problem_id int64 1 978 | question stringlengths 86 2.11k | source stringlengths 19 76 | solution stringlengths 94 14.7k | asymptote_code stringlengths 44 17.8k | solution_image_url stringclasses 16
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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 $1 ?$
$\textbf{(A) } 4 \pi \sqrt{3} \qquad\textbf{(B) } 7 \pi \qquad\textbf{(C) } \pi\left(3\sqrt{3} +2\right) \qquad\textbf{(D) } 10 \pi \left(\sqrt{3} - 1\right) \qquad\textbf{(E) } \pi\left(\sqrt{3} + 6\right)$ | 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 is $2\sqrt{3}+1$.
The area of the larger circle is thus $\left(2\sqrt{3}+1\right)^2 \pi = \left(13+4\sqrt{3}\right)\pi$, and the sum of the areas of the smaller circles is $13\pi$, so the area of the dark region is $\left(13+4\sqrt{3}\right)\pi-13\pi = \boxed{\textbf{(A) } 4 \pi \sqrt{3}}$. | // 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),white);filldraw(circle((1,-1*sqrt(3)),1),white);filldraw(circle((3,-1*sqrt(3)),1),white);filldraw(circle((-1,-1*sqrt(3)),1),white);filldraw(circle((-3,-1*sqrt(3)),1),white);filldraw(circle((0,2*sqrt(3)),1),white);filldraw(circle((0,-2*sqrt(3)),1),white);
pair O,A,B,C,H;
O=(0,0);
A=(-1,sqrt(3));
B=(1,sqrt(3));
C=(0,sqrt(3)*2);
H=(0,sqrt(3));
draw(O--A);
draw(A--C);
draw(B--C);
draw(O--B);
draw(A--B);
draw(O--C);
dot(A);
dot(B);
dot(C);
dot(O);
label("A",A, W);
label("O",O,S);
label("B",B,E);
label("C",C, N);
label("H",H, NE);
// Block 2
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);filldraw(circle((1,-1*sqrt(3)),1),white);filldraw(circle((3,-1*sqrt(3)),1),white);filldraw(circle((-1,-1*sqrt(3)),1),white);filldraw(circle((-3,-1*sqrt(3)),1),white);filldraw(circle((0,2*sqrt(3)),1),white);filldraw(circle((0,-2*sqrt(3)),1),white); pair O,A,B,C,H; O=(0,0); A=(-1,sqrt(3)); B=(1,sqrt(3)); C=(0,sqrt(3)*2); H=(0,sqrt(3)); draw(O--A); draw(A--C); draw(B--C); draw(O--B); draw(A--B); draw(O--C); dot(A); dot(B); dot(C); dot(O); label("A",A, W); label("O",O,S); label("B",B,E); label("C",C, N); label("H",H, NE); | [] |
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 $1 ?$
$\textbf{(A) } 4 \pi \sqrt{3} \qquad\textbf{(B) } 7 \pi \qquad\textbf{(C) } \pi\left(3\sqrt{3} +2\right) \qquad\textbf{(D) } 10 \pi \left(\sqrt{3} - 1\right) \qquad\textbf{(E) } \pi\left(\sqrt{3} + 6\right)$ | 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 $13\pi+4\pi\sqrt{3}-13\pi=\boxed{\textbf{(A) } 4 \pi \sqrt{3}}$. | // 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),white);filldraw(circle((1,-1*sqrt(3)),1),white);filldraw(circle((3,-1*sqrt(3)),1),white);filldraw(circle((-1,-1*sqrt(3)),1),white);filldraw(circle((-3,-1*sqrt(3)),1),white);filldraw(circle((0,2*sqrt(3)),1),white);filldraw(circle((0,-2*sqrt(3)),1),white);
pair A,B,C;
A=(0,sqrt(3)*2);
B=(-2,0);
C=(0,0);
draw(A--B);
draw(A--C);
draw(B--C);
dot(A);
dot(B);
dot(C);
label("A",A, N);
label("B",B, W);
label("C",C, S);
// Block 2
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);filldraw(circle((1,-1*sqrt(3)),1),white);filldraw(circle((3,-1*sqrt(3)),1),white);filldraw(circle((-1,-1*sqrt(3)),1),white);filldraw(circle((-3,-1*sqrt(3)),1),white);filldraw(circle((0,2*sqrt(3)),1),white);filldraw(circle((0,-2*sqrt(3)),1),white); pair A,B,C; A=(0,sqrt(3)*2); B=(-2,0); C=(0,0); draw(A--B); draw(A--C); draw(B--C); dot(A); dot(B); dot(C); label("A",A, N); label("B",B, W); label("C",C, S); | [] |
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?
$\textbf{(A) }2\sqrt{3}\qquad \textbf{(B) }4\qquad \textbf{(C) }3\sqrt{2}\qquad \textbf{(D) }2\sqrt{5}\qquad \textbf{(E) }5\qquad$ | 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 be found by drawing an altitude from the vertex between sides with length $15$ to the midpoint of the side with length $24$. The Pythagorean triple $9$ - $12$ - $15$ allows us easily to determine that the base is $24$ and the height is $9$. The formula $\frac {\text{base} \cdot \text{height}} {2}$ can also be used to find the area of the triangle as $108$, while the semiperimeter is simply $\frac {15 + 15 + 24} {2} = 27$. After plugging into the equation, we thus get $108 = \text{inradius} \cdot 27$, so the inradius is $4$. Now, let the distance between $O$ and the triangle be $x$. Choose a point on the incircle and denote it by $A$. The distance $OA$ is $6$, because it is just the radius of the sphere. The distance from point $A$ to the center of the incircle is $4$, because it is the radius of the incircle. By using the Pythagorean Theorem, we thus find $x = \sqrt{6^2-4^2}=\sqrt{20} = \boxed{\textbf {(D) } 2 \sqrt {5}}$.
Solution 2
Drop an altitude on the isosceles triangle. Let the resulting 3-4-5 right triangle $ABC$ have $AB=15$ and $BC=12$. By special triangle, $AC=9$. Let $r$ be the circle's radius. Let the circle's center be $O$ and $D$ be the closest point on $AB$ to $O$.
Then, $OD=r$. Obviously, $ODBC$ is a kite. Thus, $BC=DB=12$, and $AD=15-DB=3$. $AO=AC-r=9-r$. By Pythagoras, $AD^2+OD^2=AO^2$, so $3^2+r^2=(9-r)^2$. The $r^2$ terms cancel out, and $r=4$.
As before, using Pythagoras again, the distance is $\sqrt{6^2-4^2}=\boxed{\textbf{(D)}2\sqrt{5}}$
(Solution by BJHHar)
Solution 3
As in Solution 1, we note that by the Pythagorean Theorem, the height of the triangle is $9$, and that the three sides of the triangle are tangent to the sphere, so the circle in the cross-section of the sphere is the incenter of the triangle.
Recall that the inradius is the intersection of the angle bisectors. To find the inradius of the incircle, we use the Angle Bisector Theorem.
\[\begin{split}&\frac{AB}{BI}=\frac{AD}{DI} \\ \Rightarrow \ &\frac{15}{BI}=\frac{12}{DI} \\ \Rightarrow \ &\frac{BI}{5}=\frac{DI}{4}\end{split}\]
Since we know that $BI+DI$ (the height) is equal to $9$, $DI$ (the inradius) is $4$.
From here, the problem can be solved in the same way as in Solution 1. The answer is $\boxed{\textbf {(D) } 2 \sqrt{5}}$.
Solution 4 (similar triangles)
First, we label a few points:
We have that $\triangle{BDC}$ is a $3-4-5$ triangle, so, as in Solution 1, $BD = 9$.
From this, we know that $\overline{BI}=9-r$.
Since $AB$ is tangent to circle $I$, we also know $IEB$ is a right triangle. $\triangle{BIE}$ and $\triangle{BDA}$ share angle $DBA$, so $\triangle{BIE} \sim \triangle{BDA}$ since they have two equal angles.
Hence, by this similarity, $\dfrac{9-r}{5}=\dfrac{r}{4}$. Cross-multiplying, we get $36-4r =5r$, which gives $r=4$.
We now take another cross section of the sphere, perpendicular to the plane of the triangle.
Using the Pythagorean Theorem, we find that the distance from the center to the plane is $\boxed{\textbf {(D) } 2 \sqrt{5}}$. | // 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)); label("$15$", (6,9/2),NW); label("$12$", (6,-1),S);
// Block 2
draw((0,0)--(12,9)--(24,0)--cycle); draw((12,9)--(12,0), dashed); draw((11.5,0)--(11.5,0.5)--(12,0.5)); draw(circle((12,4),4)); draw((12,4)--(48/5, 36/5)); dot((12,4)); label("$15$", (6,9/2),NW); label("$15$", (18,9/2),NE); label("$r$",(54/5, 28/5), SW); label("$12$", (6,-1),S); label("$I$",(12,4),SE); label("$A$",(0,0),SW); label("$B$",(12,9),N); label("$C$",(24,0),SE); label("$D$",(12,-1/2),S); label("$E$",(48/5, 36/5),NW);
// Block 3
draw(circle((6,6),6)); draw((6,6)--(1.75735931,1.75735931)--(6,1.75735931)--cycle); dot((6,6)); dot((1.75735931,1.75735931)); dot((6,1.75735931)); label("$O$", (6,6),N); label("$6$", (3.87867965,3.87867965),NW); label("$4$", (3.87867965,1.75735931),SE); | [] |
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 uniformly at random from the closed interval $[0,1]$. Two random numbers $x$ and $y$ are chosen independently in this manner. What is the probability that $|x-y| > \tfrac{1}{2}$?
$\textbf{(A) } \frac{1}{3} \qquad \textbf{(B) } \frac{7}{16} \qquad \textbf{(C) } \frac{1}{2} \qquad \textbf{(D) } \frac{9}{16} \qquad \textbf{(E) } \frac{2}{3}$ | 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: $x$ is is chosen from the interval $[0,1]$, and $y$ is either $0$ or $1$.
Case 4: $x$ is is chosen from the interval $[0,1]$, and $y$ is also chosen from the interval $[0,1]$.
Each case has a $\frac{1}{4}$ chance of occurring (as it requires two coin flips).
For Case 1, we need $x$ and $y$ to be different. Therefore, the probability for success in Case 1 is $\frac{1}{2}$.
For Case 2, if $x$ is 0, we need $y$ to be in the interval $\left(\frac{1}{2}, 1\right]$. If $x$ is 1, we need $y$ to be in the interval $\left[0, \frac{1}{2}\right)$. Regardless of what $x$ is, the probability for success for Case 2 is $\frac{1}{2}$.
By symmetry, Case 3 has the same success rate as Case 2.
For Case 4, we must use geometric probability because there are an infinite number of pairs $(x, y)$ that can be selected, whether they satisfy the inequality or not. Graphing $|x-y| > \tfrac{1}{2}$ gives us the following picture where the shaded area is the set of all the points that fulfill the inequality:
The shaded area is $\frac{1}{4}$, which means the probability for success for case 4 is $\frac{1}{4}$ (since the total area of the bounding square, containing all possible pairs, is $1$).
Adding up the success rates from each case, we get:
$\left(\frac{1}{4}\right) \cdot \left(\frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{4}\right) = \boxed{\textbf{(B) }\frac{7}{16}}$. | // 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+linewidth(1)); draw((0,0)--(0,1)--(1,1)--(1,0)--cycle); | [] |
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 uniformly at random from the closed interval $[0,1]$. Two random numbers $x$ and $y$ are chosen independently in this manner. What is the probability that $|x-y| > \tfrac{1}{2}$?
$\textbf{(A) } \frac{1}{3} \qquad \textbf{(B) } \frac{7}{16} \qquad \textbf{(C) } \frac{1}{2} \qquad \textbf{(D) } \frac{9}{16} \qquad \textbf{(E) } \frac{2}{3}$ | 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 $[0,1]$. The probability that $x-y > \tfrac{1}{2}$ when both $x$ and $y$ are randomly chosen from the interval $[0,1]$ is $\frac18$, as illustrated in the geometric probability diagram below.
Therefore, the probability for Case $1$ is $\frac12 \cdot \frac12 \cdot \frac18 = \frac{1}{32}$
Case $2$: $x$ flips $HT$ and $y$ flips $T$
When $x$ flips $HT$ and $y$ flips $T$, $x=1$, and $y$ is randomly chosen from the interval $[0,1]$. The probability that $x-y > \tfrac{1}{2}$ when $x=1$, and $y$ is randomly chosen from the interval $[0,1]$ is $\frac12$ ($y$ has to be between $[0,\frac12]$).
Therefore, the probability for Case $2$ is $\frac12 \cdot \frac12 \cdot \frac12 \cdot \frac12 = \frac{1}{16}$
Case $3$: $x$ flips $T$ and $y$ flips $HH$
When $x$ flips $T$ and $y$ flips $HH$, $x$ is randomly chosen from the interval $[0,1]$, and $y = 0$. The probability that $x-y > \tfrac{1}{2}$ when $x$ is randomly chosen from the interval $[0,1]$, and $y = 0$is $\frac12$ ($x$ has to be between $[\frac12,1]$).
Therefore, the probability for Case $3$ is $\frac12 \cdot \frac12 \cdot \frac12 \cdot \frac12 = \frac{1}{16}$
Case $4$: $x$ flips $HT$ and $y$ flips $HH$
When $x$ flips $HT$ and $y$ flips $HH$, $x=1$, and $y = 0$. The probability that $x-y > \frac{1}{2}$ when $x=1$, and $y = 0$ is $1$.
Therefore, the probability for Case $4$ is $\frac12 \cdot \frac12 \cdot \frac12 \cdot \frac12 \cdot 1= \frac{1}{16}$
Hence, the answer is $2(\frac{1}{32} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16}) = \boxed{\textbf{(B)}\frac{7}{16}}$
~isabelchen | // 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 know that we cannot have 7 points of intersection or else our answer would be greater than the given answer choices. Our answer is given by the sum $0+1+3+4+5+6=\boxed{\textbf{(D)} 19}$. | unitsize(2cm); real d = 2.5; draw((-1,.6)--(1,.6),Arrows); draw((-1,.2)--(1,.2),Arrows); draw((-1,-.2)--(1,-.2),Arrows); draw((-1,-.6)--(1,-.6),Arrows); draw((-1+d,0)--(1+d,0),Arrows); draw((0+d,1)--(0+d,-1),Arrows); draw(dir(45)+(d,0)--dir(45+180)+(d,0),Arrows); draw(dir(135)+(d,0)--dir(135+180)+(d,0),Arrows); dot((0+d,0)); draw((-1+2*d,sqrt(3)/3)--(1+2*d,sqrt(3)/3),Arrows); draw((-1/4-1/2+2*d, sqrt(3)/12-sqrt(3)/2)--(-1/4+1/2+2*d,sqrt(3)/12+sqrt(3)/2),Arrows); draw((1/4+1/2+2*d, sqrt(3)/12-sqrt(3)/2)--(1/4-1/2+2*d,sqrt(3)/12+sqrt(3)/2),Arrows); draw((-1+2*d,-sqrt(3)/6)--(1+2*d,-sqrt(3)/6),Arrows); dot((0+2*d,sqrt(3)/3)); dot((-1/2+2*d,-sqrt(3)/6)); dot((1/2+2*d,-sqrt(3)/6)); draw((-1/3,1-d)--(-1/3,-1-d),Arrows); draw((1/3,1-d)--(1/3,-1-d),Arrows); draw((-1,-1/3-d)--(1,-1/3-d),Arrows); draw((-1,1/3-d)--(1,1/3-d),Arrows); dot((1/3,1/3-d)); dot((-1/3,1/3-d)); dot((1/3,-1/3-d)); dot((-1/3,-1/3-d)); draw((-1+d,sqrt(3)/12-d)--(1+d,sqrt(3)/12-d),Arrows); draw((-1/4-1/2+d, sqrt(3)/12-sqrt(3)/2-d)--(-1/4+1/2+d,sqrt(3)/12+sqrt(3)/2-d),Arrows); draw((1/4+1/2+d, sqrt(3)/12-sqrt(3)/2-d)--(1/4-1/2+d,sqrt(3)/12+sqrt(3)/2-d),Arrows); draw((-1+d,-sqrt(3)/6-d)--(1+d,-sqrt(3)/6-d),Arrows); dot((0+d,sqrt(3)/3-d)); dot((-1/2+d,-sqrt(3)/6-d)); dot((1/2+d,-sqrt(3)/6-d)); dot((-1/4+d,sqrt(3)/12-d)); dot((1/4+d,sqrt(3)/12-d)); draw((-1/4-1/2+2*d, sqrt(3)/12-sqrt(3)/2-d)--(-1/4+1/2+2*d,sqrt(3)/12+sqrt(3)/2-d),Arrows); draw((1/4+1/2+2*d, sqrt(3)/12-sqrt(3)/2-d)--(1/4-1/2+2*d,sqrt(3)/12+sqrt(3)/2-d),Arrows); draw(dir(30)+(2*d,-d)--dir(30+180)+(2*d,-d),Arrows); draw(dir(150)+(2*d,-d)--dir(-30)+(2*d,-d),Arrows); dot((0+2*d,0-d)); dot((0+2*d,sqrt(3)/3-d)); dot((-1/2+2*d,-sqrt(3)/6-d)); dot((1/2+2*d,-sqrt(3)/6-d)); dot((-1/4+2*d,sqrt(3)/12-d)); dot((1/4+2*d,sqrt(3)/12-d)); | [] |
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 $1 ?$
$\textbf{(A) } 4 \pi \sqrt{3} \qquad\textbf{(B) } 7 \pi \qquad\textbf{(C) } \pi\left(3\sqrt{3} +2\right) \qquad\textbf{(D) } 10 \pi \left(\sqrt{3} - 1\right) \qquad\textbf{(E) } \pi\left(\sqrt{3} + 6\right)$ | 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 is $2\sqrt{3}+1$.
The area of the larger circle is thus $\left(2\sqrt{3}+1\right)^2 \pi = \left(13+4\sqrt{3}\right)\pi$, and the sum of the areas of the smaller circles is $13\pi$, so the area of the dark region is $\left(13+4\sqrt{3}\right)\pi-13\pi = \boxed{\textbf{(A) } 4 \pi \sqrt{3}}$. | 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);filldraw(circle((1,-1*sqrt(3)),1),white);filldraw(circle((3,-1*sqrt(3)),1),white);filldraw(circle((-1,-1*sqrt(3)),1),white);filldraw(circle((-3,-1*sqrt(3)),1),white);filldraw(circle((0,2*sqrt(3)),1),white);filldraw(circle((0,-2*sqrt(3)),1),white); pair O,A,B,C,H; O=(0,0); A=(-1,sqrt(3)); B=(1,sqrt(3)); C=(0,sqrt(3)*2); H=(0,sqrt(3)); draw(O--A); draw(A--C); draw(B--C); draw(O--B); draw(A--B); draw(O--C); dot(A); dot(B); dot(C); dot(O); label("A",A, W); label("O",O,S); label("B",B,E); label("C",C, N); label("H",H, NE); | [] |
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 $1 ?$
$\textbf{(A) } 4 \pi \sqrt{3} \qquad\textbf{(B) } 7 \pi \qquad\textbf{(C) } \pi\left(3\sqrt{3} +2\right) \qquad\textbf{(D) } 10 \pi \left(\sqrt{3} - 1\right) \qquad\textbf{(E) } \pi\left(\sqrt{3} + 6\right)$ | 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 $13\pi+4\pi\sqrt{3}-13\pi=\boxed{\textbf{(A) } 4 \pi \sqrt{3}}$. | 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);filldraw(circle((1,-1*sqrt(3)),1),white);filldraw(circle((3,-1*sqrt(3)),1),white);filldraw(circle((-1,-1*sqrt(3)),1),white);filldraw(circle((-3,-1*sqrt(3)),1),white);filldraw(circle((0,2*sqrt(3)),1),white);filldraw(circle((0,-2*sqrt(3)),1),white); pair A,B,C; A=(0,sqrt(3)*2); B=(-2,0); C=(0,0); draw(A--B); draw(A--C); draw(B--C); dot(A); dot(B); dot(C); label("A",A, N); label("B",B, W); label("C",C, S); | [] |
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?
$\textbf{(A) }2\sqrt{3}\qquad \textbf{(B) }4\qquad \textbf{(C) }3\sqrt{2}\qquad \textbf{(D) }2\sqrt{5}\qquad \textbf{(E) }5\qquad$ | 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 be found by drawing an altitude from the vertex between sides with length $15$ to the midpoint of the side with length $24$. The Pythagorean triple $9$ - $12$ - $15$ allows us easily to determine that the base is $24$ and the height is $9$. The formula $\frac {\text{base} \cdot \text{height}} {2}$ can also be used to find the area of the triangle as $108$, while the semiperimeter is simply $\frac {15 + 15 + 24} {2} = 27$. After plugging into the equation, we thus get $108 = \text{inradius} \cdot 27$, so the inradius is $4$. Now, let the distance between $O$ and the triangle be $x$. Choose a point on the incircle and denote it by $A$. The distance $OA$ is $6$, because it is just the radius of the sphere. The distance from point $A$ to the center of the incircle is $4$, because it is the radius of the incircle. By using the Pythagorean Theorem, we thus find $x = \sqrt{6^2-4^2}=\sqrt{20} = \boxed{\textbf {(D) } 2 \sqrt {5}}$.
Solution 2
Drop an altitude on the isosceles triangle. Let the resulting 3-4-5 right triangle $ABC$ have $AB=15$ and $BC=12$. By special triangle, $AC=9$. Let $r$ be the circle's radius. Let the circle's center be $O$ and $D$ be the closest point on $AB$ to $O$.
Then, $OD=r$. Obviously, $ODBC$ is a kite. Thus, $BC=DB=12$, and $AD=15-DB=3$. $AO=AC-r=9-r$. By Pythagoras, $AD^2+OD^2=AO^2$, so $3^2+r^2=(9-r)^2$. The $r^2$ terms cancel out, and $r=4$.
As before, using Pythagoras again, the distance is $\sqrt{6^2-4^2}=\boxed{\textbf{(D)}2\sqrt{5}}$
(Solution by BJHHar)
Solution 3
As in Solution 1, we note that by the Pythagorean Theorem, the height of the triangle is $9$, and that the three sides of the triangle are tangent to the sphere, so the circle in the cross-section of the sphere is the incenter of the triangle.
Recall that the inradius is the intersection of the angle bisectors. To find the inradius of the incircle, we use the Angle Bisector Theorem.
\[\begin{split}&\frac{AB}{BI}=\frac{AD}{DI} \\ \Rightarrow \ &\frac{15}{BI}=\frac{12}{DI} \\ \Rightarrow \ &\frac{BI}{5}=\frac{DI}{4}\end{split}\]
Since we know that $BI+DI$ (the height) is equal to $9$, $DI$ (the inradius) is $4$.
From here, the problem can be solved in the same way as in Solution 1. The answer is $\boxed{\textbf {(D) } 2 \sqrt{5}}$.
Solution 4 (similar triangles)
First, we label a few points:
We have that $\triangle{BDC}$ is a $3-4-5$ triangle, so, as in Solution 1, $BD = 9$.
From this, we know that $\overline{BI}=9-r$.
Since $AB$ is tangent to circle $I$, we also know $IEB$ is a right triangle. $\triangle{BIE}$ and $\triangle{BDA}$ share angle $DBA$, so $\triangle{BIE} \sim \triangle{BDA}$ since they have two equal angles.
Hence, by this similarity, $\dfrac{9-r}{5}=\dfrac{r}{4}$. Cross-multiplying, we get $36-4r =5r$, which gives $r=4$.
We now take another cross section of the sphere, perpendicular to the plane of the triangle.
Using the Pythagorean Theorem, we find that the distance from the center to the plane is $\boxed{\textbf {(D) } 2 \sqrt{5}}$. | // 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)); label("$15$", (6,9/2),NW); label("$12$", (6,-1),S);
// Block 2
draw((0,0)--(12,9)--(24,0)--cycle); draw((12,9)--(12,0), dashed); draw((11.5,0)--(11.5,0.5)--(12,0.5)); draw(circle((12,4),4)); draw((12,4)--(48/5, 36/5)); dot((12,4)); label("$15$", (6,9/2),NW); label("$15$", (18,9/2),NE); label("$r$",(54/5, 28/5), SW); label("$12$", (6,-1),S); label("$I$",(12,4),SE); label("$A$",(0,0),SW); label("$B$",(12,9),N); label("$C$",(24,0),SE); label("$D$",(12,-1/2),S); label("$E$",(48/5, 36/5),NW);
// Block 3
draw(circle((6,6),6)); draw((6,6)--(1.75735931,1.75735931)--(6,1.75735931)--cycle); dot((6,6)); dot((1.75735931,1.75735931)); dot((6,1.75735931)); label("$O$", (6,6),N); label("$6$", (3.87867965,3.87867965),NW); label("$4$", (3.87867965,1.75735931),SE); | [] |
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 uniformly at random from the closed interval $[0,1]$. Two random numbers $x$ and $y$ are chosen independently in this manner. What is the probability that $|x-y| > \tfrac{1}{2}$?
$\textbf{(A) } \frac{1}{3} \qquad \textbf{(B) } \frac{7}{16} \qquad \textbf{(C) } \frac{1}{2} \qquad \textbf{(D) } \frac{9}{16} \qquad \textbf{(E) } \frac{2}{3}$ | 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: $x$ is is chosen from the interval $[0,1]$, and $y$ is either $0$ or $1$.
Case 4: $x$ is is chosen from the interval $[0,1]$, and $y$ is also chosen from the interval $[0,1]$.
Each case has a $\frac{1}{4}$ chance of occurring (as it requires two coin flips).
For Case 1, we need $x$ and $y$ to be different. Therefore, the probability for success in Case 1 is $\frac{1}{2}$.
For Case 2, if $x$ is 0, we need $y$ to be in the interval $\left(\frac{1}{2}, 1\right]$. If $x$ is 1, we need $y$ to be in the interval $\left[0, \frac{1}{2}\right)$. Regardless of what $x$ is, the probability for success for Case 2 is $\frac{1}{2}$.
By symmetry, Case 3 has the same success rate as Case 2.
For Case 4, we must use geometric probability because there are an infinite number of pairs $(x, y)$ that can be selected, whether they satisfy the inequality or not. Graphing $|x-y| > \tfrac{1}{2}$ gives us the following picture where the shaded area is the set of all the points that fulfill the inequality:
The shaded area is $\frac{1}{4}$, which means the probability for success for case 4 is $\frac{1}{4}$ (since the total area of the bounding square, containing all possible pairs, is $1$).
Adding up the success rates from each case, we get:
$\left(\frac{1}{4}\right) \cdot \left(\frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{4}\right) = \boxed{\textbf{(B) }\frac{7}{16}}$. | 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 uniformly at random from the closed interval $[0,1]$. Two random numbers $x$ and $y$ are chosen independently in this manner. What is the probability that $|x-y| > \tfrac{1}{2}$?
$\textbf{(A) } \frac{1}{3} \qquad \textbf{(B) } \frac{7}{16} \qquad \textbf{(C) } \frac{1}{2} \qquad \textbf{(D) } \frac{9}{16} \qquad \textbf{(E) } \frac{2}{3}$ | 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 $[0,1]$. The probability that $x-y > \tfrac{1}{2}$ when both $x$ and $y$ are randomly chosen from the interval $[0,1]$ is $\frac18$, as illustrated in the geometric probability diagram below.
Therefore, the probability for Case $1$ is $\frac12 \cdot \frac12 \cdot \frac18 = \frac{1}{32}$
Case $2$: $x$ flips $HT$ and $y$ flips $T$
When $x$ flips $HT$ and $y$ flips $T$, $x=1$, and $y$ is randomly chosen from the interval $[0,1]$. The probability that $x-y > \tfrac{1}{2}$ when $x=1$, and $y$ is randomly chosen from the interval $[0,1]$ is $\frac12$ ($y$ has to be between $[0,\frac12]$).
Therefore, the probability for Case $2$ is $\frac12 \cdot \frac12 \cdot \frac12 \cdot \frac12 = \frac{1}{16}$
Case $3$: $x$ flips $T$ and $y$ flips $HH$
When $x$ flips $T$ and $y$ flips $HH$, $x$ is randomly chosen from the interval $[0,1]$, and $y = 0$. The probability that $x-y > \tfrac{1}{2}$ when $x$ is randomly chosen from the interval $[0,1]$, and $y = 0$is $\frac12$ ($x$ has to be between $[\frac12,1]$).
Therefore, the probability for Case $3$ is $\frac12 \cdot \frac12 \cdot \frac12 \cdot \frac12 = \frac{1}{16}$
Case $4$: $x$ flips $HT$ and $y$ flips $HH$
When $x$ flips $HT$ and $y$ flips $HH$, $x=1$, and $y = 0$. The probability that $x-y > \frac{1}{2}$ when $x=1$, and $y = 0$ is $1$.
Therefore, the probability for Case $4$ is $\frac12 \cdot \frac12 \cdot \frac12 \cdot \frac12 \cdot 1= \frac{1}{16}$
Hence, the answer is $2(\frac{1}{32} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16}) = \boxed{\textbf{(B)}\frac{7}{16}}$
~isabelchen | 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 $\overline{AO}$ and $\overline{BC}$ intersect at $P$, and $\dfrac{BP}{CP} = 3$. Then $AB$ can be written in the form $\dfrac{m}{\sqrt{n}} - \dfrac{p}{\sqrt{q}}$ for positive integers $m$, $n$, $p$, $q$ with $\text{gcd}(m,n) = \text{gcd}(p,q) = 1$. What is $m+n+p+q$?
$\phantom{}$
$\textbf{(A) } 42 \qquad \textbf{(B) }86 \qquad \textbf{(C) } 92 \qquad \textbf{(D) } 114 \qquad \textbf{(E) } 130$ | 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 $\triangle ABC$, then it follows that $AM = \frac{\sqrt{3}}{2}s$ and $PM = \frac{s}{4}$. This implies that $AP = \frac{\sqrt{13}}{4}s$, so $\frac{AM}{AP} = \frac{2\sqrt{3}}{\sqrt{13}}$. Furthermore, $\frac{AM + SO}{AO} = \frac{AM}{AP}$ (because $\triangle PMA \sim \triangle PSO$) so this gives us the equation
\[\frac{\frac{\sqrt{3}}{2}s + 17}{20} = \frac{2\sqrt{3}}{\sqrt{13}}\]
to solve for the side length $s$, or $AB$. Thus,
\[\frac{\sqrt{39}}{2}s + 17\sqrt{13} = 40\sqrt{3}\]
\[\frac{\sqrt{39}}{2}s = 40\sqrt{3} - 17\sqrt{13}\]
\[s = \frac{80}{\sqrt{13}} - \frac{34}{\sqrt{3}} = AB\]
The problem asks for $m + n + p + q = 80 + 13 + 34 + 3 = \boxed{\textbf{(E) } 130}$. | // 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(3))*(sqrt(13)/sqrt(12))*(sqrt(3)/2));
real ans = (80/sqrt(13) - 34/sqrt(3));
dot((a,0));
label("$P$", (a, 0), dir(290)*0.58);
pair s = ((12*a + 0)/13, 0-sqrt(12)*a/13);
dot(s);
label("$S$", s, dir(135));
pair c = (a + 1/4*ans*1/sqrt(13), 0 + 1/4*ans*sqrt(12)/sqrt(13));
dot(c);
label("$C$", c, dir(110));
pair m = (a - 1/4*ans*1/sqrt(13), 0 - 1/4*ans*sqrt(12)/sqrt(13));
dot(m);
label("$M$", m, dir(285));
pair b = (a - 3/4*ans*1/sqrt(13), 0 - 3/4*ans*sqrt(12)/sqrt(13));
dot(b);
label("$B$", b, S);
draw(b--s);
draw(s--(0,0));
draw(aa--b);
draw(aa--c);
draw(aa--m);
markscalefactor=0.1;
draw(rightanglemark(s,m,aa,3.4));
draw(rightanglemark((0,0),s,m,3.4));
// Block 2
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(3))*(sqrt(13)/sqrt(12))*(sqrt(3)/2)); real ans = (80/sqrt(13) - 34/sqrt(3)); dot((a,0)); label("$P$", (a, 0), dir(290)*0.58); pair s = ((12*a + 0)/13, 0-sqrt(12)*a/13); dot(s); label("$S$", s, dir(135)); pair c = (a + 1/4*ans*1/sqrt(13), 0 + 1/4*ans*sqrt(12)/sqrt(13)); dot(c); label("$C$", c, dir(110)); pair m = (a - 1/4*ans*1/sqrt(13), 0 - 1/4*ans*sqrt(12)/sqrt(13)); dot(m); label("$M$", m, dir(285)); pair b = (a - 3/4*ans*1/sqrt(13), 0 - 3/4*ans*sqrt(12)/sqrt(13)); dot(b); label("$B$", b, S); draw(b--s); draw(s--(0,0)); draw(aa--b); draw(aa--c); draw(aa--m); markscalefactor=0.1; draw(rightanglemark(s,m,aa,3.4)); draw(rightanglemark((0,0),s,m,3.4)); | [] |
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 many different routes can Paula take?
$\textbf{(A) } 0 \qquad\textbf{(B) } 1 \qquad\textbf{(C) } 2 \qquad\textbf{(D) } 3 \qquad\textbf{(E) } 4$ | 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 end points, since we don't want to return to them. Thus there are $6+2=8$ vertices that we know have $1$ unused edge, and we have $17-13=4$ unused edges to work with (since there are $17$ edges in total, and we must use exactly $13$ of them). It is not hard to find that there is only one configuration satisfying these conditions:
Note: $\text{X}$s represent unused edges.
Observe that at each of the $2$ cities marked with an $\text{O}$ on a path, there are $2$ possibilities: we can either continue straight and cross back over the path later, or we can make a left turn, then turn right when we approach the junction again. This gives us a total of $2\cdot 2 = \boxed{\textbf{(E) } 4}$ valid paths. | // 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);
label("X", (0, 1.5));
label("X", (1.5, 2));
label("X", (1.5, 0));
label("X", (3, 0.5));
label("O", (1,1), NE);
label("O", (2,1), NE);
// Block 2
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); label("X", (0, 1.5)); label("X", (1.5, 2)); label("X", (1.5, 0)); label("X", (3, 0.5)); label("O", (1,1), NE); label("O", (2,1), NE); | [] |
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 many different routes can Paula take?
$\textbf{(A) } 0 \qquad\textbf{(B) } 1 \qquad\textbf{(C) } 2 \qquad\textbf{(D) } 3 \qquad\textbf{(E) } 4$ | 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 $\text{deg}(V)$ is, as usual, the degree of $V$ (i.e. the number of edges coming into/out of $V$). This is because to visit any of these vertices, we would have to enter and exit it by two different edges, in order to avoid using any portion of a road more than once. (Those who know some graph theory will recognise this a well-known principle.) We will label each vertex with this number.
Additionally, notice that if we visit $n$ vertices (not necessarily distinct, i.e. counting a vertex which is visited twice as two vertices) along our path, we must traverse $n-1$ edges (this is quite easy to prove). Thus, if we want to traverse $13$ edges in total, we have to visit $14$ vertices. We must visit $A$ and $L$, leaving $12$ vertices from the rest of the graph to visit.
If we sum the maximum numbers of visits to each vertex, we find that this is exactly equal to the $12$ found above. This means that we have to visit each vertex $M(V)$ times, and must traverse $2\cdot M(V)$ edges connected to each vertex. Specifically, we must traverse all $4$ of the edges connected to the two central vertices at $(1,1)$ and $(1,2)$, as well as both edges connected to the $2$ corner vertices (excluding $A$ and $L$), and any $2$ edges connected to the other vertices along the outside edge of the rectangle.
With this information, we can now proceed by dividing into cases.
Case 1: We first move down from $A$.
In this case, we see that we must immediately move right to $(1,1)$ in order to traverse the edge from $(0,1)$ to $(1,1)$, since we can never revisit $(0,1)$. However, by the same logic, we must immediately move to $(0,0)$. This is a contradiction, so there are no possible paths in this case.
Case 2: We first move right from $A$.
Similar to the last case, we see that we must immediately move to $(1,1)$. By symmetry, we conclude that our last two moves must be $(2,1) \rightarrow (2,0) \rightarrow (3,0)$. With this information, we have reduced the problem to traveling from $(1,1)$ to $(2,1)$ with the same constraints as before. We redraw the graph, removing the edges we have already used and updating our labels (note that $(1,1)$ and $(2,1)$ are still labeled with $2$ since we will pass through them twice, at the start and the end). Then, we remove the vertices with label $0$ and the edges we know we can never traverse, giving:
Now, it is clear that we must complete a cycle in the lower left square, return to $(1,1)$, go to $(2,1)$, and complete a cycle in the top right square, returning to $(2,1)$.
There are two ways to traverse each cycle (clockwise and anti-clockwise), giving us a total of $2\cdot 2 = \boxed{\textbf{(E) } 4}$ paths of length 13 from $A$ to $L$. | // 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);
label("1", (1,2), N);
label("1", (2,2), N);
label("1", (3,2), N);
label("1", (0,1), W);
label("2", (1,1), NW);
label("2", (2,1), NW);
label("1", (3,1), E);
label("1", (0,0),S);
label("1", (1,0),S);
label("1", (2,0),S);
// Block 2
import olympiad;
unitsize(50);
for (int i = 0; i < 3; ++i) {
for (int j = 0; j < 4; ++j) {
if(j==1 && i==2) {continue;}
if(j==2 && i==0) {continue;}
pair A = (j,i);
dot(A);
}
}
draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);
draw((2,1)--(3,1)--(3,2)--(2,2)--cycle);
draw((1,1)--(2,1));
label("$A$", (0,2), W);
label("$L$", (3,0), E);
label("1", (2,2), N);
label("1", (3,2), N);
label("1", (0,1), W);
label("2", (1,1), NW);
label("2", (2,1), NW);
label("1", (3,1), E);
label("1", (0,0),S);
label("1", (1,0),S);
// Block 3
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); label("1", (1,2), N); label("1", (2,2), N); label("1", (3,2), N); label("1", (0,1), W); label("2", (1,1), NW); label("2", (2,1), NW); label("1", (3,1), E); label("1", (0,0),S); label("1", (1,0),S); label("1", (2,0),S);
// Block 4
import olympiad; unitsize(50); for (int i = 0; i < 3; ++i) { for (int j = 0; j < 4; ++j) { if(j==1 && i==2) {continue;} if(j==2 && i==0) {continue;} pair A = (j,i); dot(A); } } draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw((2,1)--(3,1)--(3,2)--(2,2)--cycle); draw((1,1)--(2,1)); label("$A$", (0,2), W); label("$L$", (3,0), E); label("1", (2,2), N); label("1", (3,2), N); label("1", (0,1), W); label("2", (1,1), NW); label("2", (2,1), NW); label("1", (3,1), E); label("1", (0,0),S); label("1", (1,0),S); | [] |
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 parallel (in other words, they cannot be skew lines). If all $12$ line segments are extended to lines, the first (arbitrarily chosen) edge's line intersects $4$ lines and is parallel to another $3$. Thus $4+3=7$ of the $12-1=11$ remaining line segments (which could be chosen for the second edge) give a pair of lines determining a common plane. To see this, observe that in the diagram below, the red edge is parallel to the $3$ green edges and intersects with the $4$ blue edges.
This means that the probability that a randomly-selected pair of edges determine a plane is $\frac{7}{11}$, and we calculate that there are ${12 \choose 2} = 66$ total pairs of edges that could be chosen (without the restriction). Thus the answer is $\frac{7}{11} \cdot 66 =\boxed{\textbf{(D) }42}$. | // 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)--(0,0,1)--(1,0,1), blue);
draw((1,0,1)--(1,1,1)--(0,1,1));
// Block 2
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)--(0,0,1)--(1,0,1), blue); draw((1,0,1)--(1,1,1)--(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 probability that Fiona reaches pad $10$ without landing on either pad $3$ or pad $6$?
$\textbf{(A) } \frac{15}{256} \qquad \textbf{(B) } \frac{1}{16} \qquad \textbf{(C) } \frac{15}{128}\qquad \textbf{(D) } \frac{1}{8} \qquad \textbf{(E) } \frac14$ | 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 $n$, and the same applies for a $1$-jump to reach lily pad $n-1$. We will now compute the values of $p_n$ recursively, but we will skip over $3$ and $6$. That is, we will not consider any jumps from lily pads $3$ or $6$ when considering the probabilities. We obtain the following chart, where an X represents an unused/uncomputed value:
Hence the answer is $\boxed{\textbf{(A) } \frac{15}{256}}$.
Note: If we let $p_n$ be the probability of surviving if the frog is on lily pad $n$, using $p_{10}$ = 1, we can solve backwards and obtain the following chart: | // 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 = {"$15/256$", "$5/128$", "$5/64$", "X", "$5/32$", "$5/16$", "X", "$5/8$", "$3/4$", "$1/2$", "1", "X"};
for(int i =0; i<= 11; ++i) {
draw((i,0)--(i+1,0)--(i+1,1)--(i,1)--cycle);
label((string) (11-i), (i+0.5,0), S);
label(vals[(11-i)], (i+0.5, 0.5));
}
// Block 3
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 4
unitsize(40); string[] vals = {"$15/256$", "$5/128$", "$5/64$", "X", "$5/32$", "$5/16$", "X", "$5/8$", "$3/4$", "$1/2$", "1", "X"}; for(int i =0; i<= 11; ++i) { draw((i,0)--(i+1,0)--(i+1,1)--(i,1)--cycle); label((string) (11-i), (i+0.5,0), S); label(vals[(11-i)], (i+0.5, 0.5)); } | [] |
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 $C$ to $E$. What is the area, in square centimeters, of $\triangle{PQR}$?
$\textbf{(A) } \frac{3\sqrt2}{2} \qquad\textbf{(B) } \frac{3\sqrt3}{2} \qquad\textbf{(C) } 2\sqrt2 \qquad\textbf{(D) } 2\sqrt3 \qquad\textbf{(E) } 3\sqrt2$ | 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 bird's-eye view, looking down upon $P$, $Q$, and $R$ projected onto $ABCD$:
Additionally, we know that $PQ$ is parallel to the plane determined by $ABCD$, since $P$ and $Q$ have the same $z$-coordinate. Hence the height of $\triangle PQR$ is equal to the $z$-coordinate of $R$ minus the $z$-coordinate of $P$, giving $4-2= 2$. By the distance formula, $\overline{PQ} = 2\sqrt{2}$, so the area of $\triangle PQR$ is $\frac{1}{2} \cdot 2\sqrt{2} \cdot 2 = \boxed{\textbf{(C) } 2\sqrt{2}}$. | // 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));
// Block 2
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 $S$?
$\textbf{(A) } \frac{1}{2}\sqrt3 \qquad\textbf{(B) } \frac{3}{4}\sqrt3 \qquad\textbf{(C) } \frac{3}{2}\sqrt3\qquad\textbf{(D) } \frac{1}{2}\pi\sqrt3 \qquad\textbf{(E) } \pi$ | 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 $a+b\omega+c\omega^2$ unchanged. Therefore, assume that at least one of $a,b,c$ is equal to $0$. If exactly one of them is $0$, we can form an equilateral triangle of side length $1$ using the remaining terms. A similar argument works if exactly two of them are $0$. In total, we get $3+{3 \choose 2} = 6$ equilateral triangles, whose total area is $6 \cdot \frac{\sqrt{3}}{4} = \boxed{\textbf{(C) } \frac{3}{2}\sqrt3}$.
Note: A diagram of the six equilateral triangles is shown below. | // 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);
draw((-2,0)--(2,0));
draw((0,-2)--(0,2));
// Block 2
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); draw((-2,0)--(2,0)); draw((0,-2)--(0,2)); | [] |
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 $S$?
$\textbf{(A) } \frac{1}{2}\sqrt3 \qquad\textbf{(B) } \frac{3}{4}\sqrt3 \qquad\textbf{(C) } \frac{3}{2}\sqrt3\qquad\textbf{(D) } \frac{1}{2}\pi\sqrt3 \qquad\textbf{(E) } \pi$ | 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 $1$:
When $a=0, b=1$ we have $-\frac{1}{2}, \frac{\sqrt{3}}{2}$:
When $a=1, b=1$ we have $\frac{1}{2}, \frac{\sqrt{3}}{2}$:
The area of this is $\frac{\sqrt{3}}{2}$. Multiply this by $3$ to get $C$. | // 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), gray);
draw((0,0)--(-1/2,sqrt(3)/2), red);
draw((0,0)--(1,0), red);
draw((-1/2,sqrt(3)/2)--(1/2,sqrt(3)/2), red);
draw((1,0)--(1/2,sqrt(3)/2), red);
// Block 4
size(100,100); draw((-2,0)--(2,0), gray); draw((0,-2)--(0,2), gray); draw((0,0)--(1,0), red);
// Block 5
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 6
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); draw((-1/2,sqrt(3)/2)--(1/2,sqrt(3)/2), red); draw((1,0)--(1/2,sqrt(3)/2), red); | [] |
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 $S$?
$\textbf{(A) } \frac{1}{2}\sqrt3 \qquad\textbf{(B) } \frac{3}{4}\sqrt3 \qquad\textbf{(C) } \frac{3}{2}\sqrt3\qquad\textbf{(D) } \frac{1}{2}\pi\sqrt3 \qquad\textbf{(E) } \pi$ | 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
by "moving" the blue line along the red line. Finally, we can add $a$ to every point, giving
by "moving" the previous area along the green line. This leaves us with a regular hexagon with side length $1$, so, as in Solution 1, the total area is $\boxed{\textbf{(C) } \frac{3}{2}\sqrt{3}}$. | // 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)--(-1/2,-sqrt(3)/2), blue);
draw((-2,0)--(2,0));
draw((0,-2)--(0,2));
// Block 3
size(100,100);
fill((-1/2,sqrt(3)/2)--(-1,0)--(-1/2,-sqrt(3)/2)--(1/2,-sqrt(3)/2)--(1,0)--(1/2,sqrt(3)/2)--cycle, lightgray);
draw((-1/2,sqrt(3)/2)--(-1,0)--(-1/2,-sqrt(3)/2)--(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)--(-1/2,-sqrt(3)/2), blue);
draw((-2,0)--(2,0));
draw((0,-2)--(0,2));
draw((0,0)--(1,0), heavygreen);
// Block 4
size(100,100); draw((0,0)--(-1/2,-sqrt(3)/2), blue); draw((-2,0)--(2,0)); draw((0,-2)--(0,2));
// Block 5
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)--(-1/2,-sqrt(3)/2), blue); draw((-2,0)--(2,0)); draw((0,-2)--(0,2));
// Block 6
size(100,100); fill((-1/2,sqrt(3)/2)--(-1,0)--(-1/2,-sqrt(3)/2)--(1/2,-sqrt(3)/2)--(1,0)--(1/2,sqrt(3)/2)--cycle, lightgray); draw((-1/2,sqrt(3)/2)--(-1,0)--(-1/2,-sqrt(3)/2)--(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)--(-1/2,-sqrt(3)/2), blue); draw((-2,0)--(2,0)); draw((0,-2)--(0,2)); draw((0,0)--(1,0), heavygreen); | [] |
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 call the intersection of $\overline{UC}$ and $\overline{MV}$ $P$. Let $\overline{UP}=x$. Then $\overline{PC}=12-x$. Since $\overline{UC} \perp \overline{MV}$, $\overline{UP}$ and $\overline{CP}$ are heights of triangles $\triangle MUV$ and $\triangle MCV$, respectively. Both of these triangles have base $12$.
Area of $\triangle MUV = \frac{x\cdot12}{2}=6x$
Area of $\triangle MCV = \frac{(12-x)\cdot12}{2}=72-6x$
Adding these two gives us the area of trapezoid $MUVC$, which is $6x+(72-6x)=72$.
This is $\frac{3}{4}$ of the triangle, so the area of the triangle is $\frac{4}{3}\cdot{72}=\boxed{\textbf{(C) } 96}$ ~quacker88, diagram by programjames1 | // 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)); 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); | [] |
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 the two segments sum to $12$, $\overline{PC}$ and $\overline{UP}$ are $8$ and $4$, respectively.
Finally knowing that the medians divide the triangle into $6$ sections of equal area, finding the area of $\triangle PUM$ is enough. $\overline{PC} = \overline{MP} = 8$.
The area of $\triangle PUM = \frac{4\cdot8}{2}=16$. Multiplying this by $6$ gives us $6\cdot16=\boxed{\textbf{(C) }96}$
~quacker88 | // 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); 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); | [] |
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 trapezoid $MUVC$ is 72.
As said in solution 1, $[\triangle AMC] = 72 / \frac{3}{4} = \boxed{\textbf{(C) } 96}$.
-QuadraticFunctions, solution 1 by ??? | // 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)); 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); | [] |
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 $\triangle{MPC}$, \[MC^2=MP^2+PC^2=8^2+8^2=128,\] which implies $MC=\sqrt{128}=8\sqrt{2}$. By symmetry $MB=\dfrac{8\sqrt{2}}{2}=4\sqrt{2}$.
Applying the Pythagorean Theorem to right triangle $\triangle{MAB}$ gives $AB^2=AM^2-MB^2=8\sqrt{5}^2-4\sqrt{2}^2=288$, so $AB=\sqrt{288}=12\sqrt{2}$. Then the area of $\triangle{AMC}$ is \[\dfrac{AB \cdot MC}{2}=\dfrac{8\sqrt{2} \cdot 12\sqrt{2}}{2}=\dfrac{96 \cdot 2}{2}=\boxed{\textbf{(C) }96}\] | // 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); 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); | [] |
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 Pythagorean Theorem to triangle $MPC$ and get $MC=\sqrt{128}=8\sqrt{2}$. The area of triangle $MPC$ is $\dfrac{MP\cdot CP}{2}=32$, and that must equal to $\dfrac{MC\cdot BP}{2}$, so $BP=\dfrac{8}{\sqrt{2}}=4\cdot\sqrt{2}$. $BP=\dfrac{1}{3}BA$, so $BA=12\sqrt{2}$. The area of triangle $AMC$ is equal to $\dfrac{MC\cdot BA}{2}=\dfrac{8 \cdot \sqrt{2} \cdot 12 \cdot \sqrt{2}}{2}=\boxed{\textbf{(C)}\ 96}$.
-SmileKat32 | // 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); 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); | [] |
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 $\left(\frac{1}{2}\right)^2=\frac{1}{4}$. Therefore,
\[[AMC] = [MUVC] + \frac{1}{4} \cdot [AMC]\]
Simplifying gives us $\frac{3}{4} \cdot [AMC] = 72$, so $[AMC] = 72 \cdot \frac{4}{3} = \boxed{\textbf{(C)}\ 96}$
~MrThinker | // 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)); 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); | [] |
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 $\triangle MVC'$ = $\triangle MVC + \triangle MUV$ (since the latter has the same base and height as the sub-triangle $\triangle CVC'$ inside $\triangle MVC'$).
From this, we can deduce that $\textbf{(B)}$ cannot be true, since an incomplete part of $\triangle AMC$ is equal to it. We can also deduce that $\textbf{(D)}$ also cannot be true, since the unknown triangle $\triangle AUV = \triangle MUV$, and $\triangle MUV = \triangle CVC' < \triangle MVC'$. Therefore, the answer must be between $72$ and $144$, leaving $\boxed{\textbf{(C)}\ 96}$ as the only possible correct answer.
~DifrractSliver | // 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);
// Block 2
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); | [] |
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 nearest tenth$?$
$\textbf{(A) } 0.3 \qquad \textbf{(B) } 0.4 \qquad \textbf{(C) } 0.5 \qquad \textbf{(D) } 0.6 \qquad \textbf{(E) } 0.7$ | 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 this, and the fact that there are four circles, we write
\[4 \cdot \frac{1}{4} \cdot \pi d^2 = \frac{1}{2}\]
Solving for $d$, we obtain $d = \frac{1}{\sqrt{2\pi}}$, where with $\pi \approx 3$, we get $d \approx \frac{1}{\sqrt{6}} \approx \dfrac{1}{2.5} = \dfrac{10}{25} = \dfrac{2}{5}$, and from here, we see that $d \approx 0.4 \implies \boxed{\textbf{(B) } 0.4}.$
~Crypthes
~ Minor Edits by BakedPotato66
$\textbf{Note:}$ To be more rigorous, note that $d<0.5$ since if $d\geq0.5$ then clearly the probability is greater than $\frac{1}{2}$. This would make sure the above solution works, as if $d\geq0.5$ there is overlap with the quartercircles. $\textbf{- Emathmaster}$
Solution 2
As in the previous solution, we obtain the equation $4 \cdot \frac{1}{4} \cdot \pi d^2 = \frac{1}{2}$, which simplifies to $\pi d^2 = \frac{1}{2} = 0.5$. Since $\pi$ is slightly more than $3$, $d^2$ is slightly less than $\frac{0.5}{3} = 0.1\bar{6}$. We notice that $0.1\bar{6}$ is slightly more than $0.4^2 = 0.16$, so $d$ is roughly $\boxed{\textbf{(B) } 0.4}.$ ~emerald_block
Solution 3 (Estimating)
As above, we find that we need to estimate $d = \frac{1}{\sqrt{2\pi}}$.
Note that we can approximate $2\pi \approx 6.28318 \approx 6.25$ and so $\frac{1}{\sqrt{2\pi}}$ $\approx \frac{1}{\sqrt{6.25}}=\frac{1}{2.5}=0.4$.
And so our answer is $\boxed{\textbf{(B) } 0.4}$.
~Silverdragon
Solution 4 (Estimating but a bit different)
We only need to figure out the probability for a unit square, as it will scale up to the $2020\times 2020$ square. Since we want to find the probability that a point inside a unit square that is $d$ units away from a lattice point (a corner of the square) is $\frac{1}{2}$, we can find which answer will come the closest to covering $\frac{1}{2}$ of the area.
Since the closest is $0.4$ which turns out to be $(0.4)^2\times \pi = 0.16 \times \pi$ which is about $0.502$, we find that the answer rounded to the nearest tenth is $0.4$ or $\boxed{\textbf{(B)}}$.
~RuiyangWu
Solution 5 (Estimating but differently again)
As per the above diagram, realize that $\pi d^2 = \frac{1}{2}$, so $d = \frac{1}{(\sqrt{2})(\sqrt{\pi})}$.
$\sqrt{2} \approx 1.4 = \frac{7}{5}$.
$\sqrt{\pi}$ is between $1.7$ and $1.8$ $((1.7)^2 = 2.89$ and $(1.8)^2 = 3.24)$, so we can say $\sqrt{\pi} \approx 1.75 = \frac{7}{4}$.
So $d \approx \frac{1}{(\frac{7}{5})(\frac{7}{4})} = \frac{1}{\frac{49}{20}} = \frac{20}{49}$. This is slightly above $\boxed{\textbf{(B) } 0.4}$, since $\frac{20}{49} \approx \frac{2}{5}$.
-Solution by Joeya
Solution 6 (Estimating but differently again, again)
As above, we have the equation $\pi d^2 = \frac{1}{2}$, and we want to find the most accurate value of $d$. We resort to the answer choices and can plug those values of $d$ in and see which value of $d$ will lead to the most accurate value of $\pi$.
Starting off in the middle, we try option C with $d=0.5$. Plugging this in, we get $\pi \left(\frac{1}{2}\right)^2 = \frac{1}{2},$ and after simplifying we get $\pi = \frac{1}{2} \cdot 4 = 2.$
That's not very good. We know $\pi \approx 3.14.$
Let's see if we can do better. Trying option A with $d = 0.3,$ we get $\pi = \frac{1}{2} \cdot \frac{100}{9} = \frac{50}{9} = 5 \frac{5}{9}.$
Hm, let's try option B with $d = 0.4.$ We get $\pi = \frac{1}{2} \cdot \frac{25}{4} = \frac{25}{8} = 3 \frac{1}{8}$. This is very close to $\pi$ and is the best estimate for $\pi$ of the 5 options.
Therefore, the answer is $\boxed{\textbf{(B) } 0.4}.$
~ epiconan | 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), 0.3989, 270, 360)); filldraw(arc((0,1), 0.3989, 270, 360)--(0,1)--cycle, gray); | [] |
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 nearest tenth$?$
$\textbf{(A) } 0.3 \qquad \textbf{(B) } 0.4 \qquad \textbf{(C) } 0.5 \qquad \textbf{(D) } 0.6 \qquad \textbf{(E) } 0.7$ | 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 shaded region (since there are $k$ unit squares).
Let $n$ be the side length of a square. When $n=1$, the shaded areas represent half of the total area:
When $n=2$:
For $n=3$:
We can calculate the total number of shaded circles given some $n$. There are $(n-1)^2$ full circles on the inside, $4(n-1)$ semicircles on the sides, and $4$ quarter circles for the corners.
Full circles are, of course, worth one circle. Semicircles are worth half a circle each, and quarter circles are worth $\dfrac14$ of a circle. Thus, weighing our sum gives $(n-1)^2+\dfrac{4(n-1)}2+\dfrac44=n^2-2n+1+2(n-1)+1=n^2-2n+2n+2-2=n^2.$ Thus, there is $n^2\cdot\pi r^2$ worth of the shaded area for any $n$, and since the area of each circle is $\pi r^2$ if $r$ is the radius of each.
We want the ratio of this shaded area to the entire to be $\dfrac12$. The area of the entire square is $n^2$, so dividing, we see that $\dfrac{n^2\cdot\pi r^2}{n^2}=\pi r^2=\dfrac12$.
The rest is the same as solution $1$. | // 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);
draw(arc((0,1), 0.3989, 270, 360));
filldraw(arc((0,1), 0.3989, 270, 360)--(0,1)--cycle, gray);
// Block 2
size(10cm);
filldraw((arc((0,0), 0.1994, 0, 90))--(0,0)--cycle, gray);
draw(arc((1,0), 0.1994, 90, 180));
filldraw((arc((1,0), 0.1994, 90, 180))--(1,0)--cycle, gray);
draw(arc((1,1), 0.1994, 180, 270));
filldraw((arc((1,1), 0.1994, 180, 270))--(1,1)--cycle, gray);
draw(arc((0,1), 0.1994, 270, 360));
filldraw(arc((0,1), 0.1994, 270, 360)--(0,1)--cycle, gray);
draw(arc((0.5,0.5), 0.1994,0,360));
filldraw(arc((0.5,0.5), 0.1994,0,360)--(0.5,0.5)--cycle, gray);
draw(arc((0.5,0), 0.1994,0,180));
filldraw(arc((0.5,0), 0.1994,0,180)--(0.5,0)--cycle, gray);
draw(arc((0,0.5), 0.1994,-90,90));
filldraw(arc((0,0.5), 0.1994,-90,90)--(0,0.5)--cycle, gray);
filldraw(arc((1,0.5), 0.1994,90,270)--(1,0.5)--cycle, gray);
filldraw(arc((0.5,1), 0.1994,0,-180)--(0.5,1)--cycle, gray);
draw((0,0)--(0.5,0)--(0.5,1)--(0,1)--(0,0)--(1,0)--(1,1)--(0,1)--(0,0.5)--(1,0.5));
// Block 3
size(10cm);
filldraw(arc((0,0),0.1330,0,90)--(0,0)--cycle, gray);
filldraw(arc((0,1),0.1330,-90,0)--(0,1)--cycle, gray);
filldraw(arc((1,0),0.1330,90,180)--(1,0)--cycle, gray);
filldraw(arc((1,1),0.1330,-180,-90)--(1,1)--cycle, gray);
filldraw(arc((0.333,0.333),0.133,0,360)--(0.333,0.333)--cycle, gray);
filldraw(arc((0.667,0.333),0.133,0,360)--(0.667,0.333)--cycle, gray);
filldraw(arc((0.333,0.667),0.133,0,360)--(0.333,0.667)--cycle, gray);
filldraw(arc((0.667,0.667),0.133,0,360)--(0.667,0.667)--cycle, gray);
filldraw(arc((0.333,0),0.133,0,180)--(0.333,0)--cycle, gray);
filldraw(arc((0.667,0),0.133,0,180)--(0.667,0)--cycle, gray);
filldraw(arc((0.333,1),0.133,-180,0)--(0.333,1)--cycle, gray);
filldraw(arc((0.666,1),0.133,-180,0)--(0.666,1)--cycle, gray);
filldraw(arc((0,0.333),0.133,-90,90)--(0,0.333)--cycle, gray);
filldraw(arc((0,0.667),0.133,-90,90)--(0,0.667)--cycle, gray);
filldraw(arc((1,0.333),0.133,90,270)--(1,0.333)--cycle, gray);
filldraw(arc((1,0.667),0.133,90,270)--(1,0.667)--cycle, gray);
draw((0,0)--(0,1)--(0.333,1)--(0.333,0)--(0.667,0)--(0.667,1)--(1,1)--(1,0)--(0,0)--(0,0.333)--(1,0.333)--(1,0.667)--(0,0.667));
draw((0.333,1)--(0.667,1));
// Block 4
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); draw(arc((0,1), 0.3989, 270, 360)); filldraw(arc((0,1), 0.3989, 270, 360)--(0,1)--cycle, gray);
// Block 5
size(10cm); filldraw((arc((0,0), 0.1994, 0, 90))--(0,0)--cycle, gray); draw(arc((1,0), 0.1994, 90, 180)); filldraw((arc((1,0), 0.1994, 90, 180))--(1,0)--cycle, gray); draw(arc((1,1), 0.1994, 180, 270)); filldraw((arc((1,1), 0.1994, 180, 270))--(1,1)--cycle, gray); draw(arc((0,1), 0.1994, 270, 360)); filldraw(arc((0,1), 0.1994, 270, 360)--(0,1)--cycle, gray); draw(arc((0.5,0.5), 0.1994,0,360)); filldraw(arc((0.5,0.5), 0.1994,0,360)--(0.5,0.5)--cycle, gray); draw(arc((0.5,0), 0.1994,0,180)); filldraw(arc((0.5,0), 0.1994,0,180)--(0.5,0)--cycle, gray); draw(arc((0,0.5), 0.1994,-90,90)); filldraw(arc((0,0.5), 0.1994,-90,90)--(0,0.5)--cycle, gray); filldraw(arc((1,0.5), 0.1994,90,270)--(1,0.5)--cycle, gray); filldraw(arc((0.5,1), 0.1994,0,-180)--(0.5,1)--cycle, gray); draw((0,0)--(0.5,0)--(0.5,1)--(0,1)--(0,0)--(1,0)--(1,1)--(0,1)--(0,0.5)--(1,0.5));
// Block 6
size(10cm); filldraw(arc((0,0),0.1330,0,90)--(0,0)--cycle, gray); filldraw(arc((0,1),0.1330,-90,0)--(0,1)--cycle, gray); filldraw(arc((1,0),0.1330,90,180)--(1,0)--cycle, gray); filldraw(arc((1,1),0.1330,-180,-90)--(1,1)--cycle, gray); filldraw(arc((0.333,0.333),0.133,0,360)--(0.333,0.333)--cycle, gray); filldraw(arc((0.667,0.333),0.133,0,360)--(0.667,0.333)--cycle, gray); filldraw(arc((0.333,0.667),0.133,0,360)--(0.333,0.667)--cycle, gray); filldraw(arc((0.667,0.667),0.133,0,360)--(0.667,0.667)--cycle, gray); filldraw(arc((0.333,0),0.133,0,180)--(0.333,0)--cycle, gray); filldraw(arc((0.667,0),0.133,0,180)--(0.667,0)--cycle, gray); filldraw(arc((0.333,1),0.133,-180,0)--(0.333,1)--cycle, gray); filldraw(arc((0.666,1),0.133,-180,0)--(0.666,1)--cycle, gray); filldraw(arc((0,0.333),0.133,-90,90)--(0,0.333)--cycle, gray); filldraw(arc((0,0.667),0.133,-90,90)--(0,0.667)--cycle, gray); filldraw(arc((1,0.333),0.133,90,270)--(1,0.333)--cycle, gray); filldraw(arc((1,0.667),0.133,90,270)--(1,0.667)--cycle, gray); draw((0,0)--(0,1)--(0.333,1)--(0.333,0)--(0.667,0)--(0.667,1)--(1,1)--(1,0)--(0,0)--(0,0.333)--(1,0.333)--(1,0.667)--(0,0.667)); draw((0.333,1)--(0.667,1)); | [] |
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 $P(x).$
~jasmineLOVER7254 | // 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) {
if (i != 0)
draw((i, -0.1) -- (i, 0.1)); // Tick mark
// label(string(i), (i, -0.5), S); // Label each tick below the line
}
label("$1^2$", (-4, -0.5), S); // Label each tick below the line
label("$2^2$", (-3, -0.5), S); // Label each tick below the line
label("$3^2$", (-2, -0.5), S); // Label each tick below the line
label("$4^2$", (-1, -0.5), S); // Label each tick below the line
label("\dots", (0, -0.5), S); // Label each tick below the line
label("$97^2$", (1, -0.5), S); // Label each tick below the line
label("$98^2$", (2, -0.5), S); // Label each tick below the line
label("$99^2$", (3, -0.5), S); // Label each tick below the line
label("$100^2$", (4, -0.5), S); // Label each tick below the line
label("$0$", (-4, 0.1), N); // Label each tick below the line
label("$0$", (-3, 0.1), N); // Label each tick below the line
label("$0$", (-2, 0.1), N); // Label each tick below the line
label("$0$", (-1, 0.1), N); // Label each tick below the line
label("$0$", (1, 0.1), N); // Label each tick below the line
label("$0$", (2, 0.1), N); // Label each tick below the line
label("$0$", (3, 0.1), N); // Label each tick below the line
label("$0$", (4, 0.1), N); // Label each tick below the line
label("$+$", (-4.5, 0), N); // Label each tick below the line
label("$-$", (-3.5, 0), N); // Label each tick below the line
label("$+$", (-2.5, 0), N); // Label each tick below the line
label("$-$", (-1.5, 0), N); // Label each tick below the line
label("$+$", (4.5, 0), N); // Label each tick below the line
label("$-$", (3.5, 0), N); // Label each tick below the line
label("$+$", (2.5, 0), N); // Label each tick below the line
label("$-$", (1.5, 0), N); // Label each tick below the line
// Block 2
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) { if (i != 0) draw((i, -0.1) -- (i, 0.1)); // Tick mark // label(string(i), (i, -0.5), S); // Label each tick below the line } label("$1^2$", (-4, -0.5), S); // Label each tick below the line label("$2^2$", (-3, -0.5), S); // Label each tick below the line label("$3^2$", (-2, -0.5), S); // Label each tick below the line label("$4^2$", (-1, -0.5), S); // Label each tick below the line label("\dots", (0, -0.5), S); // Label each tick below the line label("$97^2$", (1, -0.5), S); // Label each tick below the line label("$98^2$", (2, -0.5), S); // Label each tick below the line label("$99^2$", (3, -0.5), S); // Label each tick below the line label("$100^2$", (4, -0.5), S); // Label each tick below the line label("$0$", (-4, 0.1), N); // Label each tick below the line label("$0$", (-3, 0.1), N); // Label each tick below the line label("$0$", (-2, 0.1), N); // Label each tick below the line label("$0$", (-1, 0.1), N); // Label each tick below the line label("$0$", (1, 0.1), N); // Label each tick below the line label("$0$", (2, 0.1), N); // Label each tick below the line label("$0$", (3, 0.1), N); // Label each tick below the line label("$0$", (4, 0.1), N); // Label each tick below the line label("$+$", (-4.5, 0), N); // Label each tick below the line label("$-$", (-3.5, 0), N); // Label each tick below the line label("$+$", (-2.5, 0), N); // Label each tick below the line label("$-$", (-1.5, 0), N); // Label each tick below the line label("$+$", (4.5, 0), N); // Label each tick below the line label("$-$", (3.5, 0), N); // Label each tick below the line label("$+$", (2.5, 0), N); // Label each tick below the line label("$-$", (1.5, 0), N); // Label each tick below the line | [] |
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 $ABCD?$
$\textbf{(A) } 330 \qquad \textbf{(B) } 340 \qquad \textbf{(C) } 350 \qquad \textbf{(D) } 360 \qquad \textbf{(E) } 370$ | 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 also see two similar triangles, $\triangle BFE \sim \triangle DCE$. Since $EC$ is $20-5=15$, and $DC=30$, we get that $BF=2x$.
Now, if we redraw another diagram just of $ABC$, we get that $(2x)^2=(5-x)(15+x)$ because of the altitude geometric mean theorem which states that in any right triangle, the altitude squared is equal to the product of the two lengths that it divides the base into.
Expanding, simplifying, and dividing by the GCF, we get $x^2+2x-15=0$. This factors to $(x+5)(x-3)$, which has roots of $x=-5, 3$. Since lengths cannot be negative, $x=3$. Since $x=3$, that means the altitude $BF=2\cdot3=6$, or $[ABC]=60$. Thus $[ABCD]=[ACD]+[ABC]=300+60=\boxed {\textbf{(D) }360}$
~ Solution by Ultraman
~ Diagram by ciceronii | // 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("20", (0,2)--(4,0), SW);
label("30", (4,0)--(6,4), SE);
label("$x$", (1,1.5)--(1.714,1.143), NE);
label("5$-$$x$", (1,1.5)--(0,2), NE);
draw(rightanglemark((0,2),(0,0),(4,0)));
draw(rightanglemark((0,2),(4,0),(6,4)));
draw(rightanglemark((0,0),(1,1.5),(0,2)));
// Block 2
size(15cm,0); import olympiad; draw((0,0)--(0,2)--(6,4)--(4,0)--cycle); label("A", (0,2), NW); label("B", (0,0), SW); label("C", (4,0), SE); label("D", (6,4), NE); label("E", (1.714,1.143), N); label("F", (1,1.5), N); draw((0,2)--(4,0), dashed); draw((0,0)--(6,4), dashed); draw((0,0)--(1,1.5), dashed); label("20", (0,2)--(4,0), SW); label("30", (4,0)--(6,4), SE); label("$x$", (1,1.5)--(1.714,1.143), NE); label("5$-$$x$", (1,1.5)--(0,2), NE); draw(rightanglemark((0,2),(0,0),(4,0))); draw(rightanglemark((0,2),(4,0),(6,4))); draw(rightanglemark((0,0),(1,1.5),(0,2))); | [] |
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 $ABCD?$
$\textbf{(A) } 330 \qquad \textbf{(B) } 340 \qquad \textbf{(C) } 350 \qquad \textbf{(D) } 360 \qquad \textbf{(E) } 370$ | 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 solutions $(0,10)$ and $(-8,-6)$. We immediately discard the $(0,10)$ solution as $y$ should be negative. Thus, we conclude that $[ABCD]=[ACD]+[ABC]=\frac{20\cdot30}{2}+\frac{20\cdot6}{2}=\boxed{\textbf{(D)}\:360}$. | // 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);
label("15",(-5,0)--(10,0), N);
label("30",(10,0)--(10,30), E);
dot((-5,0));
dot((-10,0));
dot((-8,-6));
dot((10,0));
dot((10,30));
// Block 2
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",(-5,0)--(10,0), N); label("30",(10,0)--(10,30), E); dot((-5,0)); dot((-10,0)); dot((-8,-6)); dot((10,0)); dot((10,30)); | [] |
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 $ABCD?$
$\textbf{(A) } 330 \qquad \textbf{(B) } 340 \qquad \textbf{(C) } 350 \qquad \textbf{(D) } 360 \qquad \textbf{(E) } 370$ | 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:
\[[AEB] + [BEC] = \frac{1}{2}(x)(5)\sin(\angle DEC) + \frac{1}{2}(x)(15)\sin(180-\angle DEC) = (3)(20) = 60\].
Thus the desired area is $\frac{1}{2}(30)(20) + 60 = \boxed{\textbf{(D) } 360}$
~qwertysri987 | // 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 = (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); | [] |
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 $ABCD?$
$\textbf{(A) } 330 \qquad \textbf{(B) } 340 \qquad \textbf{(C) } 350 \qquad \textbf{(D) } 360 \qquad \textbf{(E) } 370$ | 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 $\angle EFG=90^\circ, EG=\sqrt{5^2+10^2}=5\sqrt{5} \implies BE=\frac{AE\cdot EC}{EG}=\frac{5\cdot 15}{5\sqrt{5}}=3\sqrt{5}$ via Power of a Point.
The altitude from $B$ to $AC$ is then equal to $GF\cdot \frac{BE}{GE}=10\cdot \frac{3\sqrt 5}{5 \sqrt 5}=6$.
Finally, the total area of $ABCD$ is equal to $\frac 12 \cdot 20 \left(30+6 \right) =\boxed{\text{(D) } 360}.$
~asops | // 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, dotted);
// Block 2
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); | [] |
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 $ABCD?$
$\textbf{(A) } 330 \qquad \textbf{(B) } 340 \qquad \textbf{(C) } 350 \qquad \textbf{(D) } 360 \qquad \textbf{(E) } 370$ | 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 $\triangle CEF \sim \triangle CAB$, $EF = AB \cdot \frac{CE}{CA} = x \cdot \frac{15}{20} = \frac{3x}{4}$
$BF = BC - CF = BC - BC \cdot \frac{CE}{CA} = \frac{1}{4} \cdot BC = \frac{y}{4}$
$BE = \sqrt{ \left( \frac{3x}{4} \right) ^2 + \left( \frac{y}{4} \right) ^2 } = \frac{ \sqrt{9x^2 + y^2} }{4}$ , substituting $x^2 + y^2 = 400$, we get $BE = \frac{ \sqrt{8x^2 + 400} }{4} = \frac{ \sqrt{2x^2 + 100} }{2}$
$[ABC] = \frac{1}{2} \cdot x \cdot y$
Because $\triangle ABC$ and $\triangle ACD$ share the same base, $\frac{[ABC]}{[ACD]} = \frac{BE}{DE}$
$[ABC] = [ACD] \cdot \frac{BE}{DE} = 300 \cdot \frac{ \frac{ \sqrt{2x^2 + 100} } {2} }{ 15 \sqrt{5} }$
$\frac{1}{2} \cdot x \cdot y = 20 \cdot \frac{ \frac{ \sqrt{2x^2 + 100} } {2} }{ \sqrt{5} }$
$xy = 4 \sqrt{10x^2 + 500}$
By $x^2 + y^2 = 400$, $y = \sqrt{400 - x^2}$. So, $x \cdot \sqrt{400 - x^2} = 4 \sqrt{10x^2 + 500}$
$x^2 (400 - x^2) = 16 (10x^2 + 500)$
Let $x^2 = a$, $a (400 - a) = 16 (10a + 500)$, $400a - a^2 = 160a + 8000$, $a^2 - 240a + 8000 = 0$, $(a-200)(a-40) = 0$
Because $x < 20$, $a$ can only equal 40. $a = 40$, $x = 2 \sqrt{10}$, $y = 6 \sqrt{10}$
$[ABC] = \frac{1}{2} \cdot 2 \sqrt{10} \cdot 6 \sqrt{10} = 60$
$[ABCD] = [ABC] + [ACD] = 60 + 300 = \boxed{\text{(D) } 360}$
~isabelchen | // 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), dashed);
label("20", (0,2)--(4,0), SW);
label("30", (4,0)--(6,4), SE);
label("$x$", (-0.3,2)--(-0.3,0), N);
label("$y$", (0,-0.3)--(4,-0.3), E);
draw(rightanglemark((1.714,2),(1.714,0),(5.714,0)));
draw(rightanglemark((0,2),(0,0),(4,0)));
draw(rightanglemark((0,2),(4,0),(6,4)));
// Block 2
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); label("20", (0,2)--(4,0), SW); label("30", (4,0)--(6,4), SE); label("$x$", (-0.3,2)--(-0.3,0), N); label("$y$", (0,-0.3)--(4,-0.3), E); draw(rightanglemark((1.714,2),(1.714,0),(5.714,0))); draw(rightanglemark((0,2),(0,0),(4,0))); draw(rightanglemark((0,2),(4,0),(6,4))); | [] |
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 $ABCD?$
$\textbf{(A) } 330 \qquad \textbf{(B) } 340 \qquad \textbf{(C) } 350 \qquad \textbf{(D) } 360 \qquad \textbf{(E) } 370$ | 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}. \end{align*}
~luckuso | // 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((6,0)--(6,4), dashed);
draw((1.714,1.143)--(1.714,0), dashed);
label("20", (0,2)--(4,0), SW);
label("30", (4,0)--(6,4), SE);
label("$x$", (-0.3,2)--(-0.3,0), N);
label("$y$", (0,-0.3)--(4,-0.3), E);
label( "$X$", (6,0), SE);
label("5", (0,2)--(1.714,1.143), NE);
label("15",(1.714,1.143)--(4,0),NE);
label("5$C$", (0,0)--(1.714,0),S);
label("15$C$", (1.714,0)--(4,0),S);
label("30$S$", (4,0)--(6,0),S);
label("30$C$", (6,0)--(6,4),E);
draw(anglemark((1.714,1.143),(4,0),(1.714,0)));
draw(anglemark((4,0),(6,4),(6,0)));
draw(rightanglemark((1.714,2),(1.714,0),(5.714,0)));
draw(rightanglemark((0,2),(0,0),(4,0)));
draw(rightanglemark((0,2),(4,0),(6,4)));
// Block 2
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)--(6,4), dashed); draw((1.714,1.143)--(1.714,0), dashed); label("20", (0,2)--(4,0), SW); label("30", (4,0)--(6,4), SE); label("$x$", (-0.3,2)--(-0.3,0), N); label("$y$", (0,-0.3)--(4,-0.3), E); label( "$X$", (6,0), SE); label("5", (0,2)--(1.714,1.143), NE); label("15",(1.714,1.143)--(4,0),NE); label("5$C$", (0,0)--(1.714,0),S); label("15$C$", (1.714,0)--(4,0),S); label("30$S$", (4,0)--(6,0),S); label("30$C$", (6,0)--(6,4),E); draw(anglemark((1.714,1.143),(4,0),(1.714,0))); draw(anglemark((4,0),(6,4),(6,0))); draw(rightanglemark((1.714,2),(1.714,0),(5.714,0))); draw(rightanglemark((0,2),(0,0),(4,0))); draw(rightanglemark((0,2),(4,0),(6,4))); | [] |
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 $y$-axis. How many of the $125$ sequences of three of these transformations (not necessarily distinct) will return $T$ to its original position? (For example, a $180^{\circ}$ rotation, followed by a reflection across the $x$-axis, followed by a reflection across the $y$-axis will return $T$ to its original position, but a $90^{\circ}$ rotation, followed by a reflection across the $x$-axis, followed by another reflection across the $x$-axis will not return $T$ to its original position.)
$\textbf{(A) } 12 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 17 \qquad \textbf{(D) } 20 \qquad \textbf{(E) } 25$ | 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 the original orientation. Since we are only allowed $3$ transformations and an even number of them must be reflections, we either reflect $T$ $0$ times or $2$ times.
Case 1: $0$ reflections on $T$.
In this case, we must use $3$ rotations to return $T$ to its original position. Notice that our set of rotations, $\{90^\circ,180^\circ,270^\circ\}$, contains every multiple of $90^\circ$ except for $0^\circ$. We can start with any two rotations $a,b$ in $\{90^\circ,180^\circ,270^\circ\}$ and there must be exactly one $c \equiv -a - b \pmod{360^\circ}$ such that we can use the three rotations $(a,b,c)$ which ensures that $a + b + c \equiv 0^\circ \pmod{360^\circ}$. That way, the composition of rotations $a,b,c$ yields a full rotation. For example, if $a = b = 90^\circ$, then $c \equiv -90^\circ - 90^\circ = -180^\circ \pmod{360^\circ}$, so $c = 180^\circ$ and the rotations $(90^\circ,90^\circ,180^\circ)$ yields a full rotation.
The only case in which this fails is when $c$ would have to equal $0^\circ$. This happens when $(a,b)$ is already a full rotation, namely, $(a,b) = (90^\circ,270^\circ),(180^\circ,180^\circ),$ or $(270^\circ,90^\circ)$. However, we can simply subtract these three cases from the total. Selecting $(a,b)$ from $\{90^\circ,180^\circ,270^\circ\}$ yields $3 \cdot 3 = 9$ choices, and with $3$ that fail, we are left with $6$ combinations for case $1$.
Case 2: $2$ reflections on $T$.
In this case, we first eliminate the possibility of having two of the same reflection. Since two reflections across the x-axis maps $T$ back to itself, inserting a rotation before, between, or after these two reflections would change $T$'s final location, meaning that any combination involving two reflections across the x-axis would not map $T$ back to itself. The same applies to two reflections across the y-axis.
Therefore, we must use one reflection about the x-axis, one reflection about the y-axis, and one rotation. Since a reflection about the x-axis changes the sign of the y component, a reflection about the y-axis changes the sign of the x component, and a $180^\circ$ rotation changes both signs, these three transformation composed (in any order) will suffice. It is therefore only a question of arranging the three, giving us $3! = 6$ combinations for case 2.
Combining both cases we get $6+6=\boxed{\textbf{(A)} 12}$. | // 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, gray, black+linewidth(1)); | [] |
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));
filldraw((2,1)--(2,2)--(3,2)--(3,1)--cycle, gray(0.7));
filldraw((3,0)--(4,0)--(4,1)--(3,1)--cycle, gray(0.9));
filldraw((1,0)--(2,0)--(2,1)--(1,1)--cycle, gray(0.9));
filldraw((4,0)--(5,0)--(5,1)--(4,1)--cycle, gray(0.7));
filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray(0.9));
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) {
for (int j = 0; j < 6; ++j) {
if (j != 5) {
draw((j,i)--(j+1,i));
}
if (i != 4) {
draw((j,i)--(j,i+1));
}
}
}
// Block 2
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)); filldraw((2,1)--(2,2)--(3,2)--(3,1)--cycle, gray(0.7)); filldraw((3,0)--(4,0)--(4,1)--(3,1)--cycle, gray(0.9)); filldraw((1,0)--(2,0)--(2,1)--(1,1)--cycle, gray(0.9)); filldraw((4,0)--(5,0)--(5,1)--(4,1)--cycle, gray(0.7)); filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray(0.9)); 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) { for (int j = 0; j < 6; ++j) { if (j != 5) { draw((j,i)--(j+1,i)); } if (i != 4) { draw((j,i)--(j,i+1)); } } } | [] |
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 on one line of symmetry, $B$ contributes $2$ shaded unit squares to the resulting figure.
The shaded unit squares contributed by $A,B,$ and $C$ are all distinct, so we need to shade at least $4+4+2-3=\boxed{\textbf{(D) } 7}$ unit squares in addition, as shown below:
~MRENTHUSIASM | // 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) {
for (int j = 0; j < 6; ++j) {
if (j != 5) {
draw((j,i)--(j+1,i));
}
if (i != 4) {
draw((j,i)--(j,i+1));
}
}
}
draw((-1,2)--(6,2),linewidth(2)+red);
draw((2.5,-1)--(2.5,5),linewidth(2)+red);
label("$A$",(1.5,3.5));
label("$B$",(2.5,1.5));
label("$C$",(4.5,0.5));
// Block 2
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));
filldraw((2,1)--(2,2)--(3,2)--(3,1)--cycle, gray(0.7));
filldraw((3,0)--(4,0)--(4,1)--(3,1)--cycle, gray(0.9));
filldraw((1,0)--(2,0)--(2,1)--(1,1)--cycle, gray(0.9));
filldraw((4,0)--(5,0)--(5,1)--(4,1)--cycle, gray(0.7));
filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray(0.9));
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) {
for (int j = 0; j < 6; ++j) {
if (j != 5) {
draw((j,i)--(j+1,i));
}
if (i != 4) {
draw((j,i)--(j,i+1));
}
}
}
draw((-1,2)--(6,2),linewidth(2)+red);
draw((2.5,-1)--(2.5,5),linewidth(2)+red);
label("$A$",(1.5,3.5));
label("$B$",(2.5,1.5));
label("$C$",(4.5,0.5));
label("$A'$",(3.5,3.5));
label("$A'$",(1.5,0.5));
label("$A'$",(3.5,0.5));
label("$B'$",(2.5,2.5));
label("$C'$",(0.5,0.5));
label("$C'$",(0.5,3.5));
label("$C'$",(4.5,3.5));
// Block 3
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) { for (int j = 0; j < 6; ++j) { if (j != 5) { draw((j,i)--(j+1,i)); } if (i != 4) { draw((j,i)--(j,i+1)); } } } draw((-1,2)--(6,2),linewidth(2)+red); draw((2.5,-1)--(2.5,5),linewidth(2)+red); label("$A$",(1.5,3.5)); label("$B$",(2.5,1.5)); label("$C$",(4.5,0.5));
// Block 4
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)); filldraw((2,1)--(2,2)--(3,2)--(3,1)--cycle, gray(0.7)); filldraw((3,0)--(4,0)--(4,1)--(3,1)--cycle, gray(0.9)); filldraw((1,0)--(2,0)--(2,1)--(1,1)--cycle, gray(0.9)); filldraw((4,0)--(5,0)--(5,1)--(4,1)--cycle, gray(0.7)); filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray(0.9)); 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) { for (int j = 0; j < 6; ++j) { if (j != 5) { draw((j,i)--(j+1,i)); } if (i != 4) { draw((j,i)--(j,i+1)); } } } draw((-1,2)--(6,2),linewidth(2)+red); draw((2.5,-1)--(2.5,5),linewidth(2)+red); label("$A$",(1.5,3.5)); label("$B$",(2.5,1.5)); label("$C$",(4.5,0.5)); label("$A'$",(3.5,3.5)); label("$A'$",(1.5,0.5)); label("$A'$",(3.5,0.5)); label("$B'$",(2.5,2.5)); label("$C'$",(0.5,0.5)); label("$C'$",(0.5,3.5)); label("$C'$",(4.5,3.5)); | [] |
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: \[\biggl[0,\frac{\pi}{4}\biggr),\left(\frac{\pi}{4},\frac{3\pi}{4}\right),\left(\frac{3\pi}{4},\frac{5\pi}{4}\right),\left(\frac{5\pi}{4},\frac{7\pi}{4}\right),\left(\frac{7\pi}{4},2\pi\right].\]
Note that $\tan(2x)\in[0,\infty)$ for the first branch, $\tan(2x)\in(-\infty,\infty)$ for the three middle branches, and $\tan(2x)\in(-\infty,0]$ for the last branch. Moreover, all branches are strictly increasing.
The graph of $y=\cos\left(\frac x2\right)$ has a period of $4\pi$ and zeros at $x=\pi+2k\pi$ for some integer $k.$
On the interval $[0,2\pi],$ note that $\cos\left(\frac x2\right)\in[-1,1].$ Moreover, the graph is strictly decreasing.
The graphs of $y=\tan(2x)$ and $y=\cos\left(\frac x2\right)$ intersect once on each of the five branches of $y=\tan(2x),$ as shown below:
Therefore, the answer is $\boxed{\textbf{(E)}\ 5}.$
~MRENTHUSIASM ~lopkiloinm ~hi13 ~annabelle0913 ~codecow | // 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)/2+3*pi/2),red);
draw(graph(f,-atan(3)/2+4*pi/2,2*pi),red);
draw(graph(g,0,2pi),blue,"$y=\cos\left(\frac x2\right)$");
real xMin = 0;
real xMax = 9/4*pi;
real yMin = -3;
real yMax = 3;
//Draws the horizontal gridlines
void horizontalLines()
{
for (real i = yMin+1; i < yMax; ++i)
{
draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4));
}
}
//Draws the vertical gridlines
void verticalLines()
{
for (real i = xMin+pi/2; i < xMax; i+=pi/2)
{
draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4));
}
}
//Draws the horizontal ticks
void horizontalTicks()
{
for (real i = yMin+1; i < yMax; ++i)
{
draw((-1/8,i)--(1/8,i), black+linewidth(1));
}
}
//Draws the vertical ticks
void verticalTicks()
{
for (real i = xMin+pi/2; i < xMax; i+=pi/2)
{
draw((i,-1/8)--(i,1/8), black+linewidth(1));
}
}
horizontalLines();
verticalLines();
horizontalTicks();
verticalTicks();
draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5));
draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5));
label("$x$",(xMax,0),(2,0));
label("$y$",(0,yMax),(0,2));
pair A[], B[];
A[0] = (2pi,0);
A[1] = (0,2);
A[2] = (0,0);
A[3] = (0,-2);
B[0] = intersectionpoints(graph(f,0,atan(3)/2),graph(g,0,2pi))[0];
B[1] = intersectionpoints(graph(f,-atan(3)/2+pi/2,atan(3)/2+pi/2),graph(g,0,2pi))[0];
B[2] = intersectionpoints(graph(f,-atan(3)/2+2*pi/2,atan(3)/2+2*pi/2),graph(g,0,2pi))[0];
B[3] = intersectionpoints(graph(f,-atan(3)/2+3*pi/2,atan(3)/2+3*pi/2),graph(g,0,2pi))[0];
B[4] = intersectionpoints(graph(f,-atan(3)/2+4*pi/2,atan(3)/2+4*pi/2),graph(g,0,2pi))[0];
label("$2\pi$",A[0],(0,-2.5));
label("$2$",A[1],(-2.5,0));
label("$0$",A[2],(-2.5,0));
label("$-2$",A[3],(-2.5,0));
for (int i = 0; i < 5; ++i)
{
dot(B[i],black+linewidth(5));
}
add(legend(),point(E),60E,UnFill);
// Block 2
/* 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)/2+3*pi/2),red); draw(graph(f,-atan(3)/2+4*pi/2,2*pi),red); draw(graph(g,0,2pi),blue,"$y=\cos\left(\frac x2\right)$"); real xMin = 0; real xMax = 9/4*pi; real yMin = -3; real yMax = 3; //Draws the horizontal gridlines void horizontalLines() { for (real i = yMin+1; i < yMax; ++i) { draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4)); } } //Draws the vertical gridlines void verticalLines() { for (real i = xMin+pi/2; i < xMax; i+=pi/2) { draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4)); } } //Draws the horizontal ticks void horizontalTicks() { for (real i = yMin+1; i < yMax; ++i) { draw((-1/8,i)--(1/8,i), black+linewidth(1)); } } //Draws the vertical ticks void verticalTicks() { for (real i = xMin+pi/2; i < xMax; i+=pi/2) { draw((i,-1/8)--(i,1/8), black+linewidth(1)); } } horizontalLines(); verticalLines(); horizontalTicks(); verticalTicks(); draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("$x$",(xMax,0),(2,0)); label("$y$",(0,yMax),(0,2)); pair A[], B[]; A[0] = (2pi,0); A[1] = (0,2); A[2] = (0,0); A[3] = (0,-2); B[0] = intersectionpoints(graph(f,0,atan(3)/2),graph(g,0,2pi))[0]; B[1] = intersectionpoints(graph(f,-atan(3)/2+pi/2,atan(3)/2+pi/2),graph(g,0,2pi))[0]; B[2] = intersectionpoints(graph(f,-atan(3)/2+2*pi/2,atan(3)/2+2*pi/2),graph(g,0,2pi))[0]; B[3] = intersectionpoints(graph(f,-atan(3)/2+3*pi/2,atan(3)/2+3*pi/2),graph(g,0,2pi))[0]; B[4] = intersectionpoints(graph(f,-atan(3)/2+4*pi/2,atan(3)/2+4*pi/2),graph(g,0,2pi))[0]; label("$2\pi$",A[0],(0,-2.5)); label("$2$",A[1],(-2.5,0)); label("$0$",A[2],(-2.5,0)); label("$-2$",A[3],(-2.5,0)); for (int i = 0; i < 5; ++i) { dot(B[i],black+linewidth(5)); } add(legend(),point(E),60E,UnFill); | [] |
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 $k?$
$\textbf{(A) } 10 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 25 \qquad \textbf{(E) } 30$ | 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 \[BX = 20 \tan{\angle{XPB}} = 20 \cdot \frac{1 - 3/5}{1 + 3/5} = 5\] by tangent addition formula. Since $BX$ is 5 and the sidelength of the square is 20 the answer is $20 - 5 \implies \boxed{\textbf{B}}.$ | // 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)); 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)); | [] |
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 determine the length of its diagonals.
In order to do this, we first determine the area of the octagon. Letting the side length be $a$, we can create a square of length $s$ around it (see figure).
Creating a small square of side length $a$ from the corners of this figure gives us an area of $a^2$. Thus, $s^2 - a^2 = n$ where $n$ is the area of the octagon. We know from the Pythagorean Theorem that $s = a + \frac{a}{\sqrt{2}}$, meaning that $n = (a + \frac{a}{\sqrt{2}})^2 - a^2 = 2a^2(1+\sqrt{2})$.
Dividing this by $8$ gives us the area of each triangular segment which makes up the octagon. Further dividing by $2$ gives us the area of a smaller segment consisting of the right triangle with legs of the apothem and $\frac{a}{2}$. Using the area of a triangle as $\frac{1}{2}bh$, we can determine the length of apothem $r$ from
\[\frac{2a^2(1+\sqrt{2})}{8 \times 2} = \frac{\frac{a}{2}\times r}{2}\]
\[\frac{4a^2(1+\sqrt{2})}{8 } = ar\]
\[\frac{a(1+\sqrt{2})}{2} = r\].
From the apothem, we can once again use the Pythagorean Theorem, giving us the length of the circumradius $R$.
\[R^2 = (\frac{a(1+\sqrt{2})}{2})^2 + (\frac{a}{2})^2\]
\[R^2 = \frac{a^2(1+\sqrt{2})^2}{4} + \frac{a^2}{4}\]
\[R^2 = \frac{a^2(3+2\sqrt{2})}{4} + \frac{a^2}{4}\]
\[R^2 = \frac{a^2(4+2\sqrt{2})}{4}\]
\[R = \sqrt{\frac{a^2(4+2\sqrt{2})}{4}} = \frac{a\sqrt{4 + 2\sqrt2}}{2}\].
Doubling this gives us the diagonal of both the square and the octagon. From here, we can use the formula $A = \frac{1}{2}d^2$ for the area of the square:
\[A = \frac{(a\sqrt{4 + 2\sqrt2})^2}{2}\]
\[A = \frac{a^2(4+2\sqrt{2})}{2} = m\].
Thus we now only need to find the ratio $\frac{m}{n}$. This can be easily done through some algebra:
\[\frac{m}{n} = \frac{a^2(4+2\sqrt{2})}{2(2a^2(1+\sqrt{2})}\]
\[\frac{m}{n} = \frac{4+2\sqrt{2}}{4+4\sqrt{2}}\]
\[\frac{m}{n} = \frac{2+\sqrt{2}}{2+2\sqrt{2}}\]
Rationalizing the denominator by multiplying by the conjugate, we get $\frac{m}{n} = \boxed{\textbf{(B) } \frac{\sqrt{2}}{2}}$. ~ciceronii
$\textbf{Note:}$ this can more easily be done if you know any of these formulas. This was an entire derivation of the area of an octagon, it's apothem, and it's circumradius, but it can be much simpler if you have any of these memorized.
$\textbf{Note No.2:}$ The formula $[ABCDEFG]=(\sqrt{2})[ACEG]$ is one from the Intro to Geometry Textbook: Problem 16.46.
~Note No.2 by athreyay | // 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);
label("C",(2.41421356, 1),NE);
label("D",(2.41421356, -1),SE);
label("E",(1,-2.41421356),SE);
label("F",(-1,-2.41421356),SW);
label("G",(-2.41421356,-1),SW);
label("H",(-2.41421356,1),NW);
// Block 2
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);
label("a",(1, 2.41421356)--(2.41421356, 1),S);
draw((-2.41421356, 2.41421356)--(2.41421356, 2.41421356)--(2.41421356, -2.41421356)--(-2.41421356, -2.41421356)--cycle);
label("s",(-2.41421356, 2.41421356)--(2.41421356, 2.41421356),N);
// Block 3
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); label("C",(2.41421356, 1),NE); label("D",(2.41421356, -1),SE); label("E",(1,-2.41421356),SE); label("F",(-1,-2.41421356),SW); label("G",(-2.41421356,-1),SW); label("H",(-2.41421356,1),NW);
// Block 4
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); label("a",(1, 2.41421356)--(2.41421356, 1),S); draw((-2.41421356, 2.41421356)--(2.41421356, 2.41421356)--(2.41421356, -2.41421356)--(-2.41421356, -2.41421356)--cycle); label("s",(-2.41421356, 2.41421356)--(2.41421356, 2.41421356),N); | [] |
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,-2.41421356));
draw((2.41421356, 1)--(-2.41421356,-1));
draw((2.41421356, -1)--(-2.41421356,1));
label("A",(-1, 2.41421356),NW);
label("B",(1, 2.41421356),NE);
label("C",(2.41421356, 1),NE);
label("D",(2.41421356, -1),SE);
label("E",(1,-2.41421356),SE);
label("F",(-1,-2.41421356),SW);
label("G",(-2.41421356,-1),SW);
label("H",(-2.41421356,1),NW);
label("O",(.1,.1),NE);
label("P",(.60710678,1.60710678),SE);
// Block 2
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,-2.41421356)); draw((2.41421356, 1)--(-2.41421356,-1)); draw((2.41421356, -1)--(-2.41421356,1)); label("A",(-1, 2.41421356),NW); label("B",(1, 2.41421356),NE); label("C",(2.41421356, 1),NE); label("D",(2.41421356, -1),SE); label("E",(1,-2.41421356),SE); label("F",(-1,-2.41421356),SW); label("G",(-2.41421356,-1),SW); label("H",(-2.41421356,1),NW); label("O",(.1,.1),NE); label("P",(.60710678,1.60710678),SE); | [] |
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 $B?$
$\textbf{(A) } 2\sqrt{3} \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 9 \qquad \textbf{(D) } 2\sqrt{21} \qquad \textbf{(E) } 9+\sqrt{3}$ | 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 \[z^3=r^3\operatorname{cis}(3\theta)=8(1),\] from which
$r^3=8,$ so $r=2.$
$\begin{cases} \begin{aligned} \cos(3\theta) &= 1 \\ \sin(3\theta) &= 0 \end{aligned}, \end{cases}$ so $3\theta=0,2\pi,4\pi,$ or $\theta=0,\frac{2\pi}{3},\frac{4\pi}{3}.$
The set of solutions to $z^{3}-8=0$ is $\boldsymbol{A=\left\{2,-1+\sqrt{3}i,-1-\sqrt{3}i\right\}}.$ In the complex plane, the solutions form the vertices of an equilateral triangle whose circumcircle has center $0$ and radius $2.$
We solve $z^{3}-8z^{2}-8z+64=0$ by factoring by grouping.
We have
\begin{align*} z^2(z-8)-8(z-8)&=0 \\ \bigl(z^2-8\bigr)(z-8)&=0. \end{align*}
The set of solutions to $z^{3}-8z^{2}-8z+64=0$ is $\boldsymbol{B=\left\{2\sqrt{2},-2\sqrt{2},8\right\}}.$
In the graph below, the points in set $A$ are shown in red, and the points in set $B$ are shown in blue. The greatest distance between a point of $A$ and a point of $B$ is the distance between $-1\pm\sqrt{3}i$ to $8,$ as shown in the dashed line segments.
By the Distance Formula, the answer is \[\sqrt{(-1-8)^2+\left(\pm\sqrt{3}-0\right)^2}=\sqrt{84}=\boxed{\textbf{(D) } 2\sqrt{21}}.\]
~lopkiloinm ~MRENTHUSIASM | // 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; ++i)
{
draw(Circle((0,0),i), gray+linewidth(0.4));
}
for(int i=0;i<numRays;++i)
draw(rotate(i*360/numRays)*((-big,0)--(big,0)), gray+linewidth(0.4));
}
//Draws the horizontal gridlines
void horizontalLines()
{
for (int i = yMin+1; i < yMax; ++i)
{
draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4));
}
}
//Draws the vertical gridlines
void verticalLines()
{
for (int i = xMin+1; i < xMax; ++i)
{
draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4));
}
}
horizontalLines();
verticalLines();
polarGrid(xMax,numRays);
draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5));
draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5));
label("Re",(xMax,0),(2,0));
label("Im",(0,yMax),(0,2));
//The n such that we're taking the nth roots of unity multiplied by 2.
int n = 3;
pair A[];
for(int i = 0; i <= n-1; i+=1) {
A[i] = rotate(360*i/n)*(2,0);
}
draw(Circle((0,0),2),red);
draw(A[1]--(8,0),dashed);
draw(A[2]--(8,0),dashed);
for(int i = 0; i< n; ++i) dot(A[i],red+linewidth(4.5));
dot((2*sqrt(2),0),blue+linewidth(4.5));
dot((-2*sqrt(2),0),blue+linewidth(4.5));
dot((8,0),blue+linewidth(4.5));
// Block 2
/* 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; ++i) { draw(Circle((0,0),i), gray+linewidth(0.4)); } for(int i=0;i<numRays;++i) draw(rotate(i*360/numRays)*((-big,0)--(big,0)), gray+linewidth(0.4)); } //Draws the horizontal gridlines void horizontalLines() { for (int i = yMin+1; i < yMax; ++i) { draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4)); } } //Draws the vertical gridlines void verticalLines() { for (int i = xMin+1; i < xMax; ++i) { draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4)); } } horizontalLines(); verticalLines(); polarGrid(xMax,numRays); draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("Re",(xMax,0),(2,0)); label("Im",(0,yMax),(0,2)); //The n such that we're taking the nth roots of unity multiplied by 2. int n = 3; pair A[]; for(int i = 0; i <= n-1; i+=1) { A[i] = rotate(360*i/n)*(2,0); } draw(Circle((0,0),2),red); draw(A[1]--(8,0),dashed); draw(A[2]--(8,0),dashed); for(int i = 0; i< n; ++i) dot(A[i],red+linewidth(4.5)); dot((2*sqrt(2),0),blue+linewidth(4.5)); dot((-2*sqrt(2),0),blue+linewidth(4.5)); dot((8,0),blue+linewidth(4.5)); | [] |
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 nearest tenth$?$
$\textbf{(A) } 0.3 \qquad \textbf{(B) } 0.4 \qquad \textbf{(C) } 0.5 \qquad \textbf{(D) } 0.6 \qquad \textbf{(E) } 0.7$ | 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 this, and the fact that there are four circles, we write
\[4 \cdot \frac{1}{4} \cdot \pi d^2 = \frac{1}{2}\]
Solving for $d$, we obtain $d = \frac{1}{\sqrt{2\pi}}$, where with $\pi \approx 3$, we get $d \approx \frac{1}{\sqrt{6}} \approx \dfrac{1}{2.5} = \dfrac{10}{25} = \dfrac{2}{5}$, and from here, we see that $d \approx 0.4 \implies \boxed{\textbf{(B) } 0.4}.$
~Crypthes
~ Minor Edits by BakedPotato66
$\textbf{Note:}$ To be more rigorous, note that $d<0.5$ since if $d\geq0.5$ then clearly the probability is greater than $\frac{1}{2}$. This would make sure the above solution works, as if $d\geq0.5$ there is overlap with the quartercircles. $\textbf{- Emathmaster}$
Solution 2
As in the previous solution, we obtain the equation $4 \cdot \frac{1}{4} \cdot \pi d^2 = \frac{1}{2}$, which simplifies to $\pi d^2 = \frac{1}{2} = 0.5$. Since $\pi$ is slightly more than $3$, $d^2$ is slightly less than $\frac{0.5}{3} = 0.1\bar{6}$. We notice that $0.1\bar{6}$ is slightly more than $0.4^2 = 0.16$, so $d$ is roughly $\boxed{\textbf{(B) } 0.4}.$ ~emerald_block
Solution 3 (Estimating)
As above, we find that we need to estimate $d = \frac{1}{\sqrt{2\pi}}$.
Note that we can approximate $2\pi \approx 6.28318 \approx 6.25$ and so $\frac{1}{\sqrt{2\pi}}$ $\approx \frac{1}{\sqrt{6.25}}=\frac{1}{2.5}=0.4$.
And so our answer is $\boxed{\textbf{(B) } 0.4}$.
~Silverdragon
Solution 4 (Estimating but a bit different)
We only need to figure out the probability for a unit square, as it will scale up to the $2020\times 2020$ square. Since we want to find the probability that a point inside a unit square that is $d$ units away from a lattice point (a corner of the square) is $\frac{1}{2}$, we can find which answer will come the closest to covering $\frac{1}{2}$ of the area.
Since the closest is $0.4$ which turns out to be $(0.4)^2\times \pi = 0.16 \times \pi$ which is about $0.502$, we find that the answer rounded to the nearest tenth is $0.4$ or $\boxed{\textbf{(B)}}$.
~RuiyangWu
Solution 5 (Estimating but differently again)
As per the above diagram, realize that $\pi d^2 = \frac{1}{2}$, so $d = \frac{1}{(\sqrt{2})(\sqrt{\pi})}$.
$\sqrt{2} \approx 1.4 = \frac{7}{5}$.
$\sqrt{\pi}$ is between $1.7$ and $1.8$ $((1.7)^2 = 2.89$ and $(1.8)^2 = 3.24)$, so we can say $\sqrt{\pi} \approx 1.75 = \frac{7}{4}$.
So $d \approx \frac{1}{(\frac{7}{5})(\frac{7}{4})} = \frac{1}{\frac{49}{20}} = \frac{20}{49}$. This is slightly above $\boxed{\textbf{(B) } 0.4}$, since $\frac{20}{49} \approx \frac{2}{5}$.
-Solution by Joeya
Solution 6 (Estimating but differently again, again)
As above, we have the equation $\pi d^2 = \frac{1}{2}$, and we want to find the most accurate value of $d$. We resort to the answer choices and can plug those values of $d$ in and see which value of $d$ will lead to the most accurate value of $\pi$.
Starting off in the middle, we try option C with $d=0.5$. Plugging this in, we get $\pi \left(\frac{1}{2}\right)^2 = \frac{1}{2},$ and after simplifying we get $\pi = \frac{1}{2} \cdot 4 = 2.$
That's not very good. We know $\pi \approx 3.14.$
Let's see if we can do better. Trying option A with $d = 0.3,$ we get $\pi = \frac{1}{2} \cdot \frac{100}{9} = \frac{50}{9} = 5 \frac{5}{9}.$
Hm, let's try option B with $d = 0.4.$ We get $\pi = \frac{1}{2} \cdot \frac{25}{4} = \frac{25}{8} = 3 \frac{1}{8}$. This is very close to $\pi$ and is the best estimate for $\pi$ of the 5 options.
Therefore, the answer is $\boxed{\textbf{(B) } 0.4}.$
~ epiconan | 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), 0.3989, 270, 360)); filldraw(arc((0,1), 0.3989, 270, 360)--(0,1)--cycle, gray); | [] |
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 nearest tenth$?$
$\textbf{(A) } 0.3 \qquad \textbf{(B) } 0.4 \qquad \textbf{(C) } 0.5 \qquad \textbf{(D) } 0.6 \qquad \textbf{(E) } 0.7$ | 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 shaded region (since there are $k$ unit squares).
Let $n$ be the side length of a square. When $n=1$, the shaded areas represent half of the total area:
When $n=2$:
For $n=3$:
We can calculate the total number of shaded circles given some $n$. There are $(n-1)^2$ full circles on the inside, $4(n-1)$ semicircles on the sides, and $4$ quarter circles for the corners.
Full circles are, of course, worth one circle. Semicircles are worth half a circle each, and quarter circles are worth $\dfrac14$ of a circle. Thus, weighing our sum gives $(n-1)^2+\dfrac{4(n-1)}2+\dfrac44=n^2-2n+1+2(n-1)+1=n^2-2n+2n+2-2=n^2.$ Thus, there is $n^2\cdot\pi r^2$ worth of the shaded area for any $n$, and since the area of each circle is $\pi r^2$ if $r$ is the radius of each.
We want the ratio of this shaded area to the entire to be $\dfrac12$. The area of the entire square is $n^2$, so dividing, we see that $\dfrac{n^2\cdot\pi r^2}{n^2}=\pi r^2=\dfrac12$.
The rest is the same as solution $1$. | // 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); draw(arc((0,1), 0.3989, 270, 360)); filldraw(arc((0,1), 0.3989, 270, 360)--(0,1)--cycle, gray);
// Block 2
size(10cm); filldraw((arc((0,0), 0.1994, 0, 90))--(0,0)--cycle, gray); draw(arc((1,0), 0.1994, 90, 180)); filldraw((arc((1,0), 0.1994, 90, 180))--(1,0)--cycle, gray); draw(arc((1,1), 0.1994, 180, 270)); filldraw((arc((1,1), 0.1994, 180, 270))--(1,1)--cycle, gray); draw(arc((0,1), 0.1994, 270, 360)); filldraw(arc((0,1), 0.1994, 270, 360)--(0,1)--cycle, gray); draw(arc((0.5,0.5), 0.1994,0,360)); filldraw(arc((0.5,0.5), 0.1994,0,360)--(0.5,0.5)--cycle, gray); draw(arc((0.5,0), 0.1994,0,180)); filldraw(arc((0.5,0), 0.1994,0,180)--(0.5,0)--cycle, gray); draw(arc((0,0.5), 0.1994,-90,90)); filldraw(arc((0,0.5), 0.1994,-90,90)--(0,0.5)--cycle, gray); filldraw(arc((1,0.5), 0.1994,90,270)--(1,0.5)--cycle, gray); filldraw(arc((0.5,1), 0.1994,0,-180)--(0.5,1)--cycle, gray); draw((0,0)--(0.5,0)--(0.5,1)--(0,1)--(0,0)--(1,0)--(1,1)--(0,1)--(0,0.5)--(1,0.5));
// Block 3
size(10cm); filldraw(arc((0,0),0.1330,0,90)--(0,0)--cycle, gray); filldraw(arc((0,1),0.1330,-90,0)--(0,1)--cycle, gray); filldraw(arc((1,0),0.1330,90,180)--(1,0)--cycle, gray); filldraw(arc((1,1),0.1330,-180,-90)--(1,1)--cycle, gray); filldraw(arc((0.333,0.333),0.133,0,360)--(0.333,0.333)--cycle, gray); filldraw(arc((0.667,0.333),0.133,0,360)--(0.667,0.333)--cycle, gray); filldraw(arc((0.333,0.667),0.133,0,360)--(0.333,0.667)--cycle, gray); filldraw(arc((0.667,0.667),0.133,0,360)--(0.667,0.667)--cycle, gray); filldraw(arc((0.333,0),0.133,0,180)--(0.333,0)--cycle, gray); filldraw(arc((0.667,0),0.133,0,180)--(0.667,0)--cycle, gray); filldraw(arc((0.333,1),0.133,-180,0)--(0.333,1)--cycle, gray); filldraw(arc((0.666,1),0.133,-180,0)--(0.666,1)--cycle, gray); filldraw(arc((0,0.333),0.133,-90,90)--(0,0.333)--cycle, gray); filldraw(arc((0,0.667),0.133,-90,90)--(0,0.667)--cycle, gray); filldraw(arc((1,0.333),0.133,90,270)--(1,0.333)--cycle, gray); filldraw(arc((1,0.667),0.133,90,270)--(1,0.667)--cycle, gray); draw((0,0)--(0,1)--(0.333,1)--(0.333,0)--(0.667,0)--(0.667,1)--(1,1)--(1,0)--(0,0)--(0,0.333)--(1,0.333)--(1,0.667)--(0,0.667)); draw((0.333,1)--(0.667,1)); | [] |
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 $ABCD?$
$\textbf{(A) } 330 \qquad \textbf{(B) } 340 \qquad \textbf{(C) } 350 \qquad \textbf{(D) } 360 \qquad \textbf{(E) } 370$ | 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 also see two similar triangles, $\triangle BFE \sim \triangle DCE$. Since $EC$ is $20-5=15$, and $DC=30$, we get that $BF=2x$.
Now, if we redraw another diagram just of $ABC$, we get that $(2x)^2=(5-x)(15+x)$ because of the altitude geometric mean theorem which states that in any right triangle, the altitude squared is equal to the product of the two lengths that it divides the base into.
Expanding, simplifying, and dividing by the GCF, we get $x^2+2x-15=0$. This factors to $(x+5)(x-3)$, which has roots of $x=-5, 3$. Since lengths cannot be negative, $x=3$. Since $x=3$, that means the altitude $BF=2\cdot3=6$, or $[ABC]=60$. Thus $[ABCD]=[ACD]+[ABC]=300+60=\boxed {\textbf{(D) }360}$
~ Solution by Ultraman
~ Diagram by ciceronii | 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)--(4,0), SW); label("30", (4,0)--(6,4), SE); label("$x$", (1,1.5)--(1.714,1.143), NE); label("5$-$$x$", (1,1.5)--(0,2), NE); draw(rightanglemark((0,2),(0,0),(4,0))); draw(rightanglemark((0,2),(4,0),(6,4))); draw(rightanglemark((0,0),(1,1.5),(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 $ABCD?$
$\textbf{(A) } 330 \qquad \textbf{(B) } 340 \qquad \textbf{(C) } 350 \qquad \textbf{(D) } 360 \qquad \textbf{(E) } 370$ | 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 solutions $(0,10)$ and $(-8,-6)$. We immediately discard the $(0,10)$ solution as $y$ should be negative. Thus, we conclude that $[ABCD]=[ACD]+[ABC]=\frac{20\cdot30}{2}+\frac{20\cdot6}{2}=\boxed{\textbf{(D)}\:360}$. | 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",(-5,0)--(10,0), N); label("30",(10,0)--(10,30), E); dot((-5,0)); dot((-10,0)); dot((-8,-6)); dot((10,0)); dot((10,30)); | [] |
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 $ABCD?$
$\textbf{(A) } 330 \qquad \textbf{(B) } 340 \qquad \textbf{(C) } 350 \qquad \textbf{(D) } 360 \qquad \textbf{(E) } 370$ | 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:
\[[AEB] + [BEC] = \frac{1}{2}(x)(5)\sin(\angle DEC) + \frac{1}{2}(x)(15)\sin(180-\angle DEC) = (3)(20) = 60\].
Thus the desired area is $\frac{1}{2}(30)(20) + 60 = \boxed{\textbf{(D) } 360}$
~qwertysri987 | 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 $ABCD?$
$\textbf{(A) } 330 \qquad \textbf{(B) } 340 \qquad \textbf{(C) } 350 \qquad \textbf{(D) } 360 \qquad \textbf{(E) } 370$ | 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 $\angle EFG=90^\circ, EG=\sqrt{5^2+10^2}=5\sqrt{5} \implies BE=\frac{AE\cdot EC}{EG}=\frac{5\cdot 15}{5\sqrt{5}}=3\sqrt{5}$ via Power of a Point.
The altitude from $B$ to $AC$ is then equal to $GF\cdot \frac{BE}{GE}=10\cdot \frac{3\sqrt 5}{5 \sqrt 5}=6$.
Finally, the total area of $ABCD$ is equal to $\frac 12 \cdot 20 \left(30+6 \right) =\boxed{\text{(D) } 360}.$
~asops | 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 $ABCD?$
$\textbf{(A) } 330 \qquad \textbf{(B) } 340 \qquad \textbf{(C) } 350 \qquad \textbf{(D) } 360 \qquad \textbf{(E) } 370$ | 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 $\triangle CEF \sim \triangle CAB$, $EF = AB \cdot \frac{CE}{CA} = x \cdot \frac{15}{20} = \frac{3x}{4}$
$BF = BC - CF = BC - BC \cdot \frac{CE}{CA} = \frac{1}{4} \cdot BC = \frac{y}{4}$
$BE = \sqrt{ \left( \frac{3x}{4} \right) ^2 + \left( \frac{y}{4} \right) ^2 } = \frac{ \sqrt{9x^2 + y^2} }{4}$ , substituting $x^2 + y^2 = 400$, we get $BE = \frac{ \sqrt{8x^2 + 400} }{4} = \frac{ \sqrt{2x^2 + 100} }{2}$
$[ABC] = \frac{1}{2} \cdot x \cdot y$
Because $\triangle ABC$ and $\triangle ACD$ share the same base, $\frac{[ABC]}{[ACD]} = \frac{BE}{DE}$
$[ABC] = [ACD] \cdot \frac{BE}{DE} = 300 \cdot \frac{ \frac{ \sqrt{2x^2 + 100} } {2} }{ 15 \sqrt{5} }$
$\frac{1}{2} \cdot x \cdot y = 20 \cdot \frac{ \frac{ \sqrt{2x^2 + 100} } {2} }{ \sqrt{5} }$
$xy = 4 \sqrt{10x^2 + 500}$
By $x^2 + y^2 = 400$, $y = \sqrt{400 - x^2}$. So, $x \cdot \sqrt{400 - x^2} = 4 \sqrt{10x^2 + 500}$
$x^2 (400 - x^2) = 16 (10x^2 + 500)$
Let $x^2 = a$, $a (400 - a) = 16 (10a + 500)$, $400a - a^2 = 160a + 8000$, $a^2 - 240a + 8000 = 0$, $(a-200)(a-40) = 0$
Because $x < 20$, $a$ can only equal 40. $a = 40$, $x = 2 \sqrt{10}$, $y = 6 \sqrt{10}$
$[ABC] = \frac{1}{2} \cdot 2 \sqrt{10} \cdot 6 \sqrt{10} = 60$
$[ABCD] = [ABC] + [ACD] = 60 + 300 = \boxed{\text{(D) } 360}$
~isabelchen | 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); label("20", (0,2)--(4,0), SW); label("30", (4,0)--(6,4), SE); label("$x$", (-0.3,2)--(-0.3,0), N); label("$y$", (0,-0.3)--(4,-0.3), E); draw(rightanglemark((1.714,2),(1.714,0),(5.714,0))); draw(rightanglemark((0,2),(0,0),(4,0))); draw(rightanglemark((0,2),(4,0),(6,4))); | [] |
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 $ABCD?$
$\textbf{(A) } 330 \qquad \textbf{(B) } 340 \qquad \textbf{(C) } 350 \qquad \textbf{(D) } 360 \qquad \textbf{(E) } 370$ | 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}. \end{align*}
~luckuso | 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)--(6,4), dashed); draw((1.714,1.143)--(1.714,0), dashed); label("20", (0,2)--(4,0), SW); label("30", (4,0)--(6,4), SE); label("$x$", (-0.3,2)--(-0.3,0), N); label("$y$", (0,-0.3)--(4,-0.3), E); label( "$X$", (6,0), SE); label("5", (0,2)--(1.714,1.143), NE); label("15",(1.714,1.143)--(4,0),NE); label("5$C$", (0,0)--(1.714,0),S); label("15$C$", (1.714,0)--(4,0),S); label("30$S$", (4,0)--(6,0),S); label("30$C$", (6,0)--(6,4),E); draw(anglemark((1.714,1.143),(4,0),(1.714,0))); draw(anglemark((4,0),(6,4),(6,0))); draw(rightanglemark((1.714,2),(1.714,0),(5.714,0))); draw(rightanglemark((0,2),(0,0),(4,0))); draw(rightanglemark((0,2),(4,0),(6,4))); | [] |
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 $y$-axis. How many of the $125$ sequences of three of these transformations (not necessarily distinct) will return $T$ to its original position? (For example, a $180^{\circ}$ rotation, followed by a reflection across the $x$-axis, followed by a reflection across the $y$-axis will return $T$ to its original position, but a $90^{\circ}$ rotation, followed by a reflection across the $x$-axis, followed by another reflection across the $x$-axis will not return $T$ to its original position.)
$\textbf{(A) } 12 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 17 \qquad \textbf{(D) } 20 \qquad \textbf{(E) } 25$ | 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 the original orientation. Since we are only allowed $3$ transformations and an even number of them must be reflections, we either reflect $T$ $0$ times or $2$ times.
Case 1: $0$ reflections on $T$.
In this case, we must use $3$ rotations to return $T$ to its original position. Notice that our set of rotations, $\{90^\circ,180^\circ,270^\circ\}$, contains every multiple of $90^\circ$ except for $0^\circ$. We can start with any two rotations $a,b$ in $\{90^\circ,180^\circ,270^\circ\}$ and there must be exactly one $c \equiv -a - b \pmod{360^\circ}$ such that we can use the three rotations $(a,b,c)$ which ensures that $a + b + c \equiv 0^\circ \pmod{360^\circ}$. That way, the composition of rotations $a,b,c$ yields a full rotation. For example, if $a = b = 90^\circ$, then $c \equiv -90^\circ - 90^\circ = -180^\circ \pmod{360^\circ}$, so $c = 180^\circ$ and the rotations $(90^\circ,90^\circ,180^\circ)$ yields a full rotation.
The only case in which this fails is when $c$ would have to equal $0^\circ$. This happens when $(a,b)$ is already a full rotation, namely, $(a,b) = (90^\circ,270^\circ),(180^\circ,180^\circ),$ or $(270^\circ,90^\circ)$. However, we can simply subtract these three cases from the total. Selecting $(a,b)$ from $\{90^\circ,180^\circ,270^\circ\}$ yields $3 \cdot 3 = 9$ choices, and with $3$ that fail, we are left with $6$ combinations for case $1$.
Case 2: $2$ reflections on $T$.
In this case, we first eliminate the possibility of having two of the same reflection. Since two reflections across the x-axis maps $T$ back to itself, inserting a rotation before, between, or after these two reflections would change $T$'s final location, meaning that any combination involving two reflections across the x-axis would not map $T$ back to itself. The same applies to two reflections across the y-axis.
Therefore, we must use one reflection about the x-axis, one reflection about the y-axis, and one rotation. Since a reflection about the x-axis changes the sign of the y component, a reflection about the y-axis changes the sign of the x component, and a $180^\circ$ rotation changes both signs, these three transformation composed (in any order) will suffice. It is therefore only a question of arranging the three, giving us $3! = 6$ combinations for case 2.
Combining both cases we get $6+6=\boxed{\textbf{(A)} 12}$. | 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 $s$?
$\textbf{(A) } 1+\sqrt{2} \qquad \textbf{(B) } \sqrt{7} \qquad \textbf{(C) } \frac{8}{3} \qquad \textbf{(D) } \sqrt{5+\sqrt{5}} \qquad \textbf{(E) } 2\sqrt{2}$ | 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, meaning that $\angle CPQ= 60^{\circ}$. Thus, by adding the two together, we see that $\angle APC = 120^{\circ}$.
We can now use the law of cosines as following:
\begin{align*} s^2 &= (AP)^2 + (CP)^2 - 2\cdot AP\cdot CP\cdot \cos{\angle{APC}} \\ &= 1 + 4 - 2\cdot 1\cdot 2\cdot \cos{120^{\circ}} \\ &= 5 - 4\left(-\frac{1}{2}\right) \\ &= 7, \end{align*}
giving us that $s = \boxed{ \sqrt{7} \ \textbf{(B)}}$.
~ciceronii
Solution 1(b)
Rotate $\triangle CPA$ counterclockwise $60^\circ$ around point $C$ to $\triangle CQB$. Then $CP=CQ, \angle PCQ=60^\circ$, so $\triangle CPQ$ is an equilateral triangle.
Note that $\triangle PQB$ is a $30^\circ$-$60^\circ$-$90^\circ$ triangle, hence $\angle BPQ=30^\circ$, and $\angle BPC=90^\circ$, so \[BC^2=PC^2+PB^2=2^2+3=7,\] and the answer is $\boxed{\textbf{(B) } \sqrt{7}}$.
~szhangmath
Solution 1(c)
Rotate $\triangle BPC$ counterclockwise by $60^\circ$ around point $B$ to $\triangle BQA$. Then $BP=BQ$, and $\angle PBQ=60^\circ$, so $\triangle BPQ$ is an equilateral triangle.
Note that $\triangle QAP$ is a $30^\circ$-$60^\circ$-$90^\circ$ triangle, hence $\angle PQA=30^\circ$, and $\angle BQA=90^\circ$, so \[AB^2=PQ^2+AQ^2=2^2+3=7,\] and the answer is $\boxed{\textbf{(B) } \sqrt{7}}$. | // 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); label("$B$", B, 0.5*(B-P), q); label("$C$", C, 0.5*(C-P), q); label("$P$", P, dir(180), q); label("$Q$", Q, 0.25*(Q-B), q); label("$\sqrt{3}$",B--P, right, p); label("$2$",C--P, 2*left, p); label("$1$",A--P, 1.5**dir(-10), p); label("$1$", A--Q, dir(250), p); label("$1$",P--Q, down, p); label("$\sqrt{3}$",C--Q, right, p);
// Block 2
size(200); pen p = fontsize(10pt)+gray+0.4; pen q = fontsize(13pt); pair A,B,C,D,P,Q; real s=sqrt(7); A=origin; B=s*right; C=s*dir(60); P=IP(CR(A,1),CR(C,2)); Q=rotate(60,C)*P; draw(A--B--C--A, black+0.8); draw(A--P--B^^P--C, p); draw(B--Q--C^^P--Q, p+dashed); label("$A$", A, A-P, q); label("$B$", B, B-P, q); label("$C$", C, up, q); label("$P$", P, dir(140), q); label("$Q$", Q, 0.25*(Q-P), q); label("$\sqrt{3}$",P--B, dir(60), p); label("$2$",C--P, 2*left, p); label("$1$",A--P, up, p); label("$1$", B--Q, dir(-30), p); label("$2$",P--Q, down, p); label("$2$",C--Q, dir(45), p);
// Block 3
size(200); pen p = fontsize(10pt)+gray+0.4; pen q = fontsize(13pt); pair A,B,C,D,P,Q,R,X,Y,Z; real s=sqrt(7); C=origin; A=s*right; B=s*dir(60); P=IP(CR(A,1),CR(C,2)); Q=rotate(60,B)*P; draw(A--B--C--A, black+0.8); draw(A--P--B^^P--C, p); draw(B--Q--A^^P--Q, p+dashed); label("$A$", A, A-P, q); label("$B$", B, B-P, q); label("$C$", C, 0.6*(C-P), q); label("$P$", P, dir(250), q); label("$Q$", Q, 0.25*(Q-P), q); label("$\sqrt{3}$",B--Q, dir(60), p); label("$\sqrt{3}$",P--B, dir(210), p); label("$\sqrt{3}$",P--Q, dir(135), p); label("$2$",P--C, dir(120), p); label("$2$",A--Q, dir(-20), p); label("$1$",A--P, dir(210), p); | [] |
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 $s$?
$\textbf{(A) } 1+\sqrt{2} \qquad \textbf{(B) } \sqrt{7} \qquad \textbf{(C) } \frac{8}{3} \qquad \textbf{(D) } \sqrt{5+\sqrt{5}} \qquad \textbf{(E) } 2\sqrt{2}$ | 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 above). Clearly $s > 2$ and the triangle defined by these intersection points will be equilateral (pictured by the blue segments).
Take this equilateral triangle to have side length $1$. The portions of each segment outside this triangle (in red) have length $1$. Take $P'$ to be the intersection of the segments emanating from $A$ and $C$. By Law of Cosines, \[BP' = \sqrt{1 + 1 - 2\cos{120^\circ}} = \sqrt{3}.\] So, $P'$ actually satisfies the conditions of the problem, and we can obtain again by Law of Cosines \[s = \sqrt{4 + 1 - 4\cos{120^\circ}} = \boxed{\textbf{(B)} \sqrt{7}}.\]
~ hnkevin42 | // 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), SE, p);
label("$B$", (1.323,2.291), N, p);
label("$P'$", (0.756,0.655), NW, p);
label("$1$", (2.174,0.164), N, p);
label("$1$", (1.228,0.491), N, p);
D(A--P, red);
D(B--Q, red);
D(C--R, red);
D(P--Q--R--cycle, blue);
D(B--P, magenta);
// Block 2
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), SE, p); label("$B$", (1.323,2.291), N, p); label("$P'$", (0.756,0.655), NW, p); label("$1$", (2.174,0.164), N, p); label("$1$", (1.228,0.491), N, p); D(A--P, red); D(B--Q, red); D(C--R, red); D(P--Q--R--cycle, blue); D(B--P, magenta); | [] |
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 $s$?
$\textbf{(A) } 1+\sqrt{2} \qquad \textbf{(B) } \sqrt{7} \qquad \textbf{(C) } \frac{8}{3} \qquad \textbf{(D) } \sqrt{5+\sqrt{5}} \qquad \textbf{(E) } 2\sqrt{2}$ | 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:
\[\triangle{APC} = \frac{s\cdot PF}{2} = \sqrt{\frac{s+3}{2}\left(\frac{s+3}{2}-s\right)\left(\frac{s+3}{2}-1\right)\left(\frac{s+3}{2}-2\right)}\]
\[\implies PF = \frac{\sqrt{10s^2-s^4-9}}{2s}\]
Similarly, we obtain:
\[PE = \frac{\sqrt{8s^2-s^4-4}}{2s}\]
\[PG = \frac{\sqrt{14s^2-s^4-1}}{2s}\]
By Viviani's theorem,
\[\frac{\sqrt{10s^2-s^4-9}}{2s}+\frac{\sqrt{8s^2-s^4-4}}{2s}+\frac{\sqrt{14s^2-s^4-1}}{2s} = \frac{\sqrt{3}}{2}s\]
\[\sqrt{10s^2-s^4-9}+\sqrt{8s^2-s^4-4}+\sqrt{14s^2-s^4-1} = \sqrt{3}s^2\]
Note that from now on, the algebra will get extremely ugly and almost impossible to do by hand within the time frame. However, we do see that it's extremely easy to check the answer choices with the equation in this form. Testing $s = \sqrt{7}$, We obtain $7\sqrt{3}$ on both sides, revealing that our answer is in fact $\boxed{\textbf{(B) } \sqrt{7}}$
~ siluweston
~ edits by aopspandy | // 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((0,0)--(3.5,3.5));
label("$\sqrt{3}$",(0,0)--(3.5,3.5), SE,p);
draw((8,0)--(3.5,3.5));
label("$2$",(8,0)--(3.5,3.5), SW,p);
draw((4,5.65)--(3.5,3.5));
label("$1$",(4,5.65)--(3.5,3.5), E,p);
draw((3.5,3.5)--(2.8191,3.982));
draw((3.5,3.5)--(4.848,4.452));
draw((3.5,3.5)--(3.5,0));
// Block 2
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((0,0)--(3.5,3.5)); label("$\sqrt{3}$",(0,0)--(3.5,3.5), SE,p); draw((8,0)--(3.5,3.5)); label("$2$",(8,0)--(3.5,3.5), SW,p); draw((4,5.65)--(3.5,3.5)); label("$1$",(4,5.65)--(3.5,3.5), E,p); draw((3.5,3.5)--(2.8191,3.982)); draw((3.5,3.5)--(4.848,4.452)); draw((3.5,3.5)--(3.5,0)); | [] |
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 $s$?
$\textbf{(A) } 1+\sqrt{2} \qquad \textbf{(B) } \sqrt{7} \qquad \textbf{(C) } \frac{8}{3} \qquad \textbf{(D) } \sqrt{5+\sqrt{5}} \qquad \textbf{(E) } 2\sqrt{2}$ | 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.
Observe that the area of the equilateral triangle $ABC$ is half that of the hexagon $AP''CP'BP'''$.
Note that $AP=AP''=AP'''$. The same goes for the other vertices. This means that $AP''P'''$ is isosceles. Using either the Law of Cosines or simply observing that $AP''P'''$ is comprised of two 30-60-90 triangles, we find that $P''P'''= \sqrt{3}$. Similarly (pun intended), $P'P'''=3$ and $P'P''=2\sqrt{3}$. Using the previous observation that $AP''P'''$ is two 30-60-90 triangles (as are the others) we find the areas of $AP''P''$ to be $\frac{\sqrt{3}}{4}$. Again, using similarity we find the area of $BP'P'''$ to be $\frac{3\sqrt{3}}{4}$ and the area of $CP'P''$ to be $\sqrt{3}$.
Next, observe that $P'P''P'''$ is a 30-60-90 right triangle. This right triangle therefore has an area of $\frac{3\sqrt{3}}{2}$.
Adding these areas together, we get the area of the hexagon as $\frac{7\sqrt{3}}{2}$. This means that the area of $ABC$ is $\frac{7\sqrt{3}}{4}$.
The formula for the area of an equilateral triangle with side length $s$ is $\frac{s^2\sqrt{3}}{4}$ (if you don't have this memorized it's not hard to derive). Comparing this formula to the area of $ABC$, we can easily find that $s^2=7$, which means that the side length of $ABC$ is $\boxed{\textbf{(B) } \sqrt{7}}$.
While this approach feels rather convoluted in comparison to Solution 1 (which only works for isosceles triangles), it is more flexible and can actually be generalized for any point in a general triangle (although that requires use of Heron's, and potentially Law of Sines and Cosines).
~IAmTheHazard | // 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));
label("$P'$", (3.5, 0), S, p = fontsize(10pt));
draw((8,3.5)--(3.5,0));
draw((0,3.5)--(3.5,0));
label("$P''$",(6,8.5), NE, p = fontsize(10pt));
draw((4,9.15)--(6,8.5));
draw((8,3.5)--(6,8.5));
label("$P'''$",(2.25,8), NW, p = fontsize(10pt));
draw((0,3.5)--(2.25,8));
draw((4,9.15)--(2.25,8));
draw((3.5,0)--(6,8.5)--(2.25,8)--cycle);
// Block 2
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)); label("$P'$", (3.5, 0), S, p = fontsize(10pt)); draw((8,3.5)--(3.5,0)); draw((0,3.5)--(3.5,0)); label("$P''$",(6,8.5), NE, p = fontsize(10pt)); draw((4,9.15)--(6,8.5)); draw((8,3.5)--(6,8.5)); label("$P'''$",(2.25,8), NW, p = fontsize(10pt)); draw((0,3.5)--(2.25,8)); draw((4,9.15)--(2.25,8)); draw((3.5,0)--(6,8.5)--(2.25,8)--cycle); | [] |
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 $R\in\{R_1,R_5\}$ by the tangent.
If $\angle Q=90^\circ,$ then $R\in\{R_4,R_8\}$ by the tangent.
If $\angle R=90^\circ,$ then $R\in\{R_2,R_3,R_6,R_7\}$ by the Inscribed Angle Theorem.
Together, there are $\boxed{\textbf{(D)}\ 8}$ such locations for $R.$
Remarks
The reflections of $R_1,R_2,R_3,R_4$ about $\overleftrightarrow{PQ}$ are $R_5,R_6,R_7,R_8,$ respectively.
The reflections of $R_1,R_2,R_5,R_6$ about the perpendicular bisector of $\overline{PQ}$ are $R_4,R_3,R_8,R_7,$ respectively.
~MRENTHUSIASM | // 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--R8)[0];
R7 = intersectionpoints(C,R5--R8)[1];
I1 = intersectionpoint(R2--R6,P--Q);
I2 = intersectionpoint(R3--R7,P--Q);
markscalefactor=0.0375;
draw(rightanglemark(R1,P,Q)^^rightanglemark(R2,I1,Q)^^rightanglemark(R3,I2,P)^^rightanglemark(R4,Q,P),red);
draw(Circle(O,4),dashed);
draw(R1--R5^^R4--R8^^R2--R6^^R3--R7^^P--Q);
dot(O,linewidth(4));
dot("$P$",P,1.5W,linewidth(4));
dot("$Q$",Q,1.5E,linewidth(4));
dot("$R_1$",R1,1.5NW,blue+linewidth(4));
dot("$R_4$",R4,1.5NE,blue+linewidth(4));
dot("$R_5$",R5,1.5SW,blue+linewidth(4));
dot("$R_8$",R8,1.5SE,blue+linewidth(4));
dot("$R_2$",R2,1.5NW,blue+linewidth(4));
dot("$R_3$",R3,1.5NE,blue+linewidth(4));
dot("$R_6$",R6,1.5SW,blue+linewidth(4));
dot("$R_7$",R7,1.5SE,blue+linewidth(4));
dot(I1,linewidth(4));
dot(I2,linewidth(4));
Label L1 = Label("$8$", align=(0,0), position=MidPoint, filltype=Fill(3,0,white));
Label L2 = Label("$3$", align=(0,0), position=MidPoint, filltype=Fill(0,3,white));
draw(P-(0,5)--Q-(0,5), L=L1, arrow=Arrows(),bar=Bars(15));
draw(R4+(2,0)--Q+(2,0), L=L2, arrow=Arrows(),bar=Bars(15));
draw(Q+(2,0)--R8+(2,0), L=L2, arrow=Arrows(),bar=Bars(15));
// Block 2
/* 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--R8)[0]; R7 = intersectionpoints(C,R5--R8)[1]; I1 = intersectionpoint(R2--R6,P--Q); I2 = intersectionpoint(R3--R7,P--Q); markscalefactor=0.0375; draw(rightanglemark(R1,P,Q)^^rightanglemark(R2,I1,Q)^^rightanglemark(R3,I2,P)^^rightanglemark(R4,Q,P),red); draw(Circle(O,4),dashed); draw(R1--R5^^R4--R8^^R2--R6^^R3--R7^^P--Q); dot(O,linewidth(4)); dot("$P$",P,1.5W,linewidth(4)); dot("$Q$",Q,1.5E,linewidth(4)); dot("$R_1$",R1,1.5NW,blue+linewidth(4)); dot("$R_4$",R4,1.5NE,blue+linewidth(4)); dot("$R_5$",R5,1.5SW,blue+linewidth(4)); dot("$R_8$",R8,1.5SE,blue+linewidth(4)); dot("$R_2$",R2,1.5NW,blue+linewidth(4)); dot("$R_3$",R3,1.5NE,blue+linewidth(4)); dot("$R_6$",R6,1.5SW,blue+linewidth(4)); dot("$R_7$",R7,1.5SE,blue+linewidth(4)); dot(I1,linewidth(4)); dot(I2,linewidth(4)); Label L1 = Label("$8$", align=(0,0), position=MidPoint, filltype=Fill(3,0,white)); Label L2 = Label("$3$", align=(0,0), position=MidPoint, filltype=Fill(0,3,white)); draw(P-(0,5)--Q-(0,5), L=L1, arrow=Arrows(),bar=Bars(15)); draw(R4+(2,0)--Q+(2,0), L=L2, arrow=Arrows(),bar=Bars(15)); draw(Q+(2,0)--R8+(2,0), L=L2, arrow=Arrows(),bar=Bars(15)); | [] |
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?
$\textbf{(A) } 6\sqrt3 - 3\pi \qquad \textbf{(B) } \frac{9\sqrt3}{2} - 2\pi \qquad \textbf{(C) } \frac{3\sqrt3}{2} - \frac{\pi}{3} \qquad \textbf{(D) } 3\sqrt3 - \pi \qquad \textbf{(E) } \frac{9\sqrt3}{2} - \pi$ | 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 $C$ lies on, $AB = 1$, since $B$ is the center of the semicircle with radius $1$ that $A$ lies on, and $\angle BAC = 60^\circ$, as a regular hexagon has angles of 120$^\circ$, and $\angle BAC$ is half of any angle in this hexagon. Now, since $AB = BC$, $\angle ACB = \angle BAC = 60^\circ$. Since the angles in a triangle sum to 180$^\circ$, $\angle ABC$ is also 60$^\circ$. Therefore, $\triangle ABC$ is an equilateral triangle with side lengths of $1$.
Since the area of a regular hexagon can be found with the formula $\frac{3\sqrt{3}s^2}{2}$, where $s$ is the side length of the hexagon, the area of this hexagon is $\frac{3\sqrt{3}(2^2)}{2} = 6\sqrt{3}$. Since the area of an equilateral triangle can be found with the formula $\frac{\sqrt{3}}{4}s^2$, where $s$ is the side length of the equilateral triangle, the area of an equilateral triangle with side lengths of $1$ is $\frac{\sqrt{3}}{4}\left(1^2\right) = \frac{\sqrt{3}}{4}$. Since the area of a circle can be found with the formula $\pi r^2$, the area of a sixth of a circle with radius $1$ is $\frac{\pi(1^2)}{6} = \frac{\pi}{6}$. In each sixth of the hexagon, there are two equilateral triangles colored white, each with an area of $\frac{\sqrt{3}}{4}$, and one-sixth of a circle with radius $1$ colored white, with an area of $\frac{\pi}{6}$. The rest of the sixth is colored gray. Therefore, the total area that is colored white in each sixth of the hexagon is $2\left(\frac{\sqrt{3}}{4}\right) + \frac{\pi}{6}$, which equals $\frac{\sqrt{3}}{2} + \frac{\pi}{6}$, and the total area colored white is $6\left(\frac{\sqrt{3}}{2} + \frac{\pi}{6}\right)$, which equals $3\sqrt{3} + \pi$. Since the area colored gray equals the total area of the hexagon minus the area colored white, the area colored gray is $6\sqrt{3} - (3\sqrt{3} + \pi)$, which equals $\boxed{\textbf{(D) }3\sqrt{3} - \pi}$. | // 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),1,60,240)--cycle,white);
fill(arc((b,a),1,300,480)--cycle,white);
fill(arc((b,c),1,240,420)--cycle,white);
fill(arc((z,c),1,120,300)--cycle,white);
draw(A--B--C--D--E--F--A);
draw(arc((z,c),1,120,300));
draw(arc((b,c),1,240,420));
draw(arc((b,a),1,300,480));
draw(arc((z,a),1,60,240));
draw(arc((2,y),1,180,360));
draw(arc((2,0),1,0,180));
label("2",(z,c),NE);
pair X,Y,Z;
X = (1,0);
Y = (2,0);
Z = (1.5,a);
pair d = 1.9*dir(7);
dot(X);
dot(Y);
dot(Z);
label("A",X,SW);
label("B",Y,SE);
label("C",Z,N);
draw(X--Y--Z--A);
label("1",(1.5,0),S);
label("1",(1.75,a/2),dir(30));
draw(anglemark(Y,X,Z,8),blue);
label("$60^\circ$",anglemark(Y,X,Z),d);
// Block 2
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),1,60,240)--cycle,white);
fill(arc((b,a),1,300,480)--cycle,white);
fill(arc((b,c),1,240,420)--cycle,white);
fill(arc((z,c),1,120,300)--cycle,white);
draw(A--B--C--D--E--F--A);
draw(arc((z,c),1,120,300));
draw(arc((b,c),1,240,420));
draw(arc((b,a),1,300,480));
draw(arc((z,a),1,60,240));
draw(arc((2,y),1,180,360));
draw(arc((2,0),1,0,180));
pair G,H,I,J,K;
G = (2,0);
H = (2.5,a);
I = (1.5,a);
J = (1,0);
K = (3,0);
pair d = 2.7*dir(78);
dot(G);
dot(H);
dot(I);
dot(J);
dot(K);
label("2",(z,c),NE);
label("1",(1.5,0),S);
label("1",(2.5,0),S);
add(pathticks(G--J,1,0.5,0,3,red));
add(pathticks(I--J,1,0.5,0,3,red));
add(pathticks(G--I,1,0.5,0,3,red));
add(pathticks(G--H,1,0.5,0,3,red));
add(pathticks(G--K,1,0.5,0,3,red));
add(pathticks(K--H,1,0.5,0,3,red));
label("$60^\circ$",anglemark(H,G,I),d);
draw(anglemark(H,G,I,8),blue);
label("$60^\circ$",G,2*dir(146));
draw(anglemark(I,G,J,8),blue);
label("$60^\circ$",G,2.8*dir(28));
draw(anglemark(K,G,H,8),blue);
draw(G--J--I--G);
draw(G--H--K--G);
// Block 3
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),1,60,240)--cycle,white); fill(arc((b,a),1,300,480)--cycle,white); fill(arc((b,c),1,240,420)--cycle,white); fill(arc((z,c),1,120,300)--cycle,white); draw(A--B--C--D--E--F--A); draw(arc((z,c),1,120,300)); draw(arc((b,c),1,240,420)); draw(arc((b,a),1,300,480)); draw(arc((z,a),1,60,240)); draw(arc((2,y),1,180,360)); draw(arc((2,0),1,0,180)); label("2",(z,c),NE); pair X,Y,Z; X = (1,0); Y = (2,0); Z = (1.5,a); pair d = 1.9*dir(7); dot(X); dot(Y); dot(Z); label("A",X,SW); label("B",Y,SE); label("C",Z,N); draw(X--Y--Z--A); label("1",(1.5,0),S); label("1",(1.75,a/2),dir(30)); draw(anglemark(Y,X,Z,8),blue); label("$60^\circ$",anglemark(Y,X,Z),d);
// Block 4
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),1,60,240)--cycle,white); fill(arc((b,a),1,300,480)--cycle,white); fill(arc((b,c),1,240,420)--cycle,white); fill(arc((z,c),1,120,300)--cycle,white); draw(A--B--C--D--E--F--A); draw(arc((z,c),1,120,300)); draw(arc((b,c),1,240,420)); draw(arc((b,a),1,300,480)); draw(arc((z,a),1,60,240)); draw(arc((2,y),1,180,360)); draw(arc((2,0),1,0,180)); pair G,H,I,J,K; G = (2,0); H = (2.5,a); I = (1.5,a); J = (1,0); K = (3,0); pair d = 2.7*dir(78); dot(G); dot(H); dot(I); dot(J); dot(K); label("2",(z,c),NE); label("1",(1.5,0),S); label("1",(2.5,0),S); add(pathticks(G--J,1,0.5,0,3,red)); add(pathticks(I--J,1,0.5,0,3,red)); add(pathticks(G--I,1,0.5,0,3,red)); add(pathticks(G--H,1,0.5,0,3,red)); add(pathticks(G--K,1,0.5,0,3,red)); add(pathticks(K--H,1,0.5,0,3,red)); label("$60^\circ$",anglemark(H,G,I),d); draw(anglemark(H,G,I,8),blue); label("$60^\circ$",G,2*dir(146)); draw(anglemark(I,G,J,8),blue); label("$60^\circ$",G,2.8*dir(28)); draw(anglemark(K,G,H,8),blue); draw(G--J--I--G); draw(G--H--K--G); | [] |
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?
$\textbf{(A) } 6\sqrt3 - 3\pi \qquad \textbf{(B) } \frac{9\sqrt3}{2} - 2\pi \qquad \textbf{(C) } \frac{3\sqrt3}{2} - \frac{\pi}{3} \qquad \textbf{(D) } 3\sqrt3 - \pi \qquad \textbf{(E) } \frac{9\sqrt3}{2} - \pi$ | 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 included has an area of:
\[\frac16 \cdot\pi \cdot1^2 = \frac{\pi}{6}\]
Hence, the area of the shaded region in that section is \[\frac{\sqrt{3}}{2}-\frac{\pi}{6}\]
For a final area of:
\[6\left(\frac{\sqrt{3}}{2}-\frac{\pi}{6}\right)=3\sqrt{3}-\pi\Rightarrow \boxed{\mathrm{(D)}}\]
~N828335 | // 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)),1,180,360)--(2,2sqrt(3))--cycle,white); fill(arc((0.5,3sqrt(3)/2),1,240,420)--(0.5,3sqrt(3)/2)--cycle,white); fill(arc((0.5,sqrt(3)/2),1,300,480)--(0.5,sqrt(3)/2)--cycle,white); draw((1,0)--(3,0)--(4,sqrt(3))--(3,2sqrt(3))--(1,2sqrt(3))--(0,sqrt(3))--(1,0)); draw(arc((2,0),1,180,0)--(2,0)--cycle); draw(arc((3.5,sqrt(3)/2),1,60,240)--(3.5,sqrt(3)/2)--cycle); draw(arc((3.5,3sqrt(3)/2),1,120,300)--(3.5,3sqrt(3)/2)--cycle); draw(arc((2,2sqrt(3)),1,180,360)--(2,2sqrt(3))--cycle); draw(arc((0.5,3sqrt(3)/2),1,240,420)--(0.5,3sqrt(3)/2)--cycle); draw(arc((0.5,sqrt(3)/2),1,300,480)--(0.5,sqrt(3)/2)--cycle); label("$2$",(3.5,3sqrt(3)/2),NE);
draw((1,0)--(3,2sqrt(3)));
draw((3,0)--(1,2sqrt(3)));
draw((4,sqrt(3))--(0,sqrt(3)));
draw((2,0)--(3.5,3sqrt(3)/2));
draw((3.5,sqrt(3)/2)--(2,2sqrt(3)));
draw((3.5,3sqrt(3)/2)--(0.5,3sqrt(3)/2));
draw((2,2sqrt(3))--(0.5,sqrt(3)/2));
draw((2,0)--(0.5,3sqrt(3)/2));
draw((3.5,sqrt(3)/2)--(0.5,sqrt(3)/2));
// Block 2
size(100);
fill((2,sqrt(3))--(2.5,3sqrt(3)/2)--(2,2sqrt(3))--(1.5,3sqrt(3)/2)--cycle,gray(0.4));
fill(arc((2,2sqrt(3)),1,240,300)--(2,2sqrt(3))--cycle,white);
draw(arc((2,2sqrt(3)),1,240,300)--(2,2sqrt(3))--cycle);
label("$1$",(2.25,7sqrt(3)/4),NE);
draw((2,sqrt(3))--(2.5,3sqrt(3)/2)--(2,2sqrt(3))--(1.5,3sqrt(3)/2)--cycle);
draw((2.5,3sqrt(3)/2)--(1.5,3sqrt(3)/2));
// Block 3
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)--(2,2sqrt(3))--cycle,white); fill(arc((0.5,3sqrt(3)/2),1,240,420)--(0.5,3sqrt(3)/2)--cycle,white); fill(arc((0.5,sqrt(3)/2),1,300,480)--(0.5,sqrt(3)/2)--cycle,white); draw((1,0)--(3,0)--(4,sqrt(3))--(3,2sqrt(3))--(1,2sqrt(3))--(0,sqrt(3))--(1,0)); draw(arc((2,0),1,180,0)--(2,0)--cycle); draw(arc((3.5,sqrt(3)/2),1,60,240)--(3.5,sqrt(3)/2)--cycle); draw(arc((3.5,3sqrt(3)/2),1,120,300)--(3.5,3sqrt(3)/2)--cycle); draw(arc((2,2sqrt(3)),1,180,360)--(2,2sqrt(3))--cycle); draw(arc((0.5,3sqrt(3)/2),1,240,420)--(0.5,3sqrt(3)/2)--cycle); draw(arc((0.5,sqrt(3)/2),1,300,480)--(0.5,sqrt(3)/2)--cycle); label("$2$",(3.5,3sqrt(3)/2),NE); draw((1,0)--(3,2sqrt(3))); draw((3,0)--(1,2sqrt(3))); draw((4,sqrt(3))--(0,sqrt(3))); draw((2,0)--(3.5,3sqrt(3)/2)); draw((3.5,sqrt(3)/2)--(2,2sqrt(3))); draw((3.5,3sqrt(3)/2)--(0.5,3sqrt(3)/2)); draw((2,2sqrt(3))--(0.5,sqrt(3)/2)); draw((2,0)--(0.5,3sqrt(3)/2)); draw((3.5,sqrt(3)/2)--(0.5,sqrt(3)/2));
// Block 4
size(100); fill((2,sqrt(3))--(2.5,3sqrt(3)/2)--(2,2sqrt(3))--(1.5,3sqrt(3)/2)--cycle,gray(0.4)); fill(arc((2,2sqrt(3)),1,240,300)--(2,2sqrt(3))--cycle,white); draw(arc((2,2sqrt(3)),1,240,300)--(2,2sqrt(3))--cycle); label("$1$",(2.25,7sqrt(3)/4),NE); draw((2,sqrt(3))--(2.5,3sqrt(3)/2)--(2,2sqrt(3))--(1.5,3sqrt(3)/2)--cycle); draw((2.5,3sqrt(3)/2)--(1.5,3sqrt(3)/2)); | [] |
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?
$\textbf{(A) } 6\sqrt3 - 3\pi \qquad \textbf{(B) } \frac{9\sqrt3}{2} - 2\pi \qquad \textbf{(C) } \frac{3\sqrt3}{2} - \frac{\pi}{3} \qquad \textbf{(D) } 3\sqrt3 - \pi \qquad \textbf{(E) } \frac{9\sqrt3}{2} - \pi$ | 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 represented by $S$.
Then the area of the shaded region can be expressed in the following two ways:
$[1]$. $\textbf{Area}$ of hexagon - ($\textbf{Area}$ of six semicircles $-$ $\textbf{Area}$ of six $\textbf{old}$ "leaves") $=$ $\frac{\sqrt{3}}{4} \cdot 4 \cdot 6 - (3\pi - S);$
$[2]$. $\textbf{Area}$ of dashed circle $-$ $\textbf{Area}$ of six $\textbf{new}$ "leaves" = $\pi - S;$
Then we have $\frac{\sqrt{3}}{4} \cdot 4 \cdot 6 - (3\pi - S) = \pi - S$
$S = 2\pi - 3\sqrt{3}$
Plugging $S$ back in to either $[1]$ or $[2]$, we can solve for the shaded area $= 3\sqrt{3} - \pi$ $\rightarrow\boxed{\mathrm{(D)}}$
--RyanZ@BRS
--Extremelysupercooldude (Minor latex and grammar edits)
~Yelechi (LaTeX edits) | // 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)),1,180,360)--(2,2sqrt(3))--cycle,white);
fill(arc((0.5,3sqrt(3)/2),1,240,420)--(0.5,3sqrt(3)/2)--cycle,white);
fill(arc((0.5,sqrt(3)/2),1,300,480)--(0.5,sqrt(3)/2)--cycle,white);
draw((1,0)--(3,0)--(4,sqrt(3))--(3,2sqrt(3))--(1,2sqrt(3))--(0,sqrt(3))--(1,0));
draw(arc((2,0),1,180,0)--(2,0)--cycle);
draw(arc((3.5,sqrt(3)/2),1,60,240)--(3.5,sqrt(3)/2)--cycle);
draw(arc((3.5,3sqrt(3)/2),1,120,300)--(3.5,3sqrt(3)/2)--cycle);
draw(arc((2,2sqrt(3)),1,180,360)--(2,2sqrt(3))--cycle);
draw(arc((0.5,3sqrt(3)/2),1,240,420)--(0.5,3sqrt(3)/2)--cycle);
draw(arc((0.5,sqrt(3)/2),1,300,480)--(0.5,sqrt(3)/2)--cycle);
label("$2$",(3.5,3sqrt(3)/2),NE);
draw(Arc((2,sqrt(3)), 1, 0, 360),dashed+linewidth(.5));
// Block 2
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)--(2,2sqrt(3))--cycle,white); fill(arc((0.5,3sqrt(3)/2),1,240,420)--(0.5,3sqrt(3)/2)--cycle,white); fill(arc((0.5,sqrt(3)/2),1,300,480)--(0.5,sqrt(3)/2)--cycle,white); draw((1,0)--(3,0)--(4,sqrt(3))--(3,2sqrt(3))--(1,2sqrt(3))--(0,sqrt(3))--(1,0)); draw(arc((2,0),1,180,0)--(2,0)--cycle); draw(arc((3.5,sqrt(3)/2),1,60,240)--(3.5,sqrt(3)/2)--cycle); draw(arc((3.5,3sqrt(3)/2),1,120,300)--(3.5,3sqrt(3)/2)--cycle); draw(arc((2,2sqrt(3)),1,180,360)--(2,2sqrt(3))--cycle); draw(arc((0.5,3sqrt(3)/2),1,240,420)--(0.5,3sqrt(3)/2)--cycle); draw(arc((0.5,sqrt(3)/2),1,300,480)--(0.5,sqrt(3)/2)--cycle); label("$2$",(3.5,3sqrt(3)/2),NE); draw(Arc((2,sqrt(3)), 1, 0, 360),dashed+linewidth(.5)); | [] |
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?
$\textbf{(A) } 6\sqrt3 - 3\pi \qquad \textbf{(B) } \frac{9\sqrt3}{2} - 2\pi \qquad \textbf{(C) } \frac{3\sqrt3}{2} - \frac{\pi}{3} \qquad \textbf{(D) } 3\sqrt3 - \pi \qquad \textbf{(E) } \frac{9\sqrt3}{2} - \pi$ | 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 is half of $y$, their total area is $\frac{6y}{2} = 3y$. The smaller hexagon minus this region would be $x$. So,
\[x= \frac{3\sqrt{3}}{2} \cdot 1^{2} - 3y\]
Also, the area of the larger hexagon minus the semicircles then add back $6y$ would also be $x$. (Note that by just subtracting the semicircles from the larger hexagon would result in an overlap of $6y$.) So,
\[x= \frac{3\sqrt{3}}{2} \cdot 2^{2} - 3\pi + 6y\]
Equating the equations, we have
\begin{align*} \frac{3\sqrt{3}}{2} \cdot 1^{2} - 3y &= \frac{3\sqrt{3}}{2} \cdot 2^{2} - 3\pi + 6y \\ \frac{3\sqrt{3}}{2} - 3y &= 6\sqrt{3} - 3\pi + 6y \\ \end{align*}
Solving for $y$, we have
\begin{align*} 9y &= \frac{3\sqrt{3}}{2} - 6\sqrt{3} + 3\pi \\ &=-\frac{9\sqrt{3}}{2} + 3\pi \\ \end{align*}
Then, we can plug this back into
\[x= \frac{3\sqrt{3}}{2} \cdot 1^{2} - 3y\]
Since $9y = -\frac{9\sqrt{3}}{2} + 3\pi$, dividing by $3$ gives
\begin{align*} 3y &= -\frac{9\sqrt{3}}{2} \times \frac{1}{3} + \pi \\ &=-\frac{3\sqrt{3}}{2} + \pi \\ \end{align*}
Therefore, $x = \frac{3\sqrt{3}}{2} - (-\frac{3\sqrt{3}}{2} + \pi) \implies x= \boxed{\textbf{(D)} ~3\sqrt{3} - \pi}$.
~MrThinker | // 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)),1,180,360)--(2,2sqrt(3))--cycle,white);
fill(arc((0.5,3sqrt(3)/2),1,240,420)--(0.5,3sqrt(3)/2)--cycle,white);
fill(arc((0.5,sqrt(3)/2),1,300,480)--(0.5,sqrt(3)/2)--cycle,white);
draw((1,0)--(3,0)--(4,sqrt(3))--(3,2sqrt(3))--(1,2sqrt(3))--(0,sqrt(3))--(1,0));
label("$2$",(3.5,3sqrt(3)/2),NE);
label("$x$",(2,1.6),N);
label("$y$",(1.3,3.2),S);
draw(arc((2,0),1,180,0)--(2,0)--cycle);
draw(arc((3.5,sqrt(3)/2),1,60,240)--(3.5,sqrt(3)/2)--cycle);
draw(arc((3.5,3sqrt(3)/2),1,120,300)--(3.5,3sqrt(3)/2)--cycle);
draw(arc((2,2sqrt(3)),1,180,360)--(2,2sqrt(3))--cycle);
draw(arc((0.5,3sqrt(3)/2),1,240,420)--(0.5,3sqrt(3)/2)--cycle);
draw(arc((0.5,sqrt(3)/2),1,300,480)--(0.5,sqrt(3)/2)--cycle);
draw((2.5,3sqrt(3)/2)--(1.5,3sqrt(3)/2));
draw((2.5,0.87)--(3,2sqrt(3)/2));
draw((3,2sqrt(3)/2)--(2.5,3sqrt(3)/2));
draw((1,1.73)--(1.5,2.6));
draw((1.5,0.87)--(1,1.73));
draw((2.5,sqrt(3)/2)--(1.5,sqrt(3)/2));
label("$1$", (1.2,2.2),W);
// Block 2
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)--(2,2sqrt(3))--cycle,white); fill(arc((0.5,3sqrt(3)/2),1,240,420)--(0.5,3sqrt(3)/2)--cycle,white); fill(arc((0.5,sqrt(3)/2),1,300,480)--(0.5,sqrt(3)/2)--cycle,white); draw((1,0)--(3,0)--(4,sqrt(3))--(3,2sqrt(3))--(1,2sqrt(3))--(0,sqrt(3))--(1,0)); label("$2$",(3.5,3sqrt(3)/2),NE); label("$x$",(2,1.6),N); label("$y$",(1.3,3.2),S); draw(arc((2,0),1,180,0)--(2,0)--cycle); draw(arc((3.5,sqrt(3)/2),1,60,240)--(3.5,sqrt(3)/2)--cycle); draw(arc((3.5,3sqrt(3)/2),1,120,300)--(3.5,3sqrt(3)/2)--cycle); draw(arc((2,2sqrt(3)),1,180,360)--(2,2sqrt(3))--cycle); draw(arc((0.5,3sqrt(3)/2),1,240,420)--(0.5,3sqrt(3)/2)--cycle); draw(arc((0.5,sqrt(3)/2),1,300,480)--(0.5,sqrt(3)/2)--cycle); draw((2.5,3sqrt(3)/2)--(1.5,3sqrt(3)/2)); draw((2.5,0.87)--(3,2sqrt(3)/2)); draw((3,2sqrt(3)/2)--(2.5,3sqrt(3)/2)); draw((1,1.73)--(1.5,2.6)); draw((1.5,0.87)--(1,1.73)); draw((2.5,sqrt(3)/2)--(1.5,sqrt(3)/2)); label("$1$", (1.2,2.2),W); | [] |
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?
$\textbf{(A) } 6\sqrt3 - 3\pi \qquad \textbf{(B) } \frac{9\sqrt3}{2} - 2\pi \qquad \textbf{(C) } \frac{3\sqrt3}{2} - \frac{\pi}{3} \qquad \textbf{(D) } 3\sqrt3 - \pi \qquad \textbf{(E) } \frac{9\sqrt3}{2} - \pi$ | 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 by the combined area of six equilateral triangles with side length $1$, or $\frac{\sqrt{3}}{4}\cdot 6 = \frac{3\sqrt{3}}{2}.$
Now, we have to subtract off the six mini-sectors between the triangle and the shaded area. The area of one of these mini-sectors is the difference between a $60^{\circ}$ sector of the semicircle and the equilateral we just drew. The area of a sector is just $\frac{\pi}{6}$, and the area of a triangle is $\frac{\sqrt{3}}{4}.$ Therefore our combined area of the mini-sectors is
\[6\left(\frac{\pi}{6} - \frac{\sqrt{3}}{4}\right) = \pi-\frac{3\sqrt{3}}{2}.\]
Subtracting this from the area of the (smaller) hexagon, the desired area is
\[\frac{3\sqrt{3}}{2} -\left(\pi - \frac{3\sqrt{3}}{2}\right) = \boxed{\textbf{(D)} ~3\sqrt{3} - \pi}\]
-Benedict T (countmath1) | // 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)),1,180,360)--(2,2sqrt(3))--cycle,white);
fill(arc((0.5,3sqrt(3)/2),1,240,420)--(0.5,3sqrt(3)/2)--cycle,white);
fill(arc((0.5,sqrt(3)/2),1,300,480)--(0.5,sqrt(3)/2)--cycle,white);
draw((1,0)--(3,0)--(4,sqrt(3))--(3,2sqrt(3))--(1,2sqrt(3))--(0,sqrt(3))--(1,0));
label("$2$",(3.5,3sqrt(3)/2),NE);
draw(arc((2,0),1,180,0)--(2,0)--cycle);
draw(arc((3.5,sqrt(3)/2),1,60,240)--(3.5,sqrt(3)/2)--cycle);
draw(arc((3.5,3sqrt(3)/2),1,120,300)--(3.5,3sqrt(3)/2)--cycle);
draw(arc((2,2sqrt(3)),1,180,360)--(2,2sqrt(3))--cycle);
draw(arc((0.5,3sqrt(3)/2),1,240,420)--(0.5,3sqrt(3)/2)--cycle);
draw(arc((0.5,sqrt(3)/2),1,300,480)--(0.5,sqrt(3)/2)--cycle);
draw((2.5,3sqrt(3)/2)--(1.5,3sqrt(3)/2));
draw((2.5,0.87)--(3,2sqrt(3)/2));
draw((3,2sqrt(3)/2)--(2.5,3sqrt(3)/2));
draw((1,1.73)--(1.5,2.6));
draw((1.5,0.87)--(1,1.73));
draw((2.5,sqrt(3)/2)--(1.5,sqrt(3)/2));
label("$1$", (2, 0.9), S);
draw((1.43, 1)--(2, 0)--(2.57, 1));
// Block 2
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)--(2,2sqrt(3))--cycle,white); fill(arc((0.5,3sqrt(3)/2),1,240,420)--(0.5,3sqrt(3)/2)--cycle,white); fill(arc((0.5,sqrt(3)/2),1,300,480)--(0.5,sqrt(3)/2)--cycle,white); draw((1,0)--(3,0)--(4,sqrt(3))--(3,2sqrt(3))--(1,2sqrt(3))--(0,sqrt(3))--(1,0)); label("$2$",(3.5,3sqrt(3)/2),NE); draw(arc((2,0),1,180,0)--(2,0)--cycle); draw(arc((3.5,sqrt(3)/2),1,60,240)--(3.5,sqrt(3)/2)--cycle); draw(arc((3.5,3sqrt(3)/2),1,120,300)--(3.5,3sqrt(3)/2)--cycle); draw(arc((2,2sqrt(3)),1,180,360)--(2,2sqrt(3))--cycle); draw(arc((0.5,3sqrt(3)/2),1,240,420)--(0.5,3sqrt(3)/2)--cycle); draw(arc((0.5,sqrt(3)/2),1,300,480)--(0.5,sqrt(3)/2)--cycle); draw((2.5,3sqrt(3)/2)--(1.5,3sqrt(3)/2)); draw((2.5,0.87)--(3,2sqrt(3)/2)); draw((3,2sqrt(3)/2)--(2.5,3sqrt(3)/2)); draw((1,1.73)--(1.5,2.6)); draw((1.5,0.87)--(1,1.73)); draw((2.5,sqrt(3)/2)--(1.5,sqrt(3)/2)); label("$1$", (2, 0.9), S); draw((1.43, 1)--(2, 0)--(2.57, 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 resulting list (that is, the $4$th, $8$th, $12$th, $\ldots$ digits from the left in what remained), and then erased every fifth digit from what remained at that point. What is the sum of the three digits that were then in the positions $2019, 2020, 2021$?
$\textbf{(A)}\ 7 \qquad\textbf{(B)}\ 9 \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ 11 \qquad\textbf{(E)}\ 12$ | 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 erasing, one cycle has length $L\left(1-\frac13\right)=\frac23L,$ from which $L$ must be divisible by $3.$
After the second round of erasing, one cycle has length $L\left(1-\frac13\right)\left(1-\frac14\right)=\frac12L,$ from which $L$ must be divisible by $2.$
After the third round of erasing, one cycle has length $L\left(1-\frac13\right)\left(1-\frac14\right)\left(1-\frac15\right)=\frac25L,$ from which $L$ must be divisible by $5.$
The least such positive integer $L$ is $\operatorname{lcm}(5,3,2)=30.$ So, there is a repeating pattern for every $30$ digits on the original list. As shown below, the digits erased in the first, second, and third rounds are colored in red, yellow, and green, respectively:
As indicated by the white squares, each group of $30$ digits on the original list has $\frac25\cdot30=12$ digits remaining on the final list.
Since $2019,2020,2021$ are congruent to $3,4,5$ modulo $12,$ respectively, the three digits in the final positions $2019,2020,2021$ are $4,2,5,$ respectively:
Therefore, the answer is $4+2+5=\boxed{\textbf{(D)}\ 11}.$
~MRENTHUSIASM | // 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)--(19,4)--(19,0)--cycle,green);
fill((27,0)--(27,4)--(28,4)--(28,0)--cycle,green);
for (real i=1; i<5; ++i) {
for (real j=0; j<30; ++j) {
label("$"+string(1+j%5)+"$",(j+0.5,i-0.5));
}
}
for (real i=0; i<30; ++i) {
label("$\vdots$",(i+0.5,-1/3));
}
add(grid(30,4,linewidth(1.25)));
// Block 2
/* Made by MRENTHUSIASM */
size(20cm);
for (real i=2.5; i<30; i+=3) {
fill((i-0.5,0)--(i-0.5,1)--(i+0.5,1)--(i+0.5,0)--cycle,red);
}
for (real i=4.5; i<30; i+=6) {
fill((i-0.5,0)--(i-0.5,1)--(i+0.5,1)--(i+0.5,0)--cycle,yellow);
}
fill((7,0)--(7,1)--(8,1)--(8,0)--cycle,green);
fill((18,0)--(18,1)--(19,1)--(19,0)--cycle,green);
fill((27,0)--(27,1)--(28,1)--(28,0)--cycle,green);
for (real j=0; j<30; ++j) {
label("$"+string(1+j%5)+"$",(j+0.5,0.5));
}
draw((3.5,-1.25)--(3.5,-0.2),linewidth(1.25),EndArrow);
draw((6.5,-1.25)--(6.5,-0.2),linewidth(1.25),EndArrow);
draw((9.5,-1.25)--(9.5,-0.2),linewidth(1.25),EndArrow);
add(grid(30,1,linewidth(1.25)));
// Block 3
/* 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)--(19,4)--(19,0)--cycle,green); fill((27,0)--(27,4)--(28,4)--(28,0)--cycle,green); for (real i=1; i<5; ++i) { for (real j=0; j<30; ++j) { label("$"+string(1+j%5)+"$",(j+0.5,i-0.5)); } } for (real i=0; i<30; ++i) { label("$\vdots$",(i+0.5,-1/3)); } add(grid(30,4,linewidth(1.25)));
// Block 4
/* Made by MRENTHUSIASM */ size(20cm); for (real i=2.5; i<30; i+=3) { fill((i-0.5,0)--(i-0.5,1)--(i+0.5,1)--(i+0.5,0)--cycle,red); } for (real i=4.5; i<30; i+=6) { fill((i-0.5,0)--(i-0.5,1)--(i+0.5,1)--(i+0.5,0)--cycle,yellow); } fill((7,0)--(7,1)--(8,1)--(8,0)--cycle,green); fill((18,0)--(18,1)--(19,1)--(19,0)--cycle,green); fill((27,0)--(27,1)--(28,1)--(28,0)--cycle,green); for (real j=0; j<30; ++j) { label("$"+string(1+j%5)+"$",(j+0.5,0.5)); } draw((3.5,-1.25)--(3.5,-0.2),linewidth(1.25),EndArrow); draw((6.5,-1.25)--(6.5,-0.2),linewidth(1.25),EndArrow); draw((9.5,-1.25)--(9.5,-0.2),linewidth(1.25),EndArrow); add(grid(30,1,linewidth(1.25))); | [] |
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 $\overline{GJ} \perp \overline{EH}$. See the figure below. Triangle $AEH$, quadrilateral $BFIE$, quadrilateral $DHJG$, and pentagon $FCGJI$ each has area $1.$ What is $FI^2$?
$\textbf{(A) } \frac{7}{3} \qquad \textbf{(B) } 8-4\sqrt2 \qquad \textbf{(C) } 1+\sqrt2 \qquad \textbf{(D) } \frac{7}{4}\sqrt2 \qquad \textbf{(E) } 2\sqrt2$ | 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 $AH$ = $AE$ = $\sqrt{2}$, $EB$ = $2-\sqrt{2}$, $EK$ = $2\sqrt{2}-2$.
$FI^2=\text{area of} \: BFIE+\text{area of} \:FJ'H'B+\text{area of} \:EH'K \\= 1 + 1 + \frac{1}{2}(2\sqrt{2}-2)^2=8-4\sqrt{2}\rightarrow \boxed{\textbf{(B)}\ 8-4\sqrt{2}}$
--Ryan Zhang @BRS
--Minor edit by WhySean38 | // 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 = (4,l);
draw(A--B--C--D--A);
draw(H--E);
draw(J--G^^F--I);
draw(F--M);
draw(M--L);
draw(E--K,dashed+linewidth(.5));
draw(K--L,dashed+linewidth(.5));
draw(B--L);
draw(rightanglemark(G, J, I), linewidth(.5));
draw(rightanglemark(F, I, E), linewidth(.5));
draw(rightanglemark(F, M, L), linewidth(.5));
fill((4,0)--(k,0)--M--(4,y)--cycle, gray);
dot("$A$", A, S);
dot("$C$", C, dir(90));
dot("$D$", D, dir(90));
dot("$E$", E, S);
dot("$G$", G, N);
dot("$H$", H, W);
dot("$I$", I, SW);
dot("$J$", J, SW);
dot("$K$", K, S);
dot("$F(G)$", F, E);
dot("$H'$", L, S);
dot("$B(D)$", B, S);
// Block 2
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); draw(A--B--C--D--A); draw(H--E); draw(J--G^^F--I); draw(F--M); draw(M--L); draw(E--K,dashed+linewidth(.5)); draw(K--L,dashed+linewidth(.5)); draw(B--L); draw(rightanglemark(G, J, I), linewidth(.5)); draw(rightanglemark(F, I, E), linewidth(.5)); draw(rightanglemark(F, M, L), linewidth(.5)); fill((4,0)--(k,0)--M--(4,y)--cycle, gray); dot("$A$", A, S); dot("$C$", C, dir(90)); dot("$D$", D, dir(90)); dot("$E$", E, S); dot("$G$", G, N); dot("$H$", H, W); dot("$I$", I, SW); dot("$J$", J, SW); dot("$K$", K, S); dot("$F(G)$", F, E); dot("$H'$", L, S); dot("$B(D)$", B, S); | [] |
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 $\overline{GJ} \perp \overline{EH}$. See the figure below. Triangle $AEH$, quadrilateral $BFIE$, quadrilateral $DHJG$, and pentagon $FCGJI$ each has area $1.$ What is $FI^2$?
$\textbf{(A) } \frac{7}{3} \qquad \textbf{(B) } 8-4\sqrt2 \qquad \textbf{(C) } 1+\sqrt2 \qquad \textbf{(D) } \frac{7}{4}\sqrt2 \qquad \textbf{(E) } 2\sqrt2$ | 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 $FI = GJ = x$, $KL = LG = GJ - LJ = x - 2 \sqrt{2} + 2$
$[DHK] = \frac{(2 - \sqrt{2})^2}{2} = 3 - 2 \sqrt {2}$
$[GKL] = \frac{(x - 2 \sqrt{2} + 2)^2}{2} = \frac{x^2}{2} + 2x - 2x \sqrt{2} - 4 \sqrt{2} + 6$
$[HKJL] = (x - 2 \sqrt{2} + 2) \cdot (2 \sqrt{2} - 2) = 2x \sqrt{2} - 2x + 8 \sqrt{2} -12$
\[[DHK] + [GKL] + [HKLJ] = [DHJG]\]
\[3 - 2 \sqrt {2} + \frac{x^2}{2} + 2x - 2x \sqrt{2} - 4 \sqrt{2} + 6 + 2x \sqrt{2} - 2x + 8 \sqrt{2} -12= 1\]
\[\frac{x^2}{2} + 2 \sqrt{2} - 4 = 0\]
\[x^2 = 8 - 4 \sqrt{2}\]
\[FI^2 = \boxed{\textbf{(B)}\ 8-4\sqrt{2}}\]
~isabelchen | // 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--A);
draw(H--E);
draw(J--G^^F--I);
draw(H--K,dashed+linewidth(.5));
draw(L--K,dashed+linewidth(.5));
draw(rightanglemark(G, J, I), linewidth(.5));
draw(rightanglemark(F, I, E), linewidth(.5));
draw(rightanglemark(H, K, L), linewidth(.5));
draw(rightanglemark(K, L, G), linewidth(.5));
dot("$A$", A, S);
dot("$B$", B, S);
dot("$C$", C, dir(90));
dot("$D$", D, dir(90));
dot("$E$", E, S);
dot("$F$", F, dir(0));
dot("$G$", G, N);
dot("$H$", H, W);
dot("$I$", I, SW);
dot("$J$", J, SW);
dot("$K$", K, N);
dot("$L$", L, S);
// Block 2
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--E); draw(J--G^^F--I); draw(H--K,dashed+linewidth(.5)); draw(L--K,dashed+linewidth(.5)); draw(rightanglemark(G, J, I), linewidth(.5)); draw(rightanglemark(F, I, E), linewidth(.5)); draw(rightanglemark(H, K, L), linewidth(.5)); draw(rightanglemark(K, L, G), linewidth(.5)); dot("$A$", A, S); dot("$B$", B, S); dot("$C$", C, dir(90)); dot("$D$", D, dir(90)); dot("$E$", E, S); dot("$F$", F, dir(0)); dot("$G$", G, N); dot("$H$", H, W); dot("$I$", I, SW); dot("$J$", J, SW); dot("$K$", K, N); dot("$L$", L, S); | [] |
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 origin;
$\quad\bullet\qquad$ $H,$ a reflection across the $x$-axis; and
$\quad\bullet\qquad$ $V,$ a reflection across the $y$-axis.
Each of these transformations maps the squares onto itself, but the positions of the labeled vertices will change. For example, applying $R$ and then $V$ would send the vertex $A$ at $(1,1)$ to $(-1,-1)$ and would send the vertex $B$ at $(-1,1)$ to itself. How many sequences of $20$ transformations chosen from $\{L, R, H, V\}$ will send all of the labeled vertices back to their original positions? (For example, $R, R, V, H$ is one sequence of $4$ transformations that will send the vertices back to their original positions.)
$\textbf{(A)}\ 2^{37} \qquad\textbf{(B)}\ 3\cdot 2^{36} \qquad\textbf{(C)}\ 2^{38} \qquad\textbf{(D)}\ 3\cdot 2^{37} \qquad\textbf{(E)}\ 2^{39}$ | 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 vertices.
After the $19$th transformation, vertex $A$ will be at either $(1,-1)$ or $(-1,1).$ All possible configurations of the labeled vertices are shown below:
Each sequence of $19$ transformations generates one valid sequence of $20$ transformations. Therefore, the answer is $4^{19}=\boxed{\textbf{(C)}\ 2^{38}}.$
~MRENTHUSIASM | // 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));
label("$B$",(13,1));
label("$A$",(12,1));
label("$D$",(12,0));
label("\textbf{Configuration}",(-5,0.5));
label("\textbf{The 20th}",(-5,-1.5));
label("\textbf{Transformation}",(-5,-2.25));
label("$L$",(0.5,-1.875));
label("$R$",(4.5,-1.875));
label("$H$",(8.5,-1.875));
label("$V$",(12.5,-1.875));
// Block 2
/* 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("$B$",(13,1)); label("$A$",(12,1)); label("$D$",(12,0)); label("\textbf{Configuration}",(-5,0.5)); label("\textbf{The 20th}",(-5,-1.5)); label("\textbf{Transformation}",(-5,-2.25)); label("$L$",(0.5,-1.875)); label("$R$",(4.5,-1.875)); label("$H$",(8.5,-1.875)); label("$V$",(12.5,-1.875)); | [] |
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 counted as distinct). For example, the number $6$ can be written as $6$, $2\cdot 3$, and $3\cdot2$, so $D(6) = 3$. What is $D(96)$?
$\textbf{(A) } 112 \qquad\textbf{(B) } 128 \qquad\textbf{(C) } 144 \qquad\textbf{(D) } 172 \qquad\textbf{(E) } 184$ | 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 its left and below it. We label the points to obtain number of paths.
The result is $\boxed{\textbf{(A) }112}$
~reda_mandymath | // 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);
// Block 2
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); label("1", (0, 0), NW); label("1", (1, 0), NW); label("2", (2, 0), NW); label("4", (3, 0), NW); label("8", (4, 0), NW); label("16", (5, 0), NW); label("1", (0, 1), NW); label("3", (1, 1), NW); label("8", (2, 1), NW); label("20", (3, 1), NW); label("48", (4, 1), NW); label("112", (5, 1), NW); | [] |
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 $AP?$
$\textbf{(A) } \frac{\sqrt5}{12} \qquad \textbf{(B) } \frac{\sqrt5}{10} \qquad \textbf{(C) } \frac{\sqrt5}{9} \qquad \textbf{(D) } \frac{\sqrt5}{8} \qquad \textbf{(E) } \frac{2\sqrt5}{15}$ | 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{AP}{AN}&=\frac{AN}{AM} \\ \frac{AP}{1/2}&=\frac{1/2}{\sqrt5/2} \\ AP&=\boxed{\textbf{(B) } \frac{\sqrt5}{10}}. \end{align*}
~QIDb602 | // 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,1.5*(0,1),linewidth(4));
dot("$A$",A,1.5*SW,linewidth(4));
dot("$B$",B,1.5*NW,linewidth(4));
dot("$C$",C,1.5*NE,linewidth(4));
dot("$D$",D,1.5*SE,linewidth(4));
dot("$M$",M,1.5*E,linewidth(4));
dot("$N$",N,1.5*W,linewidth(4));
dot("$P$",P,1.5*dir(60),linewidth(4));
markscalefactor=0.00625;
draw(rightanglemark(A,N,M),red);
draw(rightanglemark(N,P,A),red);
draw(A--B--C--D--cycle^^A--M^^P--N--M^^p);
// Block 2
/* 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,1.5*(0,1),linewidth(4)); dot("$A$",A,1.5*SW,linewidth(4)); dot("$B$",B,1.5*NW,linewidth(4)); dot("$C$",C,1.5*NE,linewidth(4)); dot("$D$",D,1.5*SE,linewidth(4)); dot("$M$",M,1.5*E,linewidth(4)); dot("$N$",N,1.5*W,linewidth(4)); dot("$P$",P,1.5*dir(60),linewidth(4)); markscalefactor=0.00625; draw(rightanglemark(A,N,M),red); draw(rightanglemark(N,P,A),red); draw(A--B--C--D--cycle^^A--M^^P--N--M^^p); | [] |
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 $AP?$
$\textbf{(A) } \frac{\sqrt5}{12} \qquad \textbf{(B) } \frac{\sqrt5}{10} \qquad \textbf{(C) } \frac{\sqrt5}{9} \qquad \textbf{(D) } \frac{\sqrt5}{8} \qquad \textbf{(E) } \frac{2\sqrt5}{15}$ | 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 $PM=\frac{\sqrt5}{2}-x.$ Applying the Pythagorean Theorem to right $\triangle ANP$ gives $NP^2=\left(\frac12\right)^2-x^2,$ and applying the Pythagorean Theorem to right $\triangle MNP$ gives $NP^2=1^2-\left(\frac{\sqrt5}{2}-x\right)^2.$ Equating the expressions for $NP^2$ produces
\begin{align*} \left(\frac12\right)^2-x^2&=1^2-\left(\frac{\sqrt5}{2}-x\right)^2 \\ \frac14-x^2&=1-\frac54+\sqrt5x-x^2 \\ \frac12&=\sqrt5x. \end{align*}
Finally, dividing both sides by $\sqrt5$ and then rationalizing the denominator, we obtain $x=\frac{1}{2\sqrt5}=\boxed{\textbf{(B) } \frac{\sqrt5}{10}}.$
~MRENTHUSIASM | // 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("$\omega$",O,1.5*(0,1),linewidth(4));
dot("$A$",A,1.5*SW,linewidth(4));
dot("$B$",B,1.5*NW,linewidth(4));
dot("$C$",C,1.5*NE,linewidth(4));
dot("$D$",D,1.5*SE,linewidth(4));
dot("$M$",M,1.5*E,linewidth(4));
dot("$N$",N,1.5*W,linewidth(4));
dot("$P$",P,1.5*dir(60),linewidth(4));
markscalefactor=0.00625;
draw(rightanglemark(A,N,M),red);
draw(rightanglemark(N,P,A),red);
draw(A--B--C--D--cycle^^A--M^^P--N--M^^p);
// Block 2
/* 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("$\omega$",O,1.5*(0,1),linewidth(4)); dot("$A$",A,1.5*SW,linewidth(4)); dot("$B$",B,1.5*NW,linewidth(4)); dot("$C$",C,1.5*NE,linewidth(4)); dot("$D$",D,1.5*SE,linewidth(4)); dot("$M$",M,1.5*E,linewidth(4)); dot("$N$",N,1.5*W,linewidth(4)); dot("$P$",P,1.5*dir(60),linewidth(4)); markscalefactor=0.00625; draw(rightanglemark(A,N,M),red); draw(rightanglemark(N,P,A),red); draw(A--B--C--D--cycle^^A--M^^P--N--M^^p); | [] |
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?
$\textbf{(A) } 6\sqrt3 - 3\pi \qquad \textbf{(B) } \frac{9\sqrt3}{2} - 2\pi \qquad \textbf{(C) } \frac{3\sqrt3}{2} - \frac{\pi}{3} \qquad \textbf{(D) } 3\sqrt3 - \pi \qquad \textbf{(E) } \frac{9\sqrt3}{2} - \pi$ | 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 $C$ lies on, $AB = 1$, since $B$ is the center of the semicircle with radius $1$ that $A$ lies on, and $\angle BAC = 60^\circ$, as a regular hexagon has angles of 120$^\circ$, and $\angle BAC$ is half of any angle in this hexagon. Now, since $AB = BC$, $\angle ACB = \angle BAC = 60^\circ$. Since the angles in a triangle sum to 180$^\circ$, $\angle ABC$ is also 60$^\circ$. Therefore, $\triangle ABC$ is an equilateral triangle with side lengths of $1$.
Since the area of a regular hexagon can be found with the formula $\frac{3\sqrt{3}s^2}{2}$, where $s$ is the side length of the hexagon, the area of this hexagon is $\frac{3\sqrt{3}(2^2)}{2} = 6\sqrt{3}$. Since the area of an equilateral triangle can be found with the formula $\frac{\sqrt{3}}{4}s^2$, where $s$ is the side length of the equilateral triangle, the area of an equilateral triangle with side lengths of $1$ is $\frac{\sqrt{3}}{4}\left(1^2\right) = \frac{\sqrt{3}}{4}$. Since the area of a circle can be found with the formula $\pi r^2$, the area of a sixth of a circle with radius $1$ is $\frac{\pi(1^2)}{6} = \frac{\pi}{6}$. In each sixth of the hexagon, there are two equilateral triangles colored white, each with an area of $\frac{\sqrt{3}}{4}$, and one-sixth of a circle with radius $1$ colored white, with an area of $\frac{\pi}{6}$. The rest of the sixth is colored gray. Therefore, the total area that is colored white in each sixth of the hexagon is $2\left(\frac{\sqrt{3}}{4}\right) + \frac{\pi}{6}$, which equals $\frac{\sqrt{3}}{2} + \frac{\pi}{6}$, and the total area colored white is $6\left(\frac{\sqrt{3}}{2} + \frac{\pi}{6}\right)$, which equals $3\sqrt{3} + \pi$. Since the area colored gray equals the total area of the hexagon minus the area colored white, the area colored gray is $6\sqrt{3} - (3\sqrt{3} + \pi)$, which equals $\boxed{\textbf{(D) }3\sqrt{3} - \pi}$. | // 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),1,60,240)--cycle,white); fill(arc((b,a),1,300,480)--cycle,white); fill(arc((b,c),1,240,420)--cycle,white); fill(arc((z,c),1,120,300)--cycle,white); draw(A--B--C--D--E--F--A); draw(arc((z,c),1,120,300)); draw(arc((b,c),1,240,420)); draw(arc((b,a),1,300,480)); draw(arc((z,a),1,60,240)); draw(arc((2,y),1,180,360)); draw(arc((2,0),1,0,180)); label("2",(z,c),NE); pair X,Y,Z; X = (1,0); Y = (2,0); Z = (1.5,a); pair d = 1.9*dir(7); dot(X); dot(Y); dot(Z); label("A",X,SW); label("B",Y,SE); label("C",Z,N); draw(X--Y--Z--A); label("1",(1.5,0),S); label("1",(1.75,a/2),dir(30)); draw(anglemark(Y,X,Z,8),blue); label("$60^\circ$",anglemark(Y,X,Z),d);
// Block 2
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),1,60,240)--cycle,white); fill(arc((b,a),1,300,480)--cycle,white); fill(arc((b,c),1,240,420)--cycle,white); fill(arc((z,c),1,120,300)--cycle,white); draw(A--B--C--D--E--F--A); draw(arc((z,c),1,120,300)); draw(arc((b,c),1,240,420)); draw(arc((b,a),1,300,480)); draw(arc((z,a),1,60,240)); draw(arc((2,y),1,180,360)); draw(arc((2,0),1,0,180)); pair G,H,I,J,K; G = (2,0); H = (2.5,a); I = (1.5,a); J = (1,0); K = (3,0); pair d = 2.7*dir(78); dot(G); dot(H); dot(I); dot(J); dot(K); label("2",(z,c),NE); label("1",(1.5,0),S); label("1",(2.5,0),S); add(pathticks(G--J,1,0.5,0,3,red)); add(pathticks(I--J,1,0.5,0,3,red)); add(pathticks(G--I,1,0.5,0,3,red)); add(pathticks(G--H,1,0.5,0,3,red)); add(pathticks(G--K,1,0.5,0,3,red)); add(pathticks(K--H,1,0.5,0,3,red)); label("$60^\circ$",anglemark(H,G,I),d); draw(anglemark(H,G,I,8),blue); label("$60^\circ$",G,2*dir(146)); draw(anglemark(I,G,J,8),blue); label("$60^\circ$",G,2.8*dir(28)); draw(anglemark(K,G,H,8),blue); draw(G--J--I--G); draw(G--H--K--G); | [] |
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?
$\textbf{(A) } 6\sqrt3 - 3\pi \qquad \textbf{(B) } \frac{9\sqrt3}{2} - 2\pi \qquad \textbf{(C) } \frac{3\sqrt3}{2} - \frac{\pi}{3} \qquad \textbf{(D) } 3\sqrt3 - \pi \qquad \textbf{(E) } \frac{9\sqrt3}{2} - \pi$ | 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 included has an area of:
\[\frac16 \cdot\pi \cdot1^2 = \frac{\pi}{6}\]
Hence, the area of the shaded region in that section is \[\frac{\sqrt{3}}{2}-\frac{\pi}{6}\]
For a final area of:
\[6\left(\frac{\sqrt{3}}{2}-\frac{\pi}{6}\right)=3\sqrt{3}-\pi\Rightarrow \boxed{\mathrm{(D)}}\]
~N828335 | // 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)),1,180,360)--(2,2sqrt(3))--cycle,white); fill(arc((0.5,3sqrt(3)/2),1,240,420)--(0.5,3sqrt(3)/2)--cycle,white); fill(arc((0.5,sqrt(3)/2),1,300,480)--(0.5,sqrt(3)/2)--cycle,white); draw((1,0)--(3,0)--(4,sqrt(3))--(3,2sqrt(3))--(1,2sqrt(3))--(0,sqrt(3))--(1,0)); draw(arc((2,0),1,180,0)--(2,0)--cycle); draw(arc((3.5,sqrt(3)/2),1,60,240)--(3.5,sqrt(3)/2)--cycle); draw(arc((3.5,3sqrt(3)/2),1,120,300)--(3.5,3sqrt(3)/2)--cycle); draw(arc((2,2sqrt(3)),1,180,360)--(2,2sqrt(3))--cycle); draw(arc((0.5,3sqrt(3)/2),1,240,420)--(0.5,3sqrt(3)/2)--cycle); draw(arc((0.5,sqrt(3)/2),1,300,480)--(0.5,sqrt(3)/2)--cycle); label("$2$",(3.5,3sqrt(3)/2),NE); draw((1,0)--(3,2sqrt(3))); draw((3,0)--(1,2sqrt(3))); draw((4,sqrt(3))--(0,sqrt(3))); draw((2,0)--(3.5,3sqrt(3)/2)); draw((3.5,sqrt(3)/2)--(2,2sqrt(3))); draw((3.5,3sqrt(3)/2)--(0.5,3sqrt(3)/2)); draw((2,2sqrt(3))--(0.5,sqrt(3)/2)); draw((2,0)--(0.5,3sqrt(3)/2)); draw((3.5,sqrt(3)/2)--(0.5,sqrt(3)/2));
// Block 2
size(100); fill((2,sqrt(3))--(2.5,3sqrt(3)/2)--(2,2sqrt(3))--(1.5,3sqrt(3)/2)--cycle,gray(0.4)); fill(arc((2,2sqrt(3)),1,240,300)--(2,2sqrt(3))--cycle,white); draw(arc((2,2sqrt(3)),1,240,300)--(2,2sqrt(3))--cycle); label("$1$",(2.25,7sqrt(3)/4),NE); draw((2,sqrt(3))--(2.5,3sqrt(3)/2)--(2,2sqrt(3))--(1.5,3sqrt(3)/2)--cycle); draw((2.5,3sqrt(3)/2)--(1.5,3sqrt(3)/2)); | [] |
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?
$\textbf{(A) } 6\sqrt3 - 3\pi \qquad \textbf{(B) } \frac{9\sqrt3}{2} - 2\pi \qquad \textbf{(C) } \frac{3\sqrt3}{2} - \frac{\pi}{3} \qquad \textbf{(D) } 3\sqrt3 - \pi \qquad \textbf{(E) } \frac{9\sqrt3}{2} - \pi$ | 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 represented by $S$.
Then the area of the shaded region can be expressed in the following two ways:
$[1]$. $\textbf{Area}$ of hexagon - ($\textbf{Area}$ of six semicircles $-$ $\textbf{Area}$ of six $\textbf{old}$ "leaves") $=$ $\frac{\sqrt{3}}{4} \cdot 4 \cdot 6 - (3\pi - S);$
$[2]$. $\textbf{Area}$ of dashed circle $-$ $\textbf{Area}$ of six $\textbf{new}$ "leaves" = $\pi - S;$
Then we have $\frac{\sqrt{3}}{4} \cdot 4 \cdot 6 - (3\pi - S) = \pi - S$
$S = 2\pi - 3\sqrt{3}$
Plugging $S$ back in to either $[1]$ or $[2]$, we can solve for the shaded area $= 3\sqrt{3} - \pi$ $\rightarrow\boxed{\mathrm{(D)}}$
--RyanZ@BRS
--Extremelysupercooldude (Minor latex and grammar edits)
~Yelechi (LaTeX edits) | 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)--(2,2sqrt(3))--cycle,white); fill(arc((0.5,3sqrt(3)/2),1,240,420)--(0.5,3sqrt(3)/2)--cycle,white); fill(arc((0.5,sqrt(3)/2),1,300,480)--(0.5,sqrt(3)/2)--cycle,white); draw((1,0)--(3,0)--(4,sqrt(3))--(3,2sqrt(3))--(1,2sqrt(3))--(0,sqrt(3))--(1,0)); draw(arc((2,0),1,180,0)--(2,0)--cycle); draw(arc((3.5,sqrt(3)/2),1,60,240)--(3.5,sqrt(3)/2)--cycle); draw(arc((3.5,3sqrt(3)/2),1,120,300)--(3.5,3sqrt(3)/2)--cycle); draw(arc((2,2sqrt(3)),1,180,360)--(2,2sqrt(3))--cycle); draw(arc((0.5,3sqrt(3)/2),1,240,420)--(0.5,3sqrt(3)/2)--cycle); draw(arc((0.5,sqrt(3)/2),1,300,480)--(0.5,sqrt(3)/2)--cycle); label("$2$",(3.5,3sqrt(3)/2),NE); draw(Arc((2,sqrt(3)), 1, 0, 360),dashed+linewidth(.5)); | [] |
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?
$\textbf{(A) } 6\sqrt3 - 3\pi \qquad \textbf{(B) } \frac{9\sqrt3}{2} - 2\pi \qquad \textbf{(C) } \frac{3\sqrt3}{2} - \frac{\pi}{3} \qquad \textbf{(D) } 3\sqrt3 - \pi \qquad \textbf{(E) } \frac{9\sqrt3}{2} - \pi$ | 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 is half of $y$, their total area is $\frac{6y}{2} = 3y$. The smaller hexagon minus this region would be $x$. So,
\[x= \frac{3\sqrt{3}}{2} \cdot 1^{2} - 3y\]
Also, the area of the larger hexagon minus the semicircles then add back $6y$ would also be $x$. (Note that by just subtracting the semicircles from the larger hexagon would result in an overlap of $6y$.) So,
\[x= \frac{3\sqrt{3}}{2} \cdot 2^{2} - 3\pi + 6y\]
Equating the equations, we have
\begin{align*} \frac{3\sqrt{3}}{2} \cdot 1^{2} - 3y &= \frac{3\sqrt{3}}{2} \cdot 2^{2} - 3\pi + 6y \\ \frac{3\sqrt{3}}{2} - 3y &= 6\sqrt{3} - 3\pi + 6y \\ \end{align*}
Solving for $y$, we have
\begin{align*} 9y &= \frac{3\sqrt{3}}{2} - 6\sqrt{3} + 3\pi \\ &=-\frac{9\sqrt{3}}{2} + 3\pi \\ \end{align*}
Then, we can plug this back into
\[x= \frac{3\sqrt{3}}{2} \cdot 1^{2} - 3y\]
Since $9y = -\frac{9\sqrt{3}}{2} + 3\pi$, dividing by $3$ gives
\begin{align*} 3y &= -\frac{9\sqrt{3}}{2} \times \frac{1}{3} + \pi \\ &=-\frac{3\sqrt{3}}{2} + \pi \\ \end{align*}
Therefore, $x = \frac{3\sqrt{3}}{2} - (-\frac{3\sqrt{3}}{2} + \pi) \implies x= \boxed{\textbf{(D)} ~3\sqrt{3} - \pi}$.
~MrThinker | 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)--(2,2sqrt(3))--cycle,white); fill(arc((0.5,3sqrt(3)/2),1,240,420)--(0.5,3sqrt(3)/2)--cycle,white); fill(arc((0.5,sqrt(3)/2),1,300,480)--(0.5,sqrt(3)/2)--cycle,white); draw((1,0)--(3,0)--(4,sqrt(3))--(3,2sqrt(3))--(1,2sqrt(3))--(0,sqrt(3))--(1,0)); label("$2$",(3.5,3sqrt(3)/2),NE); label("$x$",(2,1.6),N); label("$y$",(1.3,3.2),S); draw(arc((2,0),1,180,0)--(2,0)--cycle); draw(arc((3.5,sqrt(3)/2),1,60,240)--(3.5,sqrt(3)/2)--cycle); draw(arc((3.5,3sqrt(3)/2),1,120,300)--(3.5,3sqrt(3)/2)--cycle); draw(arc((2,2sqrt(3)),1,180,360)--(2,2sqrt(3))--cycle); draw(arc((0.5,3sqrt(3)/2),1,240,420)--(0.5,3sqrt(3)/2)--cycle); draw(arc((0.5,sqrt(3)/2),1,300,480)--(0.5,sqrt(3)/2)--cycle); draw((2.5,3sqrt(3)/2)--(1.5,3sqrt(3)/2)); draw((2.5,0.87)--(3,2sqrt(3)/2)); draw((3,2sqrt(3)/2)--(2.5,3sqrt(3)/2)); draw((1,1.73)--(1.5,2.6)); draw((1.5,0.87)--(1,1.73)); draw((2.5,sqrt(3)/2)--(1.5,sqrt(3)/2)); label("$1$", (1.2,2.2),W); | [] |
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?
$\textbf{(A) } 6\sqrt3 - 3\pi \qquad \textbf{(B) } \frac{9\sqrt3}{2} - 2\pi \qquad \textbf{(C) } \frac{3\sqrt3}{2} - \frac{\pi}{3} \qquad \textbf{(D) } 3\sqrt3 - \pi \qquad \textbf{(E) } \frac{9\sqrt3}{2} - \pi$ | 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 by the combined area of six equilateral triangles with side length $1$, or $\frac{\sqrt{3}}{4}\cdot 6 = \frac{3\sqrt{3}}{2}.$
Now, we have to subtract off the six mini-sectors between the triangle and the shaded area. The area of one of these mini-sectors is the difference between a $60^{\circ}$ sector of the semicircle and the equilateral we just drew. The area of a sector is just $\frac{\pi}{6}$, and the area of a triangle is $\frac{\sqrt{3}}{4}.$ Therefore our combined area of the mini-sectors is
\[6\left(\frac{\pi}{6} - \frac{\sqrt{3}}{4}\right) = \pi-\frac{3\sqrt{3}}{2}.\]
Subtracting this from the area of the (smaller) hexagon, the desired area is
\[\frac{3\sqrt{3}}{2} -\left(\pi - \frac{3\sqrt{3}}{2}\right) = \boxed{\textbf{(D)} ~3\sqrt{3} - \pi}\]
-Benedict T (countmath1) | 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)--(2,2sqrt(3))--cycle,white); fill(arc((0.5,3sqrt(3)/2),1,240,420)--(0.5,3sqrt(3)/2)--cycle,white); fill(arc((0.5,sqrt(3)/2),1,300,480)--(0.5,sqrt(3)/2)--cycle,white); draw((1,0)--(3,0)--(4,sqrt(3))--(3,2sqrt(3))--(1,2sqrt(3))--(0,sqrt(3))--(1,0)); label("$2$",(3.5,3sqrt(3)/2),NE); draw(arc((2,0),1,180,0)--(2,0)--cycle); draw(arc((3.5,sqrt(3)/2),1,60,240)--(3.5,sqrt(3)/2)--cycle); draw(arc((3.5,3sqrt(3)/2),1,120,300)--(3.5,3sqrt(3)/2)--cycle); draw(arc((2,2sqrt(3)),1,180,360)--(2,2sqrt(3))--cycle); draw(arc((0.5,3sqrt(3)/2),1,240,420)--(0.5,3sqrt(3)/2)--cycle); draw(arc((0.5,sqrt(3)/2),1,300,480)--(0.5,sqrt(3)/2)--cycle); draw((2.5,3sqrt(3)/2)--(1.5,3sqrt(3)/2)); draw((2.5,0.87)--(3,2sqrt(3)/2)); draw((3,2sqrt(3)/2)--(2.5,3sqrt(3)/2)); draw((1,1.73)--(1.5,2.6)); draw((1.5,0.87)--(1,1.73)); draw((2.5,sqrt(3)/2)--(1.5,sqrt(3)/2)); label("$1$", (2, 0.9), S); draw((1.43, 1)--(2, 0)--(2.57, 1)); | [] |
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 $CE^2+DE^2?$
$\textbf{(A)}\ 96 \qquad\textbf{(B)}\ 98 \qquad\textbf{(C)}\ 44\sqrt5 \qquad\textbf{(D)}\ 70\sqrt2 \qquad\textbf{(E)}\ 100$ | 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 $\angle AEC=\frac{\overarc{AC}+\overarc{BD}}{2}=45$, so $\overarc{AC}+\overarc{BD}=90$. Since $\overarc{BD'}=\overarc{BD}$, and $\overarc{AC}+\overarc{BD'}+\overarc{CD'}=180$, $\overarc{CD'}=90$. Thus, $\angle COD'=90$.
Since $OC=OD'=5\sqrt{2}$, by the Pythagorean Theorem, the desired $CD'^2= \boxed{\textbf{(E)}\ 100}$.
~sofas103 | // 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);
dot("$O$",O,1.5*S,linewidth(4));
dot("$A$",A,1.5*dir(A),linewidth(4));
dot("$B$",B,1.5*dir(B),linewidth(4));
dot("$E$",E,1.5*dir(180+135/2),linewidth(4));
dot("$C$",C,1.5*dir(C),linewidth(4));
dot("$D$",D,1.5*dir(D),linewidth(4));
dot("$D'$",D1,1.5*dir(D1),linewidth(4));
draw(A--B^^C--D^^C--D1--O--cycle^^D1--E);
// Block 2
/* 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); dot("$O$",O,1.5*S,linewidth(4)); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*dir(B),linewidth(4)); dot("$E$",E,1.5*dir(180+135/2),linewidth(4)); dot("$C$",C,1.5*dir(C),linewidth(4)); dot("$D$",D,1.5*dir(D),linewidth(4)); dot("$D'$",D1,1.5*dir(D1),linewidth(4)); draw(A--B^^C--D^^C--D1--O--cycle^^D1--E); | [] |
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 $CE^2+DE^2?$
$\textbf{(A)}\ 96 \qquad\textbf{(B)}\ 98 \qquad\textbf{(C)}\ 44\sqrt5 \qquad\textbf{(D)}\ 70\sqrt2 \qquad\textbf{(E)}\ 100$ | 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 $BE = BP = 2\sqrt{5}$ and $PE=2\sqrt{10}$. Hence, $PD = 2\sqrt{10}-x.$
By Power of a Point, we have
\[PB^2 = PD\cdot PC\]
\[20 = \left(2\sqrt{10}-x\right)\left(y + 2\sqrt{10}\right).\]
Isolating the variables after expanding gives $x-y = 2\sqrt{10}-10.$
Using Power of a Point again, we have
\[DE\cdot EC = BE\cdot EA\]
\[xy = 2\sqrt{5}\left(10\sqrt{2}-2\sqrt{5} \right)\]
\[xy = 20\sqrt{10}-20.\]
To get $x^2 + y^2$, we can perform the operation $(x-y)^2 + 2xy$. Plugging these values in,
\[(x-y)^2 + 2xy = \left(2\sqrt{10}-10\right)^2 + 2\left(20\sqrt{10}-20\right)\]
\[= 40 + 100 - 40\sqrt{10} + 40\sqrt{10} - 40 = \boxed{\textbf{(E)}\ 100}\]
-Benedict T (countmath1) | // 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 = intersectionpoint(p,E--E+10*dir(-45));
draw(p);
dot("$A$",A,1.5*dir(A),linewidth(4));
dot("$B$",B,1.5*dir(B),linewidth(4));
dot("$E$",E,1.5*dir(180+135/2),linewidth(4));
dot("$C$",C,1.5*dir(C),linewidth(4));
dot("$D$",D,1.5*dir(D),linewidth(4));
dot("$P$",P,1.5*dir(P),linewidth(4));
label("$45^{\circ}$",Q,NW);
draw(A--B^^C--P--B);
draw((5*sqrt(2), 0)--(5*sqrt(2)-0.4, 0)--(5*sqrt(2)-0.4, -0.4)--(5*sqrt(2), -0.4));
// Block 2
/* 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 = intersectionpoint(p,E--E+10*dir(-45)); draw(p); dot("$A$",A,1.5*dir(A),linewidth(4)); dot("$B$",B,1.5*dir(B),linewidth(4)); dot("$E$",E,1.5*dir(180+135/2),linewidth(4)); dot("$C$",C,1.5*dir(C),linewidth(4)); dot("$D$",D,1.5*dir(D),linewidth(4)); dot("$P$",P,1.5*dir(P),linewidth(4)); label("$45^{\circ}$",Q,NW); draw(A--B^^C--P--B); draw((5*sqrt(2), 0)--(5*sqrt(2)-0.4, 0)--(5*sqrt(2)-0.4, -0.4)--(5*sqrt(2), -0.4)); | [] |
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 $i=\sqrt{-1}$)
$\textbf{(A) } 0 \qquad \textbf{(B) }1 \qquad \textbf{(C) } 2 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } 4$ | 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 $6$ roots from $2$ groups, and we cannot have $0$ roots, so the $3$ distinct roots, when drawn (with $2$ with multiplicities not shown), must look like the figure above.
As for the magnitudes of the roots, as all roots pictured above have the same magnitude, there is $1$ possible magnitude for each root, which is $\sqrt[5]{2020}$, meaning we don't have to care about the scale of the diagram.
To satisfy the complex conjugate root theorem (that if the coefficients are real, any root of a+bi must have a root of a-bi), the figure depicting all $5$ roots must be symmetrical across the horizontal axis. Without consideration for roots with multiplicities, there are $3$ axes of symmetry for this shape, and two unique ways to align one of the axes of symmetry such that it's parallel to the horizontal axis, the other such case created by reflecting across the vertical axis.
However, not only must the magnitudes multiply to $2020$, but the product of the roots must be *positive* $2020$. Only one of the above cases can satisfy this condition (If there exists a configuration that multiplies to $-2020$, then reflecting across the vertical axis flips the sign creating the $+2020$ case, with the converse also being true. As each case guarantees the existence of the other, and while at least one case is guaranteed to exist in the first place, we know there is exactly $1$ of each case that remain, with $1$ prevailing $+2020$ case to use.
Now, when taking roots with multiplicities back into consideration, the singular case has $2$ ways to assign multiplicities while preserving symmetry: either both collating to the horizontal root, or through one being distributed to each of the angled roots (This corresponds to the $(2,2,1)$ and $(3,1,1)$ cases discussed earlier).
Therefore, the final answer is $\boxed{\mathbf{(C)} 2}$. | // 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 $\overline{GJ} \perp \overline{EH}$. See the figure below. Triangle $AEH$, quadrilateral $BFIE$, quadrilateral $DHJG$, and pentagon $FCGJI$ each has area $1.$ What is $FI^2$?
$\textbf{(A) } \frac{7}{3} \qquad \textbf{(B) } 8-4\sqrt2 \qquad \textbf{(C) } 1+\sqrt2 \qquad \textbf{(D) } \frac{7}{4}\sqrt2 \qquad \textbf{(E) } 2\sqrt2$ | 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 $AH$ = $AE$ = $\sqrt{2}$, $EB$ = $2-\sqrt{2}$, $EK$ = $2\sqrt{2}-2$.
$FI^2=\text{area of} \: BFIE+\text{area of} \:FJ'H'B+\text{area of} \:EH'K \\= 1 + 1 + \frac{1}{2}(2\sqrt{2}-2)^2=8-4\sqrt{2}\rightarrow \boxed{\textbf{(B)}\ 8-4\sqrt{2}}$
--Ryan Zhang @BRS
--Minor edit by WhySean38 | 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); draw(A--B--C--D--A); draw(H--E); draw(J--G^^F--I); draw(F--M); draw(M--L); draw(E--K,dashed+linewidth(.5)); draw(K--L,dashed+linewidth(.5)); draw(B--L); draw(rightanglemark(G, J, I), linewidth(.5)); draw(rightanglemark(F, I, E), linewidth(.5)); draw(rightanglemark(F, M, L), linewidth(.5)); fill((4,0)--(k,0)--M--(4,y)--cycle, gray); dot("$A$", A, S); dot("$C$", C, dir(90)); dot("$D$", D, dir(90)); dot("$E$", E, S); dot("$G$", G, N); dot("$H$", H, W); dot("$I$", I, SW); dot("$J$", J, SW); dot("$K$", K, S); dot("$F(G)$", F, E); dot("$H'$", L, S); dot("$B(D)$", B, S); | [] |
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 $\overline{GJ} \perp \overline{EH}$. See the figure below. Triangle $AEH$, quadrilateral $BFIE$, quadrilateral $DHJG$, and pentagon $FCGJI$ each has area $1.$ What is $FI^2$?
$\textbf{(A) } \frac{7}{3} \qquad \textbf{(B) } 8-4\sqrt2 \qquad \textbf{(C) } 1+\sqrt2 \qquad \textbf{(D) } \frac{7}{4}\sqrt2 \qquad \textbf{(E) } 2\sqrt2$ | 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 $FI = GJ = x$, $KL = LG = GJ - LJ = x - 2 \sqrt{2} + 2$
$[DHK] = \frac{(2 - \sqrt{2})^2}{2} = 3 - 2 \sqrt {2}$
$[GKL] = \frac{(x - 2 \sqrt{2} + 2)^2}{2} = \frac{x^2}{2} + 2x - 2x \sqrt{2} - 4 \sqrt{2} + 6$
$[HKJL] = (x - 2 \sqrt{2} + 2) \cdot (2 \sqrt{2} - 2) = 2x \sqrt{2} - 2x + 8 \sqrt{2} -12$
\[[DHK] + [GKL] + [HKLJ] = [DHJG]\]
\[3 - 2 \sqrt {2} + \frac{x^2}{2} + 2x - 2x \sqrt{2} - 4 \sqrt{2} + 6 + 2x \sqrt{2} - 2x + 8 \sqrt{2} -12= 1\]
\[\frac{x^2}{2} + 2 \sqrt{2} - 4 = 0\]
\[x^2 = 8 - 4 \sqrt{2}\]
\[FI^2 = \boxed{\textbf{(B)}\ 8-4\sqrt{2}}\]
~isabelchen | 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--E); draw(J--G^^F--I); draw(H--K,dashed+linewidth(.5)); draw(L--K,dashed+linewidth(.5)); draw(rightanglemark(G, J, I), linewidth(.5)); draw(rightanglemark(F, I, E), linewidth(.5)); draw(rightanglemark(H, K, L), linewidth(.5)); draw(rightanglemark(K, L, G), linewidth(.5)); dot("$A$", A, S); dot("$B$", B, S); dot("$C$", C, dir(90)); dot("$D$", D, dir(90)); dot("$E$", E, S); dot("$F$", F, dir(0)); dot("$G$", G, N); dot("$H$", H, W); dot("$I$", I, SW); dot("$J$", J, SW); dot("$K$", K, N); dot("$L$", L, S); | [] |
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 origin;
$\quad\bullet\qquad$ $H,$ a reflection across the $x$-axis; and
$\quad\bullet\qquad$ $V,$ a reflection across the $y$-axis.
Each of these transformations maps the squares onto itself, but the positions of the labeled vertices will change. For example, applying $R$ and then $V$ would send the vertex $A$ at $(1,1)$ to $(-1,-1)$ and would send the vertex $B$ at $(-1,1)$ to itself. How many sequences of $20$ transformations chosen from $\{L, R, H, V\}$ will send all of the labeled vertices back to their original positions? (For example, $R, R, V, H$ is one sequence of $4$ transformations that will send the vertices back to their original positions.)
$\textbf{(A)}\ 2^{37} \qquad\textbf{(B)}\ 3\cdot 2^{36} \qquad\textbf{(C)}\ 2^{38} \qquad\textbf{(D)}\ 3\cdot 2^{37} \qquad\textbf{(E)}\ 2^{39}$ | 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 vertices.
After the $19$th transformation, vertex $A$ will be at either $(1,-1)$ or $(-1,1).$ All possible configurations of the labeled vertices are shown below:
Each sequence of $19$ transformations generates one valid sequence of $20$ transformations. Therefore, the answer is $4^{19}=\boxed{\textbf{(C)}\ 2^{38}}.$
~MRENTHUSIASM | /* 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("$B$",(13,1)); label("$A$",(12,1)); label("$D$",(12,0)); label("\textbf{Configuration}",(-5,0.5)); label("\textbf{The 20th}",(-5,-1.5)); label("\textbf{Transformation}",(-5,-2.25)); label("$L$",(0.5,-1.875)); label("$R$",(4.5,-1.875)); label("$H$",(8.5,-1.875)); label("$V$",(12.5,-1.875)); | [] |
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 counted as distinct). For example, the number $6$ can be written as $6$, $2\cdot 3$, and $3\cdot2$, so $D(6) = 3$. What is $D(96)$?
$\textbf{(A) } 112 \qquad\textbf{(B) } 128 \qquad\textbf{(C) } 144 \qquad\textbf{(D) } 172 \qquad\textbf{(E) } 184$ | 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 its left and below it. We label the points to obtain number of paths.
The result is $\boxed{\textbf{(A) }112}$
~reda_mandymath | // 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);
// Block 2
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); label("1", (0, 0), NW); label("1", (1, 0), NW); label("2", (2, 0), NW); label("4", (3, 0), NW); label("8", (4, 0), NW); label("16", (5, 0), NW); label("1", (0, 1), NW); label("3", (1, 1), NW); label("8", (2, 1), NW); label("20", (3, 1), NW); label("48", (4, 1), NW); label("112", (5, 1), NW); | [] |
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 $P(a)?$
$\textbf{(A)}\ \frac{7}{12} \qquad\textbf{(B)}\ 2 - \sqrt{2} \qquad\textbf{(C)}\ \frac{1+\sqrt{2}}{4} \qquad\textbf{(D)}\ \frac{\sqrt{5}-1}{2} \qquad\textbf{(E)}\ \frac{5}{8}$ | 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 case gives us \[y=\tfrac{1}{2}-x \quad \textrm{and} \quad y=x-\tfrac{1}{2}.\] Solving the second case gives us \[y=x+\tfrac{1}{2}\quad \textrm{and} \quad y=\tfrac{3}{2}-x.\] If we graph these equations in $[0,1]\times[0,1]$, we get a rhombus shape.
Testing points in each section tells us that the inside of the rhombus satisfies the inequality in the problem statement.
From the region graph, notice that in order to maximize $P(a)$, $a\geq\tfrac{1}{2}$. We can solve the rest with geometric probability.
Instead of maximizing $P(a)$, we minimize $Q(a)=1-P(a)$. $Q(a)$ consists of two squares (each broken into two triangles), one of area $\tfrac{1}{4}$ and another of area $(a-\tfrac 12)^2$. To calculate $Q(a)$, we divide this area by $a$, so \[Q(a) = \frac{1}{a}\left(\frac{1}{4}+(a-\tfrac 12)^2\right) = \frac{1}{a}\left(\frac{1}{2}+a^2-a\right)= a+\frac 1{2a}-1.\]
By AM-GM, $a+\frac{1}{2a}\geq 2\sqrt{\frac{a}{2a}}=\sqrt{2}$, which we can achieve by setting $a=\frac{\sqrt{2}}{2}$.
Therefore, the maximum value of $P(a)$ is $1-\min(Q(a))$, which is $1-(\sqrt{2}-1) =\boxed{\textbf{(B)}\ 2 - \sqrt{2}}$ | // 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("",extension(A,D,(0.5,0),(0,0.5)),2*left,p); dot(A2);
B1=MP("",extension(B,C,(0.5,0),(0,-0.5)),2*right,p); dot(B1);
B2=MP("",extension(C,D,(0.5,1),(0,0.5)),2*up,p); dot(B2);
real a=0.7;
draw(A1--B1--B2--A2--cycle, gray+0.6);
draw(a*right--a*right+up, royalblue);
draw(A1--B2, royalblue+dashed);
draw(A--B--C--D--A, black+1.2);
dot("$(a,0)$",(a,0),down); dot("$(a,1)$",(a,1),up);
// Block 2
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("",extension(A,D,(0.5,0),(0,0.5)),2*left,p); dot(A2); B1=MP("",extension(B,C,(0.5,0),(0,-0.5)),2*right,p); dot(B1); B2=MP("",extension(C,D,(0.5,1),(0,0.5)),2*up,p); dot(B2); real a=0.7; draw(A1--B1--B2--A2--cycle, gray+0.6); draw(a*right--a*right+up, royalblue); draw(A1--B2, royalblue+dashed); draw(A--B--C--D--A, black+1.2); dot("$(a,0)$",(a,0),down); dot("$(a,1)$",(a,1),up); | [] |
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?
$\textbf{(A)} ~100.5 \qquad\textbf{(B)} ~134 \qquad\textbf{(C)} ~142 \qquad\textbf{(D)} ~150.5 \qquad\textbf{(E)} ~167$ | 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 $200.$
Let $x$ be the side length such that both sides of the triangle have the same area. The desired answer is then around $x$ because about half of the numbers in the list fall on each side.
Solving for $x$ yields:
\begin{align*} \frac{x^2}{2} =& \:\frac{1}{2} \cdot \frac{200^2}{2} \\ x^2 =& \:\frac{1}{2}\cdot 200^2 \\ x =& \:\frac{200}{\sqrt{2}} = \: 100\sqrt{2} \approx 141. \end{align*}
We see that $\boxed{(C) \: 142}$ is the closest to $x$ by far, and thus, can be relatively certain this is the answer.
~thinker123 | 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 again, we have $x^2+y^2-6y=0$, i.e. $x^2+(y-3)^2=3^2$. This gives us a circle of radius $3$ centered at $(0,3)$.
Case 3: $|x-y|=x-y, |x+y|=-x-y$
Doing the same process as before, we have $x^2+y^2+6y=0$, i.e. $x^2+(y+3)^2=3^2$. This gives us a circle of radius $3$ centered at $(0,-3)$.
Case 4: $|x-y|=y-x, |x+y|=-x-y$
One last time: we have $x^2+y^2+6x=0$, i.e. $(x+3)^2+y^2=3^2$. This gives us a circle of radius $3$ centered at $(-3,0)$.
After combining all the cases and drawing them on the Cartesian Plane, this is what the diagram looks like:
Now, the area of the shaded region is just a square with side length $6$ with four semicircles of radius $3$.
The area is $6\cdot6+4\cdot \frac{9\pi}{2} = 36+18\pi$. The answer is $36+18$ which is $\boxed{\textbf{(E) }54}$
~Bryguy | // 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, grey);
// Block 2
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, grey); | [] |
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 symmetry, there are $16$ valid sequences that start with $5,4,35,$ or $34.$ So, the answer is $16+16=\boxed{\textbf{(D)} ~32}.$
~MRENTHUSIASM (inspired by Snowfan) | // 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),red,EndArrow);
draw((2.25,-3)--(3.75,-3.75),red,EndArrow);
draw((4.25,3)--(5.75,3),red,EndArrow);
draw((4.25,0.75)--(5.75,0.75),red,EndArrow);
draw((4.25,-0.75)--(5.75,-0.75),red,EndArrow);
draw((4.25,-2.25)--(5.75,-2.25),red,EndArrow);
draw((4.25,-3.75)--(5.75,-3.75),red,EndArrow);
draw((6.25,3)--(7.75,3),red,EndArrow);
draw((6.25,0.75)--(7.75,0.75),red,EndArrow);
draw((6.25,-0.75)--(7.75,-0.75),red,EndArrow);
draw((6.25,-2.25)--(7.75,-2.25),red,EndArrow);
draw((6.25,-3.75)--(7.75,-3.75),red,EndArrow);
label("$1$",(0,0));
label("$3$",(2,3));
label("$2$",(4,3));
label("$5$",(6,3));
label("$4$",(8,3));
label("$4$",(2,0));
label("$2$",(4,0.75));
label("$5$",(6,0.75));
label("$3$",(8,0.75));
label("$3$",(4,-0.75));
label("$5$",(6,-0.75));
label("$2$",(8,-0.75));
label("$5$",(2,-3));
label("$2$",(4,-2.25));
label("$4$",(6,-2.25));
label("$3$",(8,-2.25));
label("$3$",(4,-3.75));
label("$4$",(6,-3.75));
label("$2$",(8,-3.75));
draw((12.75,0)--(14.25,4.5),red,EndArrow);
draw((12.75,0)--(14.25,1.5),red,EndArrow);
draw((12.75,0)--(14.25,-1.5),red,EndArrow);
draw((12.75,0)--(14.25,-4.5),red,EndArrow);
draw((14.75,4.5)--(16.25,5.25),red,EndArrow);
draw((14.75,4.5)--(16.25,3.75),red,EndArrow);
draw((14.75,1.5)--(16.25,1.5),red,EndArrow);
draw((14.75,-1.5)--(16.25,-0.75),red,EndArrow);
draw((14.75,-1.5)--(16.25,-2.25),red,EndArrow);
draw((14.75,-4.5)--(16.25,-3.75),red,EndArrow);
draw((14.75,-4.5)--(16.25,-5.25),red,EndArrow);
draw((16.75,5.25)--(18.25,5.25),red,EndArrow);
draw((16.75,3.75)--(18.25,3.75),red,EndArrow);
draw((16.75,1.5)--(18.25,1.5),red,EndArrow);
draw((16.75,-0.75)--(18.25,-0.75),red,EndArrow);
draw((16.75,-2.25)--(18.25,-2.25),red,EndArrow);
draw((16.75,-3.75)--(18.25,-3.75),red,EndArrow);
draw((16.75,-5.25)--(18.25,-5.25),red,EndArrow);
draw((18.75,5.25)--(20.25,5.25),red,EndArrow);
draw((18.75,3.75)--(20.25,3.75),red,EndArrow);
draw((18.75,1.5)--(20.25,1.5),red,EndArrow);
draw((18.75,-0.75)--(20.25,-0.75),red,EndArrow);
draw((18.75,-2.25)--(20.25,-2.25),red,EndArrow);
draw((18.75,-3.75)--(20.25,-3.75),red,EndArrow);
draw((18.75,-5.25)--(20.25,-5.25),red,EndArrow);
label("$2$",(12.5,0));
label("$1$",(14.5,4.5));
label("$3$",(14.5,1.5));
label("$4$",(14.5,-1.5));
label("$5$",(14.5,-4.5));
label("$4$",(16.5,5.25));
label("$5$",(16.5,3.75));
label("$1$",(16.5,1.5));
label("$1$",(16.5,-0.75));
label("$3$",(16.5,-2.25));
label("$1$",(16.5,-3.75));
label("$3$",(16.5,-5.25));
label("$3$",(18.5,5.25));
label("$3$",(18.5,3.75));
label("$5$",(18.5,1.5));
label("$5$",(18.5,-0.75));
label("$5$",(18.5,-2.25));
label("$4$",(18.5,-3.75));
label("$4$",(18.5,-5.25));
label("$5$",(20.5,5.25));
label("$4$",(20.5,3.75));
label("$4$",(20.5,1.5));
label("$3$",(20.5,-0.75));
label("$1$",(20.5,-2.25));
label("$3$",(20.5,-3.75));
label("$1$",(20.5,-5.25));
draw((25.25,0)--(26.75,1.5),red,EndArrow);
draw((25.25,0)--(26.75,-1.5),red,EndArrow);
draw((27.25,1.5)--(28.75,2.25),red,EndArrow);
draw((27.25,1.5)--(28.75,0.75),red,EndArrow);
draw((27.25,-1.5)--(28.75,-0.75),red,EndArrow);
draw((27.25,-1.5)--(28.75,-2.25),red,EndArrow);
draw((29.25,2.25)--(30.75,2.25),red,EndArrow);
draw((29.25,0.75)--(30.75,0.75),red,EndArrow);
draw((29.25,-0.75)--(30.75,-0.75),red,EndArrow);
draw((29.25,-2.25)--(30.75,-2.25),red,EndArrow);
draw((31.25,2.25)--(32.75,2.25),red,EndArrow);
draw((31.25,0.75)--(32.75,0.75),red,EndArrow);
draw((31.25,-0.75)--(32.75,-0.75),red,EndArrow);
draw((31.25,-2.25)--(32.75,-2.25),red,EndArrow);
label("$3$",(25,0));
label("$1$",(27,1.5));
label("$2$",(27,-1.5));
label("$4$",(29,2.25));
label("$5$",(29,0.75));
label("$4$",(29,-0.75));
label("$5$",(29,-2.25));
label("$2$",(31,2.25));
label("$2$",(31,0.75));
label("$1$",(31,-0.75));
label("$1$",(31,-2.25));
label("$5$",(33,2.25));
label("$4$",(33,0.75));
label("$5$",(33,-0.75));
label("$4$",(33,-2.25));
// Block 2
/* 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),red,EndArrow); draw((2.25,-3)--(3.75,-3.75),red,EndArrow); draw((4.25,3)--(5.75,3),red,EndArrow); draw((4.25,0.75)--(5.75,0.75),red,EndArrow); draw((4.25,-0.75)--(5.75,-0.75),red,EndArrow); draw((4.25,-2.25)--(5.75,-2.25),red,EndArrow); draw((4.25,-3.75)--(5.75,-3.75),red,EndArrow); draw((6.25,3)--(7.75,3),red,EndArrow); draw((6.25,0.75)--(7.75,0.75),red,EndArrow); draw((6.25,-0.75)--(7.75,-0.75),red,EndArrow); draw((6.25,-2.25)--(7.75,-2.25),red,EndArrow); draw((6.25,-3.75)--(7.75,-3.75),red,EndArrow); label("$1$",(0,0)); label("$3$",(2,3)); label("$2$",(4,3)); label("$5$",(6,3)); label("$4$",(8,3)); label("$4$",(2,0)); label("$2$",(4,0.75)); label("$5$",(6,0.75)); label("$3$",(8,0.75)); label("$3$",(4,-0.75)); label("$5$",(6,-0.75)); label("$2$",(8,-0.75)); label("$5$",(2,-3)); label("$2$",(4,-2.25)); label("$4$",(6,-2.25)); label("$3$",(8,-2.25)); label("$3$",(4,-3.75)); label("$4$",(6,-3.75)); label("$2$",(8,-3.75)); draw((12.75,0)--(14.25,4.5),red,EndArrow); draw((12.75,0)--(14.25,1.5),red,EndArrow); draw((12.75,0)--(14.25,-1.5),red,EndArrow); draw((12.75,0)--(14.25,-4.5),red,EndArrow); draw((14.75,4.5)--(16.25,5.25),red,EndArrow); draw((14.75,4.5)--(16.25,3.75),red,EndArrow); draw((14.75,1.5)--(16.25,1.5),red,EndArrow); draw((14.75,-1.5)--(16.25,-0.75),red,EndArrow); draw((14.75,-1.5)--(16.25,-2.25),red,EndArrow); draw((14.75,-4.5)--(16.25,-3.75),red,EndArrow); draw((14.75,-4.5)--(16.25,-5.25),red,EndArrow); draw((16.75,5.25)--(18.25,5.25),red,EndArrow); draw((16.75,3.75)--(18.25,3.75),red,EndArrow); draw((16.75,1.5)--(18.25,1.5),red,EndArrow); draw((16.75,-0.75)--(18.25,-0.75),red,EndArrow); draw((16.75,-2.25)--(18.25,-2.25),red,EndArrow); draw((16.75,-3.75)--(18.25,-3.75),red,EndArrow); draw((16.75,-5.25)--(18.25,-5.25),red,EndArrow); draw((18.75,5.25)--(20.25,5.25),red,EndArrow); draw((18.75,3.75)--(20.25,3.75),red,EndArrow); draw((18.75,1.5)--(20.25,1.5),red,EndArrow); draw((18.75,-0.75)--(20.25,-0.75),red,EndArrow); draw((18.75,-2.25)--(20.25,-2.25),red,EndArrow); draw((18.75,-3.75)--(20.25,-3.75),red,EndArrow); draw((18.75,-5.25)--(20.25,-5.25),red,EndArrow); label("$2$",(12.5,0)); label("$1$",(14.5,4.5)); label("$3$",(14.5,1.5)); label("$4$",(14.5,-1.5)); label("$5$",(14.5,-4.5)); label("$4$",(16.5,5.25)); label("$5$",(16.5,3.75)); label("$1$",(16.5,1.5)); label("$1$",(16.5,-0.75)); label("$3$",(16.5,-2.25)); label("$1$",(16.5,-3.75)); label("$3$",(16.5,-5.25)); label("$3$",(18.5,5.25)); label("$3$",(18.5,3.75)); label("$5$",(18.5,1.5)); label("$5$",(18.5,-0.75)); label("$5$",(18.5,-2.25)); label("$4$",(18.5,-3.75)); label("$4$",(18.5,-5.25)); label("$5$",(20.5,5.25)); label("$4$",(20.5,3.75)); label("$4$",(20.5,1.5)); label("$3$",(20.5,-0.75)); label("$1$",(20.5,-2.25)); label("$3$",(20.5,-3.75)); label("$1$",(20.5,-5.25)); draw((25.25,0)--(26.75,1.5),red,EndArrow); draw((25.25,0)--(26.75,-1.5),red,EndArrow); draw((27.25,1.5)--(28.75,2.25),red,EndArrow); draw((27.25,1.5)--(28.75,0.75),red,EndArrow); draw((27.25,-1.5)--(28.75,-0.75),red,EndArrow); draw((27.25,-1.5)--(28.75,-2.25),red,EndArrow); draw((29.25,2.25)--(30.75,2.25),red,EndArrow); draw((29.25,0.75)--(30.75,0.75),red,EndArrow); draw((29.25,-0.75)--(30.75,-0.75),red,EndArrow); draw((29.25,-2.25)--(30.75,-2.25),red,EndArrow); draw((31.25,2.25)--(32.75,2.25),red,EndArrow); draw((31.25,0.75)--(32.75,0.75),red,EndArrow); draw((31.25,-0.75)--(32.75,-0.75),red,EndArrow); draw((31.25,-2.25)--(32.75,-2.25),red,EndArrow); label("$3$",(25,0)); label("$1$",(27,1.5)); label("$2$",(27,-1.5)); label("$4$",(29,2.25)); label("$5$",(29,0.75)); label("$4$",(29,-0.75)); label("$5$",(29,-2.25)); label("$2$",(31,2.25)); label("$2$",(31,0.75)); label("$1$",(31,-0.75)); label("$1$",(31,-2.25)); label("$5$",(33,2.25)); label("$4$",(33,0.75)); label("$5$",(33,-0.75)); label("$4$",(33,-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. For example if Frieda begins in the center square and makes two hops "up", the first hop would place her in the top row middle square, and the second hop would cause Frieda to jump to the opposite edge, landing in the bottom row middle square. Suppose Frieda starts from the center square, makes at most four hops at random, and stops hopping if she lands on a corner square. What is the probability that she reaches a corner square on one of the four hops?
$\textbf{(A)} ~\frac{9}{16}\qquad\textbf{(B)} ~\frac{5}{8}\qquad\textbf{(C)} ~\frac{3}{4}\qquad\textbf{(D)} ~\frac{25}{32}\qquad\textbf{(E)} ~\frac{13}{16}$ | 2021 AMC 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). Since counting the complement set is easier, we'll count the number of $4$-step paths such that Frieda never reaches a corner point.
In other words, since the reachable corner points are $(\pm 1, \pm 1), (\pm 1, \pm 2), (\pm 2, \pm 1),$ and $(\pm 2, \pm 2),$ Frieda can only travel along the collection of points included in $S$, where $S$ is all points on $x=0$ and $y=0$ such that $|y|<4$ and $|x|<4$, respectively, plus all points on the big square with side length $6$ centered at $(0, 0).$ We then can proceed with casework:
Case $1$: Frieda never reaches $(0, \pm 3)$ nor $(\pm 3, 0).$
When Frieda only moves horizontally or vertically for her four moves, she can do so in $2^4 - 4 = 12$ ways for each case . Thus, $12 \cdot 2$ total paths for the subcase of staying in one direction. (For instance, all length $4$ combinations of $F$ and $B$ except $FFFF$, $BBBB$, $FFFB$, and $BBBF$ for the horizontal direction.)
There is another subcase where she changes directions during her path. There are four symmetric cases for this subcase depending on which of the four quadrants Frieda hugs. For the first quadrant, the possible paths are $FBUD$, $FBUU$, $UDFB$, and $UDFF.$ Thus, a total of $4 \cdot 4 = 16$ ways for this subcase.
Total for Case $1$: $24 + 16 = 40$
Case $2$: Frieda reaches $(0, \pm 3)$ or $(\pm 3, 0)$.
Once Frieda reaches one of the points listed above (by using three moves), she has four choices for her last move. Thus, a total of $4 \cdot 4 = 16$ paths for this case.
Our total number of paths never reaching coroners is thus $16+40=56,$ making for an answer of \[\frac{4^4-56}{4^4} = \boxed{\textbf{(D)} ~\frac{25}{32}}.\]
-fidgetboss_4000 | dot((0,0)); draw((-1,-1)--(-1,1)--(1,1)--(1,-1)--cycle); draw((-4,-1)--(-4,1)--(-2,1)--(-2,-1)--cycle); draw((2,-1)--(2,1)--(4,1)--(4,-1)--cycle); draw((-1,2)--(-1,4)--(1,4)--(1,2)--cycle); draw((-1,-4)--(-1,-2)--(1,-2)--(1,-4)--cycle); draw((-4,2)--(-2,2)--(-2,4)--(-4,4)--cycle); draw((4,-2)--(2,-2)--(2,-4)--(4,-4)--cycle); draw((4,2)--(2,2)--(2,4)--(4,4)--cycle); draw((-4,-2)--(-2,-2)--(-2,-4)--(-4,-4)--cycle); draw((-3,-3)--(3,-3)--(3,3)--(-3,3)--cycle); draw((-4,0)--(4,0)); draw((0,4)--(0,-4)); | [] |
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. For example if Frieda begins in the center square and makes two hops "up", the first hop would place her in the top row middle square, and the second hop would cause Frieda to jump to the opposite edge, landing in the bottom row middle square. Suppose Frieda starts from the center square, makes at most four hops at random, and stops hopping if she lands on a corner square. What is the probability that she reaches a corner square on one of the four hops?
$\textbf{(A)} ~\frac{9}{16}\qquad\textbf{(B)} ~\frac{5}{8}\qquad\textbf{(C)} ~\frac{3}{4}\qquad\textbf{(D)} ~\frac{25}{32}\qquad\textbf{(E)} ~\frac{13}{16}$ | 2021 AMC 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 two corners. WLOG we let the frog hop up:
Notice that in its second hop, there are two ways for it to move to a corner (left and right) out of four, and thus the probability of Frieda reaching a corner on the second hop is $\frac{1}{2}$.
There are two other cases for Frieda’s second hop: it could either wrap around to another edge square or move back to the middle. We analyze each of these cases individually, noting that each has probability $\frac{1}{4}$:
In this case, there are again $2$ choices to go to a corner on the third hop, go back to the previous square, or go to the middle. Here there is a $\frac{1}{2}$ probability that Frieda will reach a corner on the third hop. If Frieda hops back with a $\frac{1}{4}$ probability, she will again have a $\frac{1}{2}$ chance of reaching a corner on her fourth hop, but after four hops she stops hopping, so there are no more chances for her to reach a corner if she hops back. If she hops to the middle, then her fourth hop cannot reach a corner and that case yields $0$ probability of reaching a corner.
To sum up this case, the total probability of her reaching a corner is $\frac{1}{4}\left(\frac{1}{2}+\frac{1}{4}\cdot \frac{1}{2}\right)$.
Finally suppose Frieda goes back to her initial position after $2$ steps. Then her third step again doesn’t matter because whatever she does, she will be next to $2$ corners, so the probability of her reaching a corner on her fourth hop in this case is$\frac{1}{4}\cdot\frac{1}{2}$.
Summing up all the cases, we get
\[\frac{1}{2}+\frac{1}{4}\left(\frac{1}{2}+\frac{1}{4}\cdot\frac{1}{2}\right)+\frac{1}{4}\cdot\frac{1}{2} = \frac{1}{2}+\frac{1}{4}\left(\frac{1}{2}+\frac{1}{8}\right)+\frac{1}{8} = \frac{5}{4}\cdot\left(\frac{1}{2}+\frac{1}{8}\right) = \frac{5}{4} \cdot \frac{5}{8} = \boxed{\textbf{(D)} ~\frac{25}{32}}.\]
~KingRavi | // Block 1
label("$F$",(0,0)); draw((-1,-1)--(-1,1)--(1,1)--(1,-1)--cycle); draw((1,-1)--(1,1)--(3,1)--(3,-1)--cycle); draw((-1,1)--(-1,3)--(1,3)--(1,1)--cycle); draw((-3,-1)--(-3,1)--(-1,1)--(-1,-1)--cycle); draw((-1,-3)--(-1,-1)--(1,-1)--(1,-3)--cycle); draw((-3,-3)--(3,-3)--(3,3)--(-3,3)--cycle);
// Block 2
label("$F$",(0,2)); draw((-1,-1)--(-1,1)--(1,1)--(1,-1)--cycle); draw((1,-1)--(1,1)--(3,1)--(3,-1)--cycle); draw((-1,1)--(-1,3)--(1,3)--(1,1)--cycle); draw((-3,-1)--(-3,1)--(-1,1)--(-1,-1)--cycle); draw((-1,-3)--(-1,-1)--(1,-1)--(1,-3)--cycle); draw((-3,-3)--(3,-3)--(3,3)--(-3,3)--cycle);
// Block 3
label("$F$",(0,-2)); draw((-1,-1)--(-1,1)--(1,1)--(1,-1)--cycle); draw((1,-1)--(1,1)--(3,1)--(3,-1)--cycle); draw((-1,1)--(-1,3)--(1,3)--(1,1)--cycle); draw((-3,-1)--(-3,1)--(-1,1)--(-1,-1)--cycle); draw((-1,-3)--(-1,-1)--(1,-1)--(1,-3)--cycle); draw((-3,-3)--(3,-3)--(3,3)--(-3,3)--cycle); | [] |
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?
$\textbf{(A)} ~12\qquad\textbf{(B)} ~18\qquad\textbf{(C)} ~24\qquad\textbf{(D)} ~30\qquad\textbf{(E)} ~36$ | 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, Case (1) has $3\cdot6=18$ ways in total.
Case (2): The top-center and center-left chips have the same color.
There are three subcases for Case (2):
As there are $3!=6$ permutations of the three colors, each subcase has $6$ ways. So, Case (2) has $3\cdot6=18$ ways in total.
Answer
Together, the answer is $18+18=\boxed{\textbf{(E)} ~36}.$
~MRENTHUSIASM | // 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), linewidth(1.5));
draw((6,2)--(9,2), linewidth(1.5));
draw((7,0)--(7,3), linewidth(1.5));
draw((8,0)--(8,3), linewidth(1.5));
// Block 2
/* Made by MRENTHUSIASM */
unitsize(7mm);
fill((0,2)--(1,2)--(1,3)--(0,3)--cycle, red);
fill((1,2)--(2,2)--(2,3)--(1,3)--cycle, blue);
fill((0,1)--(1,1)--(1,2)--(0,2)--cycle, green);
fill((1,1)--(2,1)--(2,2)--(1,2)--cycle, red);
fill((2,2)--(3,2)--(3,3)--(2,3)--cycle, red);
fill((2,1)--(3,1)--(3,2)--(2,2)--cycle, green);
fill((2,0)--(3,0)--(3,1)--(2,1)--cycle, blue);
fill((0,0)--(1,0)--(1,1)--(0,1)--cycle, blue);
fill((1,0)--(2,0)--(2,1)--(1,1)--cycle, green);
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);
fill((8,2)--(9,2)--(9,3)--(8,3)--cycle, green);
fill((8,1)--(9,1)--(9,2)--(8,2)--cycle, blue);
fill((8,0)--(9,0)--(9,1)--(8,1)--cycle, green);
fill((6,0)--(7,0)--(7,1)--(6,1)--cycle, red);
fill((7,0)--(8,0)--(8,1)--(7,1)--cycle, blue);
fill((12,2)--(13,2)--(13,3)--(12,3)--cycle, red);
fill((13,2)--(14,2)--(14,3)--(13,3)--cycle, blue);
fill((12,1)--(13,1)--(13,2)--(12,2)--cycle, green);
fill((13,1)--(14,1)--(14,2)--(13,2)--cycle, red);
fill((14,2)--(15,2)--(15,3)--(14,3)--cycle, green);
fill((14,1)--(15,1)--(15,2)--(14,2)--cycle, blue);
fill((14,0)--(15,0)--(15,1)--(14,1)--cycle, red);
fill((12,0)--(13,0)--(13,1)--(12,1)--cycle, blue);
fill((13,0)--(14,0)--(14,1)--(13,1)--cycle, green);
draw((0,0)--(3,0)--(3,3)--(0,3)--cycle, linewidth(1.5));
draw((0,1)--(3,1), linewidth(1.5));
draw((0,2)--(3,2), linewidth(1.5));
draw((1,0)--(1,3), linewidth(1.5));
draw((2,0)--(2,3), linewidth(1.5));
draw((6,0)--(9,0)--(9,3)--(6,3)--cycle, linewidth(1.5));
draw((6,1)--(9,1), linewidth(1.5));
draw((6,2)--(9,2), linewidth(1.5));
draw((7,0)--(7,3), linewidth(1.5));
draw((8,0)--(8,3), linewidth(1.5));
draw((12,0)--(15,0)--(15,3)--(12,3)--cycle, linewidth(1.5));
draw((12,1)--(15,1), linewidth(1.5));
draw((12,2)--(15,2), linewidth(1.5));
draw((13,0)--(13,3), linewidth(1.5));
draw((14,0)--(14,3), linewidth(1.5));
// Block 3
/* 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, blue);
draw((6,0)--(9,0)--(9,3)--(6,3)--cycle, linewidth(1.5));
draw((6,1)--(9,1), linewidth(1.5));
draw((6,2)--(9,2), linewidth(1.5));
draw((7,0)--(7,3), linewidth(1.5));
draw((8,0)--(8,3), linewidth(1.5));
// Block 4
/* Made by MRENTHUSIASM */
unitsize(7mm);
fill((0,2)--(1,2)--(1,3)--(0,3)--cycle, red);
fill((1,2)--(2,2)--(2,3)--(1,3)--cycle, blue);
fill((0,1)--(1,1)--(1,2)--(0,2)--cycle, blue);
fill((1,1)--(2,1)--(2,2)--(1,2)--cycle, green);
fill((2,2)--(3,2)--(3,3)--(2,3)--cycle, red);
fill((2,1)--(3,1)--(3,2)--(2,2)--cycle, blue);
fill((2,0)--(3,0)--(3,1)--(2,1)--cycle, green);
fill((0,0)--(1,0)--(1,1)--(0,1)--cycle, green);
fill((1,0)--(2,0)--(2,1)--(1,1)--cycle, red);
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, blue);
fill((7,1)--(8,1)--(8,2)--(7,2)--cycle, green);
fill((8,2)--(9,2)--(9,3)--(8,3)--cycle, green);
fill((8,1)--(9,1)--(9,2)--(8,2)--cycle, red);
fill((8,0)--(9,0)--(9,1)--(8,1)--cycle, green);
fill((6,0)--(7,0)--(7,1)--(6,1)--cycle, red);
fill((7,0)--(8,0)--(8,1)--(7,1)--cycle, blue);
fill((12,2)--(13,2)--(13,3)--(12,3)--cycle, red);
fill((13,2)--(14,2)--(14,3)--(13,3)--cycle, blue);
fill((12,1)--(13,1)--(13,2)--(12,2)--cycle, blue);
fill((13,1)--(14,1)--(14,2)--(13,2)--cycle, green);
fill((14,2)--(15,2)--(15,3)--(14,3)--cycle, green);
fill((14,1)--(15,1)--(15,2)--(14,2)--cycle, red);
fill((14,0)--(15,0)--(15,1)--(14,1)--cycle, blue);
fill((12,0)--(13,0)--(13,1)--(12,1)--cycle, green);
fill((13,0)--(14,0)--(14,1)--(13,1)--cycle, red);
draw((0,0)--(3,0)--(3,3)--(0,3)--cycle, linewidth(1.5));
draw((0,1)--(3,1), linewidth(1.5));
draw((0,2)--(3,2), linewidth(1.5));
draw((1,0)--(1,3), linewidth(1.5));
draw((2,0)--(2,3), linewidth(1.5));
draw((6,0)--(9,0)--(9,3)--(6,3)--cycle, linewidth(1.5));
draw((6,1)--(9,1), linewidth(1.5));
draw((6,2)--(9,2), linewidth(1.5));
draw((7,0)--(7,3), linewidth(1.5));
draw((8,0)--(8,3), linewidth(1.5));
draw((12,0)--(15,0)--(15,3)--(12,3)--cycle, linewidth(1.5));
draw((12,1)--(15,1), linewidth(1.5));
draw((12,2)--(15,2), linewidth(1.5));
draw((13,0)--(13,3), linewidth(1.5));
draw((14,0)--(14,3), linewidth(1.5));
// Block 5
/* 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), linewidth(1.5)); draw((6,2)--(9,2), linewidth(1.5)); draw((7,0)--(7,3), linewidth(1.5)); draw((8,0)--(8,3), linewidth(1.5));
// Block 6
/* Made by MRENTHUSIASM */ unitsize(7mm); fill((0,2)--(1,2)--(1,3)--(0,3)--cycle, red); fill((1,2)--(2,2)--(2,3)--(1,3)--cycle, blue); fill((0,1)--(1,1)--(1,2)--(0,2)--cycle, green); fill((1,1)--(2,1)--(2,2)--(1,2)--cycle, red); fill((2,2)--(3,2)--(3,3)--(2,3)--cycle, red); fill((2,1)--(3,1)--(3,2)--(2,2)--cycle, green); fill((2,0)--(3,0)--(3,1)--(2,1)--cycle, blue); fill((0,0)--(1,0)--(1,1)--(0,1)--cycle, blue); fill((1,0)--(2,0)--(2,1)--(1,1)--cycle, green); 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); fill((8,2)--(9,2)--(9,3)--(8,3)--cycle, green); fill((8,1)--(9,1)--(9,2)--(8,2)--cycle, blue); fill((8,0)--(9,0)--(9,1)--(8,1)--cycle, green); fill((6,0)--(7,0)--(7,1)--(6,1)--cycle, red); fill((7,0)--(8,0)--(8,1)--(7,1)--cycle, blue); fill((12,2)--(13,2)--(13,3)--(12,3)--cycle, red); fill((13,2)--(14,2)--(14,3)--(13,3)--cycle, blue); fill((12,1)--(13,1)--(13,2)--(12,2)--cycle, green); fill((13,1)--(14,1)--(14,2)--(13,2)--cycle, red); fill((14,2)--(15,2)--(15,3)--(14,3)--cycle, green); fill((14,1)--(15,1)--(15,2)--(14,2)--cycle, blue); fill((14,0)--(15,0)--(15,1)--(14,1)--cycle, red); fill((12,0)--(13,0)--(13,1)--(12,1)--cycle, blue); fill((13,0)--(14,0)--(14,1)--(13,1)--cycle, green); draw((0,0)--(3,0)--(3,3)--(0,3)--cycle, linewidth(1.5)); draw((0,1)--(3,1), linewidth(1.5)); draw((0,2)--(3,2), linewidth(1.5)); draw((1,0)--(1,3), linewidth(1.5)); draw((2,0)--(2,3), linewidth(1.5)); draw((6,0)--(9,0)--(9,3)--(6,3)--cycle, linewidth(1.5)); draw((6,1)--(9,1), linewidth(1.5)); draw((6,2)--(9,2), linewidth(1.5)); draw((7,0)--(7,3), linewidth(1.5)); draw((8,0)--(8,3), linewidth(1.5)); draw((12,0)--(15,0)--(15,3)--(12,3)--cycle, linewidth(1.5)); draw((12,1)--(15,1), linewidth(1.5)); draw((12,2)--(15,2), linewidth(1.5)); draw((13,0)--(13,3), linewidth(1.5)); draw((14,0)--(14,3), linewidth(1.5));
// Block 7
/* 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, blue); draw((6,0)--(9,0)--(9,3)--(6,3)--cycle, linewidth(1.5)); draw((6,1)--(9,1), linewidth(1.5)); draw((6,2)--(9,2), linewidth(1.5)); draw((7,0)--(7,3), linewidth(1.5)); draw((8,0)--(8,3), linewidth(1.5));
// Block 8
/* Made by MRENTHUSIASM */ unitsize(7mm); fill((0,2)--(1,2)--(1,3)--(0,3)--cycle, red); fill((1,2)--(2,2)--(2,3)--(1,3)--cycle, blue); fill((0,1)--(1,1)--(1,2)--(0,2)--cycle, blue); fill((1,1)--(2,1)--(2,2)--(1,2)--cycle, green); fill((2,2)--(3,2)--(3,3)--(2,3)--cycle, red); fill((2,1)--(3,1)--(3,2)--(2,2)--cycle, blue); fill((2,0)--(3,0)--(3,1)--(2,1)--cycle, green); fill((0,0)--(1,0)--(1,1)--(0,1)--cycle, green); fill((1,0)--(2,0)--(2,1)--(1,1)--cycle, red); 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, blue); fill((7,1)--(8,1)--(8,2)--(7,2)--cycle, green); fill((8,2)--(9,2)--(9,3)--(8,3)--cycle, green); fill((8,1)--(9,1)--(9,2)--(8,2)--cycle, red); fill((8,0)--(9,0)--(9,1)--(8,1)--cycle, green); fill((6,0)--(7,0)--(7,1)--(6,1)--cycle, red); fill((7,0)--(8,0)--(8,1)--(7,1)--cycle, blue); fill((12,2)--(13,2)--(13,3)--(12,3)--cycle, red); fill((13,2)--(14,2)--(14,3)--(13,3)--cycle, blue); fill((12,1)--(13,1)--(13,2)--(12,2)--cycle, blue); fill((13,1)--(14,1)--(14,2)--(13,2)--cycle, green); fill((14,2)--(15,2)--(15,3)--(14,3)--cycle, green); fill((14,1)--(15,1)--(15,2)--(14,2)--cycle, red); fill((14,0)--(15,0)--(15,1)--(14,1)--cycle, blue); fill((12,0)--(13,0)--(13,1)--(12,1)--cycle, green); fill((13,0)--(14,0)--(14,1)--(13,1)--cycle, red); draw((0,0)--(3,0)--(3,3)--(0,3)--cycle, linewidth(1.5)); draw((0,1)--(3,1), linewidth(1.5)); draw((0,2)--(3,2), linewidth(1.5)); draw((1,0)--(1,3), linewidth(1.5)); draw((2,0)--(2,3), linewidth(1.5)); draw((6,0)--(9,0)--(9,3)--(6,3)--cycle, linewidth(1.5)); draw((6,1)--(9,1), linewidth(1.5)); draw((6,2)--(9,2), linewidth(1.5)); draw((7,0)--(7,3), linewidth(1.5)); draw((8,0)--(8,3), linewidth(1.5)); draw((12,0)--(15,0)--(15,3)--(12,3)--cycle, linewidth(1.5)); draw((12,1)--(15,1), linewidth(1.5)); draw((12,2)--(15,2), linewidth(1.5)); draw((13,0)--(13,3), linewidth(1.5)); draw((14,0)--(14,3), linewidth(1.5)); | [] |
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?
$\textbf{(A)} ~12\qquad\textbf{(B)} ~18\qquad\textbf{(C)} ~24\qquad\textbf{(D)} ~30\qquad\textbf{(E)} ~36$ | 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)--cycle, red);
fill((12,2)--(13,2)--(13,3)--(12,3)--cycle, red);
fill((14,2)--(15,2)--(15,3)--(14,3)--cycle, red);
fill((13,0)--(14,0)--(14,1)--(13,1)--cycle, red);
fill((18,2)--(19,2)--(19,3)--(18,3)--cycle, red);
fill((20,1)--(21,1)--(21,2)--(20,2)--cycle, red);
fill((19,0)--(20,0)--(20,1)--(19,1)--cycle, red);
fill((24,1)--(25,1)--(25,2)--(24,2)--cycle, red);
fill((26,1)--(27,1)--(27,2)--(26,2)--cycle, red);
fill((25,2)--(26,2)--(26,3)--(25,3)--cycle, red);
draw((0,0)--(3,0)--(3,3)--(0,3)--cycle, linewidth(1.5));
draw((0,1)--(3,1), linewidth(1.5));
draw((0,2)--(3,2), linewidth(1.5));
draw((1,0)--(1,3), linewidth(1.5));
draw((2,0)--(2,3), linewidth(1.5));
draw((6,0)--(9,0)--(9,3)--(6,3)--cycle, linewidth(1.5));
draw((6,1)--(9,1), linewidth(1.5));
draw((6,2)--(9,2), linewidth(1.5));
draw((7,0)--(7,3), linewidth(1.5));
draw((8,0)--(8,3), linewidth(1.5));
draw((12,0)--(15,0)--(15,3)--(12,3)--cycle, linewidth(1.5));
draw((12,1)--(15,1), linewidth(1.5));
draw((12,2)--(15,2), linewidth(1.5));
draw((13,0)--(13,3), linewidth(1.5));
draw((14,0)--(14,3), linewidth(1.5));
draw((18,0)--(21,0)--(21,3)--(18,3)--cycle, linewidth(1.5));
draw((18,1)--(21,1), linewidth(1.5));
draw((18,2)--(21,2), linewidth(1.5));
draw((19,0)--(19,3), linewidth(1.5));
draw((20,0)--(20,3), linewidth(1.5));
draw((24,0)--(27,0)--(27,3)--(24,3)--cycle, linewidth(1.5));
draw((24,1)--(27,1), linewidth(1.5));
draw((24,2)--(27,2), linewidth(1.5));
draw((25,0)--(25,3), linewidth(1.5));
draw((26,0)--(26,3), linewidth(1.5));
label("Rotational",(1.5,4.5));
label("Symmetry",(1.5,3.75));
label("$2$ Configurations",(1.5,-0.75));
label("$4$ Configurations",(7.5,-0.75));
label("$4$ Configurations",(13.5,-0.75));
label("$4$ Configurations",(19.5,-0.75));
label("$4$ Configurations",(25.5,-0.75));
// Block 2
/* 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)--cycle, red); fill((12,2)--(13,2)--(13,3)--(12,3)--cycle, red); fill((14,2)--(15,2)--(15,3)--(14,3)--cycle, red); fill((13,0)--(14,0)--(14,1)--(13,1)--cycle, red); fill((18,2)--(19,2)--(19,3)--(18,3)--cycle, red); fill((20,1)--(21,1)--(21,2)--(20,2)--cycle, red); fill((19,0)--(20,0)--(20,1)--(19,1)--cycle, red); fill((24,1)--(25,1)--(25,2)--(24,2)--cycle, red); fill((26,1)--(27,1)--(27,2)--(26,2)--cycle, red); fill((25,2)--(26,2)--(26,3)--(25,3)--cycle, red); draw((0,0)--(3,0)--(3,3)--(0,3)--cycle, linewidth(1.5)); draw((0,1)--(3,1), linewidth(1.5)); draw((0,2)--(3,2), linewidth(1.5)); draw((1,0)--(1,3), linewidth(1.5)); draw((2,0)--(2,3), linewidth(1.5)); draw((6,0)--(9,0)--(9,3)--(6,3)--cycle, linewidth(1.5)); draw((6,1)--(9,1), linewidth(1.5)); draw((6,2)--(9,2), linewidth(1.5)); draw((7,0)--(7,3), linewidth(1.5)); draw((8,0)--(8,3), linewidth(1.5)); draw((12,0)--(15,0)--(15,3)--(12,3)--cycle, linewidth(1.5)); draw((12,1)--(15,1), linewidth(1.5)); draw((12,2)--(15,2), linewidth(1.5)); draw((13,0)--(13,3), linewidth(1.5)); draw((14,0)--(14,3), linewidth(1.5)); draw((18,0)--(21,0)--(21,3)--(18,3)--cycle, linewidth(1.5)); draw((18,1)--(21,1), linewidth(1.5)); draw((18,2)--(21,2), linewidth(1.5)); draw((19,0)--(19,3), linewidth(1.5)); draw((20,0)--(20,3), linewidth(1.5)); draw((24,0)--(27,0)--(27,3)--(24,3)--cycle, linewidth(1.5)); draw((24,1)--(27,1), linewidth(1.5)); draw((24,2)--(27,2), linewidth(1.5)); draw((25,0)--(25,3), linewidth(1.5)); draw((26,0)--(26,3), linewidth(1.5)); label("Rotational",(1.5,4.5)); label("Symmetry",(1.5,3.75)); label("$2$ Configurations",(1.5,-0.75)); label("$4$ Configurations",(7.5,-0.75)); label("$4$ Configurations",(13.5,-0.75)); label("$4$ Configurations",(19.5,-0.75)); label("$4$ Configurations",(25.5,-0.75)); | [] |
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?
$\textbf{(A)} ~12\qquad\textbf{(B)} ~18\qquad\textbf{(C)} ~24\qquad\textbf{(D)} ~30\qquad\textbf{(E)} ~36$ | 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 of one color and one token of another color for the bottom row, so there must only be $1$ way to fill it up since the two tokens of the same color can't be next to each other. This makes the grid uniquely determined.
The $3$ cells with the same color can either be cells $1,5,3$ or $2,4,6.$ If we choose $1,3,5,$ the two cells with the same color must be cells $4$ and $6$ and cell $2$ must be the cell of the other color. If we choose $2,4,6,$ the two cells with the same color must be cells $1$ and $3,$ and cell $5$ must be the cell of the other color. So there are $2$ ways to choose which cells share $3$ colors and $1$ way to choose which cells share $2$ and $1$ color. There are $3! = 6$ ways to assign colors to the cells. We have a total of $2 \cdot 1 \cdot 3! = 12$ colorings for this case. An example coloring is shown below with cells $1,3,5$ being chosen for the $3$ cells of the same color.
Now, what about the colorings where even if the top two rows are filled out, the grid is still not uniquely determined? In other words, what about the colorings where even if we fill in the top two rows, there is still more than one possible color combination for the last row?
In this case, the $3$ x $2$ grid made up of the top two rows would have to use two tokens of each color, and the second row would be some permutation of the colors blue, red, green. Then for the last row, we would be left with one token of each color.
There are $3! = 6$ ways to permute the second row. WLOG, say we fill the second row of the grid so that cell $4$ is red, cell $5$ is blue, and cell $6$ is green. Then cells $1,2,3$ can be green, red, blue or blue, green, red, respectively. Finally, cells $7,8,9$ can also either be green, red, blue or blue, green, red, respectively. This gives us $6 \cdot 4 = 24$ colorings.
Our final answer is $12 + 24 = \boxed{\textbf{(E)} ~36}$.
~grogg007 | // 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", (1.5,1.5));
label("6", (2.5,1.5));
label("1", (0.5,2.5));
label("2", (1.5,2.5));
label("3", (2.5,2.5));
draw((0,1)--(3,1)--(3,3)--(0,3)--cycle, linewidth(1.5));
// Block 2
size(70);
fill((0,2)--(1,2)--(1,3)--(0,3)--cycle, red); // Cell 1
fill((1,2)--(2,2)--(2,3)--(1,3)--cycle, green); // Cell 2
fill((2,2)--(3,2)--(3,3)--(2,3)--cycle, red); // Cell 3
fill((0,1)--(1,1)--(1,2)--(0,2)--cycle, blue); // Cell 4
fill((1,1)--(2,1)--(2,2)--(1,2)--cycle, red); // Cell 5
fill((2,1)--(3,1)--(3,2)--(2,2)--cycle, blue); // Cell 6
fill((0,0)--(1,0)--(1,1)--(0,1)--cycle, green); // Cell 7
fill((1,0)--(2,0)--(2,1)--(1,1)--cycle, blue); // Cell 8
fill((2,0)--(3,0)--(3,1)--(2,1)--cycle, green); // Cell 9
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("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", (1.5,1.5));
label("6", (2.5,1.5));
label("1", (0.5,2.5));
label("2", (1.5,2.5));
label("3", (2.5,2.5));
label("Only 1 configuration for last row, grid is uniquely determined from top 2 rows", (1.5,-1.2));
// Block 3
size(200);
void drawGrid(pair offset, pen[] colors) {
fill(shift(offset)*((0,2)--(1,2)--(1,3)--(0,3)--cycle), colors[0]);
fill(shift(offset)*((1,2)--(2,2)--(2,3)--(1,3)--cycle), colors[1]);
fill(shift(offset)*((2,2)--(3,2)--(3,3)--(2,3)--cycle), colors[2]);
fill(shift(offset)*((0,1)--(1,1)--(1,2)--(0,2)--cycle), colors[3]);
fill(shift(offset)*((1,1)--(2,1)--(2,2)--(1,2)--cycle), colors[4]);
fill(shift(offset)*((2,1)--(3,1)--(3,2)--(2,2)--cycle), colors[5]);
fill(shift(offset)*((0,0)--(1,0)--(1,1)--(0,1)--cycle), colors[6]);
fill(shift(offset)*((1,0)--(2,0)--(2,1)--(1,1)--cycle), colors[7]);
fill(shift(offset)*((2,0)--(3,0)--(3,1)--(2,1)--cycle), colors[8]);
for (int i = 0; i <= 3; ++i) {
draw(shift(offset)*((i,0)--(i,3)));
draw(shift(offset)*((0,i)--(3,i)));
}
label("1", offset + (0.5,2.5));
label("2", offset + (1.5,2.5));
label("3", offset + (2.5,2.5));
label("4", offset + (0.5,1.5));
label("5", offset + (1.5,1.5));
label("6", offset + (2.5,1.5));
label("7", offset + (0.5,0.5));
label("8", offset + (1.5,0.5));
label("9", offset + (2.5,0.5));
}
pen[] colors1 = {green, red, blue, red, blue, green, green, red, blue};
pen[] colors2 = {blue, green, red, red, blue, green, green, red, blue};
pen[] colors3 = {green, red, blue, red, blue, green, blue, green, red};
pen[] colors4 = {blue, green, red, red, blue, green, blue, green, red};
drawGrid((0,0), colors1);
drawGrid((4,0), colors2);
drawGrid((8,0), colors3);
drawGrid((12,0), colors4);
// Block 4
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", (1.5,1.5)); label("6", (2.5,1.5)); label("1", (0.5,2.5)); label("2", (1.5,2.5)); label("3", (2.5,2.5)); draw((0,1)--(3,1)--(3,3)--(0,3)--cycle, linewidth(1.5));
// Block 5
size(70); fill((0,2)--(1,2)--(1,3)--(0,3)--cycle, red); // Cell 1 fill((1,2)--(2,2)--(2,3)--(1,3)--cycle, green); // Cell 2 fill((2,2)--(3,2)--(3,3)--(2,3)--cycle, red); // Cell 3 fill((0,1)--(1,1)--(1,2)--(0,2)--cycle, blue); // Cell 4 fill((1,1)--(2,1)--(2,2)--(1,2)--cycle, red); // Cell 5 fill((2,1)--(3,1)--(3,2)--(2,2)--cycle, blue); // Cell 6 fill((0,0)--(1,0)--(1,1)--(0,1)--cycle, green); // Cell 7 fill((1,0)--(2,0)--(2,1)--(1,1)--cycle, blue); // Cell 8 fill((2,0)--(3,0)--(3,1)--(2,1)--cycle, green); // Cell 9 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("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", (1.5,1.5)); label("6", (2.5,1.5)); label("1", (0.5,2.5)); label("2", (1.5,2.5)); label("3", (2.5,2.5)); label("Only 1 configuration for last row, grid is uniquely determined from top 2 rows", (1.5,-1.2));
// Block 6
size(200); void drawGrid(pair offset, pen[] colors) { fill(shift(offset)*((0,2)--(1,2)--(1,3)--(0,3)--cycle), colors[0]); fill(shift(offset)*((1,2)--(2,2)--(2,3)--(1,3)--cycle), colors[1]); fill(shift(offset)*((2,2)--(3,2)--(3,3)--(2,3)--cycle), colors[2]); fill(shift(offset)*((0,1)--(1,1)--(1,2)--(0,2)--cycle), colors[3]); fill(shift(offset)*((1,1)--(2,1)--(2,2)--(1,2)--cycle), colors[4]); fill(shift(offset)*((2,1)--(3,1)--(3,2)--(2,2)--cycle), colors[5]); fill(shift(offset)*((0,0)--(1,0)--(1,1)--(0,1)--cycle), colors[6]); fill(shift(offset)*((1,0)--(2,0)--(2,1)--(1,1)--cycle), colors[7]); fill(shift(offset)*((2,0)--(3,0)--(3,1)--(2,1)--cycle), colors[8]); for (int i = 0; i <= 3; ++i) { draw(shift(offset)*((i,0)--(i,3))); draw(shift(offset)*((0,i)--(3,i))); } label("1", offset + (0.5,2.5)); label("2", offset + (1.5,2.5)); label("3", offset + (2.5,2.5)); label("4", offset + (0.5,1.5)); label("5", offset + (1.5,1.5)); label("6", offset + (2.5,1.5)); label("7", offset + (0.5,0.5)); label("8", offset + (1.5,0.5)); label("9", offset + (2.5,0.5)); } pen[] colors1 = {green, red, blue, red, blue, green, green, red, blue}; pen[] colors2 = {blue, green, red, red, blue, green, green, red, blue}; pen[] colors3 = {green, red, blue, red, blue, green, blue, green, red}; pen[] colors4 = {blue, green, red, red, blue, green, blue, green, red}; drawGrid((0,0), colors1); drawGrid((4,0), colors2); drawGrid((8,0), colors3); drawGrid((12,0), colors4); | [] |
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 always true. | 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+linewidth(1.5));
// Block 2
/* 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+linewidth(1.5)); | [] |
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?
$\textbf{(A) }2\sqrt{10} \qquad \textbf{(B) }5\sqrt2 \qquad \textbf{(C) }10\sqrt2 \qquad \textbf{(D) }15\sqrt2 \qquad \textbf{(E) }10\sqrt5$ | 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 reflections:
We reflect $\overline{BC}$ about the $y$-axis to get $\overline{BC'}.$
We reflect $\overline{CD}$ about the $x$-axis to get $\overline{CD'}$ with $D'=(7,-5),$ then reflect $\overline{CD'}$ about the $y$-axis to get $\overline{C'D''}$ with $D''=(-7,-5).$
We obtain the following diagram:
The total distance that the beam will travel is
\begin{align*} AB+BC+CD&=AB+BC+CD' \\ &=AB+BC'+C'D'' \\ &=AD'' \\ &=\sqrt{((3-(-7))^2+(5-(-5))^2} \\ &=\boxed{\textbf{(C) }10\sqrt2}. \end{align*}
~MRENTHUSIASM (Solution)
~JHawk0224 (Proposal)
~crystalkqw (Minor edits) | // 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);
pair C = (2,0);
pair D = (7,5);
pair E = (-2,0);
pair F = (7,-5);
pair G = (-7,-5);
draw(A--B--C--D,red);
draw(B--E,heavygreen+dashed);
draw(C--F,heavygreen+dashed);
draw(E--G,heavygreen+dashed);
dot("$A(3,5)$",A,(0,2),linewidth(3.5));
dot("$B$",B,(-2,0),linewidth(3.5));
dot("$C$",C,(0,-2),linewidth(3.5));
dot("$D(7,5)$",D,(0,2),linewidth(3.5));
dot("$C'$",E,(0,-2),linewidth(3.5));
dot("$D'(7,-5)$",F,(0,-2),linewidth(3.5));
dot("$D''(-7,-5)$",G,(0,-2),linewidth(3.5));
// Block 2
/* 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); pair C = (2,0); pair D = (7,5); pair E = (-2,0); pair F = (7,-5); pair G = (-7,-5); draw(A--B--C--D,red); draw(B--E,heavygreen+dashed); draw(C--F,heavygreen+dashed); draw(E--G,heavygreen+dashed); dot("$A(3,5)$",A,(0,2),linewidth(3.5)); dot("$B$",B,(-2,0),linewidth(3.5)); dot("$C$",C,(0,-2),linewidth(3.5)); dot("$D(7,5)$",D,(0,2),linewidth(3.5)); dot("$C'$",E,(0,-2),linewidth(3.5)); dot("$D'(7,-5)$",F,(0,-2),linewidth(3.5)); dot("$D''(-7,-5)$",G,(0,-2),linewidth(3.5)); | [] |
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?
$\textbf{(A) }2\sqrt{10} \qquad \textbf{(B) }5\sqrt2 \qquad \textbf{(C) }10\sqrt2 \qquad \textbf{(D) }15\sqrt2 \qquad \textbf{(E) }10\sqrt5$ | 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 coordinate axis, the angle of incidence and the angle of reflection are congruent, from which $\angle ABF=\angle CBF$ and $\angle BCF=\angle ECF.$ We conclude that $\triangle BCF\cong\triangle ECF$ by ASA, so $\angle CBF=\angle CEF.$ It follows that $\angle ABF=\angle CEF$ by transitive, so $\overline{AB}\parallel\overline{CD}$ by the Converse of the Alternate Interior Angles Theorem.
Note that $\overline{AD}\parallel\overline{BE}.$ Since the opposite sides are parallel, quadrilateral $ABED$ is a parallelogram. From $\triangle BCF\cong\triangle ECF,$ we get $BF=EF=2,$ so $C=(2,0).$
Let $B=(0,b).$ We equate the slopes of $\overline{AB}$ and $\overline{DC}:$ \[\frac{5-b}{3-0}=\frac{5-0}{7-2},\] from which $b=2,$ or $B=(0,2).$
By the Distance Formula, we have $AB=3\sqrt2,BC=2\sqrt2,$ and $CD=5\sqrt2.$ The total distance that the beam will travel is \[AB+BC+CD=\boxed{\textbf{(C) }10\sqrt2}.\]
Remark
When a straight line hits and bounces off a coordinate axis at point $P,$ the ray entering $P$ and the ray leaving $P$ always have negative slopes. In this problem, $\overline{AB}$ and $\overline{BC}$ have negative slopes; $\overline{BC}$ and $\overline{CD}$ have negative slopes. So, $\overline{AB}$ and $\overline{CD}$ have the same slope, or $\overline{AB}\parallel\overline{CD}.$
~MRENTHUSIASM | // 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);
pair C = (2,0);
pair D = (7,5);
pair E = (4,2);
pair F = (2,2);
draw(A--B--C--D,red);
draw(A--D^^B--E^^C--F,heavygreen+dashed);
dot("$A(3,5)$",A,(0,2),linewidth(3.5));
dot("$B$",B,(-2,0),linewidth(3.5));
dot("$C$",C,(0,-2),linewidth(3.5));
dot("$D(7,5)$",D,(0,2),linewidth(3.5));
dot("$E$",E,(2,0),linewidth(3.5));
dot("$F$",F,(0,1.5),linewidth(3.5));
// Block 2
/* 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); pair C = (2,0); pair D = (7,5); pair E = (4,2); pair F = (2,2); draw(A--B--C--D,red); draw(A--D^^B--E^^C--F,heavygreen+dashed); dot("$A(3,5)$",A,(0,2),linewidth(3.5)); dot("$B$",B,(-2,0),linewidth(3.5)); dot("$C$",C,(0,-2),linewidth(3.5)); dot("$D(7,5)$",D,(0,2),linewidth(3.5)); dot("$E$",E,(2,0),linewidth(3.5)); dot("$F$",F,(0,1.5),linewidth(3.5)); | [] |
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?
$\textbf{(A) }2\sqrt{10} \qquad \textbf{(B) }5\sqrt2 \qquad \textbf{(C) }10\sqrt2 \qquad \textbf{(D) }15\sqrt2 \qquad \textbf{(E) }10\sqrt5$ | 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, from which $B=(0,2)$ and $C=(2,0).$ The given condition $D=(7,5)$ verifies our guess, as shown below.
Following the last paragraph of Solution 2 gives the answer $\boxed{\textbf{(C) }10\sqrt2}.$
~MRENTHUSIASM | // 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 verticalLines()
{
for (int i = xMin+1; i < xMax; ++i)
{
draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4));
}
}
//Draws the horizontal ticks
void horizontalTicks()
{
for (int i = yMin+1; i < yMax; ++i)
{
draw((-3/16,i)--(3/16,i), black+linewidth(1));
}
}
//Draws the vertical ticks
void verticalTicks()
{
for (int i = xMin+1; i < xMax; ++i)
{
draw((i,-3/16)--(i,3/16), black+linewidth(1));
}
}
horizontalLines();
verticalLines();
horizontalTicks();
verticalTicks();
draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5));
draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5));
label("$x$",(xMax,0),(2,0));
label("$y$",(0,yMax),(0,2));
pair A = (3,5);
pair B = (0,2);
pair C = (2,0);
pair D = (7,5);
draw(A--B--C--D,red);
dot(A,linewidth(3.5));
dot(B,linewidth(3.5));
dot(C,linewidth(3.5));
dot(D,linewidth(3.5));
label("$A(3,5)$",A,(0,2),UnFill);
label("$B$",B,(-2,0),UnFill);
label("$C$",C,(0,-2),UnFill);
label("$D(7,5)$",D,(0,2),UnFill);
// Block 2
/* 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 verticalLines() { for (int i = xMin+1; i < xMax; ++i) { draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4)); } } //Draws the horizontal ticks void horizontalTicks() { for (int i = yMin+1; i < yMax; ++i) { draw((-3/16,i)--(3/16,i), black+linewidth(1)); } } //Draws the vertical ticks void verticalTicks() { for (int i = xMin+1; i < xMax; ++i) { draw((i,-3/16)--(i,3/16), black+linewidth(1)); } } horizontalLines(); verticalLines(); horizontalTicks(); verticalTicks(); draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("$x$",(xMax,0),(2,0)); label("$y$",(0,yMax),(0,2)); pair A = (3,5); pair B = (0,2); pair C = (2,0); pair D = (7,5); draw(A--B--C--D,red); dot(A,linewidth(3.5)); dot(B,linewidth(3.5)); dot(C,linewidth(3.5)); dot(D,linewidth(3.5)); label("$A(3,5)$",A,(0,2),UnFill); label("$B$",B,(-2,0),UnFill); label("$C$",C,(0,-2),UnFill); label("$D(7,5)$",D,(0,2),UnFill); | [] |
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?
$\textbf{(A)} ~100.5 \qquad\textbf{(B)} ~134 \qquad\textbf{(C)} ~142 \qquad\textbf{(D)} ~150.5 \qquad\textbf{(E)} ~167$ | 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 $200.$
Let $x$ be the side length such that both sides of the triangle have the same area. The desired answer is then around $x$ because about half of the numbers in the list fall on each side.
Solving for $x$ yields:
\begin{align*} \frac{x^2}{2} =& \:\frac{1}{2} \cdot \frac{200^2}{2} \\ x^2 =& \:\frac{1}{2}\cdot 200^2 \\ x =& \:\frac{200}{\sqrt{2}} = \: 100\sqrt{2} \approx 141. \end{align*}
We see that $\boxed{(C) \: 142}$ is the closest to $x$ by far, and thus, can be relatively certain this is the answer.
~thinker123 | // 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); label("$200$",(200,0)--(200,200),E); | [] |
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:
\[\begin{array}{c|ccc} & & & \\ [-2ex] & \boldsymbol{x=0} & \boldsymbol{x=\frac{\pi}{2}} & \boldsymbol{x=\pi} \\ [1.5ex] \hline & & & \\ [-1ex] \boldsymbol{\cos x} & 1 & 0 & -1 \\ [1.5ex] \boldsymbol{\frac{\pi}{2}\cos x} & \frac{\pi}{2} & 0 & -\frac{\pi}{2} \\ [1.5ex] \boldsymbol{\sin\left(\frac{\pi}{2}\cos x\right)} & 1 & 0 & -1 \\ [1.5ex] \hline & & & \\ [-1ex] \boldsymbol{\sin x} & 0 & 1 & 0 \\ [1.5ex] \boldsymbol{\frac{\pi}{2}\sin x} & 0 & \frac{\pi}{2} & 0 \\ [1.5ex] \boldsymbol{\cos\left(\frac{\pi}{2}\sin x\right)} & 1 & 0 & 1 \\ [1ex] \end{array}\]
For $x\in[0,\pi],$ note that:
$\frac{\pi}{2}\cos x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right],$ so $\sin\left(\frac{\pi}{2}\cos x\right)\in[-1,1].$
$\frac{\pi}{2}\sin x\in\left[0,\frac{\pi}{2}\right],$ so $\cos\left(\frac{\pi}{2}\sin x\right)\in[0,1].$
For the graphs to intersect, we need $\sin\left(\frac{\pi}{2}\cos x\right)\in[0,1].$ This occurs when $\frac{\pi}{2}\cos x\in\left[0,\frac{\pi}{2}\right].$
By the Cofunction Identity $\cos\theta=\sin\left(\frac{\pi}{2}-\theta\right),$ we rewrite the given equation:
\[\sin\left(\frac{\pi}{2}\cos x\right) = \sin\left(\frac{\pi}{2}-\frac{\pi}{2}\sin x\right).\]
Since $\frac{\pi}{2}\cos x\in\left[0,\frac{\pi}{2}\right]$ and $\frac{\pi}{2}\sin x\in\left[0,\frac{\pi}{2}\right],$ it follows that $x\in\left[0,\frac{\pi}{2}\right]$ and $\frac{\pi}{2}-\frac{\pi}{2}\sin x\in\left[0,\frac{\pi}{2}\right].$
We can apply the arcsine function to both sides, then rearrange and simplify:
\begin{align*} \frac{\pi}{2}\cos x &= \frac{\pi}{2}-\frac{\pi}{2}\sin x \\ \sin x + \cos x &= 1. \end{align*}
From Case 1 in Solution 2, we conclude that $(0,1)$ and $\left(\frac{\pi}{2},0\right)$ are the only points of intersection, as shown below:
Therefore, the answer is $\boxed{\textbf{(C) }2}.$
~MRENTHUSIASM (credit given to TheAMCHub) | // 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;
real yMin = -2;
real yMax = 2;
//Draws the horizontal gridlines
void horizontalLines()
{
for (real i = yMin+1; i < yMax; ++i)
{
draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4));
}
}
//Draws the vertical gridlines
void verticalLines()
{
for (real i = xMin+pi/2; i < xMax; i+=pi/2)
{
draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4));
}
}
//Draws the horizontal ticks
void horizontalTicks()
{
for (real i = yMin+1; i < yMax; ++i)
{
draw((-1/8,i)--(1/8,i), black+linewidth(1));
}
}
//Draws the vertical ticks
void verticalTicks()
{
for (real i = xMin+pi/2; i < xMax; i+=pi/2)
{
draw((i,-1/8)--(i,1/8), black+linewidth(1));
}
}
horizontalLines();
verticalLines();
horizontalTicks();
verticalTicks();
draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5));
draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5));
label("$x$",(xMax,0),(2,0));
label("$y$",(0,yMax),(0,2));
pair A[];
A[0] = (pi/2,0);
A[1] = (pi,0);
A[2] = (0,1);
A[3] = (0,0);
A[4] = (0,-1);
label("$\frac{\pi}{2}$",A[0],(0,-2.5));
label("$\pi$",A[1],(0,-2.5));
label("$1$",A[2],(-2.5,0));
label("$0$",A[3],(-2.5,0));
label("$-1$",A[4],(-2.5,0));
dot((0,1),linewidth(5));
dot((pi/2,0),linewidth(5));
add(legend(),point(E),40E,UnFill);
// Block 2
/* 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; real yMin = -2; real yMax = 2; //Draws the horizontal gridlines void horizontalLines() { for (real i = yMin+1; i < yMax; ++i) { draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4)); } } //Draws the vertical gridlines void verticalLines() { for (real i = xMin+pi/2; i < xMax; i+=pi/2) { draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4)); } } //Draws the horizontal ticks void horizontalTicks() { for (real i = yMin+1; i < yMax; ++i) { draw((-1/8,i)--(1/8,i), black+linewidth(1)); } } //Draws the vertical ticks void verticalTicks() { for (real i = xMin+pi/2; i < xMax; i+=pi/2) { draw((i,-1/8)--(i,1/8), black+linewidth(1)); } } horizontalLines(); verticalLines(); horizontalTicks(); verticalTicks(); draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("$x$",(xMax,0),(2,0)); label("$y$",(0,yMax),(0,2)); pair A[]; A[0] = (pi/2,0); A[1] = (pi,0); A[2] = (0,1); A[3] = (0,0); A[4] = (0,-1); label("$\frac{\pi}{2}$",A[0],(0,-2.5)); label("$\pi$",A[1],(0,-2.5)); label("$1$",A[2],(-2.5,0)); label("$0$",A[3],(-2.5,0)); label("$-1$",A[4],(-2.5,0)); dot((0,1),linewidth(5)); dot((pi/2,0),linewidth(5)); add(legend(),point(E),40E,UnFill); | [] |
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 $AF < AV$, there are two possible configurations which may arise, and they are shown below.
Set $d = FV$, which by the definition of a parabola also equals $VP$. Then as $AQ = AF = 20$, we have $AY = 20 - d$ and $AX = |20 - 2d|$. Since $FXYV$ is a rectangle, $FX = VY$, so by Pythagorean Theorem on triangles $AFX$ and $AVY$,
\begin{align*} 21^2 - (20 - d)^2 &= AV^2 - AY^2 = VY^2\\ &= FX^2 = AF^2 - AX^2 = 20^2 - (20 - 2d)^2 \end{align*}
This equation simplifies to $3d^2 - 40d + 41 = 0$, which has solutions $d = \tfrac{20\pm\sqrt{277}}3$. Both values of $d$ work - the smaller solution with the bottom configuration and the larger solution with the top configuration - and so the requested answer is $\boxed{\textbf{(B)}\ \tfrac{40}{3}}$. | // 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);
draw(P--F--A--V--Y^^F--X--Q^^rightanglemark(F,P,la,4)^^rightanglemark(A,Q,lb,4)^^rightanglemark(A,X,F,4)^^rightanglemark(A,Y,V,4),red);
draw(graph(f,-2.5,2.5));
draw(la -- lb);
dot(F^^A^^V);
label("$F$",F,NW);
label("$V$",V,SW);
label("$A$",A,dir(F--A));
label("$P$",P,S,red);
label("$Q$",Q,S,red);
label("$X$",X,E,red);
label("$Y$",Y,E,red);
// Block 2
import olympiad;
size(200);
defaultpen(linewidth(0.8)+fontsize(11pt));
real d = 0.7, edge = 2.5, Ax = 1.9;
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);
draw(Q--A--F--P^^F--X^^A--V--Y^^rightanglemark(F,P,la,4)^^rightanglemark(A,Q,lb,4)^^rightanglemark(A,X,F,4)^^rightanglemark(A,Y,V,4),red);
draw(la -- lb);
draw(graph(f,-2.5,2.5));
dot(F^^A^^V);
label("$F$",F,NW);
label("$V$",V,SW);
label("$A$",A,dir(F--A));
label("$P$",P,S,red);
label("$Q$",Q,S,red);
label("$X$",X,E,red);
label("$Y$",Y,E,red);
// Block 3
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); draw(P--F--A--V--Y^^F--X--Q^^rightanglemark(F,P,la,4)^^rightanglemark(A,Q,lb,4)^^rightanglemark(A,X,F,4)^^rightanglemark(A,Y,V,4),red); draw(graph(f,-2.5,2.5)); draw(la -- lb); dot(F^^A^^V); label("$F$",F,NW); label("$V$",V,SW); label("$A$",A,dir(F--A)); label("$P$",P,S,red); label("$Q$",Q,S,red); label("$X$",X,E,red); label("$Y$",Y,E,red);
// Block 4
import olympiad; size(200); defaultpen(linewidth(0.8)+fontsize(11pt)); real d = 0.7, edge = 2.5, Ax = 1.9; 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); draw(Q--A--F--P^^F--X^^A--V--Y^^rightanglemark(F,P,la,4)^^rightanglemark(A,Q,lb,4)^^rightanglemark(A,X,F,4)^^rightanglemark(A,Y,V,4),red); draw(la -- lb); draw(graph(f,-2.5,2.5)); dot(F^^A^^V); label("$F$",F,NW); label("$V$",V,SW); label("$A$",A,dir(F--A)); label("$P$",P,S,red); label("$Q$",Q,S,red); label("$X$",X,E,red); label("$Y$",Y,E,red); | [] |
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 $\mathcal E$ can be written in the form $\sqrt{\frac mn}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$? (Recall that the eccentricity of an ellipse $\mathcal E$ is the ratio $\frac ca$, where $2a$ is the length of the major axis of $\mathcal E$ and $2c$ is the is the distance between its two foci.)
$\textbf{(A) }7 \qquad \textbf{(B) }9 \qquad \textbf{(C) }11 \qquad \textbf{(D) }13\qquad \textbf{(E) }15$ | 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 $xy$-plane. By symmetry, the center of $\mathcal E$ must be on the $x$-axis.
The formula of $\mathcal E$ is \[\frac{(x-h)^2}{a^2}+\frac{y^2}{b^2}=1, \hspace{44.5mm} (\bigstar)\] with the center $(h,0)$ and the axes' lengths $2a$ and $2b.$
Plugging the points $(1,0),\left(-1,\sqrt3\right),$ and $\left(-2,\sqrt2\right)$ into $(\bigstar),$ respectively, we have the following system of equations:
\begin{align*} \frac{(1-h)^2}{a^2}&=1, \\ \frac{(-1-h)^2}{a^2}+\frac{{\sqrt3}^2}{b^2}&=1, \\ \frac{(-2-h)^2}{a^2}+\frac{{\sqrt2}^2}{b^2}&=1. \end{align*}
Since $t^2=(-t)^2$ holds for all real numbers $t,$ we clear fractions and simplify:
\begin{align*} (1-h)^2&=a^2, \hspace{30.25mm} &(1)\\ b^2(1+h)^2 + 3a^2 &= a^2b^2, &(2)\\ b^2(2+h)^2 + 2a^2 &= a^2b^2. &(3) \end{align*}
Applying the Transitive Property to $(2)$ and $(3),$ we isolate $a^2:$
\begin{align*} b^2(1+h)^2 + 3a^2 &= b^2(2+h)^2 + 2a^2 \\ a^2 &= b^2\left((2+h)^2-(1+h)^2\right) \\ a^2 &= b^2(2h+3). \hspace{26.75mm} (*) \end{align*}
Substituting $(1)$ and $(*)$ into $(2),$ we solve for $h:$
\begin{align*} b^2(1+h)^2 + 3\underbrace{b^2(2h+3)}_{\text{by }(*)} &= \underbrace{(1-h)^2}_{\text{by }(1)}b^2 \\ (1+h)^2+3(2h+3)&=(1-h)^2 \\ 1+2h+h^2+6h+9&=1-2h+h^2 \\ 10h&=-9 \\ h&=-\frac{9}{10}. \end{align*}
Substituting this into $(1),$ we get $a^2=\frac{361}{100}.$
Substituting the current results into $(*),$ we get $b^2=\frac{361}{120}.$
Finally, we obtain
\[c^2 = a^2-b^2 = 361\left(\frac{1}{100}-\frac{1}{120}\right) = \frac{361}{600},\]
from which \[\frac{c}{a}=\sqrt{\frac{c^2}{a^2}}=\sqrt{\frac{361/600}{361/100}}=\sqrt{\frac 16}.\]
The answer is $1+6=\boxed{\textbf{(A) } 7}.$
The graph of $\mathcal E$ is shown below. Note that the foci are at $(h\pm c,0)=\left(-\frac{9}{10}\pm\frac{19\sqrt6}{60},0\right),$ as shown in the blue points.
~MRENTHUSIASM | // 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 verticalLines()
{
for (int i = xMin+1; i < xMax; ++i)
{
draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4));
}
}
//Draws the horizontal ticks
void horizontalTicks()
{
for (int i = yMin+1; i < yMax; ++i)
{
draw((-1/8,i)--(1/8,i), black+linewidth(1));
}
}
//Draws the vertical ticks
void verticalTicks()
{
for (int i = xMin+1; i < xMax; ++i)
{
draw((i,-1/8)--(i,1/8), black+linewidth(1));
}
}
horizontalLines();
verticalLines();
horizontalTicks();
verticalTicks();
draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5));
draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5));
label("$x$",(xMax,0),(2,0));
label("$y$",(0,yMax),(0,2));
draw(ellipse((-9/10,0),19/10,19/sqrt(120)),red);
pair A = (-9/10,0);
pair B = (1,0);
pair C = (-1,sqrt(3));
pair D = (-1,-sqrt(3));
pair E = (-2,sqrt(2));
pair F = (-2,-sqrt(2));
pair G = (-9/10+19/sqrt(600),0);
pair H = (-9/10-19/sqrt(600),0);
dot(A,red+linewidth(4.5));
dot(B,red+linewidth(4.5));
dot(C,red+linewidth(4.5));
dot(D,red+linewidth(4.5));
dot(E,red+linewidth(4.5));
dot(F,red+linewidth(4.5));
dot(G,blue+linewidth(4.5));
dot(H,blue+linewidth(4.5));
label("$\left(-\frac{9}{10},0\right)$",A,(0,-2),UnFill);
label("$(1,0)$",B,(1.5,-2),UnFill);
label("$\left(-1,\sqrt3\right)$",C,N,UnFill);
label("$\left(-1,-\sqrt3\right)$",D,S,UnFill);
label("$\left(-2,\sqrt2\right)$",E,NW,UnFill);
label("$\left(-2,-\sqrt2\right)$",F,SW,UnFill);
// Block 2
/* 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 verticalLines() { for (int i = xMin+1; i < xMax; ++i) { draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4)); } } //Draws the horizontal ticks void horizontalTicks() { for (int i = yMin+1; i < yMax; ++i) { draw((-1/8,i)--(1/8,i), black+linewidth(1)); } } //Draws the vertical ticks void verticalTicks() { for (int i = xMin+1; i < xMax; ++i) { draw((i,-1/8)--(i,1/8), black+linewidth(1)); } } horizontalLines(); verticalLines(); horizontalTicks(); verticalTicks(); draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("$x$",(xMax,0),(2,0)); label("$y$",(0,yMax),(0,2)); draw(ellipse((-9/10,0),19/10,19/sqrt(120)),red); pair A = (-9/10,0); pair B = (1,0); pair C = (-1,sqrt(3)); pair D = (-1,-sqrt(3)); pair E = (-2,sqrt(2)); pair F = (-2,-sqrt(2)); pair G = (-9/10+19/sqrt(600),0); pair H = (-9/10-19/sqrt(600),0); dot(A,red+linewidth(4.5)); dot(B,red+linewidth(4.5)); dot(C,red+linewidth(4.5)); dot(D,red+linewidth(4.5)); dot(E,red+linewidth(4.5)); dot(F,red+linewidth(4.5)); dot(G,blue+linewidth(4.5)); dot(H,blue+linewidth(4.5)); label("$\left(-\frac{9}{10},0\right)$",A,(0,-2),UnFill); label("$(1,0)$",B,(1.5,-2),UnFill); label("$\left(-1,\sqrt3\right)$",C,N,UnFill); label("$\left(-1,-\sqrt3\right)$",D,S,UnFill); label("$\left(-2,\sqrt2\right)$",E,NW,UnFill); label("$\left(-2,-\sqrt2\right)$",F,SW,UnFill); | [] |
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 $abc$?
$\textbf{(A) }{-}\frac{3}{49} \qquad \textbf{(B) }{-}\frac{1}{28} \qquad \textbf{(C) }\frac{\sqrt[3]7}{64} \qquad \textbf{(D) }\frac{1}{32}\qquad \textbf{(E) }\frac{1}{28}$ | 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}+\cos\frac{6\pi}{7}\right)&=-1\\ \cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}&=-\frac12. \end{align*}
Note that $\theta=\frac{2\pi}{7},\frac{4\pi}{7},\frac{6\pi}{7}$ are roots of \[\cos\theta+\cos(2\theta)+\cos(3\theta)=-\frac12, \hspace{15mm} (\bigstar)\] as they can be verified either algebraically (by the identity $\cos\theta=\cos(-\theta)=\cos(2\pi-\theta)$) or geometrically (by the graph below).
Let $x=\cos\theta.$ We apply the Double-Angle and Triple-Angle Identities to rewrite $(\bigstar)$ in terms of $x,$ and then rearrange:
\begin{align*} x+\left(2x^2-1\right)+\left(4x^3-3x\right)&=-\frac12 \\ 4x^3+2x^2-2x-\frac12&=0 \\ x^3+\frac12 x^2 - \frac12 x - \frac18 &= 0. \end{align*}
Recall that the roots are $x=\cos\frac{2\pi}{7},\cos\frac{4\pi}{7},\cos\frac{6\pi}{7}.$
Therefore, we obtain $(a,b,c)=\left(\frac12,-\frac12,-\frac18\right),$ from which $abc=\boxed{\textbf{(D) }\frac{1}{32}}.$
~MRENTHUSIASM (inspired by Peeyush Pandaya et al) | // Block 1
/* Made by MRENTHUSIASM */
size(200);
int xMin = -2;
int xMax = 2;
int yMin = -2;
int yMax = 2;
int numRays = 24;
//Draws a polar grid that goes out to a number of circles
//equal to big, with numRays specifying the number of rays:
void polarGrid(int big, int numRays)
{
for (int i = 1; i < big+1; ++i)
{
draw(Circle((0,0),i), gray+linewidth(0.4));
}
for(int i=0;i<numRays;++i)
draw(rotate(i*360/numRays)*((-big,0)--(big,0)), gray+linewidth(0.4));
}
//Draws the horizontal gridlines
void horizontalLines()
{
for (int i = yMin+1; i < yMax; ++i)
{
draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4));
}
}
//Draws the vertical gridlines
void verticalLines()
{
for (int i = xMin+1; i < xMax; ++i)
{
draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4));
}
}
horizontalLines();
verticalLines();
polarGrid(xMax,numRays);
draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5));
draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5));
label("Re",(xMax,0),(2,0));
label("Im",(0,yMax),(0,2));
//The n such that we're taking the nth roots of unity
int n = 7;
pair A[];
for(int i = 0; i <= n-1; i+=1) {
A[i] = rotate(360*i/n)*(1,0);
}
label("$1$",A[0],NE, UnFill);
for(int i =1; i < n; ++i)
{
label("$e^{\frac{" +string(2i)+"\pi i}{" + string(n) + "}}$",A[i],dir(A[i]), UnFill);
}
draw(Circle((0,0),1),red);
for(int i = 0; i< n; ++i) dot(A[i],linewidth(3.5));
// Block 2
/* Made by MRENTHUSIASM */ size(200); int xMin = -2; int xMax = 2; int yMin = -2; int yMax = 2; int numRays = 24; //Draws a polar grid that goes out to a number of circles //equal to big, with numRays specifying the number of rays: void polarGrid(int big, int numRays) { for (int i = 1; i < big+1; ++i) { draw(Circle((0,0),i), gray+linewidth(0.4)); } for(int i=0;i<numRays;++i) draw(rotate(i*360/numRays)*((-big,0)--(big,0)), gray+linewidth(0.4)); } //Draws the horizontal gridlines void horizontalLines() { for (int i = yMin+1; i < yMax; ++i) { draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4)); } } //Draws the vertical gridlines void verticalLines() { for (int i = xMin+1; i < xMax; ++i) { draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4)); } } horizontalLines(); verticalLines(); polarGrid(xMax,numRays); draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("Re",(xMax,0),(2,0)); label("Im",(0,yMax),(0,2)); //The n such that we're taking the nth roots of unity int n = 7; pair A[]; for(int i = 0; i <= n-1; i+=1) { A[i] = rotate(360*i/n)*(1,0); } label("$1$",A[0],NE, UnFill); for(int i =1; i < n; ++i) { label("$e^{\frac{" +string(2i)+"\pi i}{" + string(n) + "}}$",A[i],dir(A[i]), UnFill); } draw(Circle((0,0),1),red); for(int i = 0; i< n; ++i) dot(A[i],linewidth(3.5)); | [] |
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