{15}{4}$.
Call the point where the line through $A$ intersects $\overline{CD}$ $E$. We know that $[ADE] = \frac{15}{4} = \frac{bh}{2}$. Furthermore, we know that $b = 4$, as $AD = 4$. Thus, solving for $h$, we find that $2h = \frac{15}{4}$, so $h = \frac{15}{8}$. This gives that the y coordinate of E is $\frac{15}{8}$.
Line CD can be expressed as $y = -3x+12$, so the $x$ coordinate of E satisfies $\frac{15}{8} = -3x + 12$. Solving for $x$, we find that $x = \frac{27}{8}$.
From this, we know that $E = \left(\frac{27}{8}, \frac{15}{8}\right)$. $27 + 15 + 8 + 8 = \boxed{\textbf{(B) }58}$ | // Block 1
size(8cm);
pair A, B, C, D, E, EE;
A = (0,0);
B = (1,2);
C = (3,3);
D = (4,0);
E = (27/8,15/8);
EE = (27/8,0);
draw(A--B--C--D--A--E);
draw(E--EE,linetype("8 8"));
dot(A);
dot(B);
dot(C);
dot(D);
dot(E);
draw(rightanglemark(E,EE,D,4));
label("A",A,SW);
label("B",B,NW);
label("C",C,NE);
label("D",D,SE);
label("E",E,NE);
label("$4$",(A+D)/2,S);
label("$\frac{27}{8}$",(A+EE)/2,S);
label("$\frac{15}{8}$",(E+EE)/2,W);
// Block 2
size(8cm); pair A, B, C, D, E, EE; A = (0,0); B = (1,2); C = (3,3); D = (4,0); E = (27/8,15/8); EE = (27/8,0); draw(A--B--C--D--A--E); draw(E--EE,linetype("8 8")); dot(A); dot(B); dot(C); dot(D); dot(E); draw(rightanglemark(E,EE,D,4)); label("A",A,SW); label("B",B,NW); label("C",C,NE); label("D",D,SE); label("E",E,NE); label("$4$",(A+D)/2,S); label("$\frac{27}{8}$",(A+EE)/2,S); label("$\frac{15}{8}$",(E+EE)/2,W); | [] |
741 | Let points $A = (0, 0)$, $B = (1, 2)$, $C=(3, 3)$, and $D = (4, 0)$. Quadrilateral $ABCD$ is cut into equal area pieces by a line passing through $A$. This line intersects $\overline{CD}$ at point $\left(\frac{p}{q}, \frac{r}{s}\right)$, where these fractions are in lowest terms. What is $p+q+r+s$?
$\textbf{(A)}\ 54\qquad\textbf{(B)}\ 58\qquad\textbf{(C)}\ 62\qquad\textbf{(D)}\ 70\qquad\textbf{(E)}\ 75$ | 2013 AMC 10A Problem 18 | Let the point where the altitude from $E$ to $\overline{AD}$ be labeled $F$.
Following the steps above, you can find that the height of $\triangle ADE$ is $\frac{15}{8}$, and from there split the base into two parts, $x$, and $4-x$, such that $x$ is the segment from the origin to the point $F$, and $4-x$ is the segment from point $F$ to point $D$. Then, by the Pythagorean Theorem, $x=\frac{27}{8}$, and the answer is $\boxed{\textbf{(B) }58}$ | // Block 1
size(8cm);
pair A, B, C, D, E, F;
A = (0,0);
B = (1,2);
C = (3,3);
D = (4,0);
E = (27/8,15/8);
F = (27/8,0);
draw(A--B--C--D--A--E);
draw(E--F,linetype("8 8"));
dot(A);
dot(B);
dot(C);
dot(D);
dot(E);
draw(rightanglemark(E,F,D,4));
label("A",A,SW);
label("B",B,NW);
label("C",C,NE);
label("D",D,SE);
label("E",E,NE);
label("F",F,S);
label("$4$",(A+D)/2,S);
label("$x$",(A+F)/2,S);
label("$\frac{15}{8}$",(E+F)/2,W);
// Block 2
size(8cm); pair A, B, C, D, E, F; A = (0,0); B = (1,2); C = (3,3); D = (4,0); E = (27/8,15/8); F = (27/8,0); draw(A--B--C--D--A--E); draw(E--F,linetype("8 8")); dot(A); dot(B); dot(C); dot(D); dot(E); draw(rightanglemark(E,F,D,4)); label("A",A,SW); label("B",B,NW); label("C",C,NE); label("D",D,SE); label("E",E,NE); label("F",F,S); label("$4$",(A+D)/2,S); label("$x$",(A+F)/2,S); label("$\frac{15}{8}$",(E+F)/2,W); | [] |
742 | A unit square is rotated $45^\circ$ about its center. What is the area of the region swept out by the interior of the square?
$\textbf{(A)}\ 1 - \frac{\sqrt2}{2} + \frac{\pi}{4}\qquad\textbf{(B)}\ \frac{1}{2} + \frac{\pi}{4} \qquad\textbf{(C)}\ 2 - \sqrt2 + \frac{\pi}{4}\qquad\textbf{(D)}\ \frac{\sqrt2}{2} + \frac{\pi}{4} \qquad\textbf{(E)}\ 1 + \frac{\sqrt2}{4} + \frac{\pi}{8}$ | 2013 AMC 10A Problem 20 | First, we need to see what this looks like. Below is a diagram.
For this square with side length 1, the distance from center to vertex is $r = \frac{\sqrt{2}}{2}$, hence the area is composed of a semicircle of radius $r$, plus $4$ times a parallelogram (or a kite with diagonals of $(\sqrt{2}-1)$ and $r \text{ or} \frac{\sqrt{2}}{2}$) with height $\frac{1}{2}$ and base $\frac{\sqrt{2}}{2(1+\sqrt{2})}$. That is to say, the total area is $\frac{1}{2} \pi \left(\frac{\sqrt{2}}{2}\right)^2 + 4 \frac{\sqrt{2}}{4(1+\sqrt{2})} = \boxed{\textbf{(C) } 2 - \sqrt{2} + \frac{\pi}{4}}$.
(To turn each dart-shaped piece into a parallelogram, cut along the dashed line and flip over one half.)
Alternatively, you can move the dart-shaped piece to the other side and make a kite. | // Block 1
size(200);
defaultpen(linewidth(0.8));
path square=shift((-.5,-.5))*unitsquare,square2=rotate(45)*square;
fill(square^^square2,grey);
for(int i=0;i<=3;i=i+1)
{
path arcrot=arc(origin,sqrt(2)/2,45+90*i,90*(i+1));
draw(arcrot);
fill(arcrot--(0,0)--cycle,grey);
draw(arc(origin,sqrt(2)/2+1/8,50+90*i,90*(i+1)-10),EndArrow);
}
draw(square^^square2);
// Block 2
size(150);defaultpen(linewidth(0.8));
path square=shift((-.5,-.5))*unitsquare,square2=rotate(45)*square;//fill(square^^square2,grey);
for(int i=0;i<=3;i=i+1){path arcrot=arc(origin,sqrt(2)/2,45+90*i,90*(i+1));draw(arcrot);
fill(arcrot--(0,0)--cycle,grey);}
//draw(arc(origin,sqrt(2)/2+1/8,50+90*i,90*(i+1)-10),EndArrow);}
draw(square^^square2);
//draw((-.5,.5)--(.5,-.5)^^(0,sqrt(.5))--(0,-sqrt(.5)),dotted);draw((.5,.5)--(-.5,-.5),dotted);
// Block 3
size(150,Aspect);real r=sqrt(2);real b=2-2/r;
draw((0,0)--(-1,1)--(b-1,1)--(0,r)--cycle);draw((0,1)--(b-1,1)--(b/2-1,1-b/2));draw((0,0)--(b-1,1),dashed);
fill((2,0)--(b+1,1)--(b+2,0)--cycle,lightgray);draw((.5,.5)--(1,.5),EndArrow);
draw((2,0)--(1,1)--(b+1,1)--(b+2,0)--(2,0)^^(b+1,1)--(b/2+1,1-b/2)^^(2,0)--(2+b/2,b/2));
draw((2,0)--(b+1,1),dashed);
// Block 4
size(75,Aspect);real r=sqrt(2);real b=2-2/r;
draw((r-1,1)--(b-1,1));
draw((0,0)--(b-1,1)--(0,r)--(r-1,1)--cycle);
draw((0,r)--(0,0),dashed);
// Block 5
size(200); defaultpen(linewidth(0.8)); path square=shift((-.5,-.5))*unitsquare,square2=rotate(45)*square; fill(square^^square2,grey); for(int i=0;i<=3;i=i+1) { path arcrot=arc(origin,sqrt(2)/2,45+90*i,90*(i+1)); draw(arcrot); fill(arcrot--(0,0)--cycle,grey); draw(arc(origin,sqrt(2)/2+1/8,50+90*i,90*(i+1)-10),EndArrow); } draw(square^^square2);
// Block 6
size(150);defaultpen(linewidth(0.8)); path square=shift((-.5,-.5))*unitsquare,square2=rotate(45)*square;//fill(square^^square2,grey); for(int i=0;i<=3;i=i+1){path arcrot=arc(origin,sqrt(2)/2,45+90*i,90*(i+1));draw(arcrot); fill(arcrot--(0,0)--cycle,grey);} //draw(arc(origin,sqrt(2)/2+1/8,50+90*i,90*(i+1)-10),EndArrow);} draw(square^^square2); //draw((-.5,.5)--(.5,-.5)^^(0,sqrt(.5))--(0,-sqrt(.5)),dotted);draw((.5,.5)--(-.5,-.5),dotted);
// Block 7
size(150,Aspect);real r=sqrt(2);real b=2-2/r; draw((0,0)--(-1,1)--(b-1,1)--(0,r)--cycle);draw((0,1)--(b-1,1)--(b/2-1,1-b/2));draw((0,0)--(b-1,1),dashed); fill((2,0)--(b+1,1)--(b+2,0)--cycle,lightgray);draw((.5,.5)--(1,.5),EndArrow); draw((2,0)--(1,1)--(b+1,1)--(b+2,0)--(2,0)^^(b+1,1)--(b/2+1,1-b/2)^^(2,0)--(2+b/2,b/2)); draw((2,0)--(b+1,1),dashed);
// Block 8
size(75,Aspect);real r=sqrt(2);real b=2-2/r; draw((r-1,1)--(b-1,1)); draw((0,0)--(b-1,1)--(0,r)--(r-1,1)--cycle); draw((0,r)--(0,0),dashed); | [] |
742 | A unit square is rotated $45^\circ$ about its center. What is the area of the region swept out by the interior of the square?
$\textbf{(A)}\ 1 - \frac{\sqrt2}{2} + \frac{\pi}{4}\qquad\textbf{(B)}\ \frac{1}{2} + \frac{\pi}{4} \qquad\textbf{(C)}\ 2 - \sqrt2 + \frac{\pi}{4}\qquad\textbf{(D)}\ \frac{\sqrt2}{2} + \frac{\pi}{4} \qquad\textbf{(E)}\ 1 + \frac{\sqrt2}{4} + \frac{\pi}{8}$ | 2013 AMC 10A Problem 20 | $\textbf{(high res image; no labels)}$
[Image: images/amc/2013_AMC_10A_Problem_20_0.jpg]
Let $O$ be the center of the square and $C$ be the intersection of $OB$ and $AD$. The desired area consists of the unit square, plus $4$ regions congruent to the region bounded by arc $AB$, $\overline{AC}$, and $\overline{BC}$, plus $4$ triangular regions congruent to right triangle $BCD$. The area of the region bounded by arc $AB$, $\overline{AC}$, and $\overline{BC}$ is $\frac{\text{Area of Circle}-\text{Area of Square}}{8}$. Since the circle has radius $\dfrac{1}{\sqrt {2}}$, the area of the region is $\dfrac{\dfrac{\pi}{2}-1}{8}$, so 4 times the area of that region is $\dfrac{\pi}{4}-\dfrac{1}{2}$. Now we find the area of $\triangle BCD$. $BC=BO-OC=\dfrac{\sqrt {2}}{2}-\dfrac{1}{2}$. Since $\triangle BCD$ is a $45-45-90$ right triangle, the area of $\triangle BCD$ is $\dfrac{BC^2}{2}=\dfrac {\left (\dfrac {\sqrt {2}}{2}-\dfrac{1}{2} \right)^2}{2}$, so $4$ times the area of $\triangle BCD$ is $\dfrac{3}{2}-\sqrt {2}$. Finally, the area of the whole region is $1+ \left(\dfrac {3}{2}-\sqrt {2} \right) + \left(\dfrac{\pi}{4}-\dfrac{1}{2} \right)=\dfrac{\pi}{4}+2-\sqrt {2}$, which we can rewrite as $\boxed{\textbf{(C) } 2 - \sqrt{2} + \frac{\pi}{4}}$. | // Block 1
size(200);
defaultpen(linewidth(0.8));
path square=shift((-.5,-.5))*unitsquare,square2=rotate(45)*square;
for(int i=0;i<=6;i=i+1)
{
path arcrot=arc(origin,sqrt(2)/2,45+270*i,270*(i+1));
draw(arcrot);
}
draw(square^^square2);
// Block 2
size(200); defaultpen(linewidth(0.8)); path square=shift((-.5,-.5))*unitsquare,square2=rotate(45)*square; for(int i=0;i<=6;i=i+1) { path arcrot=arc(origin,sqrt(2)/2,45+270*i,270*(i+1)); draw(arcrot); } draw(square^^square2); | ["https://artofproblemsolving.com/wiki/images/1/1e/AMC_10A_2013_20.jpg"] |
743 | Six spheres of radius $1$ are positioned so that their centers are at the vertices of a regular hexagon of side length $2$. The six spheres are internally tangent to a larger sphere whose center is the center of the hexagon. An eighth sphere is externally tangent to the six smaller spheres and internally tangent to the larger sphere. What is the radius of this eighth sphere?
$\textbf{(A)}\ \sqrt2\qquad\textbf{(B)}\ \frac{3}{2}\qquad\textbf{(C)}\ \frac{5}{3}\qquad\textbf{(D)}\ \sqrt3\qquad\textbf{(E)}\ 2$ | 2013 AMC 10A Problem 22 | We have a regular hexagon with side length $2$ and six spheres on each vertex with radius $1$ that are internally tangent, therefore, drawing radii to the tangent points would create this regular hexagon.
Imagine a 2D overhead view. There is a larger sphere which the $6$ spheres are internally tangent to, with the center in the center of the hexagon. To find the radius of the larger sphere we must first, either by prior knowledge or by deducing from the angle sum that the hexagon can be split into $6$ equilateral triangles from its vertices, that the radius is two more than the small radius, or $3$.
diagram made by erics118
Now imagine the figure in $3$ dimensions. The 8th sphere must be sitting atop of the $6$ spheres, which is the only possibility for it to be tangent to all the $6$ small spheres externally and the larger sphere internally. The ring of $6$ small spheres is symmetrical and the 8th sphere will be resting atop it with its center aligned with the diameter of the large sphere.
We can now create a right triangle with the legs being the line from a vertex of the hexagon to the center of the hexagon and the line from the center of the 7th sphere to the center of the 8th sphere. Let the radius of our 8th sphere be $r$. As previously mentioned, the distance from the center of the hexagon to one of its vertices is $2$. The distance between the centers will be $3-r$. The hypotenuse will be $r+1$.
We now have a right triangle. Applying the Pythagorean Theorem, $2^2+(3-r)^2=(1+r)^2$. Expanding and solving for $r$, we find $r=\frac{12}{8}=\boxed{\textbf{(B)}\frac{3}{2}}$.
3D Diagram:
https://www.desmos.com/3d/nhwgeghuby (turn on translucent surfaces in more options) | // Block 1
draw(circle((0.5,0.866025404),0.5));
draw(circle((-0.5,0.866025404),0.5));
draw(circle((1,0),0.5));
draw(circle((-1,0),0.5));
draw(circle((0.5,-0.866025404),0.5));
draw(circle((-0.5,-0.866025404),0.5));
draw(circle((0,0),1.5));
draw((-0.5,0.866025404)--(0.5,0.866025404));
draw((-1,0)--(1,0));
draw((-0.5,-0.866025404)--(0.5,-0.866025404));
draw((-1,0)--(-0.5,0.866025404));
draw((-0.5,-0.866025404)--(0.5,0.866025404));
draw((0.5,-0.866025404)--(1,0));
draw((0.5,0.866025404)--(1,0));
draw((-0.5,0.866025404)--(0.5,-0.866025404));
draw((-1,0)--(-0.5,-0.866025404));
// Block 2
draw(circle((0.5,0.866025404),0.5)); draw(circle((-0.5,0.866025404),0.5)); draw(circle((1,0),0.5)); draw(circle((-1,0),0.5)); draw(circle((0.5,-0.866025404),0.5)); draw(circle((-0.5,-0.866025404),0.5)); draw(circle((0,0),1.5)); draw((-0.5,0.866025404)--(0.5,0.866025404)); draw((-1,0)--(1,0)); draw((-0.5,-0.866025404)--(0.5,-0.866025404)); draw((-1,0)--(-0.5,0.866025404)); draw((-0.5,-0.866025404)--(0.5,0.866025404)); draw((0.5,-0.866025404)--(1,0)); draw((0.5,0.866025404)--(1,0)); draw((-0.5,0.866025404)--(0.5,-0.866025404)); draw((-1,0)--(-0.5,-0.866025404)); | [] |
744 | In $\triangle ABC$, $AB = 86$, and $AC=97$. A circle with center $A$ and radius $AB$ intersects $\overline{BC}$ at points $B$ and $X$. Moreover $\overline{BX}$ and $\overline{CX}$ have integer lengths. What is $BC$?
$\textbf{(A)}\ 11\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 33\qquad\textbf{(D)}\ 61\qquad\textbf{(E)}\ 72$ | 2013 AMC 10A Problem 23 | We first draw the height of isosceles triangle $ABD$ and get two equations by the Pythagorean Theorem.
First, $h^2 + k^2 = 86^2$. Second, $h^2 + (k + m)^2 = 97^2$.
Subtracting these two equations, we get $2km + m^2 = 97^2 - 86^2 = (97 - 86)(97 + 86) = 2013$.
We then add $k^2$ to both sides to get $k^2 + 2km + m^2 = 2013 + k^2$.
We then complete the square to get $(k + m)^2 = 2013 + k^2$. Because $k$ and $m$ are both integers, we get that $2013 + k^2$ is a square number. Simple guess and check reveals that $k = 14$.
Because $k$ equals $14$, therefore $m = 33$. We want $\overline{BC} = 2k + m$, so we get that $\overline{BC} = \boxed{(D)~61}$.
$\phantom{solution and diagram by bobjoe123}$ | // Block 1
unitsize(2);
import olympiad;
import graph;
pair A,B,C,D,E;
A = (0,0);
B = (70,51);
C = (97,0);
D = (82,29);
E = (76,40);
draw(Circle((0,0),86.609));
draw(A--B--C--A);
draw(A--B--E--A);
draw(A--D);
dot(A);
dot(B,blue);
dot(C);
dot(D,blue);
dot(E);
label("A",A,S);
label("B",B,NE);
label("C",C,S);
label("D",D,NE);
label("E",E,NE);
label("86",(A+B)/2,NW);
label("86",(A+D)/2,SE);
label("97",(A+C)/2,S);
label("h",(A+E)/2,N);
label("k",(E+D)/2,NE);
label("k",(B+E)/2,NE);
label("m",(C+D)/2,NE);
fill(anglemark(A,E,D,100),black);
label("$90^\circ$",anglemark(A,E,D),3*S);
// Block 2
unitsize(2); import olympiad; import graph; pair A,B,C,D,E; A = (0,0); B = (70,51); C = (97,0); D = (82,29); E = (76,40); draw(Circle((0,0),86.609)); draw(A--B--C--A); draw(A--B--E--A); draw(A--D); dot(A); dot(B,blue); dot(C); dot(D,blue); dot(E); label("A",A,S); label("B",B,NE); label("C",C,S); label("D",D,NE); label("E",E,NE); label("86",(A+B)/2,NW); label("86",(A+D)/2,SE); label("97",(A+C)/2,S); label("h",(A+E)/2,N); label("k",(E+D)/2,NE); label("k",(B+E)/2,NE); label("m",(C+D)/2,NE); fill(anglemark(A,E,D,100),black); label("$90^\circ$",anglemark(A,E,D),3*S); | [] |
745 | All $20$ diagonals are drawn in a regular octagon. At how many distinct points in the interior
of the octagon (not on the boundary) do two or more diagonals intersect?
$\textbf{(A)}\ 49\qquad\textbf{(B)}\ 65\qquad\textbf{(C)}\ 70\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 128$ | 2013 AMC 10A Problem 25 | If you draw a clear diagram like the one below, it is easy to see that there are $\boxed{\textbf{(A) }49}$ points. | // Block 1
size(14cm);
pathpen = brown + 1.337;
// Initialize octagon
pair[] A;
for (int i=0; i<8; ++i) {
A[i] = dir(45*i);
}
D(CR( (0,0), 1));
// Draw diagonals
// choose pen colors
pen[] colors;
colors[1] = orange + 1.337;
colors[2] = purple;
colors[3] = green;
colors[4] = black;
for (int d=1; d<=4; ++d) {
pathpen = colors[d];
for (int j=0; j<8; ++j) {
D(A[j]--A[(j+d) % 8]);
}
}
pathpen = blue + 2;
// Draw all the intersections
pointpen = red + 7;
for (int x1=0; x1<8; ++x1) {
for (int x2=x1+1; x2<8; ++x2) {
for (int x3=x2+1; x3<8; ++x3) {
for (int x4=x3+1; x4<8; ++x4) {
D(IP(A[x1]--A[x2], A[x3]--A[x4]));
D(IP(A[x1]--A[x3], A[x4]--A[x2]));
D(IP(A[x1]--A[x4], A[x2]--A[x3]));
}
}
}
}
// Block 2
size(14cm); pathpen = brown + 1.337; // Initialize octagon pair[] A; for (int i=0; i<8; ++i) { A[i] = dir(45*i); } D(CR( (0,0), 1)); // Draw diagonals // choose pen colors pen[] colors; colors[1] = orange + 1.337; colors[2] = purple; colors[3] = green; colors[4] = black; for (int d=1; d<=4; ++d) { pathpen = colors[d]; for (int j=0; j<8; ++j) { D(A[j]--A[(j+d) % 8]); } } pathpen = blue + 2; // Draw all the intersections pointpen = red + 7; for (int x1=0; x1<8; ++x1) { for (int x2=x1+1; x2<8; ++x2) { for (int x3=x2+1; x3<8; ++x3) { for (int x4=x3+1; x4<8; ++x4) { D(IP(A[x1]--A[x2], A[x3]--A[x4])); D(IP(A[x1]--A[x3], A[x4]--A[x2])); D(IP(A[x1]--A[x4], A[x2]--A[x3])); } } } } | [] |
745 | All $20$ diagonals are drawn in a regular octagon. At how many distinct points in the interior
of the octagon (not on the boundary) do two or more diagonals intersect?
$\textbf{(A)}\ 49\qquad\textbf{(B)}\ 65\qquad\textbf{(C)}\ 70\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 128$ | 2013 AMC 10A Problem 25 | This problem is a counting problem of combinatoric geometry. There are 2 cases for the above diagram:
Case 1: Red Dots
The red dots are the intersection of 3 or more lines. It consists of 8 dots that make up an octagon and 1 dot in the center. Hence, there are $9$ red dots.
Case 2: Blue Dots
The blue dots are the intersection of 2 lines. Each vertex of the octagon has 2 purple lines, 2 green lines, and 1 orange line coming out of it. There are 5 dots of intersection on a purple line, 6 dots on a green line, and 5 dots on an orange line. There are $2 \cdot 5+2 \cdot 6+5=27$ dots that come out of 1 vertex, which includes 7 red dots already counted. So there are $27-7 = 20$ blue dots coming out of 1 vertex. There are 8 vertices, but each blue dot is the intersection of 2 lines, corresponding to $2 \cdot 2 = 4$ vertices. So there are $\frac{20 \cdot 8}{4} = 40$ blue dots.
The number of intersection dots are the sum of the number of red and blue dots. Hence, the answer is $40 + 9 = \boxed{\textbf{(A) }49}$.
~isabelchen | // Block 1
size(8cm);
pathpen = black;
// draw the circle
pair[] A;
for (int i=0; i<8; ++i) {
A[i] = dir(45*i);
}
D(CR((0,0), 1));
// draw the octagon and diagonals
// choose pen colors
pen[] colors;
colors[1] = yellow;
colors[2] = purple;
colors[3] = green;
colors[4] = orange;
for (int d=1; d<=4; ++d) {
pathpen = colors[d];
for (int j=0; j<8; ++j) {
D(A[j]--A[(j+d) % 8]);
}
}
// draw the 3 or more line intersections
pointpen = red + 5;
D(IP(A[0]--A[4], A[2]--A[6])); // center of the circle
for (int x1=0; x1<8; ++x1) {
for (int x2=0; x2<8; ++x2) {
int y1 = (x1 + 4)%8;
int y2 = (x2 + 3)%8;
if (x1 != x2 && y1 != y2 && x1 != y1 && x1 != y2 && x2 != y1 && x2 != y2)
D(IP(A[x1]--A[y1], A[x2]--A[y2]));
}
}
// draw the 2 line intersections
pointpen = blue + 4;
for (int x1 = 0; x1 < 8; ++x1) {
int x2 = (x1 + 1)%8;
D(IP(A[x1]--A[(x1+2)%8], A[x2]--A[(x2+2)%8]));
D(IP(A[x1]--A[(x1+2)%8], A[x2]--A[(x2+3)%8]));
D(IP(A[x1]--A[(x1+2)%8], A[x2]--A[(x2+4)%8]));
D(IP(A[x1]--A[(x1+2)%8], A[x2]--A[(x2+5)%8]));
D(IP(A[x1]--A[(x1+3)%8], A[x2]--A[(x2+5)%8]));
}
// Block 2
size(8cm); pathpen = black; // draw the circle pair[] A; for (int i=0; i<8; ++i) { A[i] = dir(45*i); } D(CR((0,0), 1)); // draw the octagon and diagonals // choose pen colors pen[] colors; colors[1] = yellow; colors[2] = purple; colors[3] = green; colors[4] = orange; for (int d=1; d<=4; ++d) { pathpen = colors[d]; for (int j=0; j<8; ++j) { D(A[j]--A[(j+d) % 8]); } } // draw the 3 or more line intersections pointpen = red + 5; D(IP(A[0]--A[4], A[2]--A[6])); // center of the circle for (int x1=0; x1<8; ++x1) { for (int x2=0; x2<8; ++x2) { int y1 = (x1 + 4)%8; int y2 = (x2 + 3)%8; if (x1 != x2 && y1 != y2 && x1 != y1 && x1 != y2 && x2 != y1 && x2 != y2) D(IP(A[x1]--A[y1], A[x2]--A[y2])); } } // draw the 2 line intersections pointpen = blue + 4; for (int x1 = 0; x1 < 8; ++x1) { int x2 = (x1 + 1)%8; D(IP(A[x1]--A[(x1+2)%8], A[x2]--A[(x2+2)%8])); D(IP(A[x1]--A[(x1+2)%8], A[x2]--A[(x2+3)%8])); D(IP(A[x1]--A[(x1+2)%8], A[x2]--A[(x2+4)%8])); D(IP(A[x1]--A[(x1+2)%8], A[x2]--A[(x2+5)%8])); D(IP(A[x1]--A[(x1+3)%8], A[x2]--A[(x2+5)%8])); } | [] |
746 | In $\bigtriangleup ABC$, $AB = 86$, and $AC = 97$. A circle with center $A$ and radius $AB$ intersects $\overline{BC}$ at points $B$ and $X$. Moreover $\overline{BX}$ and $\overline{CX}$ have integer lengths. What is $BC$?
$\textbf{(A)} \ 11 \qquad \textbf{(B)} \ 28 \qquad \textbf{(C)} \ 33 \qquad \textbf{(D)} \ 61 \qquad \textbf{(E)} \ 72$ | 2013 AMC 12A Problem 19 | Solution 1 (Diophantine PoP)
Let circle $A$ intersect $AC$ at $D$ and $E$ as shown. We apply Power of a Point on point $C$ with respect to circle $A.$ This yields the diophantine equation
\[CX \cdot CB = CD \cdot CE\]
\[CX(CX+XB) = (97-86)(97+86)\]
\[CX(CX+XB) = 3 \cdot 11 \cdot 61.\]
Since lengths cannot be negative, we must have $CX+XB \ge CX.$ This generates the four solution pairs for $(CX,CX+XB)$: \[(1,2013) \qquad (3,671) \qquad (11,183) \qquad (33,61).\]
However, by the Triangle Inequality on $\triangle ACX,$ we see that $CX>13.$ This implies that we must have $CX+XB= \boxed{\textbf{(D) }61}.$
(Solution by unknown, latex/asy modified majorly by samrocksnature)
Solution 2
Let $BX = q$, $CX = p$, and $AC$ meet the circle at $Y$ and $Z$, with $Y$ on $AC$. Then $AZ = AY = 86$. Using the Power of a Point, we get that $p(p+q) = 11(183) = 11 * 3 * 61$. We know that $p+q>p$, and that $p>13$ by the triangle inequality on $\triangle ACX$. Thus, we get that $BC = p+q = \boxed{\textbf{(D) }61}$
Solution 3
Let $x$ represent $CX$, and let $y$ represent $BX$. Since the circle goes through $B$ and $X$, $AB = AX = 86$.
Then by Stewart's Theorem,
$xy(x+y) + 86^2 (x+y) = 97^2 y + 86^2 x.$
$x^2 y + xy^2 + 86^2 x + 86^2 y = 97^2 y + 86^2 x$
$x^2 + xy + 86^2 = 97^2$
(Since $y$ cannot be equal to $0$, dividing both sides of the equation by $y$ is allowed.)
$x(x+y) = (97+86)(97-86)$
$x(x+y) = 2013$
The prime factors of $2013$ are $3$, $11$, and $61$. Obviously, $x < x+y$. In addition, by the Triangle Inequality, $BC < AB + AC$, so $x+y < 183$. Therefore, $x$ must equal $33$, and $x+y$ must equal $\boxed{\textbf{(D) }61}$
Solution 4
Motivation and general line of reasoning: we use a law of cosines condition on triangle $ABX$ and triangle $ABC$ to derive some equivalent formations of the same quantity $\cos B$, which looks promising since it involves the desired length $BC$, as well as $BX$ and $CX$.An intermediate step would be to use the integer condition and pay attention to the divisors of stuff.
Let $BX$=$x_1$ and $CX$=$x_2$.
First we have
$\cos(B)=\frac{86^2+x_{1}^2-86^2}{172x_{1}}$
by applying the law of cosines to triangle $ABX$. Another equivalent formation of $\cos B$ is
$\frac{86^2+(x_{1}+x_2)^2-97^2}{172(x_{1}+x_2)}$.
Now that we have the necessary ingredients, we can make a system of equations and deduce that
$x_1=\frac{(x_1+x_2)^2+86^2-97^2}{x_1+x_2}$.
Don't lose focus by now-we try to find $x_1+x_2$, an integer value as given in the problem. To do this, we want the quantity
$\frac{(x_1+x_2)^2+86^2-97^2}{x_1+x_2}$
to 1) be an integer and 2) smaller than $x_1+x_2$. For the sake of conciseness in notation we let $M=x_1+x_2$, then $M+\frac{86^2-97^2}{M}$ is an integer. Now recalling the fact that $(a+b)(a-b)=a^2-b^2$, we get that $\frac{183(-11)}{M}$ must be an integer.
Now the prime factor decomposition of $183 \cdot -11$ is $(61)(3)(-11)$. Trying out all the possible integer values that divide this quantity, we get that the only viable option for $M$ is 61 (verify that yourself!) Therefore the answer is $\boxed{\textbf{(D) }61}$. (Solution by CreamyCream123) | //Made by samrocksnature size(8cm); pair A,B,C,D,E,X; A=(0,0); B=(-53.4,-67.4); C=(0,-97); D=(0,-86); E=(0,86); X=(-29,-81); draw(circle(A,86)); draw(E--C--B--A--X); label("$A$",A,NE); label("$B$",B,SW); label("$C$",C,S); label("$D$",D,NE); label("$E$",E,NE); label("$X$",X,dir(250)); dot(A^^B^^C^^D^^E^^X); | [] |
747 | Six points are equally spaced around a circle of radius 1. Three of these points are the vertices of a triangle that is neither equilateral nor isosceles. What is the area of this triangle?
$\textbf{(A)}\ \frac{\sqrt{3}}{3}\qquad\textbf{(B)}\ \frac{\sqrt{3}}{2}\qquad\textbf{(C)}\ \textbf{1}\qquad\textbf{(D)}\ \sqrt{2}\qquad\textbf{(E)}\ \text{2}$ | 2013 AMC 10B Problem 7 | If there are no two points on the circle that are adjacent, then the triangle would be equilateral. If the three points are all adjacent, it would be isosceles. Thus, the only possibility is two adjacent points and one point two away. Because one of the sides of this triangle is the diameter, the opposite angle is a right angle. Also, because the two adjacent angles are one sixth of the circle apart, the angle opposite them is thirty degrees. This is a $30-60-90$ triangle.
If the original six points are connected, a regular hexagon is created. This hexagon consists of six equilateral triangles, so the radius is equal to one of its side lengths. The radius is $1$, so the side opposite the thirty degree angle in the triangle is also $1$. From the properties of $30-60-90$ triangles, the area is $\frac{1 \cdot \sqrt{3}}{2}=\boxed{\textbf{(B) } \frac{\sqrt3}{2}}$ | // Block 1
unitsize(72);
draw((0,0)--(1/2,sqrt(3)/2));
draw((1/2,sqrt(3)/2)--(3/2,sqrt(3)/2));
draw((3/2,sqrt(3)/2)--(2,0));
draw((2,0)--(3/2,-sqrt(3)/2));
draw((3/2,-sqrt(3)/2)--(1/2,-sqrt(3)/2));
draw((1/2,-sqrt(3)/2)--(0,0));
draw((3/2,sqrt(3)/2)--(1/2,-sqrt(3)/2));
draw((3/2,sqrt(3)/2)--(3/2,-sqrt(3)/2));
label("2",(3/2,sqrt(3)/2)--(1/2,-sqrt(3)/2),NW);
label("$\sqrt{3}$",(3/2,sqrt(3)/2)--(3/2,-sqrt(3)/2),E);
label("1",(3/2,-sqrt(3)/2)--(1/2,-sqrt(3)/2),S);
import olympiad;
markscalefactor=0.01;
draw(rightanglemark((3/2,sqrt(3)/2),(3/2,-sqrt(3)/2),(1/2,-sqrt(3)/2)));
// Block 2
unitsize(72); draw((0,0)--(1/2,sqrt(3)/2)); draw((1/2,sqrt(3)/2)--(3/2,sqrt(3)/2)); draw((3/2,sqrt(3)/2)--(2,0)); draw((2,0)--(3/2,-sqrt(3)/2)); draw((3/2,-sqrt(3)/2)--(1/2,-sqrt(3)/2)); draw((1/2,-sqrt(3)/2)--(0,0)); draw((3/2,sqrt(3)/2)--(1/2,-sqrt(3)/2)); draw((3/2,sqrt(3)/2)--(3/2,-sqrt(3)/2)); label("2",(3/2,sqrt(3)/2)--(1/2,-sqrt(3)/2),NW); label("$\sqrt{3}$",(3/2,sqrt(3)/2)--(3/2,-sqrt(3)/2),E); label("1",(3/2,-sqrt(3)/2)--(1/2,-sqrt(3)/2),S); import olympiad; markscalefactor=0.01; draw(rightanglemark((3/2,sqrt(3)/2),(3/2,-sqrt(3)/2),(1/2,-sqrt(3)/2))); | [] |
748 | The regular octagon $ABCDEFGH$ has its center at $J$. Each of the vertices and the center are to be associated with one of the digits $1$ through $9$, with each digit used once, in such a way that the sums of the numbers on the lines $AJE$, $BJF$, $CJG$, and $DJH$ are all equal. In how many ways can this be done?
$\textbf{(A)}\ 384 \qquad\textbf{(B)}\ 576 \qquad\textbf{(C)}\ 1152 \qquad\textbf{(D)}\ 1680 \qquad\textbf{(E)}\ 3456$ | 2013 AMC 10B Problem 22 | First of all, note that $J$ must be $1$, $5$, or $9$ to preserve symmetry, since the sum of 1 to 9 is 45, and we need the remaining 8 to be divisible by 4 (otherwise we will have uneven sums). So, we have:
We also notice that $A+E = B+F = C+G = D+H$.
WLOG, assume that $J = 1$. Thus the pairs of vertices must be $9$ and $2$, $8$ and $3$, $7$ and $4$, and $6$ and $5$. There are $4! = 24$ ways to assign these to the vertices. Furthermore, there are $2^{4} = 16$ ways to switch them (i.e. do $2$ $9$ instead of $9$ $2$).
Thus, there are $16(24) = 384$ ways for each possible J value. There are $3$ possible J values that still preserve symmetry: $384(3) = \boxed{\textbf{(C) }1152}$ | // Block 1
pair A,B,C,D,E,F,G,H,J;
A=(20,20(2+sqrt(2)));
B=(20(1+sqrt(2)),20(2+sqrt(2)));
C=(20(2+sqrt(2)),20(1+sqrt(2)));
D=(20(2+sqrt(2)),20);
E=(20(1+sqrt(2)),0);
F=(20,0);
G=(0,20);
H=(0,20(1+sqrt(2)));
J=(10(2+sqrt(2)),10(2+sqrt(2)));
draw(A--B);
draw(B--C);
draw(C--D);
draw(D--E);
draw(E--F);
draw(F--G);
draw(G--H);
draw(H--A);
dot(A);
dot(B);
dot(C);
dot(D);
dot(E);
dot(F);
dot(G);
dot(H);
dot(J);
label("A",A,NNW);
label("B",B,NNE);
label("C",C,ENE);
label("D",D,ESE);
label("E",E,SSE);
label("F",F,SSW);
label("G",G,WSW);
label("H",H,WNW);
label("J $(1, 5, 9)$",J,SE);
// Block 2
pair A,B,C,D,E,F,G,H,J; A=(20,20(2+sqrt(2))); B=(20(1+sqrt(2)),20(2+sqrt(2))); C=(20(2+sqrt(2)),20(1+sqrt(2))); D=(20(2+sqrt(2)),20); E=(20(1+sqrt(2)),0); F=(20,0); G=(0,20); H=(0,20(1+sqrt(2))); J=(10(2+sqrt(2)),10(2+sqrt(2))); draw(A--B); draw(B--C); draw(C--D); draw(D--E); draw(E--F); draw(F--G); draw(G--H); draw(H--A); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); dot(G); dot(H); dot(J); label("A",A,NNW); label("B",B,NNE); label("C",C,ENE); label("D",D,ESE); label("E",E,SSE); label("F",F,SSW); label("G",G,WSW); label("H",H,WNW); label("J $(1, 5, 9)$",J,SE); | [] |
749 | Cities $A$, $B$, $C$, $D$, and $E$ are connected by roads $\widetilde{AB}$, $\widetilde{AD}$, $\widetilde{AE}$, $\widetilde{BC}$, $\widetilde{BD}$, $\widetilde{CD}$, and $\widetilde{DE}$. How many different routes are there from $A$ to $B$ that use each road exactly once? (Such a route will necessarily visit some cities more than once.)
$\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 9 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 16 \qquad \textbf{(E)}\ 18$ | 2013 AMC 12B Problem 12 | Note that cities $C$ and $E$ can be removed when counting paths because if a path goes in to $C$ or $E$, there is only one possible path to take out of cities $C$ or $E$.
So the diagram is as follows:
Now we proceed with casework. Remember that there are two ways to travel from $A$ to $D$, $D$ to $A$, $B$ to $D$ and $D$ to $B$.:
Case 1 $A \Rightarrow D$: From $D$, if the path returns to $A$, then the next path must go to $B\Rightarrow D \Rightarrow B$. There are $2 \cdot 1 \cdot 2 = 4$ possibilities of the path $ADABDB$. If the path goes to $D$ from $B$, then the path must continue with either $BDAB$ or $BADB$. There are $2 \cdot 2 \cdot 2 = 8$ possibilities. So, this case gives $4+8=12$ different possibilities.
Case 2 $A \Rightarrow B$: The path must continue with $BDADB$. There are $2 \cdot 2 = 4$ possibilities for this case.
Putting the two cases together gives $12+4 = \boxed{\textbf{(D)} \ 16}$ | // Block 1
unitsize(10mm);
defaultpen(linewidth(1.2pt)+fontsize(10pt));
dotfactor=4;
pair A=(1,0), B=(4.24,0), C=(5.24,3.08), D=(2.62,4.98), E=(0,3.08);
dot (A);
dot (B);
dot (D);
label("$A$",A,S);
label("$B$",B,SE);
label("$D$",D,N);
draw(A--B..D..cycle);
draw(A--D);
draw(B--D);
// Block 2
unitsize(10mm); defaultpen(linewidth(1.2pt)+fontsize(10pt)); dotfactor=4; pair A=(1,0), B=(4.24,0), C=(5.24,3.08), D=(2.62,4.98), E=(0,3.08); dot (A); dot (B); dot (D); label("$A$",A,S); label("$B$",B,SE); label("$D$",D,N); draw(A--B..D..cycle); draw(A--D); draw(B--D); | [] |
750 | Equilateral $\triangle ABC$ has side length $1$, and squares $ABDE$, $BCHI$, $CAFG$ lie outside the triangle. What is the area of hexagon $DEFGHI$?
$\textbf{(A)}\ \dfrac{12+3\sqrt3}4\qquad\textbf{(B)}\ \dfrac92\qquad\textbf{(C)}\ 3+\sqrt3\qquad\textbf{(D)}\ \dfrac{6+3\sqrt3}2\qquad\textbf{(E)}\ 6$ | 2014 AMC 10A Problem 13 | As seen in the previous solution, segment $GH$ is $\sqrt{3}$. Think of the picture as one large equilateral triangle, $\triangle{JKL}$ with the sides of $2\sqrt{3}+1$, by extending $EF$, $GH$, and $DI$ to points $J$, $K$, and $L$, respectively. This makes the area of $\triangle{JKL}$ $\dfrac{\sqrt{3}}{4}(2\sqrt{3}+1)^2=\dfrac{12+13\sqrt{3}}{4}$.
Triangles $\triangle{DIJ}$, $\triangle{EFK}$, and $\triangle{GHL}$ have sides of $\sqrt{3}$, so their total area is $3(\dfrac{\sqrt{3}}{4}(\sqrt{3})^2)=\dfrac{9\sqrt{3}}{4}$.
Now, you subtract their total area from the area of $\triangle{JKL}$:
$\dfrac{12+13\sqrt{3}}{4}-\dfrac{9\sqrt{3}}{4}=\dfrac{12+13\sqrt{3}-9\sqrt{3}}{4}=\dfrac{12+4\sqrt{3}}{4}=3+\sqrt{3}\implies\boxed{\textbf{(C)}\ 3+\sqrt3}$ | // Block 1
import graph;
size(10cm);
pen dps = linewidth(0.7) + fontsize(8); defaultpen(dps);
pair B = (0,0);
pair C = (1,0);
pair A = rotate(60,B)*C;
pair E = rotate(270,A)*B;
pair D = rotate(270,E)*A;
pair F = rotate(90,A)*C;
pair G = rotate(90,F)*A;
pair I = rotate(270,B)*C;
pair H = rotate(270,I)*B;
pair J = rotate(60,I)*D;
pair K = rotate(60,E)*F;
pair L = rotate(60,G)*H;
draw(A--B--C--cycle);
draw(A--E--D--B);
draw(A--F--G--C);
draw(B--I--H--C);
draw(E--F);
draw(D--I);
draw(I--H);
draw(H--G);
draw(I--J--D);
draw(E--K--F);
draw(G--L--H);
label("$A$",A,N);
label("$B$",B,SW);
label("$C$",C,SE);
label("$D$",D,W);
label("$E$",E,W);
label("$F$",F,E);
label("$G$",G,E);
label("$H$",H,SE);
label("$I$",I,SW);
label("$J$",J,SW);
label("$K$",K,N);
label("$L$",L,SE);
// Block 2
import graph; size(10cm); pen dps = linewidth(0.7) + fontsize(8); defaultpen(dps); pair B = (0,0); pair C = (1,0); pair A = rotate(60,B)*C; pair E = rotate(270,A)*B; pair D = rotate(270,E)*A; pair F = rotate(90,A)*C; pair G = rotate(90,F)*A; pair I = rotate(270,B)*C; pair H = rotate(270,I)*B; pair J = rotate(60,I)*D; pair K = rotate(60,E)*F; pair L = rotate(60,G)*H; draw(A--B--C--cycle); draw(A--E--D--B); draw(A--F--G--C); draw(B--I--H--C); draw(E--F); draw(D--I); draw(I--H); draw(H--G); draw(I--J--D); draw(E--K--F); draw(G--L--H); label("$A$",A,N); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,W); label("$E$",E,W); label("$F$",F,E); label("$G$",G,E); label("$H$",H,SE); label("$I$",I,SW); label("$J$",J,SW); label("$K$",K,N); label("$L$",L,SE); | [] |
751 | The $y$-intercepts, $P$ and $Q$, of two perpendicular lines intersecting at the point $A(6,8)$ have a sum of zero. What is the area of $\triangle APQ$?
$\textbf{(A)}\ 45\qquad\textbf{(B)}\ 48\qquad\textbf{(C)}\ 54\qquad\textbf{(D)}\ 60\qquad\textbf{(E)}\ 72$ | 2014 AMC 10A Problem 14 | Note that if the $y$-intercepts have a sum of $0$, the distance from the origin to each of the intercepts must be the same. Call this distance $a$. Since the $\angle PAQ = 90^\circ$, the length of the median to the midpoint of the hypotenuse is equal to half the length of the hypotenuse. Since the median's length is $\sqrt{6^2+8^2} = 10$, this means $a=10$, and the length of the hypotenuse is $2a = 20$. Since the $x$-coordinate of $A$ is the same as the altitude to the hypotenuse, $[APQ] = \dfrac{20 \cdot 6}{2} = \boxed{\textbf{(D)} \: 60}$.
(https://math.stackexchange.com/questions/2098721/a-property-of-the-midpoint-of-the-hypotenuse-in-a-right-triangle) | // Block 1
//Needs refining (hmm I think it's fine --bestwillcui1)
size(12cm);
fill((0,10)--(6,8)--(0,-10)--cycle,rgb(.7,.7,.7));
for(int i=-2;i<=8;i+=1)
draw((i,-12)--(i,12),grey);
for(int j=-12;j<=12;j+=1)
draw((-2,j)--(8,j),grey);
draw((-3,0)--(9,0),linewidth(1),Arrows); //x-axis
draw((0,-13)--(0,13),linewidth(1),Arrows); //y-axis
dot((0,0));
dot((6,8));
draw((-2,10.66667)--(8,7.33333),Arrows);
draw((7.33333,12)--(-0.66667,-12),Arrows);
draw((6,8)--(0,8));
draw((6,8)--(0,0));
draw(rightanglemark((0,10),(6,8),(0,-10),20));
label("$A$",(6,8),NE);
label("$a$", (0,5),W);
label("$a$",(0,-5),W);
label("$a$",(3,4),NW);
label("$P$",(0,10),SW);
label("$Q$",(0,-10),NW);
// wanted to import graph and use xaxis/yaxis but w/e
label("$x$",(9,0),E);
label("$y$",(0,13),N);
// Block 2
//Needs refining (hmm I think it's fine --bestwillcui1) size(12cm); fill((0,10)--(6,8)--(0,-10)--cycle,rgb(.7,.7,.7)); for(int i=-2;i<=8;i+=1) draw((i,-12)--(i,12),grey); for(int j=-12;j<=12;j+=1) draw((-2,j)--(8,j),grey); draw((-3,0)--(9,0),linewidth(1),Arrows); //x-axis draw((0,-13)--(0,13),linewidth(1),Arrows); //y-axis dot((0,0)); dot((6,8)); draw((-2,10.66667)--(8,7.33333),Arrows); draw((7.33333,12)--(-0.66667,-12),Arrows); draw((6,8)--(0,8)); draw((6,8)--(0,0)); draw(rightanglemark((0,10),(6,8),(0,-10),20)); label("$A$",(6,8),NE); label("$a$", (0,5),W); label("$a$",(0,-5),W); label("$a$",(3,4),NW); label("$P$",(0,10),SW); label("$Q$",(0,-10),NW); // wanted to import graph and use xaxis/yaxis but w/e label("$x$",(9,0),E); label("$y$",(0,13),N); | [] |
751 | The $y$-intercepts, $P$ and $Q$, of two perpendicular lines intersecting at the point $A(6,8)$ have a sum of zero. What is the area of $\triangle APQ$?
$\textbf{(A)}\ 45\qquad\textbf{(B)}\ 48\qquad\textbf{(C)}\ 54\qquad\textbf{(D)}\ 60\qquad\textbf{(E)}\ 72$ | 2014 AMC 10A Problem 14 | (Not to scale)
Long solution:
By rotating the right triangle, we get the figure shown where \[CD\perp EF\] and CE=CF. We know AB=CD=6 and AD=BC=8. By the pythagorean theorem, we have AC=10, and since C is the midpoint of EF, CE=EF=10 also. By similar triangles, \[\frac{AD}{FC}=\frac{DG}{CG}~\text{so}~DG=\frac{8}{3}~\text{and}~CG=\frac{10}{3}\]
Then, by more similar triangles,
\[\frac{CG}{FG}=\frac{EA}{FA}=\frac{1}{3}~\text{so}~AF=3AE\]
Then $AE=2\sqrt{10},~AE=6\sqrt{10}$, and the area is \[\boxed{(D)~60}\]
Short solution:
By rotating the right triangle, we get the figure shown where \[CD\perp EF\] and CE=CF. We know AB=CD=6 and AD=BC=8. By the pythagorean theorem, we have AC=10, and since C is the midpoint of EF, CE=EF=10 also, so CF=20.
\[A=\frac{bh}{2}=\frac{(EF)(AB)}{2}=\frac{(20)(6)}{2}=\frac{120}{2}=\boxed{(\textbf{D})~60}\]
Afly (talk) | // Block 1
unitsize(36);
pair A = (0,0);
pair B = (5,1);
pair C = (13/5,13);
pair D = C-B;
pair E = (26/5,0);
pair F = (0,26);
pair G = intersectionpoints(C--D,A--F)[0];
draw(A--E--F--A--B--C--D--A,linewidth(1));
label(A,scale(2)*"A",dir(-135));
label(E,scale(2)*"E",dir(-45));
label(F,scale(2)*"F",dir(90));
label(B,scale(2)*"B",dir(45));
label(C,scale(2)*"C",dir(45));
label(D,scale(2)*"D",dir(180));
label(G,scale(2)*"G",dir(135));
// Block 2
unitsize(36); pair A = (0,0); pair B = (5,1); pair C = (13/5,13); pair D = C-B; pair E = (26/5,0); pair F = (0,26); pair G = intersectionpoints(C--D,A--F)[0]; draw(A--E--F--A--B--C--D--A,linewidth(1)); label(A,scale(2)*"A",dir(-135)); label(E,scale(2)*"E",dir(-45)); label(F,scale(2)*"F",dir(90)); label(B,scale(2)*"B",dir(45)); label(C,scale(2)*"C",dir(45)); label(D,scale(2)*"D",dir(180)); label(G,scale(2)*"G",dir(135)); | [] |
752 | In rectangle $ABCD$, $AB=1$, $BC=2$, and points $E$, $F$, and $G$ are midpoints of $\overline{BC}$, $\overline{CD}$, and $\overline{AD}$, respectively. Point $H$ is the midpoint of $\overline{GE}$. What is the area of the shaded region?
$\textbf{(A)}\ \dfrac1{12}\qquad\textbf{(B)}\ \dfrac{\sqrt3}{18}\qquad\textbf{(C)}\ \dfrac{\sqrt2}{12}\qquad\textbf{(D)}\ \dfrac{\sqrt3}{12}\qquad\textbf{(E)}\ \dfrac16$ | 2014 AMC 10A Problem 16 | Denote $D=(0,0)$. Then $A= (0,2), F = \left(\frac12,0\right), H = \left(\frac12,1\right)$. Let the intersection of $AF$ and $DH$ be $X$, and the intersection of $BF$ and $CH$ be $Y$. Then we want to find the coordinates of $X$ so we can find $XY$. From our points, the slope of $AF$ is $\bigg(\dfrac{-2}{\tfrac12}\bigg) = -4$, and its $y$-intercept is just $2$. Thus the equation for $AF$ is $y = -4x + 2$. We can also quickly find that the equation of $DH$ is $y = 2x$. Setting the equations equal, we have $2x = -4x +2 \implies x = \frac13$. Because of symmetry, we can see that the distance from $Y$ to $BC$ is also $\frac13$, so $XY = 1 - 2 \cdot \frac13 = \frac13$. Now the area of the kite is simply the product of the two diagonals over $2$. Since the length $HF = 1$, our answer is $\dfrac{\dfrac{1}{3} \cdot 1}{2} = \boxed{\textbf{(E)} \: \dfrac16}$. | // Block 1
import graph;
size(9cm);
pen dps = fontsize(10); defaultpen(dps);
pair D = (0,0);
pair F = (1/2,0);
pair C = (1,0);
pair G = (0,1);
pair E = (1,1);
pair A = (0,2);
pair B = (1,2);
pair H = (1/2,1);
// do not look
pair X = (1/3,2/3);
pair Y = (2/3,2/3);
draw(A--B--C--D--cycle);
draw(G--E);
draw(A--F--B);
draw(D--H--C);
filldraw(H--X--F--Y--cycle,grey);
draw(X--Y,dashed);
label("$A\: (0,2)$",A,NW);
label("$B$",B,NE);
label("$C$",C,SE);
label("$D \: (0,0)$",D,SW);
label("$E$",E,E);
label("$F\: (\frac12,0)$",F,S);
label("$G$",G,W);
label("$H \: (\frac12,1)$",H,N);
label("$Y$",Y,E);
label("$X$",X,W);
label("$\frac12$",(0.25,0),S);
label("$\frac12$",(0.75,0),S);
label("$1$",(1,0.5),E);
label("$1$",(1,1.5),E);
// Block 2
import graph; size(9cm); pen dps = fontsize(10); defaultpen(dps); pair D = (0,0); pair F = (1/2,0); pair C = (1,0); pair G = (0,1); pair E = (1,1); pair A = (0,2); pair B = (1,2); pair H = (1/2,1); // do not look pair X = (1/3,2/3); pair Y = (2/3,2/3); draw(A--B--C--D--cycle); draw(G--E); draw(A--F--B); draw(D--H--C); filldraw(H--X--F--Y--cycle,grey); draw(X--Y,dashed); label("$A\: (0,2)$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D \: (0,0)$",D,SW); label("$E$",E,E); label("$F\: (\frac12,0)$",F,S); label("$G$",G,W); label("$H \: (\frac12,1)$",H,N); label("$Y$",Y,E); label("$X$",X,W); label("$\frac12$",(0.25,0),S); label("$\frac12$",(0.75,0),S); label("$1$",(1,0.5),E); label("$1$",(1,1.5),E); | [] |
752 | In rectangle $ABCD$, $AB=1$, $BC=2$, and points $E$, $F$, and $G$ are midpoints of $\overline{BC}$, $\overline{CD}$, and $\overline{AD}$, respectively. Point $H$ is the midpoint of $\overline{GE}$. What is the area of the shaded region?
$\textbf{(A)}\ \dfrac1{12}\qquad\textbf{(B)}\ \dfrac{\sqrt3}{18}\qquad\textbf{(C)}\ \dfrac{\sqrt2}{12}\qquad\textbf{(D)}\ \dfrac{\sqrt3}{12}\qquad\textbf{(E)}\ \dfrac16$ | 2014 AMC 10A Problem 16 | The area of the shaded area is the area of $\triangle DHC$ minus the two triangles on the side.
Extend $\overline{DH}$ so that it hits point $B$. Call the intersection of $\overline{AF}$ and $\overline{DB}$ point $P$. \[\triangle APB \sim \triangle FPD\]
Drop altitudes from $P$ down to $\overline{DF}$ and $\overline{AB}$; call the intersection points $L$ and $M$ respectively.
\[\frac{DF}{AB}=\frac{LP}{PM}=\frac{\frac{1}{2}}{1}=\frac{1}{2}\]
\[LP=\frac{1}{3}\cdot LM=\frac{2}{3}\]
Thus the two triangles on the side have area $\frac{1}{2} \cdot \frac{2}{2} \cdot \frac {1}{3} = \frac{1}{6}$. Since there are two, their total area is $2 \cdot \frac{1}{6} = \frac{1}{3}$. The area of $\triangle DHC$ is $\frac{1}{2} \cdot 1 \cdot 1=\frac{1}{2}$. The shaded region is $\frac{1}{2}-\frac{1}{3} = \frac{1}{6}$ which is $\boxed{\textbf{(E)} \frac{1}{6}}$.
~JH. L :) | // Block 1
import graph;
size(9cm);
pen dps = fontsize(10); defaultpen(dps);
pair D = (0,0);
pair F = (1/2,0);
pair C = (1,0);
pair G = (0,1);
pair E = (1,1);
pair A = (0,2);
pair B = (1,2);
pair H = (1/2,1);
// do not look
pair X = (1/3,2/3);
pair Y = (2/3,2/3);
draw(A--B--C--D--cycle);
draw(G--E);
draw(A--F--B);
draw(D--H--C);
filldraw(H--X--F--Y--cycle,grey);
label("$A$",A,NW);
label("$B$",B,NE);
label("$C$",C,SE);
label("$D$",D,SW);
label("$E$",E,E);
label("$F$",F,S);
label("$G$",G,W);
label("$H$",H,N);
label("$\frac12$",(0.25,0),S);
label("$\frac12$",(0.75,0),S);
label("$1$",(1,0.5),E);
label("$1$",(1,1.5),E);
// Block 2
import graph; size(9cm); pen dps = fontsize(10); defaultpen(dps); pair D = (0,0); pair F = (1/2,0); pair C = (1,0); pair G = (0,1); pair E = (1,1); pair A = (0,2); pair B = (1,2); pair H = (1/2,1); // do not look pair X = (1/3,2/3); pair Y = (2/3,2/3); draw(A--B--C--D--cycle); draw(G--E); draw(A--F--B); draw(D--H--C); filldraw(H--X--F--Y--cycle,grey); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$E$",E,E); label("$F$",F,S); label("$G$",G,W); label("$H$",H,N); label("$\frac12$",(0.25,0),S); label("$\frac12$",(0.75,0),S); label("$1$",(1,0.5),E); label("$1$",(1,1.5),E); | [] |
753 | Four cubes with edge lengths $1$, $2$, $3$, and $4$ are stacked as shown. What is the length of the portion of $\overline{XY}$ contained in the cube with edge length $3$?
$\textbf{(A)}\ \dfrac{3\sqrt{33}}5\qquad\textbf{(B)}\ 2\sqrt3\qquad\textbf{(C)}\ \dfrac{2\sqrt{33}}3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 3\sqrt2$ | 2014 AMC 10A Problem 19 | Let's redraw the diagram, however make a 3D coordinate plane, using D as the origin.
Now we can use the distance formula in 3D, which is $\sqrt{(x_{1}-x_{2})^2+(y_{1}-y_{2})^2+(z_{1}-z_{2})^2}$ and plug it in for the distance of $XY$.
$\sqrt{(0-4)^2+(10-0)^2+(0-4)^2}$
We get the answer as $\sqrt{132} = 2\sqrt{33}$.
Continuing with solution 1, using similar triangles, we get the answer as $\textbf{(A)}\ \dfrac{3\sqrt{33}}5$
~ghfhgvghj10 | // Block 1
dotfactor = 3;
size(10cm);
dot((0, 10));
label("$X(0,10,0)$", (0,10),W,fontsize(8pt));
dot((6,2));
label("$Y(4,0,4)$", (6,2),E,fontsize(8pt));
draw((0, 0)--(0, 10)--(1, 10)--(1, 9)--(2, 9)--(2, 7)--(3, 7)--(3,4)--(4, 4)--(4, 0)--cycle);
draw((0,9)--(1, 9)--(1.5, 9.5)--(1.5, 10.5)--(0.5, 10.5)--(0, 10));
draw((1, 10)--(1.5,10.5));
draw((1.5, 10)--(3,10)--(3,8)--(2,7)--(0,7));
draw((2,9)--(3,10));
draw((3,8.5)--(4.5,8.5)--(4.5,5.5)--(3,4)--(0,4));
draw((3,7)--(4.5,8.5));
draw((4.5,6)--(6,6)--(6,2)--(4,0));
draw((4,4)--(6,6));
label("$1$", (1,9.5), W,fontsize(8pt));
label("$2$", (2,8), W,fontsize(8pt));
label("$3$", (3,5.5), W,fontsize(8pt));
label("$4$", (4,2), W,fontsize(8pt));
label("$D(0,0,0)$", (0,0), W,fontsize(8pt));
// Block 2
dotfactor = 3; size(10cm); dot((0, 10)); label("$X(0,10,0)$", (0,10),W,fontsize(8pt)); dot((6,2)); label("$Y(4,0,4)$", (6,2),E,fontsize(8pt)); draw((0, 0)--(0, 10)--(1, 10)--(1, 9)--(2, 9)--(2, 7)--(3, 7)--(3,4)--(4, 4)--(4, 0)--cycle); draw((0,9)--(1, 9)--(1.5, 9.5)--(1.5, 10.5)--(0.5, 10.5)--(0, 10)); draw((1, 10)--(1.5,10.5)); draw((1.5, 10)--(3,10)--(3,8)--(2,7)--(0,7)); draw((2,9)--(3,10)); draw((3,8.5)--(4.5,8.5)--(4.5,5.5)--(3,4)--(0,4)); draw((3,7)--(4.5,8.5)); draw((4.5,6)--(6,6)--(6,2)--(4,0)); draw((4,4)--(6,6)); label("$1$", (1,9.5), W,fontsize(8pt)); label("$2$", (2,8), W,fontsize(8pt)); label("$3$", (3,5.5), W,fontsize(8pt)); label("$4$", (4,2), W,fontsize(8pt)); label("$D(0,0,0)$", (0,0), W,fontsize(8pt)); | [] |
754 | A rectangular piece of paper whose length is $\sqrt3$ times the width has area $A$. The paper is divided into three equal sections along the opposite lengths, and then a dotted line is drawn from the first divider to the second divider on the opposite side as shown. The paper is then folded flat along this dotted line to create a new shape with area $B$. What is the ratio $\frac{B}{A}$?
$\textbf{(A)}\ \frac{1}{2}\qquad\textbf{(B)}\ \frac{3}{5}\qquad\textbf{(C)}\ \frac{2}{3}\qquad\textbf{(D)}\ \frac{3}{4}\qquad\textbf{(E)}\ \frac{4}{5}$ | 2014 AMC 10A Problem 23 | Find the midpoint of the dotted line. Draw a line perpendicular to it. From the point this line intersects the top of the paper, draw lines to each endpoint of the dotted line. These two lines plus the dotted line form a triangle which is the double-layered portion of the folded paper. WLOG, assume the width of the paper is $1$ and the length is $\sqrt{3}$. The triangle we want to find has side lengths $\dfrac{2\sqrt{3}}{3}$, $\sqrt{\left(\dfrac{\sqrt{3}}{3}\right)^{2}+1}=\dfrac{2\sqrt{3}}{3}$, and $\sqrt{\left(\dfrac{\sqrt{3}}{3}\right)^{2}+1}=\dfrac{2\sqrt{3}}{3}$. It is an equilateral triangle with height $\dfrac{\sqrt{3}}{3}\cdot\sqrt{3}=1$, and area $\dfrac{\dfrac{2\sqrt{3}}{3}\cdot1}{2}=\dfrac{\sqrt{3}}{3}$. The area of the paper is $1\cdot\sqrt{3}=\sqrt{3}$, and the folded paper has area $\sqrt{3}-\dfrac{\sqrt{3}}{3}=\dfrac{2\sqrt{3}}{3}$. The ratio of the area of the folded paper to that of the original paper is thus $\boxed{\textbf{(C)} \: 2:3}$ | // Block 1
import graph;
unitsize(3cm);
real L = 0.05;
pair A = (0,0);
pair B = (sqrt(3),0);
pair C = (sqrt(3),1);
pair D = (0,1);
pair X1 = (sqrt(3)/3,0);
pair X2= (2*sqrt(3)/3,0);
pair Y1 = (2*sqrt(3)/3,1);
pair Y2 = (sqrt(3)/3,1);
dot(X1);
dot(Y1);
draw(A--B--C--D--cycle, linewidth(2));
draw(B--D,dashed);
draw(X1--Y1,dashed);
draw(Y2--X1--D, dotted);
draw(X2--Y1--B, dotted);
// Block 2
import graph; unitsize(3cm); real L = 0.05; pair A = (0,0); pair B = (sqrt(3),0); pair C = (sqrt(3),1); pair D = (0,1); pair X1 = (sqrt(3)/3,0); pair X2= (2*sqrt(3)/3,0); pair Y1 = (2*sqrt(3)/3,1); pair Y2 = (sqrt(3)/3,1); dot(X1); dot(Y1); draw(A--B--C--D--cycle, linewidth(2)); draw(B--D,dashed); draw(X1--Y1,dashed); draw(Y2--X1--D, dotted); draw(X2--Y1--B, dotted); | [] |
754 | A rectangular piece of paper whose length is $\sqrt3$ times the width has area $A$. The paper is divided into three equal sections along the opposite lengths, and then a dotted line is drawn from the first divider to the second divider on the opposite side as shown. The paper is then folded flat along this dotted line to create a new shape with area $B$. What is the ratio $\frac{B}{A}$?
$\textbf{(A)}\ \frac{1}{2}\qquad\textbf{(B)}\ \frac{3}{5}\qquad\textbf{(C)}\ \frac{2}{3}\qquad\textbf{(D)}\ \frac{3}{4}\qquad\textbf{(E)}\ \frac{4}{5}$ | 2014 AMC 10A Problem 23 | Our original paper can be divided like this:
After the fold across the dashed line, our paper becomes:
Since our original sheet of paper has six congruent $30-60-90$ triangles and and our new one has four, the ratio of the area $B:A$ is equal to $4:6\implies \boxed{\textbf{(C)} \: 2:3}$
~CHECKMATE2021 | // Block 1
import graph;
unitsize(3cm);
real L = 0.05;
pair A = (0,0);
pair B = (sqrt(3),0);
pair C = (sqrt(3),1);
pair D = (0,1);
pair X1 = (sqrt(3)/3,0);
pair X2= (2*sqrt(3)/3,0);
pair Y1 = (2*sqrt(3)/3,1);
pair Y2 = (sqrt(3)/3,1);
dot(X1);
dot(Y1);
draw(A--B--C--D--cycle, linewidth(2));
draw(X1--Y1,dashed);
draw(Y2--X1--D, dotted);
draw(X2--Y1--B, dotted);
// Block 2
import graph;
unitsize(3cm);
real L = 0.05;
pair A = (0,0);
pair D = (0,1);
pair X1 = (sqrt(3)/3,0);
pair X2 = (sqrt(3)/6,0.5);
pair Y1 = (2*sqrt(3)/3,1);
pair Y2 = (sqrt(3)/3,1);
pair Z1 = (sqrt(3)/2,1.5);
dot(X1);
dot(Y1);
draw(X1--A--D--Z1--Y1, linewidth(2));
draw(X1--D--Y1);
draw(X1--Y1, dashed);
draw(Y2--X1,dotted);
draw(X2--((sqrt(3)/6 + L/sqrt(3)),(0.5+L/2)));
draw(Y2--(sqrt(3)/3,1-L));
// Block 3
import graph; unitsize(3cm); real L = 0.05; pair A = (0,0); pair B = (sqrt(3),0); pair C = (sqrt(3),1); pair D = (0,1); pair X1 = (sqrt(3)/3,0); pair X2= (2*sqrt(3)/3,0); pair Y1 = (2*sqrt(3)/3,1); pair Y2 = (sqrt(3)/3,1); dot(X1); dot(Y1); draw(A--B--C--D--cycle, linewidth(2)); draw(X1--Y1,dashed); draw(Y2--X1--D, dotted); draw(X2--Y1--B, dotted);
// Block 4
import graph; unitsize(3cm); real L = 0.05; pair A = (0,0); pair D = (0,1); pair X1 = (sqrt(3)/3,0); pair X2 = (sqrt(3)/6,0.5); pair Y1 = (2*sqrt(3)/3,1); pair Y2 = (sqrt(3)/3,1); pair Z1 = (sqrt(3)/2,1.5); dot(X1); dot(Y1); draw(X1--A--D--Z1--Y1, linewidth(2)); draw(X1--D--Y1); draw(X1--Y1, dashed); draw(Y2--X1,dotted); draw(X2--((sqrt(3)/6 + L/sqrt(3)),(0.5+L/2))); draw(Y2--(sqrt(3)/3,1-L)); | [] |
755 | In $\triangle BAC$, $\angle BAC=40^\circ$, $AB=10$, and $AC=6$. Points $D$ and $E$ lie on $\overline{AB}$ and $\overline{AC}$ respectively. What is the minimum possible value of $BE+DE+CD$?
$\textbf{(A) }6\sqrt 3+3\qquad \textbf{(B) }\dfrac{27}2\qquad \textbf{(C) }8\sqrt 3\qquad \textbf{(D) }14\qquad \textbf{(E) }3\sqrt 3+9\qquad$ | 2014 AMC 12A Problem 20 | (Diagram by shihan)
Reflect $C$ across $AB$ to $C'$. Similarly, reflect $B$ across $AC$ to $B'$. Clearly, $BE = B'E$ and $CD = C'D$. Thus, the sum $BE + DE + CD = B'E + DE + C'D$. This value is minimized when $B'$, $C'$, $D$ and $E$ are collinear. To finish, we use the law of cosines on the triangle $AB'C'$: $B'C' = \sqrt{6^2 + 10^2 - 2(6)(10)\cos 120^{\circ}} = \boxed{\textbf{(D) } 14}.$
Note (note by JiYang): $\quad \because \angle BAC \textless \frac{\pi}{3}$ $\qquad \therefore \angle C'AB' \textless \pi$ $\qquad \therefore$ the line C'B' would pass through $\Delta ABC$ | // Block 1
size(300);
defaultpen(linewidth(0.4)+fontsize(10));
pen s = linewidth(0.8)+fontsize(8);
pair A,B,C,D,Ep,Bp,Cp;
A = (0,0);
B = 10*dir(-110);C = 6*dir(-70);
Bp = 10*dir(-30);Cp = 6*dir(-150);
D = IP(Cp--Bp, A--B); Ep = IP(Cp--Bp, A--C);
draw(A--B--C--A--Cp--Bp--A);
draw(Cp--B);
draw(C--Bp);
draw(C--D);
draw(B--Ep);
dot("$A$", A, N);
dot("$B$", B, SW);
dot("$C$", C, SE);
dot("$B'$", Bp, E);
dot("$C'$", Cp, W);
dot("$D$", D, dir(-70));
dot("$E$", Ep, dir(60));
MA("40^\circ",Cp,A,D, 1);
MA("40^\circ",D,A,Ep, 1);
MA("40^\circ",Ep,A,Bp, 1);
label("$6$", A--Cp);
label("$10$", Bp--A);
// Block 2
size(300); defaultpen(linewidth(0.4)+fontsize(10)); pen s = linewidth(0.8)+fontsize(8); pair A,B,C,D,Ep,Bp,Cp; A = (0,0); B = 10*dir(-110);C = 6*dir(-70); Bp = 10*dir(-30);Cp = 6*dir(-150); D = IP(Cp--Bp, A--B); Ep = IP(Cp--Bp, A--C); draw(A--B--C--A--Cp--Bp--A); draw(Cp--B); draw(C--Bp); draw(C--D); draw(B--Ep); dot("$A$", A, N); dot("$B$", B, SW); dot("$C$", C, SE); dot("$B'$", Bp, E); dot("$C'$", Cp, W); dot("$D$", D, dir(-70)); dot("$E$", Ep, dir(60)); MA("40^\circ",Cp,A,D, 1); MA("40^\circ",D,A,Ep, 1); MA("40^\circ",Ep,A,Bp, 1); label("$6$", A--Cp); label("$10$", Bp--A); | [] |
756 | Let $f_0(x)=x+|x-100|-|x+100|$, and for $n\geq 1$, let $f_n(x)=|f_{n-1}(x)|-1$. For how many values of $x$ is $f_{100}(x)=0$?
$\textbf{(A) }299\qquad \textbf{(B) }300\qquad \textbf{(C) }301\qquad \textbf{(D) }302\qquad \textbf{(E) }303\qquad$ | 2014 AMC 12A Problem 24 | 1. Draw the graph of $f_0(x)$ by dividing the domain into three parts.
2. Apply the recursive rule a few times to find the pattern.
Note: $f_n(x) = |f_{n-1}(x)| - 10$ is used to enlarge the difference, but the reasoning is the same.
3. Extrapolate to $f_{100}$. Notice that the summits start $100$ away from $0$ and get $1$ closer each iteration, so they reach $0$ exactly at $f_{100}$.
$f_{100}(x)$ reaches $0$ at $x = -300$, then zigzags between $0$ and $-1$, hitting $0$ at every even $x$, before leaving $0$ at $x = 300$.
This means that $f_{100}(x) = 0$ at all even $x$ where $-300 \le x \le 300$. This is a $601$-integer odd-size range with even numbers at the endpoints, so just over half of the integers are even, or $\frac{601+1}{2} = \boxed{\textbf{(C) }301}$.
(Revised by Flamedragon & Jason,C & emerald_block) | // Block 1
unitsize(0);
int w = 250;
int h = 125;
xaxis(-w,w,Ticks(100.0),Arrows);
yaxis(-h,h,Ticks(100.0),Arrows);
draw((-100,-h)--(-100,h),dashed);
draw((100,-h)--(100,h),dashed);
real f0(real x) { return x + abs(x-100) - abs(x+100); }
draw(graph(f0,-w,w),Arrows);
label("$f_0$",(-w,f0(-w)),W);
// Block 2
unitsize(0);
int w = 350;
int h = 125;
xaxis(-w,w,Ticks(100.0),Arrows);
yaxis(-h,h,Ticks(100.0),Arrows);
draw((-100,-h)--(-100,h),dashed);
draw((100,-h)--(100,h),dashed);
int s = 10;
real f0(real x) { return x + abs(x-100) - abs(x+100); }
real f(real x, int k) { real a = abs(f0(x))-s*k; return a >= -s ? a : k%2 == 0 ? abs(x%(2*s)-s)-s : -abs(x%(2*s)-s); }
real f1(real x) { return f(x,1); }
real f2(real x) { return f(x,2); }
real f3(real x) { return f(x,3); }
real f4(real x) { return f(x,4); }
draw(graph(f0,-w,w,w*2#s),Arrows);
draw(graph(f1,-w,w,w*2#s),red,Arrows);
draw(graph(f2,-w,w,w*2#s),orange,Arrows);
draw(graph(f3,-w,w,w*2#s),lightolive,Arrows);
label("$f_0$",(-w,f0(-w)),W);
label("$f_1$",(-w,f1(-w)),NW,red);
label("$f_2$",(-w,f2(-w)),W,orange);
label("$f_3$",(-w,f3(-w)),SW,lightolive);
// Block 3
unitsize(0);
int w = 350;
int h = 125;
xaxis(-w,w,Ticks(100.0),Arrows);
yaxis(-h,h,Ticks(100.0),Arrows);
draw((-100,-h)--(-100,h),dashed);
draw((100,-h)--(100,h),dashed);
int s = 10;
real f0(real x) { return x + abs(x-100) - abs(x+100); }
real f(real x, int k) { real a = abs(f0(x))-s*k; return a >= -s ? a : k%2 == 0 ? abs(x%(2*s)-s)-s : -abs(x%(2*s)-s); }
real f1(real x) { return f(x,1); }
real f2(real x) { return f(x,2); }
real f3(real x) { return f(x,3); }
real f4(real x) { return f(x,4); }
real f98(real x) { return f(x,100#s-2); }
real f99(real x) { return f(x,100#s-1); }
real f100(real x) { return f(x,100#s); }
draw(graph(f0,-w,w,w*2#s),Arrows);
draw(graph(f1,-w,w,w*2#s),red,Arrows);
draw(graph(f2,-w,w,w*2#s),orange,Arrows);
draw(graph(f3,-w,w,w*2#s),lightolive,Arrows);
draw(graph(f98,-w,w,w*2#s),heavygreen,Arrows);
draw(graph(f99,-w,w,w*2#s),blue,Arrows);
draw(graph(f100,-w,w,w*2#s),purple,Arrows);
label("$f_0$",(-w,f0(-w)),W);
label("$f_1$",(-w,f1(-w)),NW,red);
label("$f_2$",(-w,f2(-w)),W,orange);
label("$f_3$",(-w,f3(-w)),SW,lightolive);
label("$f_{98}$",(-w,f98(-w)),NW,heavygreen);
label("$f_{99}$",(-w,f99(-w)),W,blue);
label("$f_{100}$",(-w,f100(-w)),SW,purple);
// Block 4
unitsize(0); int w = 250; int h = 125; xaxis(-w,w,Ticks(100.0),Arrows); yaxis(-h,h,Ticks(100.0),Arrows); draw((-100,-h)--(-100,h),dashed); draw((100,-h)--(100,h),dashed); real f0(real x) { return x + abs(x-100) - abs(x+100); } draw(graph(f0,-w,w),Arrows); label("$f_0$",(-w,f0(-w)),W);
// Block 5
unitsize(0); int w = 350; int h = 125; xaxis(-w,w,Ticks(100.0),Arrows); yaxis(-h,h,Ticks(100.0),Arrows); draw((-100,-h)--(-100,h),dashed); draw((100,-h)--(100,h),dashed); int s = 10; real f0(real x) { return x + abs(x-100) - abs(x+100); } real f(real x, int k) { real a = abs(f0(x))-s*k; return a >= -s ? a : k%2 == 0 ? abs(x%(2*s)-s)-s : -abs(x%(2*s)-s); } real f1(real x) { return f(x,1); } real f2(real x) { return f(x,2); } real f3(real x) { return f(x,3); } real f4(real x) { return f(x,4); } draw(graph(f0,-w,w,w*2#s),Arrows); draw(graph(f1,-w,w,w*2#s),red,Arrows); draw(graph(f2,-w,w,w*2#s),orange,Arrows); draw(graph(f3,-w,w,w*2#s),lightolive,Arrows); label("$f_0$",(-w,f0(-w)),W); label("$f_1$",(-w,f1(-w)),NW,red); label("$f_2$",(-w,f2(-w)),W,orange); label("$f_3$",(-w,f3(-w)),SW,lightolive);
// Block 6
unitsize(0); int w = 350; int h = 125; xaxis(-w,w,Ticks(100.0),Arrows); yaxis(-h,h,Ticks(100.0),Arrows); draw((-100,-h)--(-100,h),dashed); draw((100,-h)--(100,h),dashed); int s = 10; real f0(real x) { return x + abs(x-100) - abs(x+100); } real f(real x, int k) { real a = abs(f0(x))-s*k; return a >= -s ? a : k%2 == 0 ? abs(x%(2*s)-s)-s : -abs(x%(2*s)-s); } real f1(real x) { return f(x,1); } real f2(real x) { return f(x,2); } real f3(real x) { return f(x,3); } real f4(real x) { return f(x,4); } real f98(real x) { return f(x,100#s-2); } real f99(real x) { return f(x,100#s-1); } real f100(real x) { return f(x,100#s); } draw(graph(f0,-w,w,w*2#s),Arrows); draw(graph(f1,-w,w,w*2#s),red,Arrows); draw(graph(f2,-w,w,w*2#s),orange,Arrows); draw(graph(f3,-w,w,w*2#s),lightolive,Arrows); draw(graph(f98,-w,w,w*2#s),heavygreen,Arrows); draw(graph(f99,-w,w,w*2#s),blue,Arrows); draw(graph(f100,-w,w,w*2#s),purple,Arrows); label("$f_0$",(-w,f0(-w)),W); label("$f_1$",(-w,f1(-w)),NW,red); label("$f_2$",(-w,f2(-w)),W,orange); label("$f_3$",(-w,f3(-w)),SW,lightolive); label("$f_{98}$",(-w,f98(-w)),NW,heavygreen); label("$f_{99}$",(-w,f99(-w)),W,blue); label("$f_{100}$",(-w,f100(-w)),SW,purple); | [] |
756 | Let $f_0(x)=x+|x-100|-|x+100|$, and for $n\geq 1$, let $f_n(x)=|f_{n-1}(x)|-1$. For how many values of $x$ is $f_{100}(x)=0$?
$\textbf{(A) }299\qquad \textbf{(B) }300\qquad \textbf{(C) }301\qquad \textbf{(D) }302\qquad \textbf{(E) }303\qquad$ | 2014 AMC 12A Problem 24 | Note $f_{100}(x) = 0$ when $|f_{99}(x)| -1$ = 0. This occurs when $f_{99}(x) = \pm 1$.
Then, repeating this process, we note $f_{99}(x) = \pm 1 \implies |f_{98}(x)| = 0, 2$, and hence $f_{98}(x) = 0, \pm 2$.
Similarly, $f_{97}(x) = \pm 1, \pm 3$. Extrapolating this pattern, we must have $f_{0}(x) = 0, \pm 2$, $\dots$, $\pm 100$. Then, drawing the graph of $f_0$,
we note for each of $0, \pm 2$, $\dots$, $\pm 98$, there are three solutions. For $\pm 100$, there is exactly $2$ solutions.
So, the total amount of solutions is $99 \cdot 3 + 2 \cdot 2 = \boxed{\textbf{(C) }301}$ | // Block 1
unitsize(0);
int w = 250;
int h = 125;
xaxis(-w,w,Ticks(100.0),Arrows);
yaxis(-h,h,Ticks(100.0),Arrows);
real f0(real x) { return x + abs(x-100) - abs(x+100); }
draw(graph(f0,-w,w),Arrows);
label("$f_0$",(-w,f0(-w)),W);
// Block 2
unitsize(0); int w = 250; int h = 125; xaxis(-w,w,Ticks(100.0),Arrows); yaxis(-h,h,Ticks(100.0),Arrows); real f0(real x) { return x + abs(x-100) - abs(x+100); } draw(graph(f0,-w,w),Arrows); label("$f_0$",(-w,f0(-w)),W); | [] |
757 | Two concentric circles have radii $1$ and $2$. Two points on the outer circle are chosen independently and uniformly at random. What is the probability that the chord joining the two points intersects the inner circle?
$\textbf{(A)}\ \frac{1}{6}\qquad \textbf{(B)}\ \frac{1}{4}\qquad \textbf{(C)}\ \frac{2-\sqrt{2}}{2}\qquad \textbf{(D)}\ \frac{1}{3}\qquad \textbf{(E)}\ \frac{1}{2}\qquad$ | 2014 AMC 10B Problem 19 | Let the center of the two circles be $O$. Now pick an arbitrary point $A$ on the boundary of the circle with radius $2$. We want to find the range of possible places for the second point, $A'$, such that $AA'$ passes through the circle of radius $1$. To do this, first draw the tangents from $A$ to the circle of radius $1$. Let the intersection points of the tangents (when extended) with circle of radius $2$ be $B$ and $C$. Let $H$ be the foot of the altitude from $O$ to $\overline{BC}$. Then we have the following diagram.
We want to find $\angle BOC$, as the range of desired points $A'$ is the set of points on minor arc $\overarc{BC}$. This is because $B$ and $C$ are part of the tangents, which "set the boundaries" for $A'$. Since $OH = 1$ and $OB = 2$ as shown in the diagram, $\triangle OHB$ is a $30-60-90$ triangle with $\angle BOH = 60^\circ$. Thus, $\angle BOC = 120^\circ$, and the probability $A'$ lies on the minor arc $\overarc{BC}$ is thus $\dfrac{120}{360} = \boxed{\textbf{(D)}\: \dfrac13}$. | // Block 1
scale(200);
pair A,O,B,C,H;
A = (0,1);
O = (0,0);
B = (-.866,-.5);
C = (.866,-.5);
H = (0, -.5);
draw(A--C--cycle);
draw(A--O--cycle);
draw(O--C--cycle);
draw(O--H,dashed+linewidth(.7));
draw(A--B--cycle);
draw(B--C--cycle);
draw(O--B--cycle);
dot("$A$",A,N);
dot("$O$",O,NW);
dot("$B$",B,W);
dot("$C$",C,E);
dot("$H$",H,S);
label("$2$",O--(-.7,-.385),N);
label("$1$",O--H,E);
draw(circle(O,.5));
draw(circle(O,1));
// Block 2
scale(200); pair A,O,B,C,H; A = (0,1); O = (0,0); B = (-.866,-.5); C = (.866,-.5); H = (0, -.5); draw(A--C--cycle); draw(A--O--cycle); draw(O--C--cycle); draw(O--H,dashed+linewidth(.7)); draw(A--B--cycle); draw(B--C--cycle); draw(O--B--cycle); dot("$A$",A,N); dot("$O$",O,NW); dot("$B$",B,W); dot("$C$",C,E); dot("$H$",H,S); label("$2$",O--(-.7,-.385),N); label("$1$",O--H,E); draw(circle(O,.5)); draw(circle(O,1)); | [] |
758 | Trapezoid $ABCD$ has parallel sides $\overline{AB}$ of length $33$ and $\overline {CD}$ of length $21$. The other two sides are of lengths $10$ and $14$. The angles $A$ and $B$ are acute. What is the length of the shorter diagonal of $ABCD$?
$\textbf{(A) }10\sqrt{6}\qquad\textbf{(B) }25\qquad\textbf{(C) }8\sqrt{10}\qquad\textbf{(D) }18\sqrt{2}\qquad\textbf{(E) }26$ | 2014 AMC 10B Problem 21 | In the diagram, $\overline{DE} \perp \overline{AB}, \overline{FC} \perp \overline{AB}$.
Denote $\overline{AE} = x$ and $\overline{DE} = h$. In right triangle $AED$, we have from the Pythagorean theorem: $x^2+h^2=100$. Note that since $EF = DC$, we have $BF = 33-DC-x = 12-x$. Using the Pythagorean theorem in right triangle $BFC$, we have $(12-x)^2 + h^2 = 196$.
We isolate the $h^2$ term in both equations, getting $h^2= 100-x^2$ and
$h^2 = 196-(12-x)^2$.
Setting these equal, we have $100-x^2 = 196 - 144 + 24x -x^2 \implies 24x = 48 \implies x = 2$. Now, we can determine that $h^2 = 100-4 \implies h = 4\sqrt{6}$.
The two diagonals are $\overline{AC}$ and $\overline{BD}$. Using the Pythagorean theorem again on $\bigtriangleup AFC$ and $\bigtriangleup BED$, we can find these lengths to be $\sqrt{96+529} = 25$ and $\sqrt{96+961} = \sqrt{1057}$. Since $\sqrt{96+529}<\sqrt{96+961}$, $25$ is the shorter length*, so the answer is $\boxed{\textbf{(B) }25}$.
Or, alternatively, one can notice that the two triangles have the same height but $\bigtriangleup AFC$ has a shorter base than $\bigtriangleup BED$. | // Block 1
size(7cm);
pair A,B,C,D,CC,DD;
A = (-2,7);
B = (14,7);
C = (10,0);
D = (0,0);
CC = (10,7);
DD = (0,7);
draw(A--B--C--D--cycle);
//label("33",(A+B)/2,N);
label("21",(C+D)/2,S);
label("10",(A+D)/2,W);
label("14",(B+C)/2,E);
label("$A$",A,NW);
label("$B$",B,NE);
label("$C$",C,SE);
label("$D$",D,SW);
label("$E$",DD,N);
label("$F$",CC,N);
draw(C--CC); draw(D--DD);
// Block 2
size(7cm);
pair A,B,C,D,CC,DD;
A = (-2,7);
B = (14,7);
C = (10,0);
D = (0,0);
CC = (10,7);
DD = (0,7);
draw(A--B--C--D--cycle);
//label("33",(A+B)/2,N);
label("21",(C+D)/2,S);
label("10",(A+D)/2,W);
label("14",(B+C)/2,E);
label("$A$",A,NW);
label("$B$",B,NE);
label("$C$",C,SE);
label("$D$",D,SW);
label("$D$",D,SW);
label("$E$",DD,SE);
label("$F$",CC,SW);
draw(C--CC); draw(D--DD);
label("21",(CC+DD)/2,N);
label("$2$",(A+DD)/2,N);
label("$10$",(CC+B)/2,N);
label("$4\sqrt{6}$",(C+CC)/2,W);
label("$4\sqrt{6}$",(D+DD)/2,E);
pair X = (-2,0);
//draw(X--C--A--cycle,black+2bp);
// Block 3
size(7cm); pair A,B,C,D,CC,DD; A = (-2,7); B = (14,7); C = (10,0); D = (0,0); CC = (10,7); DD = (0,7); draw(A--B--C--D--cycle); //label("33",(A+B)/2,N); label("21",(C+D)/2,S); label("10",(A+D)/2,W); label("14",(B+C)/2,E); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$E$",DD,N); label("$F$",CC,N); draw(C--CC); draw(D--DD);
// Block 4
size(7cm); pair A,B,C,D,CC,DD; A = (-2,7); B = (14,7); C = (10,0); D = (0,0); CC = (10,7); DD = (0,7); draw(A--B--C--D--cycle); //label("33",(A+B)/2,N); label("21",(C+D)/2,S); label("10",(A+D)/2,W); label("14",(B+C)/2,E); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$D$",D,SW); label("$E$",DD,SE); label("$F$",CC,SW); draw(C--CC); draw(D--DD); label("21",(CC+DD)/2,N); label("$2$",(A+DD)/2,N); label("$10$",(CC+B)/2,N); label("$4\sqrt{6}$",(C+CC)/2,W); label("$4\sqrt{6}$",(D+DD)/2,E); pair X = (-2,0); //draw(X--C--A--cycle,black+2bp); | [] |
758 | Trapezoid $ABCD$ has parallel sides $\overline{AB}$ of length $33$ and $\overline {CD}$ of length $21$. The other two sides are of lengths $10$ and $14$. The angles $A$ and $B$ are acute. What is the length of the shorter diagonal of $ABCD$?
$\textbf{(A) }10\sqrt{6}\qquad\textbf{(B) }25\qquad\textbf{(C) }8\sqrt{10}\qquad\textbf{(D) }18\sqrt{2}\qquad\textbf{(E) }26$ | 2014 AMC 10B Problem 21 | The area of $\Delta AED$ is by Heron's, $4\sqrt{9(4)(3)(2)}=24\sqrt{6}$. This makes the length of the altitude from $D$ onto $\overline{AE}$ equal to $4\sqrt{6}$. One may now proceed as in Solution $1$ to obtain an answer of $\boxed{\textbf{(B) }25}$. | // Block 1
size(7cm);
pair A,B,C,D,E;
A = (-2,7);
B = (14,7);
C = (10,0);
D = (0,0);
E = (4,7);
draw(A--B--C--D--cycle);
draw(D--E);
label("21",(C+D)/2,S);
label("10",(A+D)/2,W);
label("14",(12,1),E);
label("14",(2,1),E);
label("12",(A+E)/2,N);
label("21",(E+B)/2,N);
label("$A$",A,NW);
label("$B$",B,NE);
label("$C$",C,SE);
label("$D$",D,SW);
label("$D$",D,SW);
label("$E$",E,N);
// Block 2
size(7cm); pair A,B,C,D,E; A = (-2,7); B = (14,7); C = (10,0); D = (0,0); E = (4,7); draw(A--B--C--D--cycle); draw(D--E); label("21",(C+D)/2,S); label("10",(A+D)/2,W); label("14",(12,1),E); label("14",(2,1),E); label("12",(A+E)/2,N); label("21",(E+B)/2,N); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$D$",D,SW); label("$E$",E,N); | [] |
759 | Eight semicircles line the inside of a square with side length 2 as shown. What is the radius of the circle tangent to all of these semicircles?
$\text{(A) } \dfrac{1+\sqrt2}4 \quad \text{(B) } \dfrac{\sqrt5-1}2 \quad \text{(C) } \dfrac{\sqrt3+1}4 \quad \text{(D) } \dfrac{2\sqrt3}5 \quad \text{(E) } \dfrac{\sqrt5}3$ | 2014 AMC 10B Problem 22 | We connect the centers of the circle and one of the semicircles, then draw the perpendicular from the center of the middle circle to that side, as shown.
We will start by creating an equation by the Pythagorean theorem: \[\sqrt{1^2 + \left(\frac12\right)^2} = \sqrt{\frac54} = \frac{\sqrt5}{2}.\]
Let's call $r$ as the radius of the circle that we want to find. We see that the hypotenuse of the bold right triangle is $\dfrac{1}{2}+r$, and thus $r$ is $\boxed{\textbf{(B)} \frac{\sqrt{5}-1}{2}}$.
-- LORD_ERTY09 | // Block 1
scale(200);
draw(scale(.5)*((-1,-1)--(1,-1)--(1,1)--(-1,1)--cycle));
path p = arc((.25,-.5),.25,0,180)--arc((-.25,-.5),.25,0,180);
draw(p);
p=rotate(90)*p; draw(p);
p=rotate(90)*p; draw(p);
p=rotate(90)*p; draw(p);
draw(scale((sqrt(5)-1)/4)*unitcircle);
pair OO=(0,0);
pair XX=(-.25,-.5);
pair YY=(0,-.5);
draw(YY--OO--XX--cycle,black+1bp);
label("$\frac12$",.5*(XX+YY),S);
label("$1$",.5*YY,E);
// Block 2
scale(200); draw(scale(.5)*((-1,-1)--(1,-1)--(1,1)--(-1,1)--cycle)); path p = arc((.25,-.5),.25,0,180)--arc((-.25,-.5),.25,0,180); draw(p); p=rotate(90)*p; draw(p); p=rotate(90)*p; draw(p); p=rotate(90)*p; draw(p); draw(scale((sqrt(5)-1)/4)*unitcircle); pair OO=(0,0); pair XX=(-.25,-.5); pair YY=(0,-.5); draw(YY--OO--XX--cycle,black+1bp); label("$\frac12$",.5*(XX+YY),S); label("$1$",.5*YY,E); | [] |
760 | A sphere is inscribed in a truncated right circular cone as shown. The volume of the truncated cone is twice that of the sphere. What is the ratio of the radius of the bottom base of the truncated cone to the radius of the top base of the truncated cone?
$\text{(A) } \dfrac32 \quad \text{(B) } \dfrac{1+\sqrt5}2 \quad \text{(C) } \sqrt3 \quad \text{(D) } 2 \quad \text{(E) } \dfrac{3+\sqrt5}2$ | 2014 AMC 10B Problem 23 | First, we draw the vertical cross-section passing through the middle of the frustum.
Let the top base have a diameter of $2$ and the bottom base has a diameter of $2r$.
Then using the Pythagorean theorem we have:
$(r+1)^2=(2s)^2+(r-1)^2$,
which is equivalent to:
$r^2+2r+1=4s^2+r^2-2r+1$.
Subtracting $r^2-2r+1$ from both sides,
$4r=4s^2$, and solving for $s$, we end up with
\[s=\sqrt{r}.\]
Next, we can find the volume of the frustum (truncated cone) and the sphere. Since we know $V_{\text{frustum}}=2V_{\text{sphere}}$, we can solve for $s$
using $V_{\text{frustum}}=\frac{\pi h}{3}(R^2+r^2+Rr)$
we get:
\[V_{\text{frustum}}=\frac{\pi\cdot2\sqrt{r}}{3}(r^2+r+1)\]
Using $V_{\text{sphere}}=\dfrac{4s^{3}\pi}{3}$, we get
\[V_{\text{sphere}}=\dfrac{4(\sqrt{r})^{3}\pi}{3}\]
so we have:
\[\frac{\pi\cdot2\sqrt{r}}{3}(r^2+r+1)=2\cdot\dfrac{4(\sqrt{r})^{3}\pi}{3}.\]
Dividing by $\frac{2\pi\sqrt{r}}{3}$, we get
\[r^2+r+1=4r\]
which is equivalent to \[r^2-3r+1=0\]
by the Quadratic Formula, $r=\dfrac{3\pm\sqrt{(-3)^2-4\cdot1\cdot1}}{2\cdot1}$
, so
\[r=\dfrac{3+\sqrt{5}}{2} \longrightarrow \boxed{\textbf{(E)}}\] | // Block 1
size(7cm);
pair A,B,C,D;
real r = (3+sqrt(5))/2;
real s = sqrt(r);
A = (-r,0);
B = (r,0);
C = (1,2*s);
D = (-1,2*s);
draw(A--B--C--D--cycle);
pair O = (0,s);
draw(shift(O)*scale(s)*unitcircle);
dot(O);
pair X,Y;
X = (0,0);
Y = (0,2*s);
draw(X--Y);
label("$r-1$",(r/2+1/2,0),S);
label("$1$",(Y+C)/2,N);
label("$s$",(O+Y)/2,W);
label("$s$",(O+X)/2,W);
draw(B--C--(1,0)--cycle,blue+1bp);
pair P = 0.73*C+0.27*B;
draw(O--P);
dot(P);
label("$1$",(C+P)/2,NE);
label("$r$",(B+P)/2,NE);
// Block 2
size(7cm); pair A,B,C,D; real r = (3+sqrt(5))/2; real s = sqrt(r); A = (-r,0); B = (r,0); C = (1,2*s); D = (-1,2*s); draw(A--B--C--D--cycle); pair O = (0,s); draw(shift(O)*scale(s)*unitcircle); dot(O); pair X,Y; X = (0,0); Y = (0,2*s); draw(X--Y); label("$r-1$",(r/2+1/2,0),S); label("$1$",(Y+C)/2,N); label("$s$",(O+Y)/2,W); label("$s$",(O+X)/2,W); draw(B--C--(1,0)--cycle,blue+1bp); pair P = 0.73*C+0.27*B; draw(O--P); dot(P); label("$1$",(C+P)/2,NE); label("$r$",(B+P)/2,NE); | [] |
761 | The numbers $1, 2, 3, 4, 5$ are to be arranged in a circle. An arrangement is $\textit{bad}$ if it is not true that for every $n$ from $1$ to $15$ one can find a subset of the numbers that appear consecutively on the circle that sum to $n$. Arrangements that differ only by a rotation or a reflection are considered the same. How many different bad arrangements are there?
$\textbf {(A) } 1 \qquad \textbf {(B) } 2 \qquad \textbf {(C) } 3 \qquad \textbf {(D) } 4 \qquad \textbf {(E) } 5$
. | 2014 AMC 10B Problem 24 | We see that there are $5!$ total ways to arrange the numbers. However, we can always rotate these numbers so that, for example, the number $1$ is always at the top of the circle. Thus, there are only $4!$ ways under rotation. Every case has exactly $1$ reflection, so that gives us only $4!/2$, or $12$ cases, which is not difficult to list out. We systematically list out all $12$ cases.
Now, we must examine if they satisfy the conditions. We can see that by choosing one number at a time, we can always obtain subsets with sums $1, 2, 3, 4,$ and $5$. By choosing the full circle, we can obtain $15$. By choosing everything except for $1, 2, 3, 4,$ and $5$, we can obtain subsets with sums of $10, 11, 12, 13,$ and $14$.
This means that we now only need to check for $6, 7, 8,$ and $9$. However, once we have found a set summing to $6$, we can choose all remaining numbers and obtain a set summing to $15-6=9$, and similarly for $7$ and $8$. Thus, we only need to check each case for whether or not we can obtain $6$ or $7$.
We can make $6$ by having $4, 2$, or $3, 2, 1$, or $5, 1$. We can start with the group of three. To separate $3, 2, 1$ from each other, they must be grouped two together and one separate, like this.
Now, we note that $x$ is next to both blank spots, so we can't have a number from one of the pairs. So since we can't have $1$, because it is part of the $5, 1$ pair, and we can't have $2$ there, because it's part of the $4, 2$ pair, we must have $3$ inserted into the $x$ spot. We can insert $1$ and $2$ in $y$ and $z$ interchangeably, since reflections are considered the same.
We have $4$ and $5$ left to insert. We can't place the $4$ next to the $2$ or the $5$ next to the $1$, so we must place $4$ next to the $1$ and $5$ next to the $2$.
This is the only solution to make $6$ "bad."
Next we move on to $7$, which can be made by $3, 4$, or $5, 2$, or $4, 2, 1$. We do this the same way as before. We start with the three group. Since we can't have $4$ or $2$ in the top slot, we must have one there, and $4$ and $2$ are next to each other on the bottom. When we have $3$ and $5$ left to insert, we place them such that we don't have the two pairs adjacent.
This is the only solution to make $7$ "bad."
We've covered all needed cases, and the two examples we found are distinct, therefore the answer is $\boxed{\textbf {(B) }2}$. | // Block 1
draw(circle((0, 0), 5));
pair O, A, B, C, D, E;
O=origin;
A=(0, 5);
B=rotate(72)*A;
C=rotate(144)*A;
D=rotate(216)*A;
E=rotate(288)*A;
label("$x$", A, N);
label("$y$", C, SW);
label("$z$", D, SE);
// Block 2
draw(circle((0, 0), 5));
pair O, A, B, C, D, E;
O=origin;
A=(0, 5);
B=rotate(72)*A;
C=rotate(144)*A;
D=rotate(216)*A;
E=rotate(288)*A;
label("$3$", A, N);
label("$2$", C, SW);
label("$1$", D, SE);
// Block 3
draw(circle((0, 0), 5));
pair O, A, B, C, D, E;
O=origin;
A=(0, 5);
B=rotate(72)*A;
C=rotate(144)*A;
D=rotate(216)*A;
E=rotate(288)*A;
label("$3$", A, N);
label("$5$", B, NW);
label("$2$", C, SW);
label("$1$", D, SE);
label("$4$", E, NE);
// Block 4
draw(circle((0, 0), 5));
pair O, A, B, C, D, E;
O=origin;
A=(0, 5);
B=rotate(72)*A;
C=rotate(144)*A;
D=rotate(216)*A;
E=rotate(288)*A;
label("$1$", A, N);
label("$3$", B, NW);
label("$2$", C, SW);
label("$4$", D, SE);
label("$5$", E, NE);
// Block 5
draw(circle((0, 0), 5)); pair O, A, B, C, D, E; O=origin; A=(0, 5); B=rotate(72)*A; C=rotate(144)*A; D=rotate(216)*A; E=rotate(288)*A; label("$x$", A, N); label("$y$", C, SW); label("$z$", D, SE);
// Block 6
draw(circle((0, 0), 5)); pair O, A, B, C, D, E; O=origin; A=(0, 5); B=rotate(72)*A; C=rotate(144)*A; D=rotate(216)*A; E=rotate(288)*A; label("$3$", A, N); label("$2$", C, SW); label("$1$", D, SE);
// Block 7
draw(circle((0, 0), 5)); pair O, A, B, C, D, E; O=origin; A=(0, 5); B=rotate(72)*A; C=rotate(144)*A; D=rotate(216)*A; E=rotate(288)*A; label("$3$", A, N); label("$5$", B, NW); label("$2$", C, SW); label("$1$", D, SE); label("$4$", E, NE);
// Block 8
draw(circle((0, 0), 5)); pair O, A, B, C, D, E; O=origin; A=(0, 5); B=rotate(72)*A; C=rotate(144)*A; D=rotate(216)*A; E=rotate(288)*A; label("$1$", A, N); label("$3$", B, NW); label("$2$", C, SW); label("$4$", D, SE); label("$5$", E, NE); | [] |
762 | A sphere is inscribed in a truncated right circular cone as shown. The volume of the truncated cone is twice that of the sphere. What is the ratio of the radius of the bottom base of the truncated cone to the radius of the top base of the truncated cone?
$\text{(A) } \dfrac32 \quad \text{(B) } \dfrac{1+\sqrt5}2 \quad \text{(C) } \sqrt3 \quad \text{(D) } 2 \quad \text{(E) } \dfrac{3+\sqrt5}2$ | 2014 AMC 12B Problem 19 | Solution 1
First, we draw the vertical cross-section passing through the middle of the frustum.
let the top base equal 2 and the bottom base to be equal to 2r
then using the Pythagorean theorem we have:
$(r+1)^2=(2s)^2+(r-1)^2$
which is equivalent to:
$r^2+2r+1=4s^2+r^2-2r+1$
subtracting $r^2-2r+1$ from both sides
$4r=4s^2$
solving for s we get:
\[s=\sqrt{r}\]
next we can find the volume of the frustum and of the sphere and we know $V_{\text{frustum}}=2V_{\text{sphere}}$ so we can solve for $s$
using $V_{\text{frustum}}=\frac{\pi*h}{3}(R^2+r^2+Rr)$
we get:
\[V_{\text{frustum}}=\frac{\pi*2\sqrt{r}}{3}(r^2+r+1)\]
using $V_{\text{sphere}}=\dfrac{4s^{3}\pi}{3}$
we get
\[V_{\text{sphere}}=\dfrac{4(\sqrt{r})^{3}\pi}{3}\]
so we have:
\[\frac{\pi*2\sqrt{r}}{3}(r^2+r+1)=2*\dfrac{4(\sqrt{r})^{3}\pi}{3}\]
dividing by $\frac{2\pi*\sqrt{r}}{3}$
we get \[r^2+r+1=4r\]
which is equivalent to \[r^2-3r+1=0\]
$r=\dfrac{3\pm\sqrt{(-3)^2-4*1*1}}{2*1}$
so \[r=\dfrac{3+\sqrt{5}}{2}\longrightarrow \boxed{E}\]
Solution 2(ADD DIAGRAM)
Let's once again look at the cross section of the frustum. Let the angle from the center of the sphere to a point on the circumference of the bottom circle be $\theta.$ This implies that the angle from the center of the sphere to a point on the circumference of the top circle is $90 - \theta.$ Hence the bottom radius is $r\tan{\theta}$ and the top radius is $\frac {r}{\tan {\theta}}.$ This means that the radio between the bottom radius and top radius is $(\tan {\theta})^2.$ Using the frustum volume formula, we find that the are of this figure is $\frac{2\pi r}{3}(r^2(\tan {\theta})^2 + r^2 + \frac {r^2} {(\tan {\theta})^2}).$ We can equate this to $\frac {8\pi*r^3} 3.$ Simplifying, we are left with a quadratic conveniently in $(\tan {\theta})^2.$ The quadratic is $(\tan {\theta})^4 - 3(\tan {\theta})^2 + 1 = 0.$ This gives us $(\tan {\theta})^2 = \dfrac{3+\sqrt{5}}{2}\longrightarrow \boxed{E}$
~NeeNeeMath
Remark
For problems that involve a circle inscribed into an isosceles trapezoid, the following fact is very useful. If we let the bases be $x$ and $y$, and the height be $h$, then $h = \sqrt{xy}$. Then, we could solve from solution 1 directly knowing the radius in terms of the base. By letting the upper base be $1$ and the lower base be $x$, we can find $r$ in terms of $x$, and solve like in solution one.
~Puck_0 | size(7cm); pair A,B,C,D; real r = (3+sqrt(5))/2; real s = sqrt(r); A = (-r,0); B = (r,0); C = (1,2*s); D = (-1,2*s); draw(A--B--C--D--cycle); pair O = (0,s); draw(shift(O)*scale(s)*unitcircle); dot(O); pair X,Y; X = (0,0); Y = (0,2*s); draw(X--Y); label("$r-1$",(X+B)/2,S); label("$1$",(Y+C)/2,N); label("$s$",(O+Y)/2,W); label("$s$",(O+X)/2,W); draw(B--C--(1,0)--cycle,blue+1bp); pair P = 0.73*C+0.27*B; draw(O--P); dot(P); label("$1$",(C+P)/2,NE); label("$r$",(B+P)/2,NE); | [] |
763 | Let $ABCDE$ be a pentagon inscribed in a circle such that $AB = CD = 3$, $BC = DE = 10$, and $AE= 14$. The sum of the lengths of all diagonals of $ABCDE$ is equal to $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$ ?
$\textbf{(A) }129\qquad \textbf{(B) }247\qquad \textbf{(C) }353\qquad \textbf{(D) }391\qquad \textbf{(E) }421\qquad$ | 2014 AMC 12B Problem 24 | Let $BE=a$, $AD=b$, and $AC=CE=BD=c$. Let $F$ be on $AE$ such that $CF \perp AE$.
In $\triangle CFE$ we have $\cos\theta = -\cos(\pi-\theta)=-7/c$. We use the Law of Cosines on $\triangle ABC$ to get $60\cos\theta = 109-c^2$. Eliminating $\cos\theta$ we get $c^3-109c-420=0$ which factorizes as
\[(c+7)(c+5)(c-12)=0.\]Discarding the negative roots we have $c=12$. Thus $BD=AC=CE=12$. For $BE=a$, we use Ptolemy's theorem on cyclic quadrilateral $ABCE$ to get $a=44/3$. For $AD=b$, we use Ptolemy's theorem on cyclic quadrilateral $ACDE$ to get $b=27/2$.
The sum of the lengths of the diagonals is $12+12+12+\tfrac{44}{3}+\tfrac{27}{2} = \tfrac{385}{6}$ so the answer is $385 + 6 = \fbox{\textbf{(D) }391}$ | // Block 1
size(200);
defaultpen(linewidth(0.4)+fontsize(10));
pen s = linewidth(0.8)+fontsize(8);
pair O,A,B,C,D,E0,F;
O=origin;
A= dir(198);
path c = CR(O,1);
real r = 0.13535;
B = IP(c, CR(A,3*r));
C = IP(c, CR(B,10*r));
D = IP(c, CR(C,3*r));
E0 = OP(c, CR(D,10*r));
F = foot(C,A,E0);
dot("$A$", A, A-O);
dot("$B$", B, B-O);
dot("$C$", C, C-O);
dot("$D$", D, D-O);
dot("$E$", E0, E0-O);
dot("$F$", F, F-C);
label("$c$",A--C,S);
label("$c$",E0--C,W);
label("$7$",F--E0,S);
label("$7$",F--A,S);
label("$3$",A--B,2*W);
label("$10$",B--C,2*N);
label("$3$",C--D,2*NE);
label("$10$",D--E0,E);
draw(A--B--C--D--E0--A, black+0.8);
draw(CR(O,1), s);
draw(A--C--E0, royalblue);
draw(C--F, royalblue+dashed);
draw(rightanglemark(E0,F,C,2));
MA("\theta",A,B,C,0.075);
MA("\pi-\theta",C,E0,A,0.1);
// Block 2
size(200); defaultpen(linewidth(0.4)+fontsize(10)); pen s = linewidth(0.8)+fontsize(8); pair O,A,B,C,D,E0,F; O=origin; A= dir(198); path c = CR(O,1); real r = 0.13535; B = IP(c, CR(A,3*r)); C = IP(c, CR(B,10*r)); D = IP(c, CR(C,3*r)); E0 = OP(c, CR(D,10*r)); F = foot(C,A,E0); dot("$A$", A, A-O); dot("$B$", B, B-O); dot("$C$", C, C-O); dot("$D$", D, D-O); dot("$E$", E0, E0-O); dot("$F$", F, F-C); label("$c$",A--C,S); label("$c$",E0--C,W); label("$7$",F--E0,S); label("$7$",F--A,S); label("$3$",A--B,2*W); label("$10$",B--C,2*N); label("$3$",C--D,2*NE); label("$10$",D--E0,E); draw(A--B--C--D--E0--A, black+0.8); draw(CR(O,1), s); draw(A--C--E0, royalblue); draw(C--F, royalblue+dashed); draw(rightanglemark(E0,F,C,2)); MA("\theta",A,B,C,0.075); MA("\pi-\theta",C,E0,A,0.1); | [] |
764 | The diagram below shows the circular face of a clock with radius $20$ cm and a circular disk with radius $10$ cm externally tangent to the clock face at $12$ o' clock. The disk has an arrow painted on it, initially pointing in the upward vertical direction. Let the disk roll clockwise around the clock face. At what point on the clock face will the disk be tangent when the arrow is next pointing in the upward vertical direction?
$\textbf{(A) }\text{2 o' clock} \qquad\textbf{(B) }\text{3 o' clock} \qquad\textbf{(C) }\text{4 o' clock} \qquad\textbf{(D) }\text{6 o' clock} \qquad\textbf{(E) }\text{8 o' clock}$ | 2015 AMC 10A Problem 14 | Suppose that the small disk also had a clock face on it, and that both disks were toothed wheels, free to rotate around their centers. The part of the picture where they engage would look like this:
The small cog has half the radius, and therefore half the circumference. If the large cog turns $30^\circ$ counterclockwise (i.e. 1 hour), the small cog turns $60^\circ$ clockwise (i.e. 2 hours).
However, in the original problem the large cog does not rotate; it stays where it is. Therefore we must turn the whole diagram above $30^\circ$ clockwise to see what happens when the small cog rolls around it.
It turns out that, when the point of tangency moves $30^\circ$ clockwise (one hour), from our point of view the small disk rotates $90^\circ$ clockwise (three hours) around its center. Thus, for the small disk to perform a complete rotation of $360^\circ$ (twelve hours) around its center from our point of view, the point of tangency must move round four hours. So the answer is $\boxed{\textbf{(C) }4 \ \text{o' clock}}$ | // Block 1
fill(arc((0,0),2,30,150)--cycle,lightgrey);
draw(arc((0,0),2,30,150));
draw(1.8*dir(90)--2*dir(90));
draw(1.8*dir(60)--2*dir(60));
label("12",1.56*dir(90));
label("1",1.56*dir(60));
draw(arc((0,3),1,-15,-165));
draw(0.9*dir(-90)+(0,3)--dir(-90)+(0,3));
draw(0.9*dir(-60)+(0,3)--dir(-60)+(0,3));
draw(0.9*dir(-30)+(0,3)--dir(-30)+(0,3));
label("6",.74*dir(-90)+(0,3));
label("5",.74*dir(-60)+(0,3));
label("4",.74*dir(-30)+(0,3));
// Block 2
fill(arc((0,0),2,30,150)--cycle,lightgrey);
draw(arc((0,0),2,30,150),EndArcArrow);label("$30^\circ$",2*dir(150),W);
draw(1.8*dir(120)--2*dir(120));
draw(1.8*dir(90)--2*dir(90));
label(rotate(30)*"12",1.56*dir(120));
label(rotate(30)*"1",1.56*dir(90));
draw(arc((0,3),1,-15,-165),EndArcArrow);label("$60^\circ$",dir(-165)+(0,3),W);
draw(0.9*dir(-90)+(0,3)--dir(-90)+(0,3));
draw(0.9*dir(-120)+(0,3)--dir(-120)+(0,3));
draw(0.9*dir(-150)+(0,3)--dir(-150)+(0,3));
label(rotate(-60)*"6",.74*dir(-150)+(0,3));
label(rotate(-60)*"5",.74*dir(-120)+(0,3));
label(rotate(-60)*"4",.74*dir(-90)+(0,3));
// Block 3
fill(arc((0,0),2,30,150)--cycle,lightgrey);
draw(arc((0,0),2,30,150));
draw(1.8*dir(90)--2*dir(90));
draw(1.8*dir(60)--2*dir(60));
label("12",1.56*dir(90));
label("1",1.56*dir(60));
pair c=(1.5,sqrt(27)/2);
draw(arc(c,1,0,-200),EndArcArrow);label("$90^\circ$",dir(-180)+c,W);
draw(0.9*dir(-120)+c--dir(-120)+c);
draw(0.9*dir(-150)+c--dir(-150)+c);
draw(0.9*dir(-180)+c--dir(-180)+c);
label(rotate(-90)*"6",.74*dir(-180)+c);
label(rotate(-90)*"5",.74*dir(-150)+c);
label(rotate(-90)*"4",.74*dir(-120)+c);
// Block 4
fill(arc((0,0),2,30,150)--cycle,lightgrey); draw(arc((0,0),2,30,150)); draw(1.8*dir(90)--2*dir(90)); draw(1.8*dir(60)--2*dir(60)); label("12",1.56*dir(90)); label("1",1.56*dir(60)); draw(arc((0,3),1,-15,-165)); draw(0.9*dir(-90)+(0,3)--dir(-90)+(0,3)); draw(0.9*dir(-60)+(0,3)--dir(-60)+(0,3)); draw(0.9*dir(-30)+(0,3)--dir(-30)+(0,3)); label("6",.74*dir(-90)+(0,3)); label("5",.74*dir(-60)+(0,3)); label("4",.74*dir(-30)+(0,3));
// Block 5
fill(arc((0,0),2,30,150)--cycle,lightgrey); draw(arc((0,0),2,30,150),EndArcArrow);label("$30^\circ$",2*dir(150),W); draw(1.8*dir(120)--2*dir(120)); draw(1.8*dir(90)--2*dir(90)); label(rotate(30)*"12",1.56*dir(120)); label(rotate(30)*"1",1.56*dir(90)); draw(arc((0,3),1,-15,-165),EndArcArrow);label("$60^\circ$",dir(-165)+(0,3),W); draw(0.9*dir(-90)+(0,3)--dir(-90)+(0,3)); draw(0.9*dir(-120)+(0,3)--dir(-120)+(0,3)); draw(0.9*dir(-150)+(0,3)--dir(-150)+(0,3)); label(rotate(-60)*"6",.74*dir(-150)+(0,3)); label(rotate(-60)*"5",.74*dir(-120)+(0,3)); label(rotate(-60)*"4",.74*dir(-90)+(0,3));
// Block 6
fill(arc((0,0),2,30,150)--cycle,lightgrey); draw(arc((0,0),2,30,150)); draw(1.8*dir(90)--2*dir(90)); draw(1.8*dir(60)--2*dir(60)); label("12",1.56*dir(90)); label("1",1.56*dir(60)); pair c=(1.5,sqrt(27)/2); draw(arc(c,1,0,-200),EndArcArrow);label("$90^\circ$",dir(-180)+c,W); draw(0.9*dir(-120)+c--dir(-120)+c); draw(0.9*dir(-150)+c--dir(-150)+c); draw(0.9*dir(-180)+c--dir(-180)+c); label(rotate(-90)*"6",.74*dir(-180)+c); label(rotate(-90)*"5",.74*dir(-150)+c); label(rotate(-90)*"4",.74*dir(-120)+c); | [] |
765 | Eight people are sitting around a circular table, each holding a fair coin. All eight people flip their coins and those who flip heads stand while those who flip tails remain seated. What is the probability that no two adjacent people will stand?
$\textbf{(A)}\dfrac{47}{256}\qquad\textbf{(B)}\dfrac{3}{16}\qquad\textbf{(C) }\dfrac{49}{256}\qquad\textbf{(D) }\dfrac{25}{128}\qquad\textbf{(E) }\dfrac{51}{256}$ | 2015 AMC 10A Problem 22 | Case 1: Everyone flips tails
1 way:
Case 2: One person flips heads
8 ways:
Case 3: Two people flip heads
There are $\binom{8}{2}$ ways to choose any $2$ people in the circle and $8$ ways to choose $2$ people adjacent to each other. So this case gives us $\binom{8}{2} - 8 = 20$ ways. An example of choosing two nonadjacent people is shown below.
Case 4: Three people flip heads
We can choose the first person (shaded gray) in $8$ ways. We can choose the other $2$ people in $6$ ways (purple-blue, yellow-red, green-yellow, green-blue, purple-red, green-red). We divide by $3$ for overcounting to get $\frac{8 \cdot 6}{3} = 16$ ways for this case.
Case 5: 4 people flip heads
2 ways (we can either choose the people shaded in red or blue)
It's impossible to have more than 4 people standing without anyone being next to each other, so we stop here.
The total number of ways is $2^8,$ so the answer is $\frac{1 + 8 + 20 + 16 + 2}{2^8} = \boxed{\frac{47}{256}}.$
~grogg007 | // Block 1
size(50);
import graph;
real R = 0.5; // Radius of big circle
real r = 0.05; // Radius of small circles
int n = 8; // Number of small circles
// Draw big circle
draw(Circle((0,0), R), linewidth(1));
// Place small circles on circumference
for (int i = 0; i < n; ++i) {
real angle = 2*pi*i/n;
pair center = R * dir(degrees(angle));
draw(Circle(center, r), linewidth(0.8));
}
// Block 2
size(400);
import graph;
real R = 1.5; // Radius of big circle
real r = 0.2; // Radius of small circles
int n = 8; // Number of small circles
for (int j = 0; j < 8; ++j) {
pair centerOffset = (j * 4, 0); // space between diagrams
// Draw big circle
draw(shift(centerOffset) * Circle((0,0), R), linewidth(1));
// Draw small circles
for (int i = 0; i < n; ++i) {
real angle = 2 * pi * i / n;
pair pos = R * dir(degrees(angle));
pair finalPos = centerOffset + pos;
if (i == j) {
fill(Circle(finalPos, r), black); // Shade this one
} else {
draw(Circle(finalPos, r), linewidth(0.8)); // Outline others
}
}
}
// Block 3
size(50);
import graph;
real R = 0.5; // Radius of big circle
real r = 0.05; // Radius of small circles
int n = 8; // Number of small circles
// Indices of two circles that are one circle apart
int a = 0;
int b = 2;
// Compute centers of these two circles
real angleA = 2*pi*a/n;
real angleB = 2*pi*b/n;
pair centerA = R * dir(degrees(angleA));
pair centerB = R * dir(degrees(angleB));
// Draw big circle
draw(Circle((0,0), R), linewidth(1));
// Draw small circles and shade the two chosen ones
for (int i = 0; i < n; ++i) {
real angle = 2*pi*i/n;
pair center = R * dir(degrees(angle));
if (i == a || i == b) {
fill(Circle(center, r), black);
} else {
draw(Circle(center, r), linewidth(0.8));
}
}
// Draw line connecting the two shaded circles
draw(centerA -- centerB, linewidth(1));
// Block 4
size(50);
import graph;
real R = 1.5; // Radius of big circle
real r = 0.2; // Radius of small circles
int n = 8; // Number of small circles
pair center = (0, 0);
// Draw big circle
draw(Circle(center, R), linewidth(1));
// Compute positions (clockwise, starting at top)
pair[] pts;
for (int i = 0; i < n; ++i) {
real angle = 90 - 360 * i / n; // Start at 90° and go clockwise
pts[i] = center + R * dir(angle);
}
// Draw circles with unique colors for shaded ones
for (int i = 0; i < n; ++i) {
if (i == 0) {
fill(Circle(pts[i], r), gray); // Circle 1
} else if (i == 2) {
fill(Circle(pts[i], r), red); // Circle 3
} else if (i == 3) {
fill(Circle(pts[i], r), blue); // Circle 4
} else if (i == 4) {
fill(Circle(pts[i], r), yellow); // Circle 5
} else if (i == 5) {
fill(Circle(pts[i], r), purple); // Circle 6
} else if (i == 6) {
fill(Circle(pts[i], r), green); // Circle 7
} else {
draw(Circle(pts[i], r), linewidth(0.8)); // Others outlined only
}
}
// Draw triangle connecting circles 1, 4, 6 (indices 0, 3, 5)
draw(pts[0] -- pts[3] -- pts[5] -- cycle, linewidth(1));
// Block 5
size(50);
import graph;
real R = 1.5;
real r = 0.15;
// Circle centers, clockwise starting at top
pair c0 = (0, R);
pair c1 = R * dir(45);
pair c2 = R * dir(90 - 2*45);
pair c3 = R * dir(90 - 3*45);
pair c4 = R * dir(90 - 4*45);
pair c5 = R * dir(90 - 5*45);
pair c6 = R * dir(90 - 6*45);
pair c7 = R * dir(90 - 7*45);
// Draw big circle
draw(Circle((0,0), R), linewidth(1));
// Group 1 vertices (blue)
pen bluePen = blue + linewidth(1.5);
filldraw(Circle(c0, r), blue);
filldraw(Circle(c2, r), blue);
filldraw(Circle(c4, r), blue);
filldraw(Circle(c6, r), blue);
draw(c0--c2--c4--c6--c0, bluePen);
// Group 2 vertices (red)
pen redPen = red + linewidth(1.5);
filldraw(Circle(c1, r), red);
filldraw(Circle(c3, r), red);
filldraw(Circle(c5, r), red);
filldraw(Circle(c7, r), red);
draw(c1--c3--c5--c7--c1, redPen);
// Block 6
size(50); import graph; real R = 0.5; // Radius of big circle real r = 0.05; // Radius of small circles int n = 8; // Number of small circles // Draw big circle draw(Circle((0,0), R), linewidth(1)); // Place small circles on circumference for (int i = 0; i < n; ++i) { real angle = 2*pi*i/n; pair center = R * dir(degrees(angle)); draw(Circle(center, r), linewidth(0.8)); }
// Block 7
size(400); import graph; real R = 1.5; // Radius of big circle real r = 0.2; // Radius of small circles int n = 8; // Number of small circles for (int j = 0; j < 8; ++j) { pair centerOffset = (j * 4, 0); // space between diagrams // Draw big circle draw(shift(centerOffset) * Circle((0,0), R), linewidth(1)); // Draw small circles for (int i = 0; i < n; ++i) { real angle = 2 * pi * i / n; pair pos = R * dir(degrees(angle)); pair finalPos = centerOffset + pos; if (i == j) { fill(Circle(finalPos, r), black); // Shade this one } else { draw(Circle(finalPos, r), linewidth(0.8)); // Outline others } } }
// Block 8
size(50); import graph; real R = 0.5; // Radius of big circle real r = 0.05; // Radius of small circles int n = 8; // Number of small circles // Indices of two circles that are one circle apart int a = 0; int b = 2; // Compute centers of these two circles real angleA = 2*pi*a/n; real angleB = 2*pi*b/n; pair centerA = R * dir(degrees(angleA)); pair centerB = R * dir(degrees(angleB)); // Draw big circle draw(Circle((0,0), R), linewidth(1)); // Draw small circles and shade the two chosen ones for (int i = 0; i < n; ++i) { real angle = 2*pi*i/n; pair center = R * dir(degrees(angle)); if (i == a || i == b) { fill(Circle(center, r), black); } else { draw(Circle(center, r), linewidth(0.8)); } } // Draw line connecting the two shaded circles draw(centerA -- centerB, linewidth(1));
// Block 9
size(50); import graph; real R = 1.5; // Radius of big circle real r = 0.2; // Radius of small circles int n = 8; // Number of small circles pair center = (0, 0); // Draw big circle draw(Circle(center, R), linewidth(1)); // Compute positions (clockwise, starting at top) pair[] pts; for (int i = 0; i < n; ++i) { real angle = 90 - 360 * i / n; // Start at 90° and go clockwise pts[i] = center + R * dir(angle); } // Draw circles with unique colors for shaded ones for (int i = 0; i < n; ++i) { if (i == 0) { fill(Circle(pts[i], r), gray); // Circle 1 } else if (i == 2) { fill(Circle(pts[i], r), red); // Circle 3 } else if (i == 3) { fill(Circle(pts[i], r), blue); // Circle 4 } else if (i == 4) { fill(Circle(pts[i], r), yellow); // Circle 5 } else if (i == 5) { fill(Circle(pts[i], r), purple); // Circle 6 } else if (i == 6) { fill(Circle(pts[i], r), green); // Circle 7 } else { draw(Circle(pts[i], r), linewidth(0.8)); // Others outlined only } } // Draw triangle connecting circles 1, 4, 6 (indices 0, 3, 5) draw(pts[0] -- pts[3] -- pts[5] -- cycle, linewidth(1));
// Block 10
size(50); import graph; real R = 1.5; real r = 0.15; // Circle centers, clockwise starting at top pair c0 = (0, R); pair c1 = R * dir(45); pair c2 = R * dir(90 - 2*45); pair c3 = R * dir(90 - 3*45); pair c4 = R * dir(90 - 4*45); pair c5 = R * dir(90 - 5*45); pair c6 = R * dir(90 - 6*45); pair c7 = R * dir(90 - 7*45); // Draw big circle draw(Circle((0,0), R), linewidth(1)); // Group 1 vertices (blue) pen bluePen = blue + linewidth(1.5); filldraw(Circle(c0, r), blue); filldraw(Circle(c2, r), blue); filldraw(Circle(c4, r), blue); filldraw(Circle(c6, r), blue); draw(c0--c2--c4--c6--c0, bluePen); // Group 2 vertices (red) pen redPen = red + linewidth(1.5); filldraw(Circle(c1, r), red); filldraw(Circle(c3, r), red); filldraw(Circle(c5, r), red); filldraw(Circle(c7, r), red); draw(c1--c3--c5--c7--c1, redPen); | [] |
765 | Eight people are sitting around a circular table, each holding a fair coin. All eight people flip their coins and those who flip heads stand while those who flip tails remain seated. What is the probability that no two adjacent people will stand?
$\textbf{(A)}\dfrac{47}{256}\qquad\textbf{(B)}\dfrac{3}{16}\qquad\textbf{(C) }\dfrac{49}{256}\qquad\textbf{(D) }\dfrac{25}{128}\qquad\textbf{(E) }\dfrac{51}{256}$ | 2015 AMC 10A Problem 22 | We know that the denominator of the probability is $2^8=256$. So now we only have to calculate the numerator, which is the number of arrangements for $8$ people at a round table without $2$ or more neighboring people standing.
Denote $a_n$ as number of arrangements for $n$ people at a round table without $2$ or more neighboring people standing. We can see that $a_2=3$, $a_3=4$, we are going to prove $a_n=a_{n-1}+a_{n-2}$
We use $1$ to represent standing people, and $0$ as sitting people. The problem becomes arranging $n$ numbers of $0$ or $1$ around a circle with no consecutive $1$'s.
We elect one person as the chairman, let him be $p_1$, the first element in this circular sequence.
There are $2$ cases to generate the arrangement of $n$ numbers.
$Case$ $1:$ From the arrangement of $n-1$ numbers, add $1$ more number $p_n$ counter-clockwise next to $p_1$.
If $p_1=1$, $p_{n-1}p_np_1$ is $001$, as the diagram shows:
If $p_1=0$, $p_{n-1}p_np_1$ is $X00$, $X$ could be $0$ or $1$, as the diagram shows:
$Case$ $2:$ From the arrangement of $n-2$ numbers, add $2$ more numbers $p_{n-1}$ and $p_n$ counter-clockwise next to $p_1$.
If $p_1=1$, then $p_{n-2}=0$, let $p_{n-1}=1, p_n=0$, $p_{n-1}p_np_1$ is $101$, as the diagram shows:
If $p_1=0$, then $p_{n-2}$ could be $0$ or $1$. Let $p_{n-1}=0, p_n=1$, $p_{n-1}p_np_1$ is $010$, as the diagram shows:
From the above $2$ cases and the $4$ diagrams, the arrangements of $p_{n-1}p_np_1$ are mutually exclusive collectively exhaustive, so $a_{n}=a_{n-1}+a_{n-2}$
$a_2=3$
$a_3=4$
$a_4=7$
$a_5=11$
$a_6=18$
$a_7=29$
$a_8=47$
The answer is $\boxed{\textbf{(A) } \dfrac{47}{256}}$
This sequence of numbers is called Lucas Number.
~isabelchen | // Block 1
draw(circle((0, 0), 5));
pair A, B, C;
A=(0, 5);
B=rotate(72)*A;
C=rotate(144)*A;
label("$p_1=1$", A, N);
label("$p_n=0$", B, NW);
label("$p_{n-1}=0$", C, SW);
// Block 2
draw(circle((0, 0), 5));
pair A, B, C;
A=(0, 5);
B=rotate(72)*A;
C=rotate(144)*A;
label("$p_1=0$", A, N);
label("$p_n=0$", B, NW);
label("$p_{n-1}=X$", C, SW);
// Block 3
draw(circle((0, 0), 5));
pair A, B, C;
A=(0, 5);
B=rotate(72)*A;
C=rotate(144)*A;
label("$p_1=1$", A, N);
label("$p_n=0$", B, NW);
label("$p_{n-1}=1$", C, SW);
// Block 4
draw(circle((0, 0), 5));
pair A, B, C;
A=(0, 5);
B=rotate(72)*A;
C=rotate(144)*A;
label("$p_1=0$", A, N);
label("$p_n=1$", B, NW);
label("$p_{n-1}=0$", C, SW);
// Block 5
draw(circle((0, 0), 5)); pair A, B, C; A=(0, 5); B=rotate(72)*A; C=rotate(144)*A; label("$p_1=1$", A, N); label("$p_n=0$", B, NW); label("$p_{n-1}=0$", C, SW);
// Block 6
draw(circle((0, 0), 5)); pair A, B, C; A=(0, 5); B=rotate(72)*A; C=rotate(144)*A; label("$p_1=0$", A, N); label("$p_n=0$", B, NW); label("$p_{n-1}=X$", C, SW);
// Block 7
draw(circle((0, 0), 5)); pair A, B, C; A=(0, 5); B=rotate(72)*A; C=rotate(144)*A; label("$p_1=1$", A, N); label("$p_n=0$", B, NW); label("$p_{n-1}=1$", C, SW);
// Block 8
draw(circle((0, 0), 5)); pair A, B, C; A=(0, 5); B=rotate(72)*A; C=rotate(144)*A; label("$p_1=0$", A, N); label("$p_n=1$", B, NW); label("$p_{n-1}=0$", C, SW); | [] |
766 | Eight people are sitting around a circular table, each holding a fair coin. All eight people flip their coins and those who flip heads stand while those who flip tails remain seated. What is the probability that no two adjacent people will stand?
$\textbf{(A)}\dfrac{47}{256}\qquad\textbf{(B)}\dfrac{3}{16}\qquad\textbf{(C) }\dfrac{49}{256}\qquad\textbf{(D) }\dfrac{25}{128}\qquad\textbf{(E) }\dfrac{51}{256}$ | 2015 AMC 12A Problem 17 | Case 1: Everyone flips tails
1 way:
Case 2: One person flips heads
8 ways:
Case 3: Two people flip heads
There are $\binom{8}{2}$ ways to choose any $2$ people in the circle and $8$ ways to choose $2$ people adjacent to each other. So this case gives us $\binom{8}{2} - 8 = 20$ ways. An example of choosing two nonadjacent people is shown below.
Case 4: Three people flip heads
We can choose the first person (shaded gray) in $8$ ways. We can choose the other $2$ people in $6$ ways (purple-blue, yellow-red, green-yellow, green-blue, purple-red, green-red). We divide by $3$ for overcounting to get $\frac{8 \cdot 6}{3} = 16$ ways for this case.
Case 5: 4 people flip heads
2 ways (we can either choose the people shaded in red or blue)
It's impossible to have more than 4 people standing without anyone being next to each other, so we stop here.
The total number of ways is $2^8,$ so the answer is $\frac{1 + 8 + 20 + 16 + 2}{2^8} = \boxed{\frac{47}{256}}.$
~grogg007 | // Block 1
size(50); import graph; real R = 0.5; // Radius of big circle real r = 0.05; // Radius of small circles int n = 8; // Number of small circles // Draw big circle draw(Circle((0,0), R), linewidth(1)); // Place small circles on circumference for (int i = 0; i < n; ++i) { real angle = 2*pi*i/n; pair center = R * dir(degrees(angle)); draw(Circle(center, r), linewidth(0.8)); }
// Block 2
size(400); import graph; real R = 1.5; // Radius of big circle real r = 0.2; // Radius of small circles int n = 8; // Number of small circles for (int j = 0; j < 8; ++j) { pair centerOffset = (j * 4, 0); // space between diagrams // Draw big circle draw(shift(centerOffset) * Circle((0,0), R), linewidth(1)); // Draw small circles for (int i = 0; i < n; ++i) { real angle = 2 * pi * i / n; pair pos = R * dir(degrees(angle)); pair finalPos = centerOffset + pos; if (i == j) { fill(Circle(finalPos, r), black); // Shade this one } else { draw(Circle(finalPos, r), linewidth(0.8)); // Outline others } } }
// Block 3
size(50); import graph; real R = 0.5; // Radius of big circle real r = 0.05; // Radius of small circles int n = 8; // Number of small circles // Indices of two circles that are one circle apart int a = 0; int b = 2; // Compute centers of these two circles real angleA = 2*pi*a/n; real angleB = 2*pi*b/n; pair centerA = R * dir(degrees(angleA)); pair centerB = R * dir(degrees(angleB)); // Draw big circle draw(Circle((0,0), R), linewidth(1)); // Draw small circles and shade the two chosen ones for (int i = 0; i < n; ++i) { real angle = 2*pi*i/n; pair center = R * dir(degrees(angle)); if (i == a || i == b) { fill(Circle(center, r), black); } else { draw(Circle(center, r), linewidth(0.8)); } } // Draw line connecting the two shaded circles draw(centerA -- centerB, linewidth(1));
// Block 4
size(50); import graph; real R = 1.5; // Radius of big circle real r = 0.2; // Radius of small circles int n = 8; // Number of small circles pair center = (0, 0); // Draw big circle draw(Circle(center, R), linewidth(1)); // Compute positions (clockwise, starting at top) pair[] pts; for (int i = 0; i < n; ++i) { real angle = 90 - 360 * i / n; // Start at 90° and go clockwise pts[i] = center + R * dir(angle); } // Draw circles with unique colors for shaded ones for (int i = 0; i < n; ++i) { if (i == 0) { fill(Circle(pts[i], r), gray); // Circle 1 } else if (i == 2) { fill(Circle(pts[i], r), red); // Circle 3 } else if (i == 3) { fill(Circle(pts[i], r), blue); // Circle 4 } else if (i == 4) { fill(Circle(pts[i], r), yellow); // Circle 5 } else if (i == 5) { fill(Circle(pts[i], r), purple); // Circle 6 } else if (i == 6) { fill(Circle(pts[i], r), green); // Circle 7 } else { draw(Circle(pts[i], r), linewidth(0.8)); // Others outlined only } } // Draw triangle connecting circles 1, 4, 6 (indices 0, 3, 5) draw(pts[0] -- pts[3] -- pts[5] -- cycle, linewidth(1));
// Block 5
size(50); import graph; real R = 1.5; real r = 0.15; // Circle centers, clockwise starting at top pair c0 = (0, R); pair c1 = R * dir(45); pair c2 = R * dir(90 - 2*45); pair c3 = R * dir(90 - 3*45); pair c4 = R * dir(90 - 4*45); pair c5 = R * dir(90 - 5*45); pair c6 = R * dir(90 - 6*45); pair c7 = R * dir(90 - 7*45); // Draw big circle draw(Circle((0,0), R), linewidth(1)); // Group 1 vertices (blue) pen bluePen = blue + linewidth(1.5); filldraw(Circle(c0, r), blue); filldraw(Circle(c2, r), blue); filldraw(Circle(c4, r), blue); filldraw(Circle(c6, r), blue); draw(c0--c2--c4--c6--c0, bluePen); // Group 2 vertices (red) pen redPen = red + linewidth(1.5); filldraw(Circle(c1, r), red); filldraw(Circle(c3, r), red); filldraw(Circle(c5, r), red); filldraw(Circle(c7, r), red); draw(c1--c3--c5--c7--c1, redPen); | [] |
766 | Eight people are sitting around a circular table, each holding a fair coin. All eight people flip their coins and those who flip heads stand while those who flip tails remain seated. What is the probability that no two adjacent people will stand?
$\textbf{(A)}\dfrac{47}{256}\qquad\textbf{(B)}\dfrac{3}{16}\qquad\textbf{(C) }\dfrac{49}{256}\qquad\textbf{(D) }\dfrac{25}{128}\qquad\textbf{(E) }\dfrac{51}{256}$ | 2015 AMC 12A Problem 17 | We know that the denominator of the probability is $2^8=256$. So now we only have to calculate the numerator, which is the number of arrangements for $8$ people at a round table without $2$ or more neighboring people standing.
Denote $a_n$ as number of arrangements for $n$ people at a round table without $2$ or more neighboring people standing. We can see that $a_2=3$, $a_3=4$, we are going to prove $a_n=a_{n-1}+a_{n-2}$
We use $1$ to represent standing people, and $0$ as sitting people. The problem becomes arranging $n$ numbers of $0$ or $1$ around a circle with no consecutive $1$'s.
We elect one person as the chairman, let him be $p_1$, the first element in this circular sequence.
There are $2$ cases to generate the arrangement of $n$ numbers.
$Case$ $1:$ From the arrangement of $n-1$ numbers, add $1$ more number $p_n$ counter-clockwise next to $p_1$.
If $p_1=1$, $p_{n-1}p_np_1$ is $001$, as the diagram shows:
If $p_1=0$, $p_{n-1}p_np_1$ is $X00$, $X$ could be $0$ or $1$, as the diagram shows:
$Case$ $2:$ From the arrangement of $n-2$ numbers, add $2$ more numbers $p_{n-1}$ and $p_n$ counter-clockwise next to $p_1$.
If $p_1=1$, then $p_{n-2}=0$, let $p_{n-1}=1, p_n=0$, $p_{n-1}p_np_1$ is $101$, as the diagram shows:
If $p_1=0$, then $p_{n-2}$ could be $0$ or $1$. Let $p_{n-1}=0, p_n=1$, $p_{n-1}p_np_1$ is $010$, as the diagram shows:
From the above $2$ cases and the $4$ diagrams, the arrangements of $p_{n-1}p_np_1$ are mutually exclusive collectively exhaustive, so $a_{n}=a_{n-1}+a_{n-2}$
$a_2=3$
$a_3=4$
$a_4=7$
$a_5=11$
$a_6=18$
$a_7=29$
$a_8=47$
The answer is $\boxed{\textbf{(A) } \dfrac{47}{256}}$
This sequence of numbers is called Lucas Number.
~isabelchen | // Block 1
draw(circle((0, 0), 5)); pair A, B, C; A=(0, 5); B=rotate(72)*A; C=rotate(144)*A; label("$p_1=1$", A, N); label("$p_n=0$", B, NW); label("$p_{n-1}=0$", C, SW);
// Block 2
draw(circle((0, 0), 5)); pair A, B, C; A=(0, 5); B=rotate(72)*A; C=rotate(144)*A; label("$p_1=0$", A, N); label("$p_n=0$", B, NW); label("$p_{n-1}=X$", C, SW);
// Block 3
draw(circle((0, 0), 5)); pair A, B, C; A=(0, 5); B=rotate(72)*A; C=rotate(144)*A; label("$p_1=1$", A, N); label("$p_n=0$", B, NW); label("$p_{n-1}=1$", C, SW);
// Block 4
draw(circle((0, 0), 5)); pair A, B, C; A=(0, 5); B=rotate(72)*A; C=rotate(144)*A; label("$p_1=0$", A, N); label("$p_n=1$", B, NW); label("$p_{n-1}=0$", C, SW); | [] |
767 | Johann has $64$ fair coins. He flips all the coins. Any coin that lands on tails is tossed again. Coins that land on tails on the second toss are tossed a third time. What is the expected number of coins that are now heads?
$\textbf{(A) } 32 \qquad\textbf{(B) } 40 \qquad\textbf{(C) } 48 \qquad\textbf{(D) } 56 \qquad\textbf{(E) } 64$ | 2015 AMC 10B Problem 18 | We can simplify the problem first, then apply reasoning to the original problem. Let's say that there are $8$ coins. Shaded coins flip heads, and blank coins flip tails. So, after the first flip;
Then, after the second (new heads in blue);
And after the third (new head in green);
So in total, $7$ of the $8$ coins resulted in heads. Now we have the ratio of $\frac{7}{8}$ of the total coins will end up heads. Therefore, we have $\frac{7}{8}\cdot64=\boxed{\mathbf{(D)}\ 56}$ | // Block 1
filldraw(circle((-5,0),0.35),white);
filldraw(circle((-4,0),0.35),white);
filldraw(circle((-3,0),0.35),white);
filldraw(circle((-2,0),0.35),white);
filldraw(circle((-1,0),0.35),black);
filldraw(circle((-0,0),0.35),black);
filldraw(circle((1,0),0.35),black);
filldraw(circle((2,0),0.35),black);
// Block 2
filldraw(circle((-5,0),0.35),white);
filldraw(circle((-4,0),0.35),white);
filldraw(circle((-3,0),0.35),blue);
filldraw(circle((-2,0),0.35),blue);
filldraw(circle((-1,0),0.35),black);
filldraw(circle((-0,0),0.35),black);
filldraw(circle((1,0),0.35),black);
filldraw(circle((2,0),0.35),black);
// Block 3
filldraw(circle((-5,0),0.35),white);
filldraw(circle((-4,0),0.35),green);
filldraw(circle((-3,0),0.35),blue);
filldraw(circle((-2,0),0.35),blue);
filldraw(circle((-1,0),0.35),black);
filldraw(circle((-0,0),0.35),black);
filldraw(circle((1,0),0.35),black);
filldraw(circle((2,0),0.35),black);
// Block 4
filldraw(circle((-5,0),0.35),white); filldraw(circle((-4,0),0.35),white); filldraw(circle((-3,0),0.35),white); filldraw(circle((-2,0),0.35),white); filldraw(circle((-1,0),0.35),black); filldraw(circle((-0,0),0.35),black); filldraw(circle((1,0),0.35),black); filldraw(circle((2,0),0.35),black);
// Block 5
filldraw(circle((-5,0),0.35),white); filldraw(circle((-4,0),0.35),white); filldraw(circle((-3,0),0.35),blue); filldraw(circle((-2,0),0.35),blue); filldraw(circle((-1,0),0.35),black); filldraw(circle((-0,0),0.35),black); filldraw(circle((1,0),0.35),black); filldraw(circle((2,0),0.35),black);
// Block 6
filldraw(circle((-5,0),0.35),white); filldraw(circle((-4,0),0.35),green); filldraw(circle((-3,0),0.35),blue); filldraw(circle((-2,0),0.35),blue); filldraw(circle((-1,0),0.35),black); filldraw(circle((-0,0),0.35),black); filldraw(circle((1,0),0.35),black); filldraw(circle((2,0),0.35),black); | [] |
768 | In $\triangle{ABC}$, $\angle{C} = 90^{\circ}$ and $AB = 12$. Squares $ABXY$ and $ACWZ$ are constructed outside of the triangle. The points $X, Y, Z$, and $W$ lie on a circle. What is the perimeter of the triangle?
$\textbf{(A) }12+9\sqrt{3}\qquad\textbf{(B) }18+6\sqrt{3}\qquad\textbf{(C) }12+12\sqrt{2}\qquad\textbf{(D) }30\qquad\textbf{(E) }32$ | 2015 AMC 10B Problem 19 | The center of the circle lies on the intersection between the perpendicular bisectors of chords $ZW$ and $YX$. Therefore we know the center of the circle must also be the midpoint of the hypotenuse. Let this point be $O$. Draw perpendiculars to $ZW$ and $YX$ from $O$, and connect $OZ$ and $OY$. $OY^2=6^2+12^2=180$. Let $AC=a$ and $BC=b$. Then $\left(\dfrac{a}{2}\right)^2+\left(a+\dfrac{b}{2}\right)^2=OZ^2=OY^2=180$. Simplifying this gives $\dfrac{a^2}{4}+\dfrac{b^2}{4}+a^2+ab=180$. But by Pythagorean Theorem on $\triangle ABC$, we know $a^2+b^2=144$, because $AB=12$. Thus $\dfrac{a^2}{4}+\dfrac{b^2}{4}=\dfrac{144}{4}=36$. So our equation simplifies further to $a^2+ab=144$. However $a^2+b^2=144$, so $a^2+ab=a^2+b^2$, which means $ab=b^2$, or $a=b$. Aha! This means $\triangle ABC$ is just an isosceles right triangle, so $AC=BC=\dfrac{12}{\sqrt{2}}=6\sqrt{2}$, and thus the perimeter is $\boxed{\textbf{(C)}\ 12+12\sqrt{2}}$. | // Block 1
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(11.5cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -4.3, xmax = 18.7, ymin = -5.26, ymax = 6.3; /* image dimensions */
draw((3.46,0.96)--(3.44,-3.36)--(8.02,-3.44)--cycle);
draw((3.46,0.96)--(8.02,-3.44)--(12.42,1.12)--(7.86,5.52)--cycle);
/* draw figures */
draw((3.46,0.96)--(3.44,-3.36));
draw((3.44,-3.36)--(8.02,-3.44));
draw((8.02,-3.44)--(3.46,0.96));
draw((3.46,0.96)--(-0.86,0.98));
draw((-0.86,0.98)--(-0.88,-3.34));
draw((-0.88,-3.34)--(3.44,-3.36));
draw((3.46,0.96)--(8.02,-3.44));
draw((8.02,-3.44)--(12.42,1.12));
draw((12.42,1.12)--(7.86,5.52));
draw((7.86,5.52)--(3.46,0.96));
draw((5.74,-1.24)--(-0.86,0.98));
draw((5.74,-1.24)--(-0.87,-1.18), linetype("4 4"));
draw((5.74,-1.24)--(7.86,5.52));
draw((5.74,-1.24)--(10.14,3.32), linetype("4 4"));
draw(shift((5.82,-1.21))*xscale(6.99920709795045)*yscale(6.99920709795045)*arc((0,0),1,19.44457562540183,197.63600413408128), linetype("2 2"));
/* dots and labels */
dot((3.46,0.96),dotstyle);
label("$A$", (3.2,1.06), NE * labelscalefactor);
dot((3.44,-3.36),dotstyle);
label("$C$", (3.14,-3.86), NE * labelscalefactor);
dot((8.02,-3.44),dotstyle);
label("$B$", (8.06,-3.8), NE * labelscalefactor);
dot((-0.86,0.98),dotstyle);
label("$Z$", (-1.34,1.12), NE * labelscalefactor);
dot((-0.88,-3.34),dotstyle);
label("$W$", (-1.48,-3.54), NE * labelscalefactor);
dot((12.42,1.12),dotstyle);
label("$X$", (12.5,1.24), NE * labelscalefactor);
dot((7.86,5.52),dotstyle);
label("$Y$", (7.94,5.64), NE * labelscalefactor);
dot((5.74,-1.24),dotstyle);
label("$O$", (5.52,-1.82), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
// Block 2
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(11.5cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.3, xmax = 18.7, ymin = -5.26, ymax = 6.3; /* image dimensions */ draw((3.46,0.96)--(3.44,-3.36)--(8.02,-3.44)--cycle); draw((3.46,0.96)--(8.02,-3.44)--(12.42,1.12)--(7.86,5.52)--cycle); /* draw figures */ draw((3.46,0.96)--(3.44,-3.36)); draw((3.44,-3.36)--(8.02,-3.44)); draw((8.02,-3.44)--(3.46,0.96)); draw((3.46,0.96)--(-0.86,0.98)); draw((-0.86,0.98)--(-0.88,-3.34)); draw((-0.88,-3.34)--(3.44,-3.36)); draw((3.46,0.96)--(8.02,-3.44)); draw((8.02,-3.44)--(12.42,1.12)); draw((12.42,1.12)--(7.86,5.52)); draw((7.86,5.52)--(3.46,0.96)); draw((5.74,-1.24)--(-0.86,0.98)); draw((5.74,-1.24)--(-0.87,-1.18), linetype("4 4")); draw((5.74,-1.24)--(7.86,5.52)); draw((5.74,-1.24)--(10.14,3.32), linetype("4 4")); draw(shift((5.82,-1.21))*xscale(6.99920709795045)*yscale(6.99920709795045)*arc((0,0),1,19.44457562540183,197.63600413408128), linetype("2 2")); /* dots and labels */ dot((3.46,0.96),dotstyle); label("$A$", (3.2,1.06), NE * labelscalefactor); dot((3.44,-3.36),dotstyle); label("$C$", (3.14,-3.86), NE * labelscalefactor); dot((8.02,-3.44),dotstyle); label("$B$", (8.06,-3.8), NE * labelscalefactor); dot((-0.86,0.98),dotstyle); label("$Z$", (-1.34,1.12), NE * labelscalefactor); dot((-0.88,-3.34),dotstyle); label("$W$", (-1.48,-3.54), NE * labelscalefactor); dot((12.42,1.12),dotstyle); label("$X$", (12.5,1.24), NE * labelscalefactor); dot((7.86,5.52),dotstyle); label("$Y$", (7.94,5.64), NE * labelscalefactor); dot((5.74,-1.24),dotstyle); label("$O$", (5.52,-1.82), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); | [] |
768 | In $\triangle{ABC}$, $\angle{C} = 90^{\circ}$ and $AB = 12$. Squares $ABXY$ and $ACWZ$ are constructed outside of the triangle. The points $X, Y, Z$, and $W$ lie on a circle. What is the perimeter of the triangle?
$\textbf{(A) }12+9\sqrt{3}\qquad\textbf{(B) }18+6\sqrt{3}\qquad\textbf{(C) }12+12\sqrt{2}\qquad\textbf{(D) }30\qquad\textbf{(E) }32$ | 2015 AMC 10B Problem 19 | Both solution 1 and 2 uses Pythagorean Theorem to prove $\triangle ABC$ is isosceles right triangle. I'm going to prove $\triangle ABC$ is isosceles right triangle without using Pythagorean Theorem. I will use geometry rotation.
Let $O$ be the the midpoint of $AB$. The perpendicular bisector of line $WZ$ and $XY$ will meet at $O$. Thus $O$ is the center of the circle points $X$, $Y$, $Z$, and $W$ lie on.
$\angle ZAC=\angle OAY=90^{\circ}$, $\angle ZAC+\angle BAC=\angle OAY+\angle BAC$, $\angle ZAB=\angle CAY$, and $AZ=AC$, $AB=AY$, $\triangle AZB \cong \triangle ACY$ by $SAS$, $BZ=YC$. Because $AZ \perp AC$, $\triangle ACY$ is a $90^{\circ}$ rotation about point $A$ of $\triangle AZB$. So, $BZ \perp YC$.
Because $OZ$ and $OY$ is the radius of $\odot O$, $OZ=OY$. Because $O$ is the midpoint of hypotenuse $AB$, $OB=OC$, $BZ=CY$, $\triangle OBZ \cong \triangle OCY$ by $SSS$. Because $BZ \perp CY$, $\triangle OCY$ is a $90^{\circ}$ rotation about point $O$ of $\triangle OBZ$. So, $OB \perp OC$.
$\angle COB = 90^{\circ}$, $OC=OB$, $\triangle OCB$ is isosceles right triangle, $\angle ABC=\angle OBC=45^{\circ}$. So, $\triangle ABC$ is isosceles right triangle.
Therefore, $AC=BC=\dfrac{12}{\sqrt{2}}=6\sqrt{2}$, the perimeter is $\boxed{\textbf{(C)}\ 12+12\sqrt{2}}$.
~isabelchen | // Block 1
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(11.5cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -4.3, xmax = 18.7, ymin = -5.26, ymax = 6.3; /* image dimensions */
draw((3.46,0.96)--(3.44,-3.36)--(8.02,-3.44)--cycle);
draw((3.46,0.96)--(8.02,-3.44)--(12.42,1.12)--(7.86,5.52)--cycle);
/* draw figures */
draw((3.46,0.96)--(3.44,-3.36));
draw((3.44,-3.36)--(8.02,-3.44));
draw((8.02,-3.44)--(3.46,0.96));
draw((3.46,0.96)--(-0.86,0.98));
draw((-0.86,0.98)--(-0.88,-3.34));
draw((-0.88,-3.34)--(3.44,-3.36));
draw((3.46,0.96)--(8.02,-3.44));
draw((8.02,-3.44)--(12.42,1.12));
draw((12.42,1.12)--(7.86,5.52));
draw((7.86,5.52)--(3.46,0.96));
draw((5.74,-1.24)--(-0.86,0.98));
draw((5.74,-1.24)--(-0.87,-1.18), linetype("4 4"));
draw((5.74,-1.24)--(7.86,5.52));
draw((5.74,-1.24)--(10.14,3.32), linetype("4 4"));
draw(shift((5.82,-1.21))*xscale(6.99920709795045)*yscale(6.99920709795045)*arc((0,0),1,19.44457562540183,197.63600413408128), linetype("2 2"));
draw((8.02,-3.44)--(-0.86,0.98));
draw((3.44,-3.36)--(7.86,5.52));
draw((3.44,-3.36)--(5.74,-1.24));
/* dots and labels */
dot((3.46,0.96),dotstyle);
label("$A$", (3.2,1.06), NE * labelscalefactor);
dot((3.44,-3.36),dotstyle);
label("$C$", (3.14,-3.86), NE * labelscalefactor);
dot((8.02,-3.44),dotstyle);
label("$B$", (8.06,-3.8), NE * labelscalefactor);
dot((-0.86,0.98),dotstyle);
label("$Z$", (-1.34,1.12), NE * labelscalefactor);
dot((-0.88,-3.34),dotstyle);
label("$W$", (-1.48,-3.54), NE * labelscalefactor);
dot((12.42,1.12),dotstyle);
label("$X$", (12.5,1.24), NE * labelscalefactor);
dot((7.86,5.52),dotstyle);
label("$Y$", (7.94,5.64), NE * labelscalefactor);
dot((5.74,-1.24),dotstyle);
label("$O$", (5.52,-1.82), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
// Block 2
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(11.5cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.3, xmax = 18.7, ymin = -5.26, ymax = 6.3; /* image dimensions */ draw((3.46,0.96)--(3.44,-3.36)--(8.02,-3.44)--cycle); draw((3.46,0.96)--(8.02,-3.44)--(12.42,1.12)--(7.86,5.52)--cycle); /* draw figures */ draw((3.46,0.96)--(3.44,-3.36)); draw((3.44,-3.36)--(8.02,-3.44)); draw((8.02,-3.44)--(3.46,0.96)); draw((3.46,0.96)--(-0.86,0.98)); draw((-0.86,0.98)--(-0.88,-3.34)); draw((-0.88,-3.34)--(3.44,-3.36)); draw((3.46,0.96)--(8.02,-3.44)); draw((8.02,-3.44)--(12.42,1.12)); draw((12.42,1.12)--(7.86,5.52)); draw((7.86,5.52)--(3.46,0.96)); draw((5.74,-1.24)--(-0.86,0.98)); draw((5.74,-1.24)--(-0.87,-1.18), linetype("4 4")); draw((5.74,-1.24)--(7.86,5.52)); draw((5.74,-1.24)--(10.14,3.32), linetype("4 4")); draw(shift((5.82,-1.21))*xscale(6.99920709795045)*yscale(6.99920709795045)*arc((0,0),1,19.44457562540183,197.63600413408128), linetype("2 2")); draw((8.02,-3.44)--(-0.86,0.98)); draw((3.44,-3.36)--(7.86,5.52)); draw((3.44,-3.36)--(5.74,-1.24)); /* dots and labels */ dot((3.46,0.96),dotstyle); label("$A$", (3.2,1.06), NE * labelscalefactor); dot((3.44,-3.36),dotstyle); label("$C$", (3.14,-3.86), NE * labelscalefactor); dot((8.02,-3.44),dotstyle); label("$B$", (8.06,-3.8), NE * labelscalefactor); dot((-0.86,0.98),dotstyle); label("$Z$", (-1.34,1.12), NE * labelscalefactor); dot((-0.88,-3.34),dotstyle); label("$W$", (-1.48,-3.54), NE * labelscalefactor); dot((12.42,1.12),dotstyle); label("$X$", (12.5,1.24), NE * labelscalefactor); dot((7.86,5.52),dotstyle); label("$Y$", (7.94,5.64), NE * labelscalefactor); dot((5.74,-1.24),dotstyle); label("$O$", (5.52,-1.82), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); | [] |
769 | Erin the ant starts at a given corner of a cube and crawls along exactly 7 edges in such a way that she visits every corner exactly once and then finds that she is unable to return along an edge to her starting point. How many paths are there meeting these conditions?
$\textbf{(A) }\text{6}\qquad\textbf{(B) }\text{9}\qquad\textbf{(C) }\text{12}\qquad\textbf{(D) }\text{18}\qquad\textbf{(E) }\text{24}$ | 2015 AMC 10B Problem 20 | We label the vertices of the cube as different letters and numbers shown above. We label these so that Erin can only crawl from a number to a letter or a letter to a number (this can be seen as a coloring argument). The starting point is labeled $A$.
If we define a "move" as each time Erin crawls along a single edge from one vertex to another, we see that after 7 moves, Erin must be on a numbered vertex. Since this numbered vertex cannot be one unit away from $A$ (since Erin cannot crawl back to $A$), this vertex must be $4$.
Therefore, we now just need to count the number of paths from $A$ to $4$. To count this, we can work backwards. There are 3 choices for which vertex Erin was at before she moved to $4$, and 2 choices for which vertex Erin was at 2 moves before $4$. All of Erin's previous moves were forced, so the total number of legal paths from $A$ to $4$ is $3 \cdot 2 = \boxed{\textbf{(A)}\; 6}$. | // Block 1
import three; draw((1,1,1)--(1,0,1)--(1,0,0)--(0,0,0)--(0,0,1)--(0,1,1)--(1,1,1)--(1,1,0)--(0,1,0)--(0,1,1)); draw((0,0,1)--(1,0,1)); draw((1,0,0)--(1,1,0)); draw((0,0,0)--(0,1,0));
label("2",(0,0,0),S);
label("A",(1,0,0),W);
label("B",(0,0,1),N);
label("1",(1,0,1),NW);
label("3",(1,1,0),S);
label("C",(0,1,0),E);
label("D",(1,1,1),SE);
label("4",(0,1,1),NE);
// Block 2
import three; draw((1,1,1)--(1,0,1)--(1,0,0)--(0,0,0)--(0,0,1)--(0,1,1)--(1,1,1)--(1,1,0)--(0,1,0)--(0,1,1)); draw((0,0,1)--(1,0,1)); draw((1,0,0)--(1,1,0)); draw((0,0,0)--(0,1,0)); label("2",(0,0,0),S); label("A",(1,0,0),W); label("B",(0,0,1),N); label("1",(1,0,1),NW); label("3",(1,1,0),S); label("C",(0,1,0),E); label("D",(1,1,1),SE); label("4",(0,1,1),NE); | [] |
770 | In the figure shown below, $ABCDE$ is a regular pentagon and $AG=1$. What is $FG + JH + CD$?
$\textbf{(A) } 3 \qquad\textbf{(B) } 12-4\sqrt5 \qquad\textbf{(C) } \dfrac{5+2\sqrt5}{3} \qquad\textbf{(D) } 1+\sqrt5 \qquad\textbf{(E) } \dfrac{11+11\sqrt5}{10}$ | 2015 AMC 10B Problem 22 | Triangle $AFG$ is isosceles, so $AG=AF=1$. $FJ = FG$ since $\triangle FGJ$ is also isosceles. Using the symmetry of pentagon $FGHIJ$, notice that $\triangle JHG \cong \triangle AFG$. Therefore, $JH=AF=1$.
Since $\triangle AJH \sim \triangle AFG$,
\[\frac{JH}{AF+FJ}=\frac{FG}{FA}\]
\[\frac{1}{1+FG} = \frac{FG}1\]
\[1 = FG^2 + FG\]
\[FG^2+FG-1 = 0\]
\[FG = \frac{-1 \pm \sqrt{5} }{2}\]
So, $FG=\frac{-1 + \sqrt{5}}{2}$ since $FG$ must be greater than 0.
Notice that $CD = AE = AJ = AF + FJ = 1 + \frac{-1 + \sqrt{5}}{2} = \frac{1 + \sqrt{5}}{2}$.
Therefore, $FG+JH+CD=\frac{-1+\sqrt5}2+1+\frac{1+\sqrt5}2=\boxed{\mathbf{(D)}\ 1+\sqrt{5}\ }$
Note by Fasolinka: Alternatively, extend $FI$ and call its intersection with $DC$ $K$. It is not hard to see that quadrilaterals $FGCK$ and $JHKD$ are parallelograms, so $DC=DK+KC=JH+FG=1+\frac{-1+\sqrt{5}}{2}$, and the same result is achieved. | // Block 1
pair A=(cos(pi/5)-sin(pi/10),cos(pi/10)+sin(pi/5)), B=(2*cos(pi/5)-sin(pi/10),cos(pi/10)), C=(1,0), D=(0,0), E1=(-sin(pi/10),cos(pi/10));
//(0,0) is a convenient point
//E1 to prevent conflict with direction E(ast)
pair F=intersectionpoints(D--A,E1--B)[0], G=intersectionpoints(A--C,E1--B)[0], H=intersectionpoints(B--D,A--C)[0], I=intersectionpoints(C--E1,D--B)[0], J=intersectionpoints(E1--C,D--A)[0];
draw(A--B--C--D--E1--A);
draw(A--D--B--E1--C--A);
draw(F--I--G--J--H--F);
label("$A$",A,N);
label("$B$",B,E);
label("$C$",C,SE);
label("$D$",D,SW);
label("$E$",E1,W);
label("$F$",F,NW);
label("$G$",G,NE);
label("$H$",H,E);
label("$I$",I,S);
label("$J$",J,W);
// Block 2
pair A=(cos(pi/5)-sin(pi/10),cos(pi/10)+sin(pi/5)), B=(2*cos(pi/5)-sin(pi/10),cos(pi/10)), C=(1,0), D=(0,0), E1=(-sin(pi/10),cos(pi/10)); //(0,0) is a convenient point //E1 to prevent conflict with direction E(ast) pair F=intersectionpoints(D--A,E1--B)[0], G=intersectionpoints(A--C,E1--B)[0], H=intersectionpoints(B--D,A--C)[0], I=intersectionpoints(C--E1,D--B)[0], J=intersectionpoints(E1--C,D--A)[0]; draw(A--B--C--D--E1--A); draw(A--D--B--E1--C--A); draw(F--I--G--J--H--F); label("$A$",A,N); label("$B$",B,E); label("$C$",C,SE); label("$D$",D,SW); label("$E$",E1,W); label("$F$",F,NW); label("$G$",G,NE); label("$H$",H,E); label("$I$",I,S); label("$J$",J,W); | [] |
771 | In $\triangle ABC$, $\angle C = 90^\circ$ and $AB = 12$. Squares $ABXY$ and $CBWZ$ are constructed outside of the triangle. The points $X$, $Y$, $Z$, and $W$ lie on a circle. What is the perimeter of the triangle?
$\textbf{(A)}\; 12+9\sqrt{3} \qquad\textbf{(B)}\; 18+6\sqrt{3} \qquad\textbf{(C)}\; 12+12\sqrt{2} \qquad\textbf{(D)}\; 30 \qquad\textbf{(E)}\; 32$ | 2015 AMC 12B Problem 19 | First, we should find the center and radius of this circle. We can find the center by drawing the perpendicular bisectors of $WZ$ and $XY$ and finding their intersection point. This point happens to be the midpoint of $AB$, the hypotenuse. Let this point be $M$. To find the radius, determine $MY$, where $MY^{2} = MA^2 + AY^2$, $MA = \frac{12}{2} = 6$, and $AY = AB = 12$. Thus, the radius $=r =MY = 6\sqrt5$.
Next we let $AC = b$ and $BC = a$. Consider the right triangle $ACB$ first. Using the Pythagorean theorem, we find that $a^2 + b^2 = 12^2 = 144$.
Now, we let $E$ be the midpoint of $WZ$, and we consider right triangle $ZEM$. By the Pythagorean theorem, we have that $\left(\frac{a}{2}\right)^2 + \left(a + \frac{b}{2}\right)^2 = r^2 = 180$. Expanding this equation, we get that
\[\frac{1}{4}(a^2+b^2) + a^2 + ab = 180\]
\[\frac{144}{4} + a^2 + ab = 180\]
\[a^2 + ab = 144 = a^2 + b^2\]
\[ab = b^2\]
\[b = a\]
This means that $ABC$ is a 45-45-90 triangle, so $a = b = \frac{12}{\sqrt2} = 6\sqrt2$. Thus the perimeter is $a + b + AB = 12\sqrt2 + 12$ which is answer $\boxed{\textbf{(C)}\; 12 + 12\sqrt2}$. | // Block 1
pair A,B,C,M,E,W,Z,X,Y;
A=(2,0);
B=(0,2);
C=(0,0);
M=(A+B)/2;
W=(-2,2);
Z=(-2,-0);
X=(2,4);
Y=(4,2);
E=(W+Z)/2;
draw(A--B--C--cycle);
draw(W--B--C--Z--cycle);
draw(A--B--X--Y--cycle);
dot(M);
dot(E);
label("W",W,NW);
label("Z",Z,SW);
label("C",C,S);
label("A",A,S);
label("B",B,N);
label("X",X,NE);
label("Y",Y,SE);
label("E",E,1.5*plain.W);
label("M",M,NE);
draw(circle(M,sqrt(10)));
// Block 2
pair A,B,C,M,E,W,Z,X,Y;
A=(2,0);
B=(0,2);
C=(0,0);
M=(A+B)/2;
W=(-2,2);
Z=(-2,-0);
X=(2,4);
Y=(4,2);
E=(W+Z)/2;
draw(A--B--C--cycle);
draw(W--B--C--Z--cycle);
draw(A--B--X--Y--cycle);
dot(M);
dot(E);
label("W",W,NW);
label("Z",Z,SW);
label("C",C,S);
label("A",A,S);
label("B",B,N);
label("X",X,NE);
label("Y",Y,SE);
label("E",E,1.5*plain.W);
label("M",M,NE);
draw(circle(M,sqrt(10)));
draw(E--Z--M--cycle,dashed);
// Block 3
pair A,B,C,M,E,W,Z,X,Y; A=(2,0); B=(0,2); C=(0,0); M=(A+B)/2; W=(-2,2); Z=(-2,-0); X=(2,4); Y=(4,2); E=(W+Z)/2; draw(A--B--C--cycle); draw(W--B--C--Z--cycle); draw(A--B--X--Y--cycle); dot(M); dot(E); label("W",W,NW); label("Z",Z,SW); label("C",C,S); label("A",A,S); label("B",B,N); label("X",X,NE); label("Y",Y,SE); label("E",E,1.5*plain.W); label("M",M,NE); draw(circle(M,sqrt(10)));
// Block 4
pair A,B,C,M,E,W,Z,X,Y; A=(2,0); B=(0,2); C=(0,0); M=(A+B)/2; W=(-2,2); Z=(-2,-0); X=(2,4); Y=(4,2); E=(W+Z)/2; draw(A--B--C--cycle); draw(W--B--C--Z--cycle); draw(A--B--X--Y--cycle); dot(M); dot(E); label("W",W,NW); label("Z",Z,SW); label("C",C,S); label("A",A,S); label("B",B,N); label("X",X,NE); label("Y",Y,SE); label("E",E,1.5*plain.W); label("M",M,NE); draw(circle(M,sqrt(10))); draw(E--Z--M--cycle,dashed); | [] |
772 | A bee starts flying from point $P_0$. She flies $1$ inch due east to point $P_1$. For $j \ge 1$, once the bee reaches point $P_j$, she turns $30^{\circ}$ counterclockwise and then flies $j+1$ inches straight to point $P_{j+1}$. When the bee reaches $P_{2015}$ she is exactly $a \sqrt{b} + c \sqrt{d}$ inches away from $P_0$, where $a$, $b$, $c$ and $d$ are positive integers and $b$ and $d$ are not divisible by the square of any prime. What is $a+b+c+d$ ?
$\textbf{(A)}\; 2016 \qquad\textbf{(B)}\; 2024 \qquad\textbf{(C)}\; 2032 \qquad\textbf{(D)}\; 2040 \qquad\textbf{(E)}\; 2048$ | 2015 AMC 12B Problem 25 | Suppose that the bee makes a move of distance $i$. After $6$ turns it will be facing the opposite direction and move $i+6$ units. Combining these opposite movements gives a total movement of $-6$ units in the original direction. This means that every $12$ moves, the bee will move $-6$ units in each direction of $0^\circ, 30^\circ, 60^\circ, 90^\circ, 120^\circ, 150^\circ$.
We want to find the displacement vector for every $12$ moves. Factoring out the $-6$ for now (which flips the direction), we draw a quick diagram of one unit in each direction.
Using the 30-60-90 triangles, it is clear that the displacement vector (factoring back in the $-6$) is $-6\left\langle 1, 2+\sqrt{3}\right\rangle$.
To compute the distance to $P_{2015}$, we can compute the position of $P_{2016}$ (a multiple of $12$ moves) and then subtract the vector from $P_{2015}$ to $P_{2016}$.
The bee reaches $P_{2016}$ after $\frac{2016}{12} = 168$ sets of $12$ moves, so the total displacement vector to $P_{2016}$ is $168(-6)\left\langle 1, 2+\sqrt{3}\right\rangle = \left\langle -1008, -2006-1008\sqrt{3}\right\rangle$.
The bee moves at an angle of $-30^\circ$ from $P_{2015}$ to $P_{2016}$, so subtracting it means moving an angle of $150^\circ$. Since the vector is $2016$ units long, by a 30-60-90 triangle, it is $\left\langle -1008\sqrt{3}, 1008\right\rangle$.
Therefore the total displacement vector to $P_{2015}$ is $\left\langle -1008, -2006-1008\sqrt{3}\right\rangle + \left\langle -1008\sqrt{3}, 1008\right\rangle = \left\langle -1008-1008\sqrt{3}, -1008-1008\sqrt{3}\right\rangle$. The displacement is thus $\sqrt{2}\left|-1008-1008\sqrt{3}\right| = 1008\sqrt{2}+1008\sqrt{6} \implies 1008+2+1008+6 = \boxed{\text{(B) }2024}$.
Fun fact: The displacement to $P_{2016}$ is the same as for $P_{2015}$. | // Block 1
draw((0,0)--(1,0)--(1+sqrt(3)/2, 1/2)--(3/2+sqrt(3)/2, 1/2+sqrt(3)/2)--(3/2+sqrt(3)/2, 3/2+sqrt(3)/2)--(1+sqrt(3)/2, 3/2+sqrt(3))--(1, 2+sqrt(3)), EndArrow);
draw((1,0)--(3/2+sqrt(3)/2, 0), dashed);
draw((1,2+sqrt(3))--(3/2+sqrt(3)/2, 2+sqrt(3)), dashed);
draw((3/2+sqrt(3)/2, 2+sqrt(3))--(3/2+sqrt(3)/2, 0), dashed);
draw((1+sqrt(3)/2,0)--(1+sqrt(3)/2, 1/2)--(3/2+sqrt(3)/2, 1/2), dashed);
draw((1+sqrt(3)/2,2+sqrt(3))--(1+sqrt(3)/2, 3/2+sqrt(3))--(3/2+sqrt(3)/2, 3/2+sqrt(3)), dashed);
// Block 2
draw((0,0)--(1,0)--(1+sqrt(3)/2, 1/2)--(3/2+sqrt(3)/2, 1/2+sqrt(3)/2)--(3/2+sqrt(3)/2, 3/2+sqrt(3)/2)--(1+sqrt(3)/2, 3/2+sqrt(3))--(1, 2+sqrt(3)), EndArrow); draw((1,0)--(3/2+sqrt(3)/2, 0), dashed); draw((1,2+sqrt(3))--(3/2+sqrt(3)/2, 2+sqrt(3)), dashed); draw((3/2+sqrt(3)/2, 2+sqrt(3))--(3/2+sqrt(3)/2, 0), dashed); draw((1+sqrt(3)/2,0)--(1+sqrt(3)/2, 1/2)--(3/2+sqrt(3)/2, 1/2), dashed); draw((1+sqrt(3)/2,2+sqrt(3))--(1+sqrt(3)/2, 3/2+sqrt(3))--(3/2+sqrt(3)/2, 3/2+sqrt(3)), dashed); | [] |
773 | Find the area of the shaded region.
$\textbf{(A)}\ 4\dfrac{3}{5} \qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 5\dfrac{1}{4} \qquad \textbf{(D)}\ 6\dfrac{1}{2} \qquad \textbf{(E)}\ 8$ | 2016 AMC 10A Problem 11 | The bases of these triangles are all $1$, and by symmetry, their heights are $4$, $\frac{5}{2}$, $4$, and $\frac{5}{2}$. Thus, their areas are $2$, $\frac{5}{4}$, $2$, and $\frac{5}{4}$, which add to the area of the shaded region, which is $\boxed{6\frac{1}{2}}$. | // Block 1
size(6cm);
defaultpen(fontsize(9pt));
draw((0,0)--(8,0)--(8,5)--(0,5)--cycle);
filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8));
label("$1$",(1/2,5),dir(90));
label("$7$",(9/2,5),dir(90));
label("$1$",(8,1/2),dir(0));
label("$1$",(15/2,0),dir(270));
label("$7$",(7/2,0),dir(270));
label("$1$",(0,9/2),dir(180));
label("$4$",(0,2),dir(180));
draw((0,5)--(8,0));
// Block 2
size(6cm); defaultpen(fontsize(9pt)); draw((0,0)--(8,0)--(8,5)--(0,5)--cycle); filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8)); label("$1$",(1/2,5),dir(90)); label("$7$",(9/2,5),dir(90)); label("$1$",(8,1/2),dir(0)); label("$1$",(15/2,0),dir(270)); label("$7$",(7/2,0),dir(270)); label("$1$",(0,9/2),dir(180)); label("$4$",(0,2),dir(180)); draw((0,5)--(8,0)); | [] |
773 | Find the area of the shaded region.
$\textbf{(A)}\ 4\dfrac{3}{5} \qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 5\dfrac{1}{4} \qquad \textbf{(D)}\ 6\dfrac{1}{2} \qquad \textbf{(E)}\ 8$ | 2016 AMC 10A Problem 11 | Find the area of the unshaded area by calculating the area of the triangles and rectangles outside of the shaded region. We can do this by splitting up the unshaded areas into various triangles and rectangles as shown.
Notice that the two added lines bisect each of the $4$ sides of the large rectangle.
Subtracting the unshaded area from the total area gives us $40-33\frac{1}{2}=\boxed{6\frac{1}{2}}$, so the correct answer is $\boxed{\textbf{(D)}}$. | // Block 1
size(6cm);
defaultpen(fontsize(9pt));
draw((0,0)--(8,0)--(8,5)--(0,5)--cycle);
filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8));
label("$1$",(1/2,5),dir(90));
label("$4$",(6,5),dir(90));
label("$3$",(5/2,5),dir(90));
label("$1$",(8,1/2),dir(0));
label("$5/2$",(8,15/4),dir(0));
label("$3/2$",(8,7/4),dir(0));
label("$1$",(15/2,0),dir(270));
label("$4$",(2,0),dir(270));
label("$3$",(11/2,0),dir(270));
label("$1$",(0,9/2),dir(180));
label("$5/2$",(0,5/4),dir(180));
label("$3/2$",(0,13/4),dir(180));
draw((0,5/2)--(8,5/2));
draw((4,0)--(4,5));
// Block 2
size(6cm); defaultpen(fontsize(9pt)); draw((0,0)--(8,0)--(8,5)--(0,5)--cycle); filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8)); label("$1$",(1/2,5),dir(90)); label("$4$",(6,5),dir(90)); label("$3$",(5/2,5),dir(90)); label("$1$",(8,1/2),dir(0)); label("$5/2$",(8,15/4),dir(0)); label("$3/2$",(8,7/4),dir(0)); label("$1$",(15/2,0),dir(270)); label("$4$",(2,0),dir(270)); label("$3$",(11/2,0),dir(270)); label("$1$",(0,9/2),dir(180)); label("$5/2$",(0,5/4),dir(180)); label("$3/2$",(0,13/4),dir(180)); draw((0,5/2)--(8,5/2)); draw((4,0)--(4,5)); | [] |
773 | Find the area of the shaded region.
$\textbf{(A)}\ 4\dfrac{3}{5} \qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 5\dfrac{1}{4} \qquad \textbf{(D)}\ 6\dfrac{1}{2} \qquad \textbf{(E)}\ 8$ | 2016 AMC 10A Problem 11 | Split the region into four parts by the diagonal from the top left to the bottom right. Slide the top and bottom pieces next to each other to form a parallelogram with base $1$ and height $\frac{5}{2}$, and slide the left and right pieces next to each other to form a parallelogram with base $1$ and height $\frac{8}{2}$. The total area is then $1\cdot\frac{5}{2}+1\cdot\frac{8}{2} = \frac{(5+8)(1)}{2} = \boxed{\textbf{(D)}\ 6\frac{1}{2}}$. ~emerald_block | // Block 1
size(6cm);
defaultpen(fontsize(9pt));
draw((0,0)--(8,0)--(8,5)--(0,5)--cycle);
filldraw((7,0)--(8,0)--(0,5)--(1,5)--cycle,gray(0.7));
filldraw((8,0)--(8,1)--(0,4)--(0,5)--cycle,gray(0.9));
label("$1$",(1/2,5),dir(90));
label("$7$",(9/2,5),dir(90));
label("$1$",(8,1/2),dir(0));
label("$1$",(15/2,0),dir(270));
label("$7$",(7/2,0),dir(270));
label("$1$",(0,9/2),dir(180));
label("$4$",(0,2),dir(180));
// Block 2
size(6cm); defaultpen(fontsize(9pt)); draw((0,0)--(8,0)--(8,5)--(0,5)--cycle); filldraw((7,0)--(8,0)--(0,5)--(1,5)--cycle,gray(0.7)); filldraw((8,0)--(8,1)--(0,4)--(0,5)--cycle,gray(0.9)); label("$1$",(1/2,5),dir(90)); label("$7$",(9/2,5),dir(90)); label("$1$",(8,1/2),dir(0)); label("$1$",(15/2,0),dir(270)); label("$7$",(7/2,0),dir(270)); label("$1$",(0,9/2),dir(180)); label("$4$",(0,2),dir(180)); | [] |
773 | Find the area of the shaded region.
$\textbf{(A)}\ 4\dfrac{3}{5} \qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 5\dfrac{1}{4} \qquad \textbf{(D)}\ 6\dfrac{1}{2} \qquad \textbf{(E)}\ 8$ | 2016 AMC 10A Problem 11 | We can subtract the 4 triangular areas from the overall rectangle. It can be noticed that two triangles have area $\frac{6 \cdot \frac{5}{2}}{2} = \frac{15}{2}$, and the other two triangles have area $\frac{3 \cdot \frac{8}{2}}{2} = 6$ $\Rightarrow$ the shaded area is $(8 \cdot 5) - 2(\frac{15}{2}) - 2(6) = 40 - 15 - 12 = 13$. Since the desired area is half the shaded region, our area is $\boxed{\textbf{(D)}\ 6\frac{1}{2}}$.
~Champion1234 | // Block 1
size(6cm);
defaultpen(fontsize(9pt));
draw((0,0)--(8,0)--(8,5)--(0,5)--cycle);
filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8));
filldraw((1,0)--(0,0)--(0,1)--(8,4)--(8,5)--(7,5)--cycle,gray(0.8));
label("$1$",(1/2,5),dir(90));
label("$6$",(4,5),dir(90));
label("$1$",(15/2,5),dir(90));
label("$1$",(8,1/2),dir(0));
label("$3$",(8,5/2),dir(0));
label("$1$",(8,9/2),dir(0));
label("$1$",(15/2,0),dir(270));
label("$6$",(4,0),dir(270));
label("$1$",(1/2,0),dir(270));
label("$1$",(0,9/2),dir(180));
label("$3$",(0,5/2),dir(180));
label("$1$",(0,1/2),dir(180));
// Block 2
size(6cm); defaultpen(fontsize(9pt)); draw((0,0)--(8,0)--(8,5)--(0,5)--cycle); filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8)); filldraw((1,0)--(0,0)--(0,1)--(8,4)--(8,5)--(7,5)--cycle,gray(0.8)); label("$1$",(1/2,5),dir(90)); label("$6$",(4,5),dir(90)); label("$1$",(15/2,5),dir(90)); label("$1$",(8,1/2),dir(0)); label("$3$",(8,5/2),dir(0)); label("$1$",(8,9/2),dir(0)); label("$1$",(15/2,0),dir(270)); label("$6$",(4,0),dir(270)); label("$1$",(1/2,0),dir(270)); label("$1$",(0,9/2),dir(180)); label("$3$",(0,5/2),dir(180)); label("$1$",(0,1/2),dir(180)); | [] |
774 | In rectangle $ABCD,$ $AB=6$ and $BC=3$. Point $E$ between $B$ and $C$, and point $F$ between $E$ and $C$ are such that $BE=EF=FC$. Segments $\overline{AE}$ and $\overline{AF}$ intersect $\overline{BD}$ at $P$ and $Q$, respectively. The ratio $BP:PQ:QD$ can be written as $r:s:t$ where the greatest common factor of $r,s,$ and $t$ is $1.$ What is $r+s+t$?
$\textbf{(A) } 7 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 15 \qquad \textbf{(E) } 20$ | 2016 AMC 10A Problem 19 | Use similar triangles. Our goal is to put the ratio in terms of ${BD}$. Since $\triangle APD \sim \triangle EPB,$ $\frac{DP}{PB}=\frac{AD}{BE}=3.$ Therefore, $PB=\frac{BD}{4}$. Similarly, $\frac{DQ}{QB}=\frac{3}{2}$. This means that ${DQ}=\frac{3\cdot BD}{5}$. Therefore, $r:s:t=\frac{1}{4}:\frac{2}{5}-\frac{1}{4}:\frac{3}{5}=5:3:12,$ so $r+s+t=\boxed{\textbf{(E) }20.}$ | // Block 1
size(6cm);
pair D=(0,0), C=(6,0), B=(6,3), A=(0,3);
draw(A--B--C--D--cycle);
draw(B--D);
draw(A--(6,2));
draw(A--(6,1));
label("$A$", A, dir(135));
label("$B$", B, dir(45));
label("$C$", C, dir(-45));
label("$D$", D, dir(-135));
label("$Q$", extension(A,(6,1),B,D),dir(-90));
label("$P$", extension(A,(6,2),B,D), dir(90));
label("$F$", (6,1), dir(0));
label("$E$", (6,2), dir(0));
// Block 2
size(6cm); pair D=(0,0), C=(6,0), B=(6,3), A=(0,3); draw(A--B--C--D--cycle); draw(B--D); draw(A--(6,2)); draw(A--(6,1)); label("$A$", A, dir(135)); label("$B$", B, dir(45)); label("$C$", C, dir(-45)); label("$D$", D, dir(-135)); label("$Q$", extension(A,(6,1),B,D),dir(-90)); label("$P$", extension(A,(6,2),B,D), dir(90)); label("$F$", (6,1), dir(0)); label("$E$", (6,2), dir(0)); | [] |
774 | In rectangle $ABCD,$ $AB=6$ and $BC=3$. Point $E$ between $B$ and $C$, and point $F$ between $E$ and $C$ are such that $BE=EF=FC$. Segments $\overline{AE}$ and $\overline{AF}$ intersect $\overline{BD}$ at $P$ and $Q$, respectively. The ratio $BP:PQ:QD$ can be written as $r:s:t$ where the greatest common factor of $r,s,$ and $t$ is $1.$ What is $r+s+t$?
$\textbf{(A) } 7 \qquad \textbf{(B) } 9 \qquad \textbf{(C) } 12 \qquad \textbf{(D) } 15 \qquad \textbf{(E) } 20$ | 2016 AMC 10A Problem 19 | This problem breaks down into finding $QP:PB$ and $DQ:QB$. We can find the first using mass points, and the second using similar triangles.
Draw point $G$ on $DB$ such that $FG\parallel CD$. Then, by similar triangles $FG=BF\cdot 2=4$. Again, by similar triangles $AQB$ and $FQG$, $AQ:FQ=AB:FG=6:4=3:2$. Now we begin Mass Points.
We will consider the triangle $ABF$ with center $P$, so that $E$ balances $B$ and $F$, and $Q$ balances $A$ and $F$. Assign a mass of $1$ to $B$. Then, $BE:FE=1:1$ so $F=B=1$. By mass points addition, $E=B+F=2$ since $E$ balances $B$ and $F$.
Also, $AQ:QF=3:2$ so $A=\frac{2}{3}F=\frac{2}{3}$ so $Q=\frac{5}{3}$. Then, $QP:PB=1:\frac{5}{3}=3:5$.
To calculate $DQ:BQ$, extend $AF$ past $F$ to point $H$ such that $H$ lies on $BC$. Then $AFB$ is similar to $HAD$ so $DH=3\cdot AD=9$. Also, $AQB$ is similar to $HQD$ so $DQ:BQ=9:6=3:2$
Now, we wish to get $DQ:QP:PB$. Observe that $BQ=QP+PB$. So, $DQ:BQ=DQ:QP+PB=3:2$ so (since $QP:PB=3:5$ has sum $3+5=8$), $DQ:QP+PB=3:2=12:8$. now, we may combine the two and get $DQ:QP:PB=12:3:5$ so $r+s+t=12+3+5=\boxed{\textbf{(E) }20}$.
~Firebolt360(minor edits by vadava_lx) | // Block 1
import geometry;
size(9cm);
pair D=(0,0), C=(6,0), B=(6,3), A=(0,3), G=(2,1), H=(9,0);
draw(A--B--C--D--cycle);
draw(B--D);
draw(A--(6,2));
draw(A--(6,1));
draw(A--H);
draw((6,1)--G);
draw(D--H);
label("$A$", A, dir(135));
label("$B$", B, dir(45));
label("$C$", C, dir(-45));
label("$D$", D, dir(-135));
pair Q = extension(A,(6,1),B,D);
pair P = extension(A,(6,2),B,D);
label("$Q$", Q, dir(-90));
label("$P$", P, dir(90));
label("$F$", (6,1), dir(45));
label("$E$", (6,2), dir(45));
label("$H$", H, dir(0));
label("$G$", G, dir(135));
// Block 2
import geometry; size(9cm); pair D=(0,0), C=(6,0), B=(6,3), A=(0,3), G=(2,1), H=(9,0); draw(A--B--C--D--cycle); draw(B--D); draw(A--(6,2)); draw(A--(6,1)); draw(A--H); draw((6,1)--G); draw(D--H); label("$A$", A, dir(135)); label("$B$", B, dir(45)); label("$C$", C, dir(-45)); label("$D$", D, dir(-135)); pair Q = extension(A,(6,1),B,D); pair P = extension(A,(6,2),B,D); label("$Q$", Q, dir(-90)); label("$P$", P, dir(90)); label("$F$", (6,1), dir(45)); label("$E$", (6,2), dir(45)); label("$H$", H, dir(0)); label("$G$", G, dir(135)); | [] |
775 | Circles with centers $P, Q$ and $R$, having radii $1, 2$ and $3$, respectively, lie on the same side of line $l$ and are tangent to $l$ at $P', Q'$ and $R'$, respectively, with $Q'$ between $P'$ and $R'$. The circle with center $Q$ is externally tangent to each of the other two circles. What is the area of triangle $PQR$?
$\textbf{(A) } 0\qquad \textbf{(B) } \sqrt{\frac{2}{3}}\qquad\textbf{(C) } 1\qquad\textbf{(D) } \sqrt{6}-\sqrt{2}\qquad\textbf{(E) }\sqrt{\frac{3}{2}}$ | 2016 AMC 10A Problem 21 | Notice that we can find $[P'PQRR']$ in two different ways: $[P'PQQ']+[Q'QRR']$ and $[PQR]+[P'PRR'].$ Since we want $[PQR],$ we use the latter method, so we have $[P'PQQ']+[Q'QRR']=[PQR]+[P'PRR'].$
$\break$
$P'Q'=\sqrt{PQ^2-(QQ'-PP')^2}=\sqrt{3^2-1^2}=\sqrt{8}=2\sqrt{2}$. Additionally, $Q'R'=\sqrt{QR^2-(RR'-QQ')^2}=\sqrt{5^2-1^2}=\sqrt{24}=2\sqrt{6}$. Therefore, $[P'PQQ']=\frac{P'P+Q'Q}{2}*2\sqrt{2}=\frac{1+2}{2}*2\sqrt{2}=3\sqrt{2}$. Similarly, $[Q'QRR']=5\sqrt6$. We can calculate $[P'PRR']$ easily because $P'R'=P'Q'+Q'R'=2\sqrt{2}+2\sqrt{6}$. $[P'PRR']=4\sqrt{2}+4\sqrt{6}$. $\newline$
Plugging into first equation, the two sums of areas, $3\sqrt{2}+5\sqrt{6}=4\sqrt{2}+4\sqrt{6}+[PQR],$ so therefore $[PQR]=\boxed{\textbf{(D) }\sqrt{6}-\sqrt{2}.}$ | size(250); defaultpen(linewidth(0.4)); //Variable Declarations pair P,Q,R,Pp,Qp,Rp; pair A,B; //Variable Definitions A=(-5, 0); B=(8, 0); P=(-2.828,1); Q=(0,2); R=(4.899,3); Pp=foot(P,A,B); Qp=foot(Q,A,B); Rp=foot(R,A,B); path PQR = P--Q--R--cycle; //Initial Diagram dot(P); dot(Q); dot(R); dot(Pp); dot(Qp); dot(Rp); draw(Circle(P, 1), linewidth(0.8)); draw(Circle(Q, 2), linewidth(0.8)); draw(Circle(R, 3), linewidth(0.8)); draw(A--B,Arrows); label("$P$",P,N); label("$Q$",Q,N); label("$R$",R,N); label("$P'$",Pp,S); label("$Q'$",Qp,S); label("$R'$",Rp,S); label("$l$",B,E); //Added lines draw(PQR); draw(P--Pp); draw(Q--Qp); draw(R--Rp); //Angle marks draw(rightanglemark(P,Pp,B)); draw(rightanglemark(Q,Qp,B)); draw(rightanglemark(R,Rp,B)); | [] |
775 | Circles with centers $P, Q$ and $R$, having radii $1, 2$ and $3$, respectively, lie on the same side of line $l$ and are tangent to $l$ at $P', Q'$ and $R'$, respectively, with $Q'$ between $P'$ and $R'$. The circle with center $Q$ is externally tangent to each of the other two circles. What is the area of triangle $PQR$?
$\textbf{(A) } 0\qquad \textbf{(B) } \sqrt{\frac{2}{3}}\qquad\textbf{(C) } 1\qquad\textbf{(D) } \sqrt{6}-\sqrt{2}\qquad\textbf{(E) }\sqrt{\frac{3}{2}}$ | 2016 AMC 10A Problem 21 | The above diagram can be achieved relatively simply using basic knowledge of the Pythagorean theorem and the fact that the radius from the center to the point of tangency is perpendicular to the tangent line. From there, observe that $[PQRY]$ can be calculated in two ways: $[\triangle PQX] + [QZYX] + [\triangle QRZ]$ and $[\triangle PRY] + [\triangle PQR]$. Solving, we get:
\begin{align*} [\triangle PQR] &= [PQRY] - [\triangle PRY] \\ &= [\triangle PQX] + [QZYX] + [\triangle QRZ] - [\triangle PRY] \\ &= \sqrt{2} + 2\sqrt{6} + \sqrt{6}- 2\sqrt{2}-2\sqrt{6} \\ &= \boxed{\textbf{(D)} \sqrt{6}-\sqrt{2}} \end{align*}
- ColtsFan10, diagram partially borrowed from Solution 1 | // Initial Pen Sizing size(250); defaultpen(linewidth(0.4)); defaultpen(fontsize(10pt)); // Variable Declarations pair P,Q,R,Pp,Qp,Rp,X,Y,Z,A,B; // Variable Definitions A=(-5, 0); B=(8, 0); P=(-2.828,1); Q=(0,2); R=(4.899,3); X=(0,1); Y=(4.899,1); Z=(4.899,2); Pp=foot(P,A,B); Qp=foot(Q,A,B); Rp=foot(R,A,B); path PQR = P--Q--R--cycle; //Initial Diagram dot(P); dot(Q); dot(R); dot(Pp); dot(Qp); dot(Rp); dot(X); dot(Y); dot(Z); draw(Circle(P, 1), linewidth(0.3)); draw(Circle(Q, 2), linewidth(0.3)); draw(Circle(R, 3), linewidth(0.3)); draw(A--B,Arrows); label("$P$",P,N); label("$Q$",Q,N); label("$R$",R,N); label("$P'$",Pp,S); label("$Q'$",Qp,S); label("$R'$",Rp,S); label("$l$",B,E); label("$X$",X,NE); label("$Y$",Y,E); label("$Z$",Z,E); //Added lines filldraw(PQR,gray(0.8)); draw(P--Pp,linetype("8 8")); draw(Q--Qp,linetype("8 8")); draw(R--Rp,linetype("8 8")); draw(P--Y,linetype("8 8")); draw(Q--Z,linetype("8 8")); //Angle marks draw(rightanglemark(R,Y,P)); draw(rightanglemark(Q,X,P)); draw(rightanglemark(R,Z,Q)); //Length labeling label("$2\sqrt{2}$",P--X,fontsize(8pt)); label("$2\sqrt{6}$",X--Y,fontsize(8pt)); label("$2\sqrt{6}$",Q--Z,fontsize(8pt)); label("$1$",R--Z,E,fontsize(8pt)); label("$1$",Z--Y,E,fontsize(8pt)); label("$3$",(-2.828,1.3)--Q,W,fontsize(8pt)); label("$5$",Q--R,N,fontsize(8pt)); | [] |
776 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | 2016 AMC 10A Problem 24 | Let $AD$ intersect $OB$ at $E$ and $OC$ at $F.$
$\overarc{AB}= \overarc{BC}= \overarc{CD}=\theta$
$\angle{BAD}=\frac{1}{2} \cdot \overarc{BCD}=\theta=\angle{AOB}$
From there, $\triangle{OAB} \sim \triangle{ABE}$, thus:
$\frac{OA}{AB} = \frac{AB}{BE} = \frac{OB}{AE}$
$OA = OB$ because they are both radii of $\odot{O}$. Since $\frac{OA}{AB} = \frac{OB}{AE}$, we have that $200 = AB = AE$. Similarly, $CD = DF$.
$OE = 100\sqrt{2} = \frac{OB}{2}$ and $EF=\frac{BC}{2}=100$ , so $AD=AE + EF + FD = 200 + 100 + 200 = \boxed{\textbf{(E) } 500}$
Note
The first angle chase is done by the central angle theorem. Note that $\angle BAD$ is an inscribed angle from $BD$ to $A$, which is always $\frac{1}{2}$ of the angle of $\overarc{BD}$. Hence follows.
Furthermore, we get $OE=100\sqrt{2}$ from the similar triangles mentioned. We have: $\frac{200}{200\sqrt{2}}=\frac{BE}{200}$. Hence, $BE=100\sqrt{2}$ and so $OE=OB-BE=100\sqrt{2}$.
~mathboy282 | // Block 1
size(250);
defaultpen(linewidth(0.4));
//Variable Declarations
real RADIUS;
pair A, B, C, D, E, F, O;
RADIUS=3;
//Variable Definitions
A=RADIUS*dir(148.414);
B=RADIUS*dir(109.471);
C=RADIUS*dir(70.529);
D=RADIUS*dir(31.586);
O=(0,0);
//Path Definitions
path quad= A -- B -- C -- D -- cycle;
//Initial Diagram
draw(Circle(O, RADIUS), linewidth(0.8));
draw(quad, linewidth(0.8));
label("$A$",A,W);
label("$B$",B,NW);
label("$C$",C,NE);
label("$D$",D,ENE);
label("$O$",O,S);
label("$\theta$",O,3N);
//Radii
draw(O--A);
draw(O--B);
draw(O--C);
draw(O--D);
//Construction
E=extension(B,O,A,D);
label("$E$",E,NE);
F=extension(C,O,A,D);
label("$F$",F,NE);
//Angle marks
draw(anglemark(C,O,B));
// Block 2
size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, F, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label("$A$",A,W); label("$B$",B,NW); label("$C$",C,NE); label("$D$",D,ENE); label("$O$",O,S); label("$\theta$",O,3N); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction E=extension(B,O,A,D); label("$E$",E,NE); F=extension(C,O,A,D); label("$F$",F,NE); //Angle marks draw(anglemark(C,O,B)); | [] |
776 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | 2016 AMC 10A Problem 24 | To save us from getting big numbers with lots of zeros behind them, let's divide all side lengths by $200$ for now, then multiply it back at the end of our solution.
Construct quadrilateral $ABCD$ on the circle with $AD$ being the missing side (Notice that since the side length is less than the radius, it will be very small on the top of the circle). Now, draw the radii from center $O$ to $A,B,C,$ and $D$. Let the intersection of $BD$ and $OC$ be point $E$. Notice that $BD$ and $OC$ are perpendicular because $BCDO$ is a kite.
We set lengths $BE=ED$ equal to $x$ (Solution 1.1 begins from here). By the Pythagorean Theorem,
\[\sqrt{1^2-x^2}+\sqrt{(\sqrt{2})^2-x^2}=\sqrt{2}\]
We solve for $x$:
\[1-x^2+2-x^2+2\sqrt{(1-x^2)(2-x^2)}=2\]
\[2\sqrt{(1-x^2)(2-x^2)}=2x^2-1\]
\[4(1-x^2)(2-x^2)=(2x^2-1)^2\]
\[8-12x^2+4x^4=4x^4-4x^2+1\]
\[8x^2=7\]
\[x=\frac{\sqrt{14}}{4}\]
By Ptolemy's Theorem,
\[AB \cdot CD + BC \cdot AD = AC \cdot BD = BD^2 = (2 \cdot BE)^2\]
Substituting values,
\[1^2+1 \cdot AD = 4{\left( \frac{\sqrt{14}}{4} \right)}^2\]
\[1+AD=\frac{7}{2}\]
\[AD=\frac{5}{2}\]
Finally, we multiply back the $200$ that we divided by at the beginning of the problem to get $AD=\boxed{\textbf{(E) } 500}$. | // Block 1
size(250);
defaultpen(linewidth(0.4));
//Variable Declarations
real RADIUS;
pair A, B, C, D, E, O;
RADIUS=3;
//Variable Definitions
A=RADIUS*dir(148.414);
B=RADIUS*dir(109.471);
C=RADIUS*dir(70.529);
D=RADIUS*dir(31.586);
E=extension(B,D,O,C);
O=(0,0);
//Path Definitions
path quad= A -- B -- C -- D -- cycle;
//Initial Diagram
draw(Circle(O, RADIUS), linewidth(0.8));
draw(quad, linewidth(0.8));
label("$A$",A,W);
label("$B$",B,NW);
label("$C$",C,NE);
label("$D$",D,E);
label("$E$",E,WSW);
label("$O$",O,S);
//Radii
draw(O--A);
draw(O--B);
draw(O--C);
draw(O--D);
//Construction
draw(B--D);
draw(rightanglemark(C,E,D));
// Block 2
size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); E=extension(B,D,O,C); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label("$A$",A,W); label("$B$",B,NW); label("$C$",C,NE); label("$D$",D,E); label("$E$",E,WSW); label("$O$",O,S); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction draw(B--D); draw(rightanglemark(C,E,D)); | [] |
776 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | 2016 AMC 10A Problem 24 | size(250);
defaultpen(linewidth(0.4));
//Variable Declarations
real RADIUS;
pair A, B, C, D, O;
RADIUS=3;
//Variable Definitions
A=RADIUS*dir(148.414);
B=RADIUS*dir(109.471);
C=RADIUS*dir(70.529);
D=RADIUS*dir(31.586);
P=RADIUS*dir(270);
O=(0,0);
//Path Definitions
path quad= A -- B -- C -- D -- cycle;
//Initial Diagram
draw(circle(O, RADIUS), linewidth(0.8));
draw(quad, linewidth(0.8));
label("<imath>A</imath>",A,W);
label("<imath>B</imath>",B,NW);
label("<imath>C</imath>",C,NE);
label("<imath>D</imath>",D,E);
label("<imath>O</imath>",O,S);
//Radii
draw(O--A);
draw(O--B);
draw(O--C);
draw(O--D);
draw(P--A);
draw(P--B);
draw(P--C);
draw(P--D);
Draw some point <imath>P</imath> somewhere on major arc <imath>\overarc{AD}</imath>. Let <imath>\angle AOB \cong \angle BOC \cong \angle COD = 2\theta</imath>. Then, by inscribed angle <imath>\angle APB \cong \angle BPC \cong \angle CPD = \theta</imath>. By extended law of sines, we have <imath>\frac{200}{\text{sin}(\theta)} = 2(200\sqrt{2}) \implies \text{sin}(\theta) = \frac{\sqrt{2}}{4}</imath>. Using the triple angle sine formula, we have <imath>\text{sin}(3\theta)=3\text{sin}(\theta) - 4\text{sin}^3(\theta) = \frac{5\sqrt{2}}{8}</imath>.
Then, again by extended law of sines, we have <imath>\frac{AD}{\text{sin}(3\theta)}=400\sqrt{2}</imath>. Thus, <imath>AD = \text{sin}(3\theta) 400\sqrt{2} = \boxed{500}</imath>.
==Solution 3 (HARD Algebra)== | // Block 1
size(250);
defaultpen(linewidth(0.4));
//Variable Declarations
real RADIUS;
pair A, B, C, D, O;
RADIUS=3;
//Variable Definitions
A=RADIUS*dir(148.414);
B=RADIUS*dir(109.471);
C=RADIUS*dir(70.529);
D=RADIUS*dir(31.586);
P=RADIUS*dir(270);
O=(0,0);
//Path Definitions
path quad= A -- B -- C -- D -- cycle;
//Initial Diagram
draw(circle(O, RADIUS), linewidth(0.8));
draw(quad, linewidth(0.8));
label("<imath>A</imath>",A,W);
label("<imath>B</imath>",B,NW);
label("<imath>C</imath>",C,NE);
label("<imath>D</imath>",D,E);
label("<imath>O</imath>",O,S);
//Radii
draw(O--A);
draw(O--B);
draw(O--C);
draw(O--D);
draw(P--A);
draw(P--B);
draw(P--C);
draw(P--D);
Draw some point <imath>P</imath> somewhere on major arc <imath>\overarc{AD}</imath>. Let <imath>\angle AOB \cong \angle BOC \cong \angle COD = 2\theta</imath>. Then, by inscribed angle <imath>\angle APB \cong \angle BPC \cong \angle CPD = \theta</imath>. By extended law of sines, we have <imath>\frac{200}{\text{sin}(\theta)} = 2(200\sqrt{2}) \implies \text{sin}(\theta) = \frac{\sqrt{2}}{4}</imath>. Using the triple angle sine formula, we have <imath>\text{sin}(3\theta)=3\text{sin}(\theta) - 4\text{sin}^3(\theta) = \frac{5\sqrt{2}}{8}</imath>.
Then, again by extended law of sines, we have <imath>\frac{AD}{\text{sin}(3\theta)}=400\sqrt{2}</imath>. Thus, <imath>AD = \text{sin}(3\theta) 400\sqrt{2} = \boxed{500}</imath>.
==Solution 3 (HARD Algebra)==
<asy>
size(250);
defaultpen(linewidth(0.4));
//Variable Declarations
real RADIUS;
pair A, B, C, D, O;
RADIUS=3;
//Variable Definitions
A=RADIUS*dir(148.414);
B=RADIUS*dir(109.471);
C=RADIUS*dir(70.529);
D=RADIUS*dir(31.586);
O=(0,0);
//Path Definitions
path quad= A -- B -- C -- D -- cycle;
//Initial Diagram
draw(Circle(O, RADIUS), linewidth(0.8));
draw(quad, linewidth(0.8));
label("$A$",A,W);
label("$B$",B,NW);
label("$C$",C,NE);
label("$D$",D,E);
label("$O$",O,S);
//Radii
draw(O--A);
draw(O--B);
draw(O--C);
draw(O--D);
// Block 2
size(250);
defaultpen(linewidth(0.4));
//Variable Declarations
real RADIUS;
pair A, B, C, D, O;
RADIUS=3;
//Variable Definitions
A=RADIUS*dir(148.414);
B=RADIUS*dir(109.471);
C=RADIUS*dir(70.529);
D=RADIUS*dir(31.586);
O=(0,0);
//Path Definitions
path quad= A -- B -- C -- D -- cycle;
//Initial Diagram
draw(Circle(O, RADIUS), linewidth(0.8));
draw(quad, linewidth(0.8));
label("$A$",A,W);
label("$B$",B,NW);
label("$C$",C,NE);
label("$D$",D,E);
label("$O$",O,S);
//Radii
draw(O--A);
draw(O--B);
draw(O--C);
draw(O--D);
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Let quadrilateral $ABCD$ be inscribed in circle $O$, where $AD$ is the side of unknown length. Draw the radii from center $O$ to all four vertices of the quadrilateral, and draw the altitude of $\triangle BOC$ such that it passes through side $AD$ at the point $G$ and meets side $BC$ at point $H$.
By the Pythagorean Theorem, the length of $OH$ is
\begin{align*} \sqrt{CO^2 - HC^2} &= \sqrt{(200\sqrt{2})^2 - \left(\frac{200}{2}\right)^2} \\ &= \sqrt{80000 - 10000} \\ &= \sqrt{70000} \\ &= 100\sqrt{7}. \end{align*}
Note that $[ABCDO] = [AOB] + [BOC] + [COD] = [AOD] + [ABCD].$ Let the length of $OG$ be $h$ and the length of $AD$ be $x$; then we have that
$[AOB] + [BOC] + [COD] = \frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} = \frac{x \times h}{2} + \frac{(100\sqrt{7} - h)(200 + x)}{2} = [AOD] + [ABCD].$
Furthermore,
\begin{align*} h &= \sqrt{OD^2 - GD^2} \\ &= \sqrt{(200\sqrt{2})^2 - \left(\frac{x}{2}\right)^2} \\ &= \sqrt{80000 - \frac{x^2}{4}} \end{align*}
Substituting this value of $h$ into the previous equation and evaluating for $x$, we get:
\[\frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} = \frac{x \times h}{2} + \frac{(100\sqrt{7} - h)(200 + x)}{2}\]
\[\frac{3 \times 200 \times 100\sqrt{7}}{2} = \frac{x\sqrt{80000 - \frac{x^2}{4}}}{2} + \frac{\left(100\sqrt{7} - \sqrt{80000 - \frac{x^2}{4}}\right)(200 + x)}{2}\]
\[60000\sqrt{7} = \left(x\sqrt{80000 - \frac{x^2}{4}}\right) + \left(20000\sqrt{7}\right) + \left(100x\sqrt{7}\right) - \left(200\sqrt{80000 - \frac{x^2}{4}}\right) - \left(x\sqrt{80000 - \frac{x^2}{4}}\right)\]
\[40000\sqrt{7} = 100x\sqrt{7} - 200\sqrt{80000 - \frac{x^2}{4}}\]
\[400\sqrt{7} = x\sqrt{7} - 2\sqrt{80000 - \frac{x^2}{4}}\]
\[(x - 400)\sqrt{7} = 2\sqrt{80000 - \frac{x^2}{4}}\]
\[7(x-400)^2 = 4\left(80000 - \frac{x^2}{4}\right)\]
\[7x^2 - 5600x + 1120000 = 320000 - x^2\]
\[8x^2 - 5600x + 800000 = 0\]
\[x^2 - 700x + 100000 = 0\]
The roots of this quadratic are found by using the quadratic formula:
\begin{align*} x &= \frac{-(-700) \pm \sqrt{(-700)^2 - 4 \times 1 \times 100000}}{2 \times 1} \\ &= \frac{700 \pm \sqrt{490000 - 400000}}{2} \\ &= \frac{700}{2} \pm \frac{\sqrt{90000}}{2} \\ &= 350 \pm \frac{300}{2} \\ &= 200, 500 \end{align*}
If the length of $AD$ is $200$, then $ABCD$ would be a square. Thus, the radius of the circle would be
\[\frac{\sqrt{200^2 + 200^2}}{2} = \frac{\sqrt{80000}}{2} = \frac{200\sqrt{2}}{2} = 100\sqrt{2}\]
Which is a contradiction. Therefore, our answer is $\boxed{500}.$ | [] |
776 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | 2016 AMC 10A Problem 24 | Construct quadrilateral $ABCD$ on the circle with $AD$ being the missing side (Notice that since the side length is less than the radius, it will be very small on the top of the circle). Now, draw the radii from center $O$ to $A,B,C,$ and $D$. Apply the law of cosines on $\Delta BOC$; let $\theta = \angle BOC$. We get the following equation: \[(BC)^{2}=(OB)^{2}+(OC)^{2}-2\cdot OB \cdot OC\cdot \cos\theta\] Substituting the values in, we get \[(200)^{2}=2\cdot (200)^{2}+ 2\cdot (200)^{2}- 2\cdot 2\cdot (200)^{2}\cdot \cos\theta\] Canceling out, we get \[\cos\theta=\frac{3}{4}\]
Because $\angle AOB$, $\angle BOC$, and $\angle COD$ are congruent, $\angle AOD = 3\theta$. To find the remaining side ($AD$), we simply have to apply the law of cosines to $\Delta AOD$. Now, to find $\cos 3\theta$, we can derive a formula that only uses $\cos\theta$: \[\cos 3\theta=\cos (2\theta+\theta)= \cos 2\theta \cos\theta- (2\sin\theta \cos\theta) \cdot \sin \theta\] \[\cos 3\theta= \cos\theta (\cos 2\theta-2\sin^{2}\theta)=\cos\theta (2\cos^{2}\theta-3+2\cos^{2}\theta)\] \[\Rightarrow \cos 3\theta=4\cos^{3}\theta-3\cos\theta\] It is useful to memorize the triple angle formulas ($\cos 3\theta=4\cos^{3}\theta-3\cos\theta, \sin 3\theta=3\sin\theta-4\sin^{3}\theta$). Plugging in $\cos\theta=\frac{3}{4}$, we get $\cos 3\theta= -\frac{9}{16}$. Now, applying law of cosines on triangle $OAD$, we get \[(AD)^{2}= 2\cdot (200)^{2}+ 2\cdot (200)^{2}+2\cdot 200\sqrt2 \cdot 200\sqrt2 \cdot \frac{9}{16}\] \[\Rightarrow 2\cdot (200)^{2} \cdot (1+1+ \frac{9}{8})=(200)^{2}\cdot \frac{25}{4}\] \[AD=200 \cdot \frac{5}{2}=\boxed{500}\] | // Block 1
size(250);
defaultpen(linewidth(0.4));
//Variable Declarations
real RADIUS;
pair A, B, C, D, O;
RADIUS=3;
//Variable Definitions
A=RADIUS*dir(148.414);
B=RADIUS*dir(109.471);
C=RADIUS*dir(70.529);
D=RADIUS*dir(31.586);
O=(0,0);
//Path Definitions
path quad= A -- B -- C -- D -- cycle;
//Initial Diagram
draw(Circle(O, RADIUS), linewidth(0.8));
draw(quad, linewidth(0.8));
label("$A$",A,W);
label("$B$",B,NW);
label("$C$",C,NE);
label("$D$",D,E);
label("$O$",O,S);
label("$\theta$",O,3N);
//Radii
draw(O--A);
draw(O--B);
draw(O--C);
draw(O--D);
//Angle mark for BOC
draw(anglemark(C,O,B));
// Block 2
size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label("$A$",A,W); label("$B$",B,NW); label("$C$",C,NE); label("$D$",D,E); label("$O$",O,S); label("$\theta$",O,3N); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Angle mark for BOC draw(anglemark(C,O,B)); | [] |
776 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | 2016 AMC 10A Problem 24 | Construct quadrilateral $ABCD$ on the circle $O$ with $AD$ being the desired side. Then, drop perpendiculars from $A$ and $D$ to the extended line of $\overline{BC}$ and let these points be $E$ and $F$, respectively. Also, let $\theta = \angle BOC$. From the Law of Cosines on $\triangle BOC$, we have $\cos \theta = \frac{3}{4}$.
Now, since $\triangle BOC$ is isosceles with $\overline{OB} \cong \overline{OC}$, we have that $\angle BCO = \angle CBO = 90 - \frac{\theta}{2}$. In addition, we know that $\overline{BC} \cong \overline{CD}$ as they are both equal to $200$ and $\overline{OB} \cong \overline{OC} \cong \overline{OD}$ as they are both radii of the same circle. By SSS Congruence, we have that $\triangle OBC \cong \triangle OCD$, so we have that $\angle OCD = \angle BCO = 90 - \frac{\theta}{2}$, so $\angle DCF = \theta$.
Thus, we have $\frac{FC}{DC} = \cos \theta = \frac{3}{4}$, so $FC = 150$. Similarly, $BE = 150$, and $AD = 150 + 200 + 150 = \boxed{500}$. | // Block 1
size(250);
defaultpen(linewidth(0.4));
//Variable Declarations
real RADIUS;
pair A, B, C, D, E, F, O;
RADIUS=3;
//Variable Definitions
A=RADIUS*dir(148.414);
B=RADIUS*dir(109.471);
C=RADIUS*dir(70.529);
D=RADIUS*dir(31.586);
E=foot(A,B,C);
F=foot(D,B,C);
O=(0,0);
//Path Definitions
path quad= A -- B -- C -- D -- cycle;
//Initial Diagram
draw(Circle(O, RADIUS), linewidth(0.8));
draw(quad, linewidth(0.8));
label("$A$",A,W);
label("$B$",B,NW);
label("$C$",C,NE);
label("$D$",D,ENE);
label("$O$",O,S);
label("$\theta$",O,3N);
//Radii
draw(O--A);
draw(O--B);
draw(O--C);
draw(O--D);
//Construction
draw(A--E);
draw(E--B);
draw(C--F);
draw(F--D);
label("$E$",E,NW);
label("$F$",F,NE);
//Angle marks
draw(anglemark(C,O,B));
draw(rightanglemark(A,E,B));
draw(rightanglemark(C,F,D));
// Block 2
size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, F, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); E=foot(A,B,C); F=foot(D,B,C); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label("$A$",A,W); label("$B$",B,NW); label("$C$",C,NE); label("$D$",D,ENE); label("$O$",O,S); label("$\theta$",O,3N); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction draw(A--E); draw(E--B); draw(C--F); draw(F--D); label("$E$",E,NW); label("$F$",F,NE); //Angle marks draw(anglemark(C,O,B)); draw(rightanglemark(A,E,B)); draw(rightanglemark(C,F,D)); | [] |
776 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | 2016 AMC 10A Problem 24 | Let $s = 200$. Let $O$ be the center of the circle. Then $AC$ is twice the altitude of $\triangle OBC$ to $\overline{OB}$. Since $\triangle OBC$ is isosceles we can compute its area to be $\frac{s^2 \sqrt{7}}{4}$, hence $CA = 2 \cdot \tfrac{2 \cdot s^2\sqrt7/4}{s\sqrt2} = s\sqrt{\frac{7}{2}}$.
Now by Ptolemy's Theorem we have $CA^2 = s^2 + AD \cdot s \implies AD = \left(\frac{7}{2}-1\right)s.$ This gives us: \[\boxed{\textbf{(E) } 500.}\] | // Block 1
pathpen = black; pointpen = black;
size(6cm);
draw(unitcircle);
pair A = D("A", dir(50), dir(50));
pair B = D("B", dir(90), dir(90));
pair C = D("C", dir(130), dir(130));
pair D = D("D", dir(170), dir(170));
pair O = D("O", (0,0), dir(-90));
draw(A--C, red);
draw(B--D, blue+dashed);
draw(A--B--C--D--cycle);
draw(A--O--C);
draw(O--B);
// Block 2
pathpen = black; pointpen = black; size(6cm); draw(unitcircle); pair A = D("A", dir(50), dir(50)); pair B = D("B", dir(90), dir(90)); pair C = D("C", dir(130), dir(130)); pair D = D("D", dir(170), dir(170)); pair O = D("O", (0,0), dir(-90)); draw(A--C, red); draw(B--D, blue+dashed); draw(A--B--C--D--cycle); draw(A--O--C); draw(O--B); | [] |
776 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | 2016 AMC 10A Problem 24 | Label the points as shown, and let $\angle{EOF} = \theta$. Since $\overline{OB} = \overline{OC}$, and $\triangle{OFE} \sim \triangle{OCB}$, we get that $\angle{EFO} = 90-\frac{\theta}{2}$. We assign $\alpha$ to $90-\frac{\theta}{2}$ for simplicity.
From here, by vertical angles $\angle{CFD} = \alpha$. Also, since $\triangle{OCB} \cong \triangle{ODC}$, $\angle{OCD} = \alpha$. This means that $\angle{CDF} = 180-2\alpha = \theta$, which leads to $\triangle{OCB} \sim \triangle{DCF}$.
Since we know that $\overline{CD} = 200$, $\overline{DF} = 200$, and by similar reasoning $\overline{AE} = 200$.
Finally, again using similar triangles, we get that $\overline{CF} = 100\sqrt{2}$, which means that $\overline{OF} = \overline{OC} - \overline{CF} = 200\sqrt{2} - 100\sqrt{2} = 100\sqrt{2}$. We can again apply similar triangles (or use Power of a Point) to get $\overline{EF} = 100$, and finally $\overline{AD} = \overline{AE}+\overline{EF}+\overline{FD} = 200+100+200=\boxed{\textbf{(E)}500}$ - ColtsFan10 | // Block 1
size(250);
defaultpen(linewidth(0.4));
//Variable Declarations, L is used to write alpha= statement
real RADIUS;
pair A, B, C, D, E, F, O, L;
RADIUS=3;
//Variable Definitions
A=RADIUS*dir(148.414);
B=RADIUS*dir(109.471);
C=RADIUS*dir(70.529);
D=RADIUS*dir(31.586);
E=extension(A,D,O,B);
F=extension(A,D,O,C);
L=midpoint(C--D);
O=(0,0);
//Path Definitions
path quad = A -- B -- C -- D -- cycle;
//Initial Diagram
draw(circle(O, RADIUS), linewidth(0.8));
draw(quad, linewidth(0.8));
label("$A$",A,NW);
label("$B$",B,NW);
label("$C$",C,NE);
label("$D$",D,NE);
label("$E$",E,SW);
label("$F$",F,SE);
label("$O$",O,SE);
dot(O,linewidth(5));
//Radii
draw(O--A);
draw(O--B);
draw(O--C);
draw(O--D);
//Construction
label("$\alpha = 90-\frac{\theta}{2}$",L,5NE,rgb(128, 0, 0));
draw(anglemark(C,O,B));
label("$\theta$",O,3N);
draw(anglemark(E,F,O));
label("$\alpha$",F,3SW);
draw(anglemark(D,F,C));
label("$\alpha$",F,3NE);
draw(anglemark(F,C,D));
label("$\alpha$",C,3SSE);
draw(anglemark(C,D,F));
label("$\theta$",(RADIUS-0.04)*dir(31.586),3WNW);
// Block 2
size(250); defaultpen(linewidth(0.4)); //Variable Declarations, L is used to write alpha= statement real RADIUS; pair A, B, C, D, E, F, O, L; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); E=extension(A,D,O,B); F=extension(A,D,O,C); L=midpoint(C--D); O=(0,0); //Path Definitions path quad = A -- B -- C -- D -- cycle; //Initial Diagram draw(circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label("$A$",A,NW); label("$B$",B,NW); label("$C$",C,NE); label("$D$",D,NE); label("$E$",E,SW); label("$F$",F,SE); label("$O$",O,SE); dot(O,linewidth(5)); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction label("$\alpha = 90-\frac{\theta}{2}$",L,5NE,rgb(128, 0, 0)); draw(anglemark(C,O,B)); label("$\theta$",O,3N); draw(anglemark(E,F,O)); label("$\alpha$",F,3SW); draw(anglemark(D,F,C)); label("$\alpha$",F,3NE); draw(anglemark(F,C,D)); label("$\alpha$",C,3SSE); draw(anglemark(C,D,F)); label("$\theta$",(RADIUS-0.04)*dir(31.586),3WNW); | [] |
776 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | 2016 AMC 10A Problem 24 | Claim: $[ABCD]$ is an isosceles trapezoid.
Proof: Notice that $[ABCD]$ is cyclic, triangle $BOC$ is isosceles, and triangle $AOB$ is congruent to $DOC$ by SSS congruence. Therefore, $\angle BAD = 180 - \angle BCD = 180-(\angle BCO + \angle DCO)=180-(\angle CBO+\angle ABO) = 180 - \angle ABC = \angle CDA$. Hence, $[ABCD]$ is an isosceles trapezoid.
Let $\angle CDA=\alpha$. Notice that the length of the altitude from $C$ to $AD$ is $200sin(\alpha)$. Furthermore, the length of the altitude from $O$ to $BC$ is $100\sqrt{7}$ by the Pythagorean theorem. Therefore, the length of the altitude from $O$ to $AD$ is $100\sqrt{7}-200sin\alpha$. Let $F$ the feet of the altitude from $O$ to $AD$. Then, $FD=(200+400cos(\alpha))/2=100+200cos(\alpha)$, because $AOD$ is isosceles.
Therefore, by the Pythagorean theorem, $(100+200cos(\alpha))^2+(100\sqrt{7}-200sin(\alpha))^2=80000$. Simplifying, we have $1+cos(\alpha)=sin(\alpha) \cdot sqrt{7} \implies cos^2(\alpha)+2cos(\alpha)+1=sin^2(\alpha) \cdot 7 = 7-7cos^2(\alpha) \implies 8cos^2(\alpha)+2cos(\alpha) - 6 =0$. Solving this quadratic, we have $cos(\alpha)=\frac{3}{4}, -1$, but $0<\alpha<180 \implies cos(\alpha)=3/4$. Therefore, $AD=200cos(\alpha)+200cos(\alpha)+200=\boxed{500}$
- [mathMagicOPS] | // Block 1
size(250);
defaultpen(linewidth(0.4));
//Variable Declarations
real RADIUS;
pair A, B, C, D, O;
RADIUS=3;
//Variable Definitions
A=RADIUS*dir(148.414);
B=RADIUS*dir(109.471);
C=RADIUS*dir(70.529);
D=RADIUS*dir(31.586);
O=(0,0);
//Path Definitions
path quad= A -- B -- C -- D -- cycle;
//Initial Diagram
draw(Circle(O, RADIUS), linewidth(0.8));
draw(quad, linewidth(0.8));
label("$A$",A,W);
label("$B$",B,NW);
label("$C$",C,NE);
label("$D$",D,E);
label("$O$",O,S);
//Radii
draw(O--A);
draw(O--B);
draw(O--C);
draw(O--D);
// Block 2
size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label("$A$",A,W); label("$B$",B,NW); label("$C$",C,NE); label("$D$",D,E); label("$O$",O,S); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); | [] |
776 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | 2016 AMC 10A Problem 24 | Since $AB$, $BC$, and $CD$ are congruent chords, the triangles formed by connecting these sides to the circumcenter $O$ (i.e., $\triangle AOB$, $\triangle BOC$, $\triangle COD$) are congruent isosceles triangles.
Let $\alpha = \angle OCB$. In $\triangle BOC$, the side length $BC=200$ and $OC = OB = 200\sqrt{2}$. By dropping an altitude from $O$ to $BC$, we bisect $BC$. Half the base is $\frac{200}{2} = 100$.
Using the right triangle formed, we find $\cos(\alpha)$:
\[\cos(\alpha) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{100}{200\sqrt{2}} = \frac{\sqrt{2}}{4}\]
Since $\triangle BOC \cong \triangle COD$, we have $m\angle OCB = m\angle OCD = \alpha$.
The interior angle measure $m\angle DCB$ is given by:
\[m\angle DCB = m\angle OCB + m\angle OCD = 2\alpha\]
We calculate $\cos(2\alpha)$ using the double-angle identity:
\[\cos(2\alpha) = 2\cos^2(\alpha) - 1\]
\[\cos(2\alpha) = 2\left(\frac{\sqrt{2}}{4}\right)^2 - 1 = \frac{1}{4} - 1 = -\frac{3}{4}\]
Now, apply the Law of Cosines to $\triangle BCD$ to find $BD^2$:
\[BD^2 = BC^2 + CD^2 - 2(BC)(CD)\cos(\angle DCB)\]
\[BD^2 = 200^2 + 200^2 - 2(200)(200)\left(-\frac{3}{4}\right)\]
\[BD^2 = 80000 + 60000 = 140000\]
Since $ABCD$ is a cyclic quadrilateral, the opposite angles are supplementary:
\[m\angle DAB = 180^\circ - m\angle DCB = 180^\circ - 2\alpha\]
Thus, we find $\cos(\angle DAB)$:
\[\cos(\angle DAB) = \cos(180^\circ - 2\alpha) = -\cos(2\alpha)\]
\[\cos(\angle DAB) = -\left(-\frac{3}{4}\right) = \frac{3}{4}\]
Finally, we apply the Law of Cosines to $\triangle DAB$ to find the side length $AD$. Let $AD = a$:
\[140000 = 200^2 + a^2 - 2(200)(a)\left(\frac{3}{4}\right)\]
\[140000 = 40000 + a^2 - 300a\]
Rearranging the terms yields the quadratic equation:
\[a^2 - 300a - 100000 = 0\]
We solve this equation by factoring:
\[(a - 500)(a + 200) = 0\]
The two possible solutions for $a$ are $a=500$ and $a=-200$. Since $a$ represents a side length, it must be positive, so $a = 500$, which means $AD = 500$, or $\boxed{\textbf{(E) } 500}$.
~Voidling | // Block 1
size(250);
defaultpen(linewidth(0.4));
//Variable Declarations
real RADIUS;
pair A, B, C, D, E, F, O;
RADIUS=3;
//Variable Definitions
A=RADIUS*dir(148.414);
B=RADIUS*dir(109.471);
C=RADIUS*dir(70.529);
D=RADIUS*dir(31.586);
O=(0,0);
//Path Definitions
path quad= A -- B -- C -- D -- cycle;
//Initial Diagram
draw(Circle(O, RADIUS), linewidth(0.8));
draw(quad, linewidth(0.8));
label("$A$",A,W);
label("$B$",B,NW);
label("$C$",C,NE);
label("$D$",D,ENE);
label("$O$",O,S);
label("$\theta$",O,3N);
//Radii
draw(O--A);
draw(O--B);
draw(O--C);
draw(O--D);
//Construction
E=extension(B,O,A,D);
label("$E$",E,NE);
F=extension(C,O,A,D);
label("$F$",F,NE);
//Angle marks
draw(anglemark(C,O,B));
// Block 2
size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, F, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label("$A$",A,W); label("$B$",B,NW); label("$C$",C,NE); label("$D$",D,ENE); label("$O$",O,S); label("$\theta$",O,3N); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction E=extension(B,O,A,D); label("$E$",E,NE); F=extension(C,O,A,D); label("$F$",F,NE); //Angle marks draw(anglemark(C,O,B)); | [] |
777 | How many ordered triples $(x,y,z)$ of positive integers satisfy $\text{lcm}(x,y) = 72, \text{lcm}(x,z) = 600,$ and $\text{lcm}(y,z)=900$?
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 64$ | 2016 AMC 10A Problem 25 | As said in previous solutions, start by factoring $72, 600,$ and $900$. The prime factorizations are as follows: \[72=2^3\cdot 3^2,\] \[600=2^3\cdot 3\cdot 5^2,\] \[\text{and } 900=2^2\cdot 3^2\cdot 5^2\]
To organize $x,y, \text{ and } z$ and their respective LCMs in a simpler way, we can draw a triangle as follows such that $x,y, \text{and } z$ are the vertices and the LCMs are on the edges.
Now we can split this triangle into three separate ones for each of the three different prime factors $2,3, \text{and } 5$.
Analyzing for powers of $2$, it is quite obvious that $x$ must have $2^3$ as one of its factors since neither $y \text{ nor } z$ can have a power of $2$ exceeding $2$. Turning towards the vertices $y$ and $z$, we know at least one of them must have $2^2$ as its factors. Therefore, we have $5$ ways for the powers of $2$ for $y \text{ and } z$ since the only ones that satisfy the previous conditions are for ordered pairs $(y,z) \{(2,0)(2,1)(0,2)(1,2)(2,2)\}$.
Using the same logic as we did for powers of $2$, it becomes quite easy to note that $y$ must have $3^2$ as one of its factors. Moving onto $x \text{ and } z$, we can use the same logic to find the only ordered pairs $(x,z)$ that will work are $\{(1,0)(0,1)(1,1)\}$.
The final and last case is the powers of $5$.
This is actually quite a simple case since we know $z$ must have $5^2$ as part of its factorization while $x \text{ and } y$ cannot have a factor of $5$ in their prime factorization.
Multiplying all the possible arrangements for prime factors $2,3, \text{ and } 5$, we get the answer:
\[5\cdot3\cdot1=\boxed{\textbf{(A) }15}\]
(Diagrams by ColtsFan10) | // Block 1
//Variable Declarations
defaultpen(0.45);
size(200pt);
fontsize(15pt);
pair X, Y, Z;
real R;
path tri;
//Variable Definitions
R = 1;
X = R*dir(90);
Y = R*dir(210);
Z = R*dir(-30);
tri = X--Y--Z--cycle;
//Diagram
draw(tri);
label("$x$",X,N);
label("$y$",Y,SW);
label("$z$",Z,SE);
label("$2^33^25^0$",X--Y,2W);
label("$2^33^15^2$",X--Z,2E);
label("$2^23^25^2$",Y--Z,2S);
// Block 2
//Variable Declarations
defaultpen(0.45);
size(200pt);
fontsize(15pt);
pair X, Y, Z;
real R;
path tri;
//Variable Definitions
R = 1;
X = R*dir(90);
Y = R*dir(210);
Z = R*dir(-30);
tri = X--Y--Z--cycle;
//Diagram
draw(tri);
label("$x$",X,N);
label("$y$",Y,SW);
label("$z$",Z,SE);
label("$2^3$",X--Y,2W);
label("$2^3$",X--Z,2E);
label("$2^2$",Y--Z,2S);
// Block 3
//Variable Declarations
defaultpen(0.45);
size(200pt);
fontsize(15pt);
pair X, Y, Z;
real R;
path tri;
//Variable Definitions
R = 1;
X = R*dir(90);
Y = R*dir(210);
Z = R*dir(-30);
tri = X--Y--Z--cycle;
//Diagram
draw(tri);
label("$x$",X,N);
label("$y$",Y,SW);
label("$z$",Z,SE);
label("$3^2$",X--Y,2W);
label("$3^1$",X--Z,2E);
label("$3^2$",Y--Z,2S);
// Block 4
//Variable Declarations
defaultpen(0.45);
size(200pt);
fontsize(15pt);
pair X, Y, Z;
real R;
path tri;
//Variable Definitions
R = 1;
X = R*dir(90);
Y = R*dir(210);
Z = R*dir(-30);
tri = X--Y--Z--cycle;
//Diagram
draw(tri);
label("$x$",X,N);
label("$y$",Y,SW);
label("$z$",Z,SE);
label("$5^0$",X--Y,2W);
label("$5^2$",X--Z,2E);
label("$5^2$",Y--Z,2S);
// Block 5
//Variable Declarations defaultpen(0.45); size(200pt); fontsize(15pt); pair X, Y, Z; real R; path tri; //Variable Definitions R = 1; X = R*dir(90); Y = R*dir(210); Z = R*dir(-30); tri = X--Y--Z--cycle; //Diagram draw(tri); label("$x$",X,N); label("$y$",Y,SW); label("$z$",Z,SE); label("$2^33^25^0$",X--Y,2W); label("$2^33^15^2$",X--Z,2E); label("$2^23^25^2$",Y--Z,2S);
// Block 6
//Variable Declarations defaultpen(0.45); size(200pt); fontsize(15pt); pair X, Y, Z; real R; path tri; //Variable Definitions R = 1; X = R*dir(90); Y = R*dir(210); Z = R*dir(-30); tri = X--Y--Z--cycle; //Diagram draw(tri); label("$x$",X,N); label("$y$",Y,SW); label("$z$",Z,SE); label("$2^3$",X--Y,2W); label("$2^3$",X--Z,2E); label("$2^2$",Y--Z,2S);
// Block 7
//Variable Declarations defaultpen(0.45); size(200pt); fontsize(15pt); pair X, Y, Z; real R; path tri; //Variable Definitions R = 1; X = R*dir(90); Y = R*dir(210); Z = R*dir(-30); tri = X--Y--Z--cycle; //Diagram draw(tri); label("$x$",X,N); label("$y$",Y,SW); label("$z$",Z,SE); label("$3^2$",X--Y,2W); label("$3^1$",X--Z,2E); label("$3^2$",Y--Z,2S);
// Block 8
//Variable Declarations defaultpen(0.45); size(200pt); fontsize(15pt); pair X, Y, Z; real R; path tri; //Variable Definitions R = 1; X = R*dir(90); Y = R*dir(210); Z = R*dir(-30); tri = X--Y--Z--cycle; //Diagram draw(tri); label("$x$",X,N); label("$y$",Y,SW); label("$z$",Z,SE); label("$5^0$",X--Y,2W); label("$5^2$",X--Z,2E); label("$5^2$",Y--Z,2S); | [] |
778 | Find the area of the shaded region.
$\textbf{(A)}\ 4\dfrac{3}{5} \qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 5\dfrac{1}{4} \qquad \textbf{(D)}\ 6\dfrac{1}{2} \qquad \textbf{(E)}\ 8$ | 2016 AMC 12A Problem 8 | The bases of these triangles are all $1$, and by symmetry, their heights are $4$, $\frac{5}{2}$, $4$, and $\frac{5}{2}$. Thus, their areas are $2$, $\frac{5}{4}$, $2$, and $\frac{5}{4}$, which add to the area of the shaded region, which is $\boxed{6\frac{1}{2}}$. | size(6cm); defaultpen(fontsize(9pt)); draw((0,0)--(8,0)--(8,5)--(0,5)--cycle); filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8)); label("$1$",(1/2,5),dir(90)); label("$7$",(9/2,5),dir(90)); label("$1$",(8,1/2),dir(0)); label("$1$",(15/2,0),dir(270)); label("$7$",(7/2,0),dir(270)); label("$1$",(0,9/2),dir(180)); label("$4$",(0,2),dir(180)); draw((0,5)--(8,0)); | [] |
778 | Find the area of the shaded region.
$\textbf{(A)}\ 4\dfrac{3}{5} \qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 5\dfrac{1}{4} \qquad \textbf{(D)}\ 6\dfrac{1}{2} \qquad \textbf{(E)}\ 8$ | 2016 AMC 12A Problem 8 | Find the area of the unshaded area by calculating the area of the triangles and rectangles outside of the shaded region. We can do this by splitting up the unshaded areas into various triangles and rectangles as shown.
Notice that the two added lines bisect each of the $4$ sides of the large rectangle.
Subtracting the unshaded area from the total area gives us $40-33\frac{1}{2}=\boxed{6\frac{1}{2}}$, so the correct answer is $\boxed{\textbf{(D)}}$. | size(6cm); defaultpen(fontsize(9pt)); draw((0,0)--(8,0)--(8,5)--(0,5)--cycle); filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8)); label("$1$",(1/2,5),dir(90)); label("$4$",(6,5),dir(90)); label("$3$",(5/2,5),dir(90)); label("$1$",(8,1/2),dir(0)); label("$5/2$",(8,15/4),dir(0)); label("$3/2$",(8,7/4),dir(0)); label("$1$",(15/2,0),dir(270)); label("$4$",(2,0),dir(270)); label("$3$",(11/2,0),dir(270)); label("$1$",(0,9/2),dir(180)); label("$5/2$",(0,5/4),dir(180)); label("$3/2$",(0,13/4),dir(180)); draw((0,5/2)--(8,5/2)); draw((4,0)--(4,5)); | [] |
778 | Find the area of the shaded region.
$\textbf{(A)}\ 4\dfrac{3}{5} \qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 5\dfrac{1}{4} \qquad \textbf{(D)}\ 6\dfrac{1}{2} \qquad \textbf{(E)}\ 8$ | 2016 AMC 12A Problem 8 | Split the region into four parts by the diagonal from the top left to the bottom right. Slide the top and bottom pieces next to each other to form a parallelogram with base $1$ and height $\frac{5}{2}$, and slide the left and right pieces next to each other to form a parallelogram with base $1$ and height $\frac{8}{2}$. The total area is then $1\cdot\frac{5}{2}+1\cdot\frac{8}{2} = \frac{(5+8)(1)}{2} = \boxed{\textbf{(D)}\ 6\frac{1}{2}}$. ~emerald_block | size(6cm); defaultpen(fontsize(9pt)); draw((0,0)--(8,0)--(8,5)--(0,5)--cycle); filldraw((7,0)--(8,0)--(0,5)--(1,5)--cycle,gray(0.7)); filldraw((8,0)--(8,1)--(0,4)--(0,5)--cycle,gray(0.9)); label("$1$",(1/2,5),dir(90)); label("$7$",(9/2,5),dir(90)); label("$1$",(8,1/2),dir(0)); label("$1$",(15/2,0),dir(270)); label("$7$",(7/2,0),dir(270)); label("$1$",(0,9/2),dir(180)); label("$4$",(0,2),dir(180)); | [] |
778 | Find the area of the shaded region.
$\textbf{(A)}\ 4\dfrac{3}{5} \qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 5\dfrac{1}{4} \qquad \textbf{(D)}\ 6\dfrac{1}{2} \qquad \textbf{(E)}\ 8$ | 2016 AMC 12A Problem 8 | We can subtract the 4 triangular areas from the overall rectangle. It can be noticed that two triangles have area $\frac{6 \cdot \frac{5}{2}}{2} = \frac{15}{2}$, and the other two triangles have area $\frac{3 \cdot \frac{8}{2}}{2} = 6$ $\Rightarrow$ the shaded area is $(8 \cdot 5) - 2(\frac{15}{2}) - 2(6) = 40 - 15 - 12 = 13$. Since the desired area is half the shaded region, our area is $\boxed{\textbf{(D)}\ 6\frac{1}{2}}$.
~Champion1234 | size(6cm); defaultpen(fontsize(9pt)); draw((0,0)--(8,0)--(8,5)--(0,5)--cycle); filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8)); filldraw((1,0)--(0,0)--(0,1)--(8,4)--(8,5)--(7,5)--cycle,gray(0.8)); label("$1$",(1/2,5),dir(90)); label("$6$",(4,5),dir(90)); label("$1$",(15/2,5),dir(90)); label("$1$",(8,1/2),dir(0)); label("$3$",(8,5/2),dir(0)); label("$1$",(8,9/2),dir(0)); label("$1$",(15/2,0),dir(270)); label("$6$",(4,0),dir(270)); label("$1$",(1/2,0),dir(270)); label("$1$",(0,9/2),dir(180)); label("$3$",(0,5/2),dir(180)); label("$1$",(0,1/2),dir(180)); | [] |
779 | Circles with centers $P, Q$ and $R$, having radii $1, 2$ and $3$, respectively, lie on the same side of line $l$ and are tangent to $l$ at $P', Q'$ and $R'$, respectively, with $Q'$ between $P'$ and $R'$. The circle with center $Q$ is externally tangent to each of the other two circles. What is the area of triangle $PQR$?
$\textbf{(A) } 0\qquad \textbf{(B) } \sqrt{\frac{2}{3}}\qquad\textbf{(C) } 1\qquad\textbf{(D) } \sqrt{6}-\sqrt{2}\qquad\textbf{(E) }\sqrt{\frac{3}{2}}$ | 2016 AMC 12A Problem 15 | Notice that we can find $[P'PQRR']$ in two different ways: $[P'PQQ']+[Q'QRR']$ and $[PQR]+[P'PRR'].$ Since we want $[PQR],$ we use the latter method, so we have $[P'PQQ']+[Q'QRR']=[PQR]+[P'PRR'].$
$\break$
$P'Q'=\sqrt{PQ^2-(QQ'-PP')^2}=\sqrt{3^2-1^2}=\sqrt{8}=2\sqrt{2}$. Additionally, $Q'R'=\sqrt{QR^2-(RR'-QQ')^2}=\sqrt{5^2-1^2}=\sqrt{24}=2\sqrt{6}$. Therefore, $[P'PQQ']=\frac{P'P+Q'Q}{2}*2\sqrt{2}=\frac{1+2}{2}*2\sqrt{2}=3\sqrt{2}$. Similarly, $[Q'QRR']=5\sqrt6$. We can calculate $[P'PRR']$ easily because $P'R'=P'Q'+Q'R'=2\sqrt{2}+2\sqrt{6}$. $[P'PRR']=4\sqrt{2}+4\sqrt{6}$. $\newline$
Plugging into first equation, the two sums of areas, $3\sqrt{2}+5\sqrt{6}=4\sqrt{2}+4\sqrt{6}+[PQR],$ so therefore $[PQR]=\boxed{\textbf{(D) }\sqrt{6}-\sqrt{2}.}$ | // Block 1
size(250);
defaultpen(linewidth(0.4));
//Variable Declarations
pair P,Q,R,Pp,Qp,Rp;
pair A,B;
//Variable Definitions
A=(-5, 0);
B=(8, 0);
P=(-2.828,1);
Q=(0,2);
R=(4.899,3);
Pp=foot(P,A,B);
Qp=foot(Q,A,B);
Rp=foot(R,A,B);
path PQR = P--Q--R--cycle;
//Initial Diagram
dot(P);
dot(Q);
dot(R);
dot(Pp);
dot(Qp);
dot(Rp);
draw(Circle(P, 1), linewidth(0.8));
draw(Circle(Q, 2), linewidth(0.8));
draw(Circle(R, 3), linewidth(0.8));
draw(A--B,Arrows);
label("$P$",P,N);
label("$Q$",Q,N);
label("$R$",R,N);
label("$P'$",Pp,S);
label("$Q'$",Qp,S);
label("$R'$",Rp,S);
label("$l$",B,E);
//Added lines
draw(PQR);
draw(P--Pp);
draw(Q--Qp);
draw(R--Rp);
//Angle marks
draw(rightanglemark(P,Pp,B));
draw(rightanglemark(Q,Qp,B));
draw(rightanglemark(R,Rp,B));
// Block 2
size(250); defaultpen(linewidth(0.4)); //Variable Declarations pair P,Q,R,Pp,Qp,Rp; pair A,B; //Variable Definitions A=(-5, 0); B=(8, 0); P=(-2.828,1); Q=(0,2); R=(4.899,3); Pp=foot(P,A,B); Qp=foot(Q,A,B); Rp=foot(R,A,B); path PQR = P--Q--R--cycle; //Initial Diagram dot(P); dot(Q); dot(R); dot(Pp); dot(Qp); dot(Rp); draw(Circle(P, 1), linewidth(0.8)); draw(Circle(Q, 2), linewidth(0.8)); draw(Circle(R, 3), linewidth(0.8)); draw(A--B,Arrows); label("$P$",P,N); label("$Q$",Q,N); label("$R$",R,N); label("$P'$",Pp,S); label("$Q'$",Qp,S); label("$R'$",Rp,S); label("$l$",B,E); //Added lines draw(PQR); draw(P--Pp); draw(Q--Qp); draw(R--Rp); //Angle marks draw(rightanglemark(P,Pp,B)); draw(rightanglemark(Q,Qp,B)); draw(rightanglemark(R,Rp,B)); | [] |
779 | Circles with centers $P, Q$ and $R$, having radii $1, 2$ and $3$, respectively, lie on the same side of line $l$ and are tangent to $l$ at $P', Q'$ and $R'$, respectively, with $Q'$ between $P'$ and $R'$. The circle with center $Q$ is externally tangent to each of the other two circles. What is the area of triangle $PQR$?
$\textbf{(A) } 0\qquad \textbf{(B) } \sqrt{\frac{2}{3}}\qquad\textbf{(C) } 1\qquad\textbf{(D) } \sqrt{6}-\sqrt{2}\qquad\textbf{(E) }\sqrt{\frac{3}{2}}$ | 2016 AMC 12A Problem 15 | The above diagram can be achieved relatively simply using basic knowledge of the Pythagorean theorem and the fact that the radius from the center to the point of tangency is perpendicular to the tangent line. From there, observe that $[PQRY]$ can be calculated in two ways: $[\triangle PQX] + [QZYX] + [\triangle QRZ]$ and $[\triangle PRY] + [\triangle PQR]$. Solving, we get:
\begin{align*} [\triangle PQR] &= [PQRY] - [\triangle PRY] \\ &= [\triangle PQX] + [QZYX] + [\triangle QRZ] - [\triangle PRY] \\ &= \sqrt{2} + 2\sqrt{6} + \sqrt{6}- 2\sqrt{2}-2\sqrt{6} \\ &= \boxed{\textbf{(D)} \sqrt{6}-\sqrt{2}} \end{align*}
- ColtsFan10, diagram partially borrowed from Solution 1 | // Block 1
// Initial Pen Sizing
size(250);
defaultpen(linewidth(0.4));
defaultpen(fontsize(10pt));
// Variable Declarations
pair P,Q,R,Pp,Qp,Rp,X,Y,Z,A,B;
// Variable Definitions
A=(-5, 0);
B=(8, 0);
P=(-2.828,1);
Q=(0,2);
R=(4.899,3);
X=(0,1);
Y=(4.899,1);
Z=(4.899,2);
Pp=foot(P,A,B);
Qp=foot(Q,A,B);
Rp=foot(R,A,B);
path PQR = P--Q--R--cycle;
//Initial Diagram
dot(P);
dot(Q);
dot(R);
dot(Pp);
dot(Qp);
dot(Rp);
dot(X);
dot(Y);
dot(Z);
draw(Circle(P, 1), linewidth(0.3));
draw(Circle(Q, 2), linewidth(0.3));
draw(Circle(R, 3), linewidth(0.3));
draw(A--B,Arrows);
label("$P$",P,N);
label("$Q$",Q,N);
label("$R$",R,N);
label("$P'$",Pp,S);
label("$Q'$",Qp,S);
label("$R'$",Rp,S);
label("$l$",B,E);
label("$X$",X,NE);
label("$Y$",Y,E);
label("$Z$",Z,E);
//Added lines
filldraw(PQR,gray(0.8));
draw(P--Pp,linetype("8 8"));
draw(Q--Qp,linetype("8 8"));
draw(R--Rp,linetype("8 8"));
draw(P--Y,linetype("8 8"));
draw(Q--Z,linetype("8 8"));
//Angle marks
draw(rightanglemark(R,Y,P));
draw(rightanglemark(Q,X,P));
draw(rightanglemark(R,Z,Q));
//Length labeling
label("$2\sqrt{2}$",P--X,fontsize(8pt));
label("$2\sqrt{6}$",X--Y,fontsize(8pt));
label("$2\sqrt{6}$",Q--Z,fontsize(8pt));
label("$1$",R--Z,E,fontsize(8pt));
label("$1$",Z--Y,E,fontsize(8pt));
label("$3$",(-2.828,1.3)--Q,W,fontsize(8pt));
label("$5$",Q--R,N,fontsize(8pt));
// Block 2
// Initial Pen Sizing size(250); defaultpen(linewidth(0.4)); defaultpen(fontsize(10pt)); // Variable Declarations pair P,Q,R,Pp,Qp,Rp,X,Y,Z,A,B; // Variable Definitions A=(-5, 0); B=(8, 0); P=(-2.828,1); Q=(0,2); R=(4.899,3); X=(0,1); Y=(4.899,1); Z=(4.899,2); Pp=foot(P,A,B); Qp=foot(Q,A,B); Rp=foot(R,A,B); path PQR = P--Q--R--cycle; //Initial Diagram dot(P); dot(Q); dot(R); dot(Pp); dot(Qp); dot(Rp); dot(X); dot(Y); dot(Z); draw(Circle(P, 1), linewidth(0.3)); draw(Circle(Q, 2), linewidth(0.3)); draw(Circle(R, 3), linewidth(0.3)); draw(A--B,Arrows); label("$P$",P,N); label("$Q$",Q,N); label("$R$",R,N); label("$P'$",Pp,S); label("$Q'$",Qp,S); label("$R'$",Rp,S); label("$l$",B,E); label("$X$",X,NE); label("$Y$",Y,E); label("$Z$",Z,E); //Added lines filldraw(PQR,gray(0.8)); draw(P--Pp,linetype("8 8")); draw(Q--Qp,linetype("8 8")); draw(R--Rp,linetype("8 8")); draw(P--Y,linetype("8 8")); draw(Q--Z,linetype("8 8")); //Angle marks draw(rightanglemark(R,Y,P)); draw(rightanglemark(Q,X,P)); draw(rightanglemark(R,Z,Q)); //Length labeling label("$2\sqrt{2}$",P--X,fontsize(8pt)); label("$2\sqrt{6}$",X--Y,fontsize(8pt)); label("$2\sqrt{6}$",Q--Z,fontsize(8pt)); label("$1$",R--Z,E,fontsize(8pt)); label("$1$",Z--Y,E,fontsize(8pt)); label("$3$",(-2.828,1.3)--Q,W,fontsize(8pt)); label("$5$",Q--R,N,fontsize(8pt)); | [] |
780 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | 2016 AMC 12A Problem 21 | Let $AD$ intersect $OB$ at $E$ and $OC$ at $F.$
$\overarc{AB}= \overarc{BC}= \overarc{CD}=\theta$
$\angle{BAD}=\frac{1}{2} \cdot \overarc{BCD}=\theta=\angle{AOB}$
From there, $\triangle{OAB} \sim \triangle{ABE}$, thus:
$\frac{OA}{AB} = \frac{AB}{BE} = \frac{OB}{AE}$
$OA = OB$ because they are both radii of $\odot{O}$. Since $\frac{OA}{AB} = \frac{OB}{AE}$, we have that $200 = AB = AE$. Similarly, $CD = DF$.
$OE = 100\sqrt{2} = \frac{OB}{2}$ and $EF=\frac{BC}{2}=100$ , so $AD=AE + EF + FD = 200 + 100 + 200 = \boxed{\textbf{(E) } 500}$
Note
The first angle chase is done by the central angle theorem. Note that $\angle BAD$ is an inscribed angle from $BD$ to $A$, which is always $\frac{1}{2}$ of the angle of $\overarc{BD}$. Hence follows.
Furthermore, we get $OE=100\sqrt{2}$ from the similar triangles mentioned. We have: $\frac{200}{200\sqrt{2}}=\frac{BE}{200}$. Hence, $BE=100\sqrt{2}$ and so $OE=OB-BE=100\sqrt{2}$.
~mathboy282 | size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, F, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label("$A$",A,W); label("$B$",B,NW); label("$C$",C,NE); label("$D$",D,ENE); label("$O$",O,S); label("$\theta$",O,3N); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction E=extension(B,O,A,D); label("$E$",E,NE); F=extension(C,O,A,D); label("$F$",F,NE); //Angle marks draw(anglemark(C,O,B)); | [] |
780 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | 2016 AMC 12A Problem 21 | To save us from getting big numbers with lots of zeros behind them, let's divide all side lengths by $200$ for now, then multiply it back at the end of our solution.
Construct quadrilateral $ABCD$ on the circle with $AD$ being the missing side (Notice that since the side length is less than the radius, it will be very small on the top of the circle). Now, draw the radii from center $O$ to $A,B,C,$ and $D$. Let the intersection of $BD$ and $OC$ be point $E$. Notice that $BD$ and $OC$ are perpendicular because $BCDO$ is a kite.
We set lengths $BE=ED$ equal to $x$ (Solution 1.1 begins from here). By the Pythagorean Theorem,
\[\sqrt{1^2-x^2}+\sqrt{(\sqrt{2})^2-x^2}=\sqrt{2}\]
We solve for $x$:
\[1-x^2+2-x^2+2\sqrt{(1-x^2)(2-x^2)}=2\]
\[2\sqrt{(1-x^2)(2-x^2)}=2x^2-1\]
\[4(1-x^2)(2-x^2)=(2x^2-1)^2\]
\[8-12x^2+4x^4=4x^4-4x^2+1\]
\[8x^2=7\]
\[x=\frac{\sqrt{14}}{4}\]
By Ptolemy's Theorem,
\[AB \cdot CD + BC \cdot AD = AC \cdot BD = BD^2 = (2 \cdot BE)^2\]
Substituting values,
\[1^2+1 \cdot AD = 4{\left( \frac{\sqrt{14}}{4} \right)}^2\]
\[1+AD=\frac{7}{2}\]
\[AD=\frac{5}{2}\]
Finally, we multiply back the $200$ that we divided by at the beginning of the problem to get $AD=\boxed{\textbf{(E) } 500}$. | size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); E=extension(B,D,O,C); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label("$A$",A,W); label("$B$",B,NW); label("$C$",C,NE); label("$D$",D,E); label("$E$",E,WSW); label("$O$",O,S); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction draw(B--D); draw(rightanglemark(C,E,D)); | [] |
780 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | 2016 AMC 12A Problem 21 | size(250);
defaultpen(linewidth(0.4));
//Variable Declarations
real RADIUS;
pair A, B, C, D, O;
RADIUS=3;
//Variable Definitions
A=RADIUS*dir(148.414);
B=RADIUS*dir(109.471);
C=RADIUS*dir(70.529);
D=RADIUS*dir(31.586);
P=RADIUS*dir(270);
O=(0,0);
//Path Definitions
path quad= A -- B -- C -- D -- cycle;
//Initial Diagram
draw(circle(O, RADIUS), linewidth(0.8));
draw(quad, linewidth(0.8));
label("<imath>A</imath>",A,W);
label("<imath>B</imath>",B,NW);
label("<imath>C</imath>",C,NE);
label("<imath>D</imath>",D,E);
label("<imath>O</imath>",O,S);
//Radii
draw(O--A);
draw(O--B);
draw(O--C);
draw(O--D);
draw(P--A);
draw(P--B);
draw(P--C);
draw(P--D);
Draw some point <imath>P</imath> somewhere on major arc <imath>\overarc{AD}</imath>. Let <imath>\angle AOB \cong \angle BOC \cong \angle COD = 2\theta</imath>. Then, by inscribed angle <imath>\angle APB \cong \angle BPC \cong \angle CPD = \theta</imath>. By extended law of sines, we have <imath>\frac{200}{\text{sin}(\theta)} = 2(200\sqrt{2}) \implies \text{sin}(\theta) = \frac{\sqrt{2}}{4}</imath>. Using the triple angle sine formula, we have <imath>\text{sin}(3\theta)=3\text{sin}(\theta) - 4\text{sin}^3(\theta) = \frac{5\sqrt{2}}{8}</imath>.
Then, again by extended law of sines, we have <imath>\frac{AD}{\text{sin}(3\theta)}=400\sqrt{2}</imath>. Thus, <imath>AD = \text{sin}(3\theta) 400\sqrt{2} = \boxed{500}</imath>.
==Solution 3 (HARD Algebra)== | size(250);
defaultpen(linewidth(0.4));
//Variable Declarations
real RADIUS;
pair A, B, C, D, O;
RADIUS=3;
//Variable Definitions
A=RADIUS*dir(148.414);
B=RADIUS*dir(109.471);
C=RADIUS*dir(70.529);
D=RADIUS*dir(31.586);
O=(0,0);
//Path Definitions
path quad= A -- B -- C -- D -- cycle;
//Initial Diagram
draw(Circle(O, RADIUS), linewidth(0.8));
draw(quad, linewidth(0.8));
label("$A$",A,W);
label("$B$",B,NW);
label("$C$",C,NE);
label("$D$",D,E);
label("$O$",O,S);
//Radii
draw(O--A);
draw(O--B);
draw(O--C);
draw(O--D);
(Error making remote request. Unknown error_msg)
Let quadrilateral $ABCD$ be inscribed in circle $O$, where $AD$ is the side of unknown length. Draw the radii from center $O$ to all four vertices of the quadrilateral, and draw the altitude of $\triangle BOC$ such that it passes through side $AD$ at the point $G$ and meets side $BC$ at point $H$.
By the Pythagorean Theorem, the length of $OH$ is
\begin{align*} \sqrt{CO^2 - HC^2} &= \sqrt{(200\sqrt{2})^2 - \left(\frac{200}{2}\right)^2} \\ &= \sqrt{80000 - 10000} \\ &= \sqrt{70000} \\ &= 100\sqrt{7}. \end{align*}
Note that $[ABCDO] = [AOB] + [BOC] + [COD] = [AOD] + [ABCD].$ Let the length of $OG$ be $h$ and the length of $AD$ be $x$; then we have that
$[AOB] + [BOC] + [COD] = \frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} = \frac{x \times h}{2} + \frac{(100\sqrt{7} - h)(200 + x)}{2} = [AOD] + [ABCD].$
Furthermore,
\begin{align*} h &= \sqrt{OD^2 - GD^2} \\ &= \sqrt{(200\sqrt{2})^2 - \left(\frac{x}{2}\right)^2} \\ &= \sqrt{80000 - \frac{x^2}{4}} \end{align*}
Substituting this value of $h$ into the previous equation and evaluating for $x$, we get:
\[\frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} + \frac{200 \times 100\sqrt{7}}{2} = \frac{x \times h}{2} + \frac{(100\sqrt{7} - h)(200 + x)}{2}\]
\[\frac{3 \times 200 \times 100\sqrt{7}}{2} = \frac{x\sqrt{80000 - \frac{x^2}{4}}}{2} + \frac{\left(100\sqrt{7} - \sqrt{80000 - \frac{x^2}{4}}\right)(200 + x)}{2}\]
\[60000\sqrt{7} = \left(x\sqrt{80000 - \frac{x^2}{4}}\right) + \left(20000\sqrt{7}\right) + \left(100x\sqrt{7}\right) - \left(200\sqrt{80000 - \frac{x^2}{4}}\right) - \left(x\sqrt{80000 - \frac{x^2}{4}}\right)\]
\[40000\sqrt{7} = 100x\sqrt{7} - 200\sqrt{80000 - \frac{x^2}{4}}\]
\[400\sqrt{7} = x\sqrt{7} - 2\sqrt{80000 - \frac{x^2}{4}}\]
\[(x - 400)\sqrt{7} = 2\sqrt{80000 - \frac{x^2}{4}}\]
\[7(x-400)^2 = 4\left(80000 - \frac{x^2}{4}\right)\]
\[7x^2 - 5600x + 1120000 = 320000 - x^2\]
\[8x^2 - 5600x + 800000 = 0\]
\[x^2 - 700x + 100000 = 0\]
The roots of this quadratic are found by using the quadratic formula:
\begin{align*} x &= \frac{-(-700) \pm \sqrt{(-700)^2 - 4 \times 1 \times 100000}}{2 \times 1} \\ &= \frac{700 \pm \sqrt{490000 - 400000}}{2} \\ &= \frac{700}{2} \pm \frac{\sqrt{90000}}{2} \\ &= 350 \pm \frac{300}{2} \\ &= 200, 500 \end{align*}
If the length of $AD$ is $200$, then $ABCD$ would be a square. Thus, the radius of the circle would be
\[\frac{\sqrt{200^2 + 200^2}}{2} = \frac{\sqrt{80000}}{2} = \frac{200\sqrt{2}}{2} = 100\sqrt{2}\]
Which is a contradiction. Therefore, our answer is $\boxed{500}.$ | [] |
780 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | 2016 AMC 12A Problem 21 | Construct quadrilateral $ABCD$ on the circle with $AD$ being the missing side (Notice that since the side length is less than the radius, it will be very small on the top of the circle). Now, draw the radii from center $O$ to $A,B,C,$ and $D$. Apply the law of cosines on $\Delta BOC$; let $\theta = \angle BOC$. We get the following equation: \[(BC)^{2}=(OB)^{2}+(OC)^{2}-2\cdot OB \cdot OC\cdot \cos\theta\] Substituting the values in, we get \[(200)^{2}=2\cdot (200)^{2}+ 2\cdot (200)^{2}- 2\cdot 2\cdot (200)^{2}\cdot \cos\theta\] Canceling out, we get \[\cos\theta=\frac{3}{4}\]
Because $\angle AOB$, $\angle BOC$, and $\angle COD$ are congruent, $\angle AOD = 3\theta$. To find the remaining side ($AD$), we simply have to apply the law of cosines to $\Delta AOD$. Now, to find $\cos 3\theta$, we can derive a formula that only uses $\cos\theta$: \[\cos 3\theta=\cos (2\theta+\theta)= \cos 2\theta \cos\theta- (2\sin\theta \cos\theta) \cdot \sin \theta\] \[\cos 3\theta= \cos\theta (\cos 2\theta-2\sin^{2}\theta)=\cos\theta (2\cos^{2}\theta-3+2\cos^{2}\theta)\] \[\Rightarrow \cos 3\theta=4\cos^{3}\theta-3\cos\theta\] It is useful to memorize the triple angle formulas ($\cos 3\theta=4\cos^{3}\theta-3\cos\theta, \sin 3\theta=3\sin\theta-4\sin^{3}\theta$). Plugging in $\cos\theta=\frac{3}{4}$, we get $\cos 3\theta= -\frac{9}{16}$. Now, applying law of cosines on triangle $OAD$, we get \[(AD)^{2}= 2\cdot (200)^{2}+ 2\cdot (200)^{2}+2\cdot 200\sqrt2 \cdot 200\sqrt2 \cdot \frac{9}{16}\] \[\Rightarrow 2\cdot (200)^{2} \cdot (1+1+ \frac{9}{8})=(200)^{2}\cdot \frac{25}{4}\] \[AD=200 \cdot \frac{5}{2}=\boxed{500}\] | size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label("$A$",A,W); label("$B$",B,NW); label("$C$",C,NE); label("$D$",D,E); label("$O$",O,S); label("$\theta$",O,3N); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Angle mark for BOC draw(anglemark(C,O,B)); | [] |
780 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | 2016 AMC 12A Problem 21 | Construct quadrilateral $ABCD$ on the circle $O$ with $AD$ being the desired side. Then, drop perpendiculars from $A$ and $D$ to the extended line of $\overline{BC}$ and let these points be $E$ and $F$, respectively. Also, let $\theta = \angle BOC$. From the Law of Cosines on $\triangle BOC$, we have $\cos \theta = \frac{3}{4}$.
Now, since $\triangle BOC$ is isosceles with $\overline{OB} \cong \overline{OC}$, we have that $\angle BCO = \angle CBO = 90 - \frac{\theta}{2}$. In addition, we know that $\overline{BC} \cong \overline{CD}$ as they are both equal to $200$ and $\overline{OB} \cong \overline{OC} \cong \overline{OD}$ as they are both radii of the same circle. By SSS Congruence, we have that $\triangle OBC \cong \triangle OCD$, so we have that $\angle OCD = \angle BCO = 90 - \frac{\theta}{2}$, so $\angle DCF = \theta$.
Thus, we have $\frac{FC}{DC} = \cos \theta = \frac{3}{4}$, so $FC = 150$. Similarly, $BE = 150$, and $AD = 150 + 200 + 150 = \boxed{500}$. | size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, F, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); E=foot(A,B,C); F=foot(D,B,C); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label("$A$",A,W); label("$B$",B,NW); label("$C$",C,NE); label("$D$",D,ENE); label("$O$",O,S); label("$\theta$",O,3N); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction draw(A--E); draw(E--B); draw(C--F); draw(F--D); label("$E$",E,NW); label("$F$",F,NE); //Angle marks draw(anglemark(C,O,B)); draw(rightanglemark(A,E,B)); draw(rightanglemark(C,F,D)); | [] |
780 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | 2016 AMC 12A Problem 21 | Let $s = 200$. Let $O$ be the center of the circle. Then $AC$ is twice the altitude of $\triangle OBC$ to $\overline{OB}$. Since $\triangle OBC$ is isosceles we can compute its area to be $\frac{s^2 \sqrt{7}}{4}$, hence $CA = 2 \cdot \tfrac{2 \cdot s^2\sqrt7/4}{s\sqrt2} = s\sqrt{\frac{7}{2}}$.
Now by Ptolemy's Theorem we have $CA^2 = s^2 + AD \cdot s \implies AD = \left(\frac{7}{2}-1\right)s.$ This gives us: \[\boxed{\textbf{(E) } 500.}\] | pathpen = black; pointpen = black; size(6cm); draw(unitcircle); pair A = D("A", dir(50), dir(50)); pair B = D("B", dir(90), dir(90)); pair C = D("C", dir(130), dir(130)); pair D = D("D", dir(170), dir(170)); pair O = D("O", (0,0), dir(-90)); draw(A--C, red); draw(B--D, blue+dashed); draw(A--B--C--D--cycle); draw(A--O--C); draw(O--B); | [] |
780 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | 2016 AMC 12A Problem 21 | Label the points as shown, and let $\angle{EOF} = \theta$. Since $\overline{OB} = \overline{OC}$, and $\triangle{OFE} \sim \triangle{OCB}$, we get that $\angle{EFO} = 90-\frac{\theta}{2}$. We assign $\alpha$ to $90-\frac{\theta}{2}$ for simplicity.
From here, by vertical angles $\angle{CFD} = \alpha$. Also, since $\triangle{OCB} \cong \triangle{ODC}$, $\angle{OCD} = \alpha$. This means that $\angle{CDF} = 180-2\alpha = \theta$, which leads to $\triangle{OCB} \sim \triangle{DCF}$.
Since we know that $\overline{CD} = 200$, $\overline{DF} = 200$, and by similar reasoning $\overline{AE} = 200$.
Finally, again using similar triangles, we get that $\overline{CF} = 100\sqrt{2}$, which means that $\overline{OF} = \overline{OC} - \overline{CF} = 200\sqrt{2} - 100\sqrt{2} = 100\sqrt{2}$. We can again apply similar triangles (or use Power of a Point) to get $\overline{EF} = 100$, and finally $\overline{AD} = \overline{AE}+\overline{EF}+\overline{FD} = 200+100+200=\boxed{\textbf{(E)}500}$ - ColtsFan10 | size(250); defaultpen(linewidth(0.4)); //Variable Declarations, L is used to write alpha= statement real RADIUS; pair A, B, C, D, E, F, O, L; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); E=extension(A,D,O,B); F=extension(A,D,O,C); L=midpoint(C--D); O=(0,0); //Path Definitions path quad = A -- B -- C -- D -- cycle; //Initial Diagram draw(circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label("$A$",A,NW); label("$B$",B,NW); label("$C$",C,NE); label("$D$",D,NE); label("$E$",E,SW); label("$F$",F,SE); label("$O$",O,SE); dot(O,linewidth(5)); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction label("$\alpha = 90-\frac{\theta}{2}$",L,5NE,rgb(128, 0, 0)); draw(anglemark(C,O,B)); label("$\theta$",O,3N); draw(anglemark(E,F,O)); label("$\alpha$",F,3SW); draw(anglemark(D,F,C)); label("$\alpha$",F,3NE); draw(anglemark(F,C,D)); label("$\alpha$",C,3SSE); draw(anglemark(C,D,F)); label("$\theta$",(RADIUS-0.04)*dir(31.586),3WNW); | [] |
780 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | 2016 AMC 12A Problem 21 | Claim: $[ABCD]$ is an isosceles trapezoid.
Proof: Notice that $[ABCD]$ is cyclic, triangle $BOC$ is isosceles, and triangle $AOB$ is congruent to $DOC$ by SSS congruence. Therefore, $\angle BAD = 180 - \angle BCD = 180-(\angle BCO + \angle DCO)=180-(\angle CBO+\angle ABO) = 180 - \angle ABC = \angle CDA$. Hence, $[ABCD]$ is an isosceles trapezoid.
Let $\angle CDA=\alpha$. Notice that the length of the altitude from $C$ to $AD$ is $200sin(\alpha)$. Furthermore, the length of the altitude from $O$ to $BC$ is $100\sqrt{7}$ by the Pythagorean theorem. Therefore, the length of the altitude from $O$ to $AD$ is $100\sqrt{7}-200sin\alpha$. Let $F$ the feet of the altitude from $O$ to $AD$. Then, $FD=(200+400cos(\alpha))/2=100+200cos(\alpha)$, because $AOD$ is isosceles.
Therefore, by the Pythagorean theorem, $(100+200cos(\alpha))^2+(100\sqrt{7}-200sin(\alpha))^2=80000$. Simplifying, we have $1+cos(\alpha)=sin(\alpha) \cdot sqrt{7} \implies cos^2(\alpha)+2cos(\alpha)+1=sin^2(\alpha) \cdot 7 = 7-7cos^2(\alpha) \implies 8cos^2(\alpha)+2cos(\alpha) - 6 =0$. Solving this quadratic, we have $cos(\alpha)=\frac{3}{4}, -1$, but $0<\alpha<180 \implies cos(\alpha)=3/4$. Therefore, $AD=200cos(\alpha)+200cos(\alpha)+200=\boxed{500}$
- [mathMagicOPS] | size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label("$A$",A,W); label("$B$",B,NW); label("$C$",C,NE); label("$D$",D,E); label("$O$",O,S); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); | [] |
780 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | 2016 AMC 12A Problem 21 | Since $AB$, $BC$, and $CD$ are congruent chords, the triangles formed by connecting these sides to the circumcenter $O$ (i.e., $\triangle AOB$, $\triangle BOC$, $\triangle COD$) are congruent isosceles triangles.
Let $\alpha = \angle OCB$. In $\triangle BOC$, the side length $BC=200$ and $OC = OB = 200\sqrt{2}$. By dropping an altitude from $O$ to $BC$, we bisect $BC$. Half the base is $\frac{200}{2} = 100$.
Using the right triangle formed, we find $\cos(\alpha)$:
\[\cos(\alpha) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{100}{200\sqrt{2}} = \frac{\sqrt{2}}{4}\]
Since $\triangle BOC \cong \triangle COD$, we have $m\angle OCB = m\angle OCD = \alpha$.
The interior angle measure $m\angle DCB$ is given by:
\[m\angle DCB = m\angle OCB + m\angle OCD = 2\alpha\]
We calculate $\cos(2\alpha)$ using the double-angle identity:
\[\cos(2\alpha) = 2\cos^2(\alpha) - 1\]
\[\cos(2\alpha) = 2\left(\frac{\sqrt{2}}{4}\right)^2 - 1 = \frac{1}{4} - 1 = -\frac{3}{4}\]
Now, apply the Law of Cosines to $\triangle BCD$ to find $BD^2$:
\[BD^2 = BC^2 + CD^2 - 2(BC)(CD)\cos(\angle DCB)\]
\[BD^2 = 200^2 + 200^2 - 2(200)(200)\left(-\frac{3}{4}\right)\]
\[BD^2 = 80000 + 60000 = 140000\]
Since $ABCD$ is a cyclic quadrilateral, the opposite angles are supplementary:
\[m\angle DAB = 180^\circ - m\angle DCB = 180^\circ - 2\alpha\]
Thus, we find $\cos(\angle DAB)$:
\[\cos(\angle DAB) = \cos(180^\circ - 2\alpha) = -\cos(2\alpha)\]
\[\cos(\angle DAB) = -\left(-\frac{3}{4}\right) = \frac{3}{4}\]
Finally, we apply the Law of Cosines to $\triangle DAB$ to find the side length $AD$. Let $AD = a$:
\[140000 = 200^2 + a^2 - 2(200)(a)\left(\frac{3}{4}\right)\]
\[140000 = 40000 + a^2 - 300a\]
Rearranging the terms yields the quadratic equation:
\[a^2 - 300a - 100000 = 0\]
We solve this equation by factoring:
\[(a - 500)(a + 200) = 0\]
The two possible solutions for $a$ are $a=500$ and $a=-200$. Since $a$ represents a side length, it must be positive, so $a = 500$, which means $AD = 500$, or $\boxed{\textbf{(E) } 500}$.
~Voidling | size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, F, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O, RADIUS), linewidth(0.8)); draw(quad, linewidth(0.8)); label("$A$",A,W); label("$B$",B,NW); label("$C$",C,NE); label("$D$",D,ENE); label("$O$",O,S); label("$\theta$",O,3N); //Radii draw(O--A); draw(O--B); draw(O--C); draw(O--D); //Construction E=extension(B,O,A,D); label("$E$",E,NE); F=extension(C,O,A,D); label("$F$",F,NE); //Angle marks draw(anglemark(C,O,B)); | [] |
781 | How many ordered triples $(x,y,z)$ of positive integers satisfy $\text{lcm}(x,y) = 72, \text{lcm}(x,z) = 600,$ and $\text{lcm}(y,z)=900$?
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 64$ | 2016 AMC 12A Problem 22 | As said in previous solutions, start by factoring $72, 600,$ and $900$. The prime factorizations are as follows: \[72=2^3\cdot 3^2,\] \[600=2^3\cdot 3\cdot 5^2,\] \[\text{and } 900=2^2\cdot 3^2\cdot 5^2\]
To organize $x,y, \text{ and } z$ and their respective LCMs in a simpler way, we can draw a triangle as follows such that $x,y, \text{and } z$ are the vertices and the LCMs are on the edges.
Now we can split this triangle into three separate ones for each of the three different prime factors $2,3, \text{and } 5$.
Analyzing for powers of $2$, it is quite obvious that $x$ must have $2^3$ as one of its factors since neither $y \text{ nor } z$ can have a power of $2$ exceeding $2$. Turning towards the vertices $y$ and $z$, we know at least one of them must have $2^2$ as its factors. Therefore, we have $5$ ways for the powers of $2$ for $y \text{ and } z$ since the only ones that satisfy the previous conditions are for ordered pairs $(y,z) \{(2,0)(2,1)(0,2)(1,2)(2,2)\}$.
Using the same logic as we did for powers of $2$, it becomes quite easy to note that $y$ must have $3^2$ as one of its factors. Moving onto $x \text{ and } z$, we can use the same logic to find the only ordered pairs $(x,z)$ that will work are $\{(1,0)(0,1)(1,1)\}$.
The final and last case is the powers of $5$.
This is actually quite a simple case since we know $z$ must have $5^2$ as part of its factorization while $x \text{ and } y$ cannot have a factor of $5$ in their prime factorization.
Multiplying all the possible arrangements for prime factors $2,3, \text{ and } 5$, we get the answer:
\[5\cdot3\cdot1=\boxed{\textbf{(A) }15}\]
(Diagrams by ColtsFan10) | // Block 1
//Variable Declarations defaultpen(0.45); size(200pt); fontsize(15pt); pair X, Y, Z; real R; path tri; //Variable Definitions R = 1; X = R*dir(90); Y = R*dir(210); Z = R*dir(-30); tri = X--Y--Z--cycle; //Diagram draw(tri); label("$x$",X,N); label("$y$",Y,SW); label("$z$",Z,SE); label("$2^33^25^0$",X--Y,2W); label("$2^33^15^2$",X--Z,2E); label("$2^23^25^2$",Y--Z,2S);
// Block 2
//Variable Declarations defaultpen(0.45); size(200pt); fontsize(15pt); pair X, Y, Z; real R; path tri; //Variable Definitions R = 1; X = R*dir(90); Y = R*dir(210); Z = R*dir(-30); tri = X--Y--Z--cycle; //Diagram draw(tri); label("$x$",X,N); label("$y$",Y,SW); label("$z$",Z,SE); label("$2^3$",X--Y,2W); label("$2^3$",X--Z,2E); label("$2^2$",Y--Z,2S);
// Block 3
//Variable Declarations defaultpen(0.45); size(200pt); fontsize(15pt); pair X, Y, Z; real R; path tri; //Variable Definitions R = 1; X = R*dir(90); Y = R*dir(210); Z = R*dir(-30); tri = X--Y--Z--cycle; //Diagram draw(tri); label("$x$",X,N); label("$y$",Y,SW); label("$z$",Z,SE); label("$3^2$",X--Y,2W); label("$3^1$",X--Z,2E); label("$3^2$",Y--Z,2S);
// Block 4
//Variable Declarations defaultpen(0.45); size(200pt); fontsize(15pt); pair X, Y, Z; real R; path tri; //Variable Definitions R = 1; X = R*dir(90); Y = R*dir(210); Z = R*dir(-30); tri = X--Y--Z--cycle; //Diagram draw(tri); label("$x$",X,N); label("$y$",Y,SW); label("$z$",Z,SE); label("$5^0$",X--Y,2W); label("$5^2$",X--Z,2E); label("$5^2$",Y--Z,2S); | [] |
782 | Three numbers in the interval $\left[0,1\right]$ are chosen independently and at random. What is the probability that the chosen numbers are the side lengths of a triangle with positive area?
$\textbf{(A) }\frac16\qquad\textbf{(B) }\frac13\qquad\textbf{(C) }\frac12\qquad\textbf{(D) }\frac23\qquad\textbf{(E) }\frac56$ | 2016 AMC 12A Problem 23 | Because we can let the the sides of the triangle be any variable we want, to make it easier for us when solving, let’s let the side lengths be $x,y,$ and $a$. WLOG assume $a$ is the largest. Then, $x+y>a$, meaning the solution is $\boxed{\textbf{(C)}\;1/2}$, as shown in the graph below. | // Block 1
pair A = (0,0);
pair B = (1,0);
pair C = (1,1);
pair D = (0,1);
pair E = (0,0);
draw(A--B--C--D--cycle);
draw(B--D,dashed);
fill(B--D--C--cycle,gray);
label("$0$",A,SW);
label("$a$",B,S);
label("$a$",D,W);
label("$y$",(0,.5),W);
label("$x$",(.5,0),S);
label("$x+y>a$",(5/7,5/7));
// Block 2
pair A = (0,0); pair B = (1,0); pair C = (1,1); pair D = (0,1); pair E = (0,0); draw(A--B--C--D--cycle); draw(B--D,dashed); fill(B--D--C--cycle,gray); label("$0$",A,SW); label("$a$",B,S); label("$a$",D,W); label("$y$",(0,.5),W); label("$x$",(.5,0),S); label("$x+y>a$",(5/7,5/7)); | [] |
783 | All three vertices of $\bigtriangleup ABC$ lie on the parabola defined by $y=x^2$, with $A$ at the origin and $\overline{BC}$ parallel to the $x$-axis. The area of the triangle is $64$. What is the length of $BC$?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 16$ | 2016 AMC 10B Problem 9 | The area of the triangle is $\frac{(2r)(r^2)}{2} = r^3$, so $r^3=64\implies r=4$, giving a total distance across the top of $8$, which is answer $\textbf{(C)}$. | // Block 1
import graph;size(7cm,IgnoreAspect);
real f(real x) {return x*x;}
draw((0,0)--(4,16)--(-4,16)--cycle,blue);
draw(graph(f,-5,5,operator ..),gray);
xaxis("$x$");yaxis("$y$",-1);
label("$y=x^2$",(4.5,20.25),E);
draw((4.2,0)--(4.2,16),Arrows);
label("$r^2$",(4.2,0)--(4.2,16),E);
draw((0,17)--(4,17),Arrows);
label("$r$",(0,17)--(4,17),N);
// Block 2
import graph;size(7cm,IgnoreAspect); real f(real x) {return x*x;} draw((0,0)--(4,16)--(-4,16)--cycle,blue); draw(graph(f,-5,5,operator ..),gray); xaxis("$x$");yaxis("$y$",-1); label("$y=x^2$",(4.5,20.25),E); draw((4.2,0)--(4.2,16),Arrows); label("$r^2$",(4.2,0)--(4.2,16),E); draw((0,17)--(4,17),Arrows); label("$r$",(0,17)--(4,17),N); | [] |
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