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 |
|---|---|---|---|---|---|
784 | Carl decided to fence in his rectangular garden. He bought $20$ fence posts, placed one on each of the four corners, and spaced out the rest evenly along the edges of the garden, leaving exactly $4$ yards between neighboring posts. The longer side of his garden, including the corners, has twice as many posts as the shorter side, including the corners. What is the area, in square yards, of Carl’s garden?
$\textbf{(A)}\ 256\qquad\textbf{(B)}\ 336\qquad\textbf{(C)}\ 384\qquad\textbf{(D)}\ 448\qquad\textbf{(E)}\ 512$ | 2016 AMC 10B Problem 11 | If the dimensions are $4a\times 4b$, then one side will have $a+1$ posts (including corners) and the other $b+1$ (including corners).
The total number of posts is $2(a+b)=20$.
This diagram represents the number of posts around the garden.
We solve the system $\begin{cases}b+1=2(a+1)\\a+b=10\end{cases}$ to get $a=3,\ b=7$. Then the area is $4a\cdot 4b=336$ which is $\boxed{\textbf{(B) } 336}$.
~Edits by BakedPotato66 | // Block 1
size(7cm);fill((0,0)--(5,0)--(5,7)--(0,7)--cycle,lightgreen);
for(int i=0;i<5;++i)dot((i,0),red);for(int i=0;i<7;++i)dot((5,i),blue);dot((5,7));
draw(arc((0,0),.5,-90,-270)--arc((4,0),.5,90,-90)--cycle,gray+dotted);
draw(arc((5,0),.5,-180,0)--arc((5,6),.5,0,180)--cycle,gray+dotted);
draw((0,-1)--(5,-1),Arrows);draw((6,0)--(6,7),Arrows);
label("$a+1$",(0,-1)--(5,-1),S);label("$b+1$",(6,0)--(6,7),E);
label("$a$",(1,1));label("$b$",(4,5));
// Block 2
size(7cm);fill((0,0)--(5,0)--(5,7)--(0,7)--cycle,lightgreen); for(int i=0;i<5;++i)dot((i,0),red);for(int i=0;i<7;++i)dot((5,i),blue);dot((5,7)); draw(arc((0,0),.5,-90,-270)--arc((4,0),.5,90,-90)--cycle,gray+dotted); draw(arc((5,0),.5,-180,0)--arc((5,6),.5,0,180)--cycle,gray+dotted); draw((0,-1)--(5,-1),Arrows);draw((6,0)--(6,7),Arrows); label("$a+1$",(0,-1)--(5,-1),S);label("$b+1$",(6,0)--(6,7),E); label("$a$",(1,1));label("$b$",(4,5)); | [] |
785 | How many squares whose sides are parallel to the axes and whose vertices have coordinates that are integers lie entirely within the region bounded by the line $y=\pi x$, the line $y=-0.1$ and the line $x=5.1?$
$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 41 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ 57$ | 2016 AMC 10B Problem 14 | The region is a right triangle which contains the following lattice points:
$(0,0); (1,0)\rightarrow(1,3); (2,0)\rightarrow(2,6); (3,0)\rightarrow(3,9); (4,0)\rightarrow(4,12); (5,0)\rightarrow(5,15)$
Squares $1\times 1$:
Suppose that the top-right corner is $(x,y)$, with $2\le x\le 5$. Then to include all other corners, we need $1\le y\le 3(x-1)$.
This produces $3+6+9+12=30$ squares.
Squares $2\times 2$:
Here $3\le x\le 5$. To include all other corners, we need $2\le y\le 3(x-2)$.
This produces $2+5+8=15$ squares.
Squares $3\times 3$:
Similarly, this produces $5$ squares.
No other squares will fit in the region. Therefore the answer is $\boxed{\textbf{(D) }50}$. | // Block 1
size(10cm);
for(int i=0;i<6;++i)for(int j=0;j<=3*i;++j)dot((i,j));
draw((0,-1)--(0,16),EndArrow);draw((-1,0)--(6,0),EndArrow);
draw((-.5,-pi/2)--(5.2,5.2*pi),gray);draw((-1,-.1)--(6,-.1)^^(5.1,-1)--(5.1,16),gray);
// Block 2
size(10cm); for(int i=0;i<6;++i)for(int j=0;j<=3*i;++j)dot((i,j)); draw((0,-1)--(0,16),EndArrow);draw((-1,0)--(6,0),EndArrow); draw((-.5,-pi/2)--(5.2,5.2*pi),gray);draw((-1,-.1)--(6,-.1)^^(5.1,-1)--(5.1,16),gray); | [] |
786 | All the numbers $1, 2, 3, 4, 5, 6, 7, 8, 9$ are written in a $3\times3$ array of squares, one number in each square, in such a way that if two numbers are consecutive then they occupy squares that share an edge. The numbers in the four corners add up to $18$. What is the number in the center?
$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 9$ | 2016 AMC 10B Problem 15 | Consecutive numbers share an edge. That means that it is possible to walk from $1$ to $9$ by single steps north, south, east, or west. Consequently, the squares in the diagram with different shades have different parity:
But since there are only four even numbers in the set, the five darker squares must contain the odd numbers, which sum to $1+3+5+7+9=25.$ Therefore if the sum of the numbers in the corners is $18$, the number in the center must be $\boxed {\textbf{(C) }7}$. | // Block 1
size(4cm);
for(int i=0;i<3;++i)for(int j=0;j<3;++j)filldraw(box((i,j),(i+1,j+1)),gray((i+j)%2*.2+.7));
// Block 2
size(4cm); for(int i=0;i<3;++i)for(int j=0;j<3;++j)filldraw(box((i,j),(i+1,j+1)),gray((i+j)%2*.2+.7)); | [] |
787 | Rectangle $ABCD$ has $AB=5$ and $BC=4$. Point $E$ lies on $\overline{AB}$ so that $EB=1$, point $G$ lies on $\overline{BC}$ so that $CG=1$, and point $F$ lies on $\overline{CD}$ so that $DF=2$. Segments $\overline{AG}$ and $\overline{AC}$ intersect $\overline{EF}$ at $Q$ and $P$, respectively. What is the value of $\frac{PQ}{EF}$?
$\textbf{(A)}~\frac{\sqrt{13}}{16} \qquad \textbf{(B)}~\frac{\sqrt{2}}{13} \qquad \textbf{(C)}~\frac{9}{82} \qquad \textbf{(D)}~\frac{10}{91}\qquad \textbf{(E)}~\frac19$ | 2016 AMC 10B Problem 19 | First, we will define point $D$ as the origin. Then, we will find the equations of the following three lines: $AG$, $AC$, and $EF$. The slopes of these lines are $-\frac{3}{5}$, $-\frac{4}{5}$, and $2$, respectively. Next, we will find the equations of $AG$, $AC$, and $EF$. They are as follows:
\[AG = f(x) = -\frac{3}{5}x + 4\]
\[AC = g(x) = -\frac{4}{5}x + 4\]
\[EF = h(x) = 2x - 4\]
After drawing in altitudes to $DC$ from $P$, $Q$, and $E$, we see that $\frac{PQ}{EF} = \frac{P'Q'}{E'F}$ because of similar triangles, and so we only need to find the x-coordinates of $P$ and $Q$.
Finding the intersections of $AC$ and $EF$, and $AG$ and $EF$ gives the x-coordinates of $P$ and $Q$ to be $\frac{20}{7}$ and $\frac{40}{13}$. This means that $P'Q' = DQ' - DP' = \frac{40}{13} - \frac{20}{7} = \frac{20}{91}$. Now we can find $\frac{PQ}{EF} = \frac{P'Q'}{E'F} = \frac{\frac{20}{91}}{2} = \boxed{\textbf{(D)}~\frac{10}{91}}$ | // Block 1
pair A1=(2,0),A2=(4,4);
pair B1=(0,4),B2=(5,1);
pair C1=(5,0),C2=(0,4);
pair D1=(20/7,0),D2=(20/7,12/7);
pair E1=(40/13,0),E2=(40/13,28/13);
pair F1=(4,0),F2=(4,4);
draw(A1--A2);
draw(B1--B2);
draw(C1--C2);
draw(D1--D2,dashed);
draw(E1--E2,dashed);
draw(F1--F2,dashed);
draw((0,0)--B1--(5,4)--C1--cycle);
dot((20/7,12/7));
dot((3.07692307692,2.15384615384));
dot((20/7,0));
dot((40/13,0));
dot((4,0));
label("$Q$",(3.07692307692,2.15384615384),N);
label("$P$",(20/7,12/7),W);
label("$A$",(0,4), NW);
label("$B$",(5,4), NE);
label("$C$",(5,0),SE);
label("$D$",(0,0),SW);
label("$F$",(2,0),S); label("$G$",(5,1),E);
label("$E$",(4,4),N);
label("$P'$", (20/7,0),SSW);
label("$Q'$", (40/13,0),SSE);
label("$E'$", (4,0),S);
dot(A1); dot(A2);
dot(B1); dot(B2);
dot(C1); dot(C2);
dot((0,0)); dot((5,4));
// Block 2
pair A1=(2,0),A2=(4,4); pair B1=(0,4),B2=(5,1); pair C1=(5,0),C2=(0,4); pair D1=(20/7,0),D2=(20/7,12/7); pair E1=(40/13,0),E2=(40/13,28/13); pair F1=(4,0),F2=(4,4); draw(A1--A2); draw(B1--B2); draw(C1--C2); draw(D1--D2,dashed); draw(E1--E2,dashed); draw(F1--F2,dashed); draw((0,0)--B1--(5,4)--C1--cycle); dot((20/7,12/7)); dot((3.07692307692,2.15384615384)); dot((20/7,0)); dot((40/13,0)); dot((4,0)); label("$Q$",(3.07692307692,2.15384615384),N); label("$P$",(20/7,12/7),W); label("$A$",(0,4), NW); label("$B$",(5,4), NE); label("$C$",(5,0),SE); label("$D$",(0,0),SW); label("$F$",(2,0),S); label("$G$",(5,1),E); label("$E$",(4,4),N); label("$P'$", (20/7,0),SSW); label("$Q'$", (40/13,0),SSE); label("$E'$", (4,0),S); dot(A1); dot(A2); dot(B1); dot(B2); dot(C1); dot(C2); dot((0,0)); dot((5,4)); | [] |
787 | Rectangle $ABCD$ has $AB=5$ and $BC=4$. Point $E$ lies on $\overline{AB}$ so that $EB=1$, point $G$ lies on $\overline{BC}$ so that $CG=1$, and point $F$ lies on $\overline{CD}$ so that $DF=2$. Segments $\overline{AG}$ and $\overline{AC}$ intersect $\overline{EF}$ at $Q$ and $P$, respectively. What is the value of $\frac{PQ}{EF}$?
$\textbf{(A)}~\frac{\sqrt{13}}{16} \qquad \textbf{(B)}~\frac{\sqrt{2}}{13} \qquad \textbf{(C)}~\frac{9}{82} \qquad \textbf{(D)}~\frac{10}{91}\qquad \textbf{(E)}~\frac19$ | 2016 AMC 10B Problem 19 | Extend $AG$ to intersect $CD$ at $H$. Letting $x=\overline{HC}$, we have that \[\triangle{HCG}\sim\triangle{HDA}\implies \dfrac{\overline{HC}}{\overline{CG}}=\dfrac{\overline{HD}}{\overline{AD}}\implies \dfrac{x}{1}=\dfrac{x+5}{4}\implies x=\dfrac{5}{3}.\]
Then, notice that $\triangle{AEQ}\sim\triangle{HFQ}$ and $\triangle{AEP}\sim\triangle{CFP}$. Thus, we see that \[\dfrac{AE}{HF}=\dfrac{EQ}{QF}\implies \dfrac{AE}{HF} = \dfrac{4}{3+\frac{5}{3}} = \dfrac{12}{14}=\dfrac{6}{7}\implies \dfrac{EQ}{EF}=\dfrac{6}{13}\]
and \[\dfrac{AE}{CF}=\dfrac{EP}{FP} \implies \dfrac{4}{3}=\dfrac{EP}{FP}\implies \dfrac{FP}{EF} = \dfrac{3}{7}.\]
Thus, we see that \[\dfrac{PQ}{EF} = 1-\left(\dfrac{6}{13}+\dfrac{3}{7}\right) = 1-\left(\dfrac{42+39}{91}\right) = 1-\left(\dfrac{81}{91}\right) = \boxed{\textbf{(D)}~ \dfrac{10}{91}}.\] | // Block 1
pair A1=(2,0),A2=(4,4);
pair B1=(0,4),B2=(5,1);
pair C1=(5,0),C2=(0,4);
pair H = (20/3,0);
draw(A1--A2);
draw(B1--B2);
draw(C1--C2);
draw(B1--H);
draw((0,0)--H);
draw((0,0)--B1--(5,4)--C1--cycle);
dot((20/7,12/7));
dot((3.07692307692,2.15384615384));
label("$Q$",(3.07692307692,2.15384615384),N);
label("$P$",(20/7,12/7),W);
label("$A$",(0,4), NW);
label("$B$",(5,4), NE);
label("$C$",(5,0),SE);
label("$D$",(0,0),SW);
label("$F$",(2,0),S); label("$G$",(5,1),E);
label("$E$",(4,4),N);
label("$H$",H,E);
// Block 2
pair A1=(2,0),A2=(4,4); pair B1=(0,4),B2=(5,1); pair C1=(5,0),C2=(0,4); pair H = (20/3,0); draw(A1--A2); draw(B1--B2); draw(C1--C2); draw(B1--H); draw((0,0)--H); draw((0,0)--B1--(5,4)--C1--cycle); dot((20/7,12/7)); dot((3.07692307692,2.15384615384)); label("$Q$",(3.07692307692,2.15384615384),N); label("$P$",(20/7,12/7),W); label("$A$",(0,4), NW); label("$B$",(5,4), NE); label("$C$",(5,0),SE); label("$D$",(0,0),SW); label("$F$",(2,0),S); label("$G$",(5,1),E); label("$E$",(4,4),N); label("$H$",H,E); | [] |
787 | Rectangle $ABCD$ has $AB=5$ and $BC=4$. Point $E$ lies on $\overline{AB}$ so that $EB=1$, point $G$ lies on $\overline{BC}$ so that $CG=1$, and point $F$ lies on $\overline{CD}$ so that $DF=2$. Segments $\overline{AG}$ and $\overline{AC}$ intersect $\overline{EF}$ at $Q$ and $P$, respectively. What is the value of $\frac{PQ}{EF}$?
$\textbf{(A)}~\frac{\sqrt{13}}{16} \qquad \textbf{(B)}~\frac{\sqrt{2}}{13} \qquad \textbf{(C)}~\frac{9}{82} \qquad \textbf{(D)}~\frac{10}{91}\qquad \textbf{(E)}~\frac19$ | 2016 AMC 10B Problem 19 | We will calculate $\frac{EP}{EF}$ using similar triangle, and $\frac{EQ}{EF}$ using ratio of area of $\triangle AEG$ to $\triangle AFG$.
\[\triangle AEP \sim \triangle CFP, \frac{AE}{CF}=\frac{EP}{FP}, \frac{EP}{FP}=\frac{4}{3}, \frac{EP}{EF}=\frac{4}{7}\]
\[[AEG]=\frac{1}{2} \cdot 4\cdot 3=6\]
\[[AFG]=[ABCD]-[ADF]-[CFG]-[ABG]=20-4-\frac{3}{2}-\frac{15}{2}=7\]
Because $\triangle AEG$ and $\triangle AFG$ share the same base $AG$, the ratio $\frac{[AEG]}{[AFG]}$ is equal to the ratio of the altitude of $\triangle AEG$ to $AG$ to that of $\triangle AFG$ to $AG$, which is equal to $\frac{EQ}{QF}$:
\[\frac{[AEG]}{[AFG]}=\frac{EQ}{QF}=\frac{6}{7}\]
\[\frac{EQ}{EF}=\frac{6}{13}\]
\[\frac{PQ}{EF}=\frac{EP}{EF}-\frac{EQ}{EF}=\frac{4}{7}-\frac{6}{13}=\frac{10}{91}\]
\[\frac{PQ}{EF}=\boxed{\textbf{(D)}~\frac{10}{91}}\]
~isabelchen | // Block 1
pair A1=(2,0),A2=(4,4);
pair B1=(0,4),B2=(5,1),B3=(4,4);
pair C1=(5,0),C2=(0,4),C3=(2,0);
draw(A1--A2);
draw(B1--B2);
draw(B2--B3);
draw(C1--C2);
draw(C2--C3);
draw(A1--B2);
draw((0,0)--B1--(5,4)--C1--cycle);
dot((20/7,12/7));
dot((3.07692307692,2.15384615384));
label("$Q$",(3.07692307692,2.15384615384),N);
label("$P$",(20/7,12/7),W);
label("$A$",(0,4), NW);
label("$B$",(5,4), NE);
label("$C$",(5,0),SE);
label("$D$",(0,0),SW);
label("$F$",(2,0),S);
label("$G$",(5,1),E);
label("$E$",(4,4),N);
// Block 2
pair A1=(2,0),A2=(4,4); pair B1=(0,4),B2=(5,1),B3=(4,4); pair C1=(5,0),C2=(0,4),C3=(2,0); draw(A1--A2); draw(B1--B2); draw(B2--B3); draw(C1--C2); draw(C2--C3); draw(A1--B2); draw((0,0)--B1--(5,4)--C1--cycle); dot((20/7,12/7)); dot((3.07692307692,2.15384615384)); label("$Q$",(3.07692307692,2.15384615384),N); label("$P$",(20/7,12/7),W); label("$A$",(0,4), NW); label("$B$",(5,4), NE); label("$C$",(5,0),SE); label("$D$",(0,0),SW); label("$F$",(2,0),S); label("$G$",(5,1),E); label("$E$",(4,4),N); | [] |
787 | Rectangle $ABCD$ has $AB=5$ and $BC=4$. Point $E$ lies on $\overline{AB}$ so that $EB=1$, point $G$ lies on $\overline{BC}$ so that $CG=1$, and point $F$ lies on $\overline{CD}$ so that $DF=2$. Segments $\overline{AG}$ and $\overline{AC}$ intersect $\overline{EF}$ at $Q$ and $P$, respectively. What is the value of $\frac{PQ}{EF}$?
$\textbf{(A)}~\frac{\sqrt{13}}{16} \qquad \textbf{(B)}~\frac{\sqrt{2}}{13} \qquad \textbf{(C)}~\frac{9}{82} \qquad \textbf{(D)}~\frac{10}{91}\qquad \textbf{(E)}~\frac19$ | 2016 AMC 10B Problem 19 | We will calculate $\frac{PQ}{QE}$ using the ratio of area of $\triangle APG$ to that of $\triangle AEG$.
\[[ACG]=\frac{1}{2} \cdot 5 \cdot 1 = \frac{5}{2}\]
\[[CAB]=\frac{1}{2} \cdot 5 \cdot 4=10\]
\[\triangle AEP \sim \triangle CFP\]
\[\frac{CP}{AP}=\frac{CF}{AE}=\frac{3}{4}\]
\[\frac{CP}{AC}=\frac{3}{7}\]
\[\frac{[CPG]}{[CAB]}=\frac{CP}{CA} \cdot \frac{CG}{CB}=\frac{3 \cdot 1}{7 \cdot 4}=\frac{3}{28}\]
\[[CPG]=\frac{3}{28} \cdot [CAB]=\frac{3}{28} \cdot 10=\frac{15}{14}\]
\[[APG]=[ACG]-[CPG]=\frac{5}{2}-\frac{15}{14}=\frac{35-15}{14}=\frac{20}{14}=\frac{10}{7}\]
\[[AEG]=\frac{1}{2} \cdot 4 \cdot 3=6\]
Because $\triangle APG$ and $\triangle AEG$ share the same base $AG$, the ratio $\frac{[APG]}{[AEG]}$ is equal to the ratio of altitude of $\triangle APG$ to $AG$ to that of $\triangle AEG$ to $AG$, which is equal to $\frac{PQ}{QE}$:
\[\frac{PQ}{QE}=\frac{[APG]}{[AEG]}=\frac{\frac{10}{7}}{6}=\frac{10}{42}=\frac{5}{21}\]
\[\frac{PQ}{PE}=\frac{5}{21+5}=\frac{5}{26}\]
\[\frac{PE}{PF}=\frac{AE}{CF}=\frac{4}{3}\]
\[\frac{PE}{EF}=\frac{4}{7}\]
\[\frac{PQ}{PE} \cdot \frac{PE}{EF} = \frac{5}{26} \cdot \frac{4}{7} = \frac{10}{91}\]
\[\frac{PQ}{EF}=\boxed{\textbf{(D)}~\frac{10}{91}}\]
~isabelchen | // Block 1
pair A1=(2,0),A2=(4,4);
pair B1=(0,4),B2=(5,1),B3=(20/7,12/7);
pair C1=(5,0),C2=(0,4);
draw(A1--A2);
draw(B1--B2);
draw(C1--C2);
draw(A2--B2);
draw(B2--B3);
draw((0,0)--B1--(5,4)--C1--cycle);
dot((20/7,12/7));
dot((3.07692307692,2.15384615384));
label("$Q$",(3.07692307692,2.15384615384),N);
label("$P$",(20/7,12/7),W);
label("$A$",(0,4), NW);
label("$B$",(5,4), NE);
label("$C$",(5,0),SE);
label("$D$",(0,0),SW);
label("$F$",(2,0),S); label("$G$",(5,1),E);
label("$E$",(4,4),N);
// Block 2
pair A1=(2,0),A2=(4,4); pair B1=(0,4),B2=(5,1),B3=(20/7,12/7); pair C1=(5,0),C2=(0,4); draw(A1--A2); draw(B1--B2); draw(C1--C2); draw(A2--B2); draw(B2--B3); draw((0,0)--B1--(5,4)--C1--cycle); dot((20/7,12/7)); dot((3.07692307692,2.15384615384)); label("$Q$",(3.07692307692,2.15384615384),N); label("$P$",(20/7,12/7),W); label("$A$",(0,4), NW); label("$B$",(5,4), NE); label("$C$",(5,0),SE); label("$D$",(0,0),SW); label("$F$",(2,0),S); label("$G$",(5,1),E); label("$E$",(4,4),N); | [] |
788 | A dilation of the plane—that is, a size transformation with a positive scale factor—sends the circle of radius $2$ centered at $A(2,2)$ to the circle of radius $3$ centered at $A’(5,6)$. What distance does the origin $O(0,0)$, move under this transformation?
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ \sqrt{13}\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$ | 2016 AMC 10B Problem 20 | Using analytic geometry, we find that the center of dilation is at $(-4,-6)$ and the coefficient/factor is $1.5$. Then, we see that the origin is $2\sqrt{13}$ from the center, and will be $1.5 \times 2\sqrt{13} = 3\sqrt{13}$ from it afterwards.
Thus, it will move $3\sqrt{13} - 2\sqrt{13} = \boxed{\sqrt{13}}$. | // Block 1
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
/* by adihaya */
import graph; size(13cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ real xmin = -7., xmax = 9., ymin = -7., ymax = 9.6; /* image dimensions */
pen xdxdff = rgb(0.49019607843137253,50.49019607843137253,1.); pen uuuuuu = rgb(0.666666666,0.26666666666666666,0.26666666666666666); pen qqzzff = rgb(0.,0.6,1.); pen ffwwqq = rgb(1.,0.4,0.); pen qqwuqq = rgb(0.,0.39215686274509803,0.);
pair O = (3.,0.), A = (2.,2.), B = (2.,1.), C = (4.203155585,5.592712848525), D = (5.,4.), F = (-3.999634206191805,-5.999512274922407), G = (-3.999634206191812,-5.9995122749224175);
/* by adihaya */
draw((2.482656878,0.)---(4.482568783,0.48268779)--(2.,0.48272202065687797)--B--cycle, qqwuqq);
draw((5.482722020656878,0.)--(7.4827220878,1.48277797)--(5.,0.48272687797)--(5.,0.)--cycle, qqwuqq);
Label laxis; laxis.p = fontsize(10);
xaxis(xmin, xmax, Ticks(laxis, Step = 2., Size = 2, NoZero),EndArrow(6), above = true);
yaxis(ymin, ymax, Ticks(laxis, Step = 2., Size = 2, NoZero),EndArrow(6), above = true); /* draws axes; NoZero hides '0' label */
/* draw figures */
draw(shift(A) * scale(2., 2.)*unitcircle);
draw(shift((5.,6.)) * scale(3.000060969351735, 3.000060969351735)*unitcircle);
draw((xmin, 1.3333333333333333*xmin-0.6666666666666666)--(xmax, 1.3333333333333333*xmax-0.6666666666666666)); /* line */
draw((2.,ymin)--(2.,ymax)); /* line */
draw((5.,ymin)--(5.,ymax)); /* line */
draw((xmin, 1.6666869897839114*xmin + 0.6666260204321771)--(xmax, 1.6666869897839114*xmax + 0.6666260204321771)); /* line */
draw(O--F, qqzzff);
draw(F--A, ffwwqq);
/* dots and labels */
dot(O,blue);
label("$O$", (0.08696973475182286,0.23426871275979863), NE * labelscalefactor,blue);
dot(A,blue);
label("$A$", (2.089474351594523,2.23677332960249), NE * labelscalefactor,blue);
dot((5.,6.),blue);
label("$A'$", (5.093231276858573,6.2190268290055695), NE * labelscalefactor,blue);
dot(B,xdxdff);
label("$B$", (2.089474351594523,0.23426871275979863), NE * labelscalefactor,xdxdff);
label("$c$", (0.9971991060439592,3.2607813723061394), NE * labelscalefactor);
dot(C,xdxdff);
label("$C$", (3.2955282685566036,3.829674729363722), NE * labelscalefactor,xdxdff);
label("$d$", (3.477574142815031,8.107752774436745), NE * labelscalefactor);
label("$a$", (7.255026033677397,9.404829628528034), NE * labelscalefactor);
label("$b$", (2.1804972887237364,9.404829628528034), NE * labelscalefactor);
label("$e$", (4.615360856930201,9.404829628528034), NE * labelscalefactor);
dot(D,linewidth(3.pt) + uuuuuu);
/* Solution by adihaya */
label("$D$", (2.089474351594523,4.125499275033665), NE * labelscalefactor,uuuuuu);
dot((5.,9.000060969351734),linewidth(3.pt) + uuuuuu);
label("$E$", (5.093231276858573,9.131760817140394), NE * labelscalefactor,uuuuuu);
label("$f$", (4.933941136882449,9.404829628528034), NE * labelscalefactor);
dot(F,linewidth(3.pt) + uuuuuu);
label("$\Large{(-4,-6)}$", (-3.73599362467515,-6.273871291978948), NE * labelscalefactor,uuuuuu);
label("$\Large{2\sqrt{13}}$", (-2.916787190512227,-2.0868161840351394), NE * labelscalefactor,qqzzff);
dot(G,linewidth(3.pt) + uuuuuu);
label("$G$", (-3.9180394989335774,-5.864268074897489), NE * labelscalefactor,uuuuuu);
label("$\Large{10}$", (0.2690156090102501,-0.6759606585323339), NE * labelscalefactor,ffwwqq);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* re-scale y/x */
currentpicture = yscale(0.9090909090909091) * currentpicture;
/* end of picture */
// Block 2
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ /* by adihaya */ import graph; size(13cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ real xmin = -7., xmax = 9., ymin = -7., ymax = 9.6; /* image dimensions */ pen xdxdff = rgb(0.49019607843137253,50.49019607843137253,1.); pen uuuuuu = rgb(0.666666666,0.26666666666666666,0.26666666666666666); pen qqzzff = rgb(0.,0.6,1.); pen ffwwqq = rgb(1.,0.4,0.); pen qqwuqq = rgb(0.,0.39215686274509803,0.); pair O = (3.,0.), A = (2.,2.), B = (2.,1.), C = (4.203155585,5.592712848525), D = (5.,4.), F = (-3.999634206191805,-5.999512274922407), G = (-3.999634206191812,-5.9995122749224175); /* by adihaya */ draw((2.482656878,0.)---(4.482568783,0.48268779)--(2.,0.48272202065687797)--B--cycle, qqwuqq); draw((5.482722020656878,0.)--(7.4827220878,1.48277797)--(5.,0.48272687797)--(5.,0.)--cycle, qqwuqq); Label laxis; laxis.p = fontsize(10); xaxis(xmin, xmax, Ticks(laxis, Step = 2., Size = 2, NoZero),EndArrow(6), above = true); yaxis(ymin, ymax, Ticks(laxis, Step = 2., Size = 2, NoZero),EndArrow(6), above = true); /* draws axes; NoZero hides '0' label */ /* draw figures */ draw(shift(A) * scale(2., 2.)*unitcircle); draw(shift((5.,6.)) * scale(3.000060969351735, 3.000060969351735)*unitcircle); draw((xmin, 1.3333333333333333*xmin-0.6666666666666666)--(xmax, 1.3333333333333333*xmax-0.6666666666666666)); /* line */ draw((2.,ymin)--(2.,ymax)); /* line */ draw((5.,ymin)--(5.,ymax)); /* line */ draw((xmin, 1.6666869897839114*xmin + 0.6666260204321771)--(xmax, 1.6666869897839114*xmax + 0.6666260204321771)); /* line */ draw(O--F, qqzzff); draw(F--A, ffwwqq); /* dots and labels */ dot(O,blue); label("$O$", (0.08696973475182286,0.23426871275979863), NE * labelscalefactor,blue); dot(A,blue); label("$A$", (2.089474351594523,2.23677332960249), NE * labelscalefactor,blue); dot((5.,6.),blue); label("$A'$", (5.093231276858573,6.2190268290055695), NE * labelscalefactor,blue); dot(B,xdxdff); label("$B$", (2.089474351594523,0.23426871275979863), NE * labelscalefactor,xdxdff); label("$c$", (0.9971991060439592,3.2607813723061394), NE * labelscalefactor); dot(C,xdxdff); label("$C$", (3.2955282685566036,3.829674729363722), NE * labelscalefactor,xdxdff); label("$d$", (3.477574142815031,8.107752774436745), NE * labelscalefactor); label("$a$", (7.255026033677397,9.404829628528034), NE * labelscalefactor); label("$b$", (2.1804972887237364,9.404829628528034), NE * labelscalefactor); label("$e$", (4.615360856930201,9.404829628528034), NE * labelscalefactor); dot(D,linewidth(3.pt) + uuuuuu); /* Solution by adihaya */ label("$D$", (2.089474351594523,4.125499275033665), NE * labelscalefactor,uuuuuu); dot((5.,9.000060969351734),linewidth(3.pt) + uuuuuu); label("$E$", (5.093231276858573,9.131760817140394), NE * labelscalefactor,uuuuuu); label("$f$", (4.933941136882449,9.404829628528034), NE * labelscalefactor); dot(F,linewidth(3.pt) + uuuuuu); label("$\Large{(-4,-6)}$", (-3.73599362467515,-6.273871291978948), NE * labelscalefactor,uuuuuu); label("$\Large{2\sqrt{13}}$", (-2.916787190512227,-2.0868161840351394), NE * labelscalefactor,qqzzff); dot(G,linewidth(3.pt) + uuuuuu); label("$G$", (-3.9180394989335774,-5.864268074897489), NE * labelscalefactor,uuuuuu); label("$\Large{10}$", (0.2690156090102501,-0.6759606585323339), NE * labelscalefactor,ffwwqq); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* re-scale y/x */ currentpicture = yscale(0.9090909090909091) * currentpicture; /* end of picture */ | [] |
789 | What is the area of the region enclosed by the graph of the equation $x^2+y^2=|x|+|y|?$
$\textbf{(A)}\ \pi+\sqrt{2}\qquad\textbf{(B)}\ \pi+2\qquad\textbf{(C)}\ \pi+2\sqrt{2}\qquad\textbf{(D)}\ 2\pi+\sqrt{2}\qquad\textbf{(E)}\ 2\pi+2\sqrt{2}$ | 2016 AMC 10B Problem 21 | Without loss of generality (WLOG) note that if a point in the first quadrant satisfies the equation, so do its corresponding points in the other three quadrants. Therefore, we can assume that $x, y \ge 0$, which implies that $|x|=x$ and $|y|=y$, and multiply by $4$ at the end.
We can rearrange the equation to get $x^2-x+y^2-y=0 \Rightarrow (x-\tfrac12)^2+(y-\tfrac12)^2=(\tfrac{\sqrt2}{2})^2$, which describes a circle with center $(\tfrac12, \tfrac12)$ and radius $\tfrac{\sqrt2}{2}.$ It's clear we now want to find the union of four equal areas.
There are several ways to find the area, but note that if you connect $(0, 1)$ to its other three respective points in the other three quadrants, you get a square of area $2$, along with four half-circles of diameter $\sqrt{2}$, for a total area of $2+2\cdot(\tfrac{\sqrt2}{2})^2\pi = \pi + 2$ which is $\boxed{\textbf{(B)}}$. | // Block 1
draw((0,-1.5)--(0,1.5),EndArrow);draw((-1.5,0)--(1.5,0),EndArrow);draw((0,1)--(1,0)--(0,-1)--(-1,0)--cycle,dotted);
for(int i=0;i<4;++i){draw(rotate(i*90,(0,0))*arc((1/2,1/2),sqrt(1/2),-45,135));dot(rotate(i*90,(0,0))*(1/2,1/2));}
// Block 2
draw((0,-1.5)--(0,1.5),EndArrow);draw((-1.5,0)--(1.5,0),EndArrow);draw((0,1)--(1,0)--(0,-1)--(-1,0)--cycle,dotted); for(int i=0;i<4;++i){draw(rotate(i*90,(0,0))*arc((1/2,1/2),sqrt(1/2),-45,135));dot(rotate(i*90,(0,0))*(1/2,1/2));} | [] |
789 | What is the area of the region enclosed by the graph of the equation $x^2+y^2=|x|+|y|?$
$\textbf{(A)}\ \pi+\sqrt{2}\qquad\textbf{(B)}\ \pi+2\qquad\textbf{(C)}\ \pi+2\sqrt{2}\qquad\textbf{(D)}\ 2\pi+\sqrt{2}\qquad\textbf{(E)}\ 2\pi+2\sqrt{2}$ | 2016 AMC 10B Problem 21 | We solve with cases. The cases are:
Case 1: $x\geq0, y\geq0.$
Case 2: $x\geq0, y<0.$
Case 3: $x<0, y\geq0.$
Case 4: $x<0, y<0.$
We can quickly realize that the whole figure is symmetrical; so when you figure out the first case, you get the first part is $\left(x-\dfrac12\right)^2+\left(x-\dfrac12\right)^2$ you can figure out the whole figure: (scaled 8x).
The way we figure out the area is by splitting it the following way:
We know each of the red lines is a diameter of the circle which is $\sqrt2$. So the area of the red is 2. We know that the area of each semicircle is $\dfrac14 \pi$ so the area of the semicircles combines is $\pi$. Thus we get $\boxed{\textbf{(B)} \pi+2}.$ | // Block 1
size(400);
import TrigMacros;
rr_cartesian_axes(-10,10,-10,10,complexplane=false, usegrid = true);
draw(circle((4,4), 4*1.41421356237));
draw(circle((4,-4), 4*1.41421356237));
draw(circle((-4,4), 4*1.41421356237));
draw(circle((-4,-4), 4*1.41421356237));
// Block 2
size(400);
import TrigMacros;
rr_cartesian_axes(-10,10,-10,10,complexplane=false, usegrid = true);
draw(circle((4,4), 4*1.41421356237));
draw(circle((4,-4), 4*1.41421356237));
draw(circle((-4,4), 4*1.41421356237));
draw(circle((-4,-4), 4*1.41421356237));
real f(real x)
{
return -x+8;
}
draw(graph(f,0,8), red+linewidth(1.5));
real g(real x)
{
return x+8;
}
draw(graph(g,0,-8), red+linewidth(1.5));
real f(real x)
{
return -x+8;
}
draw(graph(f,0,8), red+linewidth(1.5));
real h(real x)
{
return -x-8;
}
draw(graph(h,-8,0), red+linewidth(1.5));
real z(real x)
{
return x-8;
}
draw(graph(z,0,8), red+linewidth(1.5));
pair A,B,C,D;
A = (8,0);
B = (0,8);
C = (-8,0);
D = (0,-8);
fill(A--B--C--D--cycle, red);
// Block 3
size(400); import TrigMacros; rr_cartesian_axes(-10,10,-10,10,complexplane=false, usegrid = true); draw(circle((4,4), 4*1.41421356237)); draw(circle((4,-4), 4*1.41421356237)); draw(circle((-4,4), 4*1.41421356237)); draw(circle((-4,-4), 4*1.41421356237));
// Block 4
size(400); import TrigMacros; rr_cartesian_axes(-10,10,-10,10,complexplane=false, usegrid = true); draw(circle((4,4), 4*1.41421356237)); draw(circle((4,-4), 4*1.41421356237)); draw(circle((-4,4), 4*1.41421356237)); draw(circle((-4,-4), 4*1.41421356237)); real f(real x) { return -x+8; } draw(graph(f,0,8), red+linewidth(1.5)); real g(real x) { return x+8; } draw(graph(g,0,-8), red+linewidth(1.5)); real f(real x) { return -x+8; } draw(graph(f,0,8), red+linewidth(1.5)); real h(real x) { return -x-8; } draw(graph(h,-8,0), red+linewidth(1.5)); real z(real x) { return x-8; } draw(graph(z,0,8), red+linewidth(1.5)); pair A,B,C,D; A = (8,0); B = (0,8); C = (-8,0); D = (0,-8); fill(A--B--C--D--cycle, red); | [] |
790 | A set of teams held a round-robin tournament in which every team played every other team exactly once. Every team won $10$ games and lost $10$ games; there were no ties. How many sets of three teams $\{A, B, C\}$ were there in which $A$ beat $B$, $B$ beat $C$, and $C$ beat $A?$
$\textbf{(A)}\ 385 \qquad \textbf{(B)}\ 665 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 1140 \qquad \textbf{(E)}\ 1330$ | 2016 AMC 10B Problem 22 | There are $10 \cdot 2+1=21$ teams. Any of the $\tbinom{21}3=1330$ sets of three teams must either be a fork (in which one team beat both the others) or a cycle:
But we know that every team beat exactly $10$ other teams, so for each possible $A$ at the head of a fork, there are always exactly $\tbinom{10}2$ choices for $B$ and $C$ as $A$ beat exactly 10 teams and we are choosing 2 of them. Therefore there are $21\cdot\tbinom{10}2=945$ forks, and all the rest must be cycles.
Thus the answer is $1330-945=385$ which is $\boxed{\textbf{(A)}}$. | // Block 1
size(7cm);label("A",(5,5));label("C",(10,0));label("B",(0,0));draw((4,4)--(1,1),EndArrow);draw((6,4)--(9,1),EndArrow);
label("A",(20,5));label("C",(25,0));label("B",(15,0));draw((19,4)--(16,1),EndArrow);draw((16,0)--(24,0),EndArrow);draw((24,1)--(21,4),EndArrow);
// Block 2
size(7cm);label("A",(5,5));label("C",(10,0));label("B",(0,0));draw((4,4)--(1,1),EndArrow);draw((6,4)--(9,1),EndArrow); label("A",(20,5));label("C",(25,0));label("B",(15,0));draw((19,4)--(16,1),EndArrow);draw((16,0)--(24,0),EndArrow);draw((24,1)--(21,4),EndArrow); | [] |
790 | A set of teams held a round-robin tournament in which every team played every other team exactly once. Every team won $10$ games and lost $10$ games; there were no ties. How many sets of three teams $\{A, B, C\}$ were there in which $A$ beat $B$, $B$ beat $C$, and $C$ beat $A?$
$\textbf{(A)}\ 385 \qquad \textbf{(B)}\ 665 \qquad \textbf{(C)}\ 945 \qquad \textbf{(D)}\ 1140 \qquad \textbf{(E)}\ 1330$ | 2016 AMC 10B Problem 22 | This is a Graph Theory problem with directed graph. There are $21$ teams in total. WLOG, pick team $A$, there are $10$ teams that lost to $A$ and $10$ teams that won over $A$. Call the group of teams that lost to $A$ group $L$, and the group of teams that won over $A$ group $W$.
Any team from group $L$ that won a team from group $W$ will form a cycle with $A$. Now we need to count how many teams in group $L$ won over a team from group $W$. The total number of wins in group $L$ is $10 \cdot 10 =100$. There are $\tbinom{10}2=45$ wins among the teams inside group $L$. So group $L$ has $100-45=55$ wins over group $W$.
\[\frac{21 \cdot 55}{3}=\boxed{385, \textbf{(A)}}\]
~isabelchen | // Block 1
size(7cm);
label("A",(20,5));
label("Group W",(27,0));
label("Group L",(13,0));
draw((19,4)--(16,1),EndArrow);
draw((16,0)--(24,0),EndArrow);
draw((24,1)--(21,4),EndArrow);
// Block 2
size(7cm); label("A",(20,5)); label("Group W",(27,0)); label("Group L",(13,0)); draw((19,4)--(16,1),EndArrow); draw((16,0)--(24,0),EndArrow); draw((24,1)--(21,4),EndArrow); | [] |
791 | In regular hexagon $ABCDEF$, points $W$, $X$, $Y$, and $Z$ are chosen on sides $\overline{BC}$, $\overline{CD}$, $\overline{EF}$, and $\overline{FA}$ respectively, so lines $AB$, $ZW$, $YX$, and $ED$ are parallel and equally spaced. What is the ratio of the area of hexagon $WCXYFZ$ to the area of hexagon $ABCDEF$?
$\textbf{(A)}\ \frac{1}{3}\qquad\textbf{(B)}\ \frac{10}{27}\qquad\textbf{(C)}\ \frac{11}{27}\qquad\textbf{(D)}\ \frac{4}{9}\qquad\textbf{(E)}\ \frac{13}{27}$ | 2016 AMC 10B Problem 23 | We draw a diagram to make our work easier:
Assume that $AB$ is of length $1$. Therefore, the area of $ABCDEF$ is $\frac{3\sqrt 3}2$. To find the area of $WCXYFZ$, we draw $\overline{CF}$, and find the area of the trapezoids $WCFZ$ and $CXYF$.
From this, we know that $CF=2$. We also know that the combined heights of the trapezoids is $\frac{\sqrt 3}3$, since $\overline{ZW}$ and $\overline{YX}$ are equally spaced, and the height of each of the trapezoids is $\frac{\sqrt 3}6$. From this, we know $\overline{ZW}$ and $\overline{YX}$ are each $\frac 13$ of the way from $\overline{CF}$ to $\overline{DE}$ and $\overline{AB}$, respectively. We know that these are both equal to $\frac 53$.
We find the area of each of the trapezoids, which both happen to be $\frac{11}6 \cdot \frac{\sqrt 3}6=\frac{11\sqrt 3}{36}$, and the combined area is $\frac{11\sqrt 3}{18}$.
We find that $\dfrac{\frac{11\sqrt 3}{18}}{\frac{3\sqrt 3}2}$ is equal to $\frac{22}{54}=\boxed{\textbf{(C)}\ \frac{11}{27}}$. | // Block 1
pair A,B,C,D,E,F,W,X,Y,Z;
A=(0,0);
B=(1,0);
C=(3/2,sqrt(3)/2);
D=(1,sqrt(3));
E=(0,sqrt(3));
F=(-1/2,sqrt(3)/2);
X=(4/3,2sqrt(3)/3);
W=(4/3,sqrt(3)/3);
Z=(-1/3,sqrt(3)/3);
Y=(-1/3,2sqrt(3)/3);
draw(A--B--C--D--E--F--cycle);
draw(W--Z);
draw(X--Y);
label("$A$",A,SW);
label("$B$",B,SE);
label("$C$",C,ESE);
label("$D$",D,NE);
label("$E$",E,NW);
label("$F$",F,WSW);
label("$W$",W,ENE);
label("$X$",X,ESE);
label("$Y$",Y,WSW);
label("$Z$",Z,WNW);
// Block 2
pair A,B,C,D,E,F,W,X,Y,Z;
A=(0,0);
B=(1,0);
C=(3/2,sqrt(3)/2);
D=(1,sqrt(3));
E=(0,sqrt(3));
F=(-1/2,sqrt(3)/2);
W=(4/3,2sqrt(3)/3);
X=(4/3,sqrt(3)/3);
Y=(-1/3,sqrt(3)/3);
Z=(-1/3,2sqrt(3)/3);
draw(A--B--C--D--E--F--cycle);
draw(W--Z);
draw(X--Y);
draw(F--C--B--E--D--A);
label("$A$",A,SW);
label("$B$",B,SE);
label("$C$",C,ESE);
label("$D$",D,NE);
label("$E$",E,NW);
label("$F$",F,WSW);
label("$W$",W,ENE);
label("$X$",X,ESE);
label("$Y$",Y,WSW);
label("$Z$",Z,WNW);
// Block 3
pair A,B,C,D,E,F,W,X,Y,Z; A=(0,0); B=(1,0); C=(3/2,sqrt(3)/2); D=(1,sqrt(3)); E=(0,sqrt(3)); F=(-1/2,sqrt(3)/2); X=(4/3,2sqrt(3)/3); W=(4/3,sqrt(3)/3); Z=(-1/3,sqrt(3)/3); Y=(-1/3,2sqrt(3)/3); draw(A--B--C--D--E--F--cycle); draw(W--Z); draw(X--Y); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,ESE); label("$D$",D,NE); label("$E$",E,NW); label("$F$",F,WSW); label("$W$",W,ENE); label("$X$",X,ESE); label("$Y$",Y,WSW); label("$Z$",Z,WNW);
// Block 4
pair A,B,C,D,E,F,W,X,Y,Z; A=(0,0); B=(1,0); C=(3/2,sqrt(3)/2); D=(1,sqrt(3)); E=(0,sqrt(3)); F=(-1/2,sqrt(3)/2); W=(4/3,2sqrt(3)/3); X=(4/3,sqrt(3)/3); Y=(-1/3,sqrt(3)/3); Z=(-1/3,2sqrt(3)/3); draw(A--B--C--D--E--F--cycle); draw(W--Z); draw(X--Y); draw(F--C--B--E--D--A); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,ESE); label("$D$",D,NE); label("$E$",E,NW); label("$F$",F,WSW); label("$W$",W,ENE); label("$X$",X,ESE); label("$Y$",Y,WSW); label("$Z$",Z,WNW); | [] |
791 | In regular hexagon $ABCDEF$, points $W$, $X$, $Y$, and $Z$ are chosen on sides $\overline{BC}$, $\overline{CD}$, $\overline{EF}$, and $\overline{FA}$ respectively, so lines $AB$, $ZW$, $YX$, and $ED$ are parallel and equally spaced. What is the ratio of the area of hexagon $WCXYFZ$ to the area of hexagon $ABCDEF$?
$\textbf{(A)}\ \frac{1}{3}\qquad\textbf{(B)}\ \frac{10}{27}\qquad\textbf{(C)}\ \frac{11}{27}\qquad\textbf{(D)}\ \frac{4}{9}\qquad\textbf{(E)}\ \frac{13}{27}$ | 2016 AMC 10B Problem 23 | First, like in the first solution, split the large hexagon into 6 equilateral triangles. Each equilateral triangle can be split into three rows of smaller equilateral triangles. The first row will have one triangle, the second three, the third five. Once you have drawn these lines, it's just a matter of counting triangles. There are $22$ small triangles in hexagon $ZWCXYF$, and $9 \cdot 6 = 54$ small triangles in the whole hexagon.
Thus, the answer is $\frac{22}{54}=\boxed{\textbf{(C)}\ \frac{11}{27}}$. | // Block 1
pair A,B,C,D,E,F,W,X,Y,Z;
A=(0,0);
B=(1,0);
C=(3/2,sqrt(3)/2);
D=(1,sqrt(3));
E=(0,sqrt(3));
F=(-1/2,sqrt(3)/2);
W=(4/3,2sqrt(3)/3);
X=(4/3,sqrt(3)/3);
Y=(-1/3,sqrt(3)/3);
Z=(-1/3,2sqrt(3)/3);
draw(A--B--C--D--E--F--cycle);
draw(W--Z);
draw(X--Y);
draw(F--C--B--E--D--A);
label("$A$",A,SW);
label("$B$",B,SE);
label("$C$",C,ESE);
label("$D$",D,NE);
label("$E$",E,NW);
label("$F$",F,WSW);
label("$W$",W,ENE);
label("$X$",X,ESE);
label("$Y$",Y,WSW);
label("$Z$",Z,WNW);
// Block 2
pair A,B,C,D,E,F,W,X,Y,Z; A=(0,0); B=(1,0); C=(3/2,sqrt(3)/2); D=(1,sqrt(3)); E=(0,sqrt(3)); F=(-1/2,sqrt(3)/2); W=(4/3,2sqrt(3)/3); X=(4/3,sqrt(3)/3); Y=(-1/3,sqrt(3)/3); Z=(-1/3,2sqrt(3)/3); draw(A--B--C--D--E--F--cycle); draw(W--Z); draw(X--Y); draw(F--C--B--E--D--A); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,ESE); label("$D$",D,NE); label("$E$",E,NW); label("$F$",F,WSW); label("$W$",W,ENE); label("$X$",X,ESE); label("$Y$",Y,WSW); label("$Z$",Z,WNW); | [] |
791 | In regular hexagon $ABCDEF$, points $W$, $X$, $Y$, and $Z$ are chosen on sides $\overline{BC}$, $\overline{CD}$, $\overline{EF}$, and $\overline{FA}$ respectively, so lines $AB$, $ZW$, $YX$, and $ED$ are parallel and equally spaced. What is the ratio of the area of hexagon $WCXYFZ$ to the area of hexagon $ABCDEF$?
$\textbf{(A)}\ \frac{1}{3}\qquad\textbf{(B)}\ \frac{10}{27}\qquad\textbf{(C)}\ \frac{11}{27}\qquad\textbf{(D)}\ \frac{4}{9}\qquad\textbf{(E)}\ \frac{13}{27}$ | 2016 AMC 10B Problem 23 | Extend $\overline{EF}$ and $\overline{CD}$ to meet at point $G$, as shown in the diagram. Then $\triangle GZW \sim \triangle GFC$. Then $[GZW] = \left(\dfrac53\right)^2[GED]$ and $[GFC] = 2^2[GED]$. Subtracting $[GED]$, we find that $[EDWZ] = \dfrac{16}{9}[GED]$ and $[EDCF] = 3[GED]$. Subtracting again, we find that \[[ZWCF] = [EDCF] - [EDWZ] = \dfrac{11}{9}[GED].\]Finally, \[\dfrac{[WCXYFZ]}{[ABCDEF]} = \dfrac{[ZWCF]}{[EDCF]} = \dfrac{\dfrac{11}{9}[GED]}{3[GED]} = \textbf{(C) } \dfrac{11}{27}.\] | // Block 1
pair A,B,C,D,E,F,W,X,Y,Z;
A=(0,0);
B=(1,0);
C=(3/2,sqrt(3)/2);
D=(1,sqrt(3));
E=(0,sqrt(3));
F=(-1/2,sqrt(3)/2);
W=(4/3,2sqrt(3)/3);
X=(4/3,sqrt(3)/3);
Y=(-1/3,sqrt(3)/3);
Z=(-1/3,2sqrt(3)/3);
pair G = (0.5, sqrt(3)*3/2);
draw(A--B--C--D--E--F--cycle);
draw(W--Z);
draw(X--Y);
draw(E--G--D);
draw(F--C);
label("$A$",A,SW);
label("$B$",B,SE);
label("$C$",C,ESE);
label("$D$",D,NE);
label("$E$",E,NW);
label("$F$",F,WSW);
label("$W$",W,ENE);
label("$X$",X,ESE);
label("$Y$",Y,WSW);
label("$Z$",Z,WNW);
label("$G$",G,N);
// Block 2
pair A,B,C,D,E,F,W,X,Y,Z; A=(0,0); B=(1,0); C=(3/2,sqrt(3)/2); D=(1,sqrt(3)); E=(0,sqrt(3)); F=(-1/2,sqrt(3)/2); W=(4/3,2sqrt(3)/3); X=(4/3,sqrt(3)/3); Y=(-1/3,sqrt(3)/3); Z=(-1/3,2sqrt(3)/3); pair G = (0.5, sqrt(3)*3/2); draw(A--B--C--D--E--F--cycle); draw(W--Z); draw(X--Y); draw(E--G--D); draw(F--C); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,ESE); label("$D$",D,NE); label("$E$",E,NW); label("$F$",F,WSW); label("$W$",W,ENE); label("$X$",X,ESE); label("$Y$",Y,WSW); label("$Z$",Z,WNW); label("$G$",G,N); | [] |
792 | The ratio of the measures of two acute angles is $5:4$, and the complement of one of these two angles is twice as large as the complement of the other. What is the sum of the degree measures of the two angles?
$\textbf{(A)}\ 75\qquad\textbf{(B)}\ 90\qquad\textbf{(C)}\ 135\qquad\textbf{(D)}\ 150\qquad\textbf{(E)}\ 270$ | 2016 AMC 12B Problem 4 | We can visualize the problem like so:
\[5x+1y = 90^\circ = 4x+2y\]
Moving like terms to the same side gets $x = y$, and substituting this back gets $6x = 90^\circ \implies x = \frac{90^\circ}{6} = 15^\circ$, so the sum of the degree measures is $5x + 4x = 9x = 9(15) = \boxed{\textbf{(C)}\ 135}$. ~emerald_block | // Block 1
path b = brace((0,10),(90,10),5);
draw(b);
label("$90^\circ$",b,N);
draw("$5x$",(0,0)--(75,0),N);
draw((75,2.5)--(75,-2.5));
draw("$1y$",(75,0)--(90,0),N);
draw("$4x$",(0,-10)--(60,-10),S);
draw((60,-7.5)--(60,-12.5));
draw("$2y$",(60,-10)--(90,-10),S);
draw((0,5)--(0,-15));
draw((90,5)--(90,-15));
// Block 2
path b = brace((0,10),(90,10),5); draw(b); label("$90^\circ$",b,N); draw("$5x$",(0,0)--(75,0),N); draw((75,2.5)--(75,-2.5)); draw("$1y$",(75,0)--(90,0),N); draw("$4x$",(0,-10)--(60,-10),S); draw((60,-7.5)--(60,-12.5)); draw("$2y$",(60,-10)--(90,-10),S); draw((0,5)--(0,-15)); draw((90,5)--(90,-15)); | [] |
793 | How many squares whose sides are parallel to the axes and whose vertices have coordinates that are integers lie entirely within the region bounded by the line $y=\pi x$, the line $y=-0.1$ and the line $x=5.1?$
$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 41 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ 57$ | 2016 AMC 12B Problem 11 | Solution by e_power_pi_times_i
Revised by Kinglogic and RJ5303707
(red shows lattice points within the triangle)
If we draw a picture showing the triangle, we see that it would be easier to count the squares vertically and not horizontally. The upper bound is $16$ squares $(y=5.1*\pi)$, and the limit for the $x$-value is $5$ squares. First we count the $1*1$ squares. In the rightmost column, there are $12$ squares with length $1$ because $y=4*\pi$ generates squares from $(4,0)$ to $(4,4\pi)$, and continuing on we have $9$, $6$, and $3$ for $x$-values for $1$, $2$, and $3$ in the equation $y=\pi x$. So there are $12+9+6+3 = 30$ squares with length $1$ in the figure.
For $2*2$ squares, each square takes up $2$ units left and $2$ units up. Squares can also overlap. For $2*2$ squares, the rightmost column stretches from $(3,0)$ to $(3,3\pi)$, so there are $8$ squares with length $2$ in a $2$ by $9$ box. Repeating the process, the next column stretches from $(2,0)$ to $(2,2\pi)$, so there are $5$ squares. Continuing and adding up in the end, there are $8+5+2=15$ squares with length $2$ in the figure.
Squares with length $3$ in the rightmost column start at $(2,0)$ and end at $(2,2\pi)$, so there are $4$ such squares in the right column. As the left row starts at $(1,0)$ and ends at $(1,\pi)$ there are $4+1=5$ squares with length $3$. As squares with length $4$ would not fit in the triangle, the answer is $30+15+5$ which is $\boxed{\textbf{(D)}\ 50}$. | // Block 1
Label l;
l.p=fontsize(8);
xaxis(-1,8,Ticks(l, 1.0));
yaxis(-1,16,Ticks(l, 1.0));
real f(real x)
{
return x * pi;
}
D(graph(f,-1/pi,5.1));
D((5.1,-1)--(5.1,16));
D((-1,-0.1)--(8,-0.1));
for(int x = 0; x < 5.1; ++x) {
for(int y = 0; y < 16; ++y) {
if(x * pi > y) {
D((x,y));
}
}
}
// Block 2
Label l; l.p=fontsize(8); xaxis(-1,8,Ticks(l, 1.0)); yaxis(-1,16,Ticks(l, 1.0)); real f(real x) { return x * pi; } D(graph(f,-1/pi,5.1)); D((5.1,-1)--(5.1,16)); D((-1,-0.1)--(8,-0.1)); for(int x = 0; x < 5.1; ++x) { for(int y = 0; y < 16; ++y) { if(x * pi > y) { D((x,y)); } } } | [] |
794 | Alice and Bob live $10$ miles apart. One day Alice looks due north from her house and sees an airplane. At the same time Bob looks due west from his house and sees the same airplane. The angle of elevation of the airplane is $30^\circ$ from Alice's position and $60^\circ$ from Bob's position. Which of the following is closest to the airplane's altitude, in miles?
$\textbf{(A)}\ 3.5 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 4.5 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ 5.5$ | 2016 AMC 12B Problem 13 | ~credit to mathmaster2012 for original diagram
~BakedPotato66 added/contributed some elements | // Block 1
import graph; usepackage("amsmath"); size(7.2cm);
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 = 10.1, ymin = -4.44, ymax = 6.3; /* image dimensions */
pen qqwuqq = rgb(0.,0.39215686274509803,0.);
draw((-0.12,5.52)--(-1.34,-0.82)--(4.78,-0.88)--cycle);
draw(arc((-1.34,-0.82),0.6,41.93351449887411,79.10776933246521)--(-1.34,-0.82)--cycle, qqwuqq);
draw(arc((4.78,-0.88),0.6,127.43857157233303,167.225815720283)--(4.78,-0.88)--cycle, qqwuqq);
/* draw figures */
draw((-0.12,5.52)--(-1.34,-0.82));
draw((-1.34,-0.82)--(4.78,-0.88));
draw((4.78,-0.88)--(-0.12,5.52));
draw((-0.12,5.52)--(-0.16,0.24));
draw((-0.16,0.24)--(4.78,-0.88));
draw((-1.34,-0.82)--(-0.16,0.24));
label("$60^{\circ}$",(3.82,-0.06),SE*labelscalefactor);
label("$30^\circ$",(-1.12,0.3),SE*labelscalefactor);
draw((-0.15694767653526337,0.6429066973452366)--(0.22,0.58));
draw((0.22,0.58)--(0.21801530426333052,0.1542961253492044));
draw((-0.15694767653526337,0.6429066973452365)--(-0.44,0.4));
draw((-0.44,0.4)--(-0.42432750397456265,0.0025532591414946237));
draw((-0.42432750397456265,0.0025532591414946237)--(-0.16,0.24));
draw((-0.16,0.24)--(-0.42432750397456265,0.0025532591414946237));
draw((-0.42432750397456265,0.0025532591414946237)--(-0.02,-0.12));
draw((-0.02,-0.12)--(0.21801530426333054,0.15429612534920437));
/* dots and labels */
dot((-0.12,5.52),dotstyle);
label("$\text{Plane}$", (-0.52,5.74), NE * labelscalefactor);
dot((-1.34,-0.82),dotstyle);
label("$\text{Alice}$", (-1.8,-1.2), NE * labelscalefactor);
dot((4.78,-0.88),dotstyle);
label("$\text{Bob}$", (4.86,-0.68), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
label("$10$", (-1.8,-1.2)--(4.86,-0.68), S);
// Block 2
import graph; usepackage("amsmath"); size(7.2cm); 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 = 10.1, ymin = -4.44, ymax = 6.3; /* image dimensions */ pen qqwuqq = rgb(0.,0.39215686274509803,0.); draw((-0.12,5.52)--(-1.34,-0.82)--(4.78,-0.88)--cycle); draw(arc((-1.34,-0.82),0.6,41.93351449887411,79.10776933246521)--(-1.34,-0.82)--cycle, qqwuqq); draw(arc((4.78,-0.88),0.6,127.43857157233303,167.225815720283)--(4.78,-0.88)--cycle, qqwuqq); /* draw figures */ draw((-0.12,5.52)--(-1.34,-0.82)); draw((-1.34,-0.82)--(4.78,-0.88)); draw((4.78,-0.88)--(-0.12,5.52)); draw((-0.12,5.52)--(-0.16,0.24)); draw((-0.16,0.24)--(4.78,-0.88)); draw((-1.34,-0.82)--(-0.16,0.24)); label("$60^{\circ}$",(3.82,-0.06),SE*labelscalefactor); label("$30^\circ$",(-1.12,0.3),SE*labelscalefactor); draw((-0.15694767653526337,0.6429066973452366)--(0.22,0.58)); draw((0.22,0.58)--(0.21801530426333052,0.1542961253492044)); draw((-0.15694767653526337,0.6429066973452365)--(-0.44,0.4)); draw((-0.44,0.4)--(-0.42432750397456265,0.0025532591414946237)); draw((-0.42432750397456265,0.0025532591414946237)--(-0.16,0.24)); draw((-0.16,0.24)--(-0.42432750397456265,0.0025532591414946237)); draw((-0.42432750397456265,0.0025532591414946237)--(-0.02,-0.12)); draw((-0.02,-0.12)--(0.21801530426333054,0.15429612534920437)); /* dots and labels */ dot((-0.12,5.52),dotstyle); label("$\text{Plane}$", (-0.52,5.74), NE * labelscalefactor); dot((-1.34,-0.82),dotstyle); label("$\text{Alice}$", (-1.8,-1.2), NE * labelscalefactor); dot((4.78,-0.88),dotstyle); label("$\text{Bob}$", (4.86,-0.68), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ label("$10$", (-1.8,-1.2)--(4.86,-0.68), S); | [] |
795 | What is the area of the region enclosed by the graph of the equation $x^2+y^2=|x|+|y|?$
$\textbf{(A)}\ \pi+\sqrt{2} \qquad\textbf{(B)}\ \pi+2 \qquad\textbf{(C)}\ \pi+2\sqrt{2} \qquad\textbf{(D)}\ 2\pi+\sqrt{2} \qquad\textbf{(E)}\ 2\pi+2\sqrt{2}$ | 2016 AMC 12B Problem 18 | Consider the case when $x \geq 0$, $y \geq 0$.
\[x^2+y^2=x+y\]
\[(x - \frac{1}{2})^2+(y - \frac{1}{2})^2=\frac{1}{2}\]
Notice the circle intersects the axes at points $(0, 1)$ and $(1, 0)$. Find the area of this circle in the first quadrant. The area is made of a semi-circle with radius of $\frac{\sqrt{2}}{2}$ and a triangle:
\[A = \frac{\pi}{4} +\frac{1}{2}\]
Because of symmetry, the area is the same in all four quadrants. The answer is $\boxed{\textbf{(B)}\ \pi + 2}$ | draw((0,-1.5)--(0,1.5),EndArrow);draw((-1.5,0)--(1.5,0),EndArrow);draw((0,1)--(1,0)--(0,-1)--(-1,0)--cycle,dotted);draw(arc((1/2,1/2),sqrt(2)/2,135, 315),dotted);for(int i=0;i<4;++i){draw(rotate(i*90,(0,0))*arc((1/2,1/2),sqrt(2)/2,-45,135)); dot(rotate(i*90,(0,0))*(1/2,1/2));} | [] |
796 | A square with side length $x$ is inscribed in a right triangle with sides of length $3$, $4$, and $5$ so that one vertex of the square coincides with the right-angle vertex of the triangle. A square with side length $y$ is inscribed in another right triangle with sides of length $3$, $4$, and $5$ so that one side of the square lies on the hypotenuse of the triangle. What is $\dfrac{x}{y}$?
$\textbf{(A) } \dfrac{12}{13} \qquad \textbf{(B) } \dfrac{35}{37} \qquad \textbf{(C) } 1 \qquad \textbf{(D) } \dfrac{37}{35} \qquad \textbf{(E) } \dfrac{13}{12}$ | 2017 AMC 10A Problem 21 | Analyze the first right triangle.
Note that $\triangle ABC$ and $\triangle FBE$ are similar, so $\frac{BF}{FE} = \frac{AB}{AC}$. This can be written as $\frac{4-x}{x}=\frac{4}{3}$. Solving, $x = \frac{12}{7}$.
Now we analyze the second triangle.
Similarly, $\triangle A'B'C'$ and $\triangle RB'Q$ are similar, so $RB' = \frac{4}{3}y$, and $C'S = \frac{3}{4}y$. Thus, $C'B' = C'S + SR + RB' = \frac{4}{3}y + y + \frac{3}{4}y = 5$. Solving for $y$, we get $y = \frac{60}{37}$. Thus, $\frac{x}{y} = \boxed{\textbf{(D)}\:\frac{37}{35}}$. | // Block 1
pair A,B,C;
pair D, e, F;
A = (0,0);
B = (4,0);
C = (0,3);
D = (0, 12/7);
e = (12/7 , 12/7);
F = (12/7, 0);
draw(A--B--C--cycle);
draw(D--e--F);
label("$x$", D/2, W);
label("$A$", A, SW);
label("$B$", B, SE);
label("$C$", C, N);
label("$D$", D, W);
label("$E$", e, NE);
label("$F$", F, S);
// Block 2
pair A,B,C;
pair q, R, S, T;
A = (0,0);
B = (4,0);
C = (0,3);
q = (1.297, 0);
R = (2.27 , 1.297);
S = (0.973, 2.27);
T = (0, 0.973);
draw(A--B--C--cycle);
draw(q--R--S--T--cycle);
label("$y$", (q+R)/2, NW);
label("$A'$", A, SW);
label("$B'$", B, SE);
label("$C'$", C, N);
label("$Q$", (q-(0,0.3)));
label("$R$", R, NE);
label("$S$", S, NE);
label("$T$", T, W);
// Block 3
pair A,B,C; pair D, e, F; A = (0,0); B = (4,0); C = (0,3); D = (0, 12/7); e = (12/7 , 12/7); F = (12/7, 0); draw(A--B--C--cycle); draw(D--e--F); label("$x$", D/2, W); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, N); label("$D$", D, W); label("$E$", e, NE); label("$F$", F, S);
// Block 4
pair A,B,C; pair q, R, S, T; A = (0,0); B = (4,0); C = (0,3); q = (1.297, 0); R = (2.27 , 1.297); S = (0.973, 2.27); T = (0, 0.973); draw(A--B--C--cycle); draw(q--R--S--T--cycle); label("$y$", (q+R)/2, NW); label("$A'$", A, SW); label("$B'$", B, SE); label("$C'$", C, N); label("$Q$", (q-(0,0.3))); label("$R$", R, NE); label("$S$", S, NE); label("$T$", T, W); | [] |
797 | Sides $\overline{AB}$ and $\overline{AC}$ of equilateral triangle $ABC$ are tangent to a circle at points $B$ and $C$ respectively. What fraction of the area of $\triangle ABC$ lies outside the circle?
$\textbf{(A) } \frac{4\sqrt{3}\pi}{27}-\frac{1}{3}\qquad \textbf{(B) } \frac{\sqrt{3}}{2}-\frac{\pi}{8}\qquad \textbf{(C) } \frac{1}{2} \qquad \textbf{(D) } \sqrt{3}-\frac{2\sqrt{3}\pi}{9}\qquad \textbf{(E) } \frac{4}{3}-\frac{4\sqrt{3}\pi}{27}$ | 2017 AMC 10A Problem 22 | Let the radius of the circle be $r$, and let its center be $O$. Since $\overline{AB}$ and $\overline{AC}$ are tangent to circle $O$, then $\angle OBA = \angle OCA = 90^{\circ}$, so $\angle BOC = 120^{\circ}$. Therefore, since $\overline{OB}$ and $\overline{OC}$ are equal to $r$, then (pick your favorite method) $\overline{BC} = r\sqrt{3}$. The area of the equilateral triangle is $\frac{(r\sqrt{3})^2 \sqrt{3}}4 = \frac{3r^2 \sqrt{3}}4$, and the area of the sector we are subtracting from it is $\frac 13 \pi r^2 - \frac 12 \cdot r\sqrt{3} \cdot \frac{r}2 = \frac{\pi r^2}3 -\frac{r^2 \sqrt{3}}4$. The area outside of the circle is $\frac{3r^2 \sqrt{3}}4-\left(\frac{\pi r^2}3 -\frac{r^2 \sqrt{3}}4\right) = r^2 \sqrt{3} - \frac{\pi r^2}3$. Therefore, the answer is \[\frac{r^2 \sqrt{3} - \frac{\pi r^2}3}{\frac{3r^2 \sqrt{3}}4} = \boxed{\textbf{(E) } \frac 43 - \frac{4\sqrt 3 \pi}{27}}\]
Note
The sector angle is $120$ because $\angle B$ and $\angle C$ are both 90 degrees meaning $\angle B + \angle C = 180^\circ$, so $ABCO$ is cyclic. Thus, the angle is $180-60=120^\circ$
~mathboy282
Alternately, $\angle ABC$ is $60^\circ$ and $\angle ABO$ is $90^\circ$, making $\angle CBO$ $30^\circ$. Symmetry allows us to use the same argument to get $\angle BCO = 30^\circ$. Since the interior angles of $\triangle BCO$ must sum to $180^\circ$, that leaves $120^\circ$ for central angle $\angle BOC$.
—wescarroll
The 90 degrees comes from the tangency.
~ Shreyansh
Multiple Choice Shortcut
Assuming WLoG that the equilateral triangle's side length $s$ and therefore area $A$ are algebraic ("$\pi$-free"):
The "crust" is a circle sector minus a triangle, so its area is $a \pi - b$, where $a$ and $b$ are algebraic. Thus the answer is $(A - (a \pi -b))/A = (A+b)/A - a\pi/A$.
Once you see that $s$ is $\sqrt{3} \times$ the circle's radius, and that the circle's 30°-30°-120° triangle is two halves of an equilateral triangle, infer that the smaller circle-sector triangle's area is $A/(\sqrt 3)^2 = A/3$, and so the algebraic part of the answer $(A+b)/A = (A - (-A/3)) /A = 4/3$.
The transcendental ("$\pi$") part of the answer is $-a \pi /A$, and since $a$ and $A$ are algebraic, $\textbf{(E)}$ is the only compatible answer choice. | // Block 1
real sqrt3 = 1.73205080757;
draw(Circle((4, 4), 4));
draw((4-2*sqrt3,6)--(4,4)--(4+2*sqrt3,6)--(4-2*sqrt3,6)--(4,12)--(4+2*sqrt3,6));
label("A", (4, 12.4));
label("B", (-.3, 6.3));
label("C", (8.3, 6.3));
label("O", (4, 3.4));
// Block 2
real sqrt3 = 1.73205080757; draw(Circle((4, 4), 4)); draw((4-2*sqrt3,6)--(4,4)--(4+2*sqrt3,6)--(4-2*sqrt3,6)--(4,12)--(4+2*sqrt3,6)); label("A", (4, 12.4)); label("B", (-.3, 6.3)); label("C", (8.3, 6.3)); label("O", (4, 3.4)); | [] |
797 | Sides $\overline{AB}$ and $\overline{AC}$ of equilateral triangle $ABC$ are tangent to a circle at points $B$ and $C$ respectively. What fraction of the area of $\triangle ABC$ lies outside the circle?
$\textbf{(A) } \frac{4\sqrt{3}\pi}{27}-\frac{1}{3}\qquad \textbf{(B) } \frac{\sqrt{3}}{2}-\frac{\pi}{8}\qquad \textbf{(C) } \frac{1}{2} \qquad \textbf{(D) } \sqrt{3}-\frac{2\sqrt{3}\pi}{9}\qquad \textbf{(E) } \frac{4}{3}-\frac{4\sqrt{3}\pi}{27}$ | 2017 AMC 10A Problem 22 | (same diagram as Solution 1)
Without the Loss of Generality, let the side length of the triangle be $1$.
Then, the area of the triangle is $\frac{\sqrt{3}}{4}$. We are looking for the area of the portion inside the triangle but outside the circle divided by the area of the triangle. Since $m\angle ABO = 90^{\circ}=m\angle ACO = 90^{\circ}$, and $m\angle ABC = m\angle ACB = 60^{\circ}$, we know $m\angle OBC=m\angle OCB=30^{\circ}$, and $m\angle BOC = 120^{\circ}$. Drop an angle bisector of $O$ onto $BC$, call the point of intersection $D$. By SAS congruence, $\triangle BDO \cong \triangle CDO$, by CPCTC (Congruent Parts of Congruent Triangles are Congruent) $BD \cong DC$ and they both measure $\frac{1}{2}$. By 30-60-90 triangle, $OC = BO = \frac{\sqrt{3}}{3}$. The area of the sector bounded by arc BC is one-third the area of circle O, whose area is $\left(\frac{\sqrt{3}}{3}\right)^{2}\pi=\frac{1}{3}\pi$. Therefore, the area of the sector bounded by arc BC is $\frac{1}{3}\cdot\frac{1}{3}\pi=\frac{\pi}{9}$.
We are nearly there. By 30-60-90 triangle, we know $DO = \frac{\sqrt{3}}{6}$, so the area of $\triangle BOC$ is $\frac{1\cdot\frac{\sqrt{3}}{6}}{2}=\frac{\sqrt{3}}{12}$. The area of the region inside both the triangle and circle is the area of the sector bounded by arc BC minus the area of $\triangle BOC$: $\frac{\pi}{9}-\frac{\sqrt{3}}{12}$. The area of the region outside of the circle but inside the triangle is $\frac{\sqrt{3}}{4}-\left(\frac{\pi}{9}-\frac{\sqrt{3}}{12}\right)=\frac{\sqrt{3}}{4}-\frac{\pi}{9}+\frac{\sqrt{3}}{12}=\frac{\sqrt{3}}{3}-\frac{\pi}{9}$ and the ratio is $\frac{\frac{\sqrt{3}}{3}-\frac{\pi}{9}}{\frac{\sqrt{3}}{4}}=\left(\frac{\sqrt{3}}{3}-\frac{\pi}{9}\right)\cdot\frac{4}{\sqrt{3}}=\frac{4\sqrt{3}}{3\sqrt{3}}-\frac{4\pi}{9\sqrt{3}}=\frac{4}{3}-\frac{4\sqrt{3}\pi}{27} \Longrightarrow \boxed{\textbf{(E) } \frac 43 - \frac{4\sqrt 3 \pi}{27}}$.
~JH. L | // Block 1
real sqrt3 = 1.73205080757;
draw(Circle((4, 4), 4));
draw((4-2*sqrt3,6)--(4,4)--(4+2*sqrt3,6)--(4-2*sqrt3,6)--(4,12)--(4+2*sqrt3,6));
draw((4, 6)--(4, 4));
label("A", (4, 12.4));
label("B", (-.3, 6.3));
label("C", (8.3, 6.3));
label("O", (4, 3.4));
label("D", (4, 6.6));
// Block 2
real sqrt3 = 1.73205080757; draw(Circle((4, 4), 4)); draw((4-2*sqrt3,6)--(4,4)--(4+2*sqrt3,6)--(4-2*sqrt3,6)--(4,12)--(4+2*sqrt3,6)); draw((4, 6)--(4, 4)); label("A", (4, 12.4)); label("B", (-.3, 6.3)); label("C", (8.3, 6.3)); label("O", (4, 3.4)); label("D", (4, 6.6)); | [] |
798 | Let $f(x) = \sin{x} + 2\cos{x} + 3\tan{x}$, using radian measure for the variable $x$. In what interval does the smallest positive value of $x$ for which $f(x) = 0$ lie?
$\textbf{(A)}\ (0,1) \qquad \textbf{(B)}\ (1, 2) \qquad\textbf{(C)}\ (2, 3) \qquad\textbf{(D)}\ (3, 4) \qquad\textbf{(E)}\ (4,5)$ | 2017 AMC 12A Problem 15 | If you quickly take a moment to sketch the graphs of the three functions, you will see that between $0$ and $\frac{\pi}{2}$ everything is positive, while the positive number created by the sin does not outweigh the negative by the cos and tan function. Upon further examination, it is clear that the positive the tan function creates will balance the other two functions, and thus the first solution is a little bit after $\pi$, which is around $3.14$. Hence the answer is $\boxed{\textbf{(D)}}$.
Solution by roadchicken~
(Not original author) Here is the graph: | // Block 1
Label f;
f.p=fontsize(6);
xaxis(-5,5,Ticks(f, 1.0));
yaxis(-8,8,Ticks(f, 2.0));
real f(real x)
{
return sin(x);
}
draw(graph(f, -5,5));
real g(real x)
{
return 2*cos(x);
}
draw(graph(g, -5,5));
real h(real x)
{
return 3*tan(x);
}
draw(graph(h, -1.2,1.2));
draw(graph(h, 1.94, 4.34));
draw(graph(h, -4.34, -1.94));
// Block 2
Label f; f.p=fontsize(6); xaxis(-5,5,Ticks(f, 1.0)); yaxis(-8,8,Ticks(f, 2.0)); real f(real x) { return sin(x); } draw(graph(f, -5,5)); real g(real x) { return 2*cos(x); } draw(graph(g, -5,5)); real h(real x) { return 3*tan(x); } draw(graph(h, -1.2,1.2)); draw(graph(h, 1.94, 4.34)); draw(graph(h, -4.34, -1.94)); | [] |
799 | In the figure below, semicircles with centers at $A$ and $B$ and with radii 2 and 1, respectively, are drawn in the interior of, and sharing bases with, a semicircle with diameter $JK$. The two smaller semicircles are externally tangent to each other and internally tangent to the largest semicircle. A circle centered at $P$ is drawn externally tangent to the two smaller semicircles and internally tangent to the largest semicircle. What is the radius of the circle centered at $P$?
$\textbf{(A)}\ \frac{3}{4} \qquad \textbf{(B)}\ \frac{6}{7} \qquad\textbf{(C)}\ \frac{\sqrt{3}}{2} \qquad\textbf{(D)}\ \frac{5}{8}\sqrt{2} \qquad\textbf{(E)}\ \frac{11}{12}$ | 2017 AMC 12A Problem 16 | Connect the centers of the tangent circles! (call the center of the large circle $C$)
Notice that we don't even need the circles anymore; thus, draw triangle $\Delta ABP$ with cevian $PC$:
and use Stewart's Theorem:
\[AB \cdot AC \cdot BC + AB \cdot {CP}^2 = AC \cdot {BP}^2 + BC \cdot {AP}^2\]
From what we learned from the tangent circles, we have $AB = 3$, $AC = 1$, $BC = 2$, $AP = 2 + r$, $BP = 1 + r$, and $CP = 3 - r$, where $r$ is the radius of the circle centered at $P$ that we seek.
Thus:
\[3 \cdot 1 \cdot 2 + 3 {\left(3-r\right)}^2 = 1 {\left(1+r\right)}^2 + 2 {\left(2+r\right)}^2\]
\[6 + 3\left(9 - 6r + r^2\right) = \left(1 + 2r + r^2\right) + 2\left(4 + 4r + r^2\right)\]
\[33 - 18r + 3r^2 = 9 + 10r + 3r^2\]
\[28r = 24\]
\[r = \boxed{\frac{6}{7}} \to \boxed{\textbf{(B)}}\] | // Block 1
size(5cm);
draw(arc((0,0),3,0,180));
draw(arc((2,0),1,0,180));
draw(arc((-1,0),2,0,180));
draw((-3,0)--(3,0));
pair P = (9/7,12/7);
pair A = (-1,0);
pair C = (0,0);
pair B = (2,0);
draw(circle(P,6/7));
dot((-1,0)); dot((2,0)); dot((0,0)); dot(P);
draw((-1,0)--P);
draw((2,0)--P);
draw((0,0)--(9/5,12/5));
label("$A$",A,SW);
label("$C$",C,S);
label("$B$",B,SE);
label("$P$",P,N);
// Block 2
size(5cm);
draw((-1,0)--(2,0));
pair P = (9/7,12/7);
pair A = (-1,0);
pair C = (0,0);
pair B = (2,0);
dot((-1,0)); dot((2,0)); dot((0,0)); dot(P);
draw((-1,0)--P);
draw((2,0)--P);
draw((0,0)--P);
label("$A$",A,SW);
label("$C$",C,S);
label("$B$",B,SE);
label("$P$",P,N);
// Block 3
size(5cm); draw(arc((0,0),3,0,180)); draw(arc((2,0),1,0,180)); draw(arc((-1,0),2,0,180)); draw((-3,0)--(3,0)); pair P = (9/7,12/7); pair A = (-1,0); pair C = (0,0); pair B = (2,0); draw(circle(P,6/7)); dot((-1,0)); dot((2,0)); dot((0,0)); dot(P); draw((-1,0)--P); draw((2,0)--P); draw((0,0)--(9/5,12/5)); label("$A$",A,SW); label("$C$",C,S); label("$B$",B,SE); label("$P$",P,N);
// Block 4
size(5cm); draw((-1,0)--(2,0)); pair P = (9/7,12/7); pair A = (-1,0); pair C = (0,0); pair B = (2,0); dot((-1,0)); dot((2,0)); dot((0,0)); dot(P); draw((-1,0)--P); draw((2,0)--P); draw((0,0)--P); label("$A$",A,SW); label("$C$",C,S); label("$B$",B,SE); label("$P$",P,N); | [] |
799 | In the figure below, semicircles with centers at $A$ and $B$ and with radii 2 and 1, respectively, are drawn in the interior of, and sharing bases with, a semicircle with diameter $JK$. The two smaller semicircles are externally tangent to each other and internally tangent to the largest semicircle. A circle centered at $P$ is drawn externally tangent to the two smaller semicircles and internally tangent to the largest semicircle. What is the radius of the circle centered at $P$?
$\textbf{(A)}\ \frac{3}{4} \qquad \textbf{(B)}\ \frac{6}{7} \qquad\textbf{(C)}\ \frac{\sqrt{3}}{2} \qquad\textbf{(D)}\ \frac{5}{8}\sqrt{2} \qquad\textbf{(E)}\ \frac{11}{12}$ | 2017 AMC 12A Problem 16 | Like the solution above, connecting the centers of the circles results in triangle $\Delta ABP$ with cevian $PC$. The two triangles $\Delta APC$ and $\Delta ABP$ share angle $A$, which means we can use Law of Cosines to set up a system of 2 equations that solve for $r$ respectively:
$(2 + r)^2 + 1^2 - 2(2 + r)(1)\cos A = (3 - r)^2$ (notice that the diameter of the largest semicircle is 6, so its radius is 3 and $PC$ is 3 - r)
$(2 + r)^2 + 3^2 - 2(2 + r)(3)\cos A = (r+1)^2$
We can eliminate the extra variable of angle $A$ by multiplying the first equation by 3 and subtracting the second from it. Then, expand to find $r$:
$2(r^2 + 4r + 4) - 6 = 2r^2 - 20r + 26$
$8r + 2 = -20r + 26$
$28r = 24$, so $r$ = $6/7$ $\boxed{(B)}$ | // Block 1
size(5cm);
draw(arc((0,0),3,0,180));
draw(arc((2,0),1,0,180));
draw(arc((-1,0),2,0,180));
draw((-3,0)--(3,0));
pair P = (9/7,12/7);
draw(circle(P,6/7));
dot((-1,0)); dot((2,0)); dot((0,0)); dot(P);
draw((-1,0)--P);
draw((2,0)--P);
draw((0,0)--(9/5,12/5));
// Block 2
size(5cm); draw(arc((0,0),3,0,180)); draw(arc((2,0),1,0,180)); draw(arc((-1,0),2,0,180)); draw((-3,0)--(3,0)); pair P = (9/7,12/7); draw(circle(P,6/7)); dot((-1,0)); dot((2,0)); dot((0,0)); dot(P); draw((-1,0)--P); draw((2,0)--P); draw((0,0)--(9/5,12/5)); | [] |
799 | In the figure below, semicircles with centers at $A$ and $B$ and with radii 2 and 1, respectively, are drawn in the interior of, and sharing bases with, a semicircle with diameter $JK$. The two smaller semicircles are externally tangent to each other and internally tangent to the largest semicircle. A circle centered at $P$ is drawn externally tangent to the two smaller semicircles and internally tangent to the largest semicircle. What is the radius of the circle centered at $P$?
$\textbf{(A)}\ \frac{3}{4} \qquad \textbf{(B)}\ \frac{6}{7} \qquad\textbf{(C)}\ \frac{\sqrt{3}}{2} \qquad\textbf{(D)}\ \frac{5}{8}\sqrt{2} \qquad\textbf{(E)}\ \frac{11}{12}$ | 2017 AMC 12A Problem 16 | Let $C$ be the center of the largest semicircle and $r$ be the radius of $\circ P$. We know that $AC = 1$, $CB = 2$, $AP = r + 2$, $BP = r + 1$, and $CP = 3 - r$. Notice that $\Delta ACP$ and $\Delta CBP$ are bounded by the same two parallel lines, so these triangles have the same heights. Because the bases of these two triangles (that have the same heights) differ by a factor of 2, the area of $\Delta CBP$ must be twice that of $\Delta ACP$, since the area of a triangle is $\frac{1}{2} \text{Base} \cdot \text{Height}$.
Again, we don't need to look at the circle and the semicircles anymore; just focus on the triangles.
Let $A_1$ equal to the area of $\Delta ACP$ and $A_2$ equal to the area of $\Delta CBP$.
Heron's Formula states that the area of an triangle with sides $a$ $b$ and $c$ is
\[\sqrt{s(s-a)(s-b)(s-c)}\]
where $s$, or the semiperimeter, is $\frac{a+b+c}{2}$
The semiperimeter $s_1$ of $\Delta ACP$ is \[\frac{[(r + 2) + (3 - r) + 1]}{2} = \frac{6}{2} = 3\]
Use Heron's Formula to obtain
\[A_1 = \sqrt{3(2)(3-2-r)(3-3+r)} = \sqrt{6r(1-r)} = \sqrt{6r-6r^2}\]
Using Heron's Formula again, find the area of $\Delta CBP$ with sides $r+1$, $2$, and $3-r$.
\[s_2 = \frac{(r + 1) + 2 + (3 - r)}{2} = 3\]
\[A_2 = \sqrt{3(3-2)(3-1-r)(3-3+r)} = \sqrt{3(2r-r^2)} = \sqrt{6r-3r^2}\]
Now,
\[2 \cdot A_1 = A_2\]
\[2\sqrt{6r-6r^2} = \sqrt{6r-3r^2}\]
\[4(6r-6r^2) = 6r-3r^2\]
\[24r-24r^2 = 6r-3r^2\]
\[18r = 21r^2\]
\[r = \frac{18}{21} = \boxed{\frac{6}{7}} \to \boxed{\textbf{(B)}}\] | // Block 1
size(7.5cm);
draw(arc((0,0),3,0,180));
draw(arc((2,0),1,0,180));
draw(arc((-1,0),2,0,180));
draw((-3,0)--(3,0));
pair P = (9/7,12/7);
pair A = (-1,0);
pair C = (0,0);
pair B = (2,0);
draw(circle(P,6/7));
dot((-1,0)); dot((2,0)); dot((0,0)); dot(P);
draw((-1,0)--P);
draw((-2.45,12/7)--P);
draw((2.45,12/7)--P);
draw((2,0)--P);
draw((0,0)--(9/5,12/5));
label("$A$",A,SW);
label("$C$",C,S);
label("$B$",B,SE);
label("$P$",P,N);
// Block 2
size(7.5cm);
pair A = (-1,0);
pair C = (0,0);
pair B = (2,0);
draw((-1,0)--(2,0));
pair P = (9/7,12/7);
dot((-1,0)); dot((2,0)); dot((0,0)); dot(P);
draw((-1,0)--P);
draw((2,0)--P);
draw((0,0)--P);
label("$A$",A,SW);
label("$C$",C,S);
label("$B$",B,SE);
label("$P$",P,N);
// Block 3
size(7.5cm); draw(arc((0,0),3,0,180)); draw(arc((2,0),1,0,180)); draw(arc((-1,0),2,0,180)); draw((-3,0)--(3,0)); pair P = (9/7,12/7); pair A = (-1,0); pair C = (0,0); pair B = (2,0); draw(circle(P,6/7)); dot((-1,0)); dot((2,0)); dot((0,0)); dot(P); draw((-1,0)--P); draw((-2.45,12/7)--P); draw((2.45,12/7)--P); draw((2,0)--P); draw((0,0)--(9/5,12/5)); label("$A$",A,SW); label("$C$",C,S); label("$B$",B,SE); label("$P$",P,N);
// Block 4
size(7.5cm); pair A = (-1,0); pair C = (0,0); pair B = (2,0); draw((-1,0)--(2,0)); pair P = (9/7,12/7); dot((-1,0)); dot((2,0)); dot((0,0)); dot(P); draw((-1,0)--P); draw((2,0)--P); draw((0,0)--P); label("$A$",A,SW); label("$C$",C,S); label("$B$",B,SE); label("$P$",P,N); | [] |
799 | In the figure below, semicircles with centers at $A$ and $B$ and with radii 2 and 1, respectively, are drawn in the interior of, and sharing bases with, a semicircle with diameter $JK$. The two smaller semicircles are externally tangent to each other and internally tangent to the largest semicircle. A circle centered at $P$ is drawn externally tangent to the two smaller semicircles and internally tangent to the largest semicircle. What is the radius of the circle centered at $P$?
$\textbf{(A)}\ \frac{3}{4} \qquad \textbf{(B)}\ \frac{6}{7} \qquad\textbf{(C)}\ \frac{\sqrt{3}}{2} \qquad\textbf{(D)}\ \frac{5}{8}\sqrt{2} \qquad\textbf{(E)}\ \frac{11}{12}$ | 2017 AMC 12A Problem 16 | Let $C$, the center of the large semicircle, to be at $(0, 0)$, and $P$ to be at $(h, k)$.
Therefore $A$ is at $(-1, 0)$ and $B$ is at $(2, 0)$.
Let the radius of circle $P$ be $r$.
Using Distance Formula, we get the following system of three equations:
\[h^2+k^2=(3-r)^2, (h+1)^2+k^2=(r+2)^2, (h-2)^2+k^2=(r+1)^2\]
By simplifying, we get
\[h^2+k^2=r^2-6r+9, h^2+2r+1+k^2=r^2+4r+4, h^2-4r+4+k^2=r^2+2r+1\]
By subtracting the first equation from the second and third equations, we get
\[8r=-4h+12, 10r=2h+6\]
which simplifies to
\[2r=3-h, 5r=h+3\]
When we add these two equations, we get
\[7r=6\]
So \[r = \boxed{\frac{6}{7}} \to \boxed{\textbf{(B)}}\] $w^5$ | // Block 1
size(5cm);
draw(arc((0,0),3,0,180));
draw(arc((2,0),1,0,180));
draw(arc((-1,0),2,0,180));
draw((-3,0)--(3,0));
pair P = (9/7,12/7);
pair A = (-1,0);
pair C = (0,0);
pair B = (2,0);
draw(circle(P,6/7));
dot((-1,0)); dot((2,0)); dot((0,0)); dot(P);
draw((-1,0)--P);
draw((2,0)--P);
draw((0,0)--(9/5,12/5));
label("$A$",A,SW);
label("$C$",C,S);
label("$B$",B,SE);
label("$P$",P,N);
// Block 2
size(5cm); draw(arc((0,0),3,0,180)); draw(arc((2,0),1,0,180)); draw(arc((-1,0),2,0,180)); draw((-3,0)--(3,0)); pair P = (9/7,12/7); pair A = (-1,0); pair C = (0,0); pair B = (2,0); draw(circle(P,6/7)); dot((-1,0)); dot((2,0)); dot((0,0)); dot(P); draw((-1,0)--P); draw((2,0)--P); draw((0,0)--(9/5,12/5)); label("$A$",A,SW); label("$C$",C,S); label("$B$",B,SE); label("$P$",P,N); | [] |
799 | In the figure below, semicircles with centers at $A$ and $B$ and with radii 2 and 1, respectively, are drawn in the interior of, and sharing bases with, a semicircle with diameter $JK$. The two smaller semicircles are externally tangent to each other and internally tangent to the largest semicircle. A circle centered at $P$ is drawn externally tangent to the two smaller semicircles and internally tangent to the largest semicircle. What is the radius of the circle centered at $P$?
$\textbf{(A)}\ \frac{3}{4} \qquad \textbf{(B)}\ \frac{6}{7} \qquad\textbf{(C)}\ \frac{\sqrt{3}}{2} \qquad\textbf{(D)}\ \frac{5}{8}\sqrt{2} \qquad\textbf{(E)}\ \frac{11}{12}$ | 2017 AMC 12A Problem 16 | Let $\Omega$ be a circle with radius of $6$ and centered at the left corner of the semi-circle (O) with radius $3$. Extend the three semicircles to full circles. Label the resulting four circles as shown in the diagram:
$C_1$ has radius $3$, $C_2$ has radius $2$, and $C_3$ has radius $1$. We want to find the radius of $C_4$.
We now invert the four circles. $C_1$ inverts to a line. Given that one point is on $\Omega$, and all points on $\Omega$ invert to themselves, we know that the resulting line must intersect that intersection point. $C_2$ also inverts to a line. $C_2$ has radius $4$, and since $\Omega$ has radius of $6$, the resulting line must be $\frac{36}{4} = 9$ units away from $O$. $C_3$ inverts to a circle. By observing the diagram, we note that $C_3'$'s center must be on $\overline{OC_3}$ and be between the two inverted lines, because $C_3$ is tangent to $C_1$ and $C_2$ (Remeber that tangency still holds in inverted diagrams). Therefore, we must have a circle with radius $\frac{3}{2}$ that is $\frac{15}{2}$ units from $O$.
Now, we invert $C_4$. Note that $C_4$ is tangent to the three other original circles. So, in the inversion, $C_4'$ must be tangent to the two lines and $C_3'$. It is then quickly seen that $C_3'$ and $C_4'$ have the same radius: $\frac{3}{2}$.
Now, we can determine the radius of $C_4$ using the formula $r = \frac{k^2 \cdot r'}{\overline{OC_2}^2 - r'^2}$. $k^2 = 36$, and $r' = \frac{3}{2}$. $\overline{OC}$ is just the distance from the center of the inverted circle to the center of inversion. The center of $C_4'$ is $3$ units above the center of $C_3'$. Since $\overline{OC_3'} = \frac{15}{2}$, we use Pythagoras to learn that $\overline{OC_4'}^2 = \left(\frac{15}{2}\right)^2 + 9$. We do not take the square root because our relationship formula takes $\overline{OC_4'}^2$.
Therefore, we have: \[r = \frac{36 \cdot \frac{3}{2}}{\left(\frac{15}{2}\right)^2 - \left(\frac{3}{2}\right)^2 + 9} = \frac{54}{9 \cdot 6 + 9} = \frac{6}{7} = \boxed{B}.\]
Here is the diagram with $C_4'$. | // Block 1
size(5cm); path circle1 = Circle((3, 0), 3); path circle2 = Circle((2, 0), 2); path circle3 = Circle((5, 0), 1); pair P = (2,0)+(2+6/7)*dir(36.86989); path circle4 = Circle(P, 6/7); draw(circle1); draw(circle2); draw(circle3); draw(circle4); draw((0, 0)--(6, 0)); dot((2,0)); dot((5,0)); dot(P); dot((3, 0)); dot(origin); path inversion = arc((0,0), 6, -30, 30); draw(inversion, dashed); label("$\Omega$", (6, 0) * dir(30), NE); label("$C_1$", (3, 0), N); label("$C_2$", (2, 0), N); label("$C_3$", (5, 0), N); label("$C_4$", P, N); label("$O$", origin, W);
// Block 2
size(10cm); path circle1 = Circle((3, 0), 3); path circle2 = Circle((2, 0), 2); path circle3 = Circle((5, 0), 1); pair P = (2,0)+(2+6/7)*dir(36.86989); path circle4 = Circle(P, 6/7); draw(circle1); draw(circle2); draw(circle3); draw(circle4); draw((0, 0)--(9, 0)); dot((2,0)); dot((5,0)); dot(P); dot((3, 0)); dot(origin); path inversion = arc((0,0), 6, -30, 30); draw(inversion, dashed); label("$\Omega$", (6, 0) * dir(30), NW); label("$C_1$", (3, 0), N); label("$C_2$", (2, 0), N); label("$C_3$", (5, 0), N); label("$C_4$", P, N); label("$O$", origin, W); draw((6, 3)--(6, -3)); draw((9, 3)--(9, -3)); draw(Circle((15/2, 0), 3/2)); label("$C_1'$", (6, 3), N); label("$C_2'$", (9, 3), N); label("$C_3'$", (15/2, 0), N); dot((15/2, 0));
// Block 3
size(10cm); path circle1 = Circle((3, 0), 3); path circle2 = Circle((2, 0), 2); path circle3 = Circle((5, 0), 1); pair P = (2,0)+(2+6/7)*dir(36.86989); path circle4 = Circle(P, 6/7); draw(circle1); draw(circle2); draw(circle3); draw(circle4); draw((0, 0)--(9, 0)); dot((2,0)); dot((5,0)); dot(P); dot((3, 0)); dot(origin); path inversion = arc((0,0), 6, -45, 45); draw(inversion, dashed); label("$\Omega$", (6, 0) * dir(45), NW); label("$C_1$", (3, 0), N); label("$C_2$", (2, 0), N); label("$C_3$", (5, 0), N); label("$C_4$", P, N); label("$O$", origin, W); draw((6, 4.5)--(6, -3)); draw((9, 4.5)--(9, -3)); draw(Circle((15/2, 0), 3/2)); label("$C_1'$", (6, -3), SW); label("$C_2'$", (9, -3), SE); label("$C_3'$", (15/2, 0), NE); dot((15/2, 0)); draw(Circle((15/2, 3), 3/2)); dot((15/2, 3)); label("$C_4'$", (15/2, 3), N); draw((15/2, 0)--(15/2, 3)); draw(rightanglemark(origin, (15/2, 0), (15/2, 3))); | [] |
800 | A square with side length $x$ is inscribed in a right triangle with sides of length $3$, $4$, and $5$ so that one vertex of the square coincides with the right-angle vertex of the triangle. A square with side length $y$ is inscribed in another right triangle with sides of length $3$, $4$, and $5$ so that one side of the square lies on the hypotenuse of the triangle. What is $\frac{x}{y}$?
$\textbf{(A)}\ \frac{12}{13} \qquad \textbf{(B)}\ \frac{35}{37} \qquad\textbf{(C)}\ 1 \qquad\textbf{(D)}\ \frac{37}{35} \qquad\textbf{(E)}\ \frac{13}{12}$ | 2017 AMC 12A Problem 19 | Analyze the first right triangle.
Note that $\triangle ABC$ and $\triangle FBE$ are similar, so $\frac{BF}{FE} = \frac{AB}{AC}$. This can be written as $\frac{4-x}{x}=\frac{4}{3}$. Solving, $x = \frac{12}{7}$.
Now we analyze the second triangle.
Similarly, $\triangle A'B'C'$ and $\triangle RB'Q$ are similar, so $RB' = \frac{4}{3}y$, and $C'S = \frac{3}{4}y$. Thus, $C'B' = C'S + SR + RB' = \frac{4}{3}y + y + \frac{3}{4}y = 5$. Solving for $y$, we get $y = \frac{60}{37}$. Thus, $\frac{x}{y} = \frac{37}{35} \implies \boxed{\textbf{D}}$. | // Block 1
pair A,B,C;
pair D, e, F;
A = (0,0);
B = (4,0);
C = (0,3);
D = (0, 12/7);
e = (12/7 , 12/7);
F = (12/7, 0);
draw(A--B--C--cycle);
draw(D--e--F);
label("$x$", D/2, W);
label("$A$", A, SW);
label("$B$", B, SE);
label("$C$", C, N);
label("$D$", D, W);
label("$E$", e, NE);
label("$F$", F, S);
// Block 2
pair A,B,C;
pair q, R, S, T;
A = (0,0);
B = (4,0);
C = (0,3);
q = (1.297, 0);
R = (2.27 , 1.297);
S = (0.973, 2.27);
T = (0, 0.973);
draw(A--B--C--cycle);
draw(q--R--S--T--cycle);
label("$y$", (q+R)/2, NW);
label("$A'$", A, SW);
label("$B'$", B, SE);
label("$C'$", C, N);
label("$Q$", (q-(0,0.3)));
label("$R$", R, NE);
label("$S$", S, NE);
label("$T$", T, W);
// Block 3
pair A,B,C; pair D, e, F; A = (0,0); B = (4,0); C = (0,3); D = (0, 12/7); e = (12/7 , 12/7); F = (12/7, 0); draw(A--B--C--cycle); draw(D--e--F); label("$x$", D/2, W); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, N); label("$D$", D, W); label("$E$", e, NE); label("$F$", F, S);
// Block 4
pair A,B,C; pair q, R, S, T; A = (0,0); B = (4,0); C = (0,3); q = (1.297, 0); R = (2.27 , 1.297); S = (0.973, 2.27); T = (0, 0.973); draw(A--B--C--cycle); draw(q--R--S--T--cycle); label("$y$", (q+R)/2, NW); label("$A'$", A, SW); label("$B'$", B, SE); label("$C'$", C, N); label("$Q$", (q-(0,0.3))); label("$R$", R, NE); label("$S$", S, NE); label("$T$", T, W); | [] |
801 | Points $A(11, 9)$ and $B(2, -3)$ are vertices of $\triangle ABC$ with $AB=AC$. The altitude from $A$ meets the opposite side at $D(-1, 3)$. What are the coordinates of point $C$?
$\textbf{(A)}\ (-8, 9)\qquad\textbf{(B)}\ (-4, 8)\qquad\textbf{(C)}\ (-4, 9)\qquad\textbf{(D)}\ (-2, 3)\qquad\textbf{(E)}\ (-1, 0)$ | 2017 AMC 10B Problem 8 | Since $AB = AC$, then $\triangle ABC$ is isosceles, so $BD = CD$. Therefore, the coordinates of $C$ are $(-1 - 3, 3 + 6) = \boxed{\textbf{(C) } (-4,9)}$. | // Block 1
pair A,B,C,D;
A=(11,9);
B=(2,-3);
C=(-4,9);
D=(-1,3);
draw(A--B--C--cycle);
draw(A--D);
draw(rightanglemark(A,D,B,30));
label("$A$",A,E);
label("$B$",B,S);
label("$D$",D,W);
label("$C$",C,N);
// Block 2
pair A,B,C,D; A=(11,9); B=(2,-3); C=(-4,9); D=(-1,3); draw(A--B--C--cycle); draw(A--D); draw(rightanglemark(A,D,B,30)); label("$A$",A,E); label("$B$",B,S); label("$D$",D,W); label("$C$",C,N); | [] |
801 | Points $A(11, 9)$ and $B(2, -3)$ are vertices of $\triangle ABC$ with $AB=AC$. The altitude from $A$ meets the opposite side at $D(-1, 3)$. What are the coordinates of point $C$?
$\textbf{(A)}\ (-8, 9)\qquad\textbf{(B)}\ (-4, 8)\qquad\textbf{(C)}\ (-4, 9)\qquad\textbf{(D)}\ (-2, 3)\qquad\textbf{(E)}\ (-1, 0)$ | 2017 AMC 10B Problem 8 | As stated in solution 1, the triangle is isosceles.
This means that $D(-1, 3)$ is the midpoint of $B(2, -3)$ and $C(x,y)$. So $\frac{x+2}{2}$ $= -1$ and so $x = -4$. Similarly for $y$, we have $\frac{y-3}{2}$ $=3$ and so $y=9$. So our final answer is $\boxed{\textbf{(C) } (-4,9)}$.
- youtube.com/indianmathguy | // Block 1
pair A,B,C,D;
A=(11,9);
B=(2,-3);
C=(-4,9);
D=(-1,3);
draw(A--B--C--cycle);
draw(A--D);
draw(rightanglemark(A,D,B));
label("$A$",A,E);
label("$B$",B,S);
label("$D$",D,W);
label("$C$",C,N);
// Block 2
pair A,B,C,D; A=(11,9); B=(2,-3); C=(-4,9); D=(-1,3); draw(A--B--C--cycle); draw(A--D); draw(rightanglemark(A,D,B)); label("$A$",A,E); label("$B$",B,S); label("$D$",D,W); label("$C$",C,N); | [] |
802 | Rectangle $ABCD$ has $AB=3$ and $BC=4$. Point $E$ is the foot of the perpendicular from $B$ to diagonal $\overline{AC}$. What is the area of $\triangle AED$?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ \frac{42}{25}\qquad\textbf{(C)}\ \frac{28}{15}\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \frac{54}{25}$ | 2017 AMC 10B Problem 15 | First, note that $AC=5$ because $ABC$ is a right triangle. In addition, we have $AB\cdot BC=2[ABC]=AC\cdot BE$, so $BE=\frac{12}{5}$. Using similar triangles within $ABC$, we get that $AE=\frac{9}{5}$ and $CE=\frac{16}{5}$.
Let $F$ be the foot of the perpendicular from $E$ to $AB$. Since $EF$ and $BC$ are parallel, $\Delta AFE$ is similar to $\Delta ABC$. Therefore, we have $\frac{AF}{AB}=\frac{AE}{AC}=\frac{9}{25}$. Since $AB=3$, $AF=\frac{27}{25}$. Note that $AF$ is an altitude of $\Delta AED$ from $AD$, which has length $4$. Therefore, the area of $\Delta AED$ is $\frac{1}{2}\cdot\frac{27}{25}\cdot4=\boxed{\textbf{(E)}\frac{54}{25}}.$ | // Block 1
pair A,B,C,D,E;
A=(0,4);
B=(3,4);
C=(3,0);
D=(0,0);
draw(A--B--C--D--cycle);
label("$A$",A,N);
label("$B$",B,N);
label("$C$",C,S);
label("$D$",D,S);
E=foot(B,A,C);
draw(E--B);
draw(A--C);
draw(rightanglemark(B,E,C));
label("$E$",E,N);
draw(D--E);
label("$3$",A--B,N);
label("$4$",B--C,E);
// Block 2
pair A,B,C,D,E; A=(0,4); B=(3,4); C=(3,0); D=(0,0); draw(A--B--C--D--cycle); label("$A$",A,N); label("$B$",B,N); label("$C$",C,S); label("$D$",D,S); E=foot(B,A,C); draw(E--B); draw(A--C); draw(rightanglemark(B,E,C)); label("$E$",E,N); draw(D--E); label("$3$",A--B,N); label("$4$",B--C,E); | [] |
803 | In the figure below, $3$ of the $6$ disks are to be painted blue, $2$ are to be painted red, and $1$ is to be painted green. Two paintings that can be obtained from one another by a rotation or a reflection of the entire figure are considered the same. How many different paintings are possible?
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 15$ | 2017 AMC 10B Problem 18 | Note that the green disk has two possibilities; in a corner or on the side. WLOG, we can arrange these as
Take the first case. Now, we must pick two of the five remaining circles to fill in the red. There are $\dbinom{5}{2}=10$ of these. However, due to reflection we must divide this by two. But, in two of these cases, the reflection is itself, so we must subtract these out before dividing by 2, and add them back afterwards, giving $\frac{10-2}{2}+2=6$ arrangements in this case.
Now, look at the second case. We again must pick two of the five remaining circles, and like in the first case, two of the reflections give the same arrangement. Thus, there are also $6$ arrangements in this case.
In total, we have $6+6=\boxed{\text{\bf(D) }12}$.
~tdeng | // Block 1
filldraw(circle((0,0),1),green);
draw(circle((2,0),1));
draw(circle((4,0),1));
draw(circle((1,sqrt(3)),1));
draw(circle((3,sqrt(3)),1));
draw(circle((2,2sqrt(3)),1));
draw(circle((8,0),1));
filldraw(circle((10,0),1),green);
draw(circle((12,0),1));
draw(circle((9,sqrt(3)),1));
draw(circle((11,sqrt(3)),1));
draw(circle((10,2sqrt(3)),1));
// Block 2
filldraw(circle((0,0),1),green); draw(circle((2,0),1)); draw(circle((4,0),1)); draw(circle((1,sqrt(3)),1)); draw(circle((3,sqrt(3)),1)); draw(circle((2,2sqrt(3)),1)); draw(circle((8,0),1)); filldraw(circle((10,0),1),green); draw(circle((12,0),1)); draw(circle((9,sqrt(3)),1)); draw(circle((11,sqrt(3)),1)); draw(circle((10,2sqrt(3)),1)); | [] |
804 | Let $ABC$ be an equilateral triangle. Extend side $\overline{AB}$ beyond $B$ to a point $B'$ so that $BB'=3 \cdot AB$. Similarly, extend side $\overline{BC}$ beyond $C$ to a point $C'$ so that $CC'=3 \cdot BC$, and extend side $\overline{CA}$ beyond $A$ to a point $A'$ so that $AA'=3 \cdot CA$. What is the ratio of the area of $\triangle A'B'C'$ to the area of $\triangle ABC$?
$\textbf{(A)}\ 9:1\qquad\textbf{(B)}\ 16:1\qquad\textbf{(C)}\ 25:1\qquad\textbf{(D)}\ 36:1\qquad\textbf{(E)}\ 37:1$ | 2017 AMC 10B Problem 19 | This looks like an easy application of Routh's Theorem, except we are only given information about the ratios of the cevians, not the side lengths of $\triangle A'B'C'$.
Let's figure those out. Extend $A'A$ and $B'B$ to meet $B'C'$ and $A'C'$ at $D$ and $E$, respectively (Note we only need to look at 2 of 3 cevians to figure everything out). Call the unknown lengths $CD=AE=x$ so our new diagram showing the cevian ratios is as follows.
with the balanced mass points on the right. Now by the symmetry in the original diagram, $\dfrac{A'E}{EC'}=\dfrac{C'D}{DB'}$ so:
\[\dfrac{3-x}{x+1}=\dfrac{x}{3-x} \implies x=\dfrac{9}{7} \implies \dfrac{A'E}{EC'}=\dfrac{3}{4}\]
Finally we can apply Routh's Theorem:
\[\dfrac{[\triangle ABC]}{[\triangle A'B'C']} = \frac{(x y z-1)^2}{(x y+y+1) (x z+x+1) (y z+z+1)} = \frac{(\frac{37}{64})^2}{(\frac{37}{16})^3} = \frac{1}{37}\]
Hence $[\triangle A'B'C']:[\triangle ABC] = \boxed{\textbf{(E) } 37:1}$
~ proloto | // Block 1
size(250);
real a = 1;
pair A = (0, 0), B = rotate(60)*A + (a, 0), C = rotate(120)*B + (a, 0);
pair shiftVector = (8, 0); // define a shift vector to the right
// calculate extended points
pair A_prime = A + 3*(A - C);
pair B_prime = B + 3*(B - A);
pair C_prime = C + 3*(C - B);
pair D = extension(A, A_prime, B_prime, C_prime);
pair EE = extension(B, B_prime, A_prime, C_prime);
// draw original triangle and extended triangle
draw(A_prime--D);
draw(B_prime--EE);
draw(A_prime--B_prime--C_prime--cycle);
// label points
dot("$A$", A, SE);
dot("$D$", D, NE);
dot("$E$", EE, W);
dot("$A'$", A_prime, SW);
dot("$B'$", B_prime, NE);
dot("$C'$", C_prime, W);
// label sides
label(scale(0.8)*"$4$", (A+B_prime)/2, S);
label(scale(0.8)*"$x+1$", (A+D)/2, E);
label(scale(0.8)*"$3$", (A+A_prime)/2, E);
label(scale(0.8)*"$x$", (A+EE)/2, S);
// draw duplicated triangle and labels shifted to the right
draw(shift(shiftVector)*A_prime--shift(shiftVector)*D);
draw(shift(shiftVector)*B_prime--shift(shiftVector)*EE);
draw(shift(shiftVector)*A_prime--shift(shiftVector)*B_prime--shift(shiftVector)*C_prime--cycle);
// label cevian ratios
label(scale(0.8)*"$4$", shift(2*shiftVector)*(A+B_prime)/2, S);
label(scale(0.8)*"$x+1$", shift(2*shiftVector)*(A+D)/2, E);
label(scale(0.8)*"$3$", shift(2*shiftVector)*(A+A_prime)/2, E);
label(scale(0.8)*"$x$", shift(2*shiftVector)*(A+EE)/2, S);
// mass point labels
pair mA = shift(shiftVector)*A + scale(0.5)*SE;
label(scale(0.4)*"$x+4$", mA, UnFill);
draw(Circle(mA, .4), linewidth(1));
pair mB_prime = shift(shiftVector)*B_prime + scale(0.5)*E;
label(scale(0.4)*"$x$", mB_prime, UnFill);
draw(Circle(mB_prime, .4), linewidth(1));
pair mE = shift(shiftVector)*EE + scale(0.5)*W;
label(scale(0.4)*"$4$", mE, UnFill);
draw(Circle(mE, .4), linewidth(1));
pair mA_prime = shift(shiftVector)*A_prime + scale(0.5)*SW;
label(scale(0.4)*"$x+1$", mA_prime, UnFill);
draw(Circle(mA_prime, .4), linewidth(1));
pair mD = shift(shiftVector)*D + scale(0.5)*NE;
label(scale(0.4)*"$3$", mD, UnFill);
draw(Circle(mD, .4), linewidth(1));
pair mC_prime = shift(shiftVector)*C_prime + scale(0.5)*NW;
label(scale(0.4)*"$3-x$", mC_prime, UnFill);
draw(Circle(mC_prime, .4), linewidth(1));
// draw double right arrow between original and duplicated diagram
draw((4.5,0)--(5.3,0),EndArrow(5));
// Block 2
size(250); real a = 1; pair A = (0, 0), B = rotate(60)*A + (a, 0), C = rotate(120)*B + (a, 0); pair shiftVector = (8, 0); // define a shift vector to the right // calculate extended points pair A_prime = A + 3*(A - C); pair B_prime = B + 3*(B - A); pair C_prime = C + 3*(C - B); pair D = extension(A, A_prime, B_prime, C_prime); pair EE = extension(B, B_prime, A_prime, C_prime); // draw original triangle and extended triangle draw(A_prime--D); draw(B_prime--EE); draw(A_prime--B_prime--C_prime--cycle); // label points dot("$A$", A, SE); dot("$D$", D, NE); dot("$E$", EE, W); dot("$A'$", A_prime, SW); dot("$B'$", B_prime, NE); dot("$C'$", C_prime, W); // label sides label(scale(0.8)*"$4$", (A+B_prime)/2, S); label(scale(0.8)*"$x+1$", (A+D)/2, E); label(scale(0.8)*"$3$", (A+A_prime)/2, E); label(scale(0.8)*"$x$", (A+EE)/2, S); // draw duplicated triangle and labels shifted to the right draw(shift(shiftVector)*A_prime--shift(shiftVector)*D); draw(shift(shiftVector)*B_prime--shift(shiftVector)*EE); draw(shift(shiftVector)*A_prime--shift(shiftVector)*B_prime--shift(shiftVector)*C_prime--cycle); // label cevian ratios label(scale(0.8)*"$4$", shift(2*shiftVector)*(A+B_prime)/2, S); label(scale(0.8)*"$x+1$", shift(2*shiftVector)*(A+D)/2, E); label(scale(0.8)*"$3$", shift(2*shiftVector)*(A+A_prime)/2, E); label(scale(0.8)*"$x$", shift(2*shiftVector)*(A+EE)/2, S); // mass point labels pair mA = shift(shiftVector)*A + scale(0.5)*SE; label(scale(0.4)*"$x+4$", mA, UnFill); draw(Circle(mA, .4), linewidth(1)); pair mB_prime = shift(shiftVector)*B_prime + scale(0.5)*E; label(scale(0.4)*"$x$", mB_prime, UnFill); draw(Circle(mB_prime, .4), linewidth(1)); pair mE = shift(shiftVector)*EE + scale(0.5)*W; label(scale(0.4)*"$4$", mE, UnFill); draw(Circle(mE, .4), linewidth(1)); pair mA_prime = shift(shiftVector)*A_prime + scale(0.5)*SW; label(scale(0.4)*"$x+1$", mA_prime, UnFill); draw(Circle(mA_prime, .4), linewidth(1)); pair mD = shift(shiftVector)*D + scale(0.5)*NE; label(scale(0.4)*"$3$", mD, UnFill); draw(Circle(mD, .4), linewidth(1)); pair mC_prime = shift(shiftVector)*C_prime + scale(0.5)*NW; label(scale(0.4)*"$3-x$", mC_prime, UnFill); draw(Circle(mC_prime, .4), linewidth(1)); // draw double right arrow between original and duplicated diagram draw((4.5,0)--(5.3,0),EndArrow(5)); | [] |
805 | In $\triangle ABC$, $AB=6$, $AC=8$, $BC=10$, and $D$ is the midpoint of $\overline{BC}$. What is the sum of the radii of the circles inscribed in $\triangle ADB$ and $\triangle ADC$?
$\textbf{(A)}\ \sqrt{5}\qquad\textbf{(B)}\ \frac{11}{4}\qquad\textbf{(C)}\ 2\sqrt{2}\qquad\textbf{(D)}\ \frac{17}{6}\qquad\textbf{(E)}\ 3$ | 2017 AMC 10B Problem 21 | We have
Let $x$ be the radius of circle $F$, and let $y$ be the radius of circle $J$. We want to find $x+y$.
We form 6 kites: $GAKF$, $HFKD$, $GFHB$, $EJIA$, $LJIC$, and $JEDL$.
Since $G$ and $I$ are the midpoints of $\overline{AB}$ and $\overline{AC}$, respectively, this means that $BG = AG = \frac{6}{2} = 3$, and $AI = IC = \frac{8}{2} = 4$.
Since $AGFK$ is a kite, $GF = FK = x$, and $AG = AK = 3$. The same applies to all kites in the diagram.
Now, we see that $AK = 3$, and $KD = 2$, thus $AD$ is $5$, making $\triangle ADC$ and $\triangle ABD$ isosceles. So, $DI=3$ using the Pythagorean Theorem, and $GD=4$ also using the Theorem. Hence, we know that $[ADC] = [ABD] = 12$.
Notice that the area of the kite (if the $2$ opposite angles are right) is $\frac{s_1 \cdot s_2}{2} \cdot 2$, where $s_1$ and $s_2$ denoting each of the 2 congruent sides. This just simplifies to $s_1 \cdot s_2$.
Hence, we have
\[4b+4b+b = 12\]
and
\[3a+3a+2a = 12\]
Solving for $a$ and $b$, we find that $a = \frac{3}{2}$ and $b = \frac{4}{3}$, so $a+b = \frac{3}{2} + \frac {4}{3} = \boxed{\textbf{(D)} ~\frac{17}6}$.
~MrThinker | // Block 1
draw((0,0)--(8,0));
draw((0,0)--(0,6));
draw((8,0)--(0,6));
draw((0,0)--(4,3));
label("A",(0,0),W);
label("B",(0,6),N);
label("C",(8,0),E);
label("D",(4,3),NE);
label("H",(2.3,4.2),NE);
label("K",(2.3,1.8),S);
draw(circle((1.54,3),1.49));
draw(circle((4,1.35),1.33));
dot((4,1.35));
dot((1.54,3));
label("F",(1.54,3),S);
label("J",(4,1.35),SW);
label("G",(0,3),W);
label("$x$",(1,3),S);
label("$y$",(4,1),E);
draw((1.54,3)--(0,3));
draw((1.54,3)--(2.3,1.8));
draw((1.54,3)--(2.3,4.2));
draw((4,1.35)--(4,0));
draw((4,1.35)--(3.12,2.4));
draw((4,1.35)--(4.8,2.3));
label("L",(4.9,2.4),NE);
label("E",(3.11,2.3),S);
label("I",(4,0),S);
// Block 2
draw((0,0)--(8,0)); draw((0,0)--(0,6)); draw((8,0)--(0,6)); draw((0,0)--(4,3)); label("A",(0,0),W); label("B",(0,6),N); label("C",(8,0),E); label("D",(4,3),NE); label("H",(2.3,4.2),NE); label("K",(2.3,1.8),S); draw(circle((1.54,3),1.49)); draw(circle((4,1.35),1.33)); dot((4,1.35)); dot((1.54,3)); label("F",(1.54,3),S); label("J",(4,1.35),SW); label("G",(0,3),W); label("$x$",(1,3),S); label("$y$",(4,1),E); draw((1.54,3)--(0,3)); draw((1.54,3)--(2.3,1.8)); draw((1.54,3)--(2.3,4.2)); draw((4,1.35)--(4,0)); draw((4,1.35)--(3.12,2.4)); draw((4,1.35)--(4.8,2.3)); label("L",(4.9,2.4),NE); label("E",(3.11,2.3),S); label("I",(4,0),S); | [] |
806 | The diameter $\overline{AB}$ of a circle of radius $2$ is extended to a point $D$ outside the circle so that $BD=3$. Point $E$ is chosen so that $ED=5$ and line $ED$ is perpendicular to line $AD$. Segment $\overline{AE}$ intersects the circle at a point $C$ between $A$ and $E$. What is the area of $\triangle ABC$?
$\textbf{(A)}\ \frac{120}{37}\qquad\textbf{(B)}\ \frac{140}{39}\qquad\textbf{(C)}\ \frac{145}{39}\qquad\textbf{(D)}\ \frac{140}{37}\qquad\textbf{(E)}\ \frac{120}{31}$ | 2017 AMC 10B Problem 22 | Notice that $ADE$ and $ABC$ are right triangles. Then $AE = \sqrt{7^2+5^2} = \sqrt{74}$. $\sin{DAE} = \frac{5}{\sqrt{74}} = \sin{BAE} = \sin{BAC} = \frac{BC}{4}$, so $BC = \frac{20}{\sqrt{74}}$. We also find that $AC = \frac{28}{\sqrt{74}}$ (You can also use power of point ~MATHWIZARD2010), and thus the area of $ABC$ is $\frac{\frac{20}{\sqrt{74}}\cdot\frac{28}{\sqrt{74}}}{2} = \frac{\frac{560}{74}}{2} = \boxed{\textbf{(D) } \frac{140}{37}}$. | // Block 1
size(10cm);
pair A, B, C, D, E, O;
A = (-2,0);
B = (2,0);
C = (2*cos(1.24),2*sin(1.24));
D = (5,0);
E = (5,5);
O = (A+B)/2;
dot(A);
dot(B);
dot(C);
dot(D);
dot(E);
dot(O);
draw(Circle((A+B)/2,2));
draw(A--D--E--C--A);
draw(C--B);
draw(rightanglemark(A,C,B,5));
draw(rightanglemark(A,D,E,5));
label("$A$",A,W);
label("$B$",B,SE);
label("$D$",D,SE);
label("$E$",E,NE);
label("$C$",C,N);
label("$2$",(O+B)/2,S);
label("$3$",(B+D)/2,S);
label("$5$",(D+E)/2,NE);
// Block 2
size(10cm); pair A, B, C, D, E, O; A = (-2,0); B = (2,0); C = (2*cos(1.24),2*sin(1.24)); D = (5,0); E = (5,5); O = (A+B)/2; dot(A); dot(B); dot(C); dot(D); dot(E); dot(O); draw(Circle((A+B)/2,2)); draw(A--D--E--C--A); draw(C--B); draw(rightanglemark(A,C,B,5)); draw(rightanglemark(A,D,E,5)); label("$A$",A,W); label("$B$",B,SE); label("$D$",D,SE); label("$E$",E,NE); label("$C$",C,N); label("$2$",(O+B)/2,S); label("$3$",(B+D)/2,S); label("$5$",(D+E)/2,NE); | [] |
807 | The diameter $AB$ of a circle of radius $2$ is extended to a point $D$ outside the circle so that $BD=3$. Point $E$ is chosen so that $ED=5$ and line $ED$ is perpendicular to line $AD$. Segment $AE$ intersects the circle at a point $C$ between $A$ and $E$. What is the area of $\triangle ABC$?
$\textbf{(A)}\ \frac{120}{37}\qquad\textbf{(B)}\ \frac{140}{39}\qquad\textbf{(C)}\ \frac{145}{39}\qquad\textbf{(D)}\ \frac{140}{37}\qquad\textbf{(E)}\ \frac{120}{31}$ | 2017 AMC 12B Problem 18 | Let $O$ be the center of the circle. Note that $EC + CA = EA = \sqrt{AD^2 + DE^2} = \sqrt{(2+2+3)^2 + 5^2} = \sqrt{74}$. However, by Power of a Point, $(EC)(EC + CA) = EO^2 - R^2 = (2+3)^2 + 5^2 - 2^2 = 25 + 25 - 4 = 46 \implies EC = \frac{46}{\sqrt{74}}$, so $AC = \sqrt{74} - \frac{46}{\sqrt{74}} = \frac{28}{\sqrt{74}}$. Now $BC = \sqrt{AB^2 - AC^2} = \sqrt{4^2 - \frac{28^2}{74}} = \sqrt{\frac{16 \cdot 74 - 28^2}{74}} = \sqrt{\frac{1184 - 784}{74}} = \frac{20}{\sqrt{74}}$. Since $\angle ACB = 90^{\circ}, [ABC] = \frac{1}{2} \cdot BC \cdot AC = \frac{1}{2} \cdot \frac{20}{\sqrt{74}} \cdot \frac{28}{\sqrt{74}} = \boxed{\textbf{(D)}\ \frac{140}{37}}$. | // Block 1
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(8.865514650638614cm);
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 = -6.36927122464312, xmax = 11.361758076634109, ymin = -3.789601803155515, ymax = 7.420015026296013; /* image dimensions */
draw((-2.,0.)--(0.6486486486486486,1.8918918918918919)--(2.,0.)--cycle);
/* draw figures */
draw(circle((0.,0.), 2.));
draw((-2.,0.)--(5.,5.));
draw((5.,5.)--(5.,0.));
draw((5.,0.)--(-2.,0.));
draw((-2.,0.)--(0.6486486486486486,1.8918918918918919));
draw((0.6486486486486486,1.8918918918918919)--(2.,0.));
draw((2.,0.)--(-2.,0.));
draw((2.,0.)--(5.,5.));
draw((0.,0.)--(5.,5.));
/* dots and labels */
dot((0.,0.),dotstyle);
label("$O$", (-0.10330578512396349,-0.39365890308038826), NE * labelscalefactor);
dot((-2.,0.),dotstyle);
label("$A$", (-2.2370398196844437,-0.42371149511645134), NE * labelscalefactor);
dot((2.,0.),dotstyle);
label("$B$", (2.045454545454548,-0.36360631104432517), NE * labelscalefactor);
dot((5.,0.),dotstyle);
label("$D$", (4.900450788880542,-0.42371149511645134), NE * labelscalefactor);
dot((5.,5.),dotstyle);
label("$E$", (5.06574004507889,5.15104432757325), NE * labelscalefactor);
dot((0.6486486486486486,1.8918918918918919),linewidth(3.pt) + dotstyle);
label("$C$", (0.48271975957926694,2.100706235912847), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
// Block 2
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(8.865514650638614cm); 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 = -6.36927122464312, xmax = 11.361758076634109, ymin = -3.789601803155515, ymax = 7.420015026296013; /* image dimensions */ draw((-2.,0.)--(0.6486486486486486,1.8918918918918919)--(2.,0.)--cycle); /* draw figures */ draw(circle((0.,0.), 2.)); draw((-2.,0.)--(5.,5.)); draw((5.,5.)--(5.,0.)); draw((5.,0.)--(-2.,0.)); draw((-2.,0.)--(0.6486486486486486,1.8918918918918919)); draw((0.6486486486486486,1.8918918918918919)--(2.,0.)); draw((2.,0.)--(-2.,0.)); draw((2.,0.)--(5.,5.)); draw((0.,0.)--(5.,5.)); /* dots and labels */ dot((0.,0.),dotstyle); label("$O$", (-0.10330578512396349,-0.39365890308038826), NE * labelscalefactor); dot((-2.,0.),dotstyle); label("$A$", (-2.2370398196844437,-0.42371149511645134), NE * labelscalefactor); dot((2.,0.),dotstyle); label("$B$", (2.045454545454548,-0.36360631104432517), NE * labelscalefactor); dot((5.,0.),dotstyle); label("$D$", (4.900450788880542,-0.42371149511645134), NE * labelscalefactor); dot((5.,5.),dotstyle); label("$E$", (5.06574004507889,5.15104432757325), NE * labelscalefactor); dot((0.6486486486486486,1.8918918918918919),linewidth(3.pt) + dotstyle); label("$C$", (0.48271975957926694,2.100706235912847), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ | [] |
808 | All of the triangles in the diagram below are similar to isosceles triangle $ABC$, in which $AB=AC$. Each of the $7$ smallest triangles has area $1,$ and $\triangle ABC$ has area $40$. What is the area of trapezoid $DBCE$?
$\textbf{(A) } 16 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 22 \qquad \textbf{(E) } 24$ | 2018 AMC 10A Problem 9 | Notice $\big[DBCE\big]=\big[ABC\big]-\big[ADE\big]$.
Let the base of the small triangles of area 1 be $x$, then the base length of $\Delta ADE=4x$. Notice, $\left(\frac{DE}{BC}\right)^2=\frac{1}{40}\implies \frac{x}{BC}=\frac{1}{\sqrt{40}}$, then $4x=\frac{4BC}{\sqrt{40}}\implies \big[ADE\big]=\left(\frac{4}{\sqrt{40}}\right)^2\cdot \big[ABC\big]=\frac{2}{5}\big[ABC\big]$
Thus, $\big[DBCE\big]=\big[ABC\big]-\big[ADE\big]=\big[ABC\big]\left(1-\frac{2}{5}\right)=\frac{3}{5}\cdot 40=\boxed{24}.$ | // Block 1
unitsize(5);
dot((0,0));
dot((60,0));
dot((50,10));
dot((10,10));
dot((30,30));
draw((0,0)--(60,0)--(50,10)--(30,30)--(10,10)--(0,0));
draw((10,10)--(50,10));
label("$B$",(0,0),SW);
label("$C$",(60,0),SE);
label("$E$",(50,10),E);
label("$D$",(10,10),W);
label("$A$",(30,30),N);
draw((10,10)--(15,15)--(20,10)--(25,15)--(30,10)--(35,15)--(40,10)--(45,15)--(50,10));
draw((15,15)--(45,15));
// Block 2
unitsize(5); dot((0,0)); dot((60,0)); dot((50,10)); dot((10,10)); dot((30,30)); draw((0,0)--(60,0)--(50,10)--(30,30)--(10,10)--(0,0)); draw((10,10)--(50,10)); label("$B$",(0,0),SW); label("$C$",(60,0),SE); label("$E$",(50,10),E); label("$D$",(10,10),W); label("$A$",(30,30),N); draw((10,10)--(15,15)--(20,10)--(25,15)--(30,10)--(35,15)--(40,10)--(45,15)--(50,10)); draw((15,15)--(45,15)); | [] |
809 | How many ordered pairs of real numbers $(x,y)$ satisfy the following system of equations?
\begin{align*} x+3y&=3 \\ \big||x|-|y|\big|&=1 \end{align*}
$\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 8$ | 2018 AMC 10A Problem 12 | We can solve this by graphing the equations. The second equation looks challenging to graph, but start by graphing it in the first quadrant only (which is easy since the inner absolute value signs can be ignored), then simply reflect that graph into the other quadrants.
The graph looks something like this:
Now, it becomes clear that there are $\boxed{\textbf{(C) } 3}$ intersection points.
(not the author)(the lines in the first quadrant are y=x+1 and y=x-1 and we reflect them into all quadrants as all instances of variables are in a context where flipping the sign does not affect the overall expression) | // Block 1
draw((-3,0)--(3,0), Arrows);
draw((0,-3)--(0,3), Arrows);
draw((2,3)--(0,1)--(-2,3), blue);
draw((-3,2)--(-1,0)--(-3,-2), blue);
draw((-2,-3)--(0,-1)--(2,-3), blue);
draw((3,-2)--(1,0)--(3,2), blue);
draw((-3,2)--(3,0), red);
dot((-3,2));
dot((3/2,1/2));
dot((0,1));
// Block 2
draw((-3,0)--(3,0), Arrows); draw((0,-3)--(0,3), Arrows); draw((2,3)--(0,1)--(-2,3), blue); draw((-3,2)--(-1,0)--(-3,-2), blue); draw((-2,-3)--(0,-1)--(2,-3), blue); draw((3,-2)--(1,0)--(3,2), blue); draw((-3,2)--(3,0), red); dot((-3,2)); dot((3/2,1/2)); dot((0,1)); | [] |
810 | A paper triangle with sides of lengths $3,4,$ and $5$ inches, as shown, is folded so that point $A$ falls on point $B$. What is the length in inches of the crease?
$\textbf{(A) } 1+\frac12 \sqrt2 \qquad \textbf{(B) } \sqrt3 \qquad \textbf{(C) } \frac74 \qquad \textbf{(D) } \frac{15}{8} \qquad \textbf{(E) } 2$ | 2018 AMC 10A Problem 13 | First, we need to realize that the crease line is just the perpendicular bisector of side $AB$, the hypotenuse of right triangle $\triangle ABC$. Call the midpoint of $AB$ point $D$. Draw this line and call the intersection point with $AC$ as $E$. Now, $\triangle ACB$ is similar to $\triangle ADE$ by $AA$ similarity. Setting up the ratios, we find that
\[\frac{BC}{AC}=\frac{DE}{AD} \Rightarrow \frac{3}{4}=\frac{DE}{\frac{5}{2}} \Rightarrow DE=\frac{15}{8}.\]
Thus, our answer is $\boxed{\textbf{(D) } \frac{15}{8}}$.
Note
In general, whenever we are asked to make a crease, think about that crease as a line of reflection over which the diagram is reflected. This is why the crease must be the perpendicular bisector of $AB$, because $A$ must be reflected onto $B$. | // Block 1
draw((0,0)--(4,0)--(4,3)--(0,0)); label("$A$", (0,0), SW); label("$B$", (4,3), NE); label("$C$", (4,0), SE); label("$D$", (2,1.5), NW); label("$E$", (3.125,0), S);
draw ((2,1.5)--(3.125,0),linewidth(1));
draw(rightanglemark((0,0),(2,1.5),(3.125,0)));
// Block 2
draw((0,0)--(4,0)--(4,3)--(0,0)); label("$A$", (0,0), SW); label("$B$", (4,3), NE); label("$C$", (4,0), SE); label("$D$", (2,1.5), NW); label("$E$", (3.125,0), S); draw ((2,1.5)--(3.125,0),linewidth(1)); draw(rightanglemark((0,0),(2,1.5),(3.125,0))); | [] |
810 | A paper triangle with sides of lengths $3,4,$ and $5$ inches, as shown, is folded so that point $A$ falls on point $B$. What is the length in inches of the crease?
$\textbf{(A) } 1+\frac12 \sqrt2 \qquad \textbf{(B) } \sqrt3 \qquad \textbf{(C) } \frac74 \qquad \textbf{(D) } \frac{15}{8} \qquad \textbf{(E) } 2$ | 2018 AMC 10A Problem 13 | Since $\triangle ABC$ is a right triangle, we can see that the slope of line $AB$ is $\frac{BC}{AC}$ = $\frac{3}{4}$. We know that if we fold $\triangle ABC$ so that point $A$ meets point $B$ the crease line will be perpendicular to $AB$ and we also know that the slopes of perpendicular lines are negative reciprocals of one another. Then, we can see that the slope of our crease line is $-\frac{4}{3}$.
Let us call the midpoint of $AB$ point $D$, the midpoint of $AC$ point $E$, and the crease line $DF$. We know that $DE$ is parallel to $BC$ and that $DE$'s length is $\frac{BC}{2}=\frac{3}{2}$. Using our slope calculation from earlier, we can see that$-\frac{DE}{EF}=-\frac{\frac{3}{2}}{EF}=-\frac{4}{3}$. With this information, we can solve for $EF$:
\[-4EF=(-\frac{3}{2})(3) \Rightarrow -4EF=-\frac{9}{2} \Rightarrow 4EF=\frac{9}{2} \Rightarrow EF=\frac{9}{8}.\]
We can then use the Pythagorean Theorem to find $DF$.
\[\frac{3}{2}^2+\frac{9}{8}^2=DF^2 \Rightarrow \frac{9}{4}+\frac{9\cdot9}{8\cdot8}=DF^2 \Rightarrow \frac{9\cdot2\cdot8}{4\cdot2\cdot8}+\frac{9\cdot9}{8\cdot8}=DF^2 \Rightarrow \frac{9\cdot2\cdot8+9\cdot9}{8\cdot8}=DF^2 \Rightarrow \frac{9(2\cdot8+9)}{8\cdot8}=DF^2\]
\[\Rightarrow DF=\sqrt{\frac{9(2\cdot8+9)}{8\cdot8}} \Rightarrow DF=\frac{3\cdot5}{8} \Rightarrow DF=\frac{15}{8}\]
Thus, our answer is $\boxed{\textbf{(D) } \frac{15}{8}}$.
~ (minor edits) <B+ | // Block 1
pen dotstyle = black;
draw((0,0)--(4,0)--(4,3)--(0,0));
fill(origin--(0,0)--(4,3)--(4,0)--cycle, gray);
dot((0,0),dotstyle);
label("$A$", (0.03153837092244126,0.07822624343603715), SW);
dot((4,0),dotstyle);
label("$C$", (4.028913881471271,0.07822624343603715), SE);
dot((4,3),dotstyle);
label("$B$", (4.028913881471271,3.078221223847919), NE);
dot((2,1.5),dotstyle);
label("$D$", (2.0341528211973956,1.578223733641978), SE);
dot((2,0),dotstyle);
label("$E$", (2.0341528211973956,0.07822624343603715), NE);
dot((3.1249518689638927,0),dotstyle);
label("$F$", (3.1571875913515854,0.07822624343603715), NE);
label("$4$", (2,0), S);
label("$3$", (4,1.5), E);
label("$5$", (2,1.5), NW);
// Block 2
pen dotstyle = black; draw((0,0)--(4,0)--(4,3)--(0,0)); fill(origin--(0,0)--(4,3)--(4,0)--cycle, gray); dot((0,0),dotstyle); label("$A$", (0.03153837092244126,0.07822624343603715), SW); dot((4,0),dotstyle); label("$C$", (4.028913881471271,0.07822624343603715), SE); dot((4,3),dotstyle); label("$B$", (4.028913881471271,3.078221223847919), NE); dot((2,1.5),dotstyle); label("$D$", (2.0341528211973956,1.578223733641978), SE); dot((2,0),dotstyle); label("$E$", (2.0341528211973956,0.07822624343603715), NE); dot((3.1249518689638927,0),dotstyle); label("$F$", (3.1571875913515854,0.07822624343603715), NE); label("$4$", (2,0), S); label("$3$", (4,1.5), E); label("$5$", (2,1.5), NW); | [] |
810 | A paper triangle with sides of lengths $3,4,$ and $5$ inches, as shown, is folded so that point $A$ falls on point $B$. What is the length in inches of the crease?
$\textbf{(A) } 1+\frac12 \sqrt2 \qquad \textbf{(B) } \sqrt3 \qquad \textbf{(C) } \frac74 \qquad \textbf{(D) } \frac{15}{8} \qquad \textbf{(E) } 2$ | 2018 AMC 10A Problem 13 | Like in Solution 3, we can make use of coordinate geometry to solve this problem. Because the length of the crease is constant no matter the positioning of the triangle, reorient the triangle so it has C at the origin and B and A at $(0,3)$ and $(4,0)$ respectively.
We know that each point in the crease is equidistant from $A$ and $B$, so the crease must pass through the midpoint of $AB$, which is $({2,\frac{3}{2}})$ , and be perpendicular to hypotenuse $AB$. The crease therefore has a slope of $\frac{4}{3}$.
Plugging into point-slope we find that the equation of the crease is, $y-y_1=m(x-x_1)\implies y-\frac{3}{2}=\frac{4}{3}(x-2)\implies y=\frac{4}{3}x-\frac{7}{6}$
Using 0 for y, we see that the crease intersects $AC$ at $(\frac{7}{8},0)$.
By the Distance Formula,
$(2-\frac{7}{8})^2+({\frac{3}{2}})^2=d^2 \implies d^2=\frac{225}{64} \implies d=\boxed{\textbf{(D) }\frac{15}{8}}$ | // Block 1
draw((0,0)--(0,3)--(4,0)--(0,0));
label("$C$", (0,0), SW); label("$B$", (0,3), NW); label("$A$", (4,0), SE);
draw ((2,1.5)--(0.875,0),blue+linewidth(1));
label("$(\frac{7}{8},0)$", (0.875,0), S); label("$(2,\frac{3}{2})$", (2,1.5), NE);
draw(rightanglemark((4,0),(2,1.5),(0.875,0)));
// Block 2
draw((0,0)--(0,3)--(4,0)--(0,0)); label("$C$", (0,0), SW); label("$B$", (0,3), NW); label("$A$", (4,0), SE); draw ((2,1.5)--(0.875,0),blue+linewidth(1)); label("$(\frac{7}{8},0)$", (0.875,0), S); label("$(2,\frac{3}{2})$", (2,1.5), NE); draw(rightanglemark((4,0),(2,1.5),(0.875,0))); | [] |
811 | Two circles of radius $5$ are externally tangent to each other and are internally tangent to a circle of radius $13$ at points $A$ and $B$, as shown in the diagram. The distance $AB$ can be written in the form $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$?
$\textbf{(A) } 21 \qquad \textbf{(B) } 29 \qquad \textbf{(C) } 58 \qquad \textbf{(D) } 69 \qquad \textbf{(E) } 93$ | 2018 AMC 10A Problem 15 | Let the center of the surrounding circle be $X$. The circle that is tangent at point $A$ will have point $Y$ as the center. Similarly, the circle that is tangent at point $B$ will have point $Z$ as the center. Connect $AB$, $YZ$, $XA$, and $XB$. Now observe that $\triangle XYZ$ is similar to $\triangle XAB$ by SAS.
Writing out the ratios, we get
\[\frac{XY}{XA}=\frac{YZ}{AB} \Rightarrow \frac{13-5}{13}=\frac{5+5}{AB} \Rightarrow \frac{8}{13}=\frac{10}{AB} \Rightarrow AB=\frac{65}{4}.\]
Therefore, our answer is $65+4= \boxed{\textbf{(D) } 69}$. | // Block 1
draw(circle((0,0),13));
draw(circle((5,-6.25),5));
draw(circle((-5,-6.25),5));
label("$A$", (-8.125,-10.15), S);
label("$B$", (8.125,-10.15), S);
draw((0,0)--(-8.125,-10.15));
draw((0,0)--(8.125,-10.15));
draw((-5,-6.25)--(5,-6.25));
draw((-8.125,-10.15)--(8.125,-10.15));
label("$X$", (0,0), N);
label("$Y$", (-5,-6.25),NW);
label("$Z$", (5,-6.25),NE);
// Block 2
draw(circle((0,0),13)); draw(circle((5,-6.25),5)); draw(circle((-5,-6.25),5)); label("$A$", (-8.125,-10.15), S); label("$B$", (8.125,-10.15), S); draw((0,0)--(-8.125,-10.15)); draw((0,0)--(8.125,-10.15)); draw((-5,-6.25)--(5,-6.25)); draw((-8.125,-10.15)--(8.125,-10.15)); label("$X$", (0,0), N); label("$Y$", (-5,-6.25),NW); label("$Z$", (5,-6.25),NE); | [] |
812 | Right triangle $ABC$ has leg lengths $AB=20$ and $BC=21$. Including $\overline{AB}$ and $\overline{BC}$, how many line segments with integer length can be drawn from vertex $B$ to a point on hypotenuse $\overline{AC}$?
$\textbf{(A) }5 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }13 \qquad \textbf{(E) }15 \qquad$ | 2018 AMC 10A Problem 16 | As the problem has no diagram, we draw a diagram. The hypotenuse has length $29$. Let $P$ be the foot of the altitude from $B$ to $AC$. Note that $BP$ is the shortest possible length of any segment. Writing the area of the triangle in two ways, we can solve for $BP=\dfrac{20\cdot 21}{29}$, which is between $14$ and $15$.
Let the line segment be $BX$, with $X$ on $AC$. As you move $X$ along the hypotenuse from $A$ to $P$, the length of $BX$ strictly decreases, hitting all the integer values from $20, 19, \dots 15$ (IVT). Similarly, moving $X$ from $P$ to $C$ hits all the integer values from $15, 16, \dots, 21$. This is a total of $\boxed{(D) 13}$ distinct line segments.
(asymptote diagram added by elements2015) | // Block 1
unitsize(4);
pair A, B, C, E, P;
A=(-20, 0);
B=origin;
C=(0,21);
E=(-21, 20);
P=extension(B,E, A, C);
draw(A--B--C--cycle);
draw(B--P);
dot("$A$", A, SW);
dot("$B$", B, SE);
dot("$C$", C, NE);
dot("$P$", P, NW);
// Block 2
unitsize(4); pair A, B, C, E, P; A=(-20, 0); B=origin; C=(0,21); E=(-21, 20); P=extension(B,E, A, C); draw(A--B--C--cycle); draw(B--P); dot("$A$", A, SW); dot("$B$", B, SE); dot("$C$", C, NE); dot("$P$", P, NW); | [] |
812 | Right triangle $ABC$ has leg lengths $AB=20$ and $BC=21$. Including $\overline{AB}$ and $\overline{BC}$, how many line segments with integer length can be drawn from vertex $B$ to a point on hypotenuse $\overline{AC}$?
$\textbf{(A) }5 \qquad \textbf{(B) }8 \qquad \textbf{(C) }12 \qquad \textbf{(D) }13 \qquad \textbf{(E) }15 \qquad$ | 2018 AMC 10A Problem 16 | Note that if a circle with an integer radius $r$ centered at vertex $B$ intersects hypotenuse $\overline{AB}$, the lines drawn from $B$ to the points of intersection are integer lengths. As in the previous solution, the shortest distance $14<\overline{BP}<15$. As a result, a circle of $14$ will not reach the hypotenuse and thus does not intersect it. We also know that a circle of radius $21$ intersects the hypotenuse once and a circle of radius $\{15, 16, 17, 18, 19, 20 \}$ intersects the hypotenuse twice. Quick graphical thinking or Euclidean construction will prove this.
It follows that we can draw circles of radii $15, 16, 17, 18, 19,$ and $20,$ that each contribute two integer lengths (since these circles intersect the hypotenuse twice) from $B$ to $\overline{AC}$ and one circle of radius $21$ that contributes only one such segment. Our answer is then \[6 \cdot 2 + 1 = 13 \implies \boxed{D}\] ~samrocksnature | // Block 1
unitsize(4);
pair A, B, C, E, P;
A=(-20, 0);
B=origin;
C=(0,21);
E=(-21, 20);
P=extension(B,E, A, C);
draw(A--B--C--cycle);
draw(B--P);
dot("$A$", A, SW);
dot("$B$", B, SE);
dot("$C$", C, NE);
dot("$P$", P, S);
draw(arc((0,0),21, 90, 180));
draw(arc((0,0),20, 90, 180));
draw(arc((0,0),19, 90, 180));
draw(arc((0,0),18, 90, 180));
draw(arc((0,0),17, 90, 180));
draw(arc((0,0),16, 90, 180));
draw(arc((0,0),15, 90, 180));
// Block 2
unitsize(4); pair A, B, C, E, P; A=(-20, 0); B=origin; C=(0,21); E=(-21, 20); P=extension(B,E, A, C); draw(A--B--C--cycle); draw(B--P); dot("$A$", A, SW); dot("$B$", B, SE); dot("$C$", C, NE); dot("$P$", P, S); draw(arc((0,0),21, 90, 180)); draw(arc((0,0),20, 90, 180)); draw(arc((0,0),19, 90, 180)); draw(arc((0,0),18, 90, 180)); draw(arc((0,0),17, 90, 180)); draw(arc((0,0),16, 90, 180)); draw(arc((0,0),15, 90, 180)); | [] |
813 | A scanning code consists of a $7 \times 7$ grid of squares, with some of its squares colored black and the rest colored white. There must be at least one square of each color in this grid of $49$ squares. A scanning code is called $\textit{symmetric}$ if its look does not change when the entire square is rotated by a multiple of $90 ^{\circ}$ counterclockwise around its center, nor when it is reflected across a line joining opposite corners or a line joining midpoints of opposite sides. What is the total number of possible symmetric scanning codes?
$\textbf{(A)} \text{ 510} \qquad \textbf{(B)} \text{ 1022} \qquad \textbf{(C)} \text{ 8190} \qquad \textbf{(D)} \text{ 8192} \qquad \textbf{(E)} \text{ 65,534}$ | 2018 AMC 10A Problem 20 | Imagine folding the scanning code along its lines of symmetry. There will be $10$ regions which you have control over coloring. Since we must subtract off $2$ cases for the all-black and all-white cases, the answer is $2^{10}-2=\boxed{\textbf{(B) } 1022.}$ | // Block 1
size(100pt);
draw((1,0)--(8,0),linewidth(0.5));
draw((1,2)--(6,2),linewidth(0.5));
draw((1,4)--(4,4),linewidth(0.5));
draw((1,6)--(2,6),linewidth(0.5));
draw((2,6)--(2,0),linewidth(0.5));
draw((4,4)--(4,0),linewidth(0.5));
draw((6,2)--(6,0),linewidth(0.5));
draw((1,0)--(1,7),dashed+linewidth(0.5));
draw((1,7)--(8,0),dashed+linewidth(0.5));
// Block 2
size(100pt); draw((1,0)--(8,0),linewidth(0.5)); draw((1,2)--(6,2),linewidth(0.5)); draw((1,4)--(4,4),linewidth(0.5)); draw((1,6)--(2,6),linewidth(0.5)); draw((2,6)--(2,0),linewidth(0.5)); draw((4,4)--(4,0),linewidth(0.5)); draw((6,2)--(6,0),linewidth(0.5)); draw((1,0)--(1,7),dashed+linewidth(0.5)); draw((1,7)--(8,0),dashed+linewidth(0.5)); | [] |
814 | Which of the following describes the set of values of $a$ for which the curves $x^2+y^2=a^2$ and $y=x^2-a$ in the real $xy$-plane intersect at exactly $3$ points?
$\textbf{(A) }a=\frac14 \qquad \textbf{(B) }\frac14 < a < \frac12 \qquad \textbf{(C) }a>\frac14 \qquad \textbf{(D) }a=\frac12 \qquad \textbf{(E) }a>\frac12 \qquad$ | 2018 AMC 10A Problem 21 | Looking at a graph, it is obvious that the two curves intersect at $(0, -a)$. We also see that if the parabola goes "in" the circle, then by going out of it (as it will), it will intersect five times. This is impossible. Thus, we only look for cases where the parabola becomes externally tangent to the circle.
We have $x^2 - a = -\sqrt{a^2 - x^2}$. Squaring both sides and solving yields $x^4 - (2a - 1)x^2 = 0$. Since $x = 0$ is already accounted for, we only need to find one solution for $x^2 = 2a - 1$, where the right hand side portion is obviously increasing. Since $a = \frac{1}{2}$ begets $x = 0$ (an overcount), we have $\boxed{\textbf{(E) }a>\frac12}$ as the right answer.
Solution by JohnHankock | // Block 1
Label f;
f.p=fontsize(6);
xaxis(-2,2,Ticks(f, 0.2));
yaxis(-2,2,Ticks(f, 0.2));
real g(real x)
{
return x^2-1;
}
draw(graph(g, 1.7, -1.7));
real h(real x)
{
return sqrt(1-x^2);
}
draw(graph(h, 1, -1));
real j(real x)
{
return -sqrt(1-x^2);
}
draw(graph(j, 1, -1));
// Block 2
Label f; f.p=fontsize(6); xaxis(-2,2,Ticks(f, 0.2)); yaxis(-2,2,Ticks(f, 0.2)); real g(real x) { return x^2-1; } draw(graph(g, 1.7, -1.7)); real h(real x) { return sqrt(1-x^2); } draw(graph(h, 1, -1)); real j(real x) { return -sqrt(1-x^2); } draw(graph(j, 1, -1)); | [] |
814 | Which of the following describes the set of values of $a$ for which the curves $x^2+y^2=a^2$ and $y=x^2-a$ in the real $xy$-plane intersect at exactly $3$ points?
$\textbf{(A) }a=\frac14 \qquad \textbf{(B) }\frac14 < a < \frac12 \qquad \textbf{(C) }a>\frac14 \qquad \textbf{(D) }a=\frac12 \qquad \textbf{(E) }a>\frac12 \qquad$ | 2018 AMC 10A Problem 21 | Now, let's graph these two equations. We want the blue parabola to be inside this red circle.
Then we substitute $y$ into the first equation to get $x^2+(x^2-a)^2=a^2$. Expanding, we get $x^4-2ax^2+x^2=0$. Factoring out the $x^2$, we get $x^2(x^2-2a+1)=0$. Then we find that $x=0$ or $x=\pm\sqrt{2a-1}$. Therefore, $2a-1>0$, which means $\boxed{\textbf{(E) }a>\frac12}$.
- kante314 -
~ minor edit by junokim1011 and RJ5303707 | // Block 1
import graph;
size(6cm);
draw((0,0)--(0,10),EndArrow);
draw((0,0)--(0,-10),EndArrow);
draw((0,0)--(10,0),EndArrow);
draw((0,0)--(-10,0),EndArrow);
Label f;
f.p=fontsize(6);
xaxis(-10,10);
yaxis(-10,10);
real f(real x)
{
return x^2-5;
}
draw(graph(f,-4,4),blue+linewidth(1));
draw(circle((0,0),5),red);
dot(scale(.7)*"$a$",(0,5),NE);
dot(scale(.7)*"$-a$",(0,-5),N);
dot(scale(.7)*"$a$",(5,0),NE);
dot(scale(.7)*"$-a$",(-5,0),SE);
// Block 2
import graph; size(6cm); draw((0,0)--(0,10),EndArrow); draw((0,0)--(0,-10),EndArrow); draw((0,0)--(10,0),EndArrow); draw((0,0)--(-10,0),EndArrow); Label f; f.p=fontsize(6); xaxis(-10,10); yaxis(-10,10); real f(real x) { return x^2-5; } draw(graph(f,-4,4),blue+linewidth(1)); draw(circle((0,0),5),red); dot(scale(.7)*"$a$",(0,5),NE); dot(scale(.7)*"$-a$",(0,-5),N); dot(scale(.7)*"$a$",(5,0),NE); dot(scale(.7)*"$-a$",(-5,0),SE); | [] |
815 | Let $a, b, c,$ and $d$ be positive integers such that $\gcd(a, b)=24$, $\gcd(b, c)=36$, $\gcd(c, d)=54$, and $70<\gcd(d, a)<100$. Which of the following must be a divisor of $a$? (gcd means greatest common factor)
$\textbf{(A)} \text{ 5} \qquad \textbf{(B)} \text{ 7} \qquad \textbf{(C)} \text{ 11} \qquad \textbf{(D)} \text{ 13} \qquad \textbf{(E)} \text{ 17}$ | 2018 AMC 10A Problem 22 | The relationship of $a$, $b$, $c$, and $d$ is shown in the above diagram. $gcd(a,d)=2 \cdot 3 \cdot k$.
$70 < 6k < 100$, $12 \le k \le 16$, $k=\boxed{\textbf{(D) }13}$
Note that $gcd(b,c)=36$ is not required to solve the problem.
~isabelchen | // Block 1
//Variable Declarations
defaultpen(0.45);
size(200pt);
fontsize(15pt);
pair X, Y, Z, W;
real R;
path quad;
//Variable Definitions
R = 1;
X = R*dir(45);
Y = R*dir(135);
Z = R*dir(-135);
W = R*dir(-45);
quad = X--Y--Z--W--cycle;
//Diagram
draw(quad);
label("$b$",X,NE);
label("$a=2^3 \cdot 3 \cdot p$",Y,NW);
label("$d=2 \cdot 3^3 \cdot q$",Z,SW);
label("$c$",W,SE);
label("$2^3 \cdot 3$",X--Y);
label("$2^2 \cdot 3^2$",X--W);
label("$2 \cdot 3^3$",Z--W);
label("$2 \cdot 3 \cdot k$",Z--Y);
// Block 2
//Variable Declarations defaultpen(0.45); size(200pt); fontsize(15pt); pair X, Y, Z, W; real R; path quad; //Variable Definitions R = 1; X = R*dir(45); Y = R*dir(135); Z = R*dir(-135); W = R*dir(-45); quad = X--Y--Z--W--cycle; //Diagram draw(quad); label("$b$",X,NE); label("$a=2^3 \cdot 3 \cdot p$",Y,NW); label("$d=2 \cdot 3^3 \cdot q$",Z,SW); label("$c$",W,SE); label("$2^3 \cdot 3$",X--Y); label("$2^2 \cdot 3^2$",X--W); label("$2 \cdot 3^3$",Z--W); label("$2 \cdot 3 \cdot k$",Z--Y); | [] |
816 | Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths $3$ and $4$ units. In the corner where those sides meet at a right angle, he leaves a small unplanted square $S$ so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortest distance from $S$ to the hypotenuse is $2$ units. What fraction of the field is planted?
$\textbf{(A) } \frac{25}{27} \qquad \textbf{(B) } \frac{26}{27} \qquad \textbf{(C) } \frac{73}{75} \qquad \textbf{(D) } \frac{145}{147} \qquad \textbf{(E) } \frac{74}{75}$ | 2018 AMC 10A Problem 23 | Note that the hypotenuse of the field is $5,$ and the area of the field is $6.$ Let $x$ be the side-length of square $S.$
We partition the field into a red triangle, a yellow triangle, and a green triangle, as shown below:
Let the brackets denote areas. By area addition, we set up an equation for $x:$
\begin{align*} [\text{Red Triangle}]+[\text{Yellow Triangle}]+[\text{Green Triangle}]&=[\text{Field}] \\ \frac{3x}{2}+\frac{4x}{2}+\frac{5\cdot2}{2}&=6, \end{align*}
from which $x=\frac27.$ Therefore, the answer is \[\frac{[\text{Field}]-[S]}{[\text{Field}]}=\frac{6-x^2}{6}=\boxed{\textbf{(D) } \frac{145}{147}}.\]
~MRENTHUSIASM | // Block 1
/* Edited by MRENTHUSIASM */
size(180);
pair A, B, C, D, F;
A = origin;
B = (4,0);
C = (0,3);
D = (2/7,2/7);
F = foot(D,B,C);
fill(A--D--C--cycle, red);
fill(A--D--B--cycle, yellow);
fill(B--D--C--cycle, green);
draw(A--B--C--cycle);
label("$5$", midpoint(B--C), NE);
label("$4$", midpoint(A--B), S);
label("$3$", midpoint(A--C), W);
label("$2$", midpoint(D--F), SE);
label("$S$", midpoint(A--D));
label("$x$", midpoint((0,2/7)--D), N);
label("$x$", midpoint((2/7,0)--D), E);
draw((2/7,0)--D--(0,2/7));
draw(A--D^^B--D^^C--D, dashed);
draw(D--F, dashed);
// Block 2
/* Edited by MRENTHUSIASM */ size(180); pair A, B, C, D, F; A = origin; B = (4,0); C = (0,3); D = (2/7,2/7); F = foot(D,B,C); fill(A--D--C--cycle, red); fill(A--D--B--cycle, yellow); fill(B--D--C--cycle, green); draw(A--B--C--cycle); label("$5$", midpoint(B--C), NE); label("$4$", midpoint(A--B), S); label("$3$", midpoint(A--C), W); label("$2$", midpoint(D--F), SE); label("$S$", midpoint(A--D)); label("$x$", midpoint((0,2/7)--D), N); label("$x$", midpoint((2/7,0)--D), E); draw((2/7,0)--D--(0,2/7)); draw(A--D^^B--D^^C--D, dashed); draw(D--F, dashed); | [] |
816 | Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths $3$ and $4$ units. In the corner where those sides meet at a right angle, he leaves a small unplanted square $S$ so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortest distance from $S$ to the hypotenuse is $2$ units. What fraction of the field is planted?
$\textbf{(A) } \frac{25}{27} \qquad \textbf{(B) } \frac{26}{27} \qquad \textbf{(C) } \frac{73}{75} \qquad \textbf{(D) } \frac{145}{147} \qquad \textbf{(E) } \frac{74}{75}$ | 2018 AMC 10A Problem 23 | Let the square have side length $x$. Connect the upper-right vertex of square $S$ with the two vertices of the triangle's hypotenuse. This divides the triangle in several regions whose areas must add up to the area of the whole triangle, which is $6$.
Square $S$ has area $x^2$, and the two thin triangle regions have area $\dfrac{x(3-x)}{2}$ and $\dfrac{x(4-x)}{2}$. The final triangular region with the hypotenuse as its base and height $2$ has area $5$. Thus, we have \[x^2+\dfrac{x(3-x)}{2}+\dfrac{x(4-x)}{2}+5=6.\]
Solving gives $x=\dfrac{2}{7}$. The area of $S$ is $\dfrac{4}{49}$ and the desired ratio is $\dfrac{6-\tfrac{4}{49}}{6}=\boxed{\textbf{(D) } \frac{145}{147}}$.
Alternatively, once you get $x=\frac{2}{7}$, you can avoid computation by noticing that there is a denominator of $7$, so the answer must have a factor of $7$ in the denominator, which only $\frac{145}{147}$ does. | // Block 1
/* Edited by MRENTHUSIASM */
size(180);
pair A, B, C, D, F;
A = origin;
B = (4,0);
C = (0,3);
D = (2/7,2/7);
F = foot(D,B,C);
fill(A--(2/7,0)--D--(0,2/7)--cycle, lightgray);
draw(A--B--C--cycle);
label("$5$", midpoint(B--C), NE);
label("$4$", midpoint(A--B), S);
label("$3$", midpoint(A--C), W);
label("$2$", midpoint(D--F), SE);
label("$S$", midpoint(A--D));
label("$x$", midpoint((0,2/7)--D), N);
label("$x$", midpoint((2/7,0)--D), E);
draw((2/7,0)--D--(0,2/7));
draw(B--D^^C--D, dashed);
draw(D--F, dashed);
// Block 2
/* Edited by MRENTHUSIASM */ size(180); pair A, B, C, D, F; A = origin; B = (4,0); C = (0,3); D = (2/7,2/7); F = foot(D,B,C); fill(A--(2/7,0)--D--(0,2/7)--cycle, lightgray); draw(A--B--C--cycle); label("$5$", midpoint(B--C), NE); label("$4$", midpoint(A--B), S); label("$3$", midpoint(A--C), W); label("$2$", midpoint(D--F), SE); label("$S$", midpoint(A--D)); label("$x$", midpoint((0,2/7)--D), N); label("$x$", midpoint((2/7,0)--D), E); draw((2/7,0)--D--(0,2/7)); draw(B--D^^C--D, dashed); draw(D--F, dashed); | [] |
816 | Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths $3$ and $4$ units. In the corner where those sides meet at a right angle, he leaves a small unplanted square $S$ so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortest distance from $S$ to the hypotenuse is $2$ units. What fraction of the field is planted?
$\textbf{(A) } \frac{25}{27} \qquad \textbf{(B) } \frac{26}{27} \qquad \textbf{(C) } \frac{73}{75} \qquad \textbf{(D) } \frac{145}{147} \qquad \textbf{(E) } \frac{74}{75}$ | 2018 AMC 10A Problem 23 | On the diagram above, find two smaller triangles similar to the large one with side lengths $3$, $4$, and $5$; consequently, the segments with length $\frac{5}{2}$ and $\frac{10}{3}$.
With $\ell$ being the side length of the square, we need to find an expression for $\ell$. Using the hypotenuse, we can see that $\frac{3}{2}+\frac{8}{3}+\frac{5}{4}\ell+\frac{5}{3}\ell=5$. Simplifying, $\frac{35}{12}\ell=\frac{5}{6}$, or $\ell=\frac27$.
A different calculation would yield $\ell+\frac{3}{4}\ell+\frac{5}{2}=3$, so $\frac{7}{4}\ell=\frac{1}{2}$. In other words, $\ell=\frac{2}{7}$, while to check, $\ell+\frac{4}{3}\ell+\frac{10}{3}=4$. As such, $\frac{7}{3}\ell=\frac{2}{3}$, and $\ell=\frac{2}{7}$.
Finally, we get $A(\Square S)=\ell^2=\frac{4}{49}$, to finish. As a proportion of the triangle with area $6$, the answer would be $1-\frac{4}{49\cdot6}=1-\frac{2}{147}=\frac{145}{147}$, so $\boxed{\textbf{(D) } \frac{145}{147}}$ is correct. | // Block 1
/* Edited by MRENTHUSIASM */
size(180);
pair A, B, C, D, F;
A = origin;
B = (4,0);
C = (0,3);
D = (2/7,2/7);
F = foot(D,B,C);
fill(A--(2/7,0)--D--(0,2/7)--cycle, lightgray);
draw(A--B--C--cycle);
label("$4$", midpoint(A--B), S);
label("$3$", midpoint(A--C), W);
label("$2$", midpoint(D--F), SE);
label("$S$", midpoint(A--D));
label("$\ell$", midpoint((0,2/7)--D), N);
label("$\ell$", midpoint((2/7,0)--D), E);
label("$\ell$", midpoint((2/7,2/7+5/2)--(0,2/7+5/2)), S);
label("$\ell$", midpoint((2/7+10/3,2/7)--(2/7+10/3,0)), W);
label("$\frac{5}{2}$", midpoint((2/7,2/7+5/2)--D), E);
label("$\frac{10}{3}$", midpoint((2/7+10/3,2/7)--D), N);
draw((2/7,0)--D--(0,2/7));
draw((2/7,2/7+5/2)--D^^(2/7+10/3,2/7)--D, dashed);
draw((2/7,2/7+5/2)--(0,2/7+5/2)^^(2/7+10/3,2/7)--(2/7+10/3,0), dashed);
draw(D--F, dashed);
// Block 2
/* Edited by MRENTHUSIASM */ size(180); pair A, B, C, D, F; A = origin; B = (4,0); C = (0,3); D = (2/7,2/7); F = foot(D,B,C); fill(A--(2/7,0)--D--(0,2/7)--cycle, lightgray); draw(A--B--C--cycle); label("$4$", midpoint(A--B), S); label("$3$", midpoint(A--C), W); label("$2$", midpoint(D--F), SE); label("$S$", midpoint(A--D)); label("$\ell$", midpoint((0,2/7)--D), N); label("$\ell$", midpoint((2/7,0)--D), E); label("$\ell$", midpoint((2/7,2/7+5/2)--(0,2/7+5/2)), S); label("$\ell$", midpoint((2/7+10/3,2/7)--(2/7+10/3,0)), W); label("$\frac{5}{2}$", midpoint((2/7,2/7+5/2)--D), E); label("$\frac{10}{3}$", midpoint((2/7+10/3,2/7)--D), N); draw((2/7,0)--D--(0,2/7)); draw((2/7,2/7+5/2)--D^^(2/7+10/3,2/7)--D, dashed); draw((2/7,2/7+5/2)--(0,2/7+5/2)^^(2/7+10/3,2/7)--(2/7+10/3,0), dashed); draw(D--F, dashed); | [] |
816 | Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths $3$ and $4$ units. In the corner where those sides meet at a right angle, he leaves a small unplanted square $S$ so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortest distance from $S$ to the hypotenuse is $2$ units. What fraction of the field is planted?
$\textbf{(A) } \frac{25}{27} \qquad \textbf{(B) } \frac{26}{27} \qquad \textbf{(C) } \frac{73}{75} \qquad \textbf{(D) } \frac{145}{147} \qquad \textbf{(E) } \frac{74}{75}$ | 2018 AMC 10A Problem 23 | Let $AD=x$. Note that $\triangle DEF$ is a $3{-}4{-}5$ triangle, so $EF=\frac{5}{4}x$ and $FD=\frac{3}{4}x$. $BF=BD+FD=4-x+\frac{3}{4}x=4-\frac{1}{4}x$. We know that $GE$ is $2$ from the problem so $GF=2+\frac{5}{4}x$. $\triangle FGB$ is also a $3{-}4{-}5$ triangle with $GF:BF=3:5$. We now have $3\left(4-\frac{1}{4}x\right)=5\left(2+\frac{5}{4}x\right)$. Solving this equation, we get that $x=\frac{2}{7}$ so the area of $S$ is $\frac{4}{49}$. The area of the triangle is $\frac{3\cdot 4}{2}=6$ so the fraction of field that is unplanted is $\frac{\frac{4}{49}}{6}=\frac{2}{147}$. Thus, the fraction of the field that is planted is $1-\frac{2}{147}=\boxed{\textbf{(D) } \frac{145}{147}}$.
~Heavytoothpaste | // Block 1
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
import graph; size(180);
real labelscalefactor = 1.5; /* changes label-to-point distance */
// pen dps = linewidth(0.5) + fontsize(10);
// defaultpen(dps); /* default pen style */
// pen dotstyle = black; /* point style */
real xmin = -1.6030465381283199, xmax = 7.095084767820557, ymin = -1.3624649422453508, ymax = 4.065350676871526; /* image dimensions */
/* draw figures */
draw((0,0)--(0,3));
draw((0,0)--(4,0));
draw((4,0)--(0,3));
draw((0,0.2857142857142857)--(0.2857142857142857,0.2857142857142857));
draw((0.2857142857142857,0.2857142857142857)--(0.2857142857142857,0));
draw((0.07142857142857142,0)--(1.4857142857142858,1.885714285714286));
label("$A$",(0, 0),SW*labelscalefactor);
label("$B$",(4,0),SE*labelscalefactor);
label("$C$",(0, 3),N*labelscalefactor);
label("$D$",(0.2857142857142857,0),S*labelscalefactor);
label("$E$",(0.2857142857142857, 0.2857142857142857),E*labelscalefactor);
label("$F$",(0.0714285714, 0),S*labelscalefactor);
label("$G$", (1.49, 1.89), NE*labelscalefactor);
/* dots and labels */
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
// Block 2
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(180); real labelscalefactor = 1.5; /* changes label-to-point distance */ // pen dps = linewidth(0.5) + fontsize(10); // defaultpen(dps); /* default pen style */ // pen dotstyle = black; /* point style */ real xmin = -1.6030465381283199, xmax = 7.095084767820557, ymin = -1.3624649422453508, ymax = 4.065350676871526; /* image dimensions */ /* draw figures */ draw((0,0)--(0,3)); draw((0,0)--(4,0)); draw((4,0)--(0,3)); draw((0,0.2857142857142857)--(0.2857142857142857,0.2857142857142857)); draw((0.2857142857142857,0.2857142857142857)--(0.2857142857142857,0)); draw((0.07142857142857142,0)--(1.4857142857142858,1.885714285714286)); label("$A$",(0, 0),SW*labelscalefactor); label("$B$",(4,0),SE*labelscalefactor); label("$C$",(0, 3),N*labelscalefactor); label("$D$",(0.2857142857142857,0),S*labelscalefactor); label("$E$",(0.2857142857142857, 0.2857142857142857),E*labelscalefactor); label("$F$",(0.0714285714, 0),S*labelscalefactor); label("$G$", (1.49, 1.89), NE*labelscalefactor); /* dots and labels */ clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ | [] |
816 | Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths $3$ and $4$ units. In the corner where those sides meet at a right angle, he leaves a small unplanted square $S$ so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortest distance from $S$ to the hypotenuse is $2$ units. What fraction of the field is planted?
$\textbf{(A) } \frac{25}{27} \qquad \textbf{(B) } \frac{26}{27} \qquad \textbf{(C) } \frac{73}{75} \qquad \textbf{(D) } \frac{145}{147} \qquad \textbf{(E) } \frac{74}{75}$ | 2018 AMC 10A Problem 23 | Denote $A,B,C$ to be the three vertices of the triangular field. Also denote $A,M,D,N$ to be the vertices of the square $S$. Let $X$ be on $BC$ such that $AC\parallel DX$ and $Y$ be on $BC$ such that $AB\parallel DY$. Let $P$ and $Q$ be the foot of the altitudes from $X$ to $AC$ and from $Y$ to $AB$ respectively.
Note that $\triangle ABC \sim \triangle DYX \sim \triangle PXC \sim \triangle QBY$. Thus, $PC = x \cdot \frac34$ and $QB = x \cdot \frac43$, making
\begin{align*} DX &= 3-x-\dfrac{3}{4}x = 3-\dfrac{7}{4}x, \\ MQ &= 4-x-\dfrac{4}{3}x = 4-\dfrac{7}{3}x. \end{align*}
Also from the similarity ratio is the fact that $CX = \frac54 x$ and $BY = \frac53 x$, making
\[XY = 5 - \dfrac{5}{4}x - \dfrac{5}{3}x = 5 - \dfrac{35}{12}x.\]
Computing the area of $\triangle XDY$ in two ways gives an equation for $x$:
\begin{align*} \left(3-\dfrac{7}{4}x\right)\left(4-\dfrac{7}{3}x\right) &= 2\cdot \left(5 - \dfrac{35}{12}x\right) \\ 10-\dfrac{35}{6}x &= \dfrac{49}{12}x^2 - 14x + 12 \\ \dfrac{49}{12}x^2 - \dfrac{49}{6}x + 2 &= 0 \\ 49x^2 - 98x + 24 &= 0 \\ x&=\dfrac{2}{7} \text{ or } \dfrac{12}{7}. \end{align*}
But $x=\dfrac{12}{7}$ is extraneous. Thus, the area of square $S = x^2 = \dfrac{4}{49}$, making the portion of the field that is planted being \[1 - \dfrac{\tfrac{4}{49}}{6} = 1 - \dfrac{2}{147} = \boxed{\textbf{(D) } \frac{145}{147}}.\]
-Solution by sml1809 | // Block 1
size(240);
pair A, B, C, D, F, X, Y, P, Q, M, N;
A = origin; label(A, "$A$", SW);
B = (4,0); label(B, "$B$", S);
C = (0,3); label(C, "$C$", W);
D = (2/7,2/7); label(D, "$D$", NE);
F = foot(D,B,C); label(F, "$F$", NE);
X = (2/7,39/14); label(X, "$X$", NE, red);
Y = (76/21,2/7); label(Y, "$Y$", NE, red);
P = foot(X,A,C); label(P, "$P$", W, red);
Q = foot(Y,A,B); label(Q, "$Q$", S, red);
M = (2/7,0); label(M, "$M$", S);
N = (0,2/7); label(N, "$N$", W);
fill(A--(2/7,0)--D--(0,2/7)--cycle, lightgray);
draw(A--B--C--cycle);
draw((2/7,0)--D--(0,2/7));
label("$x$", midpoint(A--M), S);
label("$x$", midpoint(A--N), W);
label("$2$", midpoint(D--F), SE);
draw(D--F);
draw(D--X, red);
draw(D--Y, red);
draw(X--P, red);
draw(Y--Q, red);
// Block 2
size(240); pair A, B, C, D, F, X, Y, P, Q, M, N; A = origin; label(A, "$A$", SW); B = (4,0); label(B, "$B$", S); C = (0,3); label(C, "$C$", W); D = (2/7,2/7); label(D, "$D$", NE); F = foot(D,B,C); label(F, "$F$", NE); X = (2/7,39/14); label(X, "$X$", NE, red); Y = (76/21,2/7); label(Y, "$Y$", NE, red); P = foot(X,A,C); label(P, "$P$", W, red); Q = foot(Y,A,B); label(Q, "$Q$", S, red); M = (2/7,0); label(M, "$M$", S); N = (0,2/7); label(N, "$N$", W); fill(A--(2/7,0)--D--(0,2/7)--cycle, lightgray); draw(A--B--C--cycle); draw((2/7,0)--D--(0,2/7)); label("$x$", midpoint(A--M), S); label("$x$", midpoint(A--N), W); label("$2$", midpoint(D--F), SE); draw(D--F); draw(D--X, red); draw(D--Y, red); draw(X--P, red); draw(Y--Q, red); | [] |
817 | All of the triangles in the diagram below are similar to isosceles triangle $ABC$, in which $AB=AC$. Each of the $7$ smallest triangles has area $1,$ and $\triangle ABC$ has area $40$. What is the area of trapezoid $DBCE$?
$\textbf{(A) } 16 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 22 \qquad \textbf{(E) } 24$ | 2018 AMC 12A Problem 8 | Notice $\big[DBCE\big]=\big[ABC\big]-\big[ADE\big]$.
Let the base of the small triangles of area 1 be $x$, then the base length of $\Delta ADE=4x$. Notice, $\left(\frac{DE}{BC}\right)^2=\frac{1}{40}\implies \frac{x}{BC}=\frac{1}{\sqrt{40}}$, then $4x=\frac{4BC}{\sqrt{40}}\implies \big[ADE\big]=\left(\frac{4}{\sqrt{40}}\right)^2\cdot \big[ABC\big]=\frac{2}{5}\big[ABC\big]$
Thus, $\big[DBCE\big]=\big[ABC\big]-\big[ADE\big]=\big[ABC\big]\left(1-\frac{2}{5}\right)=\frac{3}{5}\cdot 40=\boxed{24}.$ | unitsize(5); dot((0,0)); dot((60,0)); dot((50,10)); dot((10,10)); dot((30,30)); draw((0,0)--(60,0)--(50,10)--(30,30)--(10,10)--(0,0)); draw((10,10)--(50,10)); label("$B$",(0,0),SW); label("$C$",(60,0),SE); label("$E$",(50,10),E); label("$D$",(10,10),W); label("$A$",(30,30),N); draw((10,10)--(15,15)--(20,10)--(25,15)--(30,10)--(35,15)--(40,10)--(45,15)--(50,10)); draw((15,15)--(45,15)); | [] |
818 | How many ordered pairs of real numbers $(x,y)$ satisfy the following system of equations?
\begin{align*} x+3y&=3 \\ \big||x|-|y|\big|&=1 \end{align*}
$\textbf{(A) } 1 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } 3 \qquad \textbf{(D) } 4 \qquad \textbf{(E) } 8$ | 2018 AMC 12A Problem 10 | We can solve this by graphing the equations. The second equation looks challenging to graph, but start by graphing it in the first quadrant only (which is easy since the inner absolute value signs can be ignored), then simply reflect that graph into the other quadrants.
The graph looks something like this:
Now, it becomes clear that there are $\boxed{\textbf{(C) } 3}$ intersection points.
(not the author)(the lines in the first quadrant are y=x+1 and y=x-1 and we reflect them into all quadrants as all instances of variables are in a context where flipping the sign does not affect the overall expression) | draw((-3,0)--(3,0), Arrows); draw((0,-3)--(0,3), Arrows); draw((2,3)--(0,1)--(-2,3), blue); draw((-3,2)--(-1,0)--(-3,-2), blue); draw((-2,-3)--(0,-1)--(2,-3), blue); draw((3,-2)--(1,0)--(3,2), blue); draw((-3,2)--(3,0), red); dot((-3,2)); dot((3/2,1/2)); dot((0,1)); | [] |
819 | A paper triangle with sides of lengths $3,4,$ and $5$ inches, as shown, is folded so that point $A$ falls on point $B$. What is the length in inches of the crease?
$\textbf{(A) } 1+\frac12 \sqrt2 \qquad \textbf{(B) } \sqrt3 \qquad \textbf{(C) } \frac74 \qquad \textbf{(D) } \frac{15}{8} \qquad \textbf{(E) } 2$ | 2018 AMC 12A Problem 11 | First, we need to realize that the crease line is just the perpendicular bisector of side $AB$, the hypotenuse of right triangle $\triangle ABC$. Call the midpoint of $AB$ point $D$. Draw this line and call the intersection point with $AC$ as $E$. Now, $\triangle ACB$ is similar to $\triangle ADE$ by $AA$ similarity. Setting up the ratios, we find that
\[\frac{BC}{AC}=\frac{DE}{AD} \Rightarrow \frac{3}{4}=\frac{DE}{\frac{5}{2}} \Rightarrow DE=\frac{15}{8}.\]
Thus, our answer is $\boxed{\textbf{(D) } \frac{15}{8}}$.
Note
In general, whenever we are asked to make a crease, think about that crease as a line of reflection over which the diagram is reflected. This is why the crease must be the perpendicular bisector of $AB$, because $A$ must be reflected onto $B$. | draw((0,0)--(4,0)--(4,3)--(0,0)); label("$A$", (0,0), SW); label("$B$", (4,3), NE); label("$C$", (4,0), SE); label("$D$", (2,1.5), NW); label("$E$", (3.125,0), S); draw ((2,1.5)--(3.125,0),linewidth(1)); draw(rightanglemark((0,0),(2,1.5),(3.125,0))); | [] |
819 | A paper triangle with sides of lengths $3,4,$ and $5$ inches, as shown, is folded so that point $A$ falls on point $B$. What is the length in inches of the crease?
$\textbf{(A) } 1+\frac12 \sqrt2 \qquad \textbf{(B) } \sqrt3 \qquad \textbf{(C) } \frac74 \qquad \textbf{(D) } \frac{15}{8} \qquad \textbf{(E) } 2$ | 2018 AMC 12A Problem 11 | Since $\triangle ABC$ is a right triangle, we can see that the slope of line $AB$ is $\frac{BC}{AC}$ = $\frac{3}{4}$. We know that if we fold $\triangle ABC$ so that point $A$ meets point $B$ the crease line will be perpendicular to $AB$ and we also know that the slopes of perpendicular lines are negative reciprocals of one another. Then, we can see that the slope of our crease line is $-\frac{4}{3}$.
Let us call the midpoint of $AB$ point $D$, the midpoint of $AC$ point $E$, and the crease line $DF$. We know that $DE$ is parallel to $BC$ and that $DE$'s length is $\frac{BC}{2}=\frac{3}{2}$. Using our slope calculation from earlier, we can see that$-\frac{DE}{EF}=-\frac{\frac{3}{2}}{EF}=-\frac{4}{3}$. With this information, we can solve for $EF$:
\[-4EF=(-\frac{3}{2})(3) \Rightarrow -4EF=-\frac{9}{2} \Rightarrow 4EF=\frac{9}{2} \Rightarrow EF=\frac{9}{8}.\]
We can then use the Pythagorean Theorem to find $DF$.
\[\frac{3}{2}^2+\frac{9}{8}^2=DF^2 \Rightarrow \frac{9}{4}+\frac{9\cdot9}{8\cdot8}=DF^2 \Rightarrow \frac{9\cdot2\cdot8}{4\cdot2\cdot8}+\frac{9\cdot9}{8\cdot8}=DF^2 \Rightarrow \frac{9\cdot2\cdot8+9\cdot9}{8\cdot8}=DF^2 \Rightarrow \frac{9(2\cdot8+9)}{8\cdot8}=DF^2\]
\[\Rightarrow DF=\sqrt{\frac{9(2\cdot8+9)}{8\cdot8}} \Rightarrow DF=\frac{3\cdot5}{8} \Rightarrow DF=\frac{15}{8}\]
Thus, our answer is $\boxed{\textbf{(D) } \frac{15}{8}}$.
~ (minor edits) <B+ | pen dotstyle = black; draw((0,0)--(4,0)--(4,3)--(0,0)); fill(origin--(0,0)--(4,3)--(4,0)--cycle, gray); dot((0,0),dotstyle); label("$A$", (0.03153837092244126,0.07822624343603715), SW); dot((4,0),dotstyle); label("$C$", (4.028913881471271,0.07822624343603715), SE); dot((4,3),dotstyle); label("$B$", (4.028913881471271,3.078221223847919), NE); dot((2,1.5),dotstyle); label("$D$", (2.0341528211973956,1.578223733641978), SE); dot((2,0),dotstyle); label("$E$", (2.0341528211973956,0.07822624343603715), NE); dot((3.1249518689638927,0),dotstyle); label("$F$", (3.1571875913515854,0.07822624343603715), NE); label("$4$", (2,0), S); label("$3$", (4,1.5), E); label("$5$", (2,1.5), NW); | [] |
819 | A paper triangle with sides of lengths $3,4,$ and $5$ inches, as shown, is folded so that point $A$ falls on point $B$. What is the length in inches of the crease?
$\textbf{(A) } 1+\frac12 \sqrt2 \qquad \textbf{(B) } \sqrt3 \qquad \textbf{(C) } \frac74 \qquad \textbf{(D) } \frac{15}{8} \qquad \textbf{(E) } 2$ | 2018 AMC 12A Problem 11 | Like in Solution 3, we can make use of coordinate geometry to solve this problem. Because the length of the crease is constant no matter the positioning of the triangle, reorient the triangle so it has C at the origin and B and A at $(0,3)$ and $(4,0)$ respectively.
We know that each point in the crease is equidistant from $A$ and $B$, so the crease must pass through the midpoint of $AB$, which is $({2,\frac{3}{2}})$ , and be perpendicular to hypotenuse $AB$. The crease therefore has a slope of $\frac{4}{3}$.
Plugging into point-slope we find that the equation of the crease is, $y-y_1=m(x-x_1)\implies y-\frac{3}{2}=\frac{4}{3}(x-2)\implies y=\frac{4}{3}x-\frac{7}{6}$
Using 0 for y, we see that the crease intersects $AC$ at $(\frac{7}{8},0)$.
By the Distance Formula,
$(2-\frac{7}{8})^2+({\frac{3}{2}})^2=d^2 \implies d^2=\frac{225}{64} \implies d=\boxed{\textbf{(D) }\frac{15}{8}}$ | draw((0,0)--(0,3)--(4,0)--(0,0)); label("$C$", (0,0), SW); label("$B$", (0,3), NW); label("$A$", (4,0), SE); draw ((2,1.5)--(0.875,0),blue+linewidth(1)); label("$(\frac{7}{8},0)$", (0.875,0), S); label("$(2,\frac{3}{2})$", (2,1.5), NE); draw(rightanglemark((4,0),(2,1.5),(0.875,0))); | [] |
820 | A scanning code consists of a $7 \times 7$ grid of squares, with some of its squares colored black and the rest colored white. There must be at least one square of each color in this grid of $49$ squares. A scanning code is called $\textit{symmetric}$ if its look does not change when the entire square is rotated by a multiple of $90 ^{\circ}$ counterclockwise around its center, nor when it is reflected across a line joining opposite corners or a line joining midpoints of opposite sides. What is the total number of possible symmetric scanning codes?
$\textbf{(A)} \text{ 510} \qquad \textbf{(B)} \text{ 1022} \qquad \textbf{(C)} \text{ 8190} \qquad \textbf{(D)} \text{ 8192} \qquad \textbf{(E)} \text{ 65,534}$ | 2018 AMC 12A Problem 15 | Imagine folding the scanning code along its lines of symmetry. There will be $10$ regions which you have control over coloring. Since we must subtract off $2$ cases for the all-black and all-white cases, the answer is $2^{10}-2=\boxed{\textbf{(B) } 1022.}$ | size(100pt); draw((1,0)--(8,0),linewidth(0.5)); draw((1,2)--(6,2),linewidth(0.5)); draw((1,4)--(4,4),linewidth(0.5)); draw((1,6)--(2,6),linewidth(0.5)); draw((2,6)--(2,0),linewidth(0.5)); draw((4,4)--(4,0),linewidth(0.5)); draw((6,2)--(6,0),linewidth(0.5)); draw((1,0)--(1,7),dashed+linewidth(0.5)); draw((1,7)--(8,0),dashed+linewidth(0.5)); | [] |
821 | Which of the following describes the set of values of $a$ for which the curves $x^2+y^2=a^2$ and $y=x^2-a$ in the real $xy$-plane intersect at exactly $3$ points?
$\textbf{(A) }a=\frac14 \qquad \textbf{(B) }\frac14 < a < \frac12 \qquad \textbf{(C) }a>\frac14 \qquad \textbf{(D) }a=\frac12 \qquad \textbf{(E) }a>\frac12 \qquad$ | 2018 AMC 12A Problem 16 | Looking at a graph, it is obvious that the two curves intersect at $(0, -a)$. We also see that if the parabola goes "in" the circle, then by going out of it (as it will), it will intersect five times. This is impossible. Thus, we only look for cases where the parabola becomes externally tangent to the circle.
We have $x^2 - a = -\sqrt{a^2 - x^2}$. Squaring both sides and solving yields $x^4 - (2a - 1)x^2 = 0$. Since $x = 0$ is already accounted for, we only need to find one solution for $x^2 = 2a - 1$, where the right hand side portion is obviously increasing. Since $a = \frac{1}{2}$ begets $x = 0$ (an overcount), we have $\boxed{\textbf{(E) }a>\frac12}$ as the right answer.
Solution by JohnHankock | Label f; f.p=fontsize(6); xaxis(-2,2,Ticks(f, 0.2)); yaxis(-2,2,Ticks(f, 0.2)); real g(real x) { return x^2-1; } draw(graph(g, 1.7, -1.7)); real h(real x) { return sqrt(1-x^2); } draw(graph(h, 1, -1)); real j(real x) { return -sqrt(1-x^2); } draw(graph(j, 1, -1)); | [] |
821 | Which of the following describes the set of values of $a$ for which the curves $x^2+y^2=a^2$ and $y=x^2-a$ in the real $xy$-plane intersect at exactly $3$ points?
$\textbf{(A) }a=\frac14 \qquad \textbf{(B) }\frac14 < a < \frac12 \qquad \textbf{(C) }a>\frac14 \qquad \textbf{(D) }a=\frac12 \qquad \textbf{(E) }a>\frac12 \qquad$ | 2018 AMC 12A Problem 16 | Now, let's graph these two equations. We want the blue parabola to be inside this red circle.
Then we substitute $y$ into the first equation to get $x^2+(x^2-a)^2=a^2$. Expanding, we get $x^4-2ax^2+x^2=0$. Factoring out the $x^2$, we get $x^2(x^2-2a+1)=0$. Then we find that $x=0$ or $x=\pm\sqrt{2a-1}$. Therefore, $2a-1>0$, which means $\boxed{\textbf{(E) }a>\frac12}$.
- kante314 -
~ minor edit by junokim1011 and RJ5303707 | import graph; size(6cm); draw((0,0)--(0,10),EndArrow); draw((0,0)--(0,-10),EndArrow); draw((0,0)--(10,0),EndArrow); draw((0,0)--(-10,0),EndArrow); Label f; f.p=fontsize(6); xaxis(-10,10); yaxis(-10,10); real f(real x) { return x^2-5; } draw(graph(f,-4,4),blue+linewidth(1)); draw(circle((0,0),5),red); dot(scale(.7)*"$a$",(0,5),NE); dot(scale(.7)*"$-a$",(0,-5),N); dot(scale(.7)*"$a$",(5,0),NE); dot(scale(.7)*"$-a$",(-5,0),SE); | [] |
822 | Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths $3$ and $4$ units. In the corner where those sides meet at a right angle, he leaves a small unplanted square $S$ so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortest distance from $S$ to the hypotenuse is $2$ units. What fraction of the field is planted?
$\textbf{(A) } \frac{25}{27} \qquad \textbf{(B) } \frac{26}{27} \qquad \textbf{(C) } \frac{73}{75} \qquad \textbf{(D) } \frac{145}{147} \qquad \textbf{(E) } \frac{74}{75}$ | 2018 AMC 12A Problem 17 | Note that the hypotenuse of the field is $5,$ and the area of the field is $6.$ Let $x$ be the side-length of square $S.$
We partition the field into a red triangle, a yellow triangle, and a green triangle, as shown below:
Let the brackets denote areas. By area addition, we set up an equation for $x:$
\begin{align*} [\text{Red Triangle}]+[\text{Yellow Triangle}]+[\text{Green Triangle}]&=[\text{Field}] \\ \frac{3x}{2}+\frac{4x}{2}+\frac{5\cdot2}{2}&=6, \end{align*}
from which $x=\frac27.$ Therefore, the answer is \[\frac{[\text{Field}]-[S]}{[\text{Field}]}=\frac{6-x^2}{6}=\boxed{\textbf{(D) } \frac{145}{147}}.\]
~MRENTHUSIASM | /* Edited by MRENTHUSIASM */ size(180); pair A, B, C, D, F; A = origin; B = (4,0); C = (0,3); D = (2/7,2/7); F = foot(D,B,C); fill(A--D--C--cycle, red); fill(A--D--B--cycle, yellow); fill(B--D--C--cycle, green); draw(A--B--C--cycle); label("$5$", midpoint(B--C), NE); label("$4$", midpoint(A--B), S); label("$3$", midpoint(A--C), W); label("$2$", midpoint(D--F), SE); label("$S$", midpoint(A--D)); label("$x$", midpoint((0,2/7)--D), N); label("$x$", midpoint((2/7,0)--D), E); draw((2/7,0)--D--(0,2/7)); draw(A--D^^B--D^^C--D, dashed); draw(D--F, dashed); | [] |
822 | Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths $3$ and $4$ units. In the corner where those sides meet at a right angle, he leaves a small unplanted square $S$ so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortest distance from $S$ to the hypotenuse is $2$ units. What fraction of the field is planted?
$\textbf{(A) } \frac{25}{27} \qquad \textbf{(B) } \frac{26}{27} \qquad \textbf{(C) } \frac{73}{75} \qquad \textbf{(D) } \frac{145}{147} \qquad \textbf{(E) } \frac{74}{75}$ | 2018 AMC 12A Problem 17 | Let the square have side length $x$. Connect the upper-right vertex of square $S$ with the two vertices of the triangle's hypotenuse. This divides the triangle in several regions whose areas must add up to the area of the whole triangle, which is $6$.
Square $S$ has area $x^2$, and the two thin triangle regions have area $\dfrac{x(3-x)}{2}$ and $\dfrac{x(4-x)}{2}$. The final triangular region with the hypotenuse as its base and height $2$ has area $5$. Thus, we have \[x^2+\dfrac{x(3-x)}{2}+\dfrac{x(4-x)}{2}+5=6.\]
Solving gives $x=\dfrac{2}{7}$. The area of $S$ is $\dfrac{4}{49}$ and the desired ratio is $\dfrac{6-\tfrac{4}{49}}{6}=\boxed{\textbf{(D) } \frac{145}{147}}$.
Alternatively, once you get $x=\frac{2}{7}$, you can avoid computation by noticing that there is a denominator of $7$, so the answer must have a factor of $7$ in the denominator, which only $\frac{145}{147}$ does. | /* Edited by MRENTHUSIASM */ size(180); pair A, B, C, D, F; A = origin; B = (4,0); C = (0,3); D = (2/7,2/7); F = foot(D,B,C); fill(A--(2/7,0)--D--(0,2/7)--cycle, lightgray); draw(A--B--C--cycle); label("$5$", midpoint(B--C), NE); label("$4$", midpoint(A--B), S); label("$3$", midpoint(A--C), W); label("$2$", midpoint(D--F), SE); label("$S$", midpoint(A--D)); label("$x$", midpoint((0,2/7)--D), N); label("$x$", midpoint((2/7,0)--D), E); draw((2/7,0)--D--(0,2/7)); draw(B--D^^C--D, dashed); draw(D--F, dashed); | [] |
822 | Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths $3$ and $4$ units. In the corner where those sides meet at a right angle, he leaves a small unplanted square $S$ so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortest distance from $S$ to the hypotenuse is $2$ units. What fraction of the field is planted?
$\textbf{(A) } \frac{25}{27} \qquad \textbf{(B) } \frac{26}{27} \qquad \textbf{(C) } \frac{73}{75} \qquad \textbf{(D) } \frac{145}{147} \qquad \textbf{(E) } \frac{74}{75}$ | 2018 AMC 12A Problem 17 | On the diagram above, find two smaller triangles similar to the large one with side lengths $3$, $4$, and $5$; consequently, the segments with length $\frac{5}{2}$ and $\frac{10}{3}$.
With $\ell$ being the side length of the square, we need to find an expression for $\ell$. Using the hypotenuse, we can see that $\frac{3}{2}+\frac{8}{3}+\frac{5}{4}\ell+\frac{5}{3}\ell=5$. Simplifying, $\frac{35}{12}\ell=\frac{5}{6}$, or $\ell=\frac27$.
A different calculation would yield $\ell+\frac{3}{4}\ell+\frac{5}{2}=3$, so $\frac{7}{4}\ell=\frac{1}{2}$. In other words, $\ell=\frac{2}{7}$, while to check, $\ell+\frac{4}{3}\ell+\frac{10}{3}=4$. As such, $\frac{7}{3}\ell=\frac{2}{3}$, and $\ell=\frac{2}{7}$.
Finally, we get $A(\Square S)=\ell^2=\frac{4}{49}$, to finish. As a proportion of the triangle with area $6$, the answer would be $1-\frac{4}{49\cdot6}=1-\frac{2}{147}=\frac{145}{147}$, so $\boxed{\textbf{(D) } \frac{145}{147}}$ is correct. | /* Edited by MRENTHUSIASM */ size(180); pair A, B, C, D, F; A = origin; B = (4,0); C = (0,3); D = (2/7,2/7); F = foot(D,B,C); fill(A--(2/7,0)--D--(0,2/7)--cycle, lightgray); draw(A--B--C--cycle); label("$4$", midpoint(A--B), S); label("$3$", midpoint(A--C), W); label("$2$", midpoint(D--F), SE); label("$S$", midpoint(A--D)); label("$\ell$", midpoint((0,2/7)--D), N); label("$\ell$", midpoint((2/7,0)--D), E); label("$\ell$", midpoint((2/7,2/7+5/2)--(0,2/7+5/2)), S); label("$\ell$", midpoint((2/7+10/3,2/7)--(2/7+10/3,0)), W); label("$\frac{5}{2}$", midpoint((2/7,2/7+5/2)--D), E); label("$\frac{10}{3}$", midpoint((2/7+10/3,2/7)--D), N); draw((2/7,0)--D--(0,2/7)); draw((2/7,2/7+5/2)--D^^(2/7+10/3,2/7)--D, dashed); draw((2/7,2/7+5/2)--(0,2/7+5/2)^^(2/7+10/3,2/7)--(2/7+10/3,0), dashed); draw(D--F, dashed); | [] |
822 | Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths $3$ and $4$ units. In the corner where those sides meet at a right angle, he leaves a small unplanted square $S$ so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortest distance from $S$ to the hypotenuse is $2$ units. What fraction of the field is planted?
$\textbf{(A) } \frac{25}{27} \qquad \textbf{(B) } \frac{26}{27} \qquad \textbf{(C) } \frac{73}{75} \qquad \textbf{(D) } \frac{145}{147} \qquad \textbf{(E) } \frac{74}{75}$ | 2018 AMC 12A Problem 17 | Let $AD=x$. Note that $\triangle DEF$ is a $3{-}4{-}5$ triangle, so $EF=\frac{5}{4}x$ and $FD=\frac{3}{4}x$. $BF=BD+FD=4-x+\frac{3}{4}x=4-\frac{1}{4}x$. We know that $GE$ is $2$ from the problem so $GF=2+\frac{5}{4}x$. $\triangle FGB$ is also a $3{-}4{-}5$ triangle with $GF:BF=3:5$. We now have $3\left(4-\frac{1}{4}x\right)=5\left(2+\frac{5}{4}x\right)$. Solving this equation, we get that $x=\frac{2}{7}$ so the area of $S$ is $\frac{4}{49}$. The area of the triangle is $\frac{3\cdot 4}{2}=6$ so the fraction of field that is unplanted is $\frac{\frac{4}{49}}{6}=\frac{2}{147}$. Thus, the fraction of the field that is planted is $1-\frac{2}{147}=\boxed{\textbf{(D) } \frac{145}{147}}$.
~Heavytoothpaste | /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(180); real labelscalefactor = 1.5; /* changes label-to-point distance */ // pen dps = linewidth(0.5) + fontsize(10); // defaultpen(dps); /* default pen style */ // pen dotstyle = black; /* point style */ real xmin = -1.6030465381283199, xmax = 7.095084767820557, ymin = -1.3624649422453508, ymax = 4.065350676871526; /* image dimensions */ /* draw figures */ draw((0,0)--(0,3)); draw((0,0)--(4,0)); draw((4,0)--(0,3)); draw((0,0.2857142857142857)--(0.2857142857142857,0.2857142857142857)); draw((0.2857142857142857,0.2857142857142857)--(0.2857142857142857,0)); draw((0.07142857142857142,0)--(1.4857142857142858,1.885714285714286)); label("$A$",(0, 0),SW*labelscalefactor); label("$B$",(4,0),SE*labelscalefactor); label("$C$",(0, 3),N*labelscalefactor); label("$D$",(0.2857142857142857,0),S*labelscalefactor); label("$E$",(0.2857142857142857, 0.2857142857142857),E*labelscalefactor); label("$F$",(0.0714285714, 0),S*labelscalefactor); label("$G$", (1.49, 1.89), NE*labelscalefactor); /* dots and labels */ clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ | [] |
822 | Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths $3$ and $4$ units. In the corner where those sides meet at a right angle, he leaves a small unplanted square $S$ so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortest distance from $S$ to the hypotenuse is $2$ units. What fraction of the field is planted?
$\textbf{(A) } \frac{25}{27} \qquad \textbf{(B) } \frac{26}{27} \qquad \textbf{(C) } \frac{73}{75} \qquad \textbf{(D) } \frac{145}{147} \qquad \textbf{(E) } \frac{74}{75}$ | 2018 AMC 12A Problem 17 | Denote $A,B,C$ to be the three vertices of the triangular field. Also denote $A,M,D,N$ to be the vertices of the square $S$. Let $X$ be on $BC$ such that $AC\parallel DX$ and $Y$ be on $BC$ such that $AB\parallel DY$. Let $P$ and $Q$ be the foot of the altitudes from $X$ to $AC$ and from $Y$ to $AB$ respectively.
Note that $\triangle ABC \sim \triangle DYX \sim \triangle PXC \sim \triangle QBY$. Thus, $PC = x \cdot \frac34$ and $QB = x \cdot \frac43$, making
\begin{align*} DX &= 3-x-\dfrac{3}{4}x = 3-\dfrac{7}{4}x, \\ MQ &= 4-x-\dfrac{4}{3}x = 4-\dfrac{7}{3}x. \end{align*}
Also from the similarity ratio is the fact that $CX = \frac54 x$ and $BY = \frac53 x$, making
\[XY = 5 - \dfrac{5}{4}x - \dfrac{5}{3}x = 5 - \dfrac{35}{12}x.\]
Computing the area of $\triangle XDY$ in two ways gives an equation for $x$:
\begin{align*} \left(3-\dfrac{7}{4}x\right)\left(4-\dfrac{7}{3}x\right) &= 2\cdot \left(5 - \dfrac{35}{12}x\right) \\ 10-\dfrac{35}{6}x &= \dfrac{49}{12}x^2 - 14x + 12 \\ \dfrac{49}{12}x^2 - \dfrac{49}{6}x + 2 &= 0 \\ 49x^2 - 98x + 24 &= 0 \\ x&=\dfrac{2}{7} \text{ or } \dfrac{12}{7}. \end{align*}
But $x=\dfrac{12}{7}$ is extraneous. Thus, the area of square $S = x^2 = \dfrac{4}{49}$, making the portion of the field that is planted being \[1 - \dfrac{\tfrac{4}{49}}{6} = 1 - \dfrac{2}{147} = \boxed{\textbf{(D) } \frac{145}{147}}.\]
-Solution by sml1809 | size(240); pair A, B, C, D, F, X, Y, P, Q, M, N; A = origin; label(A, "$A$", SW); B = (4,0); label(B, "$B$", S); C = (0,3); label(C, "$C$", W); D = (2/7,2/7); label(D, "$D$", NE); F = foot(D,B,C); label(F, "$F$", NE); X = (2/7,39/14); label(X, "$X$", NE, red); Y = (76/21,2/7); label(Y, "$Y$", NE, red); P = foot(X,A,C); label(P, "$P$", W, red); Q = foot(Y,A,B); label(Q, "$Q$", S, red); M = (2/7,0); label(M, "$M$", S); N = (0,2/7); label(N, "$N$", W); fill(A--(2/7,0)--D--(0,2/7)--cycle, lightgray); draw(A--B--C--cycle); draw((2/7,0)--D--(0,2/7)); label("$x$", midpoint(A--M), S); label("$x$", midpoint(A--N), W); label("$2$", midpoint(D--F), SE); draw(D--F); draw(D--X, red); draw(D--Y, red); draw(X--P, red); draw(Y--Q, red); | [] |
823 | The solutions to the equations $z^2=4+4\sqrt{15}i$ and $z^2=2+2\sqrt 3i,$ where $i=\sqrt{-1},$ form the vertices of a parallelogram in the complex plane. The area of this parallelogram can be written in the form $p\sqrt q-r\sqrt s,$ where $p,$ $q,$ $r,$ and $s$ are positive integers and neither $q$ nor $s$ is divisible by the square of any prime number. What is $p+q+r+s?$
$\textbf{(A) } 20 \qquad \textbf{(B) } 21 \qquad \textbf{(C) } 22 \qquad \textbf{(D) } 23 \qquad \textbf{(E) } 24$ | 2018 AMC 12A Problem 22 | Let $z_1$ and $z_2$ be the solutions to the equation $z^2=4+4\sqrt{15}i,$ and $z_3$ and $z_4$ be the solutions to the equation $z^2=2+2\sqrt 3i.$ Clearly, $z_1$ and $z_2$ are opposite complex numbers, so are $z_3$ and $z_4.$ This solution refers to the results of De Moivre's Theorem in Solution 2.
From Solution 2, let $z_1=4\operatorname{cis}\phi$ for some $0<\phi<\frac{\pi}{4}.$ It follows that $z_2=4\operatorname{cis}(\phi+\pi).$ On the other hand, we have $z_3=2\operatorname{cis}\frac{\pi}{6}$ and $z_4=2\operatorname{cis}\frac{7\pi}{6}$ without loss of generality. Since $\tan(2\phi)>\tan\frac{\pi}{3},$ we deduce that $2\phi>\frac{\pi}{3},$ from which $\phi>\frac{\pi}{6}.$
In the complex plane, the positions of $z_1,z_2,z_3,$ and $z_4$ are shown below:
Note that the diagonals of every parallelogram partition the shape into four triangles with equal areas. Therefore, to find the area of the parallelogram with vertices $z_1,z_2,z_3,$ and $z_4,$ we find the area of the triangle with vertices $0,z_1,$ and $z_3,$ then multiply by $4.$
Recall that $|z_1|=4, |z_2|=2, \sin\phi=\frac{\sqrt6}{4},$ and $\cos\phi=\frac{\sqrt{10}}{4}$ from Solution 2. The area of the parallelogram is
\begin{align*} 4\cdot\left[\frac12\cdot|z_1|\cdot|z_3|\cdot\sin\left(\phi-\frac{\pi}{6}\right)\right] &= 16\sin\left(\phi-\frac{\pi}{6}\right) \\ &= 16\left[\sin\phi\cos\frac{\pi}{6}-\cos\phi\sin\frac{\pi}{6}\right] \\ &= 16\left[\frac{\sqrt3}{2}\sin\phi-\frac12\cos\phi\right] \\ &= 6\sqrt2-2\sqrt{10}, \end{align*}
so the answer is $6+2+2+10=\boxed{\textbf{(A) } 20}.$
~MRENTHUSIASM | // Block 1
/* Made by MRENTHUSIASM */
size(200);
int xMin = -5;
int xMax = 5;
int yMin = -5;
int yMax = 5;
int numRays = 24;
//Draws a polar grid that goes out to a number of circles
//equal to big, with numRays specifying the number of rays:
void polarGrid(int big, int numRays)
{
for (int i = 1; i < big+1; ++i)
{
draw(Circle((0,0),i), gray+linewidth(0.4));
}
for(int i=0;i<numRays;++i)
draw(rotate(i*360/numRays)*((-big,0)--(big,0)), gray+linewidth(0.4));
}
//Draws the horizontal gridlines
void horizontalLines()
{
for (int i = yMin+1; i < yMax; ++i)
{
draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4));
}
}
//Draws the vertical gridlines
void verticalLines()
{
for (int i = xMin+1; i < xMax; ++i)
{
draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4));
}
}
horizontalLines();
verticalLines();
polarGrid(xMax,numRays);
draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5));
draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5));
label("Re",(xMax,0),(2,0));
label("Im",(0,yMax),(0,2));
pair Z1, Z2, Z3, Z4;
Z1 = (sqrt(10),sqrt(6));
Z2 = (-sqrt(10),-sqrt(6));
Z3 = (sqrt(3),1);
Z4 = (-sqrt(3),-1);
label("$z_1$", Z1, dir(Z1), UnFill);
label("$z_2$", Z2, dir(Z2), UnFill);
label("$z_3$", Z3, (0.75,-0.75), UnFill);
label("$z_4$", Z4, (-0.75,0.75), UnFill);
draw(Z1--Z3--Z2--Z4--cycle,red);
dot(Z1, linewidth(3.5));
dot(Z2, linewidth(3.5));
dot(Z3, linewidth(3.5));
dot(Z4, linewidth(3.5));
// Block 2
/* Made by MRENTHUSIASM */ size(200); int xMin = -5; int xMax = 5; int yMin = -5; int yMax = 5; int numRays = 24; //Draws a polar grid that goes out to a number of circles //equal to big, with numRays specifying the number of rays: void polarGrid(int big, int numRays) { for (int i = 1; i < big+1; ++i) { draw(Circle((0,0),i), gray+linewidth(0.4)); } for(int i=0;i<numRays;++i) draw(rotate(i*360/numRays)*((-big,0)--(big,0)), gray+linewidth(0.4)); } //Draws the horizontal gridlines void horizontalLines() { for (int i = yMin+1; i < yMax; ++i) { draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4)); } } //Draws the vertical gridlines void verticalLines() { for (int i = xMin+1; i < xMax; ++i) { draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4)); } } horizontalLines(); verticalLines(); polarGrid(xMax,numRays); draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("Re",(xMax,0),(2,0)); label("Im",(0,yMax),(0,2)); pair Z1, Z2, Z3, Z4; Z1 = (sqrt(10),sqrt(6)); Z2 = (-sqrt(10),-sqrt(6)); Z3 = (sqrt(3),1); Z4 = (-sqrt(3),-1); label("$z_1$", Z1, dir(Z1), UnFill); label("$z_2$", Z2, dir(Z2), UnFill); label("$z_3$", Z3, (0.75,-0.75), UnFill); label("$z_4$", Z4, (-0.75,0.75), UnFill); draw(Z1--Z3--Z2--Z4--cycle,red); dot(Z1, linewidth(3.5)); dot(Z2, linewidth(3.5)); dot(Z3, linewidth(3.5)); dot(Z4, linewidth(3.5)); | [] |
824 | In $\triangle PAT,$ $\angle P=36^{\circ},$ $\angle A=56^{\circ},$ and $PA=10.$ Points $U$ and $G$ lie on sides $\overline{TP}$ and $\overline{TA},$ respectively, so that $PU=AG=1.$ Let $M$ and $N$ be the midpoints of segments $\overline{PA}$ and $\overline{UG},$ respectively. What is the degree measure of the acute angle formed by lines $MN$ and $PA?$
$\textbf{(A) } 76 \qquad \textbf{(B) } 77 \qquad \textbf{(C) } 78 \qquad \textbf{(D) } 79 \qquad \textbf{(E) } 80$ | 2018 AMC 12A Problem 23 | We will refer to the Diagram section. In this solution, all angle measures are in degrees.
We rotate $\triangle PUM$ by $180^\circ$ about $M$ to obtain $\triangle AU'M.$ Let $H$ be the intersection of $\overline{PA}$ and $\overline{GU'},$ as shown below.
Note that $\triangle GU'A$ is an isosceles triangle with $GA=U'A=1,$ so $\angle AGU'=\angle AU'G=\frac{180-\angle GAU'}{2}=44.$ In $\triangle GHA,$ it follows that $\angle GHA=180-\angle GAH-\angle AGH=80.$
Since $\frac{UM}{UU'}=\frac{UN}{UG}=\frac12,$ we conclude that $\triangle UMN\sim\triangle UU'G$ by SAS, from which $\angle UMN=\angle UU'G$ and $\angle UNM=\angle UGU'.$ By the Converse of the Corresponding Angles Postulate, we deduce that $\overline{MN}\parallel\overline{U'G}.$
Finally, we have $\angle NMA=\angle GHA=\boxed{\textbf{(E) } 80}$ by the Corresponding Angles Postulate.
~MRENTHUSIASM | // Block 1
/* Made by MRENTHUSIASM */
size(375);
pair P, A, T, U, G, M, N, U1, H;
P = origin;
A = (10,0);
U = intersectionpoint(Circle(P,1),P--P+2*dir(36));
G = intersectionpoint(Circle(A,1),A--A+2*dir(180-56));
T = extension(P,U,A,G);
M = midpoint(P--A);
N = midpoint(U--G);
U1 = rotate(180,M)*U;
H = intersectionpoint(P--A,G--U1);
fill(U--P--M--cycle^^M--U1--A--cycle,yellow);
dot("$P$",P,1.5*SW,linewidth(4));
dot("$A$",A,1.5*SE,linewidth(4));
dot("$U$",U,1.5*(0,1),linewidth(4));
dot("$G$",G,1.5*NE,linewidth(4));
dot("$T$",T,1.5*(0,1),linewidth(4));
dot("$M$",M,1.5*S,linewidth(4));
dot("$N$",N,1.5*(0,1),linewidth(4));
dot("$U'$",U1,1.5*S,linewidth(4));
dot("$H$",H,1.5*NW,linewidth(4));
draw(P--A--T--cycle^^U--G^^M--N^^U--U1--A);
draw(G--U1,dashed);
label("$1$",midpoint(G--A),1.5*dir(30));
label("$1$",midpoint(A--U1),1.5*dir(-30));
label("$1$",midpoint(U--P),1.5*dir(150));
label("$36^\circ$",P,5*dir(18),fontsize(8));
label("$56^\circ$",A,2.5*dir(180-56/2),fontsize(8));
label("$36^\circ$",A,2.5*dir(180+25),fontsize(8));
Label L = Label("$10$", align=(0,0), position=MidPoint, filltype=Fill(3,0,white));
draw(P-(0,1.5)--A-(0,1.5), L=L, arrow=Arrows(),bar=Bars(15));
add(pathticks(U--N, 2, .5, 4, 8, red));
add(pathticks(N--G, 2, .5, 4, 8, red));
add(pathticks(U--M, 1, .5, 0, 8, red));
add(pathticks(M--U1, 1, .5, 0, 8, red));
// Block 2
/* Made by MRENTHUSIASM */ size(375); pair P, A, T, U, G, M, N, U1, H; P = origin; A = (10,0); U = intersectionpoint(Circle(P,1),P--P+2*dir(36)); G = intersectionpoint(Circle(A,1),A--A+2*dir(180-56)); T = extension(P,U,A,G); M = midpoint(P--A); N = midpoint(U--G); U1 = rotate(180,M)*U; H = intersectionpoint(P--A,G--U1); fill(U--P--M--cycle^^M--U1--A--cycle,yellow); dot("$P$",P,1.5*SW,linewidth(4)); dot("$A$",A,1.5*SE,linewidth(4)); dot("$U$",U,1.5*(0,1),linewidth(4)); dot("$G$",G,1.5*NE,linewidth(4)); dot("$T$",T,1.5*(0,1),linewidth(4)); dot("$M$",M,1.5*S,linewidth(4)); dot("$N$",N,1.5*(0,1),linewidth(4)); dot("$U'$",U1,1.5*S,linewidth(4)); dot("$H$",H,1.5*NW,linewidth(4)); draw(P--A--T--cycle^^U--G^^M--N^^U--U1--A); draw(G--U1,dashed); label("$1$",midpoint(G--A),1.5*dir(30)); label("$1$",midpoint(A--U1),1.5*dir(-30)); label("$1$",midpoint(U--P),1.5*dir(150)); label("$36^\circ$",P,5*dir(18),fontsize(8)); label("$56^\circ$",A,2.5*dir(180-56/2),fontsize(8)); label("$36^\circ$",A,2.5*dir(180+25),fontsize(8)); Label L = Label("$10$", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); draw(P-(0,1.5)--A-(0,1.5), L=L, arrow=Arrows(),bar=Bars(15)); add(pathticks(U--N, 2, .5, 4, 8, red)); add(pathticks(N--G, 2, .5, 4, 8, red)); add(pathticks(U--M, 1, .5, 0, 8, red)); add(pathticks(M--U1, 1, .5, 0, 8, red)); | [] |
825 | Alice, Bob, and Carol play a game in which each of them chooses a real number between $0$ and $1.$ The winner of the game is the one whose number is between the numbers chosen by the other two players. Alice announces that she will choose her number uniformly at random from all the numbers between $0$ and $1,$ and Bob announces that he will choose his number uniformly at random from all the numbers between $\tfrac{1}{2}$ and $\tfrac{2}{3}.$ Armed with this information, what number should Carol choose to maximize her chance of winning?
$\textbf{(A) }\frac{1}{2}\qquad \textbf{(B) }\frac{13}{24} \qquad \textbf{(C) }\frac{7}{12} \qquad \textbf{(D) }\frac{5}{8} \qquad \textbf{(E) }\frac{2}{3}\qquad$ | 2018 AMC 12A Problem 24 | Let $a,b,$ and $c$ be the numbers that Alice, Bob, and Carol choose, respectively.
Based on the value of $c,$ we construct the following table:
\[\begin{array}{c|c|c} & & \\ [-2ex] \textbf{Case} & \textbf{Conditions for }\boldsymbol{a}\textbf{ and }\boldsymbol{b} & \textbf{Carol's Probability of Winning} \\ [0.5ex] \hline & & \\ [-1.5ex] 0<c<\frac12 & 0<a<c \text{ and } \frac12<b<\frac23 & \hspace{1.25mm}\frac{c}{1}\cdot\frac{1/6}{1/6}=c \\ [1.5ex] \frac12\leq c\leq\frac23 & \left(0<a<c \text{ and } c<b<\frac23\right) \text{ or } \left(c<a<1 \text{ and } \frac12<b<c\right) & \hspace{1.25mm}\frac{c}{1}\cdot\frac{2/3-c}{1/6}+\frac{1-c}{1}\cdot\frac{c-1/2}{1/6}=-12c^2+13c-3 \\ [1.5ex] \frac23<c<1 & c<a<1 \text{ and } \frac12<b<\frac23 & \hspace{4.375mm}\frac{1-c}{1}\cdot\frac{1/6}{1/6}=1-c \\ [1.5ex] \end{array}\]
Let $P(c)$ be Carol's probability of winning when she chooses $c.$ We write $P(c)$ as a piecewise function:
\[P(c) = \begin{cases} c & \mathrm{if} \ 0<c<\frac12 \\ -12c^2+13c-3 & \mathrm{if} \ \frac12\leq c\leq\frac23 \\ 1-c & \mathrm{if} \ \frac23<c<1 \end{cases}.\]
Note that $P(c)$ is continuous in the interval $(0,1),$ increasing in the interval $\left(0,\frac12\right),$ increasing and then decreasing in the interval $\left(\frac12,\frac23\right),$ and decreasing in the interval $\left(\frac23,1\right).$ The graph of $y=P(c)$ is shown below.
Therefore, the maximum point of $P(c)$ occurs in the interval $\left[\frac12,\frac23\right],$ namely at $c=-\frac{13}{2\cdot(-12)}=\boxed{\textbf{(B) }\frac{13}{24}}.$
~MRENTHUSIASM | // Block 1
/* Made by MRENTHUSIASM */
size(200);
real f(real x) { return x; }
real g(real x) { return -12x^2+13x-3; }
real h(real x) { return 1-x; }
draw((1/2,0)--(1/2,1.25),dashed);
draw((2/3,0)--(2/3,1.25),dashed);
draw(graph(f,0,1/2),red);
draw(graph(g,1/2,2/3),red);
draw(graph(h,2/3,1),red);
real xMin = -0.25;
real xMax = 1.25;
real yMin = -0.25;
real yMax = 1.25;
draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5));
draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5));
label("$c$",(xMax,0),(2,0));
label("$y$",(0,yMax),(0,2));
pair A[];
A[0] = (0,0);
A[1] = (1/2,1/2);
A[2] = (2/3,1/3);
A[3] = (1,0);
dot(A[1],red+linewidth(3.5));
dot(A[2],red+linewidth(3.5));
label("$0$",A[0],(-1.5,-1.5));
label("$\frac12$",(1/2,0),(0,-1.5));
label("$\frac23$",(2/3,0),(0,-1.5));
label("$1$",A[3],(0,-1.5));
label("$1$",(0,1),(-1.5,0));
draw((1/2,-0.02)--(1/2,0.02),linewidth(1));
draw((2/3,-0.02)--(2/3,0.02),linewidth(1));
draw((1,-0.02)--(1,0.02),linewidth(1));
draw((-0.02,1)--(0.02,1),linewidth(1));
// Block 2
/* Made by MRENTHUSIASM */ size(200); real f(real x) { return x; } real g(real x) { return -12x^2+13x-3; } real h(real x) { return 1-x; } draw((1/2,0)--(1/2,1.25),dashed); draw((2/3,0)--(2/3,1.25),dashed); draw(graph(f,0,1/2),red); draw(graph(g,1/2,2/3),red); draw(graph(h,2/3,1),red); real xMin = -0.25; real xMax = 1.25; real yMin = -0.25; real yMax = 1.25; draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("$c$",(xMax,0),(2,0)); label("$y$",(0,yMax),(0,2)); pair A[]; A[0] = (0,0); A[1] = (1/2,1/2); A[2] = (2/3,1/3); A[3] = (1,0); dot(A[1],red+linewidth(3.5)); dot(A[2],red+linewidth(3.5)); label("$0$",A[0],(-1.5,-1.5)); label("$\frac12$",(1/2,0),(0,-1.5)); label("$\frac23$",(2/3,0),(0,-1.5)); label("$1$",A[3],(0,-1.5)); label("$1$",(0,1),(-1.5,0)); draw((1/2,-0.02)--(1/2,0.02),linewidth(1)); draw((2/3,-0.02)--(2/3,0.02),linewidth(1)); draw((1,-0.02)--(1,0.02),linewidth(1)); draw((-0.02,1)--(0.02,1),linewidth(1)); | [] |
826 | Line segment $\overline{AB}$ is a diameter of a circle with $AB = 24$. Point $C$, not equal to $A$ or $B$, lies on the circle. As point $C$ moves around the circle, the centroid (center of mass) of $\triangle ABC$ traces out a closed curve missing two points. To the nearest positive integer, what is the area of the region bounded by this curve?
$\textbf{(A) } 25 \qquad \textbf{(B) } 38 \qquad \textbf{(C) } 50 \qquad \textbf{(D) } 63 \qquad \textbf{(E) } 75$ | 2018 AMC 10B Problem 12 | By the Inscribed Angle Theorem, $\triangle ABC$ is a right triangle with $\angle C=90^{\circ}.$ So, its circumcenter is the midpoint of $\overline{AB},$ and its median from $C$ is half as long as $\overline{AB}.$ For each $\triangle ABC,$ let $O$ and $G$ be its circumcenter and centroid, respectively. It follows that $OA=OB=OC=12.$ In any triangle, since the centroid divides each median into parts in the ratio $2:1,$ with the centroid being twice as close to the midpoint of a side as it is to the opposite vertex, we have $OG=\frac13 OC=4.$
As shown below, $\triangle ABC_1$ and $\triangle ABC_2$ are two shapes of $\triangle ABC$ with centroids $G_1$ and $G_2,$ respectively:
Therefore, point $G$ traces out a circle (missing two points) with the center $O$ and the radius $\overline{OG},$ as indicated in red. To the nearest positive integer, the area of the region bounded by the red curve is $\pi\cdot OG^2=16\pi\approx\boxed{\textbf{(C) } 50}.$
~MRENTHUSIASM ~megacleverstarfish15 | // Block 1
/* Made by MRENTHUSIASM */
size(200);
pair O, A, B, C1, C2, G1, G2, M1, M2;
O = (0,0);
A = (-12,0);
B = (12,0);
C1 = (36/5,48/5);
C2 = (-96/17,-180/17);
G1 = O + 1/3 * C1;
G2 = O + 1/3 * C2;
M1 = (4,0);
M2 = (-4,0);
draw(Circle(O,12));
draw(Circle(O,4),red);
dot("$O$", O, (3/5,-4/5), linewidth(4.5));
dot("$A$", A, W, linewidth(4.5));
dot("$B$", B, E, linewidth(4.5));
dot("$C_1$", C1, dir(C1), linewidth(4.5));
dot("$C_2$", C2, dir(C2), linewidth(4.5));
dot("$G_1$", G1, 1.5*E, linewidth(4.5));
dot("$G_2$", G2, 1.5*W, linewidth(4.5));
draw(A--B^^A--C1--B^^A--C2--B);
draw(O--C1^^O--C2);
dot(M1,red+linewidth(0.8),UnFill);
dot(M2,red+linewidth(0.8),UnFill);
// Block 2
/* Made by MRENTHUSIASM */ size(200); pair O, A, B, C1, C2, G1, G2, M1, M2; O = (0,0); A = (-12,0); B = (12,0); C1 = (36/5,48/5); C2 = (-96/17,-180/17); G1 = O + 1/3 * C1; G2 = O + 1/3 * C2; M1 = (4,0); M2 = (-4,0); draw(Circle(O,12)); draw(Circle(O,4),red); dot("$O$", O, (3/5,-4/5), linewidth(4.5)); dot("$A$", A, W, linewidth(4.5)); dot("$B$", B, E, linewidth(4.5)); dot("$C_1$", C1, dir(C1), linewidth(4.5)); dot("$C_2$", C2, dir(C2), linewidth(4.5)); dot("$G_1$", G1, 1.5*E, linewidth(4.5)); dot("$G_2$", G2, 1.5*W, linewidth(4.5)); draw(A--B^^A--C1--B^^A--C2--B); draw(O--C1^^O--C2); dot(M1,red+linewidth(0.8),UnFill); dot(M2,red+linewidth(0.8),UnFill); | [] |
827 | A closed box with a square base is to be wrapped with a square sheet of wrapping paper. The box is centered on the wrapping paper with the vertices of the base lying on the midlines of the square sheet of paper, as shown in the figure on the left. The four corners of the wrapping paper are to be folded up over the sides and brought together to meet at the center of the top of the box, point $A$ in the figure on the right. The box has base length $w$ and height $h$. What is the area of the sheet of wrapping paper?
$\textbf{(A) } 2(w+h)^2 \qquad \textbf{(B) } \frac{(w+h)^2}2 \qquad \textbf{(C) } 2w^2+4wh \qquad \textbf{(D) } 2w^2 \qquad \textbf{(E) } w^2h$ | 2018 AMC 10B Problem 15 | Consider one-quarter of the image (the wrapping paper is divided up into $4$ congruent squares). The length of each dotted line is $h$. The area of the rectangle that is $w$ by $h$ is $wh$. The combined figure of the two triangles with base $h$ is a square with $h$ as its diagonal. Using the Pythagorean Theorem, each side of this square is $\frac{h}{\sqrt2}$. Thus, the area is the side length squared which is $\frac{h^2}{2}$. Similarly, the combined figure of the two triangles with base $w$ is a square with area $\frac{w^2}{2}$. Adding all of these together, we get $\frac{w^2}{2} + \frac{h^2}{2} + wh$. Since we have four of these areas in the entire wrapping paper, we multiply this by $4$, getting $4\left(\frac{w^2}{2} + \frac{h^2}{2} + wh\right) = 2\left(w^2 + h^2 + 2wh\right) = \boxed{\textbf{(A) } 2(w+h)^2}$.
The diagram for this solution is shown below:
~Hydroquantum (Solution)
~MRENTHUSIASM (Diagram) | // Block 1
/* Edited by MRENTHUSIASM */
size(180pt);
defaultpen(fontsize(10pt));
fill(((0,3)--(-3,3)--(-3,0)--(0,0)--cycle),lightgrey);
dot((-3,3));
label("$A$",(-3,3),NW);
draw((-3,0)--(-3,3)--(0,3),linewidth(.5));
draw((0,2)--(-1,3)--(-3,1)--(-2,0),dashed+linewidth(.5));
draw((0,2)--(-2,0),linewidth(.5));
draw((0,3)--(0,0),linetype("2.5 2.5")+linewidth(.5));
draw((0,0)--(-3,0),linetype("2.5 2.5")+linewidth(.5));
label("$w$",(-1,1),NW);
label("$w$",(-2,2),SE);
label("$h$",(-2.5,0.5),NE);
label("$h$",(-0.5,2.5),SW);
label("$\frac{w}{\sqrt{2}}$",(-2,3),N);
label("$\frac{h}{\sqrt{2}}$",(-0.5,3),N);
label("$\frac{h}{\sqrt{2}}$",(0,2.5),E);
label("$\frac{w}{\sqrt{2}}$",(0,1),E);
label("$\frac{w}{\sqrt{2}}$",(-1,0),S);
label("$\frac{h}{\sqrt{2}}$",(-2.5,0),S);
label("$\frac{h}{\sqrt{2}}$",(-3,0.5),W);
label("$\frac{w}{\sqrt{2}}$",(-3,2),W);
label("$wh$",(-1.5,1.5),red);
label("$\frac{w^2}{4}$",centroid((-3,3),(-1,3),(-3,1)),red);
label("$\frac{w^2}{4}$",centroid((0,0),(-2,0),(0,2)),red);
label("$\frac{h^2}{4}$",centroid((-3,0),(-2,0),(-3,1)),red);
label("$\frac{h^2}{4}$",centroid((0,3),(-1,3),(0,2)),red);
// Block 2
/* Edited by MRENTHUSIASM */ size(180pt); defaultpen(fontsize(10pt)); fill(((0,3)--(-3,3)--(-3,0)--(0,0)--cycle),lightgrey); dot((-3,3)); label("$A$",(-3,3),NW); draw((-3,0)--(-3,3)--(0,3),linewidth(.5)); draw((0,2)--(-1,3)--(-3,1)--(-2,0),dashed+linewidth(.5)); draw((0,2)--(-2,0),linewidth(.5)); draw((0,3)--(0,0),linetype("2.5 2.5")+linewidth(.5)); draw((0,0)--(-3,0),linetype("2.5 2.5")+linewidth(.5)); label("$w$",(-1,1),NW); label("$w$",(-2,2),SE); label("$h$",(-2.5,0.5),NE); label("$h$",(-0.5,2.5),SW); label("$\frac{w}{\sqrt{2}}$",(-2,3),N); label("$\frac{h}{\sqrt{2}}$",(-0.5,3),N); label("$\frac{h}{\sqrt{2}}$",(0,2.5),E); label("$\frac{w}{\sqrt{2}}$",(0,1),E); label("$\frac{w}{\sqrt{2}}$",(-1,0),S); label("$\frac{h}{\sqrt{2}}$",(-2.5,0),S); label("$\frac{h}{\sqrt{2}}$",(-3,0.5),W); label("$\frac{w}{\sqrt{2}}$",(-3,2),W); label("$wh$",(-1.5,1.5),red); label("$\frac{w^2}{4}$",centroid((-3,3),(-1,3),(-3,1)),red); label("$\frac{w^2}{4}$",centroid((0,0),(-2,0),(0,2)),red); label("$\frac{h^2}{4}$",centroid((-3,0),(-2,0),(-3,1)),red); label("$\frac{h^2}{4}$",centroid((0,3),(-1,3),(0,2)),red); | [] |
827 | A closed box with a square base is to be wrapped with a square sheet of wrapping paper. The box is centered on the wrapping paper with the vertices of the base lying on the midlines of the square sheet of paper, as shown in the figure on the left. The four corners of the wrapping paper are to be folded up over the sides and brought together to meet at the center of the top of the box, point $A$ in the figure on the right. The box has base length $w$ and height $h$. What is the area of the sheet of wrapping paper?
$\textbf{(A) } 2(w+h)^2 \qquad \textbf{(B) } \frac{(w+h)^2}2 \qquad \textbf{(C) } 2w^2+4wh \qquad \textbf{(D) } 2w^2 \qquad \textbf{(E) } w^2h$ | 2018 AMC 10B Problem 15 | The sheet of paper is made out of the surface area of the box plus the sum of the four yellow triangles, as shown below.
The surface area is $2w^2 + 2wh + 2wh$ which equals $2w^2 + 4wh$.The four triangles each have a height and a base of $h$, so they each have an area of $\frac{h^2}{2}$. There are four of them, so multiplied by four is $2h^2$. Together, paper's area is $2w^2 + 4wh + 2h^2$. This can be factored and written as $\boxed{\textbf{(A) } 2(w+h)^2} \qquad$.
~Yee2121 (Solution)
~MRENTHUSIASM (Diagram) | // Block 1
/* Edited by MRENTHUSIASM */
size(180pt);
defaultpen(fontsize(10pt));
fill(((3,3)--(-3,3)--(-3,-3)--(3,-3)--cycle),lightgrey);
fill((2,0)--(3,1)--(3,-1)--cycle,yellow);
fill((-2,0)--(-3,1)--(-3,-1)--cycle,yellow);
fill((0,2)--(1,3)--(-1,3)--cycle,yellow);
fill((0,-2)--(1,-3)--(-1,-3)--cycle,yellow);
draw(((3,3)--(-3,3)--(-3,-3)--(3,-3)--cycle),linewidth(.5));
dot((-3,3));
label("$A$",(-3,3),NW);
draw((1,3)--(-3,-1),dashed+linewidth(.5));
draw((-1,3)--(3,-1),dashed+linewidth(.5));
draw((-1,-3)--(3,1),dashed+linewidth(.5));
draw((1,-3)--(-3,1),dashed+linewidth(.5));
draw((0,2)--(2,0)--(0,-2)--(-2,0)--cycle,linewidth(.5));
draw((0,3)--(0,-3),linetype("2.5 2.5")+linewidth(.5));
draw((3,0)--(-3,0),linetype("2.5 2.5")+linewidth(.5));
label("$w$",(-1,-1),SW);
label("$w$",(1,-1),SE);
label("$w$",(1,1),NE);
label("$w$",(-1,1),NW);
label("$h$",(2.5,0.5),NW);
label("$h$",(2.5,-0.5),SW);
label("$h$",(-2.5,0.5),NE);
label("$h$",(-2.5,-0.5),SE);
label("$h$",(0.5,2.5),SE);
label("$h$",(-0.5,2.5),SW);
label("$h$",(0.5,-2.5),NE);
label("$h$",(-0.5,-2.5),NW);
// Block 2
/* Edited by MRENTHUSIASM */ size(180pt); defaultpen(fontsize(10pt)); fill(((3,3)--(-3,3)--(-3,-3)--(3,-3)--cycle),lightgrey); fill((2,0)--(3,1)--(3,-1)--cycle,yellow); fill((-2,0)--(-3,1)--(-3,-1)--cycle,yellow); fill((0,2)--(1,3)--(-1,3)--cycle,yellow); fill((0,-2)--(1,-3)--(-1,-3)--cycle,yellow); draw(((3,3)--(-3,3)--(-3,-3)--(3,-3)--cycle),linewidth(.5)); dot((-3,3)); label("$A$",(-3,3),NW); draw((1,3)--(-3,-1),dashed+linewidth(.5)); draw((-1,3)--(3,-1),dashed+linewidth(.5)); draw((-1,-3)--(3,1),dashed+linewidth(.5)); draw((1,-3)--(-3,1),dashed+linewidth(.5)); draw((0,2)--(2,0)--(0,-2)--(-2,0)--cycle,linewidth(.5)); draw((0,3)--(0,-3),linetype("2.5 2.5")+linewidth(.5)); draw((3,0)--(-3,0),linetype("2.5 2.5")+linewidth(.5)); label("$w$",(-1,-1),SW); label("$w$",(1,-1),SE); label("$w$",(1,1),NE); label("$w$",(-1,1),NW); label("$h$",(2.5,0.5),NW); label("$h$",(2.5,-0.5),SW); label("$h$",(-2.5,0.5),NE); label("$h$",(-2.5,-0.5),SE); label("$h$",(0.5,2.5),SE); label("$h$",(-0.5,2.5),SW); label("$h$",(0.5,-2.5),NE); label("$h$",(-0.5,-2.5),NW); | [] |
828 | Let $ABCDEF$ be a regular hexagon with side length $1$. Denote by $X$, $Y$, and $Z$ the midpoints of sides $\overline {AB}$, $\overline{CD}$, and $\overline{EF}$, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of $\triangle ACE$ and $\triangle XYZ$?
$\textbf{(A)}\ \frac {3}{8}\sqrt{3} \qquad \textbf{(B)}\ \frac {7}{16}\sqrt{3} \qquad \textbf{(C)}\ \frac {15}{32}\sqrt{3} \qquad \textbf{(D)}\ \frac {1}{2}\sqrt{3} \qquad \textbf{(E)}\ \frac {9}{16}\sqrt{3}$ | 2018 AMC 10B Problem 24 | The desired area (hexagon $MPNQOR$) consists of an equilateral triangle ($\triangle MNO$) and three right triangles ($\triangle MPN,\triangle NQO,$ and $\triangle ORM$).
Notice that $\overline {AD}$ (not shown) and $\overline {BC}$ are parallel. $\overline {XY}$ divides transversals $\overline {AB}$ and $\overline {CD}$ into a $1:1$ ratio (This can be shown by similar triangles.). Thus, it must also divide transversal $\overline {AC}$ and transversal $\overline {CO}$ into a $1:1$ ratio. By symmetry, the same applies for $\overline {CE}$ and $\overline {EA}$ as well as $\overline {EM}$ and $\overline {AN}.$
In $\triangle ACE,$ we see that $\frac{[MNO]}{[ACE]} = \frac{1}{4}$ and $\frac{[MPN]}{[ACE]} = \frac{1}{8}.$ Our desired area becomes \[\left(\frac{1}{4}+3 \cdot \frac{1}{8}\right) \cdot \frac{(\sqrt{3})^2 \cdot \sqrt{3}}{4} = \boxed{\textbf{(C)}\ \frac {15}{32}\sqrt{3}}.\] | // Block 1
/* Made by MRENTHUSIASM */
size(200);
draw(polygon(6));
pair A, B, C, D, E, F, X, Y, Z, M, N, O, P, Q, R;
A = dir(120);
B = dir(60);
C = dir(0);
D = dir(300);
E = dir(240);
F = dir(180);
X = midpoint(A--B);
Y = midpoint(C--D);
Z = midpoint(E--F);
M = intersectionpoint(A--E,X--Z);
N = intersectionpoint(A--C,X--Y);
O = intersectionpoint(C--E,Y--Z);
P = intersectionpoint(A--C,X--Z);
Q = intersectionpoint(C--E,X--Y);
R = intersectionpoint(A--E,Y--Z);
fill(M--P--N--Q--O--R--cycle,mediumgray);
dot("$A$",A,1.5*dir(A),linewidth(4));
dot("$B$",B,1.5*dir(B),linewidth(4));
dot("$C$",C,1.5*dir(C),linewidth(4));
dot("$D$",D,1.5*dir(D),linewidth(4));
dot("$E$",E,1.5*dir(E),linewidth(4));
dot("$F$",F,1.5*dir(F),linewidth(4));
dot("$X$",X,1.5*dir(X),linewidth(4));
dot("$Y$",Y,1.5*dir(Y),linewidth(4));
dot("$Z$",Z,1.5*dir(Z),linewidth(4));
dot("$M$",M,1.5*dir(165),linewidth(4));
dot("$N$",N,1.5*dir(45),linewidth(4));
dot("$O$",O,1.5*dir(-75),linewidth(4));
dot("$P$",P,1.5*dir(105),linewidth(4));
dot("$Q$",Q,1.5*dir(-15),linewidth(4));
dot("$R$",R,1.5*dir(-135),linewidth(4));
draw(A--C--E--cycle^^X--Y--Z--cycle);
draw(M--N--O--cycle,dashed);
// Block 2
/* Made by MRENTHUSIASM */ size(200); draw(polygon(6)); pair A, B, C, D, E, F, X, Y, Z, M, N, O, P, Q, R; A = dir(120); B = dir(60); C = dir(0); D = dir(300); E = dir(240); F = dir(180); X = midpoint(A--B); Y = midpoint(C--D); Z = midpoint(E--F); M = intersectionpoint(A--E,X--Z); N = intersectionpoint(A--C,X--Y); O = intersectionpoint(C--E,Y--Z); P = intersectionpoint(A--C,X--Z); Q = intersectionpoint(C--E,X--Y); R = intersectionpoint(A--E,Y--Z); fill(M--P--N--Q--O--R--cycle,mediumgray); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*dir(B),linewidth(4)); dot("$C$",C,1.5*dir(C),linewidth(4)); dot("$D$",D,1.5*dir(D),linewidth(4)); dot("$E$",E,1.5*dir(E),linewidth(4)); dot("$F$",F,1.5*dir(F),linewidth(4)); dot("$X$",X,1.5*dir(X),linewidth(4)); dot("$Y$",Y,1.5*dir(Y),linewidth(4)); dot("$Z$",Z,1.5*dir(Z),linewidth(4)); dot("$M$",M,1.5*dir(165),linewidth(4)); dot("$N$",N,1.5*dir(45),linewidth(4)); dot("$O$",O,1.5*dir(-75),linewidth(4)); dot("$P$",P,1.5*dir(105),linewidth(4)); dot("$Q$",Q,1.5*dir(-15),linewidth(4)); dot("$R$",R,1.5*dir(-135),linewidth(4)); draw(A--C--E--cycle^^X--Y--Z--cycle); draw(M--N--O--cycle,dashed); | [] |
828 | Let $ABCDEF$ be a regular hexagon with side length $1$. Denote by $X$, $Y$, and $Z$ the midpoints of sides $\overline {AB}$, $\overline{CD}$, and $\overline{EF}$, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of $\triangle ACE$ and $\triangle XYZ$?
$\textbf{(A)}\ \frac {3}{8}\sqrt{3} \qquad \textbf{(B)}\ \frac {7}{16}\sqrt{3} \qquad \textbf{(C)}\ \frac {15}{32}\sqrt{3} \qquad \textbf{(D)}\ \frac {1}{2}\sqrt{3} \qquad \textbf{(E)}\ \frac {9}{16}\sqrt{3}$ | 2018 AMC 10B Problem 24 | Now, if we look at the figure, we can see that the complement of the hexagon we are trying to find is composed of $3$ isosceles trapezoids (namely $AXZF,CYXB,$ and $EZYD$) and $3$ right triangles (namely $\triangle XPN,\triangle YQO,$ and $\triangle ZRM$).
Finding the trapezoid's area, we know that one base of each trapezoid is just the side length of the hexagon, which is $1,$ and the other base is $\frac{3}{2}$ (it is halfway in between the side and the longest diagonal, which has length $2$) with a height of $\frac{\sqrt{3}}{4}$ (by using the Pythagorean Theorem and the fact that it is an isosceles trapezoid) to give each trapezoid having an area of $\frac{5\sqrt{3}}{16}$ for a total area of $\frac{15\sqrt{3}}{16}.$ (Alternatively, we could have calculated the area of hexagon $ABCDEF$ and subtracted the area of $\triangle XYZ,$ which, as we showed before, had a side length of $\frac{3}{2}$).
Now, we need to find the area of each of the small triangles, which, if we look at the triangle that has a vertex on $X,$ is similar to the triangle with a base of $YC = \frac12.$ Using similar triangles, we calculate the base to be $\frac{1}{4}$ and the height to be $\frac{\sqrt{3}}{4}$ giving us an area of $\frac{\sqrt{3}}{32}$ per triangle, and a total area of $\frac{3\sqrt{3}}{32}.$ Adding the two areas together, we get $\frac{15\sqrt{3}}{16} + \frac{3\sqrt{3}}{32} = \frac{33\sqrt{3}}{32}.$ Finding the total area, we get $6 \cdot 1^2 \cdot \frac{\sqrt{3}}{4}=\frac{3\sqrt{3}}{2}.$ Taking the complement, we get $\frac{3\sqrt{3}}{2} - \frac{33\sqrt{3}}{32} = \boxed{\textbf{(C)}\ \frac {15}{32}\sqrt{3}}.$ | // Block 1
/* Made by MRENTHUSIASM */
size(200);
draw(polygon(6));
pair A, B, C, D, E, F, X, Y, Z, M, N, O, P, Q, R;
A = dir(120);
B = dir(60);
C = dir(0);
D = dir(300);
E = dir(240);
F = dir(180);
X = midpoint(A--B);
Y = midpoint(C--D);
Z = midpoint(E--F);
M = intersectionpoint(A--E,X--Z);
N = intersectionpoint(A--C,X--Y);
O = intersectionpoint(C--E,Y--Z);
P = intersectionpoint(A--C,X--Z);
Q = intersectionpoint(C--E,X--Y);
R = intersectionpoint(A--E,Y--Z);
fill(M--P--N--Q--O--R--cycle,mediumgray);
dot("$A$",A,1.5*dir(A),linewidth(4));
dot("$B$",B,1.5*dir(B),linewidth(4));
dot("$C$",C,1.5*dir(C),linewidth(4));
dot("$D$",D,1.5*dir(D),linewidth(4));
dot("$E$",E,1.5*dir(E),linewidth(4));
dot("$F$",F,1.5*dir(F),linewidth(4));
dot("$X$",X,1.5*dir(X),linewidth(4));
dot("$Y$",Y,1.5*dir(Y),linewidth(4));
dot("$Z$",Z,1.5*dir(Z),linewidth(4));
dot("$M$",M,1.5*dir(165),linewidth(4));
dot("$N$",N,1.5*dir(45),linewidth(4));
dot("$O$",O,1.5*dir(-75),linewidth(4));
dot("$P$",P,1.5*dir(105),linewidth(4));
dot("$Q$",Q,1.5*dir(-15),linewidth(4));
dot("$R$",R,1.5*dir(-135),linewidth(4));
draw(A--C--E--cycle^^X--Y--Z--cycle);
draw(M--N--O--cycle,dashed);
// Block 2
/* Made by MRENTHUSIASM */ size(200); draw(polygon(6)); pair A, B, C, D, E, F, X, Y, Z, M, N, O, P, Q, R; A = dir(120); B = dir(60); C = dir(0); D = dir(300); E = dir(240); F = dir(180); X = midpoint(A--B); Y = midpoint(C--D); Z = midpoint(E--F); M = intersectionpoint(A--E,X--Z); N = intersectionpoint(A--C,X--Y); O = intersectionpoint(C--E,Y--Z); P = intersectionpoint(A--C,X--Z); Q = intersectionpoint(C--E,X--Y); R = intersectionpoint(A--E,Y--Z); fill(M--P--N--Q--O--R--cycle,mediumgray); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*dir(B),linewidth(4)); dot("$C$",C,1.5*dir(C),linewidth(4)); dot("$D$",D,1.5*dir(D),linewidth(4)); dot("$E$",E,1.5*dir(E),linewidth(4)); dot("$F$",F,1.5*dir(F),linewidth(4)); dot("$X$",X,1.5*dir(X),linewidth(4)); dot("$Y$",Y,1.5*dir(Y),linewidth(4)); dot("$Z$",Z,1.5*dir(Z),linewidth(4)); dot("$M$",M,1.5*dir(165),linewidth(4)); dot("$N$",N,1.5*dir(45),linewidth(4)); dot("$O$",O,1.5*dir(-75),linewidth(4)); dot("$P$",P,1.5*dir(105),linewidth(4)); dot("$Q$",Q,1.5*dir(-15),linewidth(4)); dot("$R$",R,1.5*dir(-135),linewidth(4)); draw(A--C--E--cycle^^X--Y--Z--cycle); draw(M--N--O--cycle,dashed); | [] |
828 | Let $ABCDEF$ be a regular hexagon with side length $1$. Denote by $X$, $Y$, and $Z$ the midpoints of sides $\overline {AB}$, $\overline{CD}$, and $\overline{EF}$, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of $\triangle ACE$ and $\triangle XYZ$?
$\textbf{(A)}\ \frac {3}{8}\sqrt{3} \qquad \textbf{(B)}\ \frac {7}{16}\sqrt{3} \qquad \textbf{(C)}\ \frac {15}{32}\sqrt{3} \qquad \textbf{(D)}\ \frac {1}{2}\sqrt{3} \qquad \textbf{(E)}\ \frac {9}{16}\sqrt{3}$ | 2018 AMC 10B Problem 24 | We could also subtract $\triangle APM,\triangle CQN,$ and $\triangle ERO$ from $\triangle ACE.$
Since $\angle BAF = 120^{\circ}$ and $\angle BAC = \angle FAE = 30^{\circ},$ we have $\angle CAE = \angle BAF-\angle BAC-\angle FAE=60^{\circ}.$
Since $AX=BX$ and $FZ=EZ,$ we have $AF \parallel XZ,$ from which $\angle AMX= \angle FAM = 30^{\circ}.$
We can show that $\triangle APM$ is $30$-$60$-$90$ using a similar method, $\triangle CQN$ and $\triangle ERO$ are also $30$-$60$-$90.$
Since $AC=CE=AE=\sqrt{3},$ we have $[ACE]=AC^2 \cdot \frac{\sqrt{3}}{4}=3 \cdot \frac{\sqrt{3}}{4} = \frac{3 \sqrt{3}}{4}.$
Since $AX= \frac{1}{2}$ and $AP = AX \cdot \frac{\sqrt{3}}{2}= \frac{1}{2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4},$ we have $PM = AP \cdot \sqrt{3} = \frac{\sqrt{3}}{4} \cdot \sqrt{3} = \frac{3}{4}.$
Note that \[[APM]=[CQN]=[ERO]=\frac{1}{2} \cdot AP \cdot PM = \frac{1}{2} \cdot \frac{\sqrt{3}}{4} \cdot \frac{3}{4} = \frac{3 \sqrt{3}}{32}.\]
Therefore, we get
\[[PNQORM]=[ACE]-[APM]-[CQN]-[ERO]=\frac{3 \sqrt{3}}{4} - 3 \cdot \frac{3 \sqrt{3}}{32} = \frac{24 \sqrt{3}}{32} - \frac{9 \sqrt{3}}{32} = \boxed{\textbf{(C)}\ \frac {15}{32}\sqrt{3}}.\]
~isabelchen | // Block 1
/* Made by MRENTHUSIASM */
size(200);
draw(polygon(6));
pair A, B, C, D, E, F, X, Y, Z, M, N, O, P, Q, R;
A = dir(120);
B = dir(60);
C = dir(0);
D = dir(300);
E = dir(240);
F = dir(180);
X = midpoint(A--B);
Y = midpoint(C--D);
Z = midpoint(E--F);
M = intersectionpoint(A--E,X--Z);
N = intersectionpoint(A--C,X--Y);
O = intersectionpoint(C--E,Y--Z);
P = intersectionpoint(A--C,X--Z);
Q = intersectionpoint(C--E,X--Y);
R = intersectionpoint(A--E,Y--Z);
fill(M--P--N--Q--O--R--cycle,mediumgray);
dot("$A$",A,1.5*dir(A),linewidth(4));
dot("$B$",B,1.5*dir(B),linewidth(4));
dot("$C$",C,1.5*dir(C),linewidth(4));
dot("$D$",D,1.5*dir(D),linewidth(4));
dot("$E$",E,1.5*dir(E),linewidth(4));
dot("$F$",F,1.5*dir(F),linewidth(4));
dot("$X$",X,1.5*dir(X),linewidth(4));
dot("$Y$",Y,1.5*dir(Y),linewidth(4));
dot("$Z$",Z,1.5*dir(Z),linewidth(4));
dot("$M$",M,1.5*dir(165),linewidth(4));
dot("$N$",N,1.5*dir(45),linewidth(4));
dot("$O$",O,1.5*dir(-75),linewidth(4));
dot("$P$",P,1.5*dir(105),linewidth(4));
dot("$Q$",Q,1.5*dir(-15),linewidth(4));
dot("$R$",R,1.5*dir(-135),linewidth(4));
draw(A--C--E--cycle^^X--Y--Z--cycle);
// Block 2
/* Made by MRENTHUSIASM */ size(200); draw(polygon(6)); pair A, B, C, D, E, F, X, Y, Z, M, N, O, P, Q, R; A = dir(120); B = dir(60); C = dir(0); D = dir(300); E = dir(240); F = dir(180); X = midpoint(A--B); Y = midpoint(C--D); Z = midpoint(E--F); M = intersectionpoint(A--E,X--Z); N = intersectionpoint(A--C,X--Y); O = intersectionpoint(C--E,Y--Z); P = intersectionpoint(A--C,X--Z); Q = intersectionpoint(C--E,X--Y); R = intersectionpoint(A--E,Y--Z); fill(M--P--N--Q--O--R--cycle,mediumgray); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*dir(B),linewidth(4)); dot("$C$",C,1.5*dir(C),linewidth(4)); dot("$D$",D,1.5*dir(D),linewidth(4)); dot("$E$",E,1.5*dir(E),linewidth(4)); dot("$F$",F,1.5*dir(F),linewidth(4)); dot("$X$",X,1.5*dir(X),linewidth(4)); dot("$Y$",Y,1.5*dir(Y),linewidth(4)); dot("$Z$",Z,1.5*dir(Z),linewidth(4)); dot("$M$",M,1.5*dir(165),linewidth(4)); dot("$N$",N,1.5*dir(45),linewidth(4)); dot("$O$",O,1.5*dir(-75),linewidth(4)); dot("$P$",P,1.5*dir(105),linewidth(4)); dot("$Q$",Q,1.5*dir(-15),linewidth(4)); dot("$R$",R,1.5*dir(-135),linewidth(4)); draw(A--C--E--cycle^^X--Y--Z--cycle); | [] |
828 | Let $ABCDEF$ be a regular hexagon with side length $1$. Denote by $X$, $Y$, and $Z$ the midpoints of sides $\overline {AB}$, $\overline{CD}$, and $\overline{EF}$, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of $\triangle ACE$ and $\triangle XYZ$?
$\textbf{(A)}\ \frac {3}{8}\sqrt{3} \qquad \textbf{(B)}\ \frac {7}{16}\sqrt{3} \qquad \textbf{(C)}\ \frac {15}{32}\sqrt{3} \qquad \textbf{(D)}\ \frac {1}{2}\sqrt{3} \qquad \textbf{(E)}\ \frac {9}{16}\sqrt{3}$ | 2018 AMC 10B Problem 24 | We partition hexagon $ABCDEF$ into $48$ congruent $30^\circ\text{-}60^\circ\text{-}90^\circ$ triangles, as shown below:
Let the brackets denote areas. Note that the desired region contains $15$ of the $48$ small triangles, so the answer is \[\frac{15}{48}[ABCDEF]=\frac{15}{48}\cdot\frac{3\sqrt3}{2}=\boxed{\textbf{(C)}\ \frac {15}{32}\sqrt{3}}.\]
~AlexLikeMath ~MRENTHUSIASM | // Block 1
/* Made by MRENTHUSIASM */
size(200);
pair A, B, C, D, E, F, X, Y, Z, M, N, O, P, Q, R;
A = dir(120);
B = dir(60);
C = dir(0);
D = dir(300);
E = dir(240);
F = dir(180);
X = midpoint(A--B);
Y = midpoint(C--D);
Z = midpoint(E--F);
M = intersectionpoint(A--E,X--Z);
N = intersectionpoint(A--C,X--Y);
O = intersectionpoint(C--E,Y--Z);
P = intersectionpoint(A--C,X--Z);
Q = intersectionpoint(C--E,X--Y);
R = intersectionpoint(A--E,Y--Z);
fill(M--P--N--Q--O--R--cycle,mediumgray);
draw(A--D^^B--E^^C--F^^X--Y--Z--cycle^^midpoint(A--F)--midpoint(B--C)--midpoint(D--E)--cycle,red);
draw(A--C--E--cycle^^M--N--O--cycle^^M--midpoint(F--Z)^^M--F+1/4*(A-F)^^N--midpoint(X--B)^^N--B+1/4*(C-B)^^O--midpoint(Y--D)^^O--D+1/4*(E-D),blue);
draw(polygon(6));
dot("$A$",A,1.5*dir(A),linewidth(4));
dot("$B$",B,1.5*dir(B),linewidth(4));
dot("$C$",C,1.5*dir(C),linewidth(4));
dot("$D$",D,1.5*dir(D),linewidth(4));
dot("$E$",E,1.5*dir(E),linewidth(4));
dot("$F$",F,1.5*dir(F),linewidth(4));
dot("$X$",X,1.5*dir(X),linewidth(4));
dot("$Y$",Y,1.5*dir(Y),linewidth(4));
dot("$Z$",Z,1.5*dir(Z),linewidth(4));
dot(M^^N^^O^^P^^Q^^R,linewidth(4));
// Block 2
/* Made by MRENTHUSIASM */ size(200); pair A, B, C, D, E, F, X, Y, Z, M, N, O, P, Q, R; A = dir(120); B = dir(60); C = dir(0); D = dir(300); E = dir(240); F = dir(180); X = midpoint(A--B); Y = midpoint(C--D); Z = midpoint(E--F); M = intersectionpoint(A--E,X--Z); N = intersectionpoint(A--C,X--Y); O = intersectionpoint(C--E,Y--Z); P = intersectionpoint(A--C,X--Z); Q = intersectionpoint(C--E,X--Y); R = intersectionpoint(A--E,Y--Z); fill(M--P--N--Q--O--R--cycle,mediumgray); draw(A--D^^B--E^^C--F^^X--Y--Z--cycle^^midpoint(A--F)--midpoint(B--C)--midpoint(D--E)--cycle,red); draw(A--C--E--cycle^^M--N--O--cycle^^M--midpoint(F--Z)^^M--F+1/4*(A-F)^^N--midpoint(X--B)^^N--B+1/4*(C-B)^^O--midpoint(Y--D)^^O--D+1/4*(E-D),blue); draw(polygon(6)); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*dir(B),linewidth(4)); dot("$C$",C,1.5*dir(C),linewidth(4)); dot("$D$",D,1.5*dir(D),linewidth(4)); dot("$E$",E,1.5*dir(E),linewidth(4)); dot("$F$",F,1.5*dir(F),linewidth(4)); dot("$X$",X,1.5*dir(X),linewidth(4)); dot("$Y$",Y,1.5*dir(Y),linewidth(4)); dot("$Z$",Z,1.5*dir(Z),linewidth(4)); dot(M^^N^^O^^P^^Q^^R,linewidth(4)); | [] |
828 | Let $ABCDEF$ be a regular hexagon with side length $1$. Denote by $X$, $Y$, and $Z$ the midpoints of sides $\overline {AB}$, $\overline{CD}$, and $\overline{EF}$, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of $\triangle ACE$ and $\triangle XYZ$?
$\textbf{(A)}\ \frac {3}{8}\sqrt{3} \qquad \textbf{(B)}\ \frac {7}{16}\sqrt{3} \qquad \textbf{(C)}\ \frac {15}{32}\sqrt{3} \qquad \textbf{(D)}\ \frac {1}{2}\sqrt{3} \qquad \textbf{(E)}\ \frac {9}{16}\sqrt{3}$ | 2018 AMC 10B Problem 24 | If we try to coordinate bash this problem, it's going to look very ugly with a lot of radicals. However, we can alter and skew the diagram in such a way that all ratios of lengths and areas stay the same while making it a lot easier to work with. Then, we can find the ratio of the area of the wanted region to the area of $ABCDEF$ then apply it to the old diagram.
The isosceles right triangle with a leg length of $3$ in the new diagram is $\triangle XYZ$ in the old diagram. We see that if we want to take the area of the new hexagon, we must subtract $\frac{3}{4}$ from the area of $\triangle XYZ$ (the red triangles), giving us $\frac{15}{4}.$ However, we need to take the ratio of this area to the area of $ABCDEF,$ which is $\frac{\frac{15}{4}}{12}=\frac{5}{16}.$ Now we know that our answer is $\frac{5}{16} \cdot \frac{3\sqrt{3}}{2}=\boxed{\textbf{(C)}\ \frac {15}{32}\sqrt{3}}.$ | // Block 1
unitsize(1cm);
draw((0,0)--(4,0)--(6,3.464)--(2,3.464)--(0,0));
draw((2,0)--(1,1.732));
draw((5,1.732)--(4,3.464));
draw((1.5, 0.866)--(3, 3.464)--(4.5, 0.866)--cycle);
draw((2,0)--(2,3.464)--(5,1.732)--cycle);
// Block 2
unitsize(1cm);
fill((1,4)--(1,3.5)--(2,3)--cycle,red);
fill((1,1)--(1.5,1)--(1,2)--cycle,red);
fill((3,1)--(3.5,1.5)--(4,1)--cycle,red);
draw((1,0)--(1,4),gray(.7));
draw((2,0)--(2,4),gray(.7));
draw((3,0)--(3,4),gray(.7));
draw((0,1)--(4,1),gray(.7));
draw((0,2)--(4,2),gray(.7));
draw((0,3)--(4,3),gray(.7));
draw((0,0)--(4,0)--(4,4)--(0,4)--(0,0));
draw((2,0)--(0,2));
draw((4,2)--(2,4));
draw((1,1)--(1,4)--(4,1)--cycle);
draw((0,4)--(2,0)--(4,2)--cycle);
// Block 3
unitsize(1cm); draw((0,0)--(4,0)--(6,3.464)--(2,3.464)--(0,0)); draw((2,0)--(1,1.732)); draw((5,1.732)--(4,3.464)); draw((1.5, 0.866)--(3, 3.464)--(4.5, 0.866)--cycle); draw((2,0)--(2,3.464)--(5,1.732)--cycle);
// Block 4
unitsize(1cm); fill((1,4)--(1,3.5)--(2,3)--cycle,red); fill((1,1)--(1.5,1)--(1,2)--cycle,red); fill((3,1)--(3.5,1.5)--(4,1)--cycle,red); draw((1,0)--(1,4),gray(.7)); draw((2,0)--(2,4),gray(.7)); draw((3,0)--(3,4),gray(.7)); draw((0,1)--(4,1),gray(.7)); draw((0,2)--(4,2),gray(.7)); draw((0,3)--(4,3),gray(.7)); draw((0,0)--(4,0)--(4,4)--(0,4)--(0,0)); draw((2,0)--(0,2)); draw((4,2)--(2,4)); draw((1,1)--(1,4)--(4,1)--cycle); draw((0,4)--(2,0)--(4,2)--cycle); | [] |
828 | Let $ABCDEF$ be a regular hexagon with side length $1$. Denote by $X$, $Y$, and $Z$ the midpoints of sides $\overline {AB}$, $\overline{CD}$, and $\overline{EF}$, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of $\triangle ACE$ and $\triangle XYZ$?
$\textbf{(A)}\ \frac {3}{8}\sqrt{3} \qquad \textbf{(B)}\ \frac {7}{16}\sqrt{3} \qquad \textbf{(C)}\ \frac {15}{32}\sqrt{3} \qquad \textbf{(D)}\ \frac {1}{2}\sqrt{3} \qquad \textbf{(E)}\ \frac {9}{16}\sqrt{3}$ | 2018 AMC 10B Problem 24 | (~MRENTHUSIASM's diagram + modifications)
Lemma 1: $MP$ is midsegment of $\triangle ANE$.
As $ZX$ is a midsegment of trapezoid $FEBA$, we know that $FA||ZX||EB$. Furthermore, by subdividing trapezoid $FEBA$, $ZM,MP,PX$ are midsegments of $\triangle EFA, \triangle ANE, \triangle, ANB$ respectively.
Lemma 2: $[ACE] = \frac12[ABCDEF]$
Due to radial symmetry about the center, $G$, we only have to prove $[AGC] = [CBA]$. As $\triangle AGB$ and $\triangle CBG$ are equilateral triangles, $AC$ therefore divides rhombus $ABCG$ into 2 congruent halves. $\triangle AGC$ and $\triangle CBA$ thus each contain two such congruent triangles, which means that their areas are equal.
Lemma 3: $[AMP] = \frac1{16}[ABCDEF]$
From Lemma 1, $\triangle AMP \simeq \triangle AEN$ with a scale factor of $\frac12$, which means $[AMP] = \frac14[AEN]$. Furthermore, due to symmetry of $\triangle AEN$ and $\triangle CEN$ about $EN$, $[AEN] = \frac12[ACE]$. Therefore, $[AMP] = \frac18[ACE]$, or (using lemma 2) $\frac1{16}[ABCDEF]$.
Lemma 4: $[ABCDEF] = \frac{3\sqrt3}2$
Because the hexagon is regular, $[ABCDEF] = 6[ABG] = \frac{6\sqrt3}4 = \frac{3\sqrt3}2$. (For more info on where $\frac{\sqrt3}4$ came from, use the search term "area of equilateral triangle")
Final step:
To find the area of the shaded part, we seek $[ACE] - 3[AMP]$. Using lemmas 2 and 3, the expression resolves to $\frac12[ABCDEF] - 3\frac1{16}[ABCDEF] = \frac5{16} \cdot \frac{3\sqrt3}2 = \boxed{\textbf{(C)}\ \frac{15}{32}\sqrt{3}}$.
~math660 | // Block 1
/* Made by MRENTHUSIASM */
size(200);
draw(polygon(6));
pair A, B, C, D, E, F, X, Y, Z, M, N, O, P, Q, R, G;
A = dir(120);
B = dir(60);
C = dir(0);
D = dir(300);
E = dir(240);
F = dir(180);
X = midpoint(A--B);
Y = midpoint(C--D);
Z = midpoint(E--F);
G = midpoint(A--D);
M = intersectionpoint(A--E,X--Z);
N = intersectionpoint(A--C,X--Y);
O = intersectionpoint(C--E,Y--Z);
P = intersectionpoint(A--C,X--Z);
Q = intersectionpoint(C--E,X--Y);
R = intersectionpoint(A--E,Y--Z);
fill(M--P--N--Q--O--R--cycle,mediumgray);
dot("$A$",A,1.5*dir(A),linewidth(4));
dot("$B$",B,1.5*dir(B),linewidth(4));
dot("$C$",C,1.5*dir(C),linewidth(4));
dot("$D$",D,1.5*dir(D),linewidth(4));
dot("$E$",E,1.5*dir(E),linewidth(4));
dot("$F$",F,1.5*dir(F),linewidth(4));
dot("$X$",X,1.5*dir(X),linewidth(4));
dot("$Y$",Y,1.5*dir(Y),linewidth(4));
dot("$Z$",Z,1.5*dir(Z),linewidth(4));
dot(O^^Q^^R,linewidth(4));
dot("$N$",N,1.5*dir(N),linewidth(4));
dot("$M$",M,1.5*dir(M),linewidth(4));
dot("$P$",P,1.5*dir(P),linewidth(4));
dot("$G$",G,1.5*dir(G),linewidth(4));
draw(A--C--E--cycle^^X--Y--Z--cycle);
draw(A--G^^C--G);
draw(N--B);
draw(N--E--A--cycle^^M--P, red);
// Block 2
/* Made by MRENTHUSIASM */ size(200); draw(polygon(6)); pair A, B, C, D, E, F, X, Y, Z, M, N, O, P, Q, R, G; A = dir(120); B = dir(60); C = dir(0); D = dir(300); E = dir(240); F = dir(180); X = midpoint(A--B); Y = midpoint(C--D); Z = midpoint(E--F); G = midpoint(A--D); M = intersectionpoint(A--E,X--Z); N = intersectionpoint(A--C,X--Y); O = intersectionpoint(C--E,Y--Z); P = intersectionpoint(A--C,X--Z); Q = intersectionpoint(C--E,X--Y); R = intersectionpoint(A--E,Y--Z); fill(M--P--N--Q--O--R--cycle,mediumgray); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*dir(B),linewidth(4)); dot("$C$",C,1.5*dir(C),linewidth(4)); dot("$D$",D,1.5*dir(D),linewidth(4)); dot("$E$",E,1.5*dir(E),linewidth(4)); dot("$F$",F,1.5*dir(F),linewidth(4)); dot("$X$",X,1.5*dir(X),linewidth(4)); dot("$Y$",Y,1.5*dir(Y),linewidth(4)); dot("$Z$",Z,1.5*dir(Z),linewidth(4)); dot(O^^Q^^R,linewidth(4)); dot("$N$",N,1.5*dir(N),linewidth(4)); dot("$M$",M,1.5*dir(M),linewidth(4)); dot("$P$",P,1.5*dir(P),linewidth(4)); dot("$G$",G,1.5*dir(G),linewidth(4)); draw(A--C--E--cycle^^X--Y--Z--cycle); draw(A--G^^C--G); draw(N--B); draw(N--E--A--cycle^^M--P, red); | [] |
829 | Let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x$. How many real numbers $x$ satisfy the equation $x^2 + 10,000\lfloor x \rfloor = 10,000x$?
$\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201$ | 2018 AMC 10B Problem 25 | This rewrites itself to $x^2=10,000\{x\}$ where $\lfloor x \rfloor + \{x\} = x$.
Graphing $y=10,000\{x\}$ and $y=x^2$ we see that the former is a set of line segments with slope $10,000$ from $0$ to $1$ with a hole at $x=1$, then $1$ to $2$ with a hole at $x=2$ etc.
Here is a graph of $y=x^2$ and $y=16\{x\}$ for visualization.
Now notice that when $x=\pm 100$ the graph has a hole at $(\pm 100,10,000)$ which the equation $y=x^2$ passes through and then continues upwards. Thus our set of possible solutions is bounded by $(-100,100)$. We can see that $y=x^2$ intersects each of the lines once and there are $99-(-99)+1=199$ lines for an answer of $\boxed{\text{(C)}~199}$. | // Block 1
import graph;
size(400);
xaxis("$x$",Ticks(Label(fontsize(8pt)),new real[]{-5,-4,-3, -2, -1,0,1 2,3, 4,5}));
yaxis("$y$",Ticks(Label(fontsize(8pt)),new real[]{0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18}));
real y(real x) {return x^2;}
draw(circle((-4,16), 0.1));
draw(circle((-3,16), 0.1));
draw(circle((-2,16), 0.1));
draw(circle((-1,16), 0.1));
draw(circle((0,16), 0.1));
draw(circle((1,16), 0.1));
draw(circle((2,16), 0.1));
draw(circle((3,16), 0.1));
draw(circle((4,16), 0.1));
draw((-5,0)--(-4,16), black);
draw((-4,0)--(-3,16), black);
draw((-3,0)--(-2,16), black);
draw((-2,0)--(-1,16), black);
draw((-1,0)--(-0,16), black);
draw((0,0)--(1,16), black);
draw((1,0)--(2,16), black);
draw((2,0)--(3,16), black);
draw((3,0)--(4,16), black);
draw(graph(y,-4.2,4.2),green);
// Block 2
import graph; size(400); xaxis("$x$",Ticks(Label(fontsize(8pt)),new real[]{-5,-4,-3, -2, -1,0,1 2,3, 4,5})); yaxis("$y$",Ticks(Label(fontsize(8pt)),new real[]{0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18})); real y(real x) {return x^2;} draw(circle((-4,16), 0.1)); draw(circle((-3,16), 0.1)); draw(circle((-2,16), 0.1)); draw(circle((-1,16), 0.1)); draw(circle((0,16), 0.1)); draw(circle((1,16), 0.1)); draw(circle((2,16), 0.1)); draw(circle((3,16), 0.1)); draw(circle((4,16), 0.1)); draw((-5,0)--(-4,16), black); draw((-4,0)--(-3,16), black); draw((-3,0)--(-2,16), black); draw((-2,0)--(-1,16), black); draw((-1,0)--(-0,16), black); draw((0,0)--(1,16), black); draw((1,0)--(2,16), black); draw((2,0)--(3,16), black); draw((3,0)--(4,16), black); draw(graph(y,-4.2,4.2),green); | [] |
830 | A circle has a chord of length $10$, and the distance from the center of the circle to the chord is $5$. What is the area of the circle?
$\textbf{(A) }25\pi \qquad \textbf{(B) }50\pi \qquad \textbf{(C) }75\pi \qquad \textbf{(D) }100\pi \qquad \textbf{(E) }125\pi \qquad$ | 2018 AMC 12B Problem 4 | Let $O$ be the center of the circle, $\overline{AB}$ be the chord, and $M$ be the midpoint of $\overline{AB},$ as shown below.
Note that $\overline{OM}\perp\overline{AB}.$ Since $OM=AM=BM=5,$ we conclude that $\triangle OMA$ and $\triangle OMB$ are congruent isosceles right triangles. It follows that $r=5\sqrt2,$ so the area of $\odot O$ is $\pi r^2=\boxed{\textbf{(B) }50\pi}$.
~MRENTHUSIASM | // Block 1
/* Made by MRENTHUSIASM */
size(200);
pair O, A, B, M;
O = (0,0);
A = (-5,5);
B = (5,5);
M = midpoint(A--B);
draw(Circle(O,5sqrt(2)));
dot("$O$", O, 1.5*S, linewidth(4.5));
dot("$A$", A, 1.5*NW, linewidth(4.5));
dot("$B$", B, 1.5*NE, linewidth(4.5));
dot("$M$", M, 1.5*N, linewidth(4.5));
draw(A--B^^M--O^^A--O^^M--O^^B--O);
label("$5$", midpoint(A--M), 1.5*N);
label("$5$", midpoint(B--M), 1.5*N);
label("$5$", midpoint(O--M), 1.5*E);
label("$r$", midpoint(O--A), 1.5*SW);
label("$r$", midpoint(O--B), 1.5*SE);
// Block 2
/* Made by MRENTHUSIASM */ size(200); pair O, A, B, M; O = (0,0); A = (-5,5); B = (5,5); M = midpoint(A--B); draw(Circle(O,5sqrt(2))); dot("$O$", O, 1.5*S, linewidth(4.5)); dot("$A$", A, 1.5*NW, linewidth(4.5)); dot("$B$", B, 1.5*NE, linewidth(4.5)); dot("$M$", M, 1.5*N, linewidth(4.5)); draw(A--B^^M--O^^A--O^^M--O^^B--O); label("$5$", midpoint(A--M), 1.5*N); label("$5$", midpoint(B--M), 1.5*N); label("$5$", midpoint(O--M), 1.5*E); label("$r$", midpoint(O--A), 1.5*SW); label("$r$", midpoint(O--B), 1.5*SE); | [] |
831 | Line segment $\overline{AB}$ is a diameter of a circle with $AB = 24$. Point $C$, not equal to $A$ or $B$, lies on the circle. As point $C$ moves around the circle, the centroid (center of mass) of $\triangle ABC$ traces out a closed curve missing two points. To the nearest positive integer, what is the area of the region bounded by this curve?
$\textbf{(A) } 25 \qquad \textbf{(B) } 38 \qquad \textbf{(C) } 50 \qquad \textbf{(D) } 63 \qquad \textbf{(E) } 75$ | 2018 AMC 12B Problem 8 | By the Inscribed Angle Theorem, $\triangle ABC$ is a right triangle with $\angle C=90^{\circ}.$ So, its circumcenter is the midpoint of $\overline{AB},$ and its median from $C$ is half as long as $\overline{AB}.$ For each $\triangle ABC,$ let $O$ and $G$ be its circumcenter and centroid, respectively. It follows that $OA=OB=OC=12.$ In any triangle, since the centroid divides each median into parts in the ratio $2:1,$ with the centroid being twice as close to the midpoint of a side as it is to the opposite vertex, we have $OG=\frac13 OC=4.$
As shown below, $\triangle ABC_1$ and $\triangle ABC_2$ are two shapes of $\triangle ABC$ with centroids $G_1$ and $G_2,$ respectively:
Therefore, point $G$ traces out a circle (missing two points) with the center $O$ and the radius $\overline{OG},$ as indicated in red. To the nearest positive integer, the area of the region bounded by the red curve is $\pi\cdot OG^2=16\pi\approx\boxed{\textbf{(C) } 50}.$
~MRENTHUSIASM ~megacleverstarfish15 | /* Made by MRENTHUSIASM */ size(200); pair O, A, B, C1, C2, G1, G2, M1, M2; O = (0,0); A = (-12,0); B = (12,0); C1 = (36/5,48/5); C2 = (-96/17,-180/17); G1 = O + 1/3 * C1; G2 = O + 1/3 * C2; M1 = (4,0); M2 = (-4,0); draw(Circle(O,12)); draw(Circle(O,4),red); dot("$O$", O, (3/5,-4/5), linewidth(4.5)); dot("$A$", A, W, linewidth(4.5)); dot("$B$", B, E, linewidth(4.5)); dot("$C_1$", C1, dir(C1), linewidth(4.5)); dot("$C_2$", C2, dir(C2), linewidth(4.5)); dot("$G_1$", G1, 1.5*E, linewidth(4.5)); dot("$G_2$", G2, 1.5*W, linewidth(4.5)); draw(A--B^^A--C1--B^^A--C2--B); draw(O--C1^^O--C2); dot(M1,red+linewidth(0.8),UnFill); dot(M2,red+linewidth(0.8),UnFill); | [] |
832 | A closed box with a square base is to be wrapped with a square sheet of wrapping paper. The box is centered on the wrapping paper with the vertices of the base lying on the midlines of the square sheet of paper, as shown in the figure on the left. The four corners of the wrapping paper are to be folded up over the sides and brought together to meet at the center of the top of the box, point $A$ in the figure on the right. The box has base length $w$ and height $h$. What is the area of the sheet of wrapping paper?
$\textbf{(A) } 2(w+h)^2 \qquad \textbf{(B) } \frac{(w+h)^2}2 \qquad \textbf{(C) } 2w^2+4wh \qquad \textbf{(D) } 2w^2 \qquad \textbf{(E) } w^2h$ | 2018 AMC 12B Problem 11 | Consider one-quarter of the image (the wrapping paper is divided up into $4$ congruent squares). The length of each dotted line is $h$. The area of the rectangle that is $w$ by $h$ is $wh$. The combined figure of the two triangles with base $h$ is a square with $h$ as its diagonal. Using the Pythagorean Theorem, each side of this square is $\frac{h}{\sqrt2}$. Thus, the area is the side length squared which is $\frac{h^2}{2}$. Similarly, the combined figure of the two triangles with base $w$ is a square with area $\frac{w^2}{2}$. Adding all of these together, we get $\frac{w^2}{2} + \frac{h^2}{2} + wh$. Since we have four of these areas in the entire wrapping paper, we multiply this by $4$, getting $4\left(\frac{w^2}{2} + \frac{h^2}{2} + wh\right) = 2\left(w^2 + h^2 + 2wh\right) = \boxed{\textbf{(A) } 2(w+h)^2}$.
The diagram for this solution is shown below:
~Hydroquantum (Solution)
~MRENTHUSIASM (Diagram) | /* Edited by MRENTHUSIASM */ size(180pt); defaultpen(fontsize(10pt)); fill(((0,3)--(-3,3)--(-3,0)--(0,0)--cycle),lightgrey); dot((-3,3)); label("$A$",(-3,3),NW); draw((-3,0)--(-3,3)--(0,3),linewidth(.5)); draw((0,2)--(-1,3)--(-3,1)--(-2,0),dashed+linewidth(.5)); draw((0,2)--(-2,0),linewidth(.5)); draw((0,3)--(0,0),linetype("2.5 2.5")+linewidth(.5)); draw((0,0)--(-3,0),linetype("2.5 2.5")+linewidth(.5)); label("$w$",(-1,1),NW); label("$w$",(-2,2),SE); label("$h$",(-2.5,0.5),NE); label("$h$",(-0.5,2.5),SW); label("$\frac{w}{\sqrt{2}}$",(-2,3),N); label("$\frac{h}{\sqrt{2}}$",(-0.5,3),N); label("$\frac{h}{\sqrt{2}}$",(0,2.5),E); label("$\frac{w}{\sqrt{2}}$",(0,1),E); label("$\frac{w}{\sqrt{2}}$",(-1,0),S); label("$\frac{h}{\sqrt{2}}$",(-2.5,0),S); label("$\frac{h}{\sqrt{2}}$",(-3,0.5),W); label("$\frac{w}{\sqrt{2}}$",(-3,2),W); label("$wh$",(-1.5,1.5),red); label("$\frac{w^2}{4}$",centroid((-3,3),(-1,3),(-3,1)),red); label("$\frac{w^2}{4}$",centroid((0,0),(-2,0),(0,2)),red); label("$\frac{h^2}{4}$",centroid((-3,0),(-2,0),(-3,1)),red); label("$\frac{h^2}{4}$",centroid((0,3),(-1,3),(0,2)),red); | [] |
832 | A closed box with a square base is to be wrapped with a square sheet of wrapping paper. The box is centered on the wrapping paper with the vertices of the base lying on the midlines of the square sheet of paper, as shown in the figure on the left. The four corners of the wrapping paper are to be folded up over the sides and brought together to meet at the center of the top of the box, point $A$ in the figure on the right. The box has base length $w$ and height $h$. What is the area of the sheet of wrapping paper?
$\textbf{(A) } 2(w+h)^2 \qquad \textbf{(B) } \frac{(w+h)^2}2 \qquad \textbf{(C) } 2w^2+4wh \qquad \textbf{(D) } 2w^2 \qquad \textbf{(E) } w^2h$ | 2018 AMC 12B Problem 11 | The sheet of paper is made out of the surface area of the box plus the sum of the four yellow triangles, as shown below.
The surface area is $2w^2 + 2wh + 2wh$ which equals $2w^2 + 4wh$.The four triangles each have a height and a base of $h$, so they each have an area of $\frac{h^2}{2}$. There are four of them, so multiplied by four is $2h^2$. Together, paper's area is $2w^2 + 4wh + 2h^2$. This can be factored and written as $\boxed{\textbf{(A) } 2(w+h)^2} \qquad$.
~Yee2121 (Solution)
~MRENTHUSIASM (Diagram) | /* Edited by MRENTHUSIASM */ size(180pt); defaultpen(fontsize(10pt)); fill(((3,3)--(-3,3)--(-3,-3)--(3,-3)--cycle),lightgrey); fill((2,0)--(3,1)--(3,-1)--cycle,yellow); fill((-2,0)--(-3,1)--(-3,-1)--cycle,yellow); fill((0,2)--(1,3)--(-1,3)--cycle,yellow); fill((0,-2)--(1,-3)--(-1,-3)--cycle,yellow); draw(((3,3)--(-3,3)--(-3,-3)--(3,-3)--cycle),linewidth(.5)); dot((-3,3)); label("$A$",(-3,3),NW); draw((1,3)--(-3,-1),dashed+linewidth(.5)); draw((-1,3)--(3,-1),dashed+linewidth(.5)); draw((-1,-3)--(3,1),dashed+linewidth(.5)); draw((1,-3)--(-3,1),dashed+linewidth(.5)); draw((0,2)--(2,0)--(0,-2)--(-2,0)--cycle,linewidth(.5)); draw((0,3)--(0,-3),linetype("2.5 2.5")+linewidth(.5)); draw((3,0)--(-3,0),linetype("2.5 2.5")+linewidth(.5)); label("$w$",(-1,-1),SW); label("$w$",(1,-1),SE); label("$w$",(1,1),NE); label("$w$",(-1,1),NW); label("$h$",(2.5,0.5),NW); label("$h$",(2.5,-0.5),SW); label("$h$",(-2.5,0.5),NE); label("$h$",(-2.5,-0.5),SE); label("$h$",(0.5,2.5),SE); label("$h$",(-0.5,2.5),SW); label("$h$",(0.5,-2.5),NE); label("$h$",(-0.5,-2.5),NW); | [] |
833 | Square $ABCD$ has side length $30$. Point $P$ lies inside the square so that $AP = 12$ and $BP = 26$. The centroids of $\triangle{ABP}$, $\triangle{BCP}$, $\triangle{CDP}$, and $\triangle{DAP}$ are the vertices of a convex quadrilateral. What is the area of that quadrilateral?
$\textbf{(A) }100\sqrt{2}\qquad\textbf{(B) }100\sqrt{3}\qquad\textbf{(C) }200\qquad\textbf{(D) }200\sqrt{2}\qquad\textbf{(E) }200\sqrt{3}$ | 2018 AMC 12B Problem 13 | As shown below, let $M_1,M_2,M_3,M_4$ be the midpoints of $\overline{AB},\overline{BC},\overline{CD},\overline{DA},$ respectively, and $G_1,G_2,G_3,G_4$ be the centroids of $\triangle{ABP},\triangle{BCP},\triangle{CDP},\triangle{DAP},$ respectively.
By SAS, we conclude that $\triangle G_1G_2P\sim\triangle M_1M_2P, \triangle G_2G_3P\sim\triangle M_2M_3P, \triangle G_3G_4P\sim\triangle M_3M_4P,$ and $\triangle G_4G_1P\sim\triangle M_4M_1P.$ By the properties of centroids, the ratio of similitude for each pair of triangles is $\frac{2}{3}.$
Note that quadrilateral $M_1M_2M_3M_4$ is a square of side-length $15\sqrt2.$ It follows that:
Since $\overline{G_1G_2}\parallel\overline{M_1M_2},\overline{G_2G_3}\parallel\overline{M_2M_3},\overline{G_3G_4}\parallel\overline{M_3M_4},$ and $\overline{G_4G_1}\parallel\overline{M_4M_1}$ by the Converse of the Corresponding Angles Postulate, we have $\angle G_1G_2G_3=\angle G_2G_3G_4=\angle G_3G_4G_1=\angle G_4G_1G_2=90^\circ.$
Since $G_1G_2=\frac23M_1M_2, G_2G_3=\frac23M_2M_3, G_3G_4=\frac23M_3M_4,$ and $G_4G_1=\frac23M_4M_1$ by the ratio of similitude, we have $G_1G_2=G_2G_3=G_3G_4=G_4G_1=10\sqrt2.$
Together, quadrilateral $G_1G_2G_3G_4$ is a square of side-length $10\sqrt2,$ so its area is $\left(10\sqrt2\right)^2=\boxed{\textbf{(C) }200}.$
Remark
This solution shows that, if point $P$ is within square $ABCD,$ then the shape and the area of quadrilateral $G_1G_2G_3G_4$ are independent of the location of $P.$ Let the brackets denote areas. More generally, $G_1G_2G_3G_4$ is always a square of area \[[G_1G_2G_3G_4]=\left(\frac23\right)^2[M_1M_2M_3M_4]=\frac49[M_1M_2M_3M_4]=\frac29[ABCD].\] On the other hand, the location of $G_1G_2G_3G_4$ is dependent on the location of $P.$
~RandomPieKevin ~Kyriegon ~MRENTHUSIASM | // Block 1
/* Made by MRENTHUSIASM */
unitsize(210);
pair B = (0, 0), A = (0, 1), D = (1, 1), C = (1, 0), P = (1/4, 2/3);
pair M1 = midpoint(A--B);
pair M2 = midpoint(B--C);
pair M3 = midpoint(C--D);
pair M4 = midpoint(D--A);
pair G1 = centroid(A,B,P);
pair G2 = centroid(B,C,P);
pair G3 = centroid(C,D,P);
pair G4 = centroid(D,A,P);
filldraw(M1--M2--P--cycle,red);
filldraw(M2--M3--P--cycle,yellow);
filldraw(M3--M4--P--cycle,green);
filldraw(M4--M1--P--cycle,lightblue);
draw(A--B--C--D--cycle);
draw(M1--M2--M3--M4--cycle);
draw(G1--G2--G3--G4--cycle);
dot(P);
defaultpen(fontsize(10pt));
draw(A--P--B);
draw(C--P--D);
label("$A$", A, W);
label("$B$", B, W);
label("$C$", C, E);
label("$D$", D, E);
label("$P$", P, N);
label("$M_1$", M1, W);
label("$M_2$", M2, S);
label("$M_3$", M3, E);
label("$M_4$", M4, N);
label("$G_1$", G1, 1.5S);
label("$G_2$", G2, 1.5E);
label("$G_3$", G3, 1.5NE);
label("$G_4$", G4, 1.5E);
dot(A);
dot(B);
dot(C);
dot(D);
dot(M1);
dot(M2);
dot(M3);
dot(M4);
dot(G1);
dot(G2);
dot(G3);
dot(G4);
// Block 2
/* Made by MRENTHUSIASM */ unitsize(210); pair B = (0, 0), A = (0, 1), D = (1, 1), C = (1, 0), P = (1/4, 2/3); pair M1 = midpoint(A--B); pair M2 = midpoint(B--C); pair M3 = midpoint(C--D); pair M4 = midpoint(D--A); pair G1 = centroid(A,B,P); pair G2 = centroid(B,C,P); pair G3 = centroid(C,D,P); pair G4 = centroid(D,A,P); filldraw(M1--M2--P--cycle,red); filldraw(M2--M3--P--cycle,yellow); filldraw(M3--M4--P--cycle,green); filldraw(M4--M1--P--cycle,lightblue); draw(A--B--C--D--cycle); draw(M1--M2--M3--M4--cycle); draw(G1--G2--G3--G4--cycle); dot(P); defaultpen(fontsize(10pt)); draw(A--P--B); draw(C--P--D); label("$A$", A, W); label("$B$", B, W); label("$C$", C, E); label("$D$", D, E); label("$P$", P, N); label("$M_1$", M1, W); label("$M_2$", M2, S); label("$M_3$", M3, E); label("$M_4$", M4, N); label("$G_1$", G1, 1.5S); label("$G_2$", G2, 1.5E); label("$G_3$", G3, 1.5NE); label("$G_4$", G4, 1.5E); dot(A); dot(B); dot(C); dot(D); dot(M1); dot(M2); dot(M3); dot(M4); dot(G1); dot(G2); dot(G3); dot(G4); | [] |
834 | The solutions to the equation $(z+6)^8=81$ are connected in the complex plane to form a convex regular polygon, three of whose vertices are labeled $A,B,$ and $C$. What is the least possible area of $\triangle ABC?$
$\textbf{(A) } \frac{1}{6}\sqrt{6} \qquad \textbf{(B) } \frac{3}{2}\sqrt{2}-\frac{3}{2} \qquad \textbf{(C) } 2\sqrt3-3\sqrt2 \qquad \textbf{(D) } \frac{1}{2}\sqrt{2} \qquad \textbf{(E) } \sqrt 3-1$ | 2018 AMC 12B Problem 16 | Recall that translations preserve the shapes and the sizes for all objects. We translate the solutions to the given equation $6$ units right, so they become the solutions to the equation $z^8=81.$
We rewrite $z$ to the polar form \[z=r(\cos\theta+i\sin\theta)=r\operatorname{cis}\theta,\] where $r$ is the magnitude of $z$ such that $r\geq0,$ and $\theta$ is the argument of $z$ such that $0\leq\theta<2\pi.$
By De Moivre's Theorem, we have \[z^8=r^8\operatorname{cis}(8\theta)={\sqrt3}^8(1),\] from which
$r^8={\sqrt3}^8,$ so $r=\sqrt3.$
$\begin{cases} \begin{aligned} \cos(8\theta) &= 1 \\ \sin(8\theta) &= 0 \end{aligned}, \end{cases}$ so $\theta=0,\frac{\pi}{4},\frac{\pi}{2},\frac{3\pi}{4},\pi,\frac{5\pi}{4},\frac{3\pi}{2},\frac{7\pi}{4}.$
In the complex plane, the solutions to the equation $z^8=81$ are the vertices of a regular octagon with center $0$ and radius $\sqrt3.$
The least possible area of $\triangle ABC$ occurs when $A,B,$ and $C$ are the consecutive vertices of the octagon. For simplicity purposes, let $A=\sqrt3\operatorname{cis}\frac{\pi}{4}=\frac{\sqrt6}{2}+\frac{\sqrt6}{2}i, B=\sqrt3\operatorname{cis}\frac{\pi}{2}=\sqrt3i,$ and $C=\sqrt3\operatorname{cis}\frac{3\pi}{4}=-\frac{\sqrt6}{2}+\frac{\sqrt6}{2}i,$ as shown below.
Note that $\triangle ABC$ has base $AC=\sqrt6$ and height $\sqrt3-\frac{\sqrt6}{2},$ so its area is \[\frac12\cdot\sqrt6\cdot\left(\sqrt3-\frac{\sqrt6}{2}\right)=\boxed{\textbf{(B) } \frac{3}{2}\sqrt{2}-\frac{3}{2}}.\]
~MRENTHUSIASM | // Block 1
/* Made by MRENTHUSIASM */
size(200);
int xMin = -2;
int xMax = 2;
int yMin = -2;
int yMax = 2;
int numRays = 24;
//Draws a polar grid that goes out to a number of circles
//equal to big, with numRays specifying the number of rays:
void polarGrid(int big, int numRays)
{
for (int i = 1; i < big+1; ++i)
{
draw(Circle((0,0),i), gray+linewidth(0.4));
}
for (int i=0;i<numRays;++i)
draw(rotate(i*360/numRays)*((-big,0)--(big,0)), gray+linewidth(0.4));
}
//Draws the horizontal gridlines
void horizontalLines()
{
for (int i = yMin+1; i < yMax; ++i)
{
draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4));
}
}
//Draws the vertical gridlines
void verticalLines()
{
for (int i = xMin+1; i < xMax; ++i)
{
draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4));
}
}
horizontalLines();
verticalLines();
polarGrid(xMax,numRays);
draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5));
draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5));
label("Re",(xMax,0),2*E);
label("Im",(0,yMax),2*N);
//The n such that we're taking the nth roots of unity multiplied by 2.
int n = 8;
pair A[];
for(int i = 0; i <= n-1; i+=1) {
A[i] = rotate(360*i/n)*(sqrt(3),0);
}
label("$A$",A[1],1.5*NE,UnFill);
label("$B$",A[2],1.5*NE,UnFill);
label("$C$",A[3],1.5*NW,UnFill);
fill(A[1]--A[2]--A[3]--cycle,green);
draw(A[1]--A[3]^^A[0]--A[1]--A[2]--A[3]--A[4]--A[5]--A[6]--A[7]--cycle,red);
for(int i = 0; i< n; ++i) dot(A[i],red+linewidth(4.5));
// Block 2
/* Made by MRENTHUSIASM */ size(200); int xMin = -2; int xMax = 2; int yMin = -2; int yMax = 2; int numRays = 24; //Draws a polar grid that goes out to a number of circles //equal to big, with numRays specifying the number of rays: void polarGrid(int big, int numRays) { for (int i = 1; i < big+1; ++i) { draw(Circle((0,0),i), gray+linewidth(0.4)); } for (int i=0;i<numRays;++i) draw(rotate(i*360/numRays)*((-big,0)--(big,0)), gray+linewidth(0.4)); } //Draws the horizontal gridlines void horizontalLines() { for (int i = yMin+1; i < yMax; ++i) { draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4)); } } //Draws the vertical gridlines void verticalLines() { for (int i = xMin+1; i < xMax; ++i) { draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4)); } } horizontalLines(); verticalLines(); polarGrid(xMax,numRays); draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("Re",(xMax,0),2*E); label("Im",(0,yMax),2*N); //The n such that we're taking the nth roots of unity multiplied by 2. int n = 8; pair A[]; for(int i = 0; i <= n-1; i+=1) { A[i] = rotate(360*i/n)*(sqrt(3),0); } label("$A$",A[1],1.5*NE,UnFill); label("$B$",A[2],1.5*NE,UnFill); label("$C$",A[3],1.5*NW,UnFill); fill(A[1]--A[2]--A[3]--cycle,green); draw(A[1]--A[3]^^A[0]--A[1]--A[2]--A[3]--A[4]--A[5]--A[6]--A[7]--cycle,red); for(int i = 0; i< n; ++i) dot(A[i],red+linewidth(4.5)); | [] |
835 | Let $ABCDEF$ be a regular hexagon with side length $1$. Denote by $X$, $Y$, and $Z$ the midpoints of sides $\overline {AB}$, $\overline{CD}$, and $\overline{EF}$, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of $\triangle ACE$ and $\triangle XYZ$?
$\textbf{(A)}\ \frac {3}{8}\sqrt{3} \qquad \textbf{(B)}\ \frac {7}{16}\sqrt{3} \qquad \textbf{(C)}\ \frac {15}{32}\sqrt{3} \qquad \textbf{(D)}\ \frac {1}{2}\sqrt{3} \qquad \textbf{(E)}\ \frac {9}{16}\sqrt{3}$ | 2018 AMC 12B Problem 20 | The desired area (hexagon $MPNQOR$) consists of an equilateral triangle ($\triangle MNO$) and three right triangles ($\triangle MPN,\triangle NQO,$ and $\triangle ORM$).
Notice that $\overline {AD}$ (not shown) and $\overline {BC}$ are parallel. $\overline {XY}$ divides transversals $\overline {AB}$ and $\overline {CD}$ into a $1:1$ ratio (This can be shown by similar triangles.). Thus, it must also divide transversal $\overline {AC}$ and transversal $\overline {CO}$ into a $1:1$ ratio. By symmetry, the same applies for $\overline {CE}$ and $\overline {EA}$ as well as $\overline {EM}$ and $\overline {AN}.$
In $\triangle ACE,$ we see that $\frac{[MNO]}{[ACE]} = \frac{1}{4}$ and $\frac{[MPN]}{[ACE]} = \frac{1}{8}.$ Our desired area becomes \[\left(\frac{1}{4}+3 \cdot \frac{1}{8}\right) \cdot \frac{(\sqrt{3})^2 \cdot \sqrt{3}}{4} = \boxed{\textbf{(C)}\ \frac {15}{32}\sqrt{3}}.\] | /* Made by MRENTHUSIASM */ size(200); draw(polygon(6)); pair A, B, C, D, E, F, X, Y, Z, M, N, O, P, Q, R; A = dir(120); B = dir(60); C = dir(0); D = dir(300); E = dir(240); F = dir(180); X = midpoint(A--B); Y = midpoint(C--D); Z = midpoint(E--F); M = intersectionpoint(A--E,X--Z); N = intersectionpoint(A--C,X--Y); O = intersectionpoint(C--E,Y--Z); P = intersectionpoint(A--C,X--Z); Q = intersectionpoint(C--E,X--Y); R = intersectionpoint(A--E,Y--Z); fill(M--P--N--Q--O--R--cycle,mediumgray); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*dir(B),linewidth(4)); dot("$C$",C,1.5*dir(C),linewidth(4)); dot("$D$",D,1.5*dir(D),linewidth(4)); dot("$E$",E,1.5*dir(E),linewidth(4)); dot("$F$",F,1.5*dir(F),linewidth(4)); dot("$X$",X,1.5*dir(X),linewidth(4)); dot("$Y$",Y,1.5*dir(Y),linewidth(4)); dot("$Z$",Z,1.5*dir(Z),linewidth(4)); dot("$M$",M,1.5*dir(165),linewidth(4)); dot("$N$",N,1.5*dir(45),linewidth(4)); dot("$O$",O,1.5*dir(-75),linewidth(4)); dot("$P$",P,1.5*dir(105),linewidth(4)); dot("$Q$",Q,1.5*dir(-15),linewidth(4)); dot("$R$",R,1.5*dir(-135),linewidth(4)); draw(A--C--E--cycle^^X--Y--Z--cycle); draw(M--N--O--cycle,dashed); | [] |
835 | Let $ABCDEF$ be a regular hexagon with side length $1$. Denote by $X$, $Y$, and $Z$ the midpoints of sides $\overline {AB}$, $\overline{CD}$, and $\overline{EF}$, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of $\triangle ACE$ and $\triangle XYZ$?
$\textbf{(A)}\ \frac {3}{8}\sqrt{3} \qquad \textbf{(B)}\ \frac {7}{16}\sqrt{3} \qquad \textbf{(C)}\ \frac {15}{32}\sqrt{3} \qquad \textbf{(D)}\ \frac {1}{2}\sqrt{3} \qquad \textbf{(E)}\ \frac {9}{16}\sqrt{3}$ | 2018 AMC 12B Problem 20 | Now, if we look at the figure, we can see that the complement of the hexagon we are trying to find is composed of $3$ isosceles trapezoids (namely $AXZF,CYXB,$ and $EZYD$) and $3$ right triangles (namely $\triangle XPN,\triangle YQO,$ and $\triangle ZRM$).
Finding the trapezoid's area, we know that one base of each trapezoid is just the side length of the hexagon, which is $1,$ and the other base is $\frac{3}{2}$ (it is halfway in between the side and the longest diagonal, which has length $2$) with a height of $\frac{\sqrt{3}}{4}$ (by using the Pythagorean Theorem and the fact that it is an isosceles trapezoid) to give each trapezoid having an area of $\frac{5\sqrt{3}}{16}$ for a total area of $\frac{15\sqrt{3}}{16}.$ (Alternatively, we could have calculated the area of hexagon $ABCDEF$ and subtracted the area of $\triangle XYZ,$ which, as we showed before, had a side length of $\frac{3}{2}$).
Now, we need to find the area of each of the small triangles, which, if we look at the triangle that has a vertex on $X,$ is similar to the triangle with a base of $YC = \frac12.$ Using similar triangles, we calculate the base to be $\frac{1}{4}$ and the height to be $\frac{\sqrt{3}}{4}$ giving us an area of $\frac{\sqrt{3}}{32}$ per triangle, and a total area of $\frac{3\sqrt{3}}{32}.$ Adding the two areas together, we get $\frac{15\sqrt{3}}{16} + \frac{3\sqrt{3}}{32} = \frac{33\sqrt{3}}{32}.$ Finding the total area, we get $6 \cdot 1^2 \cdot \frac{\sqrt{3}}{4}=\frac{3\sqrt{3}}{2}.$ Taking the complement, we get $\frac{3\sqrt{3}}{2} - \frac{33\sqrt{3}}{32} = \boxed{\textbf{(C)}\ \frac {15}{32}\sqrt{3}}.$ | /* Made by MRENTHUSIASM */ size(200); draw(polygon(6)); pair A, B, C, D, E, F, X, Y, Z, M, N, O, P, Q, R; A = dir(120); B = dir(60); C = dir(0); D = dir(300); E = dir(240); F = dir(180); X = midpoint(A--B); Y = midpoint(C--D); Z = midpoint(E--F); M = intersectionpoint(A--E,X--Z); N = intersectionpoint(A--C,X--Y); O = intersectionpoint(C--E,Y--Z); P = intersectionpoint(A--C,X--Z); Q = intersectionpoint(C--E,X--Y); R = intersectionpoint(A--E,Y--Z); fill(M--P--N--Q--O--R--cycle,mediumgray); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*dir(B),linewidth(4)); dot("$C$",C,1.5*dir(C),linewidth(4)); dot("$D$",D,1.5*dir(D),linewidth(4)); dot("$E$",E,1.5*dir(E),linewidth(4)); dot("$F$",F,1.5*dir(F),linewidth(4)); dot("$X$",X,1.5*dir(X),linewidth(4)); dot("$Y$",Y,1.5*dir(Y),linewidth(4)); dot("$Z$",Z,1.5*dir(Z),linewidth(4)); dot("$M$",M,1.5*dir(165),linewidth(4)); dot("$N$",N,1.5*dir(45),linewidth(4)); dot("$O$",O,1.5*dir(-75),linewidth(4)); dot("$P$",P,1.5*dir(105),linewidth(4)); dot("$Q$",Q,1.5*dir(-15),linewidth(4)); dot("$R$",R,1.5*dir(-135),linewidth(4)); draw(A--C--E--cycle^^X--Y--Z--cycle); draw(M--N--O--cycle,dashed); | [] |
835 | Let $ABCDEF$ be a regular hexagon with side length $1$. Denote by $X$, $Y$, and $Z$ the midpoints of sides $\overline {AB}$, $\overline{CD}$, and $\overline{EF}$, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of $\triangle ACE$ and $\triangle XYZ$?
$\textbf{(A)}\ \frac {3}{8}\sqrt{3} \qquad \textbf{(B)}\ \frac {7}{16}\sqrt{3} \qquad \textbf{(C)}\ \frac {15}{32}\sqrt{3} \qquad \textbf{(D)}\ \frac {1}{2}\sqrt{3} \qquad \textbf{(E)}\ \frac {9}{16}\sqrt{3}$ | 2018 AMC 12B Problem 20 | We could also subtract $\triangle APM,\triangle CQN,$ and $\triangle ERO$ from $\triangle ACE.$
Since $\angle BAF = 120^{\circ}$ and $\angle BAC = \angle FAE = 30^{\circ},$ we have $\angle CAE = \angle BAF-\angle BAC-\angle FAE=60^{\circ}.$
Since $AX=BX$ and $FZ=EZ,$ we have $AF \parallel XZ,$ from which $\angle AMX= \angle FAM = 30^{\circ}.$
We can show that $\triangle APM$ is $30$-$60$-$90$ using a similar method, $\triangle CQN$ and $\triangle ERO$ are also $30$-$60$-$90.$
Since $AC=CE=AE=\sqrt{3},$ we have $[ACE]=AC^2 \cdot \frac{\sqrt{3}}{4}=3 \cdot \frac{\sqrt{3}}{4} = \frac{3 \sqrt{3}}{4}.$
Since $AX= \frac{1}{2}$ and $AP = AX \cdot \frac{\sqrt{3}}{2}= \frac{1}{2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4},$ we have $PM = AP \cdot \sqrt{3} = \frac{\sqrt{3}}{4} \cdot \sqrt{3} = \frac{3}{4}.$
Note that \[[APM]=[CQN]=[ERO]=\frac{1}{2} \cdot AP \cdot PM = \frac{1}{2} \cdot \frac{\sqrt{3}}{4} \cdot \frac{3}{4} = \frac{3 \sqrt{3}}{32}.\]
Therefore, we get
\[[PNQORM]=[ACE]-[APM]-[CQN]-[ERO]=\frac{3 \sqrt{3}}{4} - 3 \cdot \frac{3 \sqrt{3}}{32} = \frac{24 \sqrt{3}}{32} - \frac{9 \sqrt{3}}{32} = \boxed{\textbf{(C)}\ \frac {15}{32}\sqrt{3}}.\]
~isabelchen | /* Made by MRENTHUSIASM */ size(200); draw(polygon(6)); pair A, B, C, D, E, F, X, Y, Z, M, N, O, P, Q, R; A = dir(120); B = dir(60); C = dir(0); D = dir(300); E = dir(240); F = dir(180); X = midpoint(A--B); Y = midpoint(C--D); Z = midpoint(E--F); M = intersectionpoint(A--E,X--Z); N = intersectionpoint(A--C,X--Y); O = intersectionpoint(C--E,Y--Z); P = intersectionpoint(A--C,X--Z); Q = intersectionpoint(C--E,X--Y); R = intersectionpoint(A--E,Y--Z); fill(M--P--N--Q--O--R--cycle,mediumgray); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*dir(B),linewidth(4)); dot("$C$",C,1.5*dir(C),linewidth(4)); dot("$D$",D,1.5*dir(D),linewidth(4)); dot("$E$",E,1.5*dir(E),linewidth(4)); dot("$F$",F,1.5*dir(F),linewidth(4)); dot("$X$",X,1.5*dir(X),linewidth(4)); dot("$Y$",Y,1.5*dir(Y),linewidth(4)); dot("$Z$",Z,1.5*dir(Z),linewidth(4)); dot("$M$",M,1.5*dir(165),linewidth(4)); dot("$N$",N,1.5*dir(45),linewidth(4)); dot("$O$",O,1.5*dir(-75),linewidth(4)); dot("$P$",P,1.5*dir(105),linewidth(4)); dot("$Q$",Q,1.5*dir(-15),linewidth(4)); dot("$R$",R,1.5*dir(-135),linewidth(4)); draw(A--C--E--cycle^^X--Y--Z--cycle); | [] |
835 | Let $ABCDEF$ be a regular hexagon with side length $1$. Denote by $X$, $Y$, and $Z$ the midpoints of sides $\overline {AB}$, $\overline{CD}$, and $\overline{EF}$, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of $\triangle ACE$ and $\triangle XYZ$?
$\textbf{(A)}\ \frac {3}{8}\sqrt{3} \qquad \textbf{(B)}\ \frac {7}{16}\sqrt{3} \qquad \textbf{(C)}\ \frac {15}{32}\sqrt{3} \qquad \textbf{(D)}\ \frac {1}{2}\sqrt{3} \qquad \textbf{(E)}\ \frac {9}{16}\sqrt{3}$ | 2018 AMC 12B Problem 20 | We partition hexagon $ABCDEF$ into $48$ congruent $30^\circ\text{-}60^\circ\text{-}90^\circ$ triangles, as shown below:
Let the brackets denote areas. Note that the desired region contains $15$ of the $48$ small triangles, so the answer is \[\frac{15}{48}[ABCDEF]=\frac{15}{48}\cdot\frac{3\sqrt3}{2}=\boxed{\textbf{(C)}\ \frac {15}{32}\sqrt{3}}.\]
~AlexLikeMath ~MRENTHUSIASM | /* Made by MRENTHUSIASM */ size(200); pair A, B, C, D, E, F, X, Y, Z, M, N, O, P, Q, R; A = dir(120); B = dir(60); C = dir(0); D = dir(300); E = dir(240); F = dir(180); X = midpoint(A--B); Y = midpoint(C--D); Z = midpoint(E--F); M = intersectionpoint(A--E,X--Z); N = intersectionpoint(A--C,X--Y); O = intersectionpoint(C--E,Y--Z); P = intersectionpoint(A--C,X--Z); Q = intersectionpoint(C--E,X--Y); R = intersectionpoint(A--E,Y--Z); fill(M--P--N--Q--O--R--cycle,mediumgray); draw(A--D^^B--E^^C--F^^X--Y--Z--cycle^^midpoint(A--F)--midpoint(B--C)--midpoint(D--E)--cycle,red); draw(A--C--E--cycle^^M--N--O--cycle^^M--midpoint(F--Z)^^M--F+1/4*(A-F)^^N--midpoint(X--B)^^N--B+1/4*(C-B)^^O--midpoint(Y--D)^^O--D+1/4*(E-D),blue); draw(polygon(6)); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*dir(B),linewidth(4)); dot("$C$",C,1.5*dir(C),linewidth(4)); dot("$D$",D,1.5*dir(D),linewidth(4)); dot("$E$",E,1.5*dir(E),linewidth(4)); dot("$F$",F,1.5*dir(F),linewidth(4)); dot("$X$",X,1.5*dir(X),linewidth(4)); dot("$Y$",Y,1.5*dir(Y),linewidth(4)); dot("$Z$",Z,1.5*dir(Z),linewidth(4)); dot(M^^N^^O^^P^^Q^^R,linewidth(4)); | [] |
835 | Let $ABCDEF$ be a regular hexagon with side length $1$. Denote by $X$, $Y$, and $Z$ the midpoints of sides $\overline {AB}$, $\overline{CD}$, and $\overline{EF}$, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of $\triangle ACE$ and $\triangle XYZ$?
$\textbf{(A)}\ \frac {3}{8}\sqrt{3} \qquad \textbf{(B)}\ \frac {7}{16}\sqrt{3} \qquad \textbf{(C)}\ \frac {15}{32}\sqrt{3} \qquad \textbf{(D)}\ \frac {1}{2}\sqrt{3} \qquad \textbf{(E)}\ \frac {9}{16}\sqrt{3}$ | 2018 AMC 12B Problem 20 | If we try to coordinate bash this problem, it's going to look very ugly with a lot of radicals. However, we can alter and skew the diagram in such a way that all ratios of lengths and areas stay the same while making it a lot easier to work with. Then, we can find the ratio of the area of the wanted region to the area of $ABCDEF$ then apply it to the old diagram.
The isosceles right triangle with a leg length of $3$ in the new diagram is $\triangle XYZ$ in the old diagram. We see that if we want to take the area of the new hexagon, we must subtract $\frac{3}{4}$ from the area of $\triangle XYZ$ (the red triangles), giving us $\frac{15}{4}.$ However, we need to take the ratio of this area to the area of $ABCDEF,$ which is $\frac{\frac{15}{4}}{12}=\frac{5}{16}.$ Now we know that our answer is $\frac{5}{16} \cdot \frac{3\sqrt{3}}{2}=\boxed{\textbf{(C)}\ \frac {15}{32}\sqrt{3}}.$ | // Block 1
unitsize(1cm); draw((0,0)--(4,0)--(6,3.464)--(2,3.464)--(0,0)); draw((2,0)--(1,1.732)); draw((5,1.732)--(4,3.464)); draw((1.5, 0.866)--(3, 3.464)--(4.5, 0.866)--cycle); draw((2,0)--(2,3.464)--(5,1.732)--cycle);
// Block 2
unitsize(1cm); fill((1,4)--(1,3.5)--(2,3)--cycle,red); fill((1,1)--(1.5,1)--(1,2)--cycle,red); fill((3,1)--(3.5,1.5)--(4,1)--cycle,red); draw((1,0)--(1,4),gray(.7)); draw((2,0)--(2,4),gray(.7)); draw((3,0)--(3,4),gray(.7)); draw((0,1)--(4,1),gray(.7)); draw((0,2)--(4,2),gray(.7)); draw((0,3)--(4,3),gray(.7)); draw((0,0)--(4,0)--(4,4)--(0,4)--(0,0)); draw((2,0)--(0,2)); draw((4,2)--(2,4)); draw((1,1)--(1,4)--(4,1)--cycle); draw((0,4)--(2,0)--(4,2)--cycle); | [] |
835 | Let $ABCDEF$ be a regular hexagon with side length $1$. Denote by $X$, $Y$, and $Z$ the midpoints of sides $\overline {AB}$, $\overline{CD}$, and $\overline{EF}$, respectively. What is the area of the convex hexagon whose interior is the intersection of the interiors of $\triangle ACE$ and $\triangle XYZ$?
$\textbf{(A)}\ \frac {3}{8}\sqrt{3} \qquad \textbf{(B)}\ \frac {7}{16}\sqrt{3} \qquad \textbf{(C)}\ \frac {15}{32}\sqrt{3} \qquad \textbf{(D)}\ \frac {1}{2}\sqrt{3} \qquad \textbf{(E)}\ \frac {9}{16}\sqrt{3}$ | 2018 AMC 12B Problem 20 | (~MRENTHUSIASM's diagram + modifications)
Lemma 1: $MP$ is midsegment of $\triangle ANE$.
As $ZX$ is a midsegment of trapezoid $FEBA$, we know that $FA||ZX||EB$. Furthermore, by subdividing trapezoid $FEBA$, $ZM,MP,PX$ are midsegments of $\triangle EFA, \triangle ANE, \triangle, ANB$ respectively.
Lemma 2: $[ACE] = \frac12[ABCDEF]$
Due to radial symmetry about the center, $G$, we only have to prove $[AGC] = [CBA]$. As $\triangle AGB$ and $\triangle CBG$ are equilateral triangles, $AC$ therefore divides rhombus $ABCG$ into 2 congruent halves. $\triangle AGC$ and $\triangle CBA$ thus each contain two such congruent triangles, which means that their areas are equal.
Lemma 3: $[AMP] = \frac1{16}[ABCDEF]$
From Lemma 1, $\triangle AMP \simeq \triangle AEN$ with a scale factor of $\frac12$, which means $[AMP] = \frac14[AEN]$. Furthermore, due to symmetry of $\triangle AEN$ and $\triangle CEN$ about $EN$, $[AEN] = \frac12[ACE]$. Therefore, $[AMP] = \frac18[ACE]$, or (using lemma 2) $\frac1{16}[ABCDEF]$.
Lemma 4: $[ABCDEF] = \frac{3\sqrt3}2$
Because the hexagon is regular, $[ABCDEF] = 6[ABG] = \frac{6\sqrt3}4 = \frac{3\sqrt3}2$. (For more info on where $\frac{\sqrt3}4$ came from, use the search term "area of equilateral triangle")
Final step:
To find the area of the shaded part, we seek $[ACE] - 3[AMP]$. Using lemmas 2 and 3, the expression resolves to $\frac12[ABCDEF] - 3\frac1{16}[ABCDEF] = \frac5{16} \cdot \frac{3\sqrt3}2 = \boxed{\textbf{(C)}\ \frac{15}{32}\sqrt{3}}$.
~math660 | /* Made by MRENTHUSIASM */ size(200); draw(polygon(6)); pair A, B, C, D, E, F, X, Y, Z, M, N, O, P, Q, R, G; A = dir(120); B = dir(60); C = dir(0); D = dir(300); E = dir(240); F = dir(180); X = midpoint(A--B); Y = midpoint(C--D); Z = midpoint(E--F); G = midpoint(A--D); M = intersectionpoint(A--E,X--Z); N = intersectionpoint(A--C,X--Y); O = intersectionpoint(C--E,Y--Z); P = intersectionpoint(A--C,X--Z); Q = intersectionpoint(C--E,X--Y); R = intersectionpoint(A--E,Y--Z); fill(M--P--N--Q--O--R--cycle,mediumgray); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*dir(B),linewidth(4)); dot("$C$",C,1.5*dir(C),linewidth(4)); dot("$D$",D,1.5*dir(D),linewidth(4)); dot("$E$",E,1.5*dir(E),linewidth(4)); dot("$F$",F,1.5*dir(F),linewidth(4)); dot("$X$",X,1.5*dir(X),linewidth(4)); dot("$Y$",Y,1.5*dir(Y),linewidth(4)); dot("$Z$",Z,1.5*dir(Z),linewidth(4)); dot(O^^Q^^R,linewidth(4)); dot("$N$",N,1.5*dir(N),linewidth(4)); dot("$M$",M,1.5*dir(M),linewidth(4)); dot("$P$",P,1.5*dir(P),linewidth(4)); dot("$G$",G,1.5*dir(G),linewidth(4)); draw(A--C--E--cycle^^X--Y--Z--cycle); draw(A--G^^C--G); draw(N--B); draw(N--E--A--cycle^^M--P, red); | [] |
836 | Ajay is standing at point $A$ near Pontianak, Indonesia, $0^\circ$ latitude and $110^\circ \text{ E}$ longitude. Billy is standing at point $B$ near Big Baldy Mountain, Idaho, USA, $45^\circ \text{ N}$ latitude and $115^\circ \text{ W}$ longitude. Assume that Earth is a perfect sphere with center $C.$ What is the degree measure of $\angle ACB?$
$\textbf{(A) }105 \qquad \textbf{(B) }112\frac{1}{2} \qquad \textbf{(C) }120 \qquad \textbf{(D) }135 \qquad \textbf{(E) }150 \qquad$ | 2018 AMC 12B Problem 23 | This solution refers to the Diagram section.
Let $D$ be the orthogonal projection of $B$ onto the equator. Note that $\angle BDA = \angle BDC = 90^\circ$ and $\angle BCD = 45^\circ.$ Recall that $115^\circ \text{ W}$ longitude is the same as $245^\circ \text{ E}$ longitude, so $\angle ACD=135^\circ.$
We obtain the following diagram:
Without loss of generality, let $AC=BC=1.$ For tetrahedron $ABCD:$
Since $\triangle BCD$ is an isosceles right triangle, we have $BD=CD=\frac{\sqrt2}{2}.$
In $\triangle ACD,$ we apply the Law of Cosines to get $AD=\sqrt{AC^2+CD^2-2\cdot AC\cdot CD\cdot\cos\angle ACD}=\frac{\sqrt{10}}{2}.$
In right $\triangle ABD,$ we apply the Pythagorean Theorem to get $AB=\sqrt{AD^2+BD^2}=\sqrt{3}.$
In $\triangle ABC,$ we apply the Law of Cosines to get $\cos\angle ACB=\frac{AC^2+BC^2-AB^2}{2\cdot AC\cdot BC}=-\frac12,$ so $\angle ACB=\boxed{\textbf{(C) }120}$ degrees.
~MRENTHUSIASM | // Block 1
/* Made by MRENTHUSIASM */
size(250);
import graph3;
import solids;
currentprojection=orthographic((0.2,-0.5,0.2));
triple A, B, C, D;
A = (1,0,0);
B = (-1/2,1/2,sqrt(2)/2);
C = (0,0,0);
D = (-1/2,1/2,0);
draw(unitsphere,white,light=White);
draw(surface(A--B--C--cycle),yellow);
dot(A^^B^^C^^D,linewidth(4.5));
draw(Circle(C,1,(0,0,1))^^A--B--D--C--B^^C--A--D);
label("$A$",A,3*dir(A));
label("$B$",B,3*dir(B));
label("$C$",C,3*(0,0,-1));
label("$D$",D,3*(-1/2,-1/2,0));
// Block 2
/* Made by MRENTHUSIASM */ size(250); import graph3; import solids; currentprojection=orthographic((0.2,-0.5,0.2)); triple A, B, C, D; A = (1,0,0); B = (-1/2,1/2,sqrt(2)/2); C = (0,0,0); D = (-1/2,1/2,0); draw(unitsphere,white,light=White); draw(surface(A--B--C--cycle),yellow); dot(A^^B^^C^^D,linewidth(4.5)); draw(Circle(C,1,(0,0,1))^^A--B--D--C--B^^C--A--D); label("$A$",A,3*dir(A)); label("$B$",B,3*dir(B)); label("$C$",C,3*(0,0,-1)); label("$D$",D,3*(-1/2,-1/2,0)); | [] |
836 | Ajay is standing at point $A$ near Pontianak, Indonesia, $0^\circ$ latitude and $110^\circ \text{ E}$ longitude. Billy is standing at point $B$ near Big Baldy Mountain, Idaho, USA, $45^\circ \text{ N}$ latitude and $115^\circ \text{ W}$ longitude. Assume that Earth is a perfect sphere with center $C.$ What is the degree measure of $\angle ACB?$
$\textbf{(A) }105 \qquad \textbf{(B) }112\frac{1}{2} \qquad \textbf{(C) }120 \qquad \textbf{(D) }135 \qquad \textbf{(E) }150 \qquad$ | 2018 AMC 12B Problem 23 | This solution refers to the Diagram section.
Let $D$ be the orthogonal projection of $B$ onto the equator. Note that $\angle BDA = \angle BDC = 90^\circ$ and $\angle BCD = 45^\circ.$ Recall that $115^\circ \text{ W}$ longitude is the same as $245^\circ \text{ E}$ longitude, so $\angle ACD=135^\circ.$
Without loss of generality, let $AC=BC=1.$ As shown below, we place Earth in the $xyz$-plane with $C=(0,0,0)$ such that the positive $x$-axis runs through $A,$ the positive $y$-axis runs through $0^\circ$ latitude and $160^\circ \text{ W}$ longitude, and the positive $z$-axis runs through the North Pole.
It follows that $A=(1,0,0)$ and $D=(-t,t,0)$ for some positive number $t.$ Since $\triangle BCD$ is an isosceles right triangle, we have $B=\left(-t,t,\sqrt{2}t\right).$ By the Distance Formula, we get $(-t)^2+t^2+\left(\sqrt{2}t\right)^2=1,$ from which $t=\frac12.$
As $\vec{A} = \begin{pmatrix}1 \\ 0 \\ 0 \end{pmatrix}$ and $\vec{B} = \begin{pmatrix}-1/2 \\ 1/2 \\ \sqrt2/2 \end{pmatrix},$ we obtain \[\cos\angle ACB=\frac{\vec{A}\bullet\vec{B}}{\left\lVert\vec{A}\right\rVert\left\lVert\vec{B}\right\rVert}=-\frac12\] by the dot product, so $\angle ACB=\boxed{\textbf{(C) }120}$ degrees.
~MRENTHUSIASM | // Block 1
/* Made by MRENTHUSIASM */
size(300);
import graph3;
import solids;
currentprojection=orthographic((0.2,-0.5,0.2));
triple A, B, C, D;
A = (1,0,0);
B = (-1/2,1/2,sqrt(2)/2);
C = (0,0,0);
D = (-1/2,1/2,0);
draw(unitsphere,white,light=White);
dot(A^^B^^C^^D,linewidth(4.5));
draw(Circle(C,1,(0,0,1))^^B--C--D--cycle);
label("$A$",A,5*dir((2.5,-3,0)));
label("$B$",B,3*dir(B));
label("$C$",C,1.5*(1,0,-1));
label("$D$",D,3*(-1/2,-1/2,0));
draw((-1.5,0,0)--(1.5,0,0),linewidth(1.25),EndArrow3(10));
draw((0,-1.5,0)--(0,1.5,0),linewidth(1.25),EndArrow3(10));
draw((0,0,-1.5)--(0,0,1.5),linewidth(1.25),EndArrow3(10));
label("$x$",(1.5,0,0),2*dir((1.5,0,0)));
label("$y$",(0,1.5,0),3*dir((0,1.5,0)));
label("$z$",(0,0,1.5),2*dir((0,0,1.5)));
// Block 2
/* Made by MRENTHUSIASM */ size(300); import graph3; import solids; currentprojection=orthographic((0.2,-0.5,0.2)); triple A, B, C, D; A = (1,0,0); B = (-1/2,1/2,sqrt(2)/2); C = (0,0,0); D = (-1/2,1/2,0); draw(unitsphere,white,light=White); dot(A^^B^^C^^D,linewidth(4.5)); draw(Circle(C,1,(0,0,1))^^B--C--D--cycle); label("$A$",A,5*dir((2.5,-3,0))); label("$B$",B,3*dir(B)); label("$C$",C,1.5*(1,0,-1)); label("$D$",D,3*(-1/2,-1/2,0)); draw((-1.5,0,0)--(1.5,0,0),linewidth(1.25),EndArrow3(10)); draw((0,-1.5,0)--(0,1.5,0),linewidth(1.25),EndArrow3(10)); draw((0,0,-1.5)--(0,0,1.5),linewidth(1.25),EndArrow3(10)); label("$x$",(1.5,0,0),2*dir((1.5,0,0))); label("$y$",(0,1.5,0),3*dir((0,1.5,0))); label("$z$",(0,0,1.5),2*dir((0,0,1.5))); | [] |
837 | Let $\lfloor x \rfloor$ denote the greatest integer less than or equal to $x$. How many real numbers $x$ satisfy the equation $x^2 + 10,000\lfloor x \rfloor = 10,000x$?
$\textbf{(A) } 197 \qquad \textbf{(B) } 198 \qquad \textbf{(C) } 199 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 201$ | 2018 AMC 12B Problem 24 | This rewrites itself to $x^2=10,000\{x\}$ where $\lfloor x \rfloor + \{x\} = x$.
Graphing $y=10,000\{x\}$ and $y=x^2$ we see that the former is a set of line segments with slope $10,000$ from $0$ to $1$ with a hole at $x=1$, then $1$ to $2$ with a hole at $x=2$ etc.
Here is a graph of $y=x^2$ and $y=16\{x\}$ for visualization.
Now notice that when $x=\pm 100$ the graph has a hole at $(\pm 100,10,000)$ which the equation $y=x^2$ passes through and then continues upwards. Thus our set of possible solutions is bounded by $(-100,100)$. We can see that $y=x^2$ intersects each of the lines once and there are $99-(-99)+1=199$ lines for an answer of $\boxed{\text{(C)}~199}$. | import graph; size(400); xaxis("$x$",Ticks(Label(fontsize(8pt)),new real[]{-5,-4,-3, -2, -1,0,1 2,3, 4,5})); yaxis("$y$",Ticks(Label(fontsize(8pt)),new real[]{0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18})); real y(real x) {return x^2;} draw(circle((-4,16), 0.1)); draw(circle((-3,16), 0.1)); draw(circle((-2,16), 0.1)); draw(circle((-1,16), 0.1)); draw(circle((0,16), 0.1)); draw(circle((1,16), 0.1)); draw(circle((2,16), 0.1)); draw(circle((3,16), 0.1)); draw(circle((4,16), 0.1)); draw((-5,0)--(-4,16), black); draw((-4,0)--(-3,16), black); draw((-3,0)--(-2,16), black); draw((-2,0)--(-1,16), black); draw((-1,0)--(-0,16), black); draw((0,0)--(1,16), black); draw((1,0)--(2,16), black); draw((2,0)--(3,16), black); draw((3,0)--(4,16), black); draw(graph(y,-4.2,4.2),green); | [] |
838 | Circles $\omega_1$, $\omega_2$, and $\omega_3$ each have radius $4$ and are placed in the plane so that each circle is externally tangent to the other two. Points $P_1$, $P_2$, and $P_3$ lie on $\omega_1$, $\omega_2$, and $\omega_3$ respectively such that $P_1P_2=P_2P_3=P_3P_1$ and line $P_iP_{i+1}$ is tangent to $\omega_i$ for each $i=1,2,3$, where $P_4 = P_1$. See the figure below. The area of $\triangle P_1P_2P_3$ can be written in the form $\sqrt{a}+\sqrt{b}$ for positive integers $a$ and $b$. What is $a+b$?
$\textbf{(A) }546\qquad\textbf{(B) }548\qquad\textbf{(C) }550\qquad\textbf{(D) }552\qquad\textbf{(E) }554$ | 2018 AMC 12B Problem 25 | First, note that because the $\angle P_1=\angle P_2=\angle P_3=\pi/3$, the arcs inside the shaded equilateral triangle are each $2\pi/3$. Also, the distances between the centers of any two of the $3$ given circles are each $8$.
Draw the circle $\Gamma$ concentric with $\omega_1$ with radius $2$. Because the arc of $\omega_1$ inside said triangle is $2\pi/3$, $\Gamma$ touches $P_1P_3$, say at a point $X$. Thus, $P_1P_3$ is a common tangent of $\omega_3$ and $\Gamma$, and it can be seen from inspection of the given diagram that the line is an common internal tangent.
The length of the common internal tangent segment $XP_3$ of $\Gamma$ and $\omega_3$ is then $\sqrt{8^2-(2+4)^2}=2\sqrt{7}$, and it is easily seen that $XP_1=4\sin \pi/3=2\sqrt{3}$.
Because $P_1P_3=2(\sqrt{3}+\sqrt{7})$, the area of the shaded equilateral triangle is $\sqrt{3}(\sqrt{3}+\sqrt{7})^2=10\sqrt{3}+6\sqrt{7}$. We get $\sqrt{300}+\sqrt{252}\Rightarrow\boxed{\textbf{(D) }552}.$
~crazyeyemoody907 | // Block 1
unitsize(12);
pair A = (0, 8/sqrt(3)), B = rotate(-120)*A, C = rotate(120)*A;
real theta = 41.5;
pair P1 = rotate(theta)*(2+2*sqrt(7/3), 0), P2 = rotate(-120)*P1, P3 = rotate(120)*P1;
filldraw(P1--P2--P3--cycle, gray(0.9));
draw(Circle(A, 4));
draw(Circle(B, 4));
draw(Circle(C, 4));
dot(P1);
dot(P2);
dot(P3);
defaultpen(fontsize(10pt));
label("$P_1$", P1, E*1.5);
label("$P_2$", P2, SW*1.5);
label("$P_3$", P3, N);
label("$\omega_1$", A, W*30);
label("$\omega_2$", B, E*17);
label("$\omega_3$", C, W*17);
label("$\Gamma$", A, W*15);
draw(Circle(A,2),red);
pair X=foot(A,P1,P3);
dot(X,blue);
draw(A--X,blue);
label("$2\sqrt{7}$", X--P3);
label("$2\sqrt{3}$",X--P1);
label("$X$",X,dir(-80),blue);
// Block 2
unitsize(12); pair A = (0, 8/sqrt(3)), B = rotate(-120)*A, C = rotate(120)*A; real theta = 41.5; pair P1 = rotate(theta)*(2+2*sqrt(7/3), 0), P2 = rotate(-120)*P1, P3 = rotate(120)*P1; filldraw(P1--P2--P3--cycle, gray(0.9)); draw(Circle(A, 4)); draw(Circle(B, 4)); draw(Circle(C, 4)); dot(P1); dot(P2); dot(P3); defaultpen(fontsize(10pt)); label("$P_1$", P1, E*1.5); label("$P_2$", P2, SW*1.5); label("$P_3$", P3, N); label("$\omega_1$", A, W*30); label("$\omega_2$", B, E*17); label("$\omega_3$", C, W*17); label("$\Gamma$", A, W*15); draw(Circle(A,2),red); pair X=foot(A,P1,P3); dot(X,blue); draw(A--X,blue); label("$2\sqrt{7}$", X--P3); label("$2\sqrt{3}$",X--P1); label("$X$",X,dir(-80),blue); | [] |
838 | Circles $\omega_1$, $\omega_2$, and $\omega_3$ each have radius $4$ and are placed in the plane so that each circle is externally tangent to the other two. Points $P_1$, $P_2$, and $P_3$ lie on $\omega_1$, $\omega_2$, and $\omega_3$ respectively such that $P_1P_2=P_2P_3=P_3P_1$ and line $P_iP_{i+1}$ is tangent to $\omega_i$ for each $i=1,2,3$, where $P_4 = P_1$. See the figure below. The area of $\triangle P_1P_2P_3$ can be written in the form $\sqrt{a}+\sqrt{b}$ for positive integers $a$ and $b$. What is $a+b$?
$\textbf{(A) }546\qquad\textbf{(B) }548\qquad\textbf{(C) }550\qquad\textbf{(D) }552\qquad\textbf{(E) }554$ | 2018 AMC 12B Problem 25 | Suppose $P_1P_2 = x$, noticed that $OMP_1H$ is rectangle.
\[OO_1 = \frac{O_1O_2}{\sqrt{3}} = \frac83\sqrt{3}\]
\[OH = P_1M = \frac{1}{2}P_1P_2 = \frac12 x\]
\[O_1H = O_1P_1 - HP_1 = O_1P_1 - OM = O_1P_1 - \frac{P_1P_2}{2\sqrt{3}} = 4 - \frac{\sqrt{3}}{6}x\]
$\triangle O_1HO$ is right triangle, we can use Pythagorean theorem to establish an equation,
\[OH^2 + O_1H^2 = OO_1^2\]
\[\left(\frac12 x\right)^2 + \left(4 - \frac{\sqrt{3}}{6}x\right)^2 = \left(\frac83\sqrt{3}\right)^2\]
\[x^2 - 4\sqrt{3}x - 16 = 0\]
\[x = 2\sqrt{3} + 2\sqrt{7}\]
The side length of the equilateral triangle is $2(\sqrt{7} + \sqrt{3})$, so its area is $(\sqrt{7} + \sqrt{3})^2\sqrt{3} = 10\sqrt{3} + 6\sqrt{7} = \sqrt{300} + \sqrt{252} \implies 300 + 252 = \boxed{\textbf{(D) }552}$.
~reda_mandymath | // Block 1
import geometry;
unitsize(12);
pair A = (0, 8/sqrt(3)), B = rotate(-120)*A, C = rotate(120)*A;
real theta = 41.5;
pair P1 = rotate(theta)*(2+2*sqrt(7/3), 0), P2 = rotate(-120)*P1, P3 = rotate(120)*P1;
filldraw(P1--P2--P3--cycle, gray(0.9));
draw(Circle(A, 4));
draw(Circle(B, 4));
draw(Circle(C, 4));
dot(P1);
dot(P2);
dot(P3);
pair M = extension(P3, origin, P1, P2);
pair H = extension(A, P1, origin, rotate(90) * M);
dot(M ^^ H ^^ origin ^^ A);
draw(H -- P1 ^^ P3 -- M, dashed);
draw(A -- H -- origin -- cycle, red);
markrightangle(A, H, origin, 0.2 * markangleradius(), red);
defaultpen(fontsize(10pt));
label("$P_1$", P1, E*1.5);
label("$P_2$", P2, SW*1.5);
label("$P_3$", P3, N);
label("$\omega_1$", A, W*17);
label("$\omega_2$", B, E*17);
label("$\omega_3$", C, W*17);
label("$M$", M, SE);
label("$H$", H, NE);
label("$O_1$", A, N);
label("$O$", origin, NE);
// Block 2
import geometry; unitsize(12); pair A = (0, 8/sqrt(3)), B = rotate(-120)*A, C = rotate(120)*A; real theta = 41.5; pair P1 = rotate(theta)*(2+2*sqrt(7/3), 0), P2 = rotate(-120)*P1, P3 = rotate(120)*P1; filldraw(P1--P2--P3--cycle, gray(0.9)); draw(Circle(A, 4)); draw(Circle(B, 4)); draw(Circle(C, 4)); dot(P1); dot(P2); dot(P3); pair M = extension(P3, origin, P1, P2); pair H = extension(A, P1, origin, rotate(90) * M); dot(M ^^ H ^^ origin ^^ A); draw(H -- P1 ^^ P3 -- M, dashed); draw(A -- H -- origin -- cycle, red); markrightangle(A, H, origin, 0.2 * markangleradius(), red); defaultpen(fontsize(10pt)); label("$P_1$", P1, E*1.5); label("$P_2$", P2, SW*1.5); label("$P_3$", P3, N); label("$\omega_1$", A, W*17); label("$\omega_2$", B, E*17); label("$\omega_3$", C, W*17); label("$M$", M, SE); label("$H$", H, NE); label("$O_1$", A, N); label("$O$", origin, NE); | [] |
839 | A rectangular floor that is $10$ feet wide and $17$ feet long is tiled with $170$ one-foot square tiles. A bug walks from one corner to the opposite corner in a straight line. Including the first and the last tile, how many tiles does the bug visit?
$\textbf{(A) } 17 \qquad\textbf{(B) } 25 \qquad\textbf{(C) } 26 \qquad\textbf{(D) } 27 \qquad\textbf{(E) } 28$ | 2019 AMC 10A Problem 10 | We can also draw a diagram or scale model of the entire rectangular floor (optionally with grid paper and/or a ruler so it will be to scale), then simply count the number of tiles the path crosses. To make this slightly easier, we can divide the full grid into $4$ sections, and just draw one of these $5$ feet by $8.5$ feet sections.
Though it may appear that the line we drew comes very close to several points, we know that since $10$ and $17$ are relatively prime (numbers where the only positive integer that divides both of them is 1, a.k.a. numbers with a gcd of 1), the line will not actually pass through any of these points, so the total number of squares crossed will be the same regardless of which side we count. If we count the number of squares the line passes through using the diagram, we get $13$ squares. We can then multiply this by 2 to find out the total number of squares the bug passes through on the rectangular floor giving us a total of $2 \cdot 13 = \boxed{\textbf{(C) }26}$. | // Block 1
unitsize(20);
for(int i =0; i<= 7; ++i) {
for(int j =0; j<= 4; ++j) {
draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--cycle);
}
for(int k =0; k<= 4; ++k) {
draw((8,k)--(8.5,k)--(8.5,k+1)--(8,k+1)--cycle);
}
}
draw((0,5)--(8.5,0)--cycle);
// Block 2
unitsize(20); for(int i =0; i<= 7; ++i) { for(int j =0; j<= 4; ++j) { draw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--cycle); } for(int k =0; k<= 4; ++k) { draw((8,k)--(8.5,k)--(8.5,k+1)--(8,k+1)--cycle); } } draw((0,5)--(8.5,0)--cycle); | [] |
840 | Let $\triangle ABC$ be an isosceles triangle with $BC = AC$ and $\angle ACB = 40^{\circ}$. Construct the circle with diameter $\overline{BC}$, and let $D$ and $E$ be the other intersection points of the circle with the sides $\overline{AC}$ and $\overline{AB}$, respectively. Let $F$ be the intersection of the diagonals of the quadrilateral $BCDE$. What is the degree measure of $\angle BFC ?$
$\textbf{(A) } 90 \qquad\textbf{(B) } 100 \qquad\textbf{(C) } 105 \qquad\textbf{(D) } 110 \qquad\textbf{(E) } 120$ | 2019 AMC 10A Problem 13 | Drawing it out, we see $\angle BDC$ and $\angle BEC$ are right angles, as they are inscribed in a semicircle. Using the fact that it is an isosceles triangle, we find $\angle ABC=70^{\circ}$. We can find $\angle ECB=20^{\circ}$ and $\angle DBC=50^{\circ}$ by the triangle angle sum on $\triangle ECB$ and $\triangle DBC$.
\[\angle BDC+\angle DCB+\angle DBC=180^{\circ}\implies90^{\circ}+40^{\circ}+\angle DBC=180^{\circ}\implies\angle DBC=50^{\circ}\]
\[\angle BEC+\angle EBC+\angle ECB=180^{\circ}\implies90^{\circ}+70^{\circ}+\angle ECB=180^{\circ}\implies\angle ECB=20^{\circ}\]
Then, we take triangle $BFC$, and find $\angle BFC=180^{\circ}-50^{\circ}-20^{\circ}=\boxed{\textbf{(D) } 110^{\circ}}.$ | unitsize(40);draw((-1,0)--(1,0)--(0,2.75)--cycle);draw(circumcircle((-1,0),(0,0),(0,2.75)));label("$A$",(1,0),SE);label("$C$",(0,2.75),N);label("$B$",(-1,0),SW);label("$E$",(0,0),S);label("$D$",(0.77,0.64),E);draw((0,0)--(0,2.75));draw((-1,0)--(0.77,0.64)); | [] |
841 | For a set of four distinct lines in a plane, there are exactly $N$ distinct points that lie on two or more of the lines. What is the sum of all possible values of $N$?
$\textbf{(A) } 14 \qquad \textbf{(B) } 16 \qquad \textbf{(C) } 18 \qquad \textbf{(D) } 19 \qquad \textbf{(E) } 21$ | 2019 AMC 10A Problem 14 | It is possible to obtain $0$, $1$, $3$, $4$, $5$, and $6$ points of intersection, as demonstrated in the following figures:
It is clear that the maximum number of possible intersections is ${4 \choose 2} = 6$, since each pair of lines can intersect at most once. In addition, by looking at the answer choices, we know that we cannot have 7 points of intersection or else our answer would be greater than the given answer choices. Our answer is given by the sum $0+1+3+4+5+6=\boxed{\textbf{(D)} 19}$. | // Block 1
unitsize(2cm);
real d = 2.5;
draw((-1,.6)--(1,.6),Arrows);
draw((-1,.2)--(1,.2),Arrows);
draw((-1,-.2)--(1,-.2),Arrows);
draw((-1,-.6)--(1,-.6),Arrows);
draw((-1+d,0)--(1+d,0),Arrows);
draw((0+d,1)--(0+d,-1),Arrows);
draw(dir(45)+(d,0)--dir(45+180)+(d,0),Arrows);
draw(dir(135)+(d,0)--dir(135+180)+(d,0),Arrows);
dot((0+d,0));
draw((-1+2*d,sqrt(3)/3)--(1+2*d,sqrt(3)/3),Arrows);
draw((-1/4-1/2+2*d, sqrt(3)/12-sqrt(3)/2)--(-1/4+1/2+2*d,sqrt(3)/12+sqrt(3)/2),Arrows);
draw((1/4+1/2+2*d, sqrt(3)/12-sqrt(3)/2)--(1/4-1/2+2*d,sqrt(3)/12+sqrt(3)/2),Arrows);
draw((-1+2*d,-sqrt(3)/6)--(1+2*d,-sqrt(3)/6),Arrows);
dot((0+2*d,sqrt(3)/3));
dot((-1/2+2*d,-sqrt(3)/6));
dot((1/2+2*d,-sqrt(3)/6));
draw((-1/3,1-d)--(-1/3,-1-d),Arrows);
draw((1/3,1-d)--(1/3,-1-d),Arrows);
draw((-1,-1/3-d)--(1,-1/3-d),Arrows);
draw((-1,1/3-d)--(1,1/3-d),Arrows);
dot((1/3,1/3-d));
dot((-1/3,1/3-d));
dot((1/3,-1/3-d));
dot((-1/3,-1/3-d));
draw((-1+d,sqrt(3)/12-d)--(1+d,sqrt(3)/12-d),Arrows);
draw((-1/4-1/2+d, sqrt(3)/12-sqrt(3)/2-d)--(-1/4+1/2+d,sqrt(3)/12+sqrt(3)/2-d),Arrows);
draw((1/4+1/2+d, sqrt(3)/12-sqrt(3)/2-d)--(1/4-1/2+d,sqrt(3)/12+sqrt(3)/2-d),Arrows);
draw((-1+d,-sqrt(3)/6-d)--(1+d,-sqrt(3)/6-d),Arrows);
dot((0+d,sqrt(3)/3-d));
dot((-1/2+d,-sqrt(3)/6-d));
dot((1/2+d,-sqrt(3)/6-d));
dot((-1/4+d,sqrt(3)/12-d));
dot((1/4+d,sqrt(3)/12-d));
draw((-1/4-1/2+2*d, sqrt(3)/12-sqrt(3)/2-d)--(-1/4+1/2+2*d,sqrt(3)/12+sqrt(3)/2-d),Arrows);
draw((1/4+1/2+2*d, sqrt(3)/12-sqrt(3)/2-d)--(1/4-1/2+2*d,sqrt(3)/12+sqrt(3)/2-d),Arrows);
draw(dir(30)+(2*d,-d)--dir(30+180)+(2*d,-d),Arrows);
draw(dir(150)+(2*d,-d)--dir(-30)+(2*d,-d),Arrows);
dot((0+2*d,0-d));
dot((0+2*d,sqrt(3)/3-d));
dot((-1/2+2*d,-sqrt(3)/6-d));
dot((1/2+2*d,-sqrt(3)/6-d));
dot((-1/4+2*d,sqrt(3)/12-d));
dot((1/4+2*d,sqrt(3)/12-d));
// Block 2
unitsize(2cm); real d = 2.5; draw((-1,.6)--(1,.6),Arrows); draw((-1,.2)--(1,.2),Arrows); draw((-1,-.2)--(1,-.2),Arrows); draw((-1,-.6)--(1,-.6),Arrows); draw((-1+d,0)--(1+d,0),Arrows); draw((0+d,1)--(0+d,-1),Arrows); draw(dir(45)+(d,0)--dir(45+180)+(d,0),Arrows); draw(dir(135)+(d,0)--dir(135+180)+(d,0),Arrows); dot((0+d,0)); draw((-1+2*d,sqrt(3)/3)--(1+2*d,sqrt(3)/3),Arrows); draw((-1/4-1/2+2*d, sqrt(3)/12-sqrt(3)/2)--(-1/4+1/2+2*d,sqrt(3)/12+sqrt(3)/2),Arrows); draw((1/4+1/2+2*d, sqrt(3)/12-sqrt(3)/2)--(1/4-1/2+2*d,sqrt(3)/12+sqrt(3)/2),Arrows); draw((-1+2*d,-sqrt(3)/6)--(1+2*d,-sqrt(3)/6),Arrows); dot((0+2*d,sqrt(3)/3)); dot((-1/2+2*d,-sqrt(3)/6)); dot((1/2+2*d,-sqrt(3)/6)); draw((-1/3,1-d)--(-1/3,-1-d),Arrows); draw((1/3,1-d)--(1/3,-1-d),Arrows); draw((-1,-1/3-d)--(1,-1/3-d),Arrows); draw((-1,1/3-d)--(1,1/3-d),Arrows); dot((1/3,1/3-d)); dot((-1/3,1/3-d)); dot((1/3,-1/3-d)); dot((-1/3,-1/3-d)); draw((-1+d,sqrt(3)/12-d)--(1+d,sqrt(3)/12-d),Arrows); draw((-1/4-1/2+d, sqrt(3)/12-sqrt(3)/2-d)--(-1/4+1/2+d,sqrt(3)/12+sqrt(3)/2-d),Arrows); draw((1/4+1/2+d, sqrt(3)/12-sqrt(3)/2-d)--(1/4-1/2+d,sqrt(3)/12+sqrt(3)/2-d),Arrows); draw((-1+d,-sqrt(3)/6-d)--(1+d,-sqrt(3)/6-d),Arrows); dot((0+d,sqrt(3)/3-d)); dot((-1/2+d,-sqrt(3)/6-d)); dot((1/2+d,-sqrt(3)/6-d)); dot((-1/4+d,sqrt(3)/12-d)); dot((1/4+d,sqrt(3)/12-d)); draw((-1/4-1/2+2*d, sqrt(3)/12-sqrt(3)/2-d)--(-1/4+1/2+2*d,sqrt(3)/12+sqrt(3)/2-d),Arrows); draw((1/4+1/2+2*d, sqrt(3)/12-sqrt(3)/2-d)--(1/4-1/2+2*d,sqrt(3)/12+sqrt(3)/2-d),Arrows); draw(dir(30)+(2*d,-d)--dir(30+180)+(2*d,-d),Arrows); draw(dir(150)+(2*d,-d)--dir(-30)+(2*d,-d),Arrows); dot((0+2*d,0-d)); dot((0+2*d,sqrt(3)/3-d)); dot((-1/2+2*d,-sqrt(3)/6-d)); dot((1/2+2*d,-sqrt(3)/6-d)); dot((-1/4+2*d,sqrt(3)/12-d)); dot((1/4+2*d,sqrt(3)/12-d)); | [] |
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