problem_id int64 1 978 | question stringlengths 86 2.11k | source stringlengths 19 76 | solution stringlengths 94 14.7k | asymptote_code stringlengths 44 17.8k | solution_image_url stringclasses 16
values |
|---|---|---|---|---|---|
727 | The closed curve in the figure is made up of 9 congruent circular arcs each of length $\frac{2\pi}{3}$, where each of the centers of the corresponding circles is among the vertices of a regular hexagon of side 2. What is the area enclosed by the curve?
$\textbf{(A)}\ 2\pi+6\qquad\textbf{(B)}\ 2\pi+4\sqrt{3}\qquad\te... | 2012 AMC 12A Problem 14 | We can draw the hexagon between the centers of the circles, and compute its area. The hexagon is made of $6$ equilateral triangles each with length $2$, so the area is:
\[\frac{\sqrt{3}}{4} \cdot 2^2 \cdot 6=6 \sqrt{3}.\]
Then, we add the areas of the three sectors outside the hexagon:
\[\frac 23 \pi \cdot 3=2\pi.\]
We... | unitsize(2cm); defaultpen(fontsize(6pt)); dotfactor=4; label("$\circ$",(0,1)); label("$\circ$",(0.865,0.5)); label("$\circ$",(-0.865,0.5)); label("$\circ$",(0.865,-0.5)); label("$\circ$",(-0.865,-0.5)); label("$\circ$",(0,-1)); dot((0,1.5)); dot((-0.4325,0.75)); dot((0.4325,0.75)); dot((-0.4325,-0.75)); dot((0.4325,-0.... | [] |
727 | The closed curve in the figure is made up of 9 congruent circular arcs each of length $\frac{2\pi}{3}$, where each of the centers of the corresponding circles is among the vertices of a regular hexagon of side 2. What is the area enclosed by the curve?
$\textbf{(A)}\ 2\pi+6\qquad\textbf{(B)}\ 2\pi+4\sqrt{3}\qquad\te... | 2012 AMC 12A Problem 14 | As you can see, this diagram looks like a fidget spinner ;). Fidget spinners aside, we need to add stuff to our diagram to make it look easier. In the directions, they were talking about the centers of each arc create a hexagon, so let's add that to our diagram.
The side length of the hexagon is 2 and if we plug it... | defaultpen(fontsize(6pt)); dotfactor=4; label("$\circ$",(0,1)); label("$\circ$",(0.865,0.5)); label("$\circ$",(-0.865,0.5)); label("$\circ$",(0.865,-0.5)); label("$\circ$",(-0.865,-0.5)); label("$\circ$",(0,-1)); dot((0,1.5)); dot((-0.432,0.75)); dot((0.4325,0.75)); dot((-0.4325,-0.75)); dot((0.4325,-0.75)); dot((-0.86... | [] |
728 | Distinct planes $p_1,p_2,....,p_k$ intersect the interior of a cube $Q$. Let $S$ be the union of the faces of $Q$ and let $P =\bigcup_{j=1}^{k}p_{j}$. The intersection of $P$ and $S$ consists of the union of all segments joining the midpoints of every pair of edges belonging to the same face of $Q$. What is the differe... | 2012 AMC 12A Problem 22 | We need two different kinds of planes that only intersect $Q$ at the mentioned segments (we call them traces in this solution). These will be all the possible $p_j$'s.
First, there are two kinds of segments joining the midpoints of every pair of edges belonging to the same face of $Q$: long traces are those connecting... | // Block 1
pair A=(0,0), B=(2,0), C=(2,2), D=(0,2), E=(-1,3), F=(1,3), G=(1,1), H=(-1,1), I=(1,0), J=(2,1), K=(1,2), L=(0,1), M=(-0.5,0.5), N=(-1,2), O=(-0.5,2.5), P=(0,3), Q=(1.5,2.5), R=(1,2), S=(1.5,0.5), T=(0,1);
draw(A--B--C--D--E--F);
draw(H--A);
draw(A--D);
draw(H--E);
draw(F--C);
draw(H--G--F);
draw(G--B);
labe... | [] |
729 | Let $S$ be the square one of whose diagonals has endpoints $(1/10,7/10)$ and $(-1/10,-7/10)$. A point $v=(x,y)$ is chosen uniformly at random over all pairs of real numbers $x$ and $y$ such that $0 \le x \le 2012$ and $0\le y\le 2012$. Let $T(v)$ be a translated copy of $S$ centered at $v$. What is the probability t... | 2012 AMC 12A Problem 23 | The unit square's diagonal has a length of $\sqrt{0.2^2 + 1.4^2} = \sqrt{2}$. Because $S$ square is not parallel to the axis, the two points must be adjacent.
Now consider the unit square $U$ with vertices $(0,0), (1,0), (1,1)$ and $(0,1)$. Let us first consider only two vertices, $(0,0)$ and $(1,0)$. We want to find ... | // Block 1
pair A=(0.1,0.7), C=(-0.1,-0.7), B=(-0.7,0.1), D=(0.7,-0.1), X=(1,0), W=(-1,0), Y=(0,1), Z=(0,-1);
draw (A--B--C--D--A);
draw(A--C);
draw(B--D);
draw(W--X);
draw(Y--Z);
label("\((0.1,0.7)\)",A,NE);
label("\((-0.1,-0.7)\)",C,SW);
label("\(x\)",X,NW);
label("\(y\)",Y,NE);
// Block 2
pair A=(0.1,0.7), C=(-0.1,-... | [] |
730 | Point $B$ is due east of point $A$. Point $C$ is due north of point $B$. The distance between points $A$ and $C$ is $10\sqrt 2$, and $\angle BAC = 45^\circ$. Point $D$ is $20$ meters due north of point $C$. The distance $AD$ is between which two integers?
$\textbf{(A)}\ 30\ \text{and}\ 31 \qquad\textbf{(B)}\ 31\ \text... | 2012 AMC 10B Problem 12 | If point B is due east of point A and point C is due north of point B, $\angle CBA$ is a right angle. And if $\angle BAC = 45^\circ$, $\triangle CBA$ is a 45-45-90 triangle. Thus, the lengths of sides $CB$, $BA$, and $AC$ are in the ratio $1:1:\sqrt 2$, and $CB$ is $10 \sqrt 2 \div \sqrt 2 = 10$.
$\triangle DBA$ is cl... | // Block 1
unitsize(4);
pair A=(0,0);
label ("A",(0,0),W);
pair B=(10,0);
label ("B",(10,0),E);
pair C=(10,10);
label ("C",(10,10),E);
pair D=(10,30);
label ("D",(10,30),E);
dot(A);
dot(B);
dot(C);
dot(D);
draw(A--B);
draw(A--C);
draw(A--D);
draw(C--D);
draw(B--C);
// Block 2
unitsize(4); pair A=(0,0); label ("A",(0,0)... | [] |
731 | Two equilateral triangles are contained in square whose side length is $2\sqrt 3$. The bases of these triangles are the opposite side of the square, and their intersection is a rhombus. What is the area of the rhombus?
$\text{(A) } \frac{3}{2} \qquad \text{(B) } \sqrt 3 \qquad \text{(C) } 2\sqrt 2 - 1 \qquad \text{... | 2012 AMC 10B Problem 14 | Observe that the rhombus is made up of two congruent equilateral triangles with side length equal to GF. Since AE has length $\sqrt{3}$ and triangle AEF is a 30-60-90 triangle, it follows that EF has length 1. By symmetry, HG also has length 1. Thus GF has length $2\sqrt{3} - 2$. The formula for the area of an equilate... | // Block 1
size(8cm);
pair A, B, C, D, E, F, G, H, BF, AF;
A = (0,0);
B = (1,0);
C = (1,1);
D = (0,1);
E = (1/2,0);
H = (1/2,1);
G = (1/2,1/2^(1/2));
F = (1/2,1-(1/2^(1/2)));
AF = (3^(1/2)/2,1/2);
BF = (1-3^(1/2)/2,1/2);
draw(A--B--C--D--A--AF--D);
draw(C--BF--B);
draw(H--E,linetype("8 8"));
label("$A$",A,SW);
label("$... | [] |
732 | In rectangle $ABCD$, $AB=6$, $AD=30$, and $G$ is the midpoint of $\overline{AD}$. Segment $AB$ is extended 2 units beyond $B$ to point $E$, and $F$ is the intersection of $\overline{ED}$ and $\overline{BC}$. What is the area of quadrilateral $BFDG$?
$\textbf{(A)}\ \frac{133}{2}\qquad\textbf{(B)}\ 67\qquad\textbf{(C)}\... | 2012 AMC 10B Problem 19 | Note that the area of $BFDG$ equals the area of $ABCD-\triangle AGB-\triangle DCF$.
Since $AG=\frac{AD}{2}=15,$ $\triangle AGB=\frac{15\times 6}{2}=45$. Now, $\triangle AED\sim \triangle BEF$, so $\frac{AE}{BE}=4=\frac{AD}{BF}=\frac{30}{BF}\implies BF=7.5$ and $FC=22.5,$ so $\triangle DCF=\frac{22.5\times6}{2}=\frac{1... | // Block 1
unitsize(10);
pair B=(0,0);
pair A=(0,6);
pair C=(30,0);
pair D=(30,6);
pair G=(15,6);
pair E=(0,-2);
pair F=(15/2,0);
dot(A);
dot(B);
dot(C);
dot(D);
dot(G);
dot(E);
dot(F);
label("A",(0,6),NW);
label("B",(0,0),W);
label("C",(30,0),E);
label("D",(30,6),NE);
label("G",(15,6),N);
label("E",(0,-2),SW);
label("... | [] |
733 | Four distinct points are arranged on a plane so that the segments connecting them have lengths $a$, $a$, $a$, $a$, $2a$, and $b$. What is the ratio of $b$ to $a$?
$\textbf{(A)}\ \sqrt{3}\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ \sqrt{5}\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ \pi$ | 2012 AMC 10B Problem 21 | When you see that there are lengths a and 2a, one could think of 30-60-90 triangles. Since all of the other's lengths are a, you could think that $b=\sqrt{3}a$.
Drawing the points out, it is possible to have a diagram where $b=\sqrt{3}a$. It turns out that $a,$ $2a,$ and $b$ could be the lengths of a 30-60-90 triangle,... | // Block 1
draw((0, 0)--(1/2, sqrt(3)/2)--(1, 0)--cycle);
draw((1/2, sqrt(3)/2)--(2, 0)--(1,0));
label("$a$", (0, 0)--(1, 0), S);
label("$a$", (1, 0)--(2, 0), S);
label("$a$", (0, 0)--(1/2, sqrt(3)/2), NW);
label("$a$", (1, 0)--(1/2, sqrt(3)/2), NE);
label("$b=\sqrt{3}a$", (1/2, sqrt(3)/2)--(2, 0), NE);
// Block 2
draw... | [] |
734 | A solid tetrahedron is sliced off a solid wooden unit cube by a plane passing through two nonadjacent vertices on one face and one vertex on the opposite face not adjacent to either of the first two vertices. The tetrahedron is discarded and the remaining portion of the cube is placed on a table with the cut surface fa... | 2012 AMC 10B Problem 23 | We can approach this problem similar to the solution above, by attacking it on the Cartesian plane. As mentioned before, it has been proved that the space diagonal intersects the centroid of triangle $BGF$ (I have defined the tetrahedron as cutting through points $B$, $G$, $F$, and $E$). We can label $E$ as $(0, 0, 0)$... | // Block 1
import three; draw((1,1,1)--(1,0,1)--(1,0,0)--(0,0,0)--(0,0,1)--(0,1,1)--(1,1,1)--(1,1,0)--(0,1,0)--(0,1,1)); draw((0,0,1)--(1,0,1)); draw((1,0,0)--(1,1,0)); draw((0,0,0)--(0,1,0));
label("A",(0,0,0),S);
label("B",(1,0,0),W);
label("C",(0,0,1),N);
label("D",(1,0,1),NW);
label("E",(1,1,0),S);
label("F",(0,1,... | [] |
735 | A bug travels from A to B along the segments in the hexagonal lattice pictured below. The segments marked with an arrow can be traveled only in the direction of the arrow, and the bug never travels the same segment more than once. How many different paths are there?
$\textbf{(A)}\ 2112\qquad\textbf{(B)}\ 2304\qquad\... | 2012 AMC 10B Problem 25 | There is $1$ way to get to any of the red arrows. From the first (top) red arrow, there are $2$ ways to get to each of the first and the second (top 2) blue arrows; from the second (bottom) red arrow, there are $3$ ways to get to each of the first and the second blue arrows. So there are in total $5$ ways to get to eac... | // Block 1
size(10cm);
draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)--(10.0,6.928203230275509)--(12.0,6.928203230275509));
draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3... | [] |
735 | A bug travels from A to B along the segments in the hexagonal lattice pictured below. The segments marked with an arrow can be traveled only in the direction of the arrow, and the bug never travels the same segment more than once. How many different paths are there?
$\textbf{(A)}\ 2112\qquad\textbf{(B)}\ 2304\qquad\... | 2012 AMC 10B Problem 25 | For every blue arrow, there are $2\cdot 2=4$ ways to reach it without using the reverse arrow since the bug can choose any of $2$ red arrows to pass through and $2$ black arrows to pass through. If the bug passes through the white arrow, the red arrow that the bug travels through must be the closest to the first black... | // Block 1
size(6cm);
draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632));
draw((0.0,0.0)--(1.0,-1.7320508075688772)--(3.0,-1.7320508075688772)--(4.0,-3.4641016151377544)--(6.0,-3.4641016151377544)--(7.... | [] |
736 | Square $PQRS$ lies in the first quadrant. Points $(3,0), (5,0), (7,0),$ and $(13,0)$ lie on lines $SP, RQ, PQ$, and $SR$, respectively. What is the sum of the coordinates of the center of the square $PQRS$?
$\textbf{(A)}\ 6\qquad\textbf{(B) }\frac{31}5\qquad\textbf{(C) }\frac{32}5\qquad\textbf{(D) }\frac{33}5\qquad\te... | 2012 AMC 12B Problem 17 | Construct the midpoints $E=(4,0)$ and $F=(10,0)$ and triangle $\triangle EMF$ as in the diagram, where $M$ is the center of square $PQRS$. Also construct points $G$ and $H$ as in the diagram so that $BG\parallel PQ$ and $CH\parallel QR$.
Observe that $\triangle AGB\sim\triangle CHD$ while $PQRS$ being a square implies... | // Block 1
size(14cm);
pair A=(3,0),B=(5,0),C=(7,0),D=(13,0),EE=(4,0),F=(10,0),P=(3.4,1.2),Q=(5.2,0.6),R=(5.8,2.4),SS=(4,3),M=(4.6,1.8),G=(3.2,0.6),H=(7.6,1.8);
dot(A^^B^^C^^D^^EE^^F^^P^^Q^^R^^SS^^M^^G^^H);
draw(A--SS--D--cycle);
draw(P--Q--R^^B--Q--C);
draw(EE--M--F^^G--B^^C--H,dotted);
label("A",A,SW);
label("B",B,... | [] |
736 | Square $PQRS$ lies in the first quadrant. Points $(3,0), (5,0), (7,0),$ and $(13,0)$ lie on lines $SP, RQ, PQ$, and $SR$, respectively. What is the sum of the coordinates of the center of the square $PQRS$?
$\textbf{(A)}\ 6\qquad\textbf{(B) }\frac{31}5\qquad\textbf{(C) }\frac{32}5\qquad\textbf{(D) }\frac{33}5\qquad\te... | 2012 AMC 12B Problem 17 | Let the four points be labeled $P_1$, $P_2$, $P_3$, and $P_4$, respectively. Let the lines that go through each point be labeled $L_1$, $L_2$, $L_3$, and $L_4$, respectively. Since $L_1$ and $L_2$ go through $SP$ and $RQ$, respectively, and $SP$ and $RQ$ are opposite sides of the square, we can say that $L_1$ and $L_2$... | size(7cm); pair A=(0,0),B=(1,1.5),D=B*dir(-90),C=B+D-A; draw((-4,-2)--(8,-2), Arrows); draw(A--B--C--D--cycle); pair AB = extension(A,B,(0,-2),(1,-2)); pair BC = extension(B,C,(0,-2),(1,-2)); pair CD = extension(C,D,(0,-2),(1,-2)); pair DA = extension(D,A,(0,-2),(1,-2)); draw(A--AB--B--BC--C--CD--D--DA--A, dotted); dot... | [] |
736 | Square $PQRS$ lies in the first quadrant. Points $(3,0), (5,0), (7,0),$ and $(13,0)$ lie on lines $SP, RQ, PQ$, and $SR$, respectively. What is the sum of the coordinates of the center of the square $PQRS$?
$\textbf{(A)}\ 6\qquad\textbf{(B) }\frac{31}5\qquad\textbf{(C) }\frac{32}5\qquad\textbf{(D) }\frac{33}5\qquad\te... | 2012 AMC 12B Problem 17 | $SP: y = mx - 3m$, $RQ: y = mx-5m$, $PQ: y = -\frac{1}{m}x + \frac{7}{m}$, $SR: y = -\frac{1}{m}x + \frac{13}{m}$
Let $SP = RP = PQ = SR = a$, $\angle GAB = \angle HCD = \theta$, and the slope of $SP$ be $m$.
When the slope of $SP$ is $m$, the slope of $SR$ is $-\frac{1}{m}$, $\tan \theta = m$, $\cot \theta = -\frac{... | // Block 1
size(14cm);
pair A=(3,0),B=(5,0),C=(7,0),D=(13,0),EE=(4,0),F=(10,0),P=(3.4,1.2),Q=(5.2,0.6),R=(5.8,2.4),SS=(4,3),M=(4.6,1.8),G=(3.2,0.6),H=(7.6,1.8);
dot(A^^B^^C^^D^^EE^^F^^P^^Q^^R^^SS^^M^^G^^H);
draw(A--SS--D--cycle);
draw(P--Q--R^^B--Q--C);
draw(EE--M--F^^G--B^^C--H,dotted);
label("A",A,SW);
label("B",B,... | [] |
736 | Square $PQRS$ lies in the first quadrant. Points $(3,0), (5,0), (7,0),$ and $(13,0)$ lie on lines $SP, RQ, PQ$, and $SR$, respectively. What is the sum of the coordinates of the center of the square $PQRS$?
$\textbf{(A)}\ 6\qquad\textbf{(B) }\frac{31}5\qquad\textbf{(C) }\frac{32}5\qquad\textbf{(D) }\frac{33}5\qquad\te... | 2012 AMC 12B Problem 17 | Let $PQ=x$ and $\angle PCA=\theta.$ Draw the line $BE$ such that $E$ is on $AP$ and $BE\parallel PQ.$ Also, Draw the line $CF$ such that $F$ is on $DR$ and $CF\parallel RQ.$ Then $EB=FC=x$ and $\angle EBA=\angle FDC=\theta.$ Also, note that $AB=2$ and $CD=6.$ Hence:
\[\cos(\theta)=\frac{x}{2},\sin(\theta)=\frac{x}{6}.\... | // Block 1
unitsize(1 cm);
pair P,Q,R,S,A,B,C,D,E,F,F1,F2,F3,M;
P = (3.4,1.2);
Q = (5.2,0.6);
R = (5.8,2.4);
S = (4,3);
A = (3,0);
B = (5,0);
C = (7,0);
D = (13,0);
E = (3.2,0.6);
F = (7.6,1.8);
F1 = (3.2,0);
F2 = (3.4,0);
F3 = (5.8,0);
M = (4.6,1.8);
dot(P);
dot(Q);
dot(R);
dot(S);
dot(A);
dot(B);
dot(C);
dot(D);
dot(... | [] |
737 | Square $AXYZ$ is inscribed in equiangular hexagon $ABCDEF$ with $X$ on $\overline{BC}$, $Y$ on $\overline{DE}$, and $Z$ on $\overline{EF}$. Suppose that $AB=40$, and $EF=41(\sqrt{3}-1)$. What is the side-length of the square?
$\textbf{(A)}\ 29\sqrt{3} \qquad\textbf{(B)}\ \frac{21}{2}\sqrt{2}+\frac{41}{2}\sqrt{3} \qqua... | 2012 AMC 12B Problem 21 | We can, $\textsc{wlog}$, assume $Y$ coincides with $D$ and $CD\parallel AF$ as before. In which case, we will have $BC=EF=41(\sqrt{3}-1)$. So we have square $AXDZ$ inscribed in equiangular hexagon $ABCDEF$ with $X$ on $\overline{BC}$ and $Z$ on $\overline{EF}$.
Let $\angle BXA = \theta$; then $\angle BAX=60^\circ -\th... | size(200); defaultpen(fontsize(10)+linewidth(1)); pair A=origin,B=(2.5,0),C=B+2.5*dir(60), D=C+1.75*dir(120),E=D-(3.19,0),F=E-1.8*dir(60); pair X=waypoint(B--C,0.345),Z=rotate(90,A)*X,Y=rotate(90,Z)*A; pair Cp=extension(B,C,Y,Y+dir(-60)); draw(A--B--Cp--Y--E--F--cycle); draw(A--X--Y--Z--cycle,linewidth(0.9)+linetype("2... | [] |
738 | A bug travels from A to B along the segments in the hexagonal lattice pictured below. The segments marked with an arrow can be traveled only in the direction of the arrow, and the bug never travels the same segment more than once. How many different paths are there?
$\textbf{(A)}\ 2112\qquad\textbf{(B)}\ 2304\qquad\... | 2012 AMC 12B Problem 22 | There is $1$ way to get to any of the red arrows. From the first (top) red arrow, there are $2$ ways to get to each of the first and the second (top 2) blue arrows; from the second (bottom) red arrow, there are $3$ ways to get to each of the first and the second blue arrows. So there are in total $5$ ways to get to eac... | size(10cm); draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)--(10.0,6.928203230275509)--(12.0,6.928203230275509)); draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151... | [] |
738 | A bug travels from A to B along the segments in the hexagonal lattice pictured below. The segments marked with an arrow can be traveled only in the direction of the arrow, and the bug never travels the same segment more than once. How many different paths are there?
$\textbf{(A)}\ 2112\qquad\textbf{(B)}\ 2304\qquad\... | 2012 AMC 12B Problem 22 | For every blue arrow, there are $2\cdot 2=4$ ways to reach it without using the reverse arrow since the bug can choose any of $2$ red arrows to pass through and $2$ black arrows to pass through. If the bug passes through the white arrow, the red arrow that the bug travels through must be the closest to the first black... | size(6cm); draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)); draw((0.0,0.0)--(1.0,-1.7320508075688772)--(3.0,-1.7320508075688772)--(4.0,-3.4641016151377544)--(6.0,-3.4641016151377544)--(7.0,-5.196152... | [] |
739 | Let $S=\{(x,y) : x\in \{0,1,2,3,4\}, y\in \{0,1,2,3,4,5\},\text{ and } (x,y)\ne (0,0)\}$.
Let $T$ be the set of all right triangles whose vertices are in $S$. For every right triangle $t=\triangle{ABC}$ with vertices $A$, $B$, and $C$ in counter-clockwise order and right angle at $A$, let $f(t)=\tan(\angle{CBA})$. Wha... | 2012 AMC 12B Problem 25 | This is just another way for the reasoning of solution 1. Picture the question to be a grid of unit squares instead of a coordinate system. Note that the restriction, (This is just to make visualization easier.) Define a "cell" to be a rectangle in the set of $S.$ For example, a cell can be (labeled in a red):
Note ... | // Block 1
unitsize(0.5 cm);
draw((0,1)--(0,5),black);
draw((1,0)--(1,5),black);
draw((2,0)--(2,5),black);
draw((3,0)--(3,5),black);
draw((4,0)--(4,5),black);
draw((0,1)--(5,1),black);
draw((5,0)--(5,5),black);
draw((1,0)--(5,0),black);
draw((0,2)--(5,2),black);
draw((0,3)--(5,3),black);
draw((0,4)--(5,4),black);
draw(... | [] |
740 | A triangle with vertices $(6, 5)$, $(8, -3)$, and $(9, 1)$ is reflected about the line $x=8$ to create a second triangle. What is the area of the union of the two triangles?
$\textbf{(A)}\ 9 \qquad\textbf{(B)}\ \frac{28}{3} \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ \frac{31}{3} \qquad\textbf{(E)}\ \frac{32}{3}$ | 2013 AMC 10A Problem 16 | Let $A$ be at $(6, 5)$, B be at $(8, -3)$, and $C$ be at $(9, 1)$. Reflecting over the line $x=8$, we see that $A' = D = (10,5)$, $B' = B$ (as the x-coordinate of B is 8), and $C' = E = (7, 1)$. Line $AB$ can be represented as $y=-4x+29$, so we see that $E$ is on line $AB$.
We see that if we connect $A$ to $... | // Block 1
pair A = (6, 5), B = (8, -3), C = (9, 1), D = (10, 5), E = (7, 1), F = (8, 7/3);
draw(A--B--C--cycle^^D--E--B--cycle);
dot(A^^B^^C^^D^^E^^F);
label("$A$",A,NW);
label("$B$",B,S);
label("$C$",C,SE);
label("$D$",D,NE);
label("$E$",E,W);
label("$F$", F, N);
// Block 2
pair A = (6, 5), B = (8, -3), C = (9, 1), D... | [] |
740 | A triangle with vertices $(6, 5)$, $(8, -3)$, and $(9, 1)$ is reflected about the line $x=8$ to create a second triangle. What is the area of the union of the two triangles?
$\textbf{(A)}\ 9 \qquad\textbf{(B)}\ \frac{28}{3} \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ \frac{31}{3} \qquad\textbf{(E)}\ \frac{32}{3}$ | 2013 AMC 10A Problem 16 | Since we have to find the area on a coordinate plane, we can use the Shoelace Theorem to find the area of the intersection. When you reflect it, it makes a quadrilateral.
Since it is reflected around $x=8$, the point $(8,-3)$ remains the same on both. The top right corners are $(6,5)$, and its reflection $(10,5)$. ... | // Block 1
pair A = (6, 5), B = (8, -3), C = (9, 1), D = (10, 5), E = (7, 1), F = (8, 7/3);
draw(A--B--C--cycle^^D--E--B--cycle);
dot(A^^B^^C^^D^^E^^F);
label("$A$",A,NW);
label("$B$",B,S);
label("$C$",C,SE);
label("$D$",D,NE);
label("$E$",E,W);
label("$F$", F, N);
// Block 2
pair A = (6, 5), B = (8, -3), C = (9, 1), D... | [] |
741 | Let points $A = (0, 0)$, $B = (1, 2)$, $C=(3, 3)$, and $D = (4, 0)$. Quadrilateral $ABCD$ is cut into equal area pieces by a line passing through $A$. This line intersects $\overline{CD}$ at point $\left(\frac{p}{q}, \frac{r}{s}\right)$, where these fractions are in lowest terms. What is $p+q+r+s$?
$\textbf{(A)}\... | 2013 AMC 10A Problem 18 | First, we shall find the area of quadrilateral $ABCD$. This can be done in any of three ways:
Pick's Theorem: $[ABCD] = I + \dfrac{B}{2} - 1 = 5 + \dfrac{7}{2} - 1 = \dfrac{15}{2}.$
Splitting: Drop perpendiculars from $B$ and $C$ to the x-axis to divide the quadrilateral into triangles and trapezoids, and so the area... | // Block 1
size(8cm);
pair A, B, C, D, E, EE;
A = (0,0);
B = (1,2);
C = (3,3);
D = (4,0);
E = (27/8,15/8);
EE = (27/8,0);
draw(A--B--C--D--A--E);
draw(E--EE,linetype("8 8"));
dot(A);
dot(B);
dot(C);
dot(D);
dot(E);
draw(rightanglemark(E,EE,D,4));
label("A",A,SW);
label("B",B,NW);
label("C",C,NE);
label("D",D,SE);
label... | [] |
741 | Let points $A = (0, 0)$, $B = (1, 2)$, $C=(3, 3)$, and $D = (4, 0)$. Quadrilateral $ABCD$ is cut into equal area pieces by a line passing through $A$. This line intersects $\overline{CD}$ at point $\left(\frac{p}{q}, \frac{r}{s}\right)$, where these fractions are in lowest terms. What is $p+q+r+s$?
$\textbf{(A)}\... | 2013 AMC 10A Problem 18 | Let the point where the altitude from $E$ to $\overline{AD}$ be labeled $F$.
Following the steps above, you can find that the height of $\triangle ADE$ is $\frac{15}{8}$, and from there split the base into two parts, $x$, and $4-x$, such that $x$ is the segment from the origin to the point $F$, and $4-x$ is the segment... | // Block 1
size(8cm);
pair A, B, C, D, E, F;
A = (0,0);
B = (1,2);
C = (3,3);
D = (4,0);
E = (27/8,15/8);
F = (27/8,0);
draw(A--B--C--D--A--E);
draw(E--F,linetype("8 8"));
dot(A);
dot(B);
dot(C);
dot(D);
dot(E);
draw(rightanglemark(E,F,D,4));
label("A",A,SW);
label("B",B,NW);
label("C",C,NE);
label("D",D,SE);
label("E"... | [] |
742 | A unit square is rotated $45^\circ$ about its center. What is the area of the region swept out by the interior of the square?
$\textbf{(A)}\ 1 - \frac{\sqrt2}{2} + \frac{\pi}{4}\qquad\textbf{(B)}\ \frac{1}{2} + \frac{\pi}{4} \qquad\textbf{(C)}\ 2 - \sqrt2 + \frac{\pi}{4}\qquad\textbf{(D)}\ \frac{\sqrt2}{2} + \fra... | 2013 AMC 10A Problem 20 | First, we need to see what this looks like. Below is a diagram.
For this square with side length 1, the distance from center to vertex is $r = \frac{\sqrt{2}}{2}$, hence the area is composed of a semicircle of radius $r$, plus $4$ times a parallelogram (or a kite with diagonals of $(\sqrt{2}-1)$ and $r \text{ or} \... | // Block 1
size(200);
defaultpen(linewidth(0.8));
path square=shift((-.5,-.5))*unitsquare,square2=rotate(45)*square;
fill(square^^square2,grey);
for(int i=0;i<=3;i=i+1)
{
path arcrot=arc(origin,sqrt(2)/2,45+90*i,90*(i+1));
draw(arcrot);
fill(arcrot--(0,0)--cycle,grey);
draw(arc(origin,sqrt(2)/2+1/8,50+90*i,90*(i+1)-10)... | [] |
742 | A unit square is rotated $45^\circ$ about its center. What is the area of the region swept out by the interior of the square?
$\textbf{(A)}\ 1 - \frac{\sqrt2}{2} + \frac{\pi}{4}\qquad\textbf{(B)}\ \frac{1}{2} + \frac{\pi}{4} \qquad\textbf{(C)}\ 2 - \sqrt2 + \frac{\pi}{4}\qquad\textbf{(D)}\ \frac{\sqrt2}{2} + \fra... | 2013 AMC 10A Problem 20 | $\textbf{(high res image; no labels)}$
[Image: images/amc/2013_AMC_10A_Problem_20_0.jpg]
Let $O$ be the center of the square and $C$ be the intersection of $OB$ and $AD$. The desired area consists of the unit square, plus $4$ regions congruent to the region bounded by arc $AB$, $\overline{AC}$, and $\overline{BC}$, pl... | // Block 1
size(200);
defaultpen(linewidth(0.8));
path square=shift((-.5,-.5))*unitsquare,square2=rotate(45)*square;
for(int i=0;i<=6;i=i+1)
{
path arcrot=arc(origin,sqrt(2)/2,45+270*i,270*(i+1));
draw(arcrot);
}
draw(square^^square2);
// Block 2
size(200); defaultpen(linewidth(0.8)); path square=shift((-.5,-.5))*units... | ["https://artofproblemsolving.com/wiki/images/1/1e/AMC_10A_2013_20.jpg"] |
743 | Six spheres of radius $1$ are positioned so that their centers are at the vertices of a regular hexagon of side length $2$. The six spheres are internally tangent to a larger sphere whose center is the center of the hexagon. An eighth sphere is externally tangent to the six smaller spheres and internally tangent to th... | 2013 AMC 10A Problem 22 | We have a regular hexagon with side length $2$ and six spheres on each vertex with radius $1$ that are internally tangent, therefore, drawing radii to the tangent points would create this regular hexagon.
Imagine a 2D overhead view. There is a larger sphere which the $6$ spheres are internally tangent to, with the ce... | // Block 1
draw(circle((0.5,0.866025404),0.5));
draw(circle((-0.5,0.866025404),0.5));
draw(circle((1,0),0.5));
draw(circle((-1,0),0.5));
draw(circle((0.5,-0.866025404),0.5));
draw(circle((-0.5,-0.866025404),0.5));
draw(circle((0,0),1.5));
draw((-0.5,0.866025404)--(0.5,0.866025404));
draw((-1,0)--(1,0));
draw((-0.5,-0.... | [] |
744 | In $\triangle ABC$, $AB = 86$, and $AC=97$. A circle with center $A$ and radius $AB$ intersects $\overline{BC}$ at points $B$ and $X$. Moreover $\overline{BX}$ and $\overline{CX}$ have integer lengths. What is $BC$?
$\textbf{(A)}\ 11\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 33\qquad\textbf{(D)}\ 61\qquad\textbf{(E)... | 2013 AMC 10A Problem 23 | We first draw the height of isosceles triangle $ABD$ and get two equations by the Pythagorean Theorem.
First, $h^2 + k^2 = 86^2$. Second, $h^2 + (k + m)^2 = 97^2$.
Subtracting these two equations, we get $2km + m^2 = 97^2 - 86^2 = (97 - 86)(97 + 86) = 2013$.
We then add $k^2$ to both sides to get $k^2 + 2km + m^2 = 201... | // Block 1
unitsize(2);
import olympiad;
import graph;
pair A,B,C,D,E;
A = (0,0);
B = (70,51);
C = (97,0);
D = (82,29);
E = (76,40);
draw(Circle((0,0),86.609));
draw(A--B--C--A);
draw(A--B--E--A);
draw(A--D);
dot(A);
dot(B,blue);
dot(C);
dot(D,blue);
dot(E);
label("A",A,S);
label("B",B,NE);
label("C",C,S);
label("D",... | [] |
745 | All $20$ diagonals are drawn in a regular octagon. At how many distinct points in the interior
of the octagon (not on the boundary) do two or more diagonals intersect?
$\textbf{(A)}\ 49\qquad\textbf{(B)}\ 65\qquad\textbf{(C)}\ 70\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 128$ | 2013 AMC 10A Problem 25 | If you draw a clear diagram like the one below, it is easy to see that there are $\boxed{\textbf{(A) }49}$ points. | // Block 1
size(14cm);
pathpen = brown + 1.337;
// Initialize octagon
pair[] A;
for (int i=0; i<8; ++i) {
A[i] = dir(45*i);
}
D(CR( (0,0), 1));
// Draw diagonals
// choose pen colors
pen[] colors;
colors[1] = orange + 1.337;
colors[2] = purple;
colors[3] = green;
colors[4] = black;
for (int d=1; d<=4; ++d) {
pathpen = ... | [] |
745 | All $20$ diagonals are drawn in a regular octagon. At how many distinct points in the interior
of the octagon (not on the boundary) do two or more diagonals intersect?
$\textbf{(A)}\ 49\qquad\textbf{(B)}\ 65\qquad\textbf{(C)}\ 70\qquad\textbf{(D)}\ 96\qquad\textbf{(E)}\ 128$ | 2013 AMC 10A Problem 25 | This problem is a counting problem of combinatoric geometry. There are 2 cases for the above diagram:
Case 1: Red Dots
The red dots are the intersection of 3 or more lines. It consists of 8 dots that make up an octagon and 1 dot in the center. Hence, there are $9$ red dots.
Case 2: Blue Dots
The blue dots are the i... | // Block 1
size(8cm);
pathpen = black;
// draw the circle
pair[] A;
for (int i=0; i<8; ++i) {
A[i] = dir(45*i);
}
D(CR((0,0), 1));
// draw the octagon and diagonals
// choose pen colors
pen[] colors;
colors[1] = yellow;
colors[2] = purple;
colors[3] = green;
colors[4] = orange;
for (int d=1; d<=4; ++d) {
pathpen = colo... | [] |
746 | In $\bigtriangleup ABC$, $AB = 86$, and $AC = 97$. A circle with center $A$ and radius $AB$ intersects $\overline{BC}$ at points $B$ and $X$. Moreover $\overline{BX}$ and $\overline{CX}$ have integer lengths. What is $BC$?
$\textbf{(A)} \ 11 \qquad \textbf{(B)} \ 28 \qquad \textbf{(C)} \ 33 \qquad \textbf{(D)} \ ... | 2013 AMC 12A Problem 19 | Solution 1 (Diophantine PoP)
Let circle $A$ intersect $AC$ at $D$ and $E$ as shown. We apply Power of a Point on point $C$ with respect to circle $A.$ This yields the diophantine equation
\[CX \cdot CB = CD \cdot CE\]
\[CX(CX+XB) = (97-86)(97+86)\]
\[CX(CX+XB) = 3 \cdot 11 \cdot 61.\]
Since lengths cannot be negat... | //Made by samrocksnature size(8cm); pair A,B,C,D,E,X; A=(0,0); B=(-53.4,-67.4); C=(0,-97); D=(0,-86); E=(0,86); X=(-29,-81); draw(circle(A,86)); draw(E--C--B--A--X); label("$A$",A,NE); label("$B$",B,SW); label("$C$",C,S); label("$D$",D,NE); label("$E$",E,NE); label("$X$",X,dir(250)); dot(A^^B^^C^^D^^E^^X); | [] |
747 | Six points are equally spaced around a circle of radius 1. Three of these points are the vertices of a triangle that is neither equilateral nor isosceles. What is the area of this triangle?
$\textbf{(A)}\ \frac{\sqrt{3}}{3}\qquad\textbf{(B)}\ \frac{\sqrt{3}}{2}\qquad\textbf{(C)}\ \textbf{1}\qquad\textbf{(D)}\ \sqrt{... | 2013 AMC 10B Problem 7 | If there are no two points on the circle that are adjacent, then the triangle would be equilateral. If the three points are all adjacent, it would be isosceles. Thus, the only possibility is two adjacent points and one point two away. Because one of the sides of this triangle is the diameter, the opposite angle is a... | // Block 1
unitsize(72);
draw((0,0)--(1/2,sqrt(3)/2));
draw((1/2,sqrt(3)/2)--(3/2,sqrt(3)/2));
draw((3/2,sqrt(3)/2)--(2,0));
draw((2,0)--(3/2,-sqrt(3)/2));
draw((3/2,-sqrt(3)/2)--(1/2,-sqrt(3)/2));
draw((1/2,-sqrt(3)/2)--(0,0));
draw((3/2,sqrt(3)/2)--(1/2,-sqrt(3)/2));
draw((3/2,sqrt(3)/2)--(3/2,-sqrt(3)/2));
label("2"... | [] |
748 | The regular octagon $ABCDEFGH$ has its center at $J$. Each of the vertices and the center are to be associated with one of the digits $1$ through $9$, with each digit used once, in such a way that the sums of the numbers on the lines $AJE$, $BJF$, $CJG$, and $DJH$ are all equal. In how many ways can this be done?
$\... | 2013 AMC 10B Problem 22 | First of all, note that $J$ must be $1$, $5$, or $9$ to preserve symmetry, since the sum of 1 to 9 is 45, and we need the remaining 8 to be divisible by 4 (otherwise we will have uneven sums). So, we have:
We also notice that $A+E = B+F = C+G = D+H$.
WLOG, assume that $J = 1$. Thus the pairs of vertices must be ... | // Block 1
pair A,B,C,D,E,F,G,H,J;
A=(20,20(2+sqrt(2)));
B=(20(1+sqrt(2)),20(2+sqrt(2)));
C=(20(2+sqrt(2)),20(1+sqrt(2)));
D=(20(2+sqrt(2)),20);
E=(20(1+sqrt(2)),0);
F=(20,0);
G=(0,20);
H=(0,20(1+sqrt(2)));
J=(10(2+sqrt(2)),10(2+sqrt(2)));
draw(A--B);
draw(B--C);
draw(C--D);
draw(D--E);
draw(E--F);
draw(F--G);
draw(G--... | [] |
749 | Cities $A$, $B$, $C$, $D$, and $E$ are connected by roads $\widetilde{AB}$, $\widetilde{AD}$, $\widetilde{AE}$, $\widetilde{BC}$, $\widetilde{BD}$, $\widetilde{CD}$, and $\widetilde{DE}$. How many different routes are there from $A$ to $B$ that use each road exactly once? (Such a route will necessarily visit some citie... | 2013 AMC 12B Problem 12 | Note that cities $C$ and $E$ can be removed when counting paths because if a path goes in to $C$ or $E$, there is only one possible path to take out of cities $C$ or $E$.
So the diagram is as follows:
Now we proceed with casework. Remember that there are two ways to travel from $A$ to $D$, $D$ to $A$, $B$ to $D$ and... | // Block 1
unitsize(10mm);
defaultpen(linewidth(1.2pt)+fontsize(10pt));
dotfactor=4;
pair A=(1,0), B=(4.24,0), C=(5.24,3.08), D=(2.62,4.98), E=(0,3.08);
dot (A);
dot (B);
dot (D);
label("$A$",A,S);
label("$B$",B,SE);
label("$D$",D,N);
draw(A--B..D..cycle);
draw(A--D);
draw(B--D);
// Block 2
unitsize(10mm); defaultp... | [] |
750 | Equilateral $\triangle ABC$ has side length $1$, and squares $ABDE$, $BCHI$, $CAFG$ lie outside the triangle. What is the area of hexagon $DEFGHI$?
$\textbf{(A)}\ \dfrac{12+3\sqrt3}4\qquad\textbf{(B)}\ \dfrac92\qquad\textbf{(C)}\ 3+\sqrt3\qquad\textbf{(D)}\ \dfrac{6+3\sqrt3}2\qquad\textbf{(E)}\ 6$ | 2014 AMC 10A Problem 13 | As seen in the previous solution, segment $GH$ is $\sqrt{3}$. Think of the picture as one large equilateral triangle, $\triangle{JKL}$ with the sides of $2\sqrt{3}+1$, by extending $EF$, $GH$, and $DI$ to points $J$, $K$, and $L$, respectively. This makes the area of $\triangle{JKL}$ $\dfrac{\sqrt{3}}{4}(2\sqrt{3}+1)... | // Block 1
import graph;
size(10cm);
pen dps = linewidth(0.7) + fontsize(8); defaultpen(dps);
pair B = (0,0);
pair C = (1,0);
pair A = rotate(60,B)*C;
pair E = rotate(270,A)*B;
pair D = rotate(270,E)*A;
pair F = rotate(90,A)*C;
pair G = rotate(90,F)*A;
pair I = rotate(270,B)*C;
pair H = rotate(270,I)*B;
pair J = ro... | [] |
751 | The $y$-intercepts, $P$ and $Q$, of two perpendicular lines intersecting at the point $A(6,8)$ have a sum of zero. What is the area of $\triangle APQ$?
$\textbf{(A)}\ 45\qquad\textbf{(B)}\ 48\qquad\textbf{(C)}\ 54\qquad\textbf{(D)}\ 60\qquad\textbf{(E)}\ 72$ | 2014 AMC 10A Problem 14 | Note that if the $y$-intercepts have a sum of $0$, the distance from the origin to each of the intercepts must be the same. Call this distance $a$. Since the $\angle PAQ = 90^\circ$, the length of the median to the midpoint of the hypotenuse is equal to half the length of the hypotenuse. Since the median's length is $\... | // Block 1
//Needs refining (hmm I think it's fine --bestwillcui1)
size(12cm);
fill((0,10)--(6,8)--(0,-10)--cycle,rgb(.7,.7,.7));
for(int i=-2;i<=8;i+=1)
draw((i,-12)--(i,12),grey);
for(int j=-12;j<=12;j+=1)
draw((-2,j)--(8,j),grey);
draw((-3,0)--(9,0),linewidth(1),Arrows); //x-axis
draw((0,-13)--(0,13),linewidth(1... | [] |
751 | The $y$-intercepts, $P$ and $Q$, of two perpendicular lines intersecting at the point $A(6,8)$ have a sum of zero. What is the area of $\triangle APQ$?
$\textbf{(A)}\ 45\qquad\textbf{(B)}\ 48\qquad\textbf{(C)}\ 54\qquad\textbf{(D)}\ 60\qquad\textbf{(E)}\ 72$ | 2014 AMC 10A Problem 14 | (Not to scale)
Long solution:
By rotating the right triangle, we get the figure shown where \[CD\perp EF\] and CE=CF. We know AB=CD=6 and AD=BC=8. By the pythagorean theorem, we have AC=10, and since C is the midpoint of EF, CE=EF=10 also. By similar triangles, \[\frac{AD}{FC}=\frac{DG}{CG}~\text{so}~DG=\frac{8}{3}~\t... | // Block 1
unitsize(36);
pair A = (0,0);
pair B = (5,1);
pair C = (13/5,13);
pair D = C-B;
pair E = (26/5,0);
pair F = (0,26);
pair G = intersectionpoints(C--D,A--F)[0];
draw(A--E--F--A--B--C--D--A,linewidth(1));
label(A,scale(2)*"A",dir(-135));
label(E,scale(2)*"E",dir(-45));
label(F,scale(2)*"F",dir(90));
label(B,sca... | [] |
752 | In rectangle $ABCD$, $AB=1$, $BC=2$, and points $E$, $F$, and $G$ are midpoints of $\overline{BC}$, $\overline{CD}$, and $\overline{AD}$, respectively. Point $H$ is the midpoint of $\overline{GE}$. What is the area of the shaded region?
$\textbf{(A)}\ \dfrac1{12}\qquad\textbf{(B)}\ \dfrac{\sqrt3}{18}\qquad\textbf{(C... | 2014 AMC 10A Problem 16 | Denote $D=(0,0)$. Then $A= (0,2), F = \left(\frac12,0\right), H = \left(\frac12,1\right)$. Let the intersection of $AF$ and $DH$ be $X$, and the intersection of $BF$ and $CH$ be $Y$. Then we want to find the coordinates of $X$ so we can find $XY$. From our points, the slope of $AF$ is $\bigg(\dfrac{-2}{\tfrac12}\bigg) ... | // Block 1
import graph;
size(9cm);
pen dps = fontsize(10); defaultpen(dps);
pair D = (0,0);
pair F = (1/2,0);
pair C = (1,0);
pair G = (0,1);
pair E = (1,1);
pair A = (0,2);
pair B = (1,2);
pair H = (1/2,1);
// do not look
pair X = (1/3,2/3);
pair Y = (2/3,2/3);
draw(A--B--C--D--cycle);
draw(G--E);
draw(A--F--B);
dr... | [] |
752 | In rectangle $ABCD$, $AB=1$, $BC=2$, and points $E$, $F$, and $G$ are midpoints of $\overline{BC}$, $\overline{CD}$, and $\overline{AD}$, respectively. Point $H$ is the midpoint of $\overline{GE}$. What is the area of the shaded region?
$\textbf{(A)}\ \dfrac1{12}\qquad\textbf{(B)}\ \dfrac{\sqrt3}{18}\qquad\textbf{(C... | 2014 AMC 10A Problem 16 | The area of the shaded area is the area of $\triangle DHC$ minus the two triangles on the side.
Extend $\overline{DH}$ so that it hits point $B$. Call the intersection of $\overline{AF}$ and $\overline{DB}$ point $P$. \[\triangle APB \sim \triangle FPD\]
Drop altitudes from $P$ down to $\overline{DF}$ and $\overline{AB... | // Block 1
import graph;
size(9cm);
pen dps = fontsize(10); defaultpen(dps);
pair D = (0,0);
pair F = (1/2,0);
pair C = (1,0);
pair G = (0,1);
pair E = (1,1);
pair A = (0,2);
pair B = (1,2);
pair H = (1/2,1);
// do not look
pair X = (1/3,2/3);
pair Y = (2/3,2/3);
draw(A--B--C--D--cycle);
draw(G--E);
draw(A--F--B);
dr... | [] |
753 | Four cubes with edge lengths $1$, $2$, $3$, and $4$ are stacked as shown. What is the length of the portion of $\overline{XY}$ contained in the cube with edge length $3$?
$\textbf{(A)}\ \dfrac{3\sqrt{33}}5\qquad\textbf{(B)}\ 2\sqrt3\qquad\textbf{(C)}\ \dfrac{2\sqrt{33}}3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 3\sqr... | 2014 AMC 10A Problem 19 | Let's redraw the diagram, however make a 3D coordinate plane, using D as the origin.
Now we can use the distance formula in 3D, which is $\sqrt{(x_{1}-x_{2})^2+(y_{1}-y_{2})^2+(z_{1}-z_{2})^2}$ and plug it in for the distance of $XY$.
$\sqrt{(0-4)^2+(10-0)^2+(0-4)^2}$
We get the answer as $\sqrt{132} = 2\sqrt{33}... | // Block 1
dotfactor = 3;
size(10cm);
dot((0, 10));
label("$X(0,10,0)$", (0,10),W,fontsize(8pt));
dot((6,2));
label("$Y(4,0,4)$", (6,2),E,fontsize(8pt));
draw((0, 0)--(0, 10)--(1, 10)--(1, 9)--(2, 9)--(2, 7)--(3, 7)--(3,4)--(4, 4)--(4, 0)--cycle);
draw((0,9)--(1, 9)--(1.5, 9.5)--(1.5, 10.5)--(0.5, 10.5)--(0, 10));
draw... | [] |
754 | A rectangular piece of paper whose length is $\sqrt3$ times the width has area $A$. The paper is divided into three equal sections along the opposite lengths, and then a dotted line is drawn from the first divider to the second divider on the opposite side as shown. The paper is then folded flat along this dotted line ... | 2014 AMC 10A Problem 23 | Find the midpoint of the dotted line. Draw a line perpendicular to it. From the point this line intersects the top of the paper, draw lines to each endpoint of the dotted line. These two lines plus the dotted line form a triangle which is the double-layered portion of the folded paper. WLOG, assume the width of the... | // Block 1
import graph;
unitsize(3cm);
real L = 0.05;
pair A = (0,0);
pair B = (sqrt(3),0);
pair C = (sqrt(3),1);
pair D = (0,1);
pair X1 = (sqrt(3)/3,0);
pair X2= (2*sqrt(3)/3,0);
pair Y1 = (2*sqrt(3)/3,1);
pair Y2 = (sqrt(3)/3,1);
dot(X1);
dot(Y1);
draw(A--B--C--D--cycle, linewidth(2));
draw(B--D,dashed);
draw(X1--Y... | [] |
754 | A rectangular piece of paper whose length is $\sqrt3$ times the width has area $A$. The paper is divided into three equal sections along the opposite lengths, and then a dotted line is drawn from the first divider to the second divider on the opposite side as shown. The paper is then folded flat along this dotted line ... | 2014 AMC 10A Problem 23 | Our original paper can be divided like this:
After the fold across the dashed line, our paper becomes:
Since our original sheet of paper has six congruent $30-60-90$ triangles and and our new one has four, the ratio of the area $B:A$ is equal to $4:6\implies \boxed{\textbf{(C)} \: 2:3}$
~CHECKMATE2021 | // Block 1
import graph;
unitsize(3cm);
real L = 0.05;
pair A = (0,0);
pair B = (sqrt(3),0);
pair C = (sqrt(3),1);
pair D = (0,1);
pair X1 = (sqrt(3)/3,0);
pair X2= (2*sqrt(3)/3,0);
pair Y1 = (2*sqrt(3)/3,1);
pair Y2 = (sqrt(3)/3,1);
dot(X1);
dot(Y1);
draw(A--B--C--D--cycle, linewidth(2));
draw(X1--Y1,dashed);
draw(Y2-... | [] |
755 | In $\triangle BAC$, $\angle BAC=40^\circ$, $AB=10$, and $AC=6$. Points $D$ and $E$ lie on $\overline{AB}$ and $\overline{AC}$ respectively. What is the minimum possible value of $BE+DE+CD$?
$\textbf{(A) }6\sqrt 3+3\qquad \textbf{(B) }\dfrac{27}2\qquad \textbf{(C) }8\sqrt 3\qquad \textbf{(D) }14\qquad \textbf{(E) }3\... | 2014 AMC 12A Problem 20 | (Diagram by shihan)
Reflect $C$ across $AB$ to $C'$. Similarly, reflect $B$ across $AC$ to $B'$. Clearly, $BE = B'E$ and $CD = C'D$. Thus, the sum $BE + DE + CD = B'E + DE + C'D$. This value is minimized when $B'$, $C'$, $D$ and $E$ are collinear. To finish, we use the law of cosines on the triangle $AB'C'$: $B'C' = \s... | // Block 1
size(300);
defaultpen(linewidth(0.4)+fontsize(10));
pen s = linewidth(0.8)+fontsize(8);
pair A,B,C,D,Ep,Bp,Cp;
A = (0,0);
B = 10*dir(-110);C = 6*dir(-70);
Bp = 10*dir(-30);Cp = 6*dir(-150);
D = IP(Cp--Bp, A--B); Ep = IP(Cp--Bp, A--C);
draw(A--B--C--A--Cp--Bp--A);
draw(Cp--B);
draw(C--Bp);
draw(C--D);
draw(B... | [] |
756 | Let $f_0(x)=x+|x-100|-|x+100|$, and for $n\geq 1$, let $f_n(x)=|f_{n-1}(x)|-1$. For how many values of $x$ is $f_{100}(x)=0$?
$\textbf{(A) }299\qquad \textbf{(B) }300\qquad \textbf{(C) }301\qquad \textbf{(D) }302\qquad \textbf{(E) }303\qquad$ | 2014 AMC 12A Problem 24 | 1. Draw the graph of $f_0(x)$ by dividing the domain into three parts.
2. Apply the recursive rule a few times to find the pattern.
Note: $f_n(x) = |f_{n-1}(x)| - 10$ is used to enlarge the difference, but the reasoning is the same.
3. Extrapolate to $f_{100}$. Notice that the summits start $100$ away from $0$ and... | // Block 1
unitsize(0);
int w = 250;
int h = 125;
xaxis(-w,w,Ticks(100.0),Arrows);
yaxis(-h,h,Ticks(100.0),Arrows);
draw((-100,-h)--(-100,h),dashed);
draw((100,-h)--(100,h),dashed);
real f0(real x) { return x + abs(x-100) - abs(x+100); }
draw(graph(f0,-w,w),Arrows);
label("$f_0$",(-w,f0(-w)),W);
// Block 2
unitsize... | [] |
756 | Let $f_0(x)=x+|x-100|-|x+100|$, and for $n\geq 1$, let $f_n(x)=|f_{n-1}(x)|-1$. For how many values of $x$ is $f_{100}(x)=0$?
$\textbf{(A) }299\qquad \textbf{(B) }300\qquad \textbf{(C) }301\qquad \textbf{(D) }302\qquad \textbf{(E) }303\qquad$ | 2014 AMC 12A Problem 24 | Note $f_{100}(x) = 0$ when $|f_{99}(x)| -1$ = 0. This occurs when $f_{99}(x) = \pm 1$.
Then, repeating this process, we note $f_{99}(x) = \pm 1 \implies |f_{98}(x)| = 0, 2$, and hence $f_{98}(x) = 0, \pm 2$.
Similarly, $f_{97}(x) = \pm 1, \pm 3$. Extrapolating this pattern, we must have $f_{0}(x) = 0, \pm 2$, $\dots... | // Block 1
unitsize(0);
int w = 250;
int h = 125;
xaxis(-w,w,Ticks(100.0),Arrows);
yaxis(-h,h,Ticks(100.0),Arrows);
real f0(real x) { return x + abs(x-100) - abs(x+100); }
draw(graph(f0,-w,w),Arrows);
label("$f_0$",(-w,f0(-w)),W);
// Block 2
unitsize(0); int w = 250; int h = 125; xaxis(-w,w,Ticks(100.0),Arrows); y... | [] |
757 | Two concentric circles have radii $1$ and $2$. Two points on the outer circle are chosen independently and uniformly at random. What is the probability that the chord joining the two points intersects the inner circle?
$\textbf{(A)}\ \frac{1}{6}\qquad \textbf{(B)}\ \frac{1}{4}\qquad \textbf{(C)}\ \frac{2-\sqrt{2}}{2}\... | 2014 AMC 10B Problem 19 | Let the center of the two circles be $O$. Now pick an arbitrary point $A$ on the boundary of the circle with radius $2$. We want to find the range of possible places for the second point, $A'$, such that $AA'$ passes through the circle of radius $1$. To do this, first draw the tangents from $A$ to the circle of radius ... | // Block 1
scale(200);
pair A,O,B,C,H;
A = (0,1);
O = (0,0);
B = (-.866,-.5);
C = (.866,-.5);
H = (0, -.5);
draw(A--C--cycle);
draw(A--O--cycle);
draw(O--C--cycle);
draw(O--H,dashed+linewidth(.7));
draw(A--B--cycle);
draw(B--C--cycle);
draw(O--B--cycle);
dot("$A$",A,N);
dot("$O$",O,NW);
dot("$B$",B,W);
dot("$C$",C,E);
... | [] |
758 | Trapezoid $ABCD$ has parallel sides $\overline{AB}$ of length $33$ and $\overline {CD}$ of length $21$. The other two sides are of lengths $10$ and $14$. The angles $A$ and $B$ are acute. What is the length of the shorter diagonal of $ABCD$?
$\textbf{(A) }10\sqrt{6}\qquad\textbf{(B) }25\qquad\textbf{(C) }8\sqrt{10}\qq... | 2014 AMC 10B Problem 21 | In the diagram, $\overline{DE} \perp \overline{AB}, \overline{FC} \perp \overline{AB}$.
Denote $\overline{AE} = x$ and $\overline{DE} = h$. In right triangle $AED$, we have from the Pythagorean theorem: $x^2+h^2=100$. Note that since $EF = DC$, we have $BF = 33-DC-x = 12-x$. Using the Pythagorean theorem in right tria... | // Block 1
size(7cm);
pair A,B,C,D,CC,DD;
A = (-2,7);
B = (14,7);
C = (10,0);
D = (0,0);
CC = (10,7);
DD = (0,7);
draw(A--B--C--D--cycle);
//label("33",(A+B)/2,N);
label("21",(C+D)/2,S);
label("10",(A+D)/2,W);
label("14",(B+C)/2,E);
label("$A$",A,NW);
label("$B$",B,NE);
label("$C$",C,SE);
label("$D$",D,SW);
label("$E$"... | [] |
758 | Trapezoid $ABCD$ has parallel sides $\overline{AB}$ of length $33$ and $\overline {CD}$ of length $21$. The other two sides are of lengths $10$ and $14$. The angles $A$ and $B$ are acute. What is the length of the shorter diagonal of $ABCD$?
$\textbf{(A) }10\sqrt{6}\qquad\textbf{(B) }25\qquad\textbf{(C) }8\sqrt{10}\qq... | 2014 AMC 10B Problem 21 | The area of $\Delta AED$ is by Heron's, $4\sqrt{9(4)(3)(2)}=24\sqrt{6}$. This makes the length of the altitude from $D$ onto $\overline{AE}$ equal to $4\sqrt{6}$. One may now proceed as in Solution $1$ to obtain an answer of $\boxed{\textbf{(B) }25}$. | // Block 1
size(7cm);
pair A,B,C,D,E;
A = (-2,7);
B = (14,7);
C = (10,0);
D = (0,0);
E = (4,7);
draw(A--B--C--D--cycle);
draw(D--E);
label("21",(C+D)/2,S);
label("10",(A+D)/2,W);
label("14",(12,1),E);
label("14",(2,1),E);
label("12",(A+E)/2,N);
label("21",(E+B)/2,N);
label("$A$",A,NW);
label("$B$",B,NE);
label("$C$",C,... | [] |
759 | Eight semicircles line the inside of a square with side length 2 as shown. What is the radius of the circle tangent to all of these semicircles?
$\text{(A) } \dfrac{1+\sqrt2}4 \quad \text{(B) } \dfrac{\sqrt5-1}2 \quad \text{(C) } \dfrac{\sqrt3+1}4 \quad \text{(D) } \dfrac{2\sqrt3}5 \quad \text{(E) } \dfrac{\sqrt5}3$ | 2014 AMC 10B Problem 22 | We connect the centers of the circle and one of the semicircles, then draw the perpendicular from the center of the middle circle to that side, as shown.
We will start by creating an equation by the Pythagorean theorem: \[\sqrt{1^2 + \left(\frac12\right)^2} = \sqrt{\frac54} = \frac{\sqrt5}{2}.\]
Let's call $r$ as... | // Block 1
scale(200);
draw(scale(.5)*((-1,-1)--(1,-1)--(1,1)--(-1,1)--cycle));
path p = arc((.25,-.5),.25,0,180)--arc((-.25,-.5),.25,0,180);
draw(p);
p=rotate(90)*p; draw(p);
p=rotate(90)*p; draw(p);
p=rotate(90)*p; draw(p);
draw(scale((sqrt(5)-1)/4)*unitcircle);
pair OO=(0,0);
pair XX=(-.25,-.5);
pair YY=(0,-.5);
dra... | [] |
760 | A sphere is inscribed in a truncated right circular cone as shown. The volume of the truncated cone is twice that of the sphere. What is the ratio of the radius of the bottom base of the truncated cone to the radius of the top base of the truncated cone?
$\text{(A) } \dfrac32 \quad \text{(B) } \dfrac{1+\sqrt5}2 \quad ... | 2014 AMC 10B Problem 23 | First, we draw the vertical cross-section passing through the middle of the frustum.
Let the top base have a diameter of $2$ and the bottom base has a diameter of $2r$.
Then using the Pythagorean theorem we have:
$(r+1)^2=(2s)^2+(r-1)^2$,
which is equivalent to:
$r^2+2r+1=4s^2+r^2-2r+1$.
Subtracting $r^2-2r+1$ from... | // Block 1
size(7cm);
pair A,B,C,D;
real r = (3+sqrt(5))/2;
real s = sqrt(r);
A = (-r,0);
B = (r,0);
C = (1,2*s);
D = (-1,2*s);
draw(A--B--C--D--cycle);
pair O = (0,s);
draw(shift(O)*scale(s)*unitcircle);
dot(O);
pair X,Y;
X = (0,0);
Y = (0,2*s);
draw(X--Y);
label("$r-1$",(r/2+1/2,0),S);
label("$1$",(Y+C)/2,N);
label("... | [] |
761 | The numbers $1, 2, 3, 4, 5$ are to be arranged in a circle. An arrangement is $\textit{bad}$ if it is not true that for every $n$ from $1$ to $15$ one can find a subset of the numbers that appear consecutively on the circle that sum to $n$. Arrangements that differ only by a rotation or a reflection are considered the ... | 2014 AMC 10B Problem 24 | We see that there are $5!$ total ways to arrange the numbers. However, we can always rotate these numbers so that, for example, the number $1$ is always at the top of the circle. Thus, there are only $4!$ ways under rotation. Every case has exactly $1$ reflection, so that gives us only $4!/2$, or $12$ cases, which is n... | // Block 1
draw(circle((0, 0), 5));
pair O, A, B, C, D, E;
O=origin;
A=(0, 5);
B=rotate(72)*A;
C=rotate(144)*A;
D=rotate(216)*A;
E=rotate(288)*A;
label("$x$", A, N);
label("$y$", C, SW);
label("$z$", D, SE);
// Block 2
draw(circle((0, 0), 5));
pair O, A, B, C, D, E;
O=origin;
A=(0, 5);
B=rotate(72)*A;
C=rotate(144)*A;
... | [] |
762 | A sphere is inscribed in a truncated right circular cone as shown. The volume of the truncated cone is twice that of the sphere. What is the ratio of the radius of the bottom base of the truncated cone to the radius of the top base of the truncated cone?
$\text{(A) } \dfrac32 \quad \text{(B) } \dfrac{1+\sqrt5}2 \qua... | 2014 AMC 12B Problem 19 | Solution 1
First, we draw the vertical cross-section passing through the middle of the frustum.
let the top base equal 2 and the bottom base to be equal to 2r
then using the Pythagorean theorem we have:
$(r+1)^2=(2s)^2+(r-1)^2$
which is equivalent to:
$r^2+2r+1=4s^2+r^2-2r+1$
subtracting $r^2-2r+1$ from both sides
$4... | size(7cm); pair A,B,C,D; real r = (3+sqrt(5))/2; real s = sqrt(r); A = (-r,0); B = (r,0); C = (1,2*s); D = (-1,2*s); draw(A--B--C--D--cycle); pair O = (0,s); draw(shift(O)*scale(s)*unitcircle); dot(O); pair X,Y; X = (0,0); Y = (0,2*s); draw(X--Y); label("$r-1$",(X+B)/2,S); label("$1$",(Y+C)/2,N); label("$s$",(O+Y)/2,W)... | [] |
763 | Let $ABCDE$ be a pentagon inscribed in a circle such that $AB = CD = 3$, $BC = DE = 10$, and $AE= 14$. The sum of the lengths of all diagonals of $ABCDE$ is equal to $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$ ?
$\textbf{(A) }129\qquad \textbf{(B) }247\qquad \textbf{(C) }35... | 2014 AMC 12B Problem 24 | Let $BE=a$, $AD=b$, and $AC=CE=BD=c$. Let $F$ be on $AE$ such that $CF \perp AE$.
In $\triangle CFE$ we have $\cos\theta = -\cos(\pi-\theta)=-7/c$. We use the Law of Cosines on $\triangle ABC$ to get $60\cos\theta = 109-c^2$. Eliminating $\cos\theta$ we get $c^3-109c-420=0$ which factorizes as
\[(c+7)(c+5)(c-12)=0.... | // Block 1
size(200);
defaultpen(linewidth(0.4)+fontsize(10));
pen s = linewidth(0.8)+fontsize(8);
pair O,A,B,C,D,E0,F;
O=origin;
A= dir(198);
path c = CR(O,1);
real r = 0.13535;
B = IP(c, CR(A,3*r));
C = IP(c, CR(B,10*r));
D = IP(c, CR(C,3*r));
E0 = OP(c, CR(D,10*r));
F = foot(C,A,E0);
dot("$A$", A, A-O);
dot("$B$",... | [] |
764 | The diagram below shows the circular face of a clock with radius $20$ cm and a circular disk with radius $10$ cm externally tangent to the clock face at $12$ o' clock. The disk has an arrow painted on it, initially pointing in the upward vertical direction. Let the disk roll clockwise around the clock face. At what poi... | 2015 AMC 10A Problem 14 | Suppose that the small disk also had a clock face on it, and that both disks were toothed wheels, free to rotate around their centers. The part of the picture where they engage would look like this:
The small cog has half the radius, and therefore half the circumference. If the large cog turns $30^\circ$ counterclockw... | // Block 1
fill(arc((0,0),2,30,150)--cycle,lightgrey);
draw(arc((0,0),2,30,150));
draw(1.8*dir(90)--2*dir(90));
draw(1.8*dir(60)--2*dir(60));
label("12",1.56*dir(90));
label("1",1.56*dir(60));
draw(arc((0,3),1,-15,-165));
draw(0.9*dir(-90)+(0,3)--dir(-90)+(0,3));
draw(0.9*dir(-60)+(0,3)--dir(-60)+(0,3));
draw(0.9*dir(... | [] |
765 | Eight people are sitting around a circular table, each holding a fair coin. All eight people flip their coins and those who flip heads stand while those who flip tails remain seated. What is the probability that no two adjacent people will stand?
$\textbf{(A)}\dfrac{47}{256}\qquad\textbf{(B)}\dfrac{3}{16}\qquad\textbf... | 2015 AMC 10A Problem 22 | Case 1: Everyone flips tails
1 way:
Case 2: One person flips heads
8 ways:
Case 3: Two people flip heads
There are $\binom{8}{2}$ ways to choose any $2$ people in the circle and $8$ ways to choose $2$ people adjacent to each other. So this case gives us $\binom{8}{2} - 8 = 20$ ways. An example of choosing tw... | // Block 1
size(50);
import graph;
real R = 0.5; // Radius of big circle
real r = 0.05; // Radius of small circles
int n = 8; // Number of small circles
// Draw big circle
draw(Circle((0,0), R), linewidth(1));
// Place small circles on circumference
for (int i = 0; i < n; ++i) {
real... | [] |
765 | Eight people are sitting around a circular table, each holding a fair coin. All eight people flip their coins and those who flip heads stand while those who flip tails remain seated. What is the probability that no two adjacent people will stand?
$\textbf{(A)}\dfrac{47}{256}\qquad\textbf{(B)}\dfrac{3}{16}\qquad\textbf... | 2015 AMC 10A Problem 22 | We know that the denominator of the probability is $2^8=256$. So now we only have to calculate the numerator, which is the number of arrangements for $8$ people at a round table without $2$ or more neighboring people standing.
Denote $a_n$ as number of arrangements for $n$ people at a round table without $2$ or more n... | // Block 1
draw(circle((0, 0), 5));
pair A, B, C;
A=(0, 5);
B=rotate(72)*A;
C=rotate(144)*A;
label("$p_1=1$", A, N);
label("$p_n=0$", B, NW);
label("$p_{n-1}=0$", C, SW);
// Block 2
draw(circle((0, 0), 5));
pair A, B, C;
A=(0, 5);
B=rotate(72)*A;
C=rotate(144)*A;
label("$p_1=0$", A, N);
label("$p_n=0$", B, NW);
label("... | [] |
766 | Eight people are sitting around a circular table, each holding a fair coin. All eight people flip their coins and those who flip heads stand while those who flip tails remain seated. What is the probability that no two adjacent people will stand?
$\textbf{(A)}\dfrac{47}{256}\qquad\textbf{(B)}\dfrac{3}{16}\qquad\textbf... | 2015 AMC 12A Problem 17 | Case 1: Everyone flips tails
1 way:
Case 2: One person flips heads
8 ways:
Case 3: Two people flip heads
There are $\binom{8}{2}$ ways to choose any $2$ people in the circle and $8$ ways to choose $2$ people adjacent to each other. So this case gives us $\binom{8}{2} - 8 = 20$ ways. An example of choosing tw... | // Block 1
size(50); import graph; real R = 0.5; // Radius of big circle real r = 0.05; // Radius of small circles int n = 8; // Number of small circles // Draw big circle draw(Circle((0,0), R), linewidth(1)); // Place small circles on circumference for (int i = 0; i < n; ++i) { real... | [] |
766 | Eight people are sitting around a circular table, each holding a fair coin. All eight people flip their coins and those who flip heads stand while those who flip tails remain seated. What is the probability that no two adjacent people will stand?
$\textbf{(A)}\dfrac{47}{256}\qquad\textbf{(B)}\dfrac{3}{16}\qquad\textbf... | 2015 AMC 12A Problem 17 | We know that the denominator of the probability is $2^8=256$. So now we only have to calculate the numerator, which is the number of arrangements for $8$ people at a round table without $2$ or more neighboring people standing.
Denote $a_n$ as number of arrangements for $n$ people at a round table without $2$ or more n... | // Block 1
draw(circle((0, 0), 5)); pair A, B, C; A=(0, 5); B=rotate(72)*A; C=rotate(144)*A; label("$p_1=1$", A, N); label("$p_n=0$", B, NW); label("$p_{n-1}=0$", C, SW);
// Block 2
draw(circle((0, 0), 5)); pair A, B, C; A=(0, 5); B=rotate(72)*A; C=rotate(144)*A; label("$p_1=0$", A, N); label("$p_n=0$", B, NW); label("... | [] |
767 | Johann has $64$ fair coins. He flips all the coins. Any coin that lands on tails is tossed again. Coins that land on tails on the second toss are tossed a third time. What is the expected number of coins that are now heads?
$\textbf{(A) } 32 \qquad\textbf{(B) } 40 \qquad\textbf{(C) } 48 \qquad\textbf{(D) } 56 \qquad\t... | 2015 AMC 10B Problem 18 | We can simplify the problem first, then apply reasoning to the original problem. Let's say that there are $8$ coins. Shaded coins flip heads, and blank coins flip tails. So, after the first flip;
Then, after the second (new heads in blue);
And after the third (new head in green);
So in total, $7$ of the $8$ co... | // Block 1
filldraw(circle((-5,0),0.35),white);
filldraw(circle((-4,0),0.35),white);
filldraw(circle((-3,0),0.35),white);
filldraw(circle((-2,0),0.35),white);
filldraw(circle((-1,0),0.35),black);
filldraw(circle((-0,0),0.35),black);
filldraw(circle((1,0),0.35),black);
filldraw(circle((2,0),0.35),black);
// Block 2
fill... | [] |
768 | In $\triangle{ABC}$, $\angle{C} = 90^{\circ}$ and $AB = 12$. Squares $ABXY$ and $ACWZ$ are constructed outside of the triangle. The points $X, Y, Z$, and $W$ lie on a circle. What is the perimeter of the triangle?
$\textbf{(A) }12+9\sqrt{3}\qquad\textbf{(B) }18+6\sqrt{3}\qquad\textbf{(C) }12+12\sqrt{2}\qquad\textbf{(D... | 2015 AMC 10B Problem 19 | The center of the circle lies on the intersection between the perpendicular bisectors of chords $ZW$ and $YX$. Therefore we know the center of the circle must also be the midpoint of the hypotenuse. Let this point be $O$. Draw perpendiculars to $ZW$ and $YX$ from $O$, and connect $OZ$ and $OY$. $OY^2=6^2+12^2=180$. Let... | // Block 1
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(11.5cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = bl... | [] |
768 | In $\triangle{ABC}$, $\angle{C} = 90^{\circ}$ and $AB = 12$. Squares $ABXY$ and $ACWZ$ are constructed outside of the triangle. The points $X, Y, Z$, and $W$ lie on a circle. What is the perimeter of the triangle?
$\textbf{(A) }12+9\sqrt{3}\qquad\textbf{(B) }18+6\sqrt{3}\qquad\textbf{(C) }12+12\sqrt{2}\qquad\textbf{(D... | 2015 AMC 10B Problem 19 | Both solution 1 and 2 uses Pythagorean Theorem to prove $\triangle ABC$ is isosceles right triangle. I'm going to prove $\triangle ABC$ is isosceles right triangle without using Pythagorean Theorem. I will use geometry rotation.
Let $O$ be the the midpoint of $AB$. The perpendicular bisector of line $WZ$ and $XY$ wi... | // Block 1
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(11.5cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = bl... | [] |
769 | Erin the ant starts at a given corner of a cube and crawls along exactly 7 edges in such a way that she visits every corner exactly once and then finds that she is unable to return along an edge to her starting point. How many paths are there meeting these conditions?
$\textbf{(A) }\text{6}\qquad\textbf{(B) }\text{9}\... | 2015 AMC 10B Problem 20 | We label the vertices of the cube as different letters and numbers shown above. We label these so that Erin can only crawl from a number to a letter or a letter to a number (this can be seen as a coloring argument). The starting point is labeled $A$.
If we define a "move" as each time Erin crawls along a single edge... | // Block 1
import three; draw((1,1,1)--(1,0,1)--(1,0,0)--(0,0,0)--(0,0,1)--(0,1,1)--(1,1,1)--(1,1,0)--(0,1,0)--(0,1,1)); draw((0,0,1)--(1,0,1)); draw((1,0,0)--(1,1,0)); draw((0,0,0)--(0,1,0));
label("2",(0,0,0),S);
label("A",(1,0,0),W);
label("B",(0,0,1),N);
label("1",(1,0,1),NW);
label("3",(1,1,0),S);
label("C",(0,1,... | [] |
770 | In the figure shown below, $ABCDE$ is a regular pentagon and $AG=1$. What is $FG + JH + CD$?
$\textbf{(A) } 3 \qquad\textbf{(B) } 12-4\sqrt5 \qquad\textbf{(C) } \dfrac{5+2\sqrt5}{3} \qquad\textbf{(D) } 1+\sqrt5 \qquad\textbf{(E) } \dfrac{11+11\sqrt5}{10}$ | 2015 AMC 10B Problem 22 | Triangle $AFG$ is isosceles, so $AG=AF=1$. $FJ = FG$ since $\triangle FGJ$ is also isosceles. Using the symmetry of pentagon $FGHIJ$, notice that $\triangle JHG \cong \triangle AFG$. Therefore, $JH=AF=1$.
Since $\triangle AJH \sim \triangle AFG$,
\[\frac{JH}{AF+FJ}=\frac{FG}{FA}\]
\[\frac{1}{1+FG} = \frac{FG}1\]
\[1 ... | // Block 1
pair A=(cos(pi/5)-sin(pi/10),cos(pi/10)+sin(pi/5)), B=(2*cos(pi/5)-sin(pi/10),cos(pi/10)), C=(1,0), D=(0,0), E1=(-sin(pi/10),cos(pi/10));
//(0,0) is a convenient point
//E1 to prevent conflict with direction E(ast)
pair F=intersectionpoints(D--A,E1--B)[0], G=intersectionpoints(A--C,E1--B)[0], H=intersectionp... | [] |
771 | In $\triangle ABC$, $\angle C = 90^\circ$ and $AB = 12$. Squares $ABXY$ and $CBWZ$ are constructed outside of the triangle. The points $X$, $Y$, $Z$, and $W$ lie on a circle. What is the perimeter of the triangle?
$\textbf{(A)}\; 12+9\sqrt{3} \qquad\textbf{(B)}\; 18+6\sqrt{3} \qquad\textbf{(C)}\; 12+12\sqrt{2} \qquad\... | 2015 AMC 12B Problem 19 | First, we should find the center and radius of this circle. We can find the center by drawing the perpendicular bisectors of $WZ$ and $XY$ and finding their intersection point. This point happens to be the midpoint of $AB$, the hypotenuse. Let this point be $M$. To find the radius, determine $MY$, where $MY^{2} = MA^2 ... | // Block 1
pair A,B,C,M,E,W,Z,X,Y;
A=(2,0);
B=(0,2);
C=(0,0);
M=(A+B)/2;
W=(-2,2);
Z=(-2,-0);
X=(2,4);
Y=(4,2);
E=(W+Z)/2;
draw(A--B--C--cycle);
draw(W--B--C--Z--cycle);
draw(A--B--X--Y--cycle);
dot(M);
dot(E);
label("W",W,NW);
label("Z",Z,SW);
label("C",C,S);
label("A",A,S);
label("B",B,N);
label("X",X,NE);
label("Y",... | [] |
772 | A bee starts flying from point $P_0$. She flies $1$ inch due east to point $P_1$. For $j \ge 1$, once the bee reaches point $P_j$, she turns $30^{\circ}$ counterclockwise and then flies $j+1$ inches straight to point $P_{j+1}$. When the bee reaches $P_{2015}$ she is exactly $a \sqrt{b} + c \sqrt{d}$ inches away from $P... | 2015 AMC 12B Problem 25 | Suppose that the bee makes a move of distance $i$. After $6$ turns it will be facing the opposite direction and move $i+6$ units. Combining these opposite movements gives a total movement of $-6$ units in the original direction. This means that every $12$ moves, the bee will move $-6$ units in each direction of $0^\cir... | // Block 1
draw((0,0)--(1,0)--(1+sqrt(3)/2, 1/2)--(3/2+sqrt(3)/2, 1/2+sqrt(3)/2)--(3/2+sqrt(3)/2, 3/2+sqrt(3)/2)--(1+sqrt(3)/2, 3/2+sqrt(3))--(1, 2+sqrt(3)), EndArrow);
draw((1,0)--(3/2+sqrt(3)/2, 0), dashed);
draw((1,2+sqrt(3))--(3/2+sqrt(3)/2, 2+sqrt(3)), dashed);
draw((3/2+sqrt(3)/2, 2+sqrt(3))--(3/2+sqrt(3)/2, 0), ... | [] |
773 | Find the area of the shaded region.
$\textbf{(A)}\ 4\dfrac{3}{5} \qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 5\dfrac{1}{4} \qquad \textbf{(D)}\ 6\dfrac{1}{2} \qquad \textbf{(E)}\ 8$ | 2016 AMC 10A Problem 11 | The bases of these triangles are all $1$, and by symmetry, their heights are $4$, $\frac{5}{2}$, $4$, and $\frac{5}{2}$. Thus, their areas are $2$, $\frac{5}{4}$, $2$, and $\frac{5}{4}$, which add to the area of the shaded region, which is $\boxed{6\frac{1}{2}}$. | // Block 1
size(6cm);
defaultpen(fontsize(9pt));
draw((0,0)--(8,0)--(8,5)--(0,5)--cycle);
filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8));
label("$1$",(1/2,5),dir(90));
label("$7$",(9/2,5),dir(90));
label("$1$",(8,1/2),dir(0));
label("$1$",(15/2,0),dir(270));
label("$7$",(7/2,0),dir(270));
labe... | [] |
773 | Find the area of the shaded region.
$\textbf{(A)}\ 4\dfrac{3}{5} \qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 5\dfrac{1}{4} \qquad \textbf{(D)}\ 6\dfrac{1}{2} \qquad \textbf{(E)}\ 8$ | 2016 AMC 10A Problem 11 | Find the area of the unshaded area by calculating the area of the triangles and rectangles outside of the shaded region. We can do this by splitting up the unshaded areas into various triangles and rectangles as shown.
Notice that the two added lines bisect each of the $4$ sides of the large rectangle.
Subtracting ... | // Block 1
size(6cm);
defaultpen(fontsize(9pt));
draw((0,0)--(8,0)--(8,5)--(0,5)--cycle);
filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8));
label("$1$",(1/2,5),dir(90));
label("$4$",(6,5),dir(90));
label("$3$",(5/2,5),dir(90));
label("$1$",(8,1/2),dir(0));
label("$5/2$",(8,15/4),dir(0));
label("$3/... | [] |
773 | Find the area of the shaded region.
$\textbf{(A)}\ 4\dfrac{3}{5} \qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 5\dfrac{1}{4} \qquad \textbf{(D)}\ 6\dfrac{1}{2} \qquad \textbf{(E)}\ 8$ | 2016 AMC 10A Problem 11 | Split the region into four parts by the diagonal from the top left to the bottom right. Slide the top and bottom pieces next to each other to form a parallelogram with base $1$ and height $\frac{5}{2}$, and slide the left and right pieces next to each other to form a parallelogram with base $1$ and height $\frac{8}{2}$... | // Block 1
size(6cm);
defaultpen(fontsize(9pt));
draw((0,0)--(8,0)--(8,5)--(0,5)--cycle);
filldraw((7,0)--(8,0)--(0,5)--(1,5)--cycle,gray(0.7));
filldraw((8,0)--(8,1)--(0,4)--(0,5)--cycle,gray(0.9));
label("$1$",(1/2,5),dir(90));
label("$7$",(9/2,5),dir(90));
label("$1$",(8,1/2),dir(0));
label("$1$",(15/2,0),dir(27... | [] |
773 | Find the area of the shaded region.
$\textbf{(A)}\ 4\dfrac{3}{5} \qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 5\dfrac{1}{4} \qquad \textbf{(D)}\ 6\dfrac{1}{2} \qquad \textbf{(E)}\ 8$ | 2016 AMC 10A Problem 11 | We can subtract the 4 triangular areas from the overall rectangle. It can be noticed that two triangles have area $\frac{6 \cdot \frac{5}{2}}{2} = \frac{15}{2}$, and the other two triangles have area $\frac{3 \cdot \frac{8}{2}}{2} = 6$ $\Rightarrow$ the shaded area is $(8 \cdot 5) - 2(\frac{15}{2}) - 2(6) = 40 - 15 - 1... | // Block 1
size(6cm);
defaultpen(fontsize(9pt));
draw((0,0)--(8,0)--(8,5)--(0,5)--cycle);
filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8));
filldraw((1,0)--(0,0)--(0,1)--(8,4)--(8,5)--(7,5)--cycle,gray(0.8));
label("$1$",(1/2,5),dir(90));
label("$6$",(4,5),dir(90));
label("$1$",(15/2,5),dir(90));
lab... | [] |
774 | In rectangle $ABCD,$ $AB=6$ and $BC=3$. Point $E$ between $B$ and $C$, and point $F$ between $E$ and $C$ are such that $BE=EF=FC$. Segments $\overline{AE}$ and $\overline{AF}$ intersect $\overline{BD}$ at $P$ and $Q$, respectively. The ratio $BP:PQ:QD$ can be written as $r:s:t$ where the greatest common factor of $r,s,... | 2016 AMC 10A Problem 19 | Use similar triangles. Our goal is to put the ratio in terms of ${BD}$. Since $\triangle APD \sim \triangle EPB,$ $\frac{DP}{PB}=\frac{AD}{BE}=3.$ Therefore, $PB=\frac{BD}{4}$. Similarly, $\frac{DQ}{QB}=\frac{3}{2}$. This means that ${DQ}=\frac{3\cdot BD}{5}$. Therefore, $r:s:t=\frac{1}{4}:\frac{2}{5}-\frac{1}{4}:\frac... | // Block 1
size(6cm);
pair D=(0,0), C=(6,0), B=(6,3), A=(0,3);
draw(A--B--C--D--cycle);
draw(B--D);
draw(A--(6,2));
draw(A--(6,1));
label("$A$", A, dir(135));
label("$B$", B, dir(45));
label("$C$", C, dir(-45));
label("$D$", D, dir(-135));
label("$Q$", extension(A,(6,1),B,D),dir(-90));
label("$P$", extension(A,(6,2),B,... | [] |
774 | In rectangle $ABCD,$ $AB=6$ and $BC=3$. Point $E$ between $B$ and $C$, and point $F$ between $E$ and $C$ are such that $BE=EF=FC$. Segments $\overline{AE}$ and $\overline{AF}$ intersect $\overline{BD}$ at $P$ and $Q$, respectively. The ratio $BP:PQ:QD$ can be written as $r:s:t$ where the greatest common factor of $r,s,... | 2016 AMC 10A Problem 19 | This problem breaks down into finding $QP:PB$ and $DQ:QB$. We can find the first using mass points, and the second using similar triangles.
Draw point $G$ on $DB$ such that $FG\parallel CD$. Then, by similar triangles $FG=BF\cdot 2=4$. Again, by similar triangles $AQB$ and $FQG$, $AQ:FQ=AB:FG=6:4=3:2$. Now we begin Ma... | // Block 1
import geometry;
size(9cm);
pair D=(0,0), C=(6,0), B=(6,3), A=(0,3), G=(2,1), H=(9,0);
draw(A--B--C--D--cycle);
draw(B--D);
draw(A--(6,2));
draw(A--(6,1));
draw(A--H);
draw((6,1)--G);
draw(D--H);
label("$A$", A, dir(135));
label("$B$", B, dir(45));
label("$C$", C, dir(-45));
label("$D$", D, dir(-135));
p... | [] |
775 | Circles with centers $P, Q$ and $R$, having radii $1, 2$ and $3$, respectively, lie on the same side of line $l$ and are tangent to $l$ at $P', Q'$ and $R'$, respectively, with $Q'$ between $P'$ and $R'$. The circle with center $Q$ is externally tangent to each of the other two circles. What is the area of triangle $PQ... | 2016 AMC 10A Problem 21 | Notice that we can find $[P'PQRR']$ in two different ways: $[P'PQQ']+[Q'QRR']$ and $[PQR]+[P'PRR'].$ Since we want $[PQR],$ we use the latter method, so we have $[P'PQQ']+[Q'QRR']=[PQR]+[P'PRR'].$
$\break$
$P'Q'=\sqrt{PQ^2-(QQ'-PP')^2}=\sqrt{3^2-1^2}=\sqrt{8}=2\sqrt{2}$. Additionally, $Q'R'=\sqrt{QR^2-(RR'-QQ')^2}=\sq... | size(250); defaultpen(linewidth(0.4)); //Variable Declarations pair P,Q,R,Pp,Qp,Rp; pair A,B; //Variable Definitions A=(-5, 0); B=(8, 0); P=(-2.828,1); Q=(0,2); R=(4.899,3); Pp=foot(P,A,B); Qp=foot(Q,A,B); Rp=foot(R,A,B); path PQR = P--Q--R--cycle; //Initial Diagram dot(P); dot(Q); dot(R); dot(Pp); dot(Qp); dot(Rp); d... | [] |
775 | Circles with centers $P, Q$ and $R$, having radii $1, 2$ and $3$, respectively, lie on the same side of line $l$ and are tangent to $l$ at $P', Q'$ and $R'$, respectively, with $Q'$ between $P'$ and $R'$. The circle with center $Q$ is externally tangent to each of the other two circles. What is the area of triangle $PQ... | 2016 AMC 10A Problem 21 | The above diagram can be achieved relatively simply using basic knowledge of the Pythagorean theorem and the fact that the radius from the center to the point of tangency is perpendicular to the tangent line. From there, observe that $[PQRY]$ can be calculated in two ways: $[\triangle PQX] + [QZYX] + [\triangle QRZ]$ a... | // Initial Pen Sizing size(250); defaultpen(linewidth(0.4)); defaultpen(fontsize(10pt)); // Variable Declarations pair P,Q,R,Pp,Qp,Rp,X,Y,Z,A,B; // Variable Definitions A=(-5, 0); B=(8, 0); P=(-2.828,1); Q=(0,2); R=(4.899,3); X=(0,1); Y=(4.899,1); Z=(4.899,2); Pp=foot(P,A,B); Qp=foot(Q,A,B); Rp=foot(R,A,B); path PQR ... | [] |
776 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | 2016 AMC 10A Problem 24 | Let $AD$ intersect $OB$ at $E$ and $OC$ at $F.$
$\overarc{AB}= \overarc{BC}= \overarc{CD}=\theta$
$\angle{BAD}=\frac{1}{2} \cdot \overarc{BCD}=\theta=\angle{AOB}$
From there, $\triangle{OAB} \sim \triangle{ABE}$, thus:
$\frac{OA}{AB} = \frac{AB}{BE} = \frac{OB}{AE}$
$OA = OB$ because they are both radii of $\o... | // Block 1
size(250);
defaultpen(linewidth(0.4));
//Variable Declarations
real RADIUS;
pair A, B, C, D, E, F, O;
RADIUS=3;
//Variable Definitions
A=RADIUS*dir(148.414);
B=RADIUS*dir(109.471);
C=RADIUS*dir(70.529);
D=RADIUS*dir(31.586);
O=(0,0);
//Path Definitions
path quad= A -- B -- C -- D -- cycle;
//Initial Diagr... | [] |
776 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | 2016 AMC 10A Problem 24 | To save us from getting big numbers with lots of zeros behind them, let's divide all side lengths by $200$ for now, then multiply it back at the end of our solution.
Construct quadrilateral $ABCD$ on the circle with $AD$ being the missing side (Notice that since the side length is less than the radius, it will be ... | // Block 1
size(250);
defaultpen(linewidth(0.4));
//Variable Declarations
real RADIUS;
pair A, B, C, D, E, O;
RADIUS=3;
//Variable Definitions
A=RADIUS*dir(148.414);
B=RADIUS*dir(109.471);
C=RADIUS*dir(70.529);
D=RADIUS*dir(31.586);
E=extension(B,D,O,C);
O=(0,0);
//Path Definitions
path quad= A -- B -- C -- D -- cycl... | [] |
776 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | 2016 AMC 10A Problem 24 | size(250);
defaultpen(linewidth(0.4));
//Variable Declarations
real RADIUS;
pair A, B, C, D, O;
RADIUS=3;
//Variable Definitions
A=RADIUS*dir(148.414);
B=RADIUS*dir(109.471);
C=RADIUS*dir(70.529);
D=RADIUS*dir(31.586);
P=RADIUS*dir(270);
O=(0,0);
//Path Definitions
path quad= A -- B -- C -- D -- cycle;
//Initial Dia... | // Block 1
size(250);
defaultpen(linewidth(0.4));
//Variable Declarations
real RADIUS;
pair A, B, C, D, O;
RADIUS=3;
//Variable Definitions
A=RADIUS*dir(148.414);
B=RADIUS*dir(109.471);
C=RADIUS*dir(70.529);
D=RADIUS*dir(31.586);
P=RADIUS*dir(270);
O=(0,0);
//Path Definitions
path quad= A -- B -- C -- D -- cycle;
//... | [] |
776 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | 2016 AMC 10A Problem 24 | Construct quadrilateral $ABCD$ on the circle with $AD$ being the missing side (Notice that since the side length is less than the radius, it will be very small on the top of the circle). Now, draw the radii from center $O$ to $A,B,C,$ and $D$. Apply the law of cosines on $\Delta BOC$; let $\theta = \angle BOC$. We get ... | // Block 1
size(250);
defaultpen(linewidth(0.4));
//Variable Declarations
real RADIUS;
pair A, B, C, D, O;
RADIUS=3;
//Variable Definitions
A=RADIUS*dir(148.414);
B=RADIUS*dir(109.471);
C=RADIUS*dir(70.529);
D=RADIUS*dir(31.586);
O=(0,0);
//Path Definitions
path quad= A -- B -- C -- D -- cycle;
//Initial Diagram
dra... | [] |
776 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | 2016 AMC 10A Problem 24 | Construct quadrilateral $ABCD$ on the circle $O$ with $AD$ being the desired side. Then, drop perpendiculars from $A$ and $D$ to the extended line of $\overline{BC}$ and let these points be $E$ and $F$, respectively. Also, let $\theta = \angle BOC$. From the Law of Cosines on $\triangle BOC$, we have $\cos \theta = \fr... | // Block 1
size(250);
defaultpen(linewidth(0.4));
//Variable Declarations
real RADIUS;
pair A, B, C, D, E, F, O;
RADIUS=3;
//Variable Definitions
A=RADIUS*dir(148.414);
B=RADIUS*dir(109.471);
C=RADIUS*dir(70.529);
D=RADIUS*dir(31.586);
E=foot(A,B,C);
F=foot(D,B,C);
O=(0,0);
//Path Definitions
path quad= A -- B -- C -... | [] |
776 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | 2016 AMC 10A Problem 24 | Let $s = 200$. Let $O$ be the center of the circle. Then $AC$ is twice the altitude of $\triangle OBC$ to $\overline{OB}$. Since $\triangle OBC$ is isosceles we can compute its area to be $\frac{s^2 \sqrt{7}}{4}$, hence $CA = 2 \cdot \tfrac{2 \cdot s^2\sqrt7/4}{s\sqrt2} = s\sqrt{\frac{7}{2}}$.
Now by Ptolemy's Theo... | // Block 1
pathpen = black; pointpen = black;
size(6cm);
draw(unitcircle);
pair A = D("A", dir(50), dir(50));
pair B = D("B", dir(90), dir(90));
pair C = D("C", dir(130), dir(130));
pair D = D("D", dir(170), dir(170));
pair O = D("O", (0,0), dir(-90));
draw(A--C, red);
draw(B--D, blue+dashed);
draw(A--B--C--D--cycle);
... | [] |
776 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | 2016 AMC 10A Problem 24 | Label the points as shown, and let $\angle{EOF} = \theta$. Since $\overline{OB} = \overline{OC}$, and $\triangle{OFE} \sim \triangle{OCB}$, we get that $\angle{EFO} = 90-\frac{\theta}{2}$. We assign $\alpha$ to $90-\frac{\theta}{2}$ for simplicity.
From here, by vertical angles $\angle{CFD} = \alpha$. Also, since $\tr... | // Block 1
size(250);
defaultpen(linewidth(0.4));
//Variable Declarations, L is used to write alpha= statement
real RADIUS;
pair A, B, C, D, E, F, O, L;
RADIUS=3;
//Variable Definitions
A=RADIUS*dir(148.414);
B=RADIUS*dir(109.471);
C=RADIUS*dir(70.529);
D=RADIUS*dir(31.586);
E=extension(A,D,O,B);
F=extension(A,D,O,C);... | [] |
776 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | 2016 AMC 10A Problem 24 | Claim: $[ABCD]$ is an isosceles trapezoid.
Proof: Notice that $[ABCD]$ is cyclic, triangle $BOC$ is isosceles, and triangle $AOB$ is congruent to $DOC$ by SSS congruence. Therefore, $\angle BAD = 180 - \angle BCD = 180-(\angle BCO + \angle DCO)=180-(\angle CBO+\angle ABO) = 180 - \angle ABC = \angle CDA$. Hence, $[ABC... | // Block 1
size(250);
defaultpen(linewidth(0.4));
//Variable Declarations
real RADIUS;
pair A, B, C, D, O;
RADIUS=3;
//Variable Definitions
A=RADIUS*dir(148.414);
B=RADIUS*dir(109.471);
C=RADIUS*dir(70.529);
D=RADIUS*dir(31.586);
O=(0,0);
//Path Definitions
path quad= A -- B -- C -- D -- cycle;
//Initial Diagram
dra... | [] |
776 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | 2016 AMC 10A Problem 24 | Since $AB$, $BC$, and $CD$ are congruent chords, the triangles formed by connecting these sides to the circumcenter $O$ (i.e., $\triangle AOB$, $\triangle BOC$, $\triangle COD$) are congruent isosceles triangles.
Let $\alpha = \angle OCB$. In $\triangle BOC$, the side length $BC=200$ and $OC = OB = 200\sqrt{2}$. By dro... | // Block 1
size(250);
defaultpen(linewidth(0.4));
//Variable Declarations
real RADIUS;
pair A, B, C, D, E, F, O;
RADIUS=3;
//Variable Definitions
A=RADIUS*dir(148.414);
B=RADIUS*dir(109.471);
C=RADIUS*dir(70.529);
D=RADIUS*dir(31.586);
O=(0,0);
//Path Definitions
path quad= A -- B -- C -- D -- cycle;
//Initial Diagr... | [] |
777 | How many ordered triples $(x,y,z)$ of positive integers satisfy $\text{lcm}(x,y) = 72, \text{lcm}(x,z) = 600,$ and $\text{lcm}(y,z)=900$?
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 64$ | 2016 AMC 10A Problem 25 | As said in previous solutions, start by factoring $72, 600,$ and $900$. The prime factorizations are as follows: \[72=2^3\cdot 3^2,\] \[600=2^3\cdot 3\cdot 5^2,\] \[\text{and } 900=2^2\cdot 3^2\cdot 5^2\]
To organize $x,y, \text{ and } z$ and their respective LCMs in a simpler way, we can draw a triangle as follows suc... | // Block 1
//Variable Declarations
defaultpen(0.45);
size(200pt);
fontsize(15pt);
pair X, Y, Z;
real R;
path tri;
//Variable Definitions
R = 1;
X = R*dir(90);
Y = R*dir(210);
Z = R*dir(-30);
tri = X--Y--Z--cycle;
//Diagram
draw(tri);
label("$x$",X,N);
label("$y$",Y,SW);
label("$z$",Z,SE);
label("$2^33^25^0$",X--Y,2W)... | [] |
778 | Find the area of the shaded region.
$\textbf{(A)}\ 4\dfrac{3}{5} \qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 5\dfrac{1}{4} \qquad \textbf{(D)}\ 6\dfrac{1}{2} \qquad \textbf{(E)}\ 8$ | 2016 AMC 12A Problem 8 | The bases of these triangles are all $1$, and by symmetry, their heights are $4$, $\frac{5}{2}$, $4$, and $\frac{5}{2}$. Thus, their areas are $2$, $\frac{5}{4}$, $2$, and $\frac{5}{4}$, which add to the area of the shaded region, which is $\boxed{6\frac{1}{2}}$. | size(6cm); defaultpen(fontsize(9pt)); draw((0,0)--(8,0)--(8,5)--(0,5)--cycle); filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8)); label("$1$",(1/2,5),dir(90)); label("$7$",(9/2,5),dir(90)); label("$1$",(8,1/2),dir(0)); label("$1$",(15/2,0),dir(270)); label("$7$",(7/2,0),dir(270)); label("$1$",(0,... | [] |
778 | Find the area of the shaded region.
$\textbf{(A)}\ 4\dfrac{3}{5} \qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 5\dfrac{1}{4} \qquad \textbf{(D)}\ 6\dfrac{1}{2} \qquad \textbf{(E)}\ 8$ | 2016 AMC 12A Problem 8 | Find the area of the unshaded area by calculating the area of the triangles and rectangles outside of the shaded region. We can do this by splitting up the unshaded areas into various triangles and rectangles as shown.
Notice that the two added lines bisect each of the $4$ sides of the large rectangle.
Subtracting ... | size(6cm); defaultpen(fontsize(9pt)); draw((0,0)--(8,0)--(8,5)--(0,5)--cycle); filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8)); label("$1$",(1/2,5),dir(90)); label("$4$",(6,5),dir(90)); label("$3$",(5/2,5),dir(90)); label("$1$",(8,1/2),dir(0)); label("$5/2$",(8,15/4),dir(0)); label("$3/2$",(8,7/4)... | [] |
778 | Find the area of the shaded region.
$\textbf{(A)}\ 4\dfrac{3}{5} \qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 5\dfrac{1}{4} \qquad \textbf{(D)}\ 6\dfrac{1}{2} \qquad \textbf{(E)}\ 8$ | 2016 AMC 12A Problem 8 | Split the region into four parts by the diagonal from the top left to the bottom right. Slide the top and bottom pieces next to each other to form a parallelogram with base $1$ and height $\frac{5}{2}$, and slide the left and right pieces next to each other to form a parallelogram with base $1$ and height $\frac{8}{2}$... | size(6cm); defaultpen(fontsize(9pt)); draw((0,0)--(8,0)--(8,5)--(0,5)--cycle); filldraw((7,0)--(8,0)--(0,5)--(1,5)--cycle,gray(0.7)); filldraw((8,0)--(8,1)--(0,4)--(0,5)--cycle,gray(0.9)); label("$1$",(1/2,5),dir(90)); label("$7$",(9/2,5),dir(90)); label("$1$",(8,1/2),dir(0)); label("$1$",(15/2,0),dir(270)); label(... | [] |
778 | Find the area of the shaded region.
$\textbf{(A)}\ 4\dfrac{3}{5} \qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 5\dfrac{1}{4} \qquad \textbf{(D)}\ 6\dfrac{1}{2} \qquad \textbf{(E)}\ 8$ | 2016 AMC 12A Problem 8 | We can subtract the 4 triangular areas from the overall rectangle. It can be noticed that two triangles have area $\frac{6 \cdot \frac{5}{2}}{2} = \frac{15}{2}$, and the other two triangles have area $\frac{3 \cdot \frac{8}{2}}{2} = 6$ $\Rightarrow$ the shaded area is $(8 \cdot 5) - 2(\frac{15}{2}) - 2(6) = 40 - 15 - 1... | size(6cm); defaultpen(fontsize(9pt)); draw((0,0)--(8,0)--(8,5)--(0,5)--cycle); filldraw((7,0)--(8,0)--(8,1)--(0,4)--(0,5)--(1,5)--cycle,gray(0.8)); filldraw((1,0)--(0,0)--(0,1)--(8,4)--(8,5)--(7,5)--cycle,gray(0.8)); label("$1$",(1/2,5),dir(90)); label("$6$",(4,5),dir(90)); label("$1$",(15/2,5),dir(90)); label("$1$",(8... | [] |
779 | Circles with centers $P, Q$ and $R$, having radii $1, 2$ and $3$, respectively, lie on the same side of line $l$ and are tangent to $l$ at $P', Q'$ and $R'$, respectively, with $Q'$ between $P'$ and $R'$. The circle with center $Q$ is externally tangent to each of the other two circles. What is the area of triangle $PQ... | 2016 AMC 12A Problem 15 | Notice that we can find $[P'PQRR']$ in two different ways: $[P'PQQ']+[Q'QRR']$ and $[PQR]+[P'PRR'].$ Since we want $[PQR],$ we use the latter method, so we have $[P'PQQ']+[Q'QRR']=[PQR]+[P'PRR'].$
$\break$
$P'Q'=\sqrt{PQ^2-(QQ'-PP')^2}=\sqrt{3^2-1^2}=\sqrt{8}=2\sqrt{2}$. Additionally, $Q'R'=\sqrt{QR^2-(RR'-QQ')^2}=\sq... | // Block 1
size(250);
defaultpen(linewidth(0.4));
//Variable Declarations
pair P,Q,R,Pp,Qp,Rp;
pair A,B;
//Variable Definitions
A=(-5, 0);
B=(8, 0);
P=(-2.828,1);
Q=(0,2);
R=(4.899,3);
Pp=foot(P,A,B);
Qp=foot(Q,A,B);
Rp=foot(R,A,B);
path PQR = P--Q--R--cycle;
//Initial Diagram
dot(P);
dot(Q);
dot(R);
dot(Pp);
dot(Qp);... | [] |
779 | Circles with centers $P, Q$ and $R$, having radii $1, 2$ and $3$, respectively, lie on the same side of line $l$ and are tangent to $l$ at $P', Q'$ and $R'$, respectively, with $Q'$ between $P'$ and $R'$. The circle with center $Q$ is externally tangent to each of the other two circles. What is the area of triangle $PQ... | 2016 AMC 12A Problem 15 | The above diagram can be achieved relatively simply using basic knowledge of the Pythagorean theorem and the fact that the radius from the center to the point of tangency is perpendicular to the tangent line. From there, observe that $[PQRY]$ can be calculated in two ways: $[\triangle PQX] + [QZYX] + [\triangle QRZ]$ a... | // Block 1
// Initial Pen Sizing
size(250);
defaultpen(linewidth(0.4));
defaultpen(fontsize(10pt));
// Variable Declarations
pair P,Q,R,Pp,Qp,Rp,X,Y,Z,A,B;
// Variable Definitions
A=(-5, 0);
B=(8, 0);
P=(-2.828,1);
Q=(0,2);
R=(4.899,3);
X=(0,1);
Y=(4.899,1);
Z=(4.899,2);
Pp=foot(P,A,B);
Qp=foot(Q,A,B);
Rp=foot(R,A,B)... | [] |
780 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | 2016 AMC 12A Problem 21 | Let $AD$ intersect $OB$ at $E$ and $OC$ at $F.$
$\overarc{AB}= \overarc{BC}= \overarc{CD}=\theta$
$\angle{BAD}=\frac{1}{2} \cdot \overarc{BCD}=\theta=\angle{AOB}$
From there, $\triangle{OAB} \sim \triangle{ABE}$, thus:
$\frac{OA}{AB} = \frac{AB}{BE} = \frac{OB}{AE}$
$OA = OB$ because they are both radii of $\o... | size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, F, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Cir... | [] |
780 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | 2016 AMC 12A Problem 21 | To save us from getting big numbers with lots of zeros behind them, let's divide all side lengths by $200$ for now, then multiply it back at the end of our solution.
Construct quadrilateral $ABCD$ on the circle with $AD$ being the missing side (Notice that since the side length is less than the radius, it will be ... | size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); E=extension(B,D,O,C); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initi... | [] |
780 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | 2016 AMC 12A Problem 21 | size(250);
defaultpen(linewidth(0.4));
//Variable Declarations
real RADIUS;
pair A, B, C, D, O;
RADIUS=3;
//Variable Definitions
A=RADIUS*dir(148.414);
B=RADIUS*dir(109.471);
C=RADIUS*dir(70.529);
D=RADIUS*dir(31.586);
P=RADIUS*dir(270);
O=(0,0);
//Path Definitions
path quad= A -- B -- C -- D -- cycle;
//Initial Dia... | size(250);
defaultpen(linewidth(0.4));
//Variable Declarations
real RADIUS;
pair A, B, C, D, O;
RADIUS=3;
//Variable Definitions
A=RADIUS*dir(148.414);
B=RADIUS*dir(109.471);
C=RADIUS*dir(70.529);
D=RADIUS*dir(31.586);
O=(0,0);
//Path Definitions
path quad= A -- B -- C -- D -- cycle;
//Initial Diagram
draw(Circle(O,... | [] |
780 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | 2016 AMC 12A Problem 21 | Construct quadrilateral $ABCD$ on the circle with $AD$ being the missing side (Notice that since the side length is less than the radius, it will be very small on the top of the circle). Now, draw the radii from center $O$ to $A,B,C,$ and $D$. Apply the law of cosines on $\Delta BOC$; let $\theta = \angle BOC$. We get ... | size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O,... | [] |
780 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | 2016 AMC 12A Problem 21 | Construct quadrilateral $ABCD$ on the circle $O$ with $AD$ being the desired side. Then, drop perpendiculars from $A$ and $D$ to the extended line of $\overline{BC}$ and let these points be $E$ and $F$, respectively. Also, let $\theta = \angle BOC$. From the Law of Cosines on $\triangle BOC$, we have $\cos \theta = \fr... | size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, F, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); E=foot(A,B,C); F=foot(D,B,C); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycl... | [] |
780 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | 2016 AMC 12A Problem 21 | Let $s = 200$. Let $O$ be the center of the circle. Then $AC$ is twice the altitude of $\triangle OBC$ to $\overline{OB}$. Since $\triangle OBC$ is isosceles we can compute its area to be $\frac{s^2 \sqrt{7}}{4}$, hence $CA = 2 \cdot \tfrac{2 \cdot s^2\sqrt7/4}{s\sqrt2} = s\sqrt{\frac{7}{2}}$.
Now by Ptolemy's Theo... | pathpen = black; pointpen = black; size(6cm); draw(unitcircle); pair A = D("A", dir(50), dir(50)); pair B = D("B", dir(90), dir(90)); pair C = D("C", dir(130), dir(130)); pair D = D("D", dir(170), dir(170)); pair O = D("O", (0,0), dir(-90)); draw(A--C, red); draw(B--D, blue+dashed); draw(A--B--C--D--cycle); draw(A--O--... | [] |
780 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | 2016 AMC 12A Problem 21 | Label the points as shown, and let $\angle{EOF} = \theta$. Since $\overline{OB} = \overline{OC}$, and $\triangle{OFE} \sim \triangle{OCB}$, we get that $\angle{EFO} = 90-\frac{\theta}{2}$. We assign $\alpha$ to $90-\frac{\theta}{2}$ for simplicity.
From here, by vertical angles $\angle{CFD} = \alpha$. Also, since $\tr... | size(250); defaultpen(linewidth(0.4)); //Variable Declarations, L is used to write alpha= statement real RADIUS; pair A, B, C, D, E, F, O, L; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); E=extension(A,D,O,B); F=extension(A,D,O,C); L=midpoint... | [] |
780 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | 2016 AMC 12A Problem 21 | Claim: $[ABCD]$ is an isosceles trapezoid.
Proof: Notice that $[ABCD]$ is cyclic, triangle $BOC$ is isosceles, and triangle $AOB$ is congruent to $DOC$ by SSS congruence. Therefore, $\angle BAD = 180 - \angle BCD = 180-(\angle BCO + \angle DCO)=180-(\angle CBO+\angle ABO) = 180 - \angle ABC = \angle CDA$. Hence, $[ABC... | size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Circle(O,... | [] |
780 | A quadrilateral is inscribed in a circle of radius $200\sqrt{2}$. Three of the sides of this quadrilateral have length $200$. What is the length of the fourth side?
$\textbf{(A) }200\qquad \textbf{(B) }200\sqrt{2}\qquad\textbf{(C) }200\sqrt{3}\qquad\textbf{(D) }300\sqrt{2}\qquad\textbf{(E) } 500$ | 2016 AMC 12A Problem 21 | Since $AB$, $BC$, and $CD$ are congruent chords, the triangles formed by connecting these sides to the circumcenter $O$ (i.e., $\triangle AOB$, $\triangle BOC$, $\triangle COD$) are congruent isosceles triangles.
Let $\alpha = \angle OCB$. In $\triangle BOC$, the side length $BC=200$ and $OC = OB = 200\sqrt{2}$. By dro... | size(250); defaultpen(linewidth(0.4)); //Variable Declarations real RADIUS; pair A, B, C, D, E, F, O; RADIUS=3; //Variable Definitions A=RADIUS*dir(148.414); B=RADIUS*dir(109.471); C=RADIUS*dir(70.529); D=RADIUS*dir(31.586); O=(0,0); //Path Definitions path quad= A -- B -- C -- D -- cycle; //Initial Diagram draw(Cir... | [] |
781 | How many ordered triples $(x,y,z)$ of positive integers satisfy $\text{lcm}(x,y) = 72, \text{lcm}(x,z) = 600,$ and $\text{lcm}(y,z)=900$?
$\textbf{(A)}\ 15\qquad\textbf{(B)}\ 16\qquad\textbf{(C)}\ 24\qquad\textbf{(D)}\ 27\qquad\textbf{(E)}\ 64$ | 2016 AMC 12A Problem 22 | As said in previous solutions, start by factoring $72, 600,$ and $900$. The prime factorizations are as follows: \[72=2^3\cdot 3^2,\] \[600=2^3\cdot 3\cdot 5^2,\] \[\text{and } 900=2^2\cdot 3^2\cdot 5^2\]
To organize $x,y, \text{ and } z$ and their respective LCMs in a simpler way, we can draw a triangle as follows suc... | // Block 1
//Variable Declarations defaultpen(0.45); size(200pt); fontsize(15pt); pair X, Y, Z; real R; path tri; //Variable Definitions R = 1; X = R*dir(90); Y = R*dir(210); Z = R*dir(-30); tri = X--Y--Z--cycle; //Diagram draw(tri); label("$x$",X,N); label("$y$",Y,SW); label("$z$",Z,SE); label("$2^33^25^0$",X--Y,2W)... | [] |
782 | Three numbers in the interval $\left[0,1\right]$ are chosen independently and at random. What is the probability that the chosen numbers are the side lengths of a triangle with positive area?
$\textbf{(A) }\frac16\qquad\textbf{(B) }\frac13\qquad\textbf{(C) }\frac12\qquad\textbf{(D) }\frac23\qquad\textbf{(E) }\frac56$ | 2016 AMC 12A Problem 23 | Because we can let the the sides of the triangle be any variable we want, to make it easier for us when solving, let’s let the side lengths be $x,y,$ and $a$. WLOG assume $a$ is the largest. Then, $x+y>a$, meaning the solution is $\boxed{\textbf{(C)}\;1/2}$, as shown in the graph below. | // Block 1
pair A = (0,0);
pair B = (1,0);
pair C = (1,1);
pair D = (0,1);
pair E = (0,0);
draw(A--B--C--D--cycle);
draw(B--D,dashed);
fill(B--D--C--cycle,gray);
label("$0$",A,SW);
label("$a$",B,S);
label("$a$",D,W);
label("$y$",(0,.5),W);
label("$x$",(.5,0),S);
label("$x+y>a$",(5/7,5/7));
// Block 2
pair A = (0,0); ... | [] |
783 | All three vertices of $\bigtriangleup ABC$ lie on the parabola defined by $y=x^2$, with $A$ at the origin and $\overline{BC}$ parallel to the $x$-axis. The area of the triangle is $64$. What is the length of $BC$?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 16$ | 2016 AMC 10B Problem 9 | The area of the triangle is $\frac{(2r)(r^2)}{2} = r^3$, so $r^3=64\implies r=4$, giving a total distance across the top of $8$, which is answer $\textbf{(C)}$. | // Block 1
import graph;size(7cm,IgnoreAspect);
real f(real x) {return x*x;}
draw((0,0)--(4,16)--(-4,16)--cycle,blue);
draw(graph(f,-5,5,operator ..),gray);
xaxis("$x$");yaxis("$y$",-1);
label("$y=x^2$",(4.5,20.25),E);
draw((4.2,0)--(4.2,16),Arrows);
label("$r^2$",(4.2,0)--(4.2,16),E);
draw((0,17)--(4,17),Arrows);
labe... | [] |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.