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 |
|---|---|---|---|---|---|
842 | The figure below shows $13$ circles of radius $1$ within a larger circle. All the intersections occur at points of tangency. What is the area of the region, shaded in the figure, inside the larger circle but outside all the circles of radius $1 ?$
$\textbf{(A) } 4 \pi \sqrt{3} \qquad\textbf{(B) } 7 \pi \qquad\textbf... | 2019 AMC 10A Problem 16 | In the diagram above, notice that triangle $OAB$ and triangle $ABC$ are congruent and equilateral with side length $2$. We can see the radius of the larger circle is $2\overline{OH} + 1$. Using $30^{\circ}-60^{\circ}-90^{\circ}$ triangles, we know $\overline{OH} = \sqrt{3}$. Therefore, the radius of the larger circle i... | // Block 1
unitsize(20);filldraw(circle((0,0),2*sqrt(3)+1),rgb(0.5,0.5,0.5));filldraw(circle((-2,0),1),white);filldraw(circle((0,0),1),white);filldraw(circle((2,0),1),white);filldraw(circle((1,sqrt(3)),1),white);filldraw(circle((3,sqrt(3)),1),white);filldraw(circle((-1,sqrt(3)),1),white);filldraw(circle((-3,sqrt(3)),1)... | [] |
842 | The figure below shows $13$ circles of radius $1$ within a larger circle. All the intersections occur at points of tangency. What is the area of the region, shaded in the figure, inside the larger circle but outside all the circles of radius $1 ?$
$\textbf{(A) } 4 \pi \sqrt{3} \qquad\textbf{(B) } 7 \pi \qquad\textbf... | 2019 AMC 10A Problem 16 | In the diagram above, $AB=4$ and $BC=2$, so $AC=\sqrt{4^2-2^2}=2\sqrt{3}$. The larger circle's radius is $AC+1=2\sqrt{3}+1$, so the larger circle's area is $\pi\left(2\sqrt{3}+1\right)^2=\pi\left(13+4\sqrt{3}\right)=13\pi+4\pi\sqrt{3}$. Now, subtracting the combined area of the smaller circles gives $13\pi+4\pi\sqrt{3}... | // Block 1
unitsize(20);filldraw(circle((0,0),2*sqrt(3)+1),rgb(0.5,0.5,0.5));filldraw(circle((-2,0),1),white);filldraw(circle((0,0),1),white);filldraw(circle((2,0),1),white);filldraw(circle((1,sqrt(3)),1),white);filldraw(circle((3,sqrt(3)),1),white);filldraw(circle((-1,sqrt(3)),1),white);filldraw(circle((-3,sqrt(3)),1)... | [] |
843 | A sphere with center $O$ has radius $6$. A triangle with sides of length $15, 15,$ and $24$ is situated in space so that each of its sides is tangent to the sphere. What is the distance between $O$ and the plane determined by the triangle?
$\textbf{(A) }2\sqrt{3}\qquad \textbf{(B) }4\qquad \textbf{(C) }3\sqrt{2}\qquad... | 2019 AMC 10A Problem 21 | Solution 1
The triangle is placed on the sphere so that its three sides are tangent to the sphere. The cross-section of the sphere created by the plane of the triangle is also the incircle of the triangle. To find the inradius, use $\text{area} = \text{inradius} \cdot \text{semiperimeter}$. The area of the triangle can... | // Block 1
draw((0,0)--(12,9)--(24,0)--cycle); dot((0,0)); dot((12,9)); dot((24,0)); dot((12,0)); label("$A$",(0,0),SW); label("$B$",(12,9),N); label("$C$",(24,0),SE); label("$D$",(12,-1/2),S); label("$I$",(12,4),SE); draw((12,9)--(12,0), dashed); draw(circle((12,4),4)); draw((0,0)--(216/13,216/39)); dot((12,4)); labe... | [] |
844 | Real numbers between 0 and 1, inclusive, are chosen in the following manner. A fair coin is flipped. If it lands heads, then it is flipped again and the chosen number is 0 if the second flip is heads, and 1 if the second flip is tails. On the other hand, if the first coin flip is tails, then the number is chosen unifor... | 2019 AMC 10A Problem 22 | There are several cases depending on what the first coin flip is when determining $x$ and what the first coin flip is when determining $y$.
The four cases are:
Case 1: $x$ is either $0$ or $1$, and $y$ is either $0$ or $1$.
Case 2: $x$ is either $0$ or $1$, and $y$ is chosen from the interval $[0,1]$.
Case 3: $x$ i... | // Block 1
filldraw((0,0)--(0,1)--(1/2,1)--(0,1/2)--cycle,black+linewidth(1));
filldraw((0,0)--(1,0)--(1,1/2)--(1/2,0)--cycle,black+linewidth(1));
draw((0,0)--(0,1)--(1,1)--(1,0)--cycle);
// Block 2
filldraw((0,0)--(0,1)--(1/2,1)--(0,1/2)--cycle,black+linewidth(1)); filldraw((0,0)--(1,0)--(1,1/2)--(1/2,0)--cycle,black+... | [] |
844 | Real numbers between 0 and 1, inclusive, are chosen in the following manner. A fair coin is flipped. If it lands heads, then it is flipped again and the chosen number is 0 if the second flip is heads, and 1 if the second flip is tails. On the other hand, if the first coin flip is tails, then the number is chosen unifor... | 2019 AMC 10A Problem 22 | Notice the probability that $|x-y| > \tfrac{1}{2}$ is equal to $2$ times the probability that $x-y > \tfrac{1}{2}$.
For $x-y > \tfrac{1}{2}$ to be true; there are $4$ cases.
Case $1$: $x$ flips $T$ and $y$ flips $T$
When $x$ flips $T$ and $y$ flips $T$, both $x$ and $y$ are randomly chosen from the interval $[0,1]$.... | // Block 1
size(3cm,0);
filldraw((0,0)--(1,0)--(1,1/2)--(1/2,0)--cycle,black+linewidth(1));
draw((0,0)--(0,1)--(1,1)--(1,0)--cycle);
// Block 2
size(3cm,0); filldraw((0,0)--(1,0)--(1,1/2)--(1/2,0)--cycle,black+linewidth(1)); draw((0,0)--(0,1)--(1,1)--(1,0)--cycle); | [] |
845 | For a set of four distinct lines in a plane, there are exactly $N$ distinct points that lie on two or more of the lines. What is the sum of all possible values of $N$?
$\textbf{(A) } 14 \qquad \textbf{(B) } 16 \qquad \textbf{(C) } 18 \qquad \textbf{(D) } 19 \qquad \textbf{(E) } 21$ | 2019 AMC 12A Problem 8 | It is possible to obtain $0$, $1$, $3$, $4$, $5$, and $6$ points of intersection, as demonstrated in the following figures:
It is clear that the maximum number of possible intersections is ${4 \choose 2} = 6$, since each pair of lines can intersect at most once. In addition, by looking at the answer choices, we kn... | unitsize(2cm); real d = 2.5; draw((-1,.6)--(1,.6),Arrows); draw((-1,.2)--(1,.2),Arrows); draw((-1,-.2)--(1,-.2),Arrows); draw((-1,-.6)--(1,-.6),Arrows); draw((-1+d,0)--(1+d,0),Arrows); draw((0+d,1)--(0+d,-1),Arrows); draw(dir(45)+(d,0)--dir(45+180)+(d,0),Arrows); draw(dir(135)+(d,0)--dir(135+180)+(d,0),Arrows); dot((0... | [] |
846 | The figure below shows $13$ circles of radius $1$ within a larger circle. All the intersections occur at points of tangency. What is the area of the region, shaded in the figure, inside the larger circle but outside all the circles of radius $1 ?$
$\textbf{(A) } 4 \pi \sqrt{3} \qquad\textbf{(B) } 7 \pi \qquad\textbf... | 2019 AMC 12A Problem 10 | In the diagram above, notice that triangle $OAB$ and triangle $ABC$ are congruent and equilateral with side length $2$. We can see the radius of the larger circle is $2\overline{OH} + 1$. Using $30^{\circ}-60^{\circ}-90^{\circ}$ triangles, we know $\overline{OH} = \sqrt{3}$. Therefore, the radius of the larger circle i... | unitsize(20);filldraw(circle((0,0),2*sqrt(3)+1),rgb(0.5,0.5,0.5));filldraw(circle((-2,0),1),white);filldraw(circle((0,0),1),white);filldraw(circle((2,0),1),white);filldraw(circle((1,sqrt(3)),1),white);filldraw(circle((3,sqrt(3)),1),white);filldraw(circle((-1,sqrt(3)),1),white);filldraw(circle((-3,sqrt(3)),1),white);fil... | [] |
846 | The figure below shows $13$ circles of radius $1$ within a larger circle. All the intersections occur at points of tangency. What is the area of the region, shaded in the figure, inside the larger circle but outside all the circles of radius $1 ?$
$\textbf{(A) } 4 \pi \sqrt{3} \qquad\textbf{(B) } 7 \pi \qquad\textbf... | 2019 AMC 12A Problem 10 | In the diagram above, $AB=4$ and $BC=2$, so $AC=\sqrt{4^2-2^2}=2\sqrt{3}$. The larger circle's radius is $AC+1=2\sqrt{3}+1$, so the larger circle's area is $\pi\left(2\sqrt{3}+1\right)^2=\pi\left(13+4\sqrt{3}\right)=13\pi+4\pi\sqrt{3}$. Now, subtracting the combined area of the smaller circles gives $13\pi+4\pi\sqrt{3}... | unitsize(20);filldraw(circle((0,0),2*sqrt(3)+1),rgb(0.5,0.5,0.5));filldraw(circle((-2,0),1),white);filldraw(circle((0,0),1),white);filldraw(circle((2,0),1),white);filldraw(circle((1,sqrt(3)),1),white);filldraw(circle((3,sqrt(3)),1),white);filldraw(circle((-1,sqrt(3)),1),white);filldraw(circle((-3,sqrt(3)),1),white);fil... | [] |
847 | A sphere with center $O$ has radius $6$. A triangle with sides of length $15, 15,$ and $24$ is situated in space so that each of its sides is tangent to the sphere. What is the distance between $O$ and the plane determined by the triangle?
$\textbf{(A) }2\sqrt{3}\qquad \textbf{(B) }4\qquad \textbf{(C) }3\sqrt{2}\qquad... | 2019 AMC 12A Problem 18 | Solution 1
The triangle is placed on the sphere so that its three sides are tangent to the sphere. The cross-section of the sphere created by the plane of the triangle is also the incircle of the triangle. To find the inradius, use $\text{area} = \text{inradius} \cdot \text{semiperimeter}$. The area of the triangle can... | // Block 1
draw((0,0)--(12,9)--(24,0)--cycle); dot((0,0)); dot((12,9)); dot((24,0)); dot((12,0)); label("$A$",(0,0),SW); label("$B$",(12,9),N); label("$C$",(24,0),SE); label("$D$",(12,-1/2),S); label("$I$",(12,4),SE); draw((12,9)--(12,0), dashed); draw(circle((12,4),4)); draw((0,0)--(216/13,216/39)); dot((12,4)); labe... | [] |
848 | Real numbers between 0 and 1, inclusive, are chosen in the following manner. A fair coin is flipped. If it lands heads, then it is flipped again and the chosen number is 0 if the second flip is heads, and 1 if the second flip is tails. On the other hand, if the first coin flip is tails, then the number is chosen unifor... | 2019 AMC 12A Problem 20 | There are several cases depending on what the first coin flip is when determining $x$ and what the first coin flip is when determining $y$.
The four cases are:
Case 1: $x$ is either $0$ or $1$, and $y$ is either $0$ or $1$.
Case 2: $x$ is either $0$ or $1$, and $y$ is chosen from the interval $[0,1]$.
Case 3: $x$ i... | filldraw((0,0)--(0,1)--(1/2,1)--(0,1/2)--cycle,black+linewidth(1)); filldraw((0,0)--(1,0)--(1,1/2)--(1/2,0)--cycle,black+linewidth(1)); draw((0,0)--(0,1)--(1,1)--(1,0)--cycle); | [] |
848 | Real numbers between 0 and 1, inclusive, are chosen in the following manner. A fair coin is flipped. If it lands heads, then it is flipped again and the chosen number is 0 if the second flip is heads, and 1 if the second flip is tails. On the other hand, if the first coin flip is tails, then the number is chosen unifor... | 2019 AMC 12A Problem 20 | Notice the probability that $|x-y| > \tfrac{1}{2}$ is equal to $2$ times the probability that $x-y > \tfrac{1}{2}$.
For $x-y > \tfrac{1}{2}$ to be true; there are $4$ cases.
Case $1$: $x$ flips $T$ and $y$ flips $T$
When $x$ flips $T$ and $y$ flips $T$, both $x$ and $y$ are randomly chosen from the interval $[0,1]$.... | size(3cm,0); filldraw((0,0)--(1,0)--(1,1/2)--(1/2,0)--cycle,black+linewidth(1)); draw((0,0)--(0,1)--(1,1)--(1,0)--cycle); | [] |
849 | Circles $\omega$ and $\gamma$, both centered at $O$, have radii $20$ and $17$, respectively. Equilateral triangle $ABC$, whose interior lies in the interior of $\omega$ but in the exterior of $\gamma$, has vertex $A$ on $\omega$, and the line containing side $\overline{BC}$ is tangent to $\gamma$. Segments $\overline{A... | 2019 AMC 12A Problem 22 | Let $S$ be the point of tangency between $\overline{BC}$ and $\gamma$, and $M$ be the midpoint of $\overline{BC}$. Note that $AM \perp BS$ and $OS \perp BS$. This implies that $\angle OAM \cong \angle AOS$, and $\angle AMP \cong \angle OSP$. Thus, $\triangle PMA \sim \triangle PSO$.
If we let $s$ be the side length of... | // Block 1
size(20cm);
draw(circle((0,0), 20));
label("$\omega$", (0,0), 4.05*20*dir(149)*20/21);
draw(circle((0,0), 17));
label("$\gamma$", (0,0), 4.05*17*dir(149)*20/21);
dot((0,0));
label("$O$", (0,0), E);
pair aa = (-20, 0);
dot(aa);
label("$A$", aa, W);
draw((-20,0)--(0,0));
real a = (-20 + (80/sqrt(13) - 34/sqrt(... | [] |
850 | The figure below is a map showing $12$ cities and $17$ roads connecting certain pairs of cities. Paula wishes to travel along exactly $13$ of those roads, starting at city $A$ and ending at city $L$, without traveling along any portion of a road more than once. (Paula is allowed to visit a city more than once.)
How m... | 2019 AMC 12B Problem 10 | Note that of the $12$ cities, $6$ of them ($2$ on the top, $2$ on the bottom, and $1$ on each side) have $3$ edges coming into/out of them (i.e., in graph theory terms, they have degree $3$). Therefore, at least $1$ edge connecting to each of these cities cannot be used. Additionally, the same applies to the start and ... | // Block 1
import olympiad;
unitsize(50);
for (int i = 0; i < 3; ++i) {
for (int j = 0; j < 4; ++j) {
pair A = (j,i);
dot(A);
}
}
for (int i = 0; i < 3; ++i) {
for (int j = 0; j < 4; ++j) {
if (j != 3) {
draw((j,i)--(j+1,i));
}
if (i != 2) {
draw((j,i)--(j,i+1));
}
}
}
label("$A$", (0,2), W);
label("$L$", (3,0), E);
... | [] |
850 | The figure below is a map showing $12$ cities and $17$ roads connecting certain pairs of cities. Paula wishes to travel along exactly $13$ of those roads, starting at city $A$ and ending at city $L$, without traveling along any portion of a road more than once. (Paula is allowed to visit a city more than once.)
How m... | 2019 AMC 12B Problem 10 | Let the bottom-left vertex be $(0,0)$, and let each of the edges have length $1$, so that all of the vertices are at lattice points.
Firstly, notice that for any vertex $V$ on the graph (other than $A$ or $L$), we can visit it at most $M(V) = \left\lfloor \frac{\text{deg}(V)}{2} \right\rfloor$ times, where $\text{deg}(... | // Block 1
import olympiad;
unitsize(50);
for (int i = 0; i < 3; ++i) {
for (int j = 0; j < 4; ++j) {
pair A = (j,i);
dot(A);
}
}
for (int i = 0; i < 3; ++i) {
for (int j = 0; j < 4; ++j) {
if (j != 3) {
draw((j,i)--(j+1,i));
}
if (i != 2) {
draw((j,i)--(j,i+1));
}
}
}
label("$A$", (0,2), W);
label("$L$", (3,0), E);
l... | [] |
851 | How many unordered pairs of edges of a given cube determine a plane?
$\textbf{(A) } 12 \qquad \textbf{(B) } 28 \qquad \textbf{(C) } 36\qquad \textbf{(D) } 42 \qquad \textbf{(E) } 66$ | 2019 AMC 12B Problem 11 | Without loss of generality, choose one of the $12$ edges of the cube to be among the two selected. We now calculate the probability that a randomly-selected second edge makes the pair satisfy the condition in the problem statement.
For two lines in space to determine a common plane, they must either intersect or be pa... | // Block 1
import three;
import three;
unitsize(1cm);
size(200);
currentprojection=perspective(-6/5,-8/5,7/8);
draw((0,1,0)--(0,0,0)--(1,0,0), blue);
draw((1,0,0)--(1,1,0)--(0,1,0));
draw((0,0,0)--(0,0,1), red);
draw((0,1,0)--(0,1,1), green);
draw((1,1,0)--(1,1,1), green);
draw((1,0,0)--(1,0,1), green);
draw((0,1,1)--(... | [] |
852 | There are lily pads in a row numbered $0$ to $11$, in that order. There are predators on lily pads $3$ and $6$, and a morsel of food on lily pad $10$. Fiona the frog starts on pad $0$, and from any given lily pad, has a $\frac{1}{2}$ chance to hop to the next pad, and an equal chance to jump $2$ pads. What is the proba... | 2019 AMC 12B Problem 16 | Let $p_n$ be the probability of landing on lily pad $n$. Observe that if there are no restrictions, we would have
\[p_n = \frac{1}{2} \cdot p_{n-1} + \frac{1}{2} \cdot p_{n-2}\]
This is because, given that Fiona is at lily pad $n-2$, there is a $\frac{1}{2}$ probability that she will make a $2$-jump to reach lily pad ... | // Block 1
unitsize(40);
string[] vals = {"1", "$1/2$", "$3/4$", "X", "$3/8$", "$3/16$", "X", "$3/32$", "$3/64$", "$9/128$", "$15/256$", "X"};
for(int i =0; i<= 11; ++i) {
draw((i,0)--(i+1,0)--(i+1,1)--(i,1)--cycle);
label((string) i, (i+0.5,0), S);
label(vals[i], (i+0.5, 0.5));
}
// Block 2
unitsize(40);
string[] vals... | [] |
853 | Square pyramid $ABCDE$ has base $ABCD$, which measures $3$ cm on a side, and altitude $AE$ perpendicular to the base, which measures $6$ cm. Point $P$ lies on $BE$, one third of the way from $B$ to $E$; point $Q$ lies on $DE$, one third of the way from $D$ to $E$; and point $R$ lies on $CE$, two thirds of the way from ... | 2019 AMC 12B Problem 18 | As in Solution 1, let $A=(0, 0, 0), B=(3, 0, 0), C=(3, 3, 0), D=(0, 3, 0),$ and $E=(0, 0, 6)$, and calculate the coordinates of $P$, $Q$, and $R$ as $P=(2,0,2), Q=(0,2,2), R=(1,1,4)$. Now notice that the plane determined by $\triangle PQR$ is perpendicular to the plane determined by $ABCD$. To see this, consider the bi... | // Block 1
unitsize(40);
for(int i =0; i<=3; ++i) {
draw((0,i)--(3,i));
draw((i,0)--(i,3));
}
label("$A$", (0,0), SW);
label("$B$", (3,0), SE);
label("$C$", (3,3), NE);
label("$D$", (0,3), NW);
label("$P$", (2,0), S);
label("$Q$", (0,2), W);
label("$R$", (1,1), NE);
dot((2,0));
dot((0,2));
dot((1,1));
draw((0,2)--(2,0)... | [] |
854 | Let $\omega=-\tfrac{1}{2}+\tfrac{1}{2}i\sqrt3.$ Let $S$ denote all points in the complex plane of the form $a+b\omega+c\omega^2,$ where $0\leq a \leq 1,0\leq b\leq 1,$ and $0\leq c\leq 1.$ What is the area of $S$?
$\textbf{(A) } \frac{1}{2}\sqrt3 \qquad\textbf{(B) } \frac{3}{4}\sqrt3 \qquad\textbf{(C) } \frac{3}{2}\sq... | 2019 AMC 12B Problem 24 | Notice that $\omega=e^{\frac{2i\pi}{3}}$, which is one of the cube roots of unity. We wish to find the span of $(a+b\omega+c\omega^2)$ for reals $0\le a,b,c\le 1$.
Observe also that if $a,b,c>0$, then replacing $a$, $b$, and $c$ by $a-\min(a,b,c), b-\min(a,b,c),$ and $c-\min(a,b,c)$ leaves the value of $a+b\omega+c\ome... | // Block 1
size(200,200);
draw((0,0)--(1,0)--(1/2,sqrt(3)/2)--cycle);
draw((0,0)--(1/2,sqrt(3)/2)--(-1/2,sqrt(3)/2)--cycle);
draw((0,0)--(-1/2,sqrt(3)/2)--(-1,0)--cycle);
draw((0,0)--(-1,0)--(-1/2,-sqrt(3)/2)--cycle);
draw((0,0)--(-1/2,-sqrt(3)/2)--(1/2,-sqrt(3)/2)--cycle);
draw((0,0)--(1/2,-sqrt(3)/2)--(1,0)--cycle);
... | [] |
854 | Let $\omega=-\tfrac{1}{2}+\tfrac{1}{2}i\sqrt3.$ Let $S$ denote all points in the complex plane of the form $a+b\omega+c\omega^2,$ where $0\leq a \leq 1,0\leq b\leq 1,$ and $0\leq c\leq 1.$ What is the area of $S$?
$\textbf{(A) } \frac{1}{2}\sqrt3 \qquad\textbf{(B) } \frac{3}{4}\sqrt3 \qquad\textbf{(C) } \frac{3}{2}\sq... | 2019 AMC 12B Problem 24 | WLOG $c$ is the smallest of the $3$. Then the expression is equivalent to $a+b\omega$. To find the area of the region, we need only consider the extremities ($a=0, a=1, b=0, b=1$), as they will form a polygon which contains all points. So, when $a=0, b=0$ we have the origin (diagram omitted). When $a=1, b=0$ we have $1... | // Block 1
size(100,100);
draw((-2,0)--(2,0), gray);
draw((0,-2)--(0,2), gray);
draw((0,0)--(1,0), red);
// Block 2
size(100,100);
draw((-2,0)--(2,0), gray);
draw((0,-2)--(0,2), gray);
draw((0,0)--(-1/2,sqrt(3)/2), red);
draw((0,0)--(1,0), red);
// Block 3
size(100,100);
draw((-2,0)--(2,0), gray);
draw((0,-2)--(0,2), g... | [] |
854 | Let $\omega=-\tfrac{1}{2}+\tfrac{1}{2}i\sqrt3.$ Let $S$ denote all points in the complex plane of the form $a+b\omega+c\omega^2,$ where $0\leq a \leq 1,0\leq b\leq 1,$ and $0\leq c\leq 1.$ What is the area of $S$?
$\textbf{(A) } \frac{1}{2}\sqrt3 \qquad\textbf{(B) } \frac{3}{4}\sqrt3 \qquad\textbf{(C) } \frac{3}{2}\sq... | 2019 AMC 12B Problem 24 | We can add on each term one at a time. Firstly, the possible values of $\textstyle c\omega^2=c\left(-\frac{1}{2}-\frac{\sqrt{3}}{2}i\right)$ lie on the following line:
For each point on the line, we can add $\textstyle b\omega=b\left(-\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)$. This means that we can extend the area to... | // Block 1
size(100,100);
draw((0,0)--(-1/2,-sqrt(3)/2), blue);
draw((-2,0)--(2,0));
draw((0,-2)--(0,2));
// Block 2
size(100,100);
fill((0,0)--(-1/2,sqrt(3)/2)--(-1,0)--(-1/2,-sqrt(3)/2)--cycle, lightgray);
draw((0,0)--(-1/2,sqrt(3)/2)--(-1,0)--(-1/2,-sqrt(3)/2)--cycle);
draw((0,0)--(-1/2,sqrt(3)/2), red);
draw((0,0)-... | [] |
855 | Triangle $AMC$ is isosceles with $AM = AC$. Medians $\overline{MV}$ and $\overline{CU}$ are perpendicular to each other, and $MV=CU=12$. What is the area of $\triangle AMC?$
$\textbf{(A) } 48 \qquad \textbf{(B) } 72 \qquad \textbf{(C) } 96 \qquad \textbf{(D) } 144 \qquad \textbf{(E) } 192$ | 2020 AMC 10A Problem 12 | We know that $\triangle AUV \sim \triangle AMC$, and since the ratios of its sides are $\frac{1}{2}$, the ratio of of their areas is $\left(\frac{1}{2}\right)^2=\frac{1}{4}$.
If $\triangle AUV$ is $\frac{1}{4}$ the area of $\triangle AMC$, then trapezoid $MUVC$ is $\frac{3}{4}$ the area of $\triangle AMC$.
Let's ca... | // Block 1
draw((-4,0)--(4,0)--(0,12)--cycle);
draw((-2,6)--(4,0));
draw((2,6)--(-4,0));
draw((-2,6)--(2,6));
label("M", (-4,0), W);
label("C", (4,0), E);
label("A", (0, 12), N);
label("V", (2, 6), NE);
label("U", (-2, 6), NW);
label("P", (0, 3.6), S);
// Block 2
draw((-4,0)--(4,0)--(0,12)--cycle); draw((-2,6)--(4,0));... | [] |
855 | Triangle $AMC$ is isosceles with $AM = AC$. Medians $\overline{MV}$ and $\overline{CU}$ are perpendicular to each other, and $MV=CU=12$. What is the area of $\triangle AMC?$
$\textbf{(A) } 48 \qquad \textbf{(B) } 72 \qquad \textbf{(C) } 96 \qquad \textbf{(D) } 144 \qquad \textbf{(E) } 192$ | 2020 AMC 10A Problem 12 | Draw median $\overline{AB}$.
Since we know that all medians of a triangle intersect at the centroid, we know that $\overline{AB}$ passes through point $P$. We also know that medians of a triangle divide each other into segments of ratio $2:1$. Knowing this, we can see that $\overline{PC}:\overline{UP}=2:1$, and since... | // Block 1
draw((-4,0)--(4,0)--(0,12)--cycle);
draw((-2,6)--(4,0));
draw((2,6)--(-4,0));
draw((0,12)--(0,0));
label("M", (-4,0), W);
label("C", (4,0), E);
label("A", (0, 12), N);
label("V", (2, 6), NE);
label("U", (-2, 6), NW);
label("P", (0.5, 4), E);
label("B", (0, 0), S);
// Block 2
draw((-4,0)--(4,0)--(0,12)--cycle... | [] |
855 | Triangle $AMC$ is isosceles with $AM = AC$. Medians $\overline{MV}$ and $\overline{CU}$ are perpendicular to each other, and $MV=CU=12$. What is the area of $\triangle AMC?$
$\textbf{(A) } 48 \qquad \textbf{(B) } 72 \qquad \textbf{(C) } 96 \qquad \textbf{(D) } 144 \qquad \textbf{(E) } 192$ | 2020 AMC 10A Problem 12 | We know that $AU = UM$, $AV = VC$, so $UV = \frac{1}{2} MC$.
As $\angle UPM = \angle VPC = 90$, we can see that $\triangle UPM \cong \triangle VPC$ and $\triangle UVP \sim \triangle MPC$ with a side ratio of $1 : 2$.
So $UP = VP = 4$, $MP = PC = 8$.
With that, we can see that $[\triangle UPM] = 16$, and the area of ... | // Block 1
draw((-4,0)--(4,0)--(0,12)--cycle);
draw((-2,6)--(4,0));
draw((2,6)--(-4,0));
draw((-2,6)--(2,6));
label("M", (-4,0), W);
label("C", (4,0), E);
label("A", (0, 12), N);
label("V", (2, 6), NE);
label("U", (-2, 6), NW);
label("P", (0, 3.6), S);
// Block 2
draw((-4,0)--(4,0)--(0,12)--cycle); draw((-2,6)--(4,0));... | [] |
855 | Triangle $AMC$ is isosceles with $AM = AC$. Medians $\overline{MV}$ and $\overline{CU}$ are perpendicular to each other, and $MV=CU=12$. What is the area of $\triangle AMC?$
$\textbf{(A) } 48 \qquad \textbf{(B) } 72 \qquad \textbf{(C) } 96 \qquad \textbf{(D) } 144 \qquad \textbf{(E) } 192$ | 2020 AMC 10A Problem 12 | Let $AB$ be the height. Since medians divide each other into a $2:1$ ratio, and the medians have length 12, we have $PC=MP=8$ and $UP=VP=4$. From right triangle $\triangle{MUP}$, \[MU^2=MP^2+UP^2=8^2+4^2=80,\] so $MU=\sqrt{80}=4\sqrt{5}$. Since $CU$ is a median, $AM=8\sqrt{5}$. From right triangle $\triangle{MPC}$, \[M... | // Block 1
draw((-4,0)--(4,0)--(0,12)--cycle);
draw((-2,6)--(4,0));
draw((2,6)--(-4,0));
draw((0,12)--(0,0));
label("M", (-4,0), W);
label("C", (4,0), E);
label("A", (0, 12), N);
label("V", (2, 6), NE);
label("U", (-2, 6), NW);
label("P", (0.5, 4), E);
label("B", (0, 0), S);
// Block 2
draw((-4,0)--(4,0)--(0,12)--cycle... | [] |
855 | Triangle $AMC$ is isosceles with $AM = AC$. Medians $\overline{MV}$ and $\overline{CU}$ are perpendicular to each other, and $MV=CU=12$. What is the area of $\triangle AMC?$
$\textbf{(A) } 48 \qquad \textbf{(B) } 72 \qquad \textbf{(C) } 96 \qquad \textbf{(D) } 144 \qquad \textbf{(E) } 192$ | 2020 AMC 10A Problem 12 | Connect $AP$, and let $B$ be the point where $AP$ intersects $MC$. $MB=CB$ because all medians of a triangle intersect at one point, which in this case is $P$. $MP:PV=2:1$ because the point at which all medians intersect divides the medians into segments of ratio $2:1$, so $MP=8$ and similarly $CP=8$. We apply the Pyth... | // Block 1
draw((-4,0)--(4,0)--(0,12)--cycle);
draw((-2,6)--(4,0));
draw((2,6)--(-4,0));
draw((0,12)--(0,0));
label("M", (-4,0), W);
label("C", (4,0), E);
label("A", (0, 12), N);
label("V", (2, 6), NE);
label("U", (-2, 6), NW);
label("P", (0.5, 4), E);
label("B", (0, 0), S);
// Block 2
draw((-4,0)--(4,0)--(0,12)--cycle... | [] |
855 | Triangle $AMC$ is isosceles with $AM = AC$. Medians $\overline{MV}$ and $\overline{CU}$ are perpendicular to each other, and $MV=CU=12$. What is the area of $\triangle AMC?$
$\textbf{(A) } 48 \qquad \textbf{(B) } 72 \qquad \textbf{(C) } 96 \qquad \textbf{(D) } 144 \qquad \textbf{(E) } 192$ | 2020 AMC 10A Problem 12 | Since $\overline{MV}$ and $\overline{CU}$ intersect at a right angle, this means $MUVC$ is a kite. Hence, the area of this kite is $\frac{12 \cdot 12}{2} = 72$.
Also, notice that $\triangle AUV \sim \triangle AMC$ by AAA Similarity. Since the ratios of its sides is $\frac{1}{2}$, the ratios of the area is $\left(\frac... | // Block 1
draw((-4,0)--(4,0)--(0,12)--cycle);
draw((-2,6)--(4,0));
draw((2,6)--(-4,0));
draw((-2,6)--(2,6));
label("M", (-4,0), W);
label("C", (4,0), E);
label("A", (0, 12), N);
label("V", (2, 6), NE);
label("U", (-2, 6), NW);
label("P", (0, 3.6), S);
// Block 2
draw((-4,0)--(4,0)--(0,12)--cycle); draw((-2,6)--(4,0));... | [] |
855 | Triangle $AMC$ is isosceles with $AM = AC$. Medians $\overline{MV}$ and $\overline{CU}$ are perpendicular to each other, and $MV=CU=12$. What is the area of $\triangle AMC?$
$\textbf{(A) } 48 \qquad \textbf{(B) } 72 \qquad \textbf{(C) } 96 \qquad \textbf{(D) } 144 \qquad \textbf{(E) } 192$ | 2020 AMC 10A Problem 12 | Horizontally translate line $\overline{UC}$ until point $U$ is at point $V$, with $C$ subsequently at $C'$, and then connect up $C'$ and $C$ to create $\triangle MVC'$, which is a right triangle.
Notice that $\triangle MVC'$ = $12 \cdot 12 \cdot \frac{1}{2} = 72$, and $\triangle MVC'$ = $\triangle MVC + \triangle MU... | // Block 1
draw((-4,0)--(4,0)--(0,12)--cycle);
draw((-2,6)--(4,0));
draw((2,6)--(-4,0));
draw((-2,6)--(2,6));
draw((2,6)--(8,0));
draw((4,0)--(8,0));
label("M", (-4,0), W);
label("C", (4,0), S);
label("A", (0, 12), N);
label("V", (2, 6), NE);
label("U", (-2, 6), NW);
label("P", (0, 3.6), S);
label("C'", (8,0), E);
// B... | [] |
856 | A point is chosen at random within the square in the coordinate plane whose vertices are $(0, 0), (2020, 0), (2020, 2020),$ and $(0, 2020)$. The probability that the point is within $d$ units of a lattice point is $\tfrac{1}{2}$. (A point $(x, y)$ is a lattice point if $x$ and $y$ are both integers.) What is $d$ to the... | 2020 AMC 10A Problem 16 | Diagram
The diagram represents each unit square of the given $2020 \times 2020$ square.
Solution 1
We consider an individual one-by-one block.
If we draw a quarter of a circle from each corner (where the lattice points are located), each with radius $d$, the area covered by the circles should be $0.5$. Because of t... | size(5cm); draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); filldraw((arc((0,0), 0.3989, 0, 90))--(0,0)--cycle, gray); draw(arc((1,0), 0.3989, 90, 180)); filldraw((arc((1,0), 0.3989, 90, 180))--(1,0)--cycle, gray); draw(arc((1,1), 0.3989, 180, 270)); filldraw((arc((1,1), 0.3989, 180, 270))--(1,1)--cycle, gray); draw(arc((0,1),... | [] |
856 | A point is chosen at random within the square in the coordinate plane whose vertices are $(0, 0), (2020, 0), (2020, 2020),$ and $(0, 2020)$. The probability that the point is within $d$ units of a lattice point is $\tfrac{1}{2}$. (A point $(x, y)$ is a lattice point if $x$ and $y$ are both integers.) What is $d$ to the... | 2020 AMC 10A Problem 16 | PLEASE NOTE: Solution 1 IS rigorous. Say there are $k$ unit squares (it doesn't matter how many). There is a $\frac{1}k$ probability the point is in some unit square. There is a $\pi d^2$ probability the point is in the shaded region. So, there is a $\frac{1}k \pi d^2 \cdot k = \pi d^2$ probability the point is in any ... | // Block 1
size(10cm);
draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);
filldraw((arc((0,0), 0.3989, 0, 90))--(0,0)--cycle, gray);
draw(arc((1,0), 0.3989, 90, 180));
filldraw((arc((1,0), 0.3989, 90, 180))--(1,0)--cycle, gray);
draw(arc((1,1), 0.3989, 180, 270));
filldraw((arc((1,1), 0.3989, 180, 270))--(1,1)--cycle, gray);
dra... | [] |
857 | Define \[P(x) =(x-1^2)(x-2^2)\cdots(x-100^2).\] How many integers $n$ are there such that $P(n)\leq 0$?
$\textbf{(A) } 4900 \qquad \textbf{(B) } 4950\qquad \textbf{(C) } 5000\qquad \textbf{(D) } 5050 \qquad \textbf{(E) } 5100$ | 2020 AMC 10A Problem 17 | Visualization that makes it easier to see solutions:
Note
Clearly, there are a finite amount of integers that result in $P(x)$ being nonpositive. This is because there are an even number of factors of $P(x).$ From here, you can read other solutions that count the amount of negative numbers of $P(x).$
~jasmineLOVER7... | // Block 1
size(600);
// Draw the main number line (x-axis)
draw((-5, 0) -- (5, 0), Arrow); // Line from -10 to 10 with an arrow at the end
draw((5, 0) -- (-5, 0), Arrow); // Line from -10 to 10 with an arrow at the end
// Add ticks and labels at each integer point from -10 to 10
for (int i = -4; i <= 4; ++i) {
... | [] |
858 | Quadrilateral $ABCD$ satisfies $\angle ABC = \angle ACD = 90^{\circ}, AC=20,$ and $CD=30.$ Diagonals $\overline{AC}$ and $\overline{BD}$ intersect at point $E,$ and $AE=5.$ What is the area of quadrilateral $ABCD?$
$\textbf{(A) } 330 \qquad \textbf{(B) } 340 \qquad \textbf{(C) } 350 \qquad \textbf{(D) } 360 \qquad \te... | 2020 AMC 10A Problem 20 | It's crucial to draw a good diagram for this one. Since $AC=20$ and $CD=30$, we get $[ACD]=300$. Now we need to find $[ABC]$ to get the area of the whole quadrilateral. Drop an altitude from $B$ to $AC$ and call the point of intersection $F$. Let $FE=x$. Since $AE=5$, then $AF=5-x$.
By dropping this altitude, we can ... | // Block 1
size(15cm,0);
import olympiad;
draw((0,0)--(0,2)--(6,4)--(4,0)--cycle);
label("A", (0,2), NW);
label("B", (0,0), SW);
label("C", (4,0), SE);
label("D", (6,4), NE);
label("E", (1.714,1.143), N);
label("F", (1,1.5), N);
draw((0,2)--(4,0), dashed);
draw((0,0)--(6,4), dashed);
draw((0,0)--(1,1.5), dashed);
label... | [] |
858 | Quadrilateral $ABCD$ satisfies $\angle ABC = \angle ACD = 90^{\circ}, AC=20,$ and $CD=30.$ Diagonals $\overline{AC}$ and $\overline{BD}$ intersect at point $E,$ and $AE=5.$ What is the area of quadrilateral $ABCD?$
$\textbf{(A) } 330 \qquad \textbf{(B) } 340 \qquad \textbf{(C) } 350 \qquad \textbf{(D) } 360 \qquad \te... | 2020 AMC 10A Problem 20 | Let the points be $A(-10,0)$, $\:B(x,y)$, $\:C(10,0)$, $\:D(10,30)$,and $\:E(-5,0)$, respectively. Since $B$ lies on line $DE$, we know that $y=2x+10$. Furthermore, since $\angle{ABC}=90^\circ$, $B$ lies on the circle with diameter $AC$, so $x^2+y^2=100$. Solving for $x$ and $y$ with these equations, we get the solutio... | // Block 1
size(10cm,0);
draw((10,30)--(10,0)--(-8,-6)--(-10,0)--(10,30));
draw((-20,0)--(20,0));
draw((0,-15)--(0,35));
draw((10,30)--(-8,-6));
draw(circle((0,0),10));
label("E",(-4.05,-.25),S);
label("D",(10,30),NE);
label("C",(10,0),NE);
label("B",(-8,-6),SW);
label("A",(-10,0),NW);
label("5",(-10,0)--(-5,0), NE);
l... | [] |
858 | Quadrilateral $ABCD$ satisfies $\angle ABC = \angle ACD = 90^{\circ}, AC=20,$ and $CD=30.$ Diagonals $\overline{AC}$ and $\overline{BD}$ intersect at point $E,$ and $AE=5.$ What is the area of quadrilateral $ABCD?$
$\textbf{(A) } 330 \qquad \textbf{(B) } 340 \qquad \textbf{(C) } 350 \qquad \textbf{(D) } 360 \qquad \te... | 2020 AMC 10A Problem 20 | Denote $EB$ as $x$. By the Law of Cosines:
\[AB^2 = 25 + x^2 - 10x\cos(\angle DEC)\]
\[BC^2 = 225 + x^2 + 30x\cos(\angle DEC)\]
Adding these up yields:
\[400 = 250 + 2x^2 + 20x\cos(\angle DEC) \Longrightarrow x^2 + \frac{10x}{\sqrt{5}} - 75 = 0\]
By the quadratic formula, $x = 3\sqrt5$.
Observe:
\[[AEB] + [BEC] = \fr... | // Block 1
import olympiad;
pair A = (0, 189), B = (0,0), C = (570,0), D = (798, 798);
dot("$A$", A, W); dot("$B$", B, S); dot("$C$", C, E); dot("$D$", D, N);dot("$E$",(140, 140), N);
draw(A--B--C--D--A);
draw(A--C, dotted); draw(B--D, dotted);
// Block 2
import olympiad; pair A = (0, 189), B = (0,0), C = (570,0), D = ... | [] |
858 | Quadrilateral $ABCD$ satisfies $\angle ABC = \angle ACD = 90^{\circ}, AC=20,$ and $CD=30.$ Diagonals $\overline{AC}$ and $\overline{BD}$ intersect at point $E,$ and $AE=5.$ What is the area of quadrilateral $ABCD?$
$\textbf{(A) } 330 \qquad \textbf{(B) } 340 \qquad \textbf{(C) } 350 \qquad \textbf{(D) } 360 \qquad \te... | 2020 AMC 10A Problem 20 | Let $F$ be the midpoint of $AC$, and draw $FG // CD$ where $G$ is on $BD$. We have $EF=5,FC=10$.
$\Delta EFG \sim \Delta ECD \implies FG=10=FA=FC$. Therefore $ABCG$ is a cyclic quadrilateral.
Notice that $\angle EFG=90^\circ, EG=\sqrt{5^2+10^2}=5\sqrt{5} \implies BE=\frac{AE\cdot EC}{EG}=\frac{5\cdot 15}{5\sqrt{5}}=3... | // Block 1
import olympiad;
pair A = (0, 189), B = (0,0), C = (570,0), D = (798, 798),F=(285,94.5),G=(361.2,361.2);
dot("$A$", A, W); dot("$B$", B, S); dot("$C$", C, E); dot("$D$", D, N);dot("$E$",(140, 140), N);dot("$F$",F,N);dot("$G$",G,N);
draw(A--B--C--D--A);
draw(A--C, dotted); draw(B--D, dotted); draw(F--G, dotte... | [] |
858 | Quadrilateral $ABCD$ satisfies $\angle ABC = \angle ACD = 90^{\circ}, AC=20,$ and $CD=30.$ Diagonals $\overline{AC}$ and $\overline{BD}$ intersect at point $E,$ and $AE=5.$ What is the area of quadrilateral $ABCD?$
$\textbf{(A) } 330 \qquad \textbf{(B) } 340 \qquad \textbf{(C) } 350 \qquad \textbf{(D) } 360 \qquad \te... | 2020 AMC 10A Problem 20 | Let $AB = x$, $BC = y$
Looking at the diagram we have $x^2 + y^2 = 20^2$,
$DE = \sqrt{30^2+15^2} = 15\sqrt{5}$, $[ACD] = \frac{1}{2} \cdot 20 \cdot 30 = 300$
Because $\triangle CEF \sim \triangle CAB$, $EF = AB \cdot \frac{CE}{CA} = x \cdot \frac{15}{20} = \frac{3x}{4}$
$BF = BC - CF = BC - BC \cdot \frac{CE}{CA} = ... | // Block 1
size(15cm,0);
import olympiad;
draw((0,0)--(0,2)--(6,4)--(4,0)--cycle);
label("A", (0,2), NW);
label("B", (0,0), SW);
label("C", (4,0), SE);
label("D", (6,4), NE);
label("E", (1.714,1.143), N);
label("F", (1.714,0), SE);
draw((0,2)--(4,0), dashed);
draw((0,0)--(6,4), dashed);
draw((1.714,1.143)--(1.714,0), d... | [] |
858 | Quadrilateral $ABCD$ satisfies $\angle ABC = \angle ACD = 90^{\circ}, AC=20,$ and $CD=30.$ Diagonals $\overline{AC}$ and $\overline{BD}$ intersect at point $E,$ and $AE=5.$ What is the area of quadrilateral $ABCD?$
$\textbf{(A) } 330 \qquad \textbf{(B) } 340 \qquad \textbf{(C) } 350 \qquad \textbf{(D) } 360 \qquad \te... | 2020 AMC 10A Problem 20 | set \[\angle ACB = \theta , C= \cos(\theta), S = \sin(\theta)\]
\[\dfrac{E_y}{E_x} = \dfrac{30C} { 20C+30S} = \dfrac{15S} {20C-15C}\]
\[2SC = \dfrac35\]
\begin{align*} [ABCD] &= [ABD]+[CBD] \\ &= \dfrac12\cdot 20C\cdot 30C + \dfrac12 \cdot 20S (20C+30S) \\ &= 100\cdot 2SC + 300 \\ &= \boxed{\text{(D) } 360}. ... | // Block 1
size(15cm,0);
import olympiad;
draw((0,0)--(0,2)--(6,4)--(4,0)--cycle);
label("A", (0,2), NW);
label("B", (0,0), SW);
label("C", (4,0), SE);
label("D", (6,4), NE);
label("E", (1.714,1.143), N);
label("F", (1.714,0), SE);
draw((0,2)--(4,0), dashed);
draw((0,0)--(6,4), dashed);
draw((4,0)--(6,0), dashed);
draw... | [] |
859 | Let $T$ be the triangle in the coordinate plane with vertices $(0,0), (4,0),$ and $(0,3).$ Consider the following five isometries (rigid transformations) of the plane: rotations of $90^{\circ}, 180^{\circ},$ and $270^{\circ}$ counterclockwise around the origin, reflection across the $x$-axis, and reflection across the ... | 2020 AMC 10A Problem 23 | First, any combination of motions we can make must reflect $T$ an even number of times. This is because every time we reflect $T$, it changes orientation. Once $T$ has been flipped once, no combination of rotations will put it back in place because it is the mirror image; however, flipping it again changes it back to t... | // Block 1
size(10cm);
Label f;
f.p=fontsize(6);
xaxis(-6,6,Ticks(f, 2.0));
yaxis(-6,6,Ticks(f, 2.0));
filldraw(origin--(4,0)--(0,3)--cycle, gray, black+linewidth(1));
// Block 2
size(10cm); Label f; f.p=fontsize(6); xaxis(-6,6,Ticks(f, 2.0)); yaxis(-6,6,Ticks(f, 2.0)); filldraw(origin--(4,0)--(0,3)--cycle, gra... | [] |
860 | In the plane figure shown below, $3$ of the unit squares have been shaded. What is the least number of additional unit squares that must be shaded so that the resulting figure has two lines of symmetry?
$\textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8$ | 2020 AMC 12A Problem 6 | The two lines of symmetry must be horizontally and vertically through the middle. We can then fill the boxes in like so:
where the light gray boxes are the ones we have filled. Counting these, we get $\boxed{\textbf{(D) } 7}$ total boxes.
~ciceronii | // Block 1
import olympiad;
unitsize(25);
filldraw((1,3)--(1,4)--(2,4)--(2,3)--cycle, gray(0.7));
filldraw((0,3)--(0,4)--(1,4)--(1,3)--cycle, gray(0.9));
filldraw((3,3)--(3,4)--(4,4)--(4,3)--cycle, gray(0.9));
filldraw((4,3)--(4,4)--(5,4)--(5,3)--cycle, gray(0.9));
filldraw((2,2)--(2,3)--(3,3)--(3,2)--cycle, gray(0.9))... | [] |
860 | In the plane figure shown below, $3$ of the unit squares have been shaded. What is the least number of additional unit squares that must be shaded so that the resulting figure has two lines of symmetry?
$\textbf{(A) } 4 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 8$ | 2020 AMC 12A Problem 6 | We label the three shaded unit squares $A,B,$ and $C,$ then construct the two lines of symmetry of the resulting figure, as shown below:
Note that:
Since the centers of $A$ and $C$ are on neither line of symmetry, $A$ and $C$ each contribute $4$ shaded unit squares to the resulting figure.
Since the center of $B$ is... | // Block 1
import olympiad;
unitsize(25);
filldraw((1,3)--(1,4)--(2,4)--(2,3)--cycle, gray(0.7));
filldraw((2,1)--(2,2)--(3,2)--(3,1)--cycle, gray(0.7));
filldraw((4,0)--(5,0)--(5,1)--(4,1)--cycle, gray(0.7));
for (int i = 0; i < 5; ++i) {
for (int j = 0; j < 6; ++j) {
pair A = (j,i);
}
}
for (int i = 0; i < 5; ++i) {
... | [] |
861 | How many solutions does the equation $\tan(2x)=\cos(\tfrac{x}{2})$ have on the interval $[0,2\pi]?$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$ | 2020 AMC 12A Problem 9 | We count the intersections of the graphs of $y=\tan(2x)$ and $y=\cos\left(\frac x2\right):$
The graph of $y=\tan(2x)$ has a period of $\frac{\pi}{2},$ asymptotes at $x=\frac{\pi}{4}+\frac{k\pi}{2},$ and zeros at $x=\frac{k\pi}{2}$ for some integer $k.$
On the interval $[0,2\pi],$ the graph has five branches: \[\bigg... | // Block 1
/* Made by MRENTHUSIASM */
size(800,200);
real f(real x) { return tan(2*x); }
real g(real x) { return cos(x/2); }
draw(graph(f,0,atan(3)/2),red,"$y=\tan(2x)$");
draw(graph(f,-atan(3)/2+pi/2,atan(3)/2+pi/2),red);
draw(graph(f,-atan(3)/2+2*pi/2,atan(3)/2+2*pi/2),red);
draw(graph(f,-atan(3)/2+3*pi/2,atan(3)... | [] |
862 | Line $l$ in the coordinate plane has equation $3x-5y+40=0$. This line is rotated $45^{\circ}$ counterclockwise about the point $(20,20)$ to obtain line $k$. What is the $x$-coordinate of the $x$-intercept of line $k?$
$\textbf{(A) } 10 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 25 \qquad \tex... | 2020 AMC 12A Problem 12 | Let $P$ be $(20, 20)$ and $X, Y$ be $(20, 0)$ and $(0, 20)$ respectively. Since the slope of the line is $3/5$ we know that $\tan{\angle{YPA}} = 3/5.$ Segments $\overline{PA}$ and $\overline{PB}$ represent the before and after of rotating $l$ by 45 counterclockwise. Thus, $\angle{XPB} = 45 - \angle{YPA}$ and \[BX = 20 ... | // Block 1
draw((0,0)--(20, 0)--(20, 20)--(0, 20)--cycle);
draw((20, 20)--(0, 8));
draw((15, 0)--(20, 20));
dot("$P$", (20, 20));
dot("$A$", (0, 8), dir(75));
dot("$B$", (15, 0), dir(45));
dot("$X$", (20, 0));
dot("$Y$", (0, 20), dir(50));
// Block 2
draw((0,0)--(20, 0)--(20, 20)--(0, 20)--cycle); draw((20, 20)--(0, 8... | [] |
863 | Regular octagon $ABCDEFGH$ has area $n$. Let $m$ be the area of quadrilateral $ACEG$. What is $\tfrac{m}{n}?$
$\textbf{(A) } \frac{\sqrt{2}}{4} \qquad \textbf{(B) } \frac{\sqrt{2}}{2} \qquad \textbf{(C) } \frac{3}{4} \qquad \textbf{(D) } \frac{3\sqrt{2}}{5} \qquad \textbf{(E) } \frac{2\sqrt{2}}{3}$ | 2020 AMC 12A Problem 14 | The first thing to notice is that $ACEG$ is a square. This is because, as $\bigtriangleup ABC \cong \bigtriangleup CDE \cong \bigtriangleup EFG \cong \bigtriangleup GHE$, they all have the same base, meaning that $AC = DE = EG = GA$. Hence, we have that it is a square. To determine the area of this square, we can deter... | // Block 1
draw((1, 2.41421356)--(2.41421356, 1)--(2.41421356, -1)--(1, -2.41421356)--(-1, -2.41421356)--(-2.41421356, -1)--(-2.41421356, 1)--(-1, 2.41421356)--cycle);
draw((2.41421356, 1)--(1, -2.41421356)--(-2.41421356, -1)--(-1, 2.41421356)--cycle);
label("A",(-1, 2.41421356),NW);
label("B",(1, 2.41421356),NE);
labe... | [] |
863 | Regular octagon $ABCDEFGH$ has area $n$. Let $m$ be the area of quadrilateral $ACEG$. What is $\tfrac{m}{n}?$
$\textbf{(A) } \frac{\sqrt{2}}{4} \qquad \textbf{(B) } \frac{\sqrt{2}}{2} \qquad \textbf{(C) } \frac{3}{4} \qquad \textbf{(D) } \frac{3\sqrt{2}}{5} \qquad \textbf{(E) } \frac{2\sqrt{2}}{3}$ | 2020 AMC 12A Problem 14 | The ratio $\frac{m}{n}=\frac{[OPC]}{[OBC]}=\frac{OP}{OB}=\frac{OP}{OC}=\frac{\sqrt {2}}{2}$.
<<ERMS Coach>> | // Block 1
draw((1, 2.41421356)--(2.41421356, 1)--(2.41421356, -1)--(1, -2.41421356)--(-1, -2.41421356)--(-2.41421356, -1)--(-2.41421356, 1)--(-1, 2.41421356)--cycle);
draw((2.41421356, 1)--(1, -2.41421356)--(-2.41421356, -1)--(-1, 2.41421356)--cycle);
draw((-1, 2.41421356)--(1,-2.41421356));
draw((1, 2.41421356)--(-1,... | [] |
864 | In the complex plane, let $A$ be the set of solutions to $z^{3}-8=0$ and let $B$ be the set of solutions to $z^{3}-8z^{2}-8z+64=0.$ What is the greatest distance between a point of $A$ and a point of $B?$
$\textbf{(A) } 2\sqrt{3} \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 9 \qquad \textbf{(D) } 2\sqrt{21} \qquad \tex... | 2020 AMC 12A Problem 15 | We solve each equation separately:
We solve $z^{3}-8=0$ by De Moivre's Theorem.
Let $z=r(\cos\theta+i\sin\theta)=r\operatorname{cis}\theta,$ where $r$ is the magnitude of $z$ such that $r\geq0,$ and $\theta$ is the argument of $z$ such that $0\leq\theta<2\pi.$
We have \[z^3=r^3\operatorname{cis}(3\theta)=8(1),\] fro... | // Block 1
/* Made by MRENTHUSIASM */
size(200);
int xMin = -10;
int xMax = 10;
int yMin = -10;
int yMax = 10;
int numRays = 24;
//Draws a polar grid that goes out to a number of circles
//equal to big, with numRays specifying the number of rays:
void polarGrid(int big, int numRays)
{
for (int i = 1; i < big+1;... | [] |
865 | A point is chosen at random within the square in the coordinate plane whose vertices are $(0, 0), (2020, 0), (2020, 2020),$ and $(0, 2020)$. The probability that the point is within $d$ units of a lattice point is $\tfrac{1}{2}$. (A point $(x, y)$ is a lattice point if $x$ and $y$ are both integers.) What is $d$ to the... | 2020 AMC 12A Problem 16 | Diagram
The diagram represents each unit square of the given $2020 \times 2020$ square.
Solution 1
We consider an individual one-by-one block.
If we draw a quarter of a circle from each corner (where the lattice points are located), each with radius $d$, the area covered by the circles should be $0.5$. Because of t... | size(5cm); draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); filldraw((arc((0,0), 0.3989, 0, 90))--(0,0)--cycle, gray); draw(arc((1,0), 0.3989, 90, 180)); filldraw((arc((1,0), 0.3989, 90, 180))--(1,0)--cycle, gray); draw(arc((1,1), 0.3989, 180, 270)); filldraw((arc((1,1), 0.3989, 180, 270))--(1,1)--cycle, gray); draw(arc((0,1),... | [] |
865 | A point is chosen at random within the square in the coordinate plane whose vertices are $(0, 0), (2020, 0), (2020, 2020),$ and $(0, 2020)$. The probability that the point is within $d$ units of a lattice point is $\tfrac{1}{2}$. (A point $(x, y)$ is a lattice point if $x$ and $y$ are both integers.) What is $d$ to the... | 2020 AMC 12A Problem 16 | PLEASE NOTE: Solution 1 IS rigorous. Say there are $k$ unit squares (it doesn't matter how many). There is a $\frac{1}k$ probability the point is in some unit square. There is a $\pi d^2$ probability the point is in the shaded region. So, there is a $\frac{1}k \pi d^2 \cdot k = \pi d^2$ probability the point is in any ... | // Block 1
size(10cm); draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); filldraw((arc((0,0), 0.3989, 0, 90))--(0,0)--cycle, gray); draw(arc((1,0), 0.3989, 90, 180)); filldraw((arc((1,0), 0.3989, 90, 180))--(1,0)--cycle, gray); draw(arc((1,1), 0.3989, 180, 270)); filldraw((arc((1,1), 0.3989, 180, 270))--(1,1)--cycle, gray); dra... | [] |
866 | Quadrilateral $ABCD$ satisfies $\angle ABC = \angle ACD = 90^{\circ}, AC=20,$ and $CD=30.$ Diagonals $\overline{AC}$ and $\overline{BD}$ intersect at point $E,$ and $AE=5.$ What is the area of quadrilateral $ABCD?$
$\textbf{(A) } 330 \qquad \textbf{(B) } 340 \qquad \textbf{(C) } 350 \qquad \textbf{(D) } 360 \qquad \te... | 2020 AMC 12A Problem 18 | It's crucial to draw a good diagram for this one. Since $AC=20$ and $CD=30$, we get $[ACD]=300$. Now we need to find $[ABC]$ to get the area of the whole quadrilateral. Drop an altitude from $B$ to $AC$ and call the point of intersection $F$. Let $FE=x$. Since $AE=5$, then $AF=5-x$.
By dropping this altitude, we can ... | size(15cm,0); import olympiad; draw((0,0)--(0,2)--(6,4)--(4,0)--cycle); label("A", (0,2), NW); label("B", (0,0), SW); label("C", (4,0), SE); label("D", (6,4), NE); label("E", (1.714,1.143), N); label("F", (1,1.5), N); draw((0,2)--(4,0), dashed); draw((0,0)--(6,4), dashed); draw((0,0)--(1,1.5), dashed); label("20", (0,2... | [] |
866 | Quadrilateral $ABCD$ satisfies $\angle ABC = \angle ACD = 90^{\circ}, AC=20,$ and $CD=30.$ Diagonals $\overline{AC}$ and $\overline{BD}$ intersect at point $E,$ and $AE=5.$ What is the area of quadrilateral $ABCD?$
$\textbf{(A) } 330 \qquad \textbf{(B) } 340 \qquad \textbf{(C) } 350 \qquad \textbf{(D) } 360 \qquad \te... | 2020 AMC 12A Problem 18 | Let the points be $A(-10,0)$, $\:B(x,y)$, $\:C(10,0)$, $\:D(10,30)$,and $\:E(-5,0)$, respectively. Since $B$ lies on line $DE$, we know that $y=2x+10$. Furthermore, since $\angle{ABC}=90^\circ$, $B$ lies on the circle with diameter $AC$, so $x^2+y^2=100$. Solving for $x$ and $y$ with these equations, we get the solutio... | size(10cm,0); draw((10,30)--(10,0)--(-8,-6)--(-10,0)--(10,30)); draw((-20,0)--(20,0)); draw((0,-15)--(0,35)); draw((10,30)--(-8,-6)); draw(circle((0,0),10)); label("E",(-4.05,-.25),S); label("D",(10,30),NE); label("C",(10,0),NE); label("B",(-8,-6),SW); label("A",(-10,0),NW); label("5",(-10,0)--(-5,0), NE); label("15",(... | [] |
866 | Quadrilateral $ABCD$ satisfies $\angle ABC = \angle ACD = 90^{\circ}, AC=20,$ and $CD=30.$ Diagonals $\overline{AC}$ and $\overline{BD}$ intersect at point $E,$ and $AE=5.$ What is the area of quadrilateral $ABCD?$
$\textbf{(A) } 330 \qquad \textbf{(B) } 340 \qquad \textbf{(C) } 350 \qquad \textbf{(D) } 360 \qquad \te... | 2020 AMC 12A Problem 18 | Denote $EB$ as $x$. By the Law of Cosines:
\[AB^2 = 25 + x^2 - 10x\cos(\angle DEC)\]
\[BC^2 = 225 + x^2 + 30x\cos(\angle DEC)\]
Adding these up yields:
\[400 = 250 + 2x^2 + 20x\cos(\angle DEC) \Longrightarrow x^2 + \frac{10x}{\sqrt{5}} - 75 = 0\]
By the quadratic formula, $x = 3\sqrt5$.
Observe:
\[[AEB] + [BEC] = \fr... | import olympiad; pair A = (0, 189), B = (0,0), C = (570,0), D = (798, 798); dot("$A$", A, W); dot("$B$", B, S); dot("$C$", C, E); dot("$D$", D, N);dot("$E$",(140, 140), N); draw(A--B--C--D--A); draw(A--C, dotted); draw(B--D, dotted); | [] |
866 | Quadrilateral $ABCD$ satisfies $\angle ABC = \angle ACD = 90^{\circ}, AC=20,$ and $CD=30.$ Diagonals $\overline{AC}$ and $\overline{BD}$ intersect at point $E,$ and $AE=5.$ What is the area of quadrilateral $ABCD?$
$\textbf{(A) } 330 \qquad \textbf{(B) } 340 \qquad \textbf{(C) } 350 \qquad \textbf{(D) } 360 \qquad \te... | 2020 AMC 12A Problem 18 | Let $F$ be the midpoint of $AC$, and draw $FG // CD$ where $G$ is on $BD$. We have $EF=5,FC=10$.
$\Delta EFG \sim \Delta ECD \implies FG=10=FA=FC$. Therefore $ABCG$ is a cyclic quadrilateral.
Notice that $\angle EFG=90^\circ, EG=\sqrt{5^2+10^2}=5\sqrt{5} \implies BE=\frac{AE\cdot EC}{EG}=\frac{5\cdot 15}{5\sqrt{5}}=3... | import olympiad; pair A = (0, 189), B = (0,0), C = (570,0), D = (798, 798),F=(285,94.5),G=(361.2,361.2); dot("$A$", A, W); dot("$B$", B, S); dot("$C$", C, E); dot("$D$", D, N);dot("$E$",(140, 140), N);dot("$F$",F,N);dot("$G$",G,N); draw(A--B--C--D--A); draw(A--C, dotted); draw(B--D, dotted); draw(F--G, dotted); | [] |
866 | Quadrilateral $ABCD$ satisfies $\angle ABC = \angle ACD = 90^{\circ}, AC=20,$ and $CD=30.$ Diagonals $\overline{AC}$ and $\overline{BD}$ intersect at point $E,$ and $AE=5.$ What is the area of quadrilateral $ABCD?$
$\textbf{(A) } 330 \qquad \textbf{(B) } 340 \qquad \textbf{(C) } 350 \qquad \textbf{(D) } 360 \qquad \te... | 2020 AMC 12A Problem 18 | Let $AB = x$, $BC = y$
Looking at the diagram we have $x^2 + y^2 = 20^2$,
$DE = \sqrt{30^2+15^2} = 15\sqrt{5}$, $[ACD] = \frac{1}{2} \cdot 20 \cdot 30 = 300$
Because $\triangle CEF \sim \triangle CAB$, $EF = AB \cdot \frac{CE}{CA} = x \cdot \frac{15}{20} = \frac{3x}{4}$
$BF = BC - CF = BC - BC \cdot \frac{CE}{CA} = ... | size(15cm,0); import olympiad; draw((0,0)--(0,2)--(6,4)--(4,0)--cycle); label("A", (0,2), NW); label("B", (0,0), SW); label("C", (4,0), SE); label("D", (6,4), NE); label("E", (1.714,1.143), N); label("F", (1.714,0), SE); draw((0,2)--(4,0), dashed); draw((0,0)--(6,4), dashed); draw((1.714,1.143)--(1.714,0), dashed); lab... | [] |
866 | Quadrilateral $ABCD$ satisfies $\angle ABC = \angle ACD = 90^{\circ}, AC=20,$ and $CD=30.$ Diagonals $\overline{AC}$ and $\overline{BD}$ intersect at point $E,$ and $AE=5.$ What is the area of quadrilateral $ABCD?$
$\textbf{(A) } 330 \qquad \textbf{(B) } 340 \qquad \textbf{(C) } 350 \qquad \textbf{(D) } 360 \qquad \te... | 2020 AMC 12A Problem 18 | set \[\angle ACB = \theta , C= \cos(\theta), S = \sin(\theta)\]
\[\dfrac{E_y}{E_x} = \dfrac{30C} { 20C+30S} = \dfrac{15S} {20C-15C}\]
\[2SC = \dfrac35\]
\begin{align*} [ABCD] &= [ABD]+[CBD] \\ &= \dfrac12\cdot 20C\cdot 30C + \dfrac12 \cdot 20S (20C+30S) \\ &= 100\cdot 2SC + 300 \\ &= \boxed{\text{(D) } 360}. ... | size(15cm,0); import olympiad; draw((0,0)--(0,2)--(6,4)--(4,0)--cycle); label("A", (0,2), NW); label("B", (0,0), SW); label("C", (4,0), SE); label("D", (6,4), NE); label("E", (1.714,1.143), N); label("F", (1.714,0), SE); draw((0,2)--(4,0), dashed); draw((0,0)--(6,4), dashed); draw((4,0)--(6,0), dashed); draw((6,0)--(6,... | [] |
867 | Let $T$ be the triangle in the coordinate plane with vertices $(0,0), (4,0),$ and $(0,3).$ Consider the following five isometries (rigid transformations) of the plane: rotations of $90^{\circ}, 180^{\circ},$ and $270^{\circ}$ counterclockwise around the origin, reflection across the $x$-axis, and reflection across the ... | 2020 AMC 12A Problem 20 | First, any combination of motions we can make must reflect $T$ an even number of times. This is because every time we reflect $T$, it changes orientation. Once $T$ has been flipped once, no combination of rotations will put it back in place because it is the mirror image; however, flipping it again changes it back to t... | size(10cm); Label f; f.p=fontsize(6); xaxis(-6,6,Ticks(f, 2.0)); yaxis(-6,6,Ticks(f, 2.0)); filldraw(origin--(4,0)--(0,3)--cycle, gray, black+linewidth(1)); | [] |
868 | Suppose that $\triangle{ABC}$ is an equilateral triangle of side length $s$, with the property that there is a unique point $P$ inside the triangle such that $AP=1$, $BP=\sqrt{3}$, and $CP=2$. What is $s$?
$\textbf{(A) } 1+\sqrt{2} \qquad \textbf{(B) } \sqrt{7} \qquad \textbf{(C) } \frac{8}{3} \qquad \textbf{(D) } \sq... | 2020 AMC 12A Problem 24 | Solution 1(a)
We begin by rotating $\triangle{ APB}$ counterclockwise by $60^{\circ}$ about $A$, such that $P\mapsto Q$ and $B\mapsto C$. We see that $\triangle{ APQ}$ is equilateral with side length $1$, meaning that $\angle APQ = 60^{\circ}$. We also see that $\triangle{CPQ}$ is a $30$-$60$-$90$ right triangle, meani... | // Block 1
size(200); pen p = fontsize(10pt)+gray+0.5; pen q = fontsize(13pt); pair A,B,C,D,P,Q; real s=sqrt(7); B=origin; A=s*dir(60); C=s*right; P=IP(CR(A,1),CR(C,2)); Q=rotate(60,A)*P; draw(A--B--C--A, black+0.8); draw(A--P--B^^P--C^^A--Q--C, p); draw(P--Q, p+dashed); //draw(A--A+C--C, p); label("$A$", A, up, q); l... | [] |
868 | Suppose that $\triangle{ABC}$ is an equilateral triangle of side length $s$, with the property that there is a unique point $P$ inside the triangle such that $AP=1$, $BP=\sqrt{3}$, and $CP=2$. What is $s$?
$\textbf{(A) } 1+\sqrt{2} \qquad \textbf{(B) } \sqrt{7} \qquad \textbf{(C) } \frac{8}{3} \qquad \textbf{(D) } \sq... | 2020 AMC 12A Problem 24 | Suppose that triangle $ABC$ had three segments of length $2$, emanating from each of its vertices, making equal angles with each of its sides, and going into its interior. Suppose each of these segments intersected the segment clockwise to it precisely at its other endpoint and inside $ABC$ (as pictured in the diagram ... | // Block 1
unitsize(1inch);
pen p = fontsize(10pt);
dot((0.756,0.655));
dot((1.512,1.309));
dot((1.701,0.327));
pair A = origin, B = (1.323,2.291), C = (2.646,0), P = (0.756,0.655), Q = (1.512,1.309), R = (1.701,0.327);
draw((0,0)--(1.323,2.291)--(2.646,0)--cycle);
label("$A$", (0,0), SW, p);
label("$C$", (2.646,0), ... | [] |
868 | Suppose that $\triangle{ABC}$ is an equilateral triangle of side length $s$, with the property that there is a unique point $P$ inside the triangle such that $AP=1$, $BP=\sqrt{3}$, and $CP=2$. What is $s$?
$\textbf{(A) } 1+\sqrt{2} \qquad \textbf{(B) } \sqrt{7} \qquad \textbf{(C) } \frac{8}{3} \qquad \textbf{(D) } \sq... | 2020 AMC 12A Problem 24 | We begin by dropping altitudes from point $P$ down to all three sides of the triangle as shown above. We can therefore make equations regarding the areas of triangles $\triangle{APC}$, $\triangle{APB}$, and $\triangle{BPC}$. Let $s$ be the side of the equilateral triangle, we use the Heron's formula:
\[\triangle{APC}... | // Block 1
unitsize(0.4inch);
pen p = fontsize(10pt);
draw((0,0)--(4,5.65)--(8,0)--cycle);
label("$A$", (4,5.65), N, p);
label("$C$", (8,0), SE, p);
label("$B$", (0,0), SW, p);
label("$P$", (3.5,3.5), E, p);
label("$E$", (2.8191,3.982), NW, p);
label("$F$", (4.848,4.452), NE, p);
label("$G$", (3.5,0), down, p);
draw((... | [] |
868 | Suppose that $\triangle{ABC}$ is an equilateral triangle of side length $s$, with the property that there is a unique point $P$ inside the triangle such that $AP=1$, $BP=\sqrt{3}$, and $CP=2$. What is $s$?
$\textbf{(A) } 1+\sqrt{2} \qquad \textbf{(B) } \sqrt{7} \qquad \textbf{(C) } \frac{8}{3} \qquad \textbf{(D) } \sq... | 2020 AMC 12A Problem 24 | Instead of directly finding the side length of the equilateral triangle, we instead find the area and use it to find the side length.
Begin by reflecting $P$ over each of the sides. Label these reflected points $P', P'', P'''$. Connect these points to the vertices of the equilateral triangle, as well as to each other.
... | // Block 1
size(300);
draw((0,3.5)--(4,9.15)--(8,3.5)--cycle);
label("$A$", (4,9.15), N, p = fontsize(10pt));
label("$C$", (8,3.5), SE, p = fontsize(10pt));
label("$B$", (0,3.5), SW, p = fontsize(10pt));
label("$P$", (3.5,7), NW, p = fontsize(10pt));
draw((0,3.5)--(3.5,7));
draw((8,3.5)--(3.5,7));
draw((4,9.15)--(3.5,7... | [] |
869 | Points $P$ and $Q$ lie in a plane with $PQ=8$. How many locations for point $R$ in this plane are there such that the triangle with vertices $P$, $Q$, and $R$ is a right triangle with area $12$ square units?
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 8 \qquad\textbf{(E)}\ 12$ | 2020 AMC 10B Problem 8 | Let the brackets denote areas. We are given that \[[PQR]=\frac12\cdot PQ\cdot h_R=12.\] Since $PQ=8,$ it follows that $h_R=3.$
We construct a circle with diameter $\overline{PQ}.$ All such locations for $R$ are shown below:
We apply casework to the right angle of $\triangle PQR:$
If $\angle P=90^\circ,$ then $R\i... | // Block 1
/* Made by MRENTHUSIASM */
size(250);
pair O, P, Q, R1, R2, R3, R4, R5, R6, R7, R8, I1, I2;
O = (0,0);
P = (-4,0);
Q = (4,0);
R1 = (-4,3);
R4 = (4,3);
R5 = (-4,-3);
R8 = (4,-3);
path C;
C = Circle(O,4);
R3 = intersectionpoints(C,R1--R4)[0];
R2 = intersectionpoints(C,R1--R4)[1];
R6 = intersectionpoints(C,R5--... | [] |
870 | As shown in the figure below, six semicircles lie in the interior of a regular hexagon with side length 2 so that the diameters of the semicircles coincide with the sides of the hexagon. What is the area of the shaded region ---- inside the hexagon but outside all of the semicircles?
$\textbf{(A) } 6\sqrt3 - 3\pi \qqu... | 2020 AMC 10B Problem 14 | Let point $A$ be a vertex of the regular hexagon, let point $B$ be the midpoint of the line connecting point $A$ and a neighboring vertex, and let point $C$ be the second intersection of the two semicircles that pass through point $A$. Then, $BC = 1$, since $B$ is the center of the semicircle with radius $1$ that $C$ l... | // Block 1
real x=sqrt(3);
real y=2sqrt(3);
real z=3.5;
real a=x/2;
real b=0.5;
real c=3a;
pair A, B, C, D, E, F;
A = (1,0);
B = (3,0);
C = (4,x);
D = (3,y);
E = (1,y);
F = (0,x);
fill(A--B--C--D--E--F--A--cycle,grey);
fill(arc((2,0),1,0,180)--cycle,white);
fill(arc((2,y),1,180,360)--cycle,white);
fill(arc((z,a),1,60,... | [] |
870 | As shown in the figure below, six semicircles lie in the interior of a regular hexagon with side length 2 so that the diameters of the semicircles coincide with the sides of the hexagon. What is the area of the shaded region ---- inside the hexagon but outside all of the semicircles?
$\textbf{(A) } 6\sqrt3 - 3\pi \qqu... | 2020 AMC 10B Problem 14 | First, subdivide the hexagon into 24 equilateral triangles with side length 1:
Now note that the entire shaded region is just 6 times this part:
The entire rhombus is just 2 equilateral triangles with side lengths of 1, so it has an area of:
\[2\cdot\frac{\sqrt{3}}{4}=\frac{\sqrt{3}}{2}\]
The sector that is not inclu... | // Block 1
size(140); fill((1,0)--(3,0)--(4,sqrt(3))--(3,2sqrt(3))--(1,2sqrt(3))--(0,sqrt(3))--cycle,gray(0.4)); fill(arc((2,0),1,180,0)--(2,0)--cycle,white); fill(arc((3.5,sqrt(3)/2),1,60,240)--(3.5,sqrt(3)/2)--cycle,white); fill(arc((3.5,3sqrt(3)/2),1,120,300)--(3.5,3sqrt(3)/2)--cycle,white); fill(arc((2,2sqrt(3)),1,... | [] |
870 | As shown in the figure below, six semicircles lie in the interior of a regular hexagon with side length 2 so that the diameters of the semicircles coincide with the sides of the hexagon. What is the area of the shaded region ---- inside the hexagon but outside all of the semicircles?
$\textbf{(A) } 6\sqrt3 - 3\pi \qqu... | 2020 AMC 10B Problem 14 | We can see there are six congruent "leaves" on the original figure, and let us say the total area of them is $S$. If we make a circle with radius one in the circle (the dashed one), we can get another six congruent "leaves", and they are the same as the old six ones. So the area of the six new leaves can also be repres... | // Block 1
size(140);
fill((1,0)--(3,0)--(4,sqrt(3))--(3,2sqrt(3))--(1,2sqrt(3))--(0,sqrt(3))--cycle,gray(0.4));
fill(arc((2,0),1,180,0)--(2,0)--cycle,white);
fill(arc((3.5,sqrt(3)/2),1,60,240)--(3.5,sqrt(3)/2)--cycle,white);
fill(arc((3.5,3sqrt(3)/2),1,120,300)--(3.5,3sqrt(3)/2)--cycle,white);
fill(arc((2,2sqrt(3)),1,... | [] |
870 | As shown in the figure below, six semicircles lie in the interior of a regular hexagon with side length 2 so that the diameters of the semicircles coincide with the sides of the hexagon. What is the area of the shaded region ---- inside the hexagon but outside all of the semicircles?
$\textbf{(A) } 6\sqrt3 - 3\pi \qqu... | 2020 AMC 10B Problem 14 | We can set several variables.
Let $x$ be the area of the shaded region, and let $y$ be the area of one of the 'leaf-shaped' regions.
We then can connect the vertices of $x$, and we end up with another hexagon, with side length $1$. Since each of the regions that are outside of $x$ but inside of the smaller hexagon ... | // Block 1
size(140);
fill((1,0)--(3,0)--(4,sqrt(3))--(3,2sqrt(3))--(1,2sqrt(3))--(0,sqrt(3))--cycle,gray(0.4));
fill(arc((2,0),1,180,0)--(2,0)--cycle,white);
fill(arc((3.5,sqrt(3)/2),1,60,240)--(3.5,sqrt(3)/2)--cycle,white);
fill(arc((3.5,3sqrt(3)/2),1,120,300)--(3.5,3sqrt(3)/2)--cycle,white);
fill(arc((2,2sqrt(3)),1,... | [] |
870 | As shown in the figure below, six semicircles lie in the interior of a regular hexagon with side length 2 so that the diameters of the semicircles coincide with the sides of the hexagon. What is the area of the shaded region ---- inside the hexagon but outside all of the semicircles?
$\textbf{(A) } 6\sqrt3 - 3\pi \qqu... | 2020 AMC 10B Problem 14 | First we connect every vertex of the shaded portion to another, forming an equilateral hexagon. Each interior angle of this hexagon is $120^{\circ},$ the interior angles of the new triangle are all $60^{\circ}$. As the radius of the semicircle is $1$, we know that this hexagon has side length $1$, so its area is given ... | // Block 1
size(140);
fill((1,0)--(3,0)--(4,sqrt(3))--(3,2sqrt(3))--(1,2sqrt(3))--(0,sqrt(3))--cycle,gray(0.4));
fill(arc((2,0),1,180,0)--(2,0)--cycle,white);
fill(arc((3.5,sqrt(3)/2),1,60,240)--(3.5,sqrt(3)/2)--cycle,white);
fill(arc((3.5,3sqrt(3)/2),1,120,300)--(3.5,3sqrt(3)/2)--cycle,white);
fill(arc((2,2sqrt(3)),1,... | [] |
871 | Steve wrote the digits $1$, $2$, $3$, $4$, and $5$ in order repeatedly from left to right, forming a list of $10,000$ digits, beginning $123451234512\ldots.$ He then erased every third digit from his list (that is, the $3$rd, $6$th, $9$th, $\ldots$ digits from the left), then erased every fourth digit from the resultin... | 2020 AMC 10B Problem 15 | Note that cycles exist initially and after each round of erasing.
We will consider one cycle after all three rounds of erasing. Suppose this cycle has length $L$ before any round of erasing. It follows that:
Initially, one cycle has length $5,$ from which $L$ must be divisible by $5.$
After the first round of erasin... | // Block 1
/* Made by MRENTHUSIASM */
size(20cm);
for (real i=2.5; i<30; i+=3) {
fill((i-0.5,0)--(i-0.5,4)--(i+0.5,4)--(i+0.5,0)--cycle,red);
}
for (real i=4.5; i<30; i+=6) {
fill((i-0.5,0)--(i-0.5,4)--(i+0.5,4)--(i+0.5,0)--cycle,yellow);
}
fill((7,0)--(7,4)--(8,4)--(8,0)--cycle,green);
fill((18,0)--(18,4)--(1... | [] |
872 | In square $ABCD$, points $E$ and $H$ lie on $\overline{AB}$ and $\overline{DA}$, respectively, so that $AE=AH.$ Points $F$ and $G$ lie on $\overline{BC}$ and $\overline{CD}$, respectively, and points $I$ and $J$ lie on $\overline{EH}$ so that $\overline{FI} \perp \overline{EH}$ and $\overline{GJ} \perp \overline{EH}$. ... | 2020 AMC 10B Problem 21 | Easily, we can find that: quadrilateral $BFIE$ and $DHJG$ are congruent with each other, so we can move $DHJG$ to the shaded area ($F$ and $G$, $B$ and $D$ overlapping) to form a square $FIKJ'$ ($DG$ = $FB$, $CG$ = $FC$, ${\angle} CGF$ = ${\angle}CFG$ = $45^{\circ}$ so ${\angle} IFJ'= 90^{\circ}$). Then we can solve $A... | // Block 1
real x=2sqrt(2);
real y=2sqrt(16-8sqrt(2))-4+2sqrt(2);
real z=2sqrt(8-4sqrt(2));
real k= 8-2sqrt(2);
real l= 2sqrt(2)-4;
pair A, B, C, D, E, F, G, H, I, J, L, M, K;
A = (0,0);
B = (4,0);
C = (4,4);
D = (0,4);
E = (x,0);
F = (4,y);
G = (y,4);
H = (0,x);
I = F + z * dir(225);
L = (k,0);
M = F + z * dir(315);
K... | [] |
872 | In square $ABCD$, points $E$ and $H$ lie on $\overline{AB}$ and $\overline{DA}$, respectively, so that $AE=AH.$ Points $F$ and $G$ lie on $\overline{BC}$ and $\overline{CD}$, respectively, and points $I$ and $J$ lie on $\overline{EH}$ so that $\overline{FI} \perp \overline{EH}$ and $\overline{GJ} \perp \overline{EH}$. ... | 2020 AMC 10B Problem 21 | $[ABCD] = 4$, $AB = 2$, $[AHE] = 1$, $AH = AE = \sqrt{2}$, $DH = 2 - \sqrt{2}$, $JL = HK = \sqrt{2} \cdot DH = 2 \sqrt{2} - 2$
Because $ABCD$ is a square and $AH = AE$, $AC$ is the line of symmetry of pentagon $CDHEB$. Because $[DHJG] = [BFIE]$, $DHJG$ is the reflection of $BFIE$ about line $AC$
Let $FI = GJ = x$, $K... | // Block 1
real x=2sqrt(2);
real y=2sqrt(16-8sqrt(2))-4+2sqrt(2);
real z=2sqrt(8-4sqrt(2));
pair A, B, C, D, E, F, G, H, I, J, K, L;
A = (0,0);
B = (4,0);
C = (4,4);
D = (0,4);
E = (x,0);
F = (4,y);
G = (y,4);
H = (0,x);
I = F + z * dir(225);
J = G + z * dir(225);
K = (4-x,4);
L = J + 1.68 * dir(45);
draw(A--B--C--D--... | [] |
873 | Square $ABCD$ in the coordinate plane has vertices at the points $A(1,1), B(-1,1), C(-1,-1),$ and $D(1,-1).$ Consider the following four transformations:
$\quad\bullet\qquad$ $L,$ a rotation of $90^{\circ}$ counterclockwise around the origin;
$\quad\bullet\qquad$ $R,$ a rotation of $90^{\circ}$ clockwise around the o... | 2020 AMC 10B Problem 23 | For each transformation:
Each labeled vertex will move to an adjacent position.
The labeled vertices will maintain the consecutive order $ABCD$ in either direction (clockwise or counterclockwise).
$L$ and $R$ will retain the direction of the labeled vertices, but $H$ and $V$ will alter the direction of the labeled ve... | // Block 1
/* Made by MRENTHUSIASM */
unitsize(7mm);
label("$A$",(1,0));
label("$B$",(1,1));
label("$C$",(0,1));
label("$D$",(0,0));
label("$C$",(5,0));
label("$D$",(5,1));
label("$A$",(4,1));
label("$B$",(4,0));
label("$A$",(9,0));
label("$D$",(9,1));
label("$C$",(8,1));
label("$B$",(8,0));
label("$C$",(13,0));
lab... | [] |
874 | Let $D(n)$ denote the number of ways of writing the positive integer $n$ as a product\[n = f_1\cdot f_2\cdots f_k,\]where $k\ge1$, the $f_i$ are integers strictly greater than $1$, and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are counte... | 2020 AMC 10B Problem 25 | This question is like 2010 AMC 8 problem 25, if we treat each prime factor of $96 = 2^5 \cdot 3$ as a step, this is a recursive problem of climbing stairs.
The stair can be designed like
In this grid, we can only take any number of steps to the right or up, so the number of each point is the sum of all numbers to it... | // Block 1
size(8cm); defaultpen(fontsize(10pt)); draw((0, 1) -- (5, 1) ^^ (1, 0) -- (1, 1) ^^ (2, 0) -- (2, 1) ^^ (3, 0) -- (3, 1) ^^ (4, 0) -- (4, 1) ^^ (5, 0) -- (5, 1)); draw((0, 2) -- (0, 0) -- (6, 0), Arrows); label("2", (6, 0), E); label("3", (0, 2), N); label("start", (0, 0), SW); label("end", (5, 1), NE);
// B... | [] |
875 | In unit square $ABCD,$ the inscribed circle $\omega$ intersects $\overline{CD}$ at $M,$ and $\overline{AM}$ intersects $\omega$ at a point $P$ different from $M.$ What is $AP?$
$\textbf{(A) } \frac{\sqrt5}{12} \qquad \textbf{(B) } \frac{\sqrt5}{10} \qquad \textbf{(C) } \frac{\sqrt5}{9} \qquad \textbf{(D) } \frac{\sqrt... | 2020 AMC 12B Problem 10 | Call the midpoint of $\overline{AB}$ point $N.$ Draw in $\overline{NM}$ and $\overline{NP}.$ Note that $\angle{NPM}=90^{\circ}$ due to Thales's Theorem.
Using the Pythagorean theorem, $AM=\frac{\sqrt{5}}{2}.$ Now we just need to find $AP$ using similar triangles $\triangle APN\sim\triangle ANM:$
\begin{align*} \frac{A... | // Block 1
/* Made by QIDb602; edited by MRENTHUSIASM */
size(180);
pair A, B, C, D, M, N, O, P;
O = origin;
A = (-1/2,-1/2);
B = (-1/2,1/2);
C = (1/2,1/2);
D = (1/2,-1/2);
M = midpoint(C--D);
N = midpoint(A--B);
path p;
p = Circle(O,1/2);
P = intersectionpoints(A--M,p)[0];
fill(N--A--M--cycle,yellow);
dot("$\omega$",O... | [] |
875 | In unit square $ABCD,$ the inscribed circle $\omega$ intersects $\overline{CD}$ at $M,$ and $\overline{AM}$ intersects $\omega$ at a point $P$ different from $M.$ What is $AP?$
$\textbf{(A) } \frac{\sqrt5}{12} \qquad \textbf{(B) } \frac{\sqrt5}{10} \qquad \textbf{(C) } \frac{\sqrt5}{9} \qquad \textbf{(D) } \frac{\sqrt... | 2020 AMC 12B Problem 10 | Let $N$ be the midpoint of $\overline{AB},$ from which $\angle ANM=90^\circ.$ Note that $\angle NPM=90^\circ$ by the Inscribed Angle Theorem.
We have the following diagram:
Since $AN=\frac12$ and $NM=1,$ we get $AM=\frac{\sqrt5}{2}$ by the Pythagorean Theorem.
Let $AP=x.$ It follows that $PM=\frac{\sqrt5}{2}-x.$ Ap... | // Block 1
/* Made by MRENTHUSIASM */
size(180);
pair A, B, C, D, M, N, O, P;
O = origin;
A = (-1/2,-1/2);
B = (-1/2,1/2);
C = (1/2,1/2);
D = (1/2,-1/2);
M = midpoint(C--D);
N = midpoint(A--B);
path p;
p = Circle(O,1/2);
P = intersectionpoints(A--M,p)[0];
fill(N--P--A--cycle,yellow);
fill(N--P--M--cycle,green);
dot("$\... | [] |
876 | As shown in the figure below, six semicircles lie in the interior of a regular hexagon with side length 2 so that the diameters of the semicircles coincide with the sides of the hexagon. What is the area of the shaded region ---- inside the hexagon but outside all of the semicircles?
$\textbf{(A) } 6\sqrt3 - 3\pi \qqu... | 2020 AMC 12B Problem 11 | Let point $A$ be a vertex of the regular hexagon, let point $B$ be the midpoint of the line connecting point $A$ and a neighboring vertex, and let point $C$ be the second intersection of the two semicircles that pass through point $A$. Then, $BC = 1$, since $B$ is the center of the semicircle with radius $1$ that $C$ l... | // Block 1
real x=sqrt(3); real y=2sqrt(3); real z=3.5; real a=x/2; real b=0.5; real c=3a; pair A, B, C, D, E, F; A = (1,0); B = (3,0); C = (4,x); D = (3,y); E = (1,y); F = (0,x); fill(A--B--C--D--E--F--A--cycle,grey); fill(arc((2,0),1,0,180)--cycle,white); fill(arc((2,y),1,180,360)--cycle,white); fill(arc((z,a),1,60,... | [] |
876 | As shown in the figure below, six semicircles lie in the interior of a regular hexagon with side length 2 so that the diameters of the semicircles coincide with the sides of the hexagon. What is the area of the shaded region ---- inside the hexagon but outside all of the semicircles?
$\textbf{(A) } 6\sqrt3 - 3\pi \qqu... | 2020 AMC 12B Problem 11 | First, subdivide the hexagon into 24 equilateral triangles with side length 1:
Now note that the entire shaded region is just 6 times this part:
The entire rhombus is just 2 equilateral triangles with side lengths of 1, so it has an area of:
\[2\cdot\frac{\sqrt{3}}{4}=\frac{\sqrt{3}}{2}\]
The sector that is not inclu... | // Block 1
size(140); fill((1,0)--(3,0)--(4,sqrt(3))--(3,2sqrt(3))--(1,2sqrt(3))--(0,sqrt(3))--cycle,gray(0.4)); fill(arc((2,0),1,180,0)--(2,0)--cycle,white); fill(arc((3.5,sqrt(3)/2),1,60,240)--(3.5,sqrt(3)/2)--cycle,white); fill(arc((3.5,3sqrt(3)/2),1,120,300)--(3.5,3sqrt(3)/2)--cycle,white); fill(arc((2,2sqrt(3)),1,... | [] |
876 | As shown in the figure below, six semicircles lie in the interior of a regular hexagon with side length 2 so that the diameters of the semicircles coincide with the sides of the hexagon. What is the area of the shaded region ---- inside the hexagon but outside all of the semicircles?
$\textbf{(A) } 6\sqrt3 - 3\pi \qqu... | 2020 AMC 12B Problem 11 | We can see there are six congruent "leaves" on the original figure, and let us say the total area of them is $S$. If we make a circle with radius one in the circle (the dashed one), we can get another six congruent "leaves", and they are the same as the old six ones. So the area of the six new leaves can also be repres... | size(140); fill((1,0)--(3,0)--(4,sqrt(3))--(3,2sqrt(3))--(1,2sqrt(3))--(0,sqrt(3))--cycle,gray(0.4)); fill(arc((2,0),1,180,0)--(2,0)--cycle,white); fill(arc((3.5,sqrt(3)/2),1,60,240)--(3.5,sqrt(3)/2)--cycle,white); fill(arc((3.5,3sqrt(3)/2),1,120,300)--(3.5,3sqrt(3)/2)--cycle,white); fill(arc((2,2sqrt(3)),1,180,360)--(... | [] |
876 | As shown in the figure below, six semicircles lie in the interior of a regular hexagon with side length 2 so that the diameters of the semicircles coincide with the sides of the hexagon. What is the area of the shaded region ---- inside the hexagon but outside all of the semicircles?
$\textbf{(A) } 6\sqrt3 - 3\pi \qqu... | 2020 AMC 12B Problem 11 | We can set several variables.
Let $x$ be the area of the shaded region, and let $y$ be the area of one of the 'leaf-shaped' regions.
We then can connect the vertices of $x$, and we end up with another hexagon, with side length $1$. Since each of the regions that are outside of $x$ but inside of the smaller hexagon ... | size(140); fill((1,0)--(3,0)--(4,sqrt(3))--(3,2sqrt(3))--(1,2sqrt(3))--(0,sqrt(3))--cycle,gray(0.4)); fill(arc((2,0),1,180,0)--(2,0)--cycle,white); fill(arc((3.5,sqrt(3)/2),1,60,240)--(3.5,sqrt(3)/2)--cycle,white); fill(arc((3.5,3sqrt(3)/2),1,120,300)--(3.5,3sqrt(3)/2)--cycle,white); fill(arc((2,2sqrt(3)),1,180,360)--(... | [] |
876 | As shown in the figure below, six semicircles lie in the interior of a regular hexagon with side length 2 so that the diameters of the semicircles coincide with the sides of the hexagon. What is the area of the shaded region ---- inside the hexagon but outside all of the semicircles?
$\textbf{(A) } 6\sqrt3 - 3\pi \qqu... | 2020 AMC 12B Problem 11 | First we connect every vertex of the shaded portion to another, forming an equilateral hexagon. Each interior angle of this hexagon is $120^{\circ},$ the interior angles of the new triangle are all $60^{\circ}$. As the radius of the semicircle is $1$, we know that this hexagon has side length $1$, so its area is given ... | size(140); fill((1,0)--(3,0)--(4,sqrt(3))--(3,2sqrt(3))--(1,2sqrt(3))--(0,sqrt(3))--cycle,gray(0.4)); fill(arc((2,0),1,180,0)--(2,0)--cycle,white); fill(arc((3.5,sqrt(3)/2),1,60,240)--(3.5,sqrt(3)/2)--cycle,white); fill(arc((3.5,3sqrt(3)/2),1,120,300)--(3.5,3sqrt(3)/2)--cycle,white); fill(arc((2,2sqrt(3)),1,180,360)--(... | [] |
877 | Let $\overline{AB}$ be a diameter in a circle of radius $5\sqrt2.$ Let $\overline{CD}$ be a chord in the circle that intersects $\overline{AB}$ at a point $E$ such that $BE=2\sqrt5$ and $\angle AEC = 45^{\circ}.$ What is $CE^2+DE^2?$
$\textbf{(A)}\ 96 \qquad\textbf{(B)}\ 98 \qquad\textbf{(C)}\ 44\sqrt5 \qquad\textbf{... | 2020 AMC 12B Problem 12 | Let $O$ be the center of the circle. By reflecting $D$ across the line $AB$ to produce $D'$, we have that $\angle BED'=45$. Since $\angle AEC=45$, $\angle CED'=90$. Since $DE=ED'$, by the Pythagorean Theorem, our desired solution is just $CD'^2$.
Looking next to circle arcs, we know that $\angle AEC=\frac{\overarc{AC}+... | // Block 1
/* Made by sofas103; edited by MRENTHUSIASM */
size(250);
pair O, A, B, C, D, E, D1;
O = origin;
A = (-5*sqrt(2),0);
B = (5*sqrt(2),0);
E = (5*sqrt(2)-2*sqrt(5),0);
path p;
p = Circle(O,5*sqrt(2));
C = intersectionpoint(p,E--E+10*dir(135));
D = intersectionpoint(p,E--E+10*dir(-45));
D1 = (D.x,-D.y);
draw(p)... | [] |
877 | Let $\overline{AB}$ be a diameter in a circle of radius $5\sqrt2.$ Let $\overline{CD}$ be a chord in the circle that intersects $\overline{AB}$ at a point $E$ such that $BE=2\sqrt5$ and $\angle AEC = 45^{\circ}.$ What is $CE^2+DE^2?$
$\textbf{(A)}\ 96 \qquad\textbf{(B)}\ 98 \qquad\textbf{(C)}\ 44\sqrt5 \qquad\textbf{... | 2020 AMC 12B Problem 12 | For ease of notation, let $x = DE$ and $y=EC$.
Extend $\overline{CD}$ to point $P$ until $\overline{BP}$ is perpendicular to $AB$. It's given that $\angle AEC = 45^{\circ}$, so, by vertical angles, we have $\angle BEP = \angle EPB = 45^{\circ}$.
Since $PEB$ is a $45-45-90$ right triangle, we have $BE = BP = 2\sqrt{5... | // Block 1
/* Made by sofas103; edited by MRENTHUSIASM */
size(250);
pair O, A, B, C, D, E, P, Q;
O = origin;
A = (-5*sqrt(2),0);
B = (5*sqrt(2),0);
E = (5*sqrt(2)-2*sqrt(5),0);
P = (5*sqrt(2), -4.5);
Q=(5*sqrt(2)-2*sqrt(5)-0.5, 0);
path p;
p = Circle(O,5*sqrt(2));
C = intersectionpoint(p,E--E+10*dir(135));
D = inters... | [] |
878 | How many polynomials of the form $x^5 + ax^4 + bx^3 + cx^2 + dx + 2020$, where $a$, $b$, $c$, and $d$ are real numbers, have the property that whenever $r$ is a root, so is $\frac{-1+i\sqrt{3}}{2} \cdot r$? (Note that $i=\sqrt{-1}$)
$\textbf{(A) } 0 \qquad \textbf{(B) }1 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \... | 2020 AMC 12B Problem 17 | Due to Euler's Formula, $\frac{-1+i\sqrt3}2$ is a 120 degree rotation CCW; therefore, the problem stipulates that a root must be followed by a root rotated 120 degrees CCW, and that the rotated root must also have yet another root 240 degrees CCW. Therefore, all roots must come similar groups of $3$. We cannot have $6$... | // Block 1
import graph;
draw(Circle((0,0),10));
draw((0,0)--(-5,8.6602540378),red);
draw((0,0)--(10,0),red);
draw((0,0)--(-5,-8.6602540378),red);
// Block 2
import graph; draw(Circle((0,0),10)); draw((0,0)--(-5,8.6602540378),red); draw((0,0)--(10,0),red); draw((0,0)--(-5,-8.6602540378),red); | [] |
879 | In square $ABCD$, points $E$ and $H$ lie on $\overline{AB}$ and $\overline{DA}$, respectively, so that $AE=AH.$ Points $F$ and $G$ lie on $\overline{BC}$ and $\overline{CD}$, respectively, and points $I$ and $J$ lie on $\overline{EH}$ so that $\overline{FI} \perp \overline{EH}$ and $\overline{GJ} \perp \overline{EH}$. ... | 2020 AMC 12B Problem 18 | Easily, we can find that: quadrilateral $BFIE$ and $DHJG$ are congruent with each other, so we can move $DHJG$ to the shaded area ($F$ and $G$, $B$ and $D$ overlapping) to form a square $FIKJ'$ ($DG$ = $FB$, $CG$ = $FC$, ${\angle} CGF$ = ${\angle}CFG$ = $45^{\circ}$ so ${\angle} IFJ'= 90^{\circ}$). Then we can solve $A... | real x=2sqrt(2); real y=2sqrt(16-8sqrt(2))-4+2sqrt(2); real z=2sqrt(8-4sqrt(2)); real k= 8-2sqrt(2); real l= 2sqrt(2)-4; pair A, B, C, D, E, F, G, H, I, J, L, M, K; A = (0,0); B = (4,0); C = (4,4); D = (0,4); E = (x,0); F = (4,y); G = (y,4); H = (0,x); I = F + z * dir(225); L = (k,0); M = F + z * dir(315); K = (4,l); ... | [] |
879 | In square $ABCD$, points $E$ and $H$ lie on $\overline{AB}$ and $\overline{DA}$, respectively, so that $AE=AH.$ Points $F$ and $G$ lie on $\overline{BC}$ and $\overline{CD}$, respectively, and points $I$ and $J$ lie on $\overline{EH}$ so that $\overline{FI} \perp \overline{EH}$ and $\overline{GJ} \perp \overline{EH}$. ... | 2020 AMC 12B Problem 18 | $[ABCD] = 4$, $AB = 2$, $[AHE] = 1$, $AH = AE = \sqrt{2}$, $DH = 2 - \sqrt{2}$, $JL = HK = \sqrt{2} \cdot DH = 2 \sqrt{2} - 2$
Because $ABCD$ is a square and $AH = AE$, $AC$ is the line of symmetry of pentagon $CDHEB$. Because $[DHJG] = [BFIE]$, $DHJG$ is the reflection of $BFIE$ about line $AC$
Let $FI = GJ = x$, $K... | real x=2sqrt(2); real y=2sqrt(16-8sqrt(2))-4+2sqrt(2); real z=2sqrt(8-4sqrt(2)); pair A, B, C, D, E, F, G, H, I, J, K, L; A = (0,0); B = (4,0); C = (4,4); D = (0,4); E = (x,0); F = (4,y); G = (y,4); H = (0,x); I = F + z * dir(225); J = G + z * dir(225); K = (4-x,4); L = J + 1.68 * dir(45); draw(A--B--C--D--A); draw(H-... | [] |
880 | Square $ABCD$ in the coordinate plane has vertices at the points $A(1,1), B(-1,1), C(-1,-1),$ and $D(1,-1).$ Consider the following four transformations:
$\quad\bullet\qquad$ $L,$ a rotation of $90^{\circ}$ counterclockwise around the origin;
$\quad\bullet\qquad$ $R,$ a rotation of $90^{\circ}$ clockwise around the o... | 2020 AMC 12B Problem 19 | For each transformation:
Each labeled vertex will move to an adjacent position.
The labeled vertices will maintain the consecutive order $ABCD$ in either direction (clockwise or counterclockwise).
$L$ and $R$ will retain the direction of the labeled vertices, but $H$ and $V$ will alter the direction of the labeled ve... | /* Made by MRENTHUSIASM */ unitsize(7mm); label("$A$",(1,0)); label("$B$",(1,1)); label("$C$",(0,1)); label("$D$",(0,0)); label("$C$",(5,0)); label("$D$",(5,1)); label("$A$",(4,1)); label("$B$",(4,0)); label("$A$",(9,0)); label("$D$",(9,1)); label("$C$",(8,1)); label("$B$",(8,0)); label("$C$",(13,0)); label("$B$",(1... | [] |
881 | Let $D(n)$ denote the number of ways of writing the positive integer $n$ as a product\[n = f_1\cdot f_2\cdots f_k,\]where $k\ge1$, the $f_i$ are integers strictly greater than $1$, and the order in which the factors are listed matters (that is, two representations that differ only in the order of the factors are counte... | 2020 AMC 12B Problem 24 | This question is like 2010 AMC 8 problem 25, if we treat each prime factor of $96 = 2^5 \cdot 3$ as a step, this is a recursive problem of climbing stairs.
The stair can be designed like
In this grid, we can only take any number of steps to the right or up, so the number of each point is the sum of all numbers to it... | // Block 1
size(8cm); defaultpen(fontsize(10pt)); draw((0, 1) -- (5, 1) ^^ (1, 0) -- (1, 1) ^^ (2, 0) -- (2, 1) ^^ (3, 0) -- (3, 1) ^^ (4, 0) -- (4, 1) ^^ (5, 0) -- (5, 1)); draw((0, 2) -- (0, 0) -- (6, 0), Arrows); label("2", (6, 0), E); label("3", (0, 2), N); label("start", (0, 0), SW); label("end", (5, 1), NE);
// B... | [] |
882 | For each real number $a$ with $0 \leq a \leq 1$, let numbers $x$ and $y$ be chosen independently at random from the intervals $[0, a]$ and $[0, 1]$, respectively, and let $P(a)$ be the probability that
\[\sin^2{(\pi x)} + \sin^2{(\pi y)} > 1\]
What is the maximum value of $P(a)?$
$\textbf{(A)}\ \frac{7}{12} \qquad\te... | 2020 AMC 12B Problem 25 | Let's start first by manipulating the given inequality.
\[\sin^{2}{(\pi x)}+\sin^{2}{(\pi y)}>1\]
\[\sin^{2}{(\pi x)}>1-\sin^{2}{(\pi y)}=\cos^{2}{(\pi y)}\]
Let's consider the boundary cases: $\sin{(\pi x)}=\pm \cos{(\pi y)}$.
\[\sin{(\pi x)}=\pm \cos{(\pi y)}=\sin{(\tfrac 12 {\pi}\pm \pi y)}\]
Solving the first c... | // Block 1
defaultpen(fontsize(9)+0.8); size(200);
pen p=fontsize(10);
pair A,B,C,D,A1,A2,B1,B2,C1,C2,D1,D2,I,L;
A=MP("(0,0)",origin,down+left,p); B=MP("(1,0)",right,down+right,p); C=MP("(1,1)",right+up,up+right,p); D=MP("(0,1)",up,up+left,p);
A1=MP("",extension(A,B,(0.5,0),(0,0.5)),2*down,p); dot(A1);
A2=MP("",extens... | [] |
883 | In the following list of numbers, the integer $n$ appears $n$ times in the list for $1 \leq n \leq 200$.\[1, 2, 2, 3, 3, 3, 4, 4, 4, 4, \ldots, 200, 200, \ldots , 200\]What is the median of the numbers in this list?
$\textbf{(A)} ~100.5 \qquad\textbf{(B)} ~134 \qquad\textbf{(C)} ~142 \qquad\textbf{(D)} ~150.5 \qquad\t... | 2021 AMC 10A Problem 16 | We can arrange the numbers in the following pattern:
\[ \begin{array}{cccccc} \ &\ &\ &\ &\ 200 & \\ \ &\ &\ &\ 199 & \ 200 & \\ \ &\ &\ \iddots& \ \vdots& \ \vdots& \\ \ &\ 2& \ \cdots& \ 199& \ 200& \\ 1 & \ 2 & \ \cdots& \ 199& \ 200& \end{array} \]
We can see this as a isosceles right triangle, with legs of length... | draw((0,0)--(200,200)--(200,0)--cycle); draw((142,0)--(142,142)); label("$x$",(142,0)--(142,142),E); label("$x$",(0,0)--(142,0),S); label("$200$",(200,0)--(200,200),E); | [] |
884 | The area of the region bounded by the graph of\[x^2+y^2 = 3|x-y| + 3|x+y|\]is $m+n\pi$, where $m$ and $n$ are integers. What is $m + n$?
$\textbf{(A)} ~18\qquad\textbf{(B)} ~27\qquad\textbf{(C)} ~36\qquad\textbf{(D)} ~45\qquad\textbf{(E)} ~54$ | 2021 AMC 10A Problem 19 | In order to attack this problem, we can use casework on the sign of $|x-y|$ and $|x+y|$.
Case 1: $|x-y|=x-y, |x+y|=x+y$
Substituting and simplifying, we have $x^2-6x+y^2=0$, i.e. $(x-3)^2+y^2=3^2$, which gives us a circle of radius $3$ centered at $(3,0)$.
Case 2: $|x-y|=y-x, |x+y|=x+y$
Substituting and simplifying... | // Block 1
size(10cm);
Label f;
f.p=fontsize(7);
xaxis(-8,8,Ticks(f, 1.0));
yaxis(-8,8,Ticks(f, 1.0));
draw(arc((-3,0),3,90,270) -- cycle, gray);
draw(arc((0,3),3,0,180) -- cycle, gray);
draw(arc((3,0),3,-90,90) -- cycle, gray);
draw(arc((0,-3),3,-180,0) -- cycle, gray);
draw((-3,3)--(3,3)--(3,-3)--(-3,-3)--cycle, ... | [] |
885 | In how many ways can the sequence $1,2,3,4,5$ be rearranged so that no three consecutive terms are increasing and no three consecutive terms are decreasing?
$\textbf{(A)} ~10\qquad\textbf{(B)} ~18\qquad\textbf{(C)} ~24 \qquad\textbf{(D)} ~32 \qquad\textbf{(E)} ~44$ | 2021 AMC 10A Problem 20 | By symmetry with respect to $3,$ note that $(x_1,x_2,x_3,x_4,x_5)$ is a valid sequence if and only if $(6-x_1,6-x_2,6-x_3,6-x_4,6-x_5)$ is a valid sequence. We enumerate the valid sequences that start with $1,2,31,$ or $32,$ as shown below:
There are $16$ valid sequences that start with $1,2,31,$ or $32.$ By symmetr... | // Block 1
/* Made by MRENTHUSIASM */
size(16cm);
draw((0.25,0)--(1.75,3),red,EndArrow);
draw((0.25,0)--(1.75,0),red,EndArrow);
draw((0.25,0)--(1.75,-3),red,EndArrow);
draw((2.25,3)--(3.75,3),red,EndArrow);
draw((2.25,0)--(3.75,0.75),red,EndArrow);
draw((2.25,0)--(3.75,-0.75),red,EndArrow);
draw((2.25,-3)--(3.75,-2.25... | [] |
886 | Frieda the frog begins a sequence of hops on a $3 \times 3$ grid of squares, moving one square on each hop and choosing at random the direction of each hop-up, down, left, or right. She does not hop diagonally. When the direction of a hop would take Frieda off the grid, she "wraps around" and jumps to the opposite edge... | 2021 AMC 10A Problem 23 | Imagine an infinite grid of $2$ by $2$ squares such that there is a $2$ by $2$ square centered at $(3x, 3y)$ for all ordered pairs of integers $(x, y).$
It is easy to see that the problem is equivalent to Frieda moving left, right, up, or down on this infinite grid starting at $(0, 0)$ (minus the teleportations). Si... | dot((0,0)); draw((-1,-1)--(-1,1)--(1,1)--(1,-1)--cycle); draw((-4,-1)--(-4,1)--(-2,1)--(-2,-1)--cycle); draw((2,-1)--(2,1)--(4,1)--(4,-1)--cycle); draw((-1,2)--(-1,4)--(1,4)--(1,2)--cycle); draw((-1,-4)--(-1,-2)--(1,-2)--(1,-4)--cycle); draw((-4,2)--(-2,2)--(-2,4)--(-4,4)--cycle); draw((4,-2)--(2,-2)--(2,-4)--(4,-4)--c... | [] |
886 | Frieda the frog begins a sequence of hops on a $3 \times 3$ grid of squares, moving one square on each hop and choosing at random the direction of each hop-up, down, left, or right. She does not hop diagonally. When the direction of a hop would take Frieda off the grid, she "wraps around" and jumps to the opposite edge... | 2021 AMC 10A Problem 23 | Let us go through the frog’s jumps step by step to find the probability that it will land in a corner. Denote the frog’s position as $F$:
First notice that since the frog only moves up, down, left or right, it’s original hop will not matter by symmetry, because all first hops will bring the frog at the edge next to tw... | // Block 1
label("$F$",(0,0)); draw((-1,-1)--(-1,1)--(1,1)--(1,-1)--cycle); draw((1,-1)--(1,1)--(3,1)--(3,-1)--cycle); draw((-1,1)--(-1,3)--(1,3)--(1,1)--cycle); draw((-3,-1)--(-3,1)--(-1,1)--(-1,-1)--cycle); draw((-1,-3)--(-1,-1)--(1,-1)--(1,-3)--cycle); draw((-3,-3)--(3,-3)--(3,3)--(-3,3)--cycle);
// Block 2
label("... | [] |
887 | How many ways are there to place $3$ indistinguishable red chips, $3$ indistinguishable blue chips, and $3$ indistinguishable green chips in the squares of a $3 \times 3$ grid so that no two chips of the same color are directly adjacent to each other, either vertically or horizontally?
$\textbf{(A)} ~12\qquad\textbf{(... | 2021 AMC 10A Problem 25 | Without the loss of generality, we fix the top-left square with a red chip. We apply casework to its two adjacent chips:
Case (1): The top-center and center-left chips have different colors.
There are three subcases for Case (1):
As there are $3!=6$ permutations of the three colors, each subcase has $6$ ways. So, Ca... | // Block 1
/* Made by MRENTHUSIASM */
unitsize(7mm);
fill((6,2)--(7,2)--(7,3)--(6,3)--cycle, red);
fill((7,2)--(8,2)--(8,3)--(7,3)--cycle, blue);
fill((6,1)--(7,1)--(7,2)--(6,2)--cycle, green);
fill((7,1)--(8,1)--(8,2)--(7,2)--cycle, red);
draw((6,0)--(9,0)--(9,3)--(6,3)--cycle, linewidth(1.5));
draw((6,1)--(9,1), line... | [] |
887 | How many ways are there to place $3$ indistinguishable red chips, $3$ indistinguishable blue chips, and $3$ indistinguishable green chips in the squares of a $3 \times 3$ grid so that no two chips of the same color are directly adjacent to each other, either vertically or horizontally?
$\textbf{(A)} ~12\qquad\textbf{(... | 2021 AMC 10A Problem 25 | We consider all possible configurations of the red chips for which rotations matter:
As there are $2!=2$ permutations of blue and green for each configuration, the answer is $2\cdot(2+4+4+4+4)=\boxed{\textbf{(E)} ~36}.$
~MRENTHUSIASM (credit given to FlameKhoEmberish) | // Block 1
/* Made by MRENTHUSIASM */
unitsize(7mm);
fill((0,2)--(1,2)--(1,3)--(0,3)--cycle, red);
fill((1,1)--(2,1)--(2,2)--(1,2)--cycle, red);
fill((2,0)--(3,0)--(3,1)--(2,1)--cycle, red);
fill((6,2)--(7,2)--(7,3)--(6,3)--cycle, red);
fill((8,2)--(9,2)--(9,3)--(8,3)--cycle, red);
fill((7,1)--(8,1)--(8,2)--(7,2)--cyc... | [] |
887 | How many ways are there to place $3$ indistinguishable red chips, $3$ indistinguishable blue chips, and $3$ indistinguishable green chips in the squares of a $3 \times 3$ grid so that no two chips of the same color are directly adjacent to each other, either vertically or horizontally?
$\textbf{(A)} ~12\qquad\textbf{(... | 2021 AMC 10A Problem 25 | We will first count the colorings where if the top two rows are filled out, the grid is uniquely determined. This will only happen if there are $3$ cells of one color, $2$ cells of another color, and $1$ cell of the remaining color in the top $3$ x $2$ grid made up of cells $1-6$. With this, we are left with two tokens... | // Block 1
size(50);
draw((0,0)--(3,0));
draw((0,1)--(3,1));
draw((0,2)--(3,2));
draw((0,3)--(3,3));
draw((0,0)--(0,3));
draw((1,0)--(1,3));
draw((2,0)--(2,3));
draw((3,0)--(3,3));
// Label cells (A1 at bottom-left)
label("7",(0.5,0.5));
label("8", (1.5,0.5));
label("9",(2.5,0.5));
label("4", (0.5,1.5));
label("5", (... | [] |
888 | Under what conditions is $\sqrt{a^2+b^2}=a+b$ true, where $a$ and $b$ are real numbers?
$\textbf{(A) }$ It is never true.
$\textbf{(B) }$ It is true if and only if $ab=0$.
$\textbf{(C) }$ It is true if and only if $a+b\ge 0$.
$\textbf{(D) }$ It is true if and only if $ab=0$ and $a+b\ge 0$.
$\textbf{(E) }$ It is al... | 2021 AMC 12A Problem 2 | If we graph $\sqrt{x^2+y^2}=x+y,$ then we get the union of:
positive $x$-axis
positive $y$-axis
origin
Therefore, the answer is $\boxed{\textbf{(D)}}.$
The graph of $\sqrt{x^2+y^2}=x+y$ is shown below.
~MRENTHUSIASM (credit given to TheAMCHub) | // Block 1
/* Made by MRENTHUSIASM */
size(200);
int xMin = -5;
int xMax = 5;
int yMin = -5;
int yMax = 5;
draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5));
draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5));
label("$x$",(xMax,0),(2,0));
label("$y$",(0,yMax),(0,2));
draw((xMax,0)--(0,0)--(0,yMax),red... | [] |
889 | A laser is placed at the point $(3,5)$. The laser beam travels in a straight line. Larry wants the beam to hit and bounce off the $y$-axis, then hit and bounce off the $x$-axis, then hit the point $(7,5)$. What is the total distance the beam will travel along this path?
$\textbf{(A) }2\sqrt{10} \qquad \textbf{(B) }5\s... | 2021 AMC 12A Problem 11 | Let $A=(3,5)$ and $D=(7,5).$ Suppose that the beam hits and bounces off the $y$-axis at $B,$ then hits and bounces off the $x$-axis at $C.$
When the beam hits and bounces off a coordinate axis, the angle of incidence and the angle of reflection are congruent. Therefore, we straighten up the path of the beam by reflect... | // Block 1
/* Made by MRENTHUSIASM */
size(225);
int xMin = -9;
int xMax = 9;
int yMin = -7;
int yMax = 7;
draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5));
draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5));
label("$x$",(xMax,0),(2,0));
label("$y$",(0,yMax),(0,2));
pair A = (3,5);
pair B = (0,2);
p... | [] |
889 | A laser is placed at the point $(3,5)$. The laser beam travels in a straight line. Larry wants the beam to hit and bounce off the $y$-axis, then hit and bounce off the $x$-axis, then hit the point $(7,5)$. What is the total distance the beam will travel along this path?
$\textbf{(A) }2\sqrt{10} \qquad \textbf{(B) }5\s... | 2021 AMC 12A Problem 11 | Define points $A,B,C,$ and $D$ as Solution 1 does. Moreover, let $E$ be a point on $\overline{CD}$ such that $\overline{BE}$ is perpendicular to the $y$-axis, and $F$ be a point on $\overline{BE}$ such that $\overline{CF}$ is perpendicular to the $x$-axis, as shown below.
When the beam hits and bounces off a coordinat... | // Block 1
/* Made by MRENTHUSIASM */
size(200);
int xMin = -3;
int xMax = 9;
int yMin = -3;
int yMax = 7;
draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5));
draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5));
label("$x$",(xMax,0),(2,0));
label("$y$",(0,yMax),(0,2));
pair A = (3,5);
pair B = (0,2);
p... | [] |
889 | A laser is placed at the point $(3,5)$. The laser beam travels in a straight line. Larry wants the beam to hit and bounce off the $y$-axis, then hit and bounce off the $x$-axis, then hit the point $(7,5)$. What is the total distance the beam will travel along this path?
$\textbf{(A) }2\sqrt{10} \qquad \textbf{(B) }5\s... | 2021 AMC 12A Problem 11 | Define points $A,B,C,$ and $D$ as Solution 1 does.
Since choices $\textbf{(B)}, \textbf{(C)},$ and $\textbf{(D)}$ all involve $\sqrt2,$ we suspect that one of them is the correct answer. We take a guess in faith that $\overline{AB},\overline{BC},$ and $\overline{CD}$ all form $45^\circ$ angles with the coordinate axes... | // Block 1
/* Made by MRENTHUSIASM */
size(200);
int xMin = -3;
int xMax = 9;
int yMin = -3;
int yMax = 7;
//Draws the horizontal gridlines
void horizontalLines()
{
for (int i = yMin+1; i < yMax; ++i)
{
draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4));
}
}
//Draws the vertical gridlines
void verticalLi... | [] |
890 | In the following list of numbers, the integer $n$ appears $n$ times in the list for $1 \leq n \leq 200$.\[1, 2, 2, 3, 3, 3, 4, 4, 4, 4, \ldots, 200, 200, \ldots , 200\]What is the median of the numbers in this list?
$\textbf{(A)} ~100.5 \qquad\textbf{(B)} ~134 \qquad\textbf{(C)} ~142 \qquad\textbf{(D)} ~150.5 \qquad\t... | 2021 AMC 12A Problem 16 | We can arrange the numbers in the following pattern:
\[ \begin{array}{cccccc} \ &\ &\ &\ &\ 200 & \\ \ &\ &\ &\ 199 & \ 200 & \\ \ &\ &\ \iddots& \ \vdots& \ \vdots& \\ \ &\ 2& \ \cdots& \ 199& \ 200& \\ 1 & \ 2 & \ \cdots& \ 199& \ 200& \end{array} \]
We can see this as a isosceles right triangle, with legs of length... | // Block 1
draw((0,0)--(200,200)--(200,0)--cycle);
draw((142,0)--(142,142));
label("$x$",(142,0)--(142,142),E);
label("$x$",(0,0)--(142,0),S);
label("$200$",(200,0)--(200,200),E);
// Block 2
draw((0,0)--(200,200)--(200,0)--cycle); draw((142,0)--(142,142)); label("$x$",(142,0)--(142,142),E); label("$x$",(0,0)--(142,0),S... | [] |
891 | How many solutions does the equation $\sin \left( \frac{\pi}2 \cos x\right)=\cos \left( \frac{\pi}2 \sin x\right)$ have in the closed interval $[0,\pi]$?
$\textbf{(A) }0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }2 \qquad \textbf{(D) }3\qquad \textbf{(E) }4$ | 2021 AMC 12A Problem 19 | This problem is equivalent to counting the intersections of the graphs of $y=\sin\left(\frac{\pi}{2}\cos x\right)$ and $y=\cos\left(\frac{\pi}{2}\sin x\right)$ in the closed interval $[0,\pi].$ We construct a table of values, as shown below:
\[\begin{array}{c|ccc} & & & \\ [-2ex] & \boldsymbol{x=0} & \boldsymbol{x=\fra... | // Block 1
/* Made by MRENTHUSIASM */
size(600,200);
real f(real x) { return sin(pi/2*cos(x)); }
real g(real x) { return cos(pi/2*sin(x)); }
draw(graph(f,0,pi),red,"$y=\sin\left(\frac{\pi}{2}\cos x\right)$");
draw(graph(g,0,pi),blue,"$y=\cos\left(\frac{\pi}{2}\sin x\right)$");
real xMin = 0;
real xMax = 5/4*pi;
re... | [] |
892 | Suppose that on a parabola with vertex $V$ and a focus $F$ there exists a point $A$ such that $AF=20$ and $AV=21$. What is the sum of all possible values of the length $FV?$
$\textbf{(A) }13 \qquad \textbf{(B) }\frac{40}3 \qquad \textbf{(C) }\frac{41}3 \qquad \textbf{(D) }14\qquad \textbf{(E) }\frac{43}3$ | 2021 AMC 12A Problem 20 | Let $\ell$ be the directrix of $\mathcal P$; recall that $\mathcal P$ is the set of points $T$ such that the distance from $T$ to $\ell$ is equal to $TF$. Let $P$ and $Q$ be the orthogonal projections of $F$ and $A$ onto $\ell$, and further let $X$ and $Y$ be the orthogonal projections of $F$ and $V$ onto $AQ$. Because... | // Block 1
import olympiad;
size(230);
defaultpen(linewidth(0.8)+fontsize(11pt));
real d = 1.1, edge = 2.5, Ax = 1.6;
real f(real x)
{
return 1/(4*d) * x * x;
}
pair V = origin, F = (0,d), la = (-edge,-d), lb = (edge,-d), A = (Ax, f(Ax));
pair P = foot(F,la,lb), Q = foot(A,la,lb), X = foot(F,A,Q), Y = foot(V,A,Q);
dra... | [] |
893 | The five solutions to the equation\[(z-1)(z^2+2z+4)(z^2+4z+6)=0\] may be written in the form $x_k+y_ki$ for $1\le k\le 5,$ where $x_k$ and $y_k$ are real. Let $\mathcal E$ be the unique ellipse that passes through the points $(x_1,y_1),(x_2,y_2),(x_3,y_3),(x_4,y_4),$ and $(x_5,y_5)$. The eccentricity of $\mathcal E$ ca... | 2021 AMC 12A Problem 21 | Completing the square in the original equation, we have \[(z-1)\left((z+1)^2+3\right)\left((z+2)^2+2\right)=0,\] from which $z=1,-1\pm\sqrt{3}i,-2\pm\sqrt{2}i.$
Now, we will find the equation of an ellipse $\mathcal E$ that passes through $(1,0),\left(-1,\pm\sqrt3\right),$ and $\left(-2,\pm\sqrt2\right)$ in the $xy$-p... | // Block 1
/* Made by MRENTHUSIASM */
size(220);
int xMin = -4;
int xMax = 2;
int yMin = -3;
int yMax = 3;
//Draws the horizontal gridlines
void horizontalLines()
{
for (int i = yMin+1; i < yMax; ++i)
{
draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4));
}
}
//Draws the vertical gridlines
void verticalLi... | [] |
894 | Suppose that the roots of the polynomial $P(x)=x^3+ax^2+bx+c$ are $\cos \frac{2\pi}7,\cos \frac{4\pi}7,$ and $\cos \frac{6\pi}7$, where angles are in radians. What is $abc$?
$\textbf{(A) }{-}\frac{3}{49} \qquad \textbf{(B) }{-}\frac{1}{28} \qquad \textbf{(C) }\frac{\sqrt[3]7}{64} \qquad \textbf{(D) }\frac{1}{32}\qquad... | 2021 AMC 12A Problem 22 | Let $z=e^{\frac{2\pi i}{7}}.$ In Solution 1, we conclude that $\sum_{k=1}^{6}z^k=-1,$ so \[\sum_{k=1}^{6}\operatorname{Re}\left(z^k\right)=\sum_{k=1}^{6}\cos\frac{2k\pi}{7}=-1.\]
Since $\cos\theta=\cos(2\pi-\theta)$ holds for all $\theta,$ this sum becomes
\begin{align*} 2\left(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\co... | // Block 1
/* Made by MRENTHUSIASM */
size(200);
int xMin = -2;
int xMax = 2;
int yMin = -2;
int yMax = 2;
int numRays = 24;
//Draws a polar grid that goes out to a number of circles
//equal to big, with numRays specifying the number of rays:
void polarGrid(int big, int numRays)
{
for (int i = 1; i < big+1; ++i... | [] |
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