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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572 | An insect lives on the surface of a regular tetrahedron with edges of length 1. It wishes to travel on the surface of the tetrahedron from the midpoint of one edge to the midpoint of the opposite edge. What is the length of the shortest such trip? (Note: Two edges of a tetrahedron are opposite if they have no common endpoint.)
$\text{(A) }\frac {1}{2} \sqrt {3} \qquad \text{(B) }1 \qquad \text{(C) }\sqrt {2} \qquad \text{(D) }\frac {3}{2} \qquad \text{(E) }2$ | 2001 AMC 12 Problem 15 | Given any path on the surface, we can unfold the surface into a plane to get a path of the same length in the plane. Consider the net of a tetrahedron in the picture below. A pair of opposite points is marked by dots. It is obvious that in the plane the shortest path is just a segment that connects these two points. By symmetry (as the tetrahedron is regular, so all of its faces are equilateral triangles), its length is the same as the length of the tetrahedron's edge, i.e. $\boxed{\text{(B) }1}$. | // Block 1
unitsize(2cm);
defaultpen(0.8);
pair A=(0,0), B=(1,0), C=rotate(60)*B, D1=B+C-A, D2=A+C-B, D3=A+B-C;
draw(A--B--C--cycle);
draw(D1--B--C--cycle);
draw(D2--A--C--cycle);
draw(D3--B--A--cycle);
dot( A+0.5*(C-A) );
dot( B+0.5*(C-A) );
draw( ( A+0.5*(C-A) ) -- ( B+0.5*(C-A) ), dashed );
// Block 2
unitsize(2cm); defaultpen(0.8); pair A=(0,0), B=(1,0), C=rotate(60)*B, D1=B+C-A, D2=A+C-B, D3=A+B-C; draw(A--B--C--cycle); draw(D1--B--C--cycle); draw(D2--A--C--cycle); draw(D3--B--A--cycle); dot( A+0.5*(C-A) ); dot( B+0.5*(C-A) ); draw( ( A+0.5*(C-A) ) -- ( B+0.5*(C-A) ), dashed ); | [] |
573 | A point $P$ is selected at random from the interior of the pentagon with vertices $A = (0,2)$, $B = (4,0)$, $C = (2 \pi + 1, 0)$, $D = (2 \pi + 1,4)$, and $E=(0,4)$. What is the probability that $\angle APB$ is obtuse?
$\text{(A) }\frac {1}{5} \qquad \text{(B) }\frac {1}{4} \qquad \text{(C) }\frac {5}{16} \qquad \text{(D) }\frac {3}{8} \qquad \text{(E) }\frac {1}{2}$ | 2001 AMC 12 Problem 17 | The angle $APB$ is obtuse if and only if $P$ lies inside the circle with diameter $AB$. (This follows for example from the fact that the inscribed angle is half of the central angle for the same arc.)
The area of $AFB$ is $[AFB] = \frac {AF\cdot FB}2 = 4$, and the area of $ABCDE$ is $CD\cdot DE - [AFB] = 4\cdot (2\pi+1) - 4 = 8\pi$.
From the Pythagorean theorem the length of $AB$ is $\sqrt{2^2 + 4^2} = 2\sqrt{5}$, thus the radius of the circle is $\sqrt{5}$, and the area of the half-circle that is inside $ABCDE$ is $\frac{ 5\pi }2$.
Therefore the probability that $APB$ is obtuse is $\frac{ \frac{ 5\pi }2 }{ 8\pi } = \boxed{\text{(C) } \frac 5{16}}$. | // Block 1
defaultpen(0.8);
real pi=3.14159265359;
pair A=(0,2), B=(4,0), C=(2*pi+1, 0), D=(2*pi+1,4), E=(0,4), F=(0,0);
draw(A--B--C--D--E--cycle);
draw(circle((A+B)/2,length(B-A)/2));
label("$A$",A,W);
label("$B$",B,SE);
label("$C$",C,SE);
label("$D$",D,NE);
label("$E$",E,NW);
label("$F$",F,SW);
draw(A--F--B,dashed);
// Block 2
defaultpen(0.8); real pi=3.14159265359; pair A=(0,2), B=(4,0), C=(2*pi+1, 0), D=(2*pi+1,4), E=(0,4), F=(0,0); draw(A--B--C--D--E--cycle); draw(circle((A+B)/2,length(B-A)/2)); label("$A$",A,W); label("$B$",B,SE); label("$C$",C,SE); label("$D$",D,NE); label("$E$",E,NW); label("$F$",F,SW); draw(A--F--B,dashed); | [] |
573 | A point $P$ is selected at random from the interior of the pentagon with vertices $A = (0,2)$, $B = (4,0)$, $C = (2 \pi + 1, 0)$, $D = (2 \pi + 1,4)$, and $E=(0,4)$. What is the probability that $\angle APB$ is obtuse?
$\text{(A) }\frac {1}{5} \qquad \text{(B) }\frac {1}{4} \qquad \text{(C) }\frac {5}{16} \qquad \text{(D) }\frac {3}{8} \qquad \text{(E) }\frac {1}{2}$ | 2001 AMC 12 Problem 17 | (Alcumus Solution)
Since $\angle APB = 90^{\circ}$ if and only if $P$ lies on the semicircle with center $(2,1)$ and radius $\sqrt{5}$, the angle is obtuse if and only if the point $P$ lies inside this semicircle. The semicircle lies entirely inside the pentagon, since the distance, 3, from $(2,1)$ to $\overline{DE}$ is greater than the radius of the circle. Thus the probability that the angle is obtuse is the ratio of the area of the semicircle to the area of the pentagon.
Let $O=(0,0)$, $A=(0,2)$, $B=(4,0)$, $C=(2\pi+1,0)$, $D=(2\pi+1,4)$, and $E=(0,4)$. Then the area of the pentagon is\[[ABCDE]=[OCDE]-[OAB] = 4\cdot(2\pi+1)-\frac{1}{2}(2\cdot4) = 8\pi,\]and the area of the semicircle is\[\frac{1}{2}\pi(\sqrt{5})^2 = \frac{5}{2}\pi.\]The probability is\[\frac{\frac{5}{2}\pi}{8\pi} = \boxed{\text{(C) } \frac 5{16}}.\] | // Block 1
pair A,B,C,D,I;
A=(0,2);
B=(4,0);
C=(7.3,0);
D=(7.3,4);
I=(0,4);
draw(A--B--C--D--I--cycle);
label("$A$",A,W);
label("$B$",B,S);
label("$C$",C,E);
label("$D$",D,E);
label("$E$",I,W);
draw(A--(0,0)--B,dashed);
draw((3,3)..A--B..cycle,dashed);
dot((2,1));
// Block 2
pair A,B,C,D,I; A=(0,2); B=(4,0); C=(7.3,0); D=(7.3,4); I=(0,4); draw(A--B--C--D--I--cycle); label("$A$",A,W); label("$B$",B,S); label("$C$",C,E); label("$D$",D,E); label("$E$",I,W); draw(A--(0,0)--B,dashed); draw((3,3)..A--B..cycle,dashed); dot((2,1)); | [] |
574 | A circle centered at $A$ with a radius of 1 and a circle centered at $B$ with a radius of 4 are externally tangent. A third circle is tangent to the first two and to one of their common external tangents as shown. What is the radius of the third circle?
$\text{(A) }\frac {1}{3} \qquad \text{(B) }\frac {2}{5} \qquad \text{(C) }\frac {5}{12} \qquad \text{(D) }\frac {4}{9} \qquad \text{(E) }\frac {1}{2}$ | 2001 AMC 12 Problem 18 | Solution 1
In the triangle $ABC$ we have $AB = 1+4 = 5$ and $BC=4-1 = 3$, thus by the Pythagorean theorem we have $AC=4$.
Let $r$ be the radius of the small circle, and let $s$ be the perpendicular distance from $S$ to $\overline{AC}$. Moreover, the small circle is tangent to both other circles, hence we have $SA=1+r$ and $SB=4+r$.
We have $SA = \sqrt{s^2 + (1-r)^2}$ and $SB=\sqrt{(4-s)^2 + (4-r)^2}$. Hence we get the following two equations:
\begin{align*} s^2 + (1-r)^2 & = (1+r)^2 \\ (4-s)^2 + (4-r)^2 & = (4+r)^2 \end{align*}
Simplifying both, we get
\begin{align*} s^2 & = 4r \\ (4-s)^2 & = 16r \end{align*}
As in our case both $r$ and $s$ are positive, we can divide the second one by the first one to get $\left( \frac{4-s}s \right)^2 = 4$.
Now there are two possibilities: either $\frac{4-s}s=-2$, or $\frac{4-s}s=2$.
In the first case clearly $s<0$, which puts the center on the wrong side of $A$, so this is not the correct case.
(Note: This case corresponds to the other circle that is tangent to both given circles and the common tangent line. By coincidence, due to the $4:1$ ratio between radii of $A$ and $B$, this circle turns out to have the same radius as circle $B$, with center directly left of center $B$, and tangent to $B$ directly above center $A$.)
The second case solves to $s=\frac 43$. We then have $4r = s^2 = \frac {16}9$, hence $r = \boxed{\frac 49}$.
More generally, for two large circles of radius $a$ and $b$, the radius $c$ of the small circle is $c = \frac{ab}{\left(\sqrt{a}+\sqrt{b}\right)^2} = \frac{1}{\left(1/\sqrt{a}+1/\sqrt{b}\right)^2}$.
Equivalently, we have that $1/\sqrt{c} = 1/\sqrt{a} + 1/\sqrt{b}$.
Solution 2
The horizontal line is the equivalent of a circle of curvature $0$, thus we can apply Descartes' Circle Formula.
The four circles have curvatures $0, 1, \frac 14$, and $\frac 1r$.
We have $2\left(0^2+1^2+\frac {1}{4^2}+\frac{1}{r^2}\right)=\left(0+1+\frac 14+\frac 1r\right)^2$
Simplifying, we get $\frac{34}{16}+\frac{2}{r^2}=\frac{25}{16}+\frac{5}{2r}+\frac{1}{r^2}$
\[\frac{1}{r^2}-\frac{5}{2r}+\frac{9}{16}=0\]
\[\frac{16}{r^2}-\frac{40}{r}+9=0\]
\[\left(\frac{4}{r}-9\right)\left(\frac{4}{r}-1\right)=0\]
Obviously $r$ cannot equal $4$, therefore $r = \boxed{\frac 49}$.
Solution 3 (Basically 1 but less complicated)
As in solution 1, in triangle $ABC$ we have $AB = 1+4 = 5$ and $BC=4-1 = 3$, thus by the Pythagorean theorem or pythagorean triples in general, we have $AC=4$.
Let $r$ be the radius. Let $s$ be the perpendicular intersecting point $S$ and line $BC$. $AC=s$ because $s,$ both perpendicular radii, and $AC$ form a rectangle. We just have to find $AC$ in terms of $r$ and solve for $r$ now. From the Pythagorean theorem and subtracting to get lengths, we get $AC=s=4=\sqrt{(r+1)^2 - (1-r)^2} + \sqrt{(r+4)^2 - (4-r)^2}$, which is simply $4=\sqrt{4r}+\sqrt{16r} \implies \sqrt{r}=\frac{2}{3} \implies r= \boxed{\textbf{(D) } \frac{4}{9}}.$
~Wezzerwez7254
Video Solution
https://youtu.be/zOwYoFOUg2U | unitsize(1cm); pair A=(0,1), B=(4,4), C=(4,1), S=(12/9,4/9); dot(A); dot(B); draw( circle(A,1) ); draw( circle(B,4) ); draw( (-1.5,0)--(8.5,0) ); draw( (A+(4,0)) -- A -- (A+(0,-1)) ); draw( A -- B -- (B+(0,-4)) ); label("$A$",A,N); label("$B$",B,N); label("$C$",C,E); label("$S$",S,N); filldraw( circle(S,4/9), lightgray, black ); dot(S); draw( rightanglemark(A,C,B) ); draw( S -- A ); draw( S -- B ); | [] |
575 | Points $A = (3,9)$, $B = (1,1)$, $C = (5,3)$, and $D=(a,b)$ lie in the first quadrant and are the vertices of quadrilateral $ABCD$. The quadrilateral formed by joining the midpoints of $\overline{AB}$, $\overline{BC}$, $\overline{CD}$, and $\overline{DA}$ is a square. What is the sum of the coordinates of point $D$?
$\text{(A) }7 \qquad \text{(B) }9 \qquad \text{(C) }10 \qquad \text{(D) }12 \qquad \text{(E) }16$ | 2001 AMC 12 Problem 20 | We already know two vertices of the square: $(A+B)/2 = (2,5)$ and $(B+C)/2 = (3,2)$.
There are only two possibilities for the other vertices of the square: either they are $(6,3)$ and $(5,6)$, or they are $(0,1)$ and $(-1,4)$. The second case would give us $D$ outside the first quadrant, hence the first case is the correct one. As $(6,3)$ is the midpoint of $CD$, we can compute $D=(7,3)$, and $7+3=\boxed{10}$. | // Block 1
pair A=(3,9), B=(1,1), C=(5,3), D=(7,3);
draw(A--B--C--D--cycle);
label("$A$",A,N);
label("$B$",B,SW);
label("$C$",C,N);
label("$D$",D,E);
pair AB = (A + B)/2, BC = (B + C)/2, CD = (C + D)/2, DA = (D + A)/2;
draw(AB--BC--CD--DA--cycle);
// Block 2
pair A=(3,9), B=(1,1), C=(5,3), D=(7,3); draw(A--B--C--D--cycle); label("$A$",A,N); label("$B$",B,SW); label("$C$",C,N); label("$D$",D,E); pair AB = (A + B)/2, BC = (B + C)/2, CD = (C + D)/2, DA = (D + A)/2; draw(AB--BC--CD--DA--cycle); | [] |
576 | Contents
1 Solution 3
2 Solution 4
3 Solution 5
4 Solution 6 (Mass Points)
5 See also | 2001 AMC 12 Problem 22 | We need one more pair of ratios to fully define our mass point system. Let's use $\triangle AFH \sim \triangle CEH\implies EH:HF = 3:2$ and now do mass points on $\triangle AEG$:
Now it's just a standard "Area Reduction by Ratios"™ problem going from:
\[[ABCD]\xrightarrow[]{\frac{1}{2}}[AEB]\xrightarrow[]{\frac{2}{3}}[AEG]\xrightarrow[]{\frac{3}{7}}[AEJ]\xrightarrow[]{\frac{3}{10}}[HEJ]\]
or,
\[70 \cdot \frac{1}{2}\cdot \frac{2}{3}\cdot \frac{3}{7}\cdot \frac{3}{10} = \boxed{\textbf{(C) }3}\]
~ proloto | // Block 1
size(250);
pair A = (0,0), B = (10,0), C = (10,7), D = (0,7);
pair F = (10/3,0), G = (20/3,0), E = (5,7);
pair H = intersectionpoint(A--C, E--F);
pair J = intersectionpoint(A--C, E--G);
filldraw(A--E--G--cycle, rgb(1,1,1)+opacity(0.3), red+2bp);
draw(A--B--C--D--cycle);
draw(A--C);
draw(E--F);
draw(E--G);
draw(A--E, dashed);
draw(E--B, dashed);
dot("$A$", A, SW);
dot("$B$", B, SE);
dot("$C$", C, NE);
dot("$D$", D, NW);
dot("$E$", E, N);
dot("$F$", F, S);
dot("$G$", G, S);
dot("$H$", H, ESE);
dot("$J$", J, W);
// mass point labels
pair mass = A + SW;
label(scale(0.8)*"$3$", mass, UnFill);
draw(circle(mass, .4), linewidth(1));
pair mass = F + S;
label(scale(0.8)*"$6$", mass, UnFill);
draw(circle(mass, .4), linewidth(1));
pair mass = G + SE;
label(scale(0.8)*"$3$", mass, UnFill);
draw(circle(mass, .4), linewidth(1));
pair mass = J + .7*ESE;
label(scale(0.8)*"$7$", mass, UnFill);
draw(circle(mass, .4), linewidth(1));
pair mass = E + N;
label(scale(0.8)*"$4$", mass, UnFill);
draw(circle(mass, .4), linewidth(1));
pair mass = H + .7*WNW;
label(scale(0.8)*"$10$", mass, UnFill);
draw(circle(mass, .4), linewidth(1));
// Block 2
size(250); pair A = (0,0), B = (10,0), C = (10,7), D = (0,7); pair F = (10/3,0), G = (20/3,0), E = (5,7); pair H = intersectionpoint(A--C, E--F); pair J = intersectionpoint(A--C, E--G); filldraw(A--E--G--cycle, rgb(1,1,1)+opacity(0.3), red+2bp); draw(A--B--C--D--cycle); draw(A--C); draw(E--F); draw(E--G); draw(A--E, dashed); draw(E--B, dashed); dot("$A$", A, SW); dot("$B$", B, SE); dot("$C$", C, NE); dot("$D$", D, NW); dot("$E$", E, N); dot("$F$", F, S); dot("$G$", G, S); dot("$H$", H, ESE); dot("$J$", J, W); // mass point labels pair mass = A + SW; label(scale(0.8)*"$3$", mass, UnFill); draw(circle(mass, .4), linewidth(1)); pair mass = F + S; label(scale(0.8)*"$6$", mass, UnFill); draw(circle(mass, .4), linewidth(1)); pair mass = G + SE; label(scale(0.8)*"$3$", mass, UnFill); draw(circle(mass, .4), linewidth(1)); pair mass = J + .7*ESE; label(scale(0.8)*"$7$", mass, UnFill); draw(circle(mass, .4), linewidth(1)); pair mass = E + N; label(scale(0.8)*"$4$", mass, UnFill); draw(circle(mass, .4), linewidth(1)); pair mass = H + .7*WNW; label(scale(0.8)*"$10$", mass, UnFill); draw(circle(mass, .4), linewidth(1)); | [] |
577 | Betsy designed a flag using blue triangles, small white squares, and a red center square, as shown. Let $B$ be the total area of the blue triangles, $W$ the total area of the white squares, and $P$ the area of the red square. Which of the following is correct?
$\textbf{(A)}\ B = W \qquad \textbf{(B)}\ W = P \qquad \textbf{(C)}\ B = P \qquad \textbf{(D)}\ 3B = 2P \qquad \textbf{(E)}\ 2P = W$ | 2002 AMC 10A Problem 8 | Cut the pattern into 16 square regions and use symmetry.
This clearly shows $\boxed{\textbf{(A) }B=W}$. | unitsize(3mm); fill((-4,-4)--(-4,4)--(4,4)--(4,-4)--cycle,blue); fill((-2,-2)--(-2,2)--(2,2)--(2,-2)--cycle,red); path onewhite=(-3,3)--(-2,4)--(-1,3)--(-2,2)--(-3,3)--(-1,3)--(0,4)--(1,3)--(0,2)--(-1,3)--(1,3)--(2,4)--(3,3)--(2,2)--(1,3)--cycle; path divider=(-2,2)--(-3,3)--cycle; fill(onewhite,white); fill(rotate(90)*onewhite,white); fill(rotate(180)*onewhite,white); fill(rotate(270)*onewhite,white); draw((-4,-2)--(4,-2),black); draw((-4,0)--(4,0),black); draw((-4,2)--(4,2),black); draw((-2,-4)--(-2,4),black); draw((0,-4)--(0,4),black); draw((2,-4)--(2,4),black); | [] |
578 | Spot's doghouse has a regular hexagonal base that measures one yard on each side. He is tethered to a vertex with a two-yard rope. What is the area, in square yards, of the region outside of the doghouse that Spot can reach?
$\text{(A)}\ 2\pi/3 \qquad \text{(B)}\ 2\pi \qquad \text{(C)}\ 5\pi/2 \qquad \text{(D)}\ 8\pi/3 \qquad \text{(E)}\ 3\pi$ | 2002 AMC 10A Problem 19 | Part of what Spot can reach is $\frac{240}{360}=\frac{2}{3}$ of a circle with radius 2, which
gives him $\frac{8\pi}{3}$. He can also reach two $\frac{60}{360}$ parts of a unit circle, which combines to give $\frac{\pi}{3}$. The total area is then $3\pi$, which gives $\boxed{\text{(E)}}$.
Note
We can clearly see that the area must be more than $\frac{8\pi}{3}$, and the only such answer is $\boxed{\text{(E)}}$. | // Block 1
draw(polygon(6));
draw(Arc((1/2,sqrt(3)/2),2,-60,180));
draw(Arc((-1/2,sqrt(3)/2),1,180,240));
draw(Arc((1,0),1,240,300));
draw((-1/2,sqrt(3)/2)--(-3/2,sqrt(3)/2), dotted);
draw((1,0)--(3/2,-sqrt(3)/2),dotted);
// Block 2
draw(polygon(6)); draw(Arc((1/2,sqrt(3)/2),2,-60,180)); draw(Arc((-1/2,sqrt(3)/2),1,180,240)); draw(Arc((1,0),1,240,300)); draw((-1/2,sqrt(3)/2)--(-3/2,sqrt(3)/2), dotted); draw((1,0)--(3/2,-sqrt(3)/2),dotted); | [] |
579 | Points $A,B,C,D,E$ and $F$ lie, in that order, on $\overline{AF}$, dividing it into five segments, each of length 1. Point $G$ is not on line $AF$. Point $H$ lies on $\overline{GD}$, and point $J$ lies on $\overline{GF}$. The line segments $\overline{HC}, \overline{JE},$ and $\overline{AG}$ are parallel. Find $HC/JE$.
$\text{(A)}\ 5/4 \qquad \text{(B)}\ 4/3 \qquad \text{(C)}\ 3/2 \qquad \text{(D)}\ 5/3 \qquad \text{(E)}\ 2$ | 2002 AMC 10A Problem 20 | First we can draw an image.
Since $\overline{AG}$ and $\overline{CH}$ are parallel, triangles $\triangle GAD$ and $\triangle HCD$ are similar. Hence, $\frac{CH}{AG} = \frac{CD}{AD} = \frac{1}{3}$.
Since $\overline{AG}$ and $\overline{JE}$ are parallel, triangles $\triangle GAF$ and $\triangle JEF$ are similar. Hence, $\frac{EJ}{AG} = \frac{EF}{AF} = \frac{1}{5}$. Therefore, $\frac{CH}{EJ} = \left(\frac{CH}{AG}\right)\div\left(\frac{EJ}{AG}\right) = \left(\frac{1}{3}\right)\div\left(\frac{1}{5}\right) = \boxed{\frac{5}{3}}$. The answer is $\boxed{(D) 5/3}$. | // Block 1
unitsize(0.8 cm);
pair A, B, C, D, E, F, G, H, J;
A = (0,0);
B = (1,0);
C = (2,0);
D = (3,0);
E = (4,0);
F = (5,0);
G = (-1.5,4);
H = extension(D, G, C, C + G - A);
J = extension(F, G, E, E + G - A);
draw(A--F--G--cycle);
draw(B--G);
draw(C--G);
draw(D--G);
draw(E--G);
draw(C--H);
draw(E--J);
label("$A$", A, SW);
label("$B$", B, S);
label("$C$", C, S);
label("$D$", D, S);
label("$E$", E, S);
label("$F$", F, SE);
label("$G$", G, NW);
label("$H$", H, W);
label("$J$", J, NE);
// Block 2
unitsize(0.8 cm); pair A, B, C, D, E, F, G, H, J; A = (0,0); B = (1,0); C = (2,0); D = (3,0); E = (4,0); F = (5,0); G = (-1.5,4); H = extension(D, G, C, C + G - A); J = extension(F, G, E, E + G - A); draw(A--F--G--cycle); draw(B--G); draw(C--G); draw(D--G); draw(E--G); draw(C--H); draw(E--J); label("$A$", A, SW); label("$B$", B, S); label("$C$", C, S); label("$D$", D, S); label("$E$", E, S); label("$F$", F, SE); label("$G$", G, NW); label("$H$", H, W); label("$J$", J, NE); | [] |
580 | Points $A,B,C$ and $D$ lie on a line, in that order, with $AB = CD$ and $BC = 12$. Point $E$ is not on the line, and $BE = CE = 10$. The perimeter of $\triangle AED$ is twice the perimeter of $\triangle BEC$. Find $AB$.
$\text{(A)}\ 15/2 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 17/2 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 19/2$ | 2002 AMC 10A Problem 23 | First, we draw an altitude to $BC$ from $E$. Let it intersect at $M$. As $\triangle BEC$ is isosceles, we immediately get $MB=MC=6$, so the altitude is $8$. Now, let $AB=CD=x$. Using the Pythagorean Theorem on $\triangle EMA$, we find $AE=\sqrt{x^2+12x+100}$. From symmetry, $DE=\sqrt{x^2+12x+100}$ as well. Now, we use the fact that the perimeter of $\triangle AED$ is twice the perimeter of $\triangle BEC$.
We have $2\sqrt{x^2+12x+100}+2x+12=2(32)$ so $\sqrt{x^2+12x+100}=26-x$. Squaring both sides, we have $x^2+12x+100=676-52x+x^2$ which nicely rearranges into $64x=576\rightarrow{x=9}$. Hence, AB is 9 so our answer is $\boxed{\text{(D)}}$. | // Block 1
unitsize(0.25 cm);
pair A, B, C, D, E, M;
A = (0,0);
B = (9,0);
C = (21,0);
D = (30,0);
E = (15,-8);
M = (15,0);
draw(A--D--E--cycle);
draw(B--E);
draw(M--E);
draw(C--E);
label("$A$", A, N);
label("$B$", B, N);
label("$C$", C, N);
label("$D$", D, N);
label("$E$", E, S);
label("$M$", M, N);
// Block 2
unitsize(0.25 cm); pair A, B, C, D, E, M; A = (0,0); B = (9,0); C = (21,0); D = (30,0); E = (15,-8); M = (15,0); draw(A--D--E--cycle); draw(B--E); draw(M--E); draw(C--E); label("$A$", A, N); label("$B$", B, N); label("$C$", C, N); label("$D$", D, N); label("$E$", E, S); label("$M$", M, N); | [] |
581 | In trapezoid $ABCD$ with bases $AB$ and $CD$, we have $AB = 52$, $BC = 12$, $CD = 39$, and $DA = 5$. The area of $ABCD$ is
$\text{(A)}\ 182 \qquad \text{(B)}\ 195 \qquad \text{(C)}\ 210 \qquad \text{(D)}\ 234 \qquad \text{(E)}\ 260$ | 2002 AMC 10A Problem 25 | Solution 1
It shouldn't be hard to use trigonometry to bash this and find the height, but there is a much easier way. Extend $\overline{AD}$ and $\overline{BC}$ to meet at point $E$:
Since $\overline{AB} || \overline{CD}$ we have $\triangle AEB \sim \triangle DEC$, with the ratio of proportionality being $\frac {39}{52} = \frac {3}{4}$. Thus
\begin{align*} \frac {CE}{CE + 12} = \frac {3}{4} & \Longrightarrow CE = 36 \\ \frac {DE}{DE + 5} = \frac {3}{4} & \Longrightarrow DE = 15 \end{align*}
So the sides of $\triangle CDE$ are $15,36,39$, which we recognize to be a $5 - 12 - 13$ right triangle. Therefore (we could simplify some of the calculation using that the ratio of areas is equal to the ratio of the sides squared),
\[[ABCD] = [ABE] - [CDE] = \frac {1}{2}\cdot 20 \cdot 48 - \frac {1}{2} \cdot 15 \cdot 36 = \boxed{\mathrm{(C)}\ 210}\]
Solution 2
Draw altitudes from points $C$ and $D$:
Translate the triangle $ADD'$ so that $DD'$ coincides with $CC'$. We get the following triangle:
The length of $A'B$ in this triangle is equal to the length of the original $AB$, minus the length of $CD$.
Thus $A'B = 52 - 39 = 13$.
Therefore $A'BC$ is a well-known $(5,12,13)$ right triangle. Its area is $[A'BC]=\frac{A'C\cdot BC}2 = \frac{5\cdot 12}2 = 30$, and therefore its altitude $CC'$ is $\frac{[A'BC]}{A'B} = \frac{60}{13}$.
Now the area of the original trapezoid is $\frac{(AB+CD)\cdot CC'}2 = \frac{91 \cdot 60}{13 \cdot 2} = 7\cdot 30 = \boxed{\mathrm{(C)}\ 210}$
Solution 3
Draw altitudes from points $C$ and $D$:
Call the length of $AD'$ to be $y$, the length of $BC'$ to be $z$, and the height of the trapezoid to be $x$.
By the Pythagorean Theorem, we have:
\[z^2 + x^2 = 144\]
\[y^2 + x^2 = 25\]
Subtracting these two equation yields:
\[z^2-y^2=119 \implies (z+y)(z-y)=119\]
We also have: $z+y=52-39=13$.
We can substitute the value of $z+y$ into the equation we just obtained, so we now have:
\[(13) (z-y)=119 \implies z-y=\frac{119}{13}\].
We can add the $z+y$ and the $z-y$ equation to find the value of $z$, which simplifies down to be $\frac{144}{13}$. Finally, we can plug in $z$ and use the Pythagorean theorem to find the height of the trapezoid.
\[\frac{12^4}{13^2} + x^2 = 12^2 \implies x^2 = \frac{(12^2)(13^2)}{13^2} -\frac{12^4}{13^2} \implies x^2 = \frac{(12 \cdot 13)^2 - (144)^2}{13^2} \implies x^2 = \frac{(156+144)(156-144)}{13^2} \implies x = \sqrt{\frac{3600}{169}} = \frac{60}{13}\]
Now that we have the height of the trapezoid, we can multiply this by the median to find our answer.
The median of the trapezoid is $\frac{39+52}{2} = \frac{91}{2}$, and multiplying this and the height of the trapezoid gets us:
\[\frac{60 \cdot 91}{13 \cdot 2} = \boxed{\mathrm{(C)}\ 210}\]
Solution 4
We construct a line segment parallel to $\overline{AD}$ from point $C$ to line $\overline{AB},$ and label the intersection of this segment with line $\overline{AB}$ as point $E.$ Then quadrilateral $AECD$ is a parallelogram, so $CE=5, AE=39,$ and $EB=13.$ Triangle $EBC$ is therefore a right triangle, with area $\frac12 \cdot 5 \cdot 12 = 30.$
By continuing to split $\overline{AB}$ and $\overline{CD}$ into segments of length $13,$ we can connect these vertices in a "zig-zag," creating seven congruent right triangles, each with sides $5,12,$ and $13,$ and each with area $30.$ The total area is therefore $7 \cdot 30 = \boxed{\textbf{(C)} 210}.$
Alternative: Instead of creating seven congruent right triangles, we can find the height of parallelogram $AECD$ by drawing an altitude from $D$ to side $AE$, creating the new point $F$. By recognizing that triangle $DAF$ is similar to triangle $BEC$, we can use properties of similar triangles and find that $DE = 12 \cdot \frac{5}{13} = \frac{60}{13}$. Thus, the area of parallelogram $AECD$ is $\frac{60}{13} \cdot 39 = 180$. Finally, we add the areas of the parallelogram $AECD$ and the right triangle $BEC$ together and we get $180 + 30 = \boxed{\textbf{(C)} 210}$. ~scarletsyc
Solution 2 but quicker
From Solution $2$ we know that the the altitude of the trapezoid is $\frac{60}{13}$ and the triangle's area is $30$.
Note that once we remove the triangle we get a rectangle with length $39$ and height $\frac{60}{13}$.
The numbers multiply nicely to get $180+30=\boxed{(C) 210}$
-harsha12345
Quick Time Trouble Solution 5
First note how the answer choices are all integers.
The area of the trapezoid is $\frac{39+52}{2} \cdot h = \frac{91}{2} h$. So h divides 2. Let $x$ be $2h$. The area is now $91x$.
Trying $x=1$ and $x=2$ can easily be seen to not work. Those make the only integers possible so now you know x is a fraction.
Since the area is an integer the denominator of x must divide either 13 or 7 since $91 = 13\cdot7$.
Seeing how $39 = 3\cdot13$ and $52 = 4\cdot13$ assume that the denominator divides 13. Letting $y = \frac{x}{13}$ the area is now $7y$.
Note that (A) and (C) are the only multiples of 7. We know that A doesn't work because that would mean $h$ is $4$ which we ruled out.
So the answer is $\boxed{\textbf{(C)} 210}$. - megateleportingrobots | // Block 1
size(250); defaultpen(0.8); pair A=(0,0), B = (52,0), C=(52-144/13,60/13), D=(25/13,60/13), F=(100/13,240/13); draw(A--B--C--D--cycle); draw(D--F--C,dashed); label("\(A\)",A,S); label("\(B\)",B,S); label("\(C\)",C,NE); label("\(D\)",D,W); label("\(E\)",F,N); label("39",(C+D)/2,N); label("52",(A+B)/2,S); label("5",(A+D)/2,E); label("12",(B+C)/2,WSW);
// Block 2
unitsize(0.2cm); defaultpen(0.8); pair A=(0,0), B = (52,0), C=(52-144/13,60/13), D=(25/13,60/13), E=(52-144/13,0), F=(25/13,0); draw(A--B--C--D--cycle); draw(C--E,dashed); draw(D--F,dashed); label("\(A\)",A,SW); label("\(B\)",B,S); label("\(C\)",C,NE); label("\(D\)",D,N); label("\(D'\)",F,SSE); label("\(C'\)",E,S); label("39",(C+D)/2,N); label("52",(A+B)/2,S); label("5",(A+D)/2,W); label("12",(B+C)/2,ENE);
// Block 3
unitsize(0.2cm); defaultpen(0.8); pair A=(0,0), B = (13,0), C=(25/13,60/13), F=(25/13,0); draw(A--B--C--cycle); draw(C--F,dashed); label("\(A'\)",A,SW); label("\(B\)",B,S); label("\(C\)",C,N); label("\(C'\)",F,SE); label("5",(A+C)/2,W); label("12",(B+C)/2,ENE);
// Block 4
unitsize(0.2cm); defaultpen(0.8); pair A=(0,0), B = (52,0), C=(52-144/13,60/13), D=(25/13,60/13), E=(52-144/13,0), F=(25/13,0); draw(A--B--C--D--cycle); draw(C--E,dashed); draw(D--F,dashed); label("\(A\)",A,SW); label("\(B\)",B,S); label("\(C\)",C,NE); label("\(D\)",D,N); label("\(D'\)",F,SSE); label("\(C'\)",E,S); label("39",(C+D)/2,N); label("52",(A+B)/2,S); label("5",(A+D)/2,W); label("12",(B+C)/2,ENE); | [] |
582 | Betsy designed a flag using blue triangles, small white squares, and a red center square, as shown. Let $B$ be the total area of the blue triangles, $W$ the total area of the white squares, and $P$ the area of the red square. Which of the following is correct?
$\textbf{(A)}\ B = W \qquad \textbf{(B)}\ W = P \qquad \textbf{(C)}\ B = P \qquad \textbf{(D)}\ 3B = 2P \qquad \textbf{(E)}\ 2P = W$ | 2002 AMC 12A Problem 8 | Cut the pattern into 16 square regions and use symmetry.
This clearly shows $\boxed{\textbf{(A) }B=W}$. | // Block 1
unitsize(3mm);
fill((-4,-4)--(-4,4)--(4,4)--(4,-4)--cycle,blue);
fill((-2,-2)--(-2,2)--(2,2)--(2,-2)--cycle,red);
path onewhite=(-3,3)--(-2,4)--(-1,3)--(-2,2)--(-3,3)--(-1,3)--(0,4)--(1,3)--(0,2)--(-1,3)--(1,3)--(2,4)--(3,3)--(2,2)--(1,3)--cycle;
path divider=(-2,2)--(-3,3)--cycle;
fill(onewhite,white);
fill(rotate(90)*onewhite,white);
fill(rotate(180)*onewhite,white);
fill(rotate(270)*onewhite,white);
draw((-4,-2)--(4,-2),black);
draw((-4,0)--(4,0),black);
draw((-4,2)--(4,2),black);
draw((-2,-4)--(-2,4),black);
draw((0,-4)--(0,4),black);
draw((2,-4)--(2,4),black);
// Block 2
unitsize(3mm); fill((-4,-4)--(-4,4)--(4,4)--(4,-4)--cycle,blue); fill((-2,-2)--(-2,2)--(2,2)--(2,-2)--cycle,red); path onewhite=(-3,3)--(-2,4)--(-1,3)--(-2,2)--(-3,3)--(-1,3)--(0,4)--(1,3)--(0,2)--(-1,3)--(1,3)--(2,4)--(3,3)--(2,2)--(1,3)--cycle; path divider=(-2,2)--(-3,3)--cycle; fill(onewhite,white); fill(rotate(90)*onewhite,white); fill(rotate(180)*onewhite,white); fill(rotate(270)*onewhite,white); draw((-4,-2)--(4,-2),black); draw((-4,0)--(4,0),black); draw((-4,2)--(4,2),black); draw((-2,-4)--(-2,4),black); draw((0,-4)--(0,4),black); draw((2,-4)--(2,4),black); | [] |
583 | Let $C_1$ and $C_2$ be circles defined by $(x-10)^2 + y^2 = 36$ and $(x+15)^2 + y^2 = 81$
respectively. What is the length of the shortest line segment $PQ$ that is tangent to $C_1$ at $P$ and to $C_2$ at $Q$?
$\text{(A) }15 \qquad \text{(B) }18 \qquad \text{(C) }20 \qquad \text{(D) }21 \qquad \text{(E) }24$ | 2002 AMC 12A Problem 18 | First examine the formula $(x-10)^2+y^2=36$, for the circle $C_1$. Its center, $D_1$, is located at (10,0) and it has a radius of $\sqrt{36}$ = 6. The next circle, using the same pattern, has its center, $D_2$, at (-15,0) and has a radius of $\sqrt{81}$ = 9. So we can construct this diagram:
Line PQ is tangent to both circles, so it forms a right angle with the radii (6 and 9). This, as well as the two vertical angles near O, prove triangles $D_2QO$ and $D_1PO$ similar by AA, with a scale factor of $6:9$, or $2:3$. Next, we must subdivide the line $D_2D_1$ in a 2:3 ratio to get the length of the segments $D_2O$ and $D_1O$. The total length is $10 - (-15)$, or $25$, so applying the ratio, $D_2O$ = 15 and $D_1O$ = 10. These are the hypotenuses of the triangles. We already know the length of $D_2Q$ and $D_1P$, 9 and 6 (they're radii). So in order to find $PQ$, we must find the length of the longer legs of the two triangles and add them.
$15^2 - 9^2 = (15-9)(15+9) = 6 \times 24 = 144$
$\sqrt{144} = 12$
$10^2-6^2 = (10-6)(10+6) = 4 \times 16 = 64$
$\sqrt{64} = 8$
Finally, the length of PQ is $12+8=\boxed{20}$, or (C). | // Block 1
unitsize(0.3cm);
defaultpen(0.8);
path C1=circle((10,0),6);
path C2=circle((-15,0),9);
draw(C1); draw(C2);
draw( (-25,0) -- (17,0) );
dot((10,0)); dot((-15,0));
pair[] p1 = intersectionpoints(C1, circle((5,0),5) );
pair[] p2 = intersectionpoints(C2, circle((-7.5,0),7.5) );
dot(p1[1]); dot(p2[0]); draw((10,0)--p1[1]--p2[0]--(-15,0));
label("$C_1$",(10,0) + 6*dir(-45), SE );
label("$C_2$",(-15,0) + 9*dir(225), SW );
label("$D_1$",(10,0), SE );
label("$D_2$",(-15,0), SW );
label("$Q$", p2[0], NE );
label("$P$", p1[1], SW );
label("$O$", (0,0), SW );
// Block 2
unitsize(0.3cm); defaultpen(0.8); path C1=circle((10,0),6); path C2=circle((-15,0),9); draw(C1); draw(C2); draw( (-25,0) -- (17,0) ); dot((10,0)); dot((-15,0)); pair[] p1 = intersectionpoints(C1, circle((5,0),5) ); pair[] p2 = intersectionpoints(C2, circle((-7.5,0),7.5) ); dot(p1[1]); dot(p2[0]); draw((10,0)--p1[1]--p2[0]--(-15,0)); label("$C_1$",(10,0) + 6*dir(-45), SE ); label("$C_2$",(-15,0) + 9*dir(225), SW ); label("$D_1$",(10,0), SE ); label("$D_2$",(-15,0), SW ); label("$Q$", p2[0], NE ); label("$P$", p1[1], SW ); label("$O$", (0,0), SW ); | [] |
584 | The graph of the function $f$ is shown below. How many solutions does the equation $f(f(x))=6$ have?
$\text{(A) }2 \qquad \text{(B) }4 \qquad \text{(C) }5 \qquad \text{(D) }6 \qquad \text{(E) }7$ | 2002 AMC 12A Problem 19 | First of all, note that the equation $f(t)=6$ has two solutions: $t=-2$ and $t=1$.
Given an $x$, let $f(x)=t$. Obviously, to have $f(f(x))=6$, we need to have $f(t)=6$, and we already know when that happens. In other words, the solutions to $f(f(x))=6$ are precisely the solutions to ($f(x)=-2$ or $f(x)=1$).
Without actually computing the exact values, it is obvious from the graph that the equation $f(x)=-2$ has two and $f(x)=1$ has four different solutions, giving us a total of $2+4=\boxed{(D)6}$ solutions. | // Block 1
size(200);
defaultpen(fontsize(10pt)+linewidth(.8pt));
dotfactor=4;
pair P1=(-7,-4), P2=(-2,6), P3=(0,0), P4=(1,6), P5=(5,-6);
real[] xticks={-7,-6,-5,-4,-3,-2,-1,1,2,3,4,5,6};
real[] yticks={-6,-5,-4,-3,-2,-1,1,2,3,4,5,6};
path graph = P1--P2--P3--P4--P5;
path line1 = (-7,1)--(6,1);
path line2 = (-7,-2)--(6,-2);
draw(graph);
draw(line1, red);
draw(line2, red);
dot("(-7, -4)",P1);
dot("(-2, 6)",P2,LeftSide);
dot("(1, 6)",P4);
dot("(5, -6)",P5);
dot(intersectionpoints(graph,line1),red);
dot(intersectionpoints(graph,line2),red);
xaxis("$x$",-7.5,7,Ticks(xticks),EndArrow(6));
yaxis("$y$",-6.5,7,Ticks(yticks),EndArrow(6));
// Block 2
size(200); defaultpen(fontsize(10pt)+linewidth(.8pt)); dotfactor=4; pair P1=(-7,-4), P2=(-2,6), P3=(0,0), P4=(1,6), P5=(5,-6); real[] xticks={-7,-6,-5,-4,-3,-2,-1,1,2,3,4,5,6}; real[] yticks={-6,-5,-4,-3,-2,-1,1,2,3,4,5,6}; path graph = P1--P2--P3--P4--P5; path line1 = (-7,1)--(6,1); path line2 = (-7,-2)--(6,-2); draw(graph); draw(line1, red); draw(line2, red); dot("(-7, -4)",P1); dot("(-2, 6)",P2,LeftSide); dot("(1, 6)",P4); dot("(5, -6)",P5); dot(intersectionpoints(graph,line1),red); dot(intersectionpoints(graph,line2),red); xaxis("$x$",-7.5,7,Ticks(xticks),EndArrow(6)); yaxis("$y$",-6.5,7,Ticks(yticks),EndArrow(6)); | [] |
585 | In triangle $ABC$, side $AC$ and the perpendicular bisector of $BC$ meet in point $D$, and $BD$ bisects $\angle ABC$. If $AD=9$ and $DC=7$, what is the area of triangle $ABD$?
$\text{(A)}\ 14 \qquad \text{(B)}\ 21 \qquad \text{(C)}\ 28 \qquad \text{(D)}\ 14\sqrt5 \qquad \text{(E)}\ 28\sqrt5$ | 2002 AMC 12A Problem 23 | Looking at the triangle $BCD$, we see that its perpendicular bisector reaches the vertex, therefore implying it is isosceles. Let $x = \angle C$, so that $B=2x$ from given and the previous deducted. Then $\angle ABD=x, \angle ADB=2x$ because any exterior angle of a triangle has a measure that is the sum of the two interior angles that are not adjacent to the exterior angle. That means $\triangle ABD$ and $\triangle ACB$ are similar, so $\frac {16}{AB}=\frac {AB}{9} \Longrightarrow AB=12$.
Then by using Heron's Formula on $ABD$ (with sides $12,7,9$), we have $[\triangle ABD]= \sqrt{14(2)(7)(5)} = 14\sqrt5 \Longrightarrow \boxed{\text{D}}$. | // Block 1
unitsize(0.25 cm);
pair A, B, C, D, M;
A = (0,0);
B = (88/9, 28*sqrt(5)/9);
C = (16,0);
D = 9/16*C;
M = (B + C)/2;
draw(A--B--C--cycle);
draw(B--D--M);
label("$A$", A, SW);
label("$B$", B, N);
label("$C$", C, SE);
label("$D$", D, S);
// Block 2
unitsize(0.25 cm); pair A, B, C, D, M; A = (0,0); B = (88/9, 28*sqrt(5)/9); C = (16,0); D = 9/16*C; M = (B + C)/2; draw(A--B--C--cycle); draw(B--D--M); label("$A$", A, SW); label("$B$", B, N); label("$C$", C, SE); label("$D$", D, S); | [] |
585 | In triangle $ABC$, side $AC$ and the perpendicular bisector of $BC$ meet in point $D$, and $BD$ bisects $\angle ABC$. If $AD=9$ and $DC=7$, what is the area of triangle $ABD$?
$\text{(A)}\ 14 \qquad \text{(B)}\ 21 \qquad \text{(C)}\ 28 \qquad \text{(D)}\ 14\sqrt5 \qquad \text{(E)}\ 28\sqrt5$ | 2002 AMC 12A Problem 23 | Draw $DN$ such that $DN \bot AB$, $\triangle BND \cong \triangle BMD$
$\angle ACB = \angle DBC = \angle ABD$, $\triangle ABD \sim \triangle ACB$ by $AA$
$\frac{AB}{AD} = \frac{AC}{AB}$, $AB^2 = AD \cdot AC = 9 \cdot 16$, $AB = 12$
By the Angle Bisector Theorem, $\frac{BC}{AB} = \frac{CD}{AD}$
$\frac{BC}{12} = \frac{7}{9}$
$BC = \frac{28}{3}$, $CM = \frac{14}{3}$, $DN = DM = \sqrt{CD^2 - CM^2} = \frac{7 \sqrt{5} }{3}$
$[ABD] = \frac12 \cdot AB \cdot DN = \frac12 \cdot 12 \cdot \frac{7 \sqrt{5} }{3} = \boxed{\textbf{(D) } 14 \sqrt{5} }$
~isabelchen | // Block 1
size(12cm, 12cm);
pair A, B, C, D, M, N;
A = (0,0);
B = (88/9, 28*sqrt(5)/9);
C = (16,0);
D = 9/16*C;
M = (B + C)/2;
N = (6,4.27);
draw(A--B--C--cycle);
draw(B--D--M);
draw(D--N--B);
label("$A$", A, SW);
label("$B$", B, N);
label("$C$", C, SE);
label("$D$", D, S);
label("$M$", M, NE);
label("$N$", N, NW);
draw(rightanglemark(B, M, D), linewidth(.5));
draw(rightanglemark(A, N, D), linewidth(.5));
// Block 2
size(12cm, 12cm); pair A, B, C, D, M, N; A = (0,0); B = (88/9, 28*sqrt(5)/9); C = (16,0); D = 9/16*C; M = (B + C)/2; N = (6,4.27); draw(A--B--C--cycle); draw(B--D--M); draw(D--N--B); label("$A$", A, SW); label("$B$", B, N); label("$C$", C, SE); label("$D$", D, S); label("$M$", M, NE); label("$N$", N, NW); draw(rightanglemark(B, M, D), linewidth(.5)); draw(rightanglemark(A, N, D), linewidth(.5)); | [] |
586 | A regular octagon $ABCDEFGH$ has sides of length two. Find the area of $\triangle ADG$.
$\textbf{(A) } 4 + 2\sqrt2 \qquad \textbf{(B) } 6 + \sqrt2\qquad \textbf{(C) } 4 + 3\sqrt2 \qquad \textbf{(D) } 3 + 4\sqrt2 \qquad \textbf{(E) } 8 + \sqrt2$ | 2002 AMC 10B Problem 17 | The area of the triangle $ADG$ can be computed as $\frac{DG \cdot AP}2$. We will now find $DG$ and $AP$.
Clearly, $PFG$ is a right isosceles triangle with hypotenuse of length $2$, hence $PG=\sqrt 2$.
The same holds for triangle $QED$ and its leg $QD$. The length of $PQ$ is equal to $FE=2$.
Hence $GD = 2 + 2\sqrt 2$, and $AP = PD = 2 + \sqrt 2$.
Then the area of $ADG$ equals $\frac{DG \cdot AP}2 = \frac{(2+2\sqrt 2)(2+\sqrt 2)}2 = \frac{8+6\sqrt 2}2 = \boxed{\textbf{(C)}=4+3\sqrt 2}$. | // Block 1
unitsize(1cm);
defaultpen(0.8);
pair[] A = new pair[8];
A[0]=(0,0);
for (int i=1; i<8; ++i) A[i] = A[i-1] + 2*dir(45*(i-1));
draw( A[0]--A[1]--A[2]--A[3]--A[4]--A[5]--A[6]--A[7]--cycle );
label("$A$",A[0],SW);
label("$B$",A[1],SE);
label("$C$",A[2],SE);
label("$D$",A[3],NE);
label("$E$",A[4],NE);
label("$F$",A[5],NW);
label("$G$",A[6],NW);
label("$H$",A[7],SW);
filldraw( A[0]--A[3]--A[6]--cycle, lightgray, black );
pair P = intersectionpoint( A[3]--A[6], A[0]--A[5] );
draw( A[0]--P );
draw( P -- A[5], dashed );
label("$P$",P,NE);
draw( A[1]--A[4], dashed );
pair Q = intersectionpoint( A[3]--A[6], A[1]--A[4] );
label("$Q$",Q,NW);
// Block 2
unitsize(1cm); defaultpen(0.8); pair[] A = new pair[8]; A[0]=(0,0); for (int i=1; i<8; ++i) A[i] = A[i-1] + 2*dir(45*(i-1)); draw( A[0]--A[1]--A[2]--A[3]--A[4]--A[5]--A[6]--A[7]--cycle ); label("$A$",A[0],SW); label("$B$",A[1],SE); label("$C$",A[2],SE); label("$D$",A[3],NE); label("$E$",A[4],NE); label("$F$",A[5],NW); label("$G$",A[6],NW); label("$H$",A[7],SW); filldraw( A[0]--A[3]--A[6]--cycle, lightgray, black ); pair P = intersectionpoint( A[3]--A[6], A[0]--A[5] ); draw( A[0]--P ); draw( P -- A[5], dashed ); label("$P$",P,NE); draw( A[1]--A[4], dashed ); pair Q = intersectionpoint( A[3]--A[6], A[1]--A[4] ); label("$Q$",Q,NW); | [] |
587 | Let $\{a_k\}$ be a sequence of integers such that $a_1=1$ and $a_{m+n}=a_m+a_n+mn,$ for all positive integers $m$ and $n.$ Then $a_{12}$ is
$\mathrm{(A) \ } 45\qquad \mathrm{(B) \ } 56\qquad \mathrm{(C) \ } 67\qquad \mathrm{(D) \ } 78\qquad \mathrm{(E) \ } 89$ | 2002 AMC 10B Problem 23 | Note that the sequence of triangular numbers $T_n=1+2+3+...+n$ satisfies these conditions. It is immediately obvious that it satisfies $a_1=1$, and $a_{m+n}=a_m+a_n+mn$ can be visually proven with the diagram below.
This means that we can use the triangular number formula $T_n = \frac{n(n+1)}{2}$, so the answer is $T_{12} = \frac{12(12+1)}{2} = \boxed{\mathrm{(D) \ } 78}$.
~emerald_block | // Block 1
for(int i=5; i > 0; --i) {
for(int j=0; j < i; ++j) {
draw(circle((j+(5-i)/2,(5-i)*sqrt(3)/2),.2));
};
};
path m1 = brace((2,-.3),(0,-.3),.2);
draw(m1);
label("$m$",m1,S);
path n1 = brace((4,-.3),(3,-.3),.2);
draw(n1);
label("$n$",n1,S);
draw((-.2*sqrt(3),-.2)--(2+.2*sqrt(3),-.2)--(1,.4+sqrt(3))--cycle);
label("$T_m$",(1,1/3*sqrt(3)));
draw((3-.2*sqrt(3),-.2)--(4+.2*sqrt(3),-.2)--(3.5,.4+.5*sqrt(3))--cycle);
label("$T_n$",(3.5,.5/3*sqrt(3)));
path m2 = brace((2+.15*sqrt(3),.15+2*sqrt(3)),(3+.15*sqrt(3),.15+sqrt(3)),.2);
draw(m2);
label("$m$",m2,(.5*sqrt(3),.5));
path n2 = brace((1.5-.15*sqrt(3),.15+1.5*sqrt(3)),(2-.15*sqrt(3),.15+2*sqrt(3)),.2);
draw(n2);
label("$n$",n2,(-.5*sqrt(3),.5));
draw((2.5,-.4+.5*sqrt(3))--(3+.4/3*sqrt(3),sqrt(3))--(2,.4+2*sqrt(3))--(1.5-.4/3*sqrt(3),1.5*sqrt(3))--cycle);
label("$mn$",(2.25,1.25*sqrt(3)));
// Block 2
for(int i=5; i > 0; --i) { for(int j=0; j < i; ++j) { draw(circle((j+(5-i)/2,(5-i)*sqrt(3)/2),.2)); }; }; path m1 = brace((2,-.3),(0,-.3),.2); draw(m1); label("$m$",m1,S); path n1 = brace((4,-.3),(3,-.3),.2); draw(n1); label("$n$",n1,S); draw((-.2*sqrt(3),-.2)--(2+.2*sqrt(3),-.2)--(1,.4+sqrt(3))--cycle); label("$T_m$",(1,1/3*sqrt(3))); draw((3-.2*sqrt(3),-.2)--(4+.2*sqrt(3),-.2)--(3.5,.4+.5*sqrt(3))--cycle); label("$T_n$",(3.5,.5/3*sqrt(3))); path m2 = brace((2+.15*sqrt(3),.15+2*sqrt(3)),(3+.15*sqrt(3),.15+sqrt(3)),.2); draw(m2); label("$m$",m2,(.5*sqrt(3),.5)); path n2 = brace((1.5-.15*sqrt(3),.15+1.5*sqrt(3)),(2-.15*sqrt(3),.15+2*sqrt(3)),.2); draw(n2); label("$n$",n2,(-.5*sqrt(3),.5)); draw((2.5,-.4+.5*sqrt(3))--(3+.4/3*sqrt(3),sqrt(3))--(2,.4+2*sqrt(3))--(1.5-.4/3*sqrt(3),1.5*sqrt(3))--cycle); label("$mn$",(2.25,1.25*sqrt(3))); | [] |
588 | Riders on a Ferris wheel travel in a circle in a vertical plane. A particular wheel has radius $20$ feet and revolves at the constant rate of one revolution per minute. How many seconds does it take a rider to travel from the bottom of the wheel to a point $10$ vertical feet above the bottom?
$\mathrm{(A) \ } 5\qquad \mathrm{(B) \ } 6\qquad \mathrm{(C) \ } 7.5\qquad \mathrm{(D) \ } 10\qquad \mathrm{(E) \ } 15$ | 2002 AMC 10B Problem 24 | We can let this circle represent the ferris wheel with center $O,$ and $C$ represent the desired point $10$ feet above the bottom. Draw a diagram like the one above. We find out $\triangle OBC$ is a $30-60-90$ triangle. That means $\angle BOC = 60^\circ$ and the ferris wheel has made $\frac{60}{360} = \frac{1}{6}$ of a revolution. Therefore, the time it takes to travel that much of a distance is $\frac{1}{6}\text{th}$ of a minute, or $10$ seconds. The answer is $\boxed{\mathrm{(D) \ } 10}$. Alternatively, we could also say that $\triangle ABC$ is congruent to $\triangle OBC$ by SAS, so $AC$ is 20, and $\triangle AOC$ is equilateral, and $\angle BOC = 60^\circ$ | // Block 1
unitsize(1.5mm);
defaultpen(linewidth(.8pt)+fontsize(10pt));
dotfactor=4;
pair O=(0,0), A=(0,-20), B=(0,-10), C=(10sqrt(3),-10);
real r=20;
path ferriswheel=Circle(O,r);
draw(ferriswheel);
draw(O--A); draw(O--C); draw(B--C); draw(A--C);
pair[] ps={A,B,C,O}; dot(ps);
label("$O$",O,N); label("$A$",A,S); label("$B$",B,W); label("$C$",C,SE);
label("$10$",(O--B),W);
label("$10$",(A--B),W);
label("$20$",(O--C),NE);
// Block 2
unitsize(1.5mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair O=(0,0), A=(0,-20), B=(0,-10), C=(10sqrt(3),-10); real r=20; path ferriswheel=Circle(O,r); draw(ferriswheel); draw(O--A); draw(O--C); draw(B--C); draw(A--C); pair[] ps={A,B,C,O}; dot(ps); label("$O$",O,N); label("$A$",A,S); label("$B$",B,W); label("$C$",C,SE); label("$10$",(O--B),W); label("$10$",(A--B),W); label("$20$",(O--C),NE); | [] |
589 | In $\triangle ABC$, we have $AB = 1$ and $AC = 2$. Side $\overline{BC}$ and the median from $A$ to $\overline{BC}$ have the same length. What is $BC$?
$\mathrm{(A)}\ \frac{1+\sqrt{2}}{2} \qquad\mathrm{(B)}\ \frac{1+\sqrt{3}}2 \qquad\mathrm{(C)}\ \sqrt{2} \qquad\mathrm{(D)}\ \frac 32 \qquad\mathrm{(E)}\ \sqrt{3}$ | 2002 AMC 12B Problem 23 | Solution 1: Pythagorean Theorem
Let $D$ be the foot of the altitude from $A$ to $\overline{BC}$ extended past $B$. Let $AD = x$ and $BD = y$.
Using the Pythagorean Theorem, we obtain the equations
\begin{align*} x^2 + y^2 = 1 \hspace{0.5cm}(1)\\ x^2 + y^2 + 2ya + a^2 = 4a^2 \hspace{0.5cm}(2)\\ x^2 + y^2 + 4ya + 4a^2 = 4 \hspace{0.5cm}(3) \end{align*}
Subtracting $(1)$ equation from $(2)$ and $(3)$, we get
\begin{align*} 2ya + a^2 = 4a^2 - 1 \hspace{0.5cm}(4)\\ 4ya + 4a^2 = 3 \hspace{0.5cm}(5) \end{align*}
Then, subtracting $2 \times (4)$ from $(5)$ and rearranging, we get $10a^2 = 5$, so $BC = 2a = \sqrt{2}\Rightarrow \boxed{\mathrm{(C)}}$
~greenturtle 11/28/2017
Solution 2: Law of Cosines
[Image: images/amc/2002_AMC_12B_Problem_23_0.png]
Let $D$ be the foot of the median from $A$ to $\overline{BC}$, and we let $AD = BC = 2a$. Then by the Law of Cosines on $\triangle ABD, \triangle ACD$, we have
\begin{align*} 1^2 &= a^2 + (2a)^2 - 2(a)(2a)\cos ADB \\ 2^2 &= a^2 + (2a)^2 - 2(a)(2a)\cos ADC \end{align*}
Since $\cos ADC = \cos (180 - ADB) = -\cos ADB$, we can add these two equations and get
\[5 = 10a^2\]
Hence $a = \frac{1}{\sqrt{2}}$ and $BC = 2a = \sqrt{2} \Rightarrow \mathrm{(C)}$.
Solution 3: Stewart's Theorem
From Stewart's Theorem, we have $(2)(1/2)a(2) + (1)(1/2)a(1) = (a)(a)(a) + (1/2)a(a)(1/2)a.$ Simplifying, we get $(5/4)a^3 = (5/2)a \implies (5/4)a^2 = 5/2 \implies a^2 = 2 \implies a = \boxed{\sqrt{2}}.$
- awu2014
Solution 4: Pappus's Median Theorem
There is a theorem in geometry known as Pappus's Median Theorem. It states that if you have $\triangle{ABC}$, and you draw a median from point $A$ to side $BC$ (label this as $M$), then: $(AM)^2 = \dfrac{2(b^2) + 2(c^2) - (a^2)}{4}$. Note that $b$ is the length of side $\overline{AC}$, $c$ is the length of side $\overline{AB}$, and $a$ is length of side $\overline{BC}$. Let $MB = MC = x$. Then $AM = 2x$. Now, we can plug into the formula given above: $AM = 2x$, $b = 2$, $c = 1$, and $a = 2x$. After some simple algebra, we find $x = \dfrac{\sqrt{2}}{2}$. Then, $BC = \boxed{\sqrt{2}} \implies \boxed{C}$.
-Flames
Note: Pappus's Median Theorem is just a special case of Stewart's Theorem, with $m = n$. ~Puck_0
aka Apollonius' Theorem - Orion 2010
Video Solution by TheBeautyofMath
https://youtu.be/jEVMgWKQIW8
~IceMatrix | unitsize(4cm); pair A, B, C, D, M; A = (1.768,0.935); B = (1.414,0); C = (0,0); D = (1.768,0); M = (0.707,0); draw(A--B--C--cycle); draw(A--D); draw(D--B); draw(A--M); label("$A$",A,N); label("$B$",B,S); label("$C$",C,S); label("$D$",D,S); label("$M$",M,S); label("$x$",(A+D)/2,E); label("$y$",(B+D)/2,S); label("$a$",(C+M)/2,S); label("$a$",(M+B)/2,S); label("$2a$",(A+M)/2,SE); label("$1$",(A+B)/2,SE); label("$2$",(A+C)/2,NW); draw(rightanglemark(B,D,A,3)); | ["https://artofproblemsolving.com/wiki/images/9/98/2002_12B_AMC-23.png"] |
590 | A convex quadrilateral $ABCD$ with area $2002$ contains a point $P$ in its interior such that $PA = 24, PB = 32, PC = 28, PD = 45$. Find the perimeter of $ABCD$.
$\mathrm{(A)}\ 4\sqrt{2002} \qquad \mathrm{(B)}\ 2\sqrt{8465} \qquad \mathrm{(C)}\ 2$ $(48+\sqrt{2002}) \qquad \mathrm{(D)}\ 2\sqrt{8633} \qquad \mathrm{(E)}\ 4(36 + \sqrt{113})$ | 2002 AMC 12B Problem 24 | We have
\[[ABCD] = 2002 \le \frac 12 (AC \cdot BD)\]
(This is true for any convex quadrilateral: split the quadrilateral along $AC$ and then using the triangle area formula to evaluate $[ACB]$ and $[ACD]$), with equality only if $\overline{AC} \perp \overline{BD}$. By the triangle inequality,
\begin{align*}AC &\le PA + PC = 52\\ BD &\le PB + PD = 77\end{align*}
with equality if $P$ lies on $\overline{AC}$ and $\overline{BD}$ respectively. Thus
\[2002 \le \frac{1}{2} AC \cdot BD \le \frac 12 \cdot 52 \cdot 77 = 2002\]
Since we have the equality case, $\overline{AC} \perp \overline{BD}$ at point $P$, as shown below.
By the Pythagorean Theorem,
\begin{align*} AB = \sqrt{PA^2 + PB^2} & = \sqrt{24^2 + 32^2} = 40\\ BC = \sqrt{PB^2 + PC^2} & = \sqrt{32^2 + 28^2} = 4\sqrt{113}\\ CD = \sqrt{PC^2 + PD^2} & = \sqrt{28^2 + 45^2} = 53\\ DA = \sqrt{PD^2 + PA^2} & = \sqrt{45^2 + 24^2} = 51 \end{align*}
The perimeter of $ABCD$ is $AB + BC + CD + DA = 4(36 + \sqrt{113}) \Rightarrow \mathrm{(E)}$. | // Block 1
size(200);
defaultpen(0.6);
pair A = (0,0), B = (40,0), C = (25.6 * 52 / 24, 19.2 * 52 / 24), D = (40 - (40-25.6)*77/32,19.2*77/32), P = (25.6,19.2), Q = (25.6, 18.5);
pair E=(A+P)/2, F=(B+P)/2, G=(C+P)/2, H=(D+P)/2;
draw(A--B--C--D--cycle);
draw(A--P--B--P--C--P--D);
label("\(A\)",A,WSW);
label("\(B\)",B,ESE);
label("\(C\)",C,ESE);
label("\(D\)",D,NW);
label("\(P\)",Q,SSW);
label("24",E,WNW);
label("32",F,WSW);
label("28",G,ESE);
label("45",H,ENE);
draw(rightanglemark(C,P,D,50));
// Block 2
size(200); defaultpen(0.6); pair A = (0,0), B = (40,0), C = (25.6 * 52 / 24, 19.2 * 52 / 24), D = (40 - (40-25.6)*77/32,19.2*77/32), P = (25.6,19.2), Q = (25.6, 18.5); pair E=(A+P)/2, F=(B+P)/2, G=(C+P)/2, H=(D+P)/2; draw(A--B--C--D--cycle); draw(A--P--B--P--C--P--D); label("\(A\)",A,WSW); label("\(B\)",B,ESE); label("\(C\)",C,ESE); label("\(D\)",D,NW); label("\(P\)",Q,SSW); label("24",E,WNW); label("32",F,WSW); label("28",G,ESE); label("45",H,ENE); draw(rightanglemark(C,P,D,50)); | [] |
591 | A semicircle of diameter $1$ sits at the top of a semicircle of diameter $2$, as shown. The shaded area inside the smaller semicircle and outside the larger semicircle is called a lune. Determine the area of this lune.
$\mathrm{(A) \ } \frac{1}{6}\pi-\frac{\sqrt{3}}{4}\qquad \mathrm{(B) \ } \frac{\sqrt{3}}{4}-\frac{1}{12}\pi\qquad \mathrm{(C) \ } \frac{\sqrt{3}}{4}-\frac{1}{24}\pi\qquad \mathrm{(D) \ } \frac{\sqrt{3}}{4}+\frac{1}{24}\pi\qquad \mathrm{(E) \ } \frac{\sqrt{3}}{4}+\frac{1}{12}\pi$ | 2003 AMC 10A Problem 19 | The shaded area $[A]$ is equal to the area of the smaller semicircle $[A+B]$ minus the area of a sector of the larger circle $[B+C]$ plus the area of a triangle formed by two radii of the larger semicircle and the diameter of the smaller semicircle $[C]$.
The area of the smaller semicircle is $[A+B] = \frac{1}{2}\pi\cdot\left(\frac{1}{2}\right)^{2}=\frac{1}{8}\pi$.
Since the radius of the larger semicircle is equal to the diameter of the smaller half circle, the triangle is an equilateral triangle and the sector measures $60^\circ$.
The area of the $60^\circ$ sector of the larger semicircle is $[B+C] = \frac{60}{360}\pi\cdot\left(\frac{2}{2}\right)^{2}=\frac{1}{6}\pi$.
The area of the triangle is $[C] = \frac{1^{2}\sqrt{3}}{4}=\frac{\sqrt{3}}{4}$.
So the shaded area is $[A] = [A+B]-[B+C]+[C] = \left(\frac{1}{8}\pi\right)-\left(\frac{1}{6}\pi\right)+\left(\frac{\sqrt{3}}{4}\right)=\boxed{\mathrm{(C)}\ \frac{\sqrt{3}}{4}-\frac{1}{24}\pi}$. We have thus solved the problem. | import graph; size(150); defaultpen(fontsize(8)); pair A=(-2,0), B=(2,0); filldraw(Arc((0,sqrt(3)),1,0,180)--cycle,mediumgray); fill(Arc((0,0),2,0,180)--cycle,white); draw(Arc((0,0),2,0,180)--cycle); draw((0,0)--2*expi(2*pi/6)--2*expi(2*pi/6*2)--(0,0)); label("A",(0,2),(0,4)); label("B",(0,2),(0,-1)); label("C",(0,sqrt(3)/2),(0,2)); label("1",(-0.5,sqrt(3)/2),(-1,0)); label("1",(0.5,sqrt(3)/2),(1,0)); | [] |
592 | A semicircle of diameter $1$ sits at the top of a semicircle of diameter $2$, as shown. The shaded area inside the smaller semicircle and outside the larger semicircle is called a lune. Determine the area of this lune.
$\mathrm{(A) \ } \frac{1}{6}\pi-\frac{\sqrt{3}}{4}\qquad \mathrm{(B) \ } \frac{\sqrt{3}}{4}-\frac{1}{12}\pi\qquad \mathrm{(C) \ } \frac{\sqrt{3}}{4}-\frac{1}{24}\pi\qquad \mathrm{(D) \ } \frac{\sqrt{3}}{4}+\frac{1}{24}\pi\qquad \mathrm{(E) \ } \frac{\sqrt{3}}{4}+\frac{1}{12}\pi$ | 2003 AMC 12A Problem 15 | The shaded area $[A]$ is equal to the area of the smaller semicircle $[A+B]$ minus the area of a sector of the larger circle $[B+C]$ plus the area of a triangle formed by two radii of the larger semicircle and the diameter of the smaller semicircle $[C]$.
The area of the smaller semicircle is $[A+B] = \frac{1}{2}\pi\cdot\left(\frac{1}{2}\right)^{2}=\frac{1}{8}\pi$.
Since the radius of the larger semicircle is equal to the diameter of the smaller half circle, the triangle is an equilateral triangle and the sector measures $60^\circ$.
The area of the $60^\circ$ sector of the larger semicircle is $[B+C] = \frac{60}{360}\pi\cdot\left(\frac{2}{2}\right)^{2}=\frac{1}{6}\pi$.
The area of the triangle is $[C] = \frac{1^{2}\sqrt{3}}{4}=\frac{\sqrt{3}}{4}$.
So the shaded area is $[A] = [A+B]-[B+C]+[C] = \left(\frac{1}{8}\pi\right)-\left(\frac{1}{6}\pi\right)+\left(\frac{\sqrt{3}}{4}\right)=\boxed{\mathrm{(C)}\ \frac{\sqrt{3}}{4}-\frac{1}{24}\pi}$. We have thus solved the problem. | // Block 1
import graph;
size(150);
defaultpen(fontsize(8));
pair A=(-2,0), B=(2,0);
filldraw(Arc((0,sqrt(3)),1,0,180)--cycle,mediumgray);
fill(Arc((0,0),2,0,180)--cycle,white);
draw(Arc((0,0),2,0,180)--cycle);
draw((0,0)--2*expi(2*pi/6)--2*expi(2*pi/6*2)--(0,0));
label("A",(0,2),(0,4));
label("B",(0,2),(0,-1));
label("C",(0,sqrt(3)/2),(0,2));
label("1",(-0.5,sqrt(3)/2),(-1,0));
label("1",(0.5,sqrt(3)/2),(1,0));
// Block 2
import graph; size(150); defaultpen(fontsize(8)); pair A=(-2,0), B=(2,0); filldraw(Arc((0,sqrt(3)),1,0,180)--cycle,mediumgray); fill(Arc((0,0),2,0,180)--cycle,white); draw(Arc((0,0),2,0,180)--cycle); draw((0,0)--2*expi(2*pi/6)--2*expi(2*pi/6*2)--(0,0)); label("A",(0,2),(0,4)); label("B",(0,2),(0,-1)); label("C",(0,sqrt(3)/2),(0,2)); label("1",(-0.5,sqrt(3)/2),(-1,0)); label("1",(0.5,sqrt(3)/2),(1,0)); | [] |
593 | A point P is chosen at random in the interior of equilateral triangle $ABC$. What is the probability that $\triangle ABP$ has a greater area than each of $\triangle ACP$ and $\triangle BCP$?
$\textbf{(A)}\ \frac{1}{6}\qquad\textbf{(B)}\ \frac{1}{4}\qquad\textbf{(C)}\ \frac{1}{3}\qquad\textbf{(D)}\ \frac{1}{2}\qquad\textbf{(E)}\ \frac{2}{3}$ | 2003 AMC 12A Problem 16 | Solution 1
After we pick point $P$, we realize that $ABC$ is symmetric for this purpose, and so the probability that $ACP$ is the greatest area, or $ABP$ or $BCP$, are all the same. Since they add to $1$, the probability that $ABP$ has the greatest area is $\boxed{\mathrm{(C)}\ \dfrac{1}{3}}$
Solution 2
We will use geometric probability. Let us take point $P$, and draw the perpendiculars to $BC$, $CA$, and $AB$, and call the feet of these perpendiculars $D$, $E$, and $F$ respectively. The area of $\triangle ACP$ is simply $\frac{1}{2} * AC * PF$. Similarly we can find the area of triangles $BCP$ and $ABP$. If we add these up and realize that it equals the area of the entire triangle, we see that no matter where we choose $P, PD + PE + PF$ = the height of the triangle. Setting the area of triangle $ABP$ greater than $ACP$ and $BCP$, we want $PF$ to be the largest of $PF$, $PD$, and $PE$. We then realize that $PF = PD = PE$ when $P$ is the incenter of $\triangle ABC$. Let us call the incenter of the triangle $Q$. If we want $PF$ to be the largest of the three, by testing points we realize that $P$ must be in the interior of quadrilateral $QDCE$. So our probability (using geometric probability) is the area of $QDCE$ divided by the area of $ABC$. We will now show that the three quadrilaterals, $QDCE$, $QEAF$, and $QFBD$ are congruent. As the definition of point $Q$ yields, $QF$ = $QD$ = $QE$. Since $ABC$ is equilateral, $Q$ is also the circumcenter of $\triangle ABC$, so $QA = QB = QC$. By the Pythagorean Theorem, $BD = DC = CE = EA = AF = FB$. Also, angles $BDQ, BFQ, CEQ, CDQ, AFQ$, and $AEQ$ are all equal to $90^\circ$. Angles $DBF, FAE, ECD$ are all equal to $60$ degrees, so it is now clear that quadrilaterals $QDCE, QEAF, QFBD$ are all congruent. Summing up these areas gives us the area of $\triangle ABC$. $QDCE$ contributes to a third of that area so $\frac{[QDCE]}{[ABC]}=\boxed{\mathrm{(C)}\ \dfrac{1}{3}}$. | // Block 1
draw((0,10)--(8.660254037844385792,-5)--(-8.660254037844385792,-5)--cycle);
dot((1.2,-0.68));
label("$P$",(1.2,-0.68),N);
// Block 2
draw((0,10)--(8.660254037844385792,-5)--(-8.660254037844385792,-5)--cycle); dot((1.2,-0.68)); label("$P$",(1.2,-0.68),N); | [] |
594 | Square $ABCD$ has sides of length $4$, and $M$ is the midpoint of $\overline{CD}$. A circle with radius $2$ and center $M$ intersects a circle with radius $4$ and center $A$ at points $P$ and $D$. What is the distance from $P$ to $\overline{AD}$?
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ \frac {16}{5} \qquad \textbf{(C)}\ \frac {13}{4} \qquad \textbf{(D)}\ 2\sqrt {3} \qquad \textbf{(E)}\ \frac {7}{2}$ | 2003 AMC 12A Problem 17 | Draw $AM$, $DP$, and $PR$. $PR$ is parallel with $CD$
$[AMD] = \frac12 \cdot AD \cdot DM = 4$, $AM = \sqrt{AD^2 + DM^2} = 2 \sqrt{5}$
$\triangle ADQ \sim \triangle AMD$ by $AA$, $[ADQ] = [AMD] \cdot \left( \frac{AD}{AM} \right) ^2 = 4 \cdot \left( \frac{2 \sqrt{5}}{5} \right)^2 = \frac{16}{5}$
$\triangle ADQ \cong \triangle APQ$, $[APD] = 2 \cdot [ADQ] = 2 \cdot \frac{16}{5} = \frac{32}{5}$
$PR = \frac{2 \cdot [APD]}{AD} = \frac{2 \cdot \frac{32}{5}}{4} = \boxed{\textbf{(B) } \frac{16}{5} }$
~isabelchen | // Block 1
size(8cm, 8cm);
pair A,B,C,D,M,P,Q,R;
D = (0,0);
C = (10,0);
B = (10,10);
A = (0,10);
M = (5,0);
P = (8,4);
Q = (D+P)/2;
R = (0,4);
dot(M);
dot(P);
draw(A--B--C--D--cycle,linewidth(0.7));
draw((5,5)..D--C..cycle,linewidth(0.7));
draw((7.07,2.93)..B--A--D..cycle,linewidth(0.7));
draw(A--M,linewidth(0.7));
draw(A--P,linewidth(0.7));
draw(D--P,linewidth(0.7));
draw(R--P,linewidth(0.7));
label("$A$",A,NW);
label("$B$",B,NE);
label("$C$",C,SE);
label("$D$",D,SW);
label("$M$",M,S);
label("$P$",P,N);
label("$Q$",Q,W);
label("$R$",R,W);
draw(rightanglemark(M, Q, P), linewidth(.5));
// Block 2
size(8cm, 8cm); pair A,B,C,D,M,P,Q,R; D = (0,0); C = (10,0); B = (10,10); A = (0,10); M = (5,0); P = (8,4); Q = (D+P)/2; R = (0,4); dot(M); dot(P); draw(A--B--C--D--cycle,linewidth(0.7)); draw((5,5)..D--C..cycle,linewidth(0.7)); draw((7.07,2.93)..B--A--D..cycle,linewidth(0.7)); draw(A--M,linewidth(0.7)); draw(A--P,linewidth(0.7)); draw(D--P,linewidth(0.7)); draw(R--P,linewidth(0.7)); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$M$",M,S); label("$P$",P,N); label("$Q$",Q,W); label("$R$",R,W); draw(rightanglemark(M, Q, P), linewidth(.5)); | [] |
595 | Three semicircles of radius $1$ are constructed on diameter $\overline{AB}$ of a semicircle of radius $2$. The centers of the small semicircles divide $\overline{AB}$ into four line segments of equal length, as shown. What is the area of the shaded region that lies within the large semicircle but outside the smaller semicircles?
$\textbf{(A) } \pi - \sqrt{3} \qquad\textbf{(B) } \pi - \sqrt{2} \qquad\textbf{(C) } \frac{\pi + \sqrt{2}}{2} \qquad\textbf{(D) } \frac{\pi +\sqrt{3}}{2} \qquad\textbf{(E) } \frac{7}{6}\pi - \frac{\sqrt{3}}{2}$ | 2003 AMC 10B Problem 19 | By drawing four lines from the intersect of the semicircles to their centers, we have split the white region into $\frac{5}{6}$ of a circle with radius $1$ and two equilateral triangles with side length $1$.
This gives the area of the white region as $\frac{5}{6}\pi+\frac{2\cdot\sqrt3}{4}=\frac{5}{6}\pi+\frac{\sqrt3}{2}$.
The area of the shaded region is the area of the white region subtracted from the area of the large semicircle. This is equivalent to $2\pi-\left(\frac{5}{6}\pi+\frac{\sqrt3}{2}\right)=\frac{7}{6}\pi-\frac{\sqrt3}{2}$.
Thus the answer is $\boxed{\textbf{(E)}\ \frac{7}{6}\pi-\frac{\sqrt3}{2}}$.
Note
The reason why it is $\frac{5}{6}$ of a circle and why the triangles are equilateral are because, first, the radii are the same and they make up the equilateral triangles.
Secondly, the reason it is $\frac{5}{6}$ of a circle is because the middle sector has a degree of $180-2 \cdot 60 = 60$ and thus $\frac{60}{360}=\frac{1}{6}$ of a circle.
The other two have areas of $\frac{180-60}{360}=\frac{1}{3}$ of a triangle each.
Therefore, the total fraction of the circle(since they have the same radii) is $\frac{1}{6} + 2 \cdot \frac{1}{3} = \frac{1}{6} + \frac{4}{6} = \frac{5}{6}.$
~mathboy282 | // Block 1
import graph;
unitsize(14mm);
defaultpen(linewidth(.8pt)+fontsize(8pt));
dashed=linetype("4 4");
dotfactor=3;
pair A=(-2,0), B=(2,0);
fill(Arc((0,0),2,0,180)--cycle,mediumgray);
fill(Arc((-1,0),1,0,180)--cycle,white);
fill(Arc((0,0),1,0,180)--cycle,white);
fill(Arc((1,0),1,0,180)--cycle,white);
draw(Arc((-1,0),1,60,180));
draw(Arc((0,0),1,0,60),dashed);
draw(Arc((0,0),1,60,120));
draw(Arc((0,0),1,120,180),dashed);
draw(Arc((1,0),1,0,120));
draw(Arc((0,0),2,0,180)--cycle);
dot((0,0));
dot((-1,0));
dot((1,0));
draw((-2,-0.1)--(-2,-0.3),gray);
draw((-1,-0.1)--(-1,-0.3),gray);
draw((1,-0.1)--(1,-0.3),gray);
draw((2,-0.1)--(2,-0.3),gray);
label("$A$",A,W);
label("$B$",B,E);
label("1",(-1.5,-0.1),S);
label("2",(0,-0.1),S);
label("1",(1.5,-0.1),S);
draw((1,0)--(0.5,0.866));
draw((0,0)--(0.5,0.866));
draw((-1,0)--(-0.5,0.866));
draw((0,0)--(-0.5,0.866));
// Block 2
import graph; unitsize(14mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); dashed=linetype("4 4"); dotfactor=3; pair A=(-2,0), B=(2,0); fill(Arc((0,0),2,0,180)--cycle,mediumgray); fill(Arc((-1,0),1,0,180)--cycle,white); fill(Arc((0,0),1,0,180)--cycle,white); fill(Arc((1,0),1,0,180)--cycle,white); draw(Arc((-1,0),1,60,180)); draw(Arc((0,0),1,0,60),dashed); draw(Arc((0,0),1,60,120)); draw(Arc((0,0),1,120,180),dashed); draw(Arc((1,0),1,0,120)); draw(Arc((0,0),2,0,180)--cycle); dot((0,0)); dot((-1,0)); dot((1,0)); draw((-2,-0.1)--(-2,-0.3),gray); draw((-1,-0.1)--(-1,-0.3),gray); draw((1,-0.1)--(1,-0.3),gray); draw((2,-0.1)--(2,-0.3),gray); label("$A$",A,W); label("$B$",B,E); label("1",(-1.5,-0.1),S); label("2",(0,-0.1),S); label("1",(1.5,-0.1),S); draw((1,0)--(0.5,0.866)); draw((0,0)--(0.5,0.866)); draw((-1,0)--(-0.5,0.866)); draw((0,0)--(-0.5,0.866)); | [] |
596 | In rectangle $ABCD, AB=5$ and $BC=3$. Points $F$ and $G$ are on $\overline{CD}$ so that $DF=1$ and $GC=2$. Lines $AF$ and $BG$ intersect at $E$. Find the area of $\triangle AEB$.
$\textbf{(A) } 10 \qquad\textbf{(B) } \frac{21}{2} \qquad\textbf{(C) } 12 \qquad\textbf{(D) } \frac{25}{2} \qquad\textbf{(E) } 15$ | 2003 AMC 10B Problem 20 | Since $\Delta{ABE}\sim{\Delta{FGE}}$ then $[AFGB]\sim{[FXYG]}$, where $X$ and $Y$ are ponts on $EF$ and $EG$ respectivley which make the areas similar. This process can be done over and over again multiple times by the ratio of $\frac{FG}{AB}=\frac{2}{5}$, or something like this
\[[AEB]=[AFGB]+[FXYZ]+...\]\[[AEB]=[AFGB]+\frac{2}{5}[AFGB]+(\frac{2}{5})^2[AFGB]+...\]we have to find the ratio of the areas when the sides have shrunk by length $\frac{2}{5}l$
Let $[AFGB]'$ be the area of the shape whose length is $\frac{2}{5}l$
\[[AFGB]'=[ADCB]-[ADF]-[BCG]\]\[[AFGB]'=12/5-6/25-12/25\]\[[AFGB]'=42/25\]Now comparing the ratios of $[AFGB]'$ to $[AFGB]$ we get
\[\frac{[AFGB]'}{[AFGB]}=\frac{42}{25}/\frac{21}{2}\implies \frac{[AFGB]'}{[AFGB]}=\frac{4}{25}\]By applying an infinite summation
\[[AEB]=\sum_{n=0}^{\infty} \frac{21}{2}\cdot{(\frac{4}{25})^n}\]\[S=\frac{a_1}{1-r}\]\[\boxed{[AEB]=\frac{25}{2}}\] | // Block 1
unitsize(0.6 cm);
pair A, B, C, D, E, F, G;
A = (0,0);
B = (5,0);
C = (5,3);
D = (0,3);
F = (1,3);
G = (3,3);
E = extension(A,F,B,G);
draw(A--B--C--D--cycle);
draw(A--E--B);
label("$A$", A, SW);
label("$B$", B, SE);
label("$C$", C, NE);
label("$D$", D, NW);
label("$E$", E, N);
label("$F$", F, SE);
label("$G$", G, SW);
label("$2/5$", (D + F)/2, N);
label("$4/5$", (G + C)/2, N);
label("$6/5$", (B + C)/2, dir(0));
label("$6/5$", (A + D)/2, W);
label("$2$", (A + B)/2, S);
// Block 2
unitsize(0.6 cm); pair A, B, C, D, E, F, G; A = (0,0); B = (5,0); C = (5,3); D = (0,3); F = (1,3); G = (3,3); E = extension(A,F,B,G); draw(A--B--C--D--cycle); draw(A--E--B); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, NE); label("$D$", D, NW); label("$E$", E, N); label("$F$", F, SE); label("$G$", G, SW); label("$2/5$", (D + F)/2, N); label("$4/5$", (G + C)/2, N); label("$6/5$", (B + C)/2, dir(0)); label("$6/5$", (A + D)/2, W); label("$2$", (A + B)/2, S); | [] |
597 | A regular octagon $ABCDEFGH$ has an area of one square unit. What is the area of the rectangle $ABEF$?
$\textbf{(A)}\ 1-\frac{\sqrt2}{2}\qquad\textbf{(B)}\ \frac{\sqrt2}{4}\qquad\textbf{(C)}\ \sqrt2-1\qquad\textbf{(D)}\ \frac{1}2\qquad\textbf{(E)}\ \frac{1+\sqrt2}{4}$ | 2003 AMC 10B Problem 23 | Here is a less complicated way than that of the user below. If you draw a line segment from each vertex to the center of the octagon and draw the rectangle ABEF (in red), you can see that $2$ of the triangles (in blue) share the same base and height with $\dfrac{1}{2}$ the rectangle. Therefore, the rectangle's area is the same as $2\cdot2$ of the $8$ triangles, and is $\boxed{\textbf{(D)}\ \frac{1}{2}}$ the area of the octagon. | // Block 1
unitsize(1cm);
defaultpen(linewidth(.8pt)+fontsize(8pt));
pair C=dir(22.5), B=dir(67.5), A=dir(112.5), H=dir(157.5), G=dir(202.5), F=dir(247.5), E=dir(292.5), D=dir(337.5);
draw(A--B--C--D--E--F--G--H--cycle);
label("$A$",A,NNW);
label("$B$",B,NNE);
label("$C$",C,ENE);
label("$D$",D,ESE);
label("$E$",E,SSE);
label("$F$",F,SSW);
label("$G$",G,WSW);
label("$H$",H,WNW);
draw(A--E--F--B--C--G--H--D);
draw(A--E--F--B--A,blue);
draw(A--F--E--B--A,red);
// Block 2
unitsize(1cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); pair C=dir(22.5), B=dir(67.5), A=dir(112.5), H=dir(157.5), G=dir(202.5), F=dir(247.5), E=dir(292.5), D=dir(337.5); draw(A--B--C--D--E--F--G--H--cycle); label("$A$",A,NNW); label("$B$",B,NNE); label("$C$",C,ENE); label("$D$",D,ESE); label("$E$",E,SSE); label("$F$",F,SSW); label("$G$",G,WSW); label("$H$",H,WNW); draw(A--E--F--B--C--G--H--D); draw(A--E--F--B--A,blue); draw(A--F--E--B--A,red); | [] |
597 | A regular octagon $ABCDEFGH$ has an area of one square unit. What is the area of the rectangle $ABEF$?
$\textbf{(A)}\ 1-\frac{\sqrt2}{2}\qquad\textbf{(B)}\ \frac{\sqrt2}{4}\qquad\textbf{(C)}\ \sqrt2-1\qquad\textbf{(D)}\ \frac{1}2\qquad\textbf{(E)}\ \frac{1+\sqrt2}{4}$ | 2003 AMC 10B Problem 23 | Drawing lines $AD$, $BG$, $CF$, and $EH$, we can see that the octagon is comprised of $1$ square, $4$ rectangles, and $4$ triangles. The triangles each are $45-45-90$ triangles, and since their diagonal is length $x$, each of their sides is $\frac{\sqrt{2}}{2}x$. The area of the entire figure is, likewise, $x^2$ (the square)$+4x^2\frac{\sqrt{2}}{2}$ (the 4 rectangles)$+2\cdot(\frac{\sqrt{2}}{2}x)^2$ (the triangles), which simplifies to $2x^2 + 2\sqrt{2}x^2$. The area of $ABEF$ is just $x(x+\frac{2\sqrt{2}}{2}x)$, or $x^2$ + $x^2\sqrt{2}$, which we can see is the area of $\frac{ABCDEFGH}{2} = \boxed{\textbf{(D)}\ \frac{1}{2}}$ the area of the octagon. | // Block 1
unitsize(1cm);
defaultpen(linewidth(.8pt)+fontsize(8pt));
pair C=dir(22.5), B=dir(67.5), A=dir(112.5), H=dir(157.5), G=dir(202.5), F=dir(247.5), E=dir(292.5), D=dir(337.5);
draw(A--B--C--D--E--F--G--H--cycle);
label("$A$",A,NNW);
label("$B$",B,NNE);
label("$C$",C,ENE);
label("$D$",D,ESE);
label("$E$",E,SSE);
label("$F$",F,SSW);
label("$G$",G,WSW);
label("$H$",H,WNW);
draw(A--D--E--H--G--B--C--F);
// Block 2
unitsize(1cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); pair C=dir(22.5), B=dir(67.5), A=dir(112.5), H=dir(157.5), G=dir(202.5), F=dir(247.5), E=dir(292.5), D=dir(337.5); draw(A--B--C--D--E--F--G--H--cycle); label("$A$",A,NNW); label("$B$",B,NNE); label("$C$",C,ENE); label("$D$",D,ESE); label("$E$",E,SSE); label("$F$",F,SSW); label("$G$",G,WSW); label("$H$",H,WNW); draw(A--D--E--H--G--B--C--F); | [] |
597 | A regular octagon $ABCDEFGH$ has an area of one square unit. What is the area of the rectangle $ABEF$?
$\textbf{(A)}\ 1-\frac{\sqrt2}{2}\qquad\textbf{(B)}\ \frac{\sqrt2}{4}\qquad\textbf{(C)}\ \sqrt2-1\qquad\textbf{(D)}\ \frac{1}2\qquad\textbf{(E)}\ \frac{1+\sqrt2}{4}$ | 2003 AMC 10B Problem 23 | First, we are going to divide the diagram. Now we need to find the ratio of the area of the rectangle to the area of the trapezoid.
The area of a trapezoid is $\frac{b_1 + b_2}{2}h$
Note that the trapezoid is made out of 2 45-45-90 triangles and a rectangle, and that $AH=FG=1$. By realizing that, the area of the trapezoid is $(\frac{2+\sqrt{2}}{2})(\frac{\sqrt{2}}{2}$). To make this product easier, note there is two trapezoids, so the new product is now this, $(\frac{2+\sqrt{2}}{2})(\sqrt{2}) = \sqrt{2} + 1$
Notice how the rectangle has side lengths $\sqrt{2}+1$ and $1$, so it's area is also $\sqrt{2}+1$.
The ratio of the area of rectangle $ABEF$ to the two trapezoids is $1:1$, meaning they share half the area of the octagon. Since the area of the octagon is 1, $\therefore$ the area of the rectangle is $\boxed{\textbf{(D)}\ \frac{1}{2}}$.
~ghfhgvghj10 | unitsize(1cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); pair C=dir(22.5), B=dir(67.5), A=dir(112.5), H=dir(157.5), G=dir(202.5), F=dir(247.5), E=dir(292.5), D=dir(337.5); draw(A--B--C--D--E--F--G--H--cycle); label("$A$",A,NNW); label("$B$",B,NNE); label("$C$",C,ENE); label("$D$",D,ESE); label("$E$",E,SSE); label("$F$",F,SSW); label("$G$",G,WSW); label("$H$",H,WNW);draw(A--F);draw(B--E);draw(D--G);draw(C--H); | [] |
598 | In rectangle $ABCD, AB=5$ and $BC=3$. Points $F$ and $G$ are on $\overline{CD}$ so that $DF=1$ and $GC=2$. Lines $AF$ and $BG$ intersect at $E$. Find the area of $\triangle AEB$.
$\textbf{(A) } 10 \qquad\textbf{(B) } \frac{21}{2} \qquad\textbf{(C) } 12 \qquad\textbf{(D) } \frac{25}{2} \qquad\textbf{(E) } 15$ | 2003 AMC 12B Problem 14 | Since $\Delta{ABE}\sim{\Delta{FGE}}$ then $[AFGB]\sim{[FXYG]}$, where $X$ and $Y$ are ponts on $EF$ and $EG$ respectivley which make the areas similar. This process can be done over and over again multiple times by the ratio of $\frac{FG}{AB}=\frac{2}{5}$, or something like this
\[[AEB]=[AFGB]+[FXYZ]+...\]\[[AEB]=[AFGB]+\frac{2}{5}[AFGB]+(\frac{2}{5})^2[AFGB]+...\]we have to find the ratio of the areas when the sides have shrunk by length $\frac{2}{5}l$
Let $[AFGB]'$ be the area of the shape whose length is $\frac{2}{5}l$
\[[AFGB]'=[ADCB]-[ADF]-[BCG]\]\[[AFGB]'=12/5-6/25-12/25\]\[[AFGB]'=42/25\]Now comparing the ratios of $[AFGB]'$ to $[AFGB]$ we get
\[\frac{[AFGB]'}{[AFGB]}=\frac{42}{25}/\frac{21}{2}\implies \frac{[AFGB]'}{[AFGB]}=\frac{4}{25}\]By applying an infinite summation
\[[AEB]=\sum_{n=0}^{\infty} \frac{21}{2}\cdot{(\frac{4}{25})^n}\]\[S=\frac{a_1}{1-r}\]\[\boxed{[AEB]=\frac{25}{2}}\] | unitsize(0.6 cm); pair A, B, C, D, E, F, G; A = (0,0); B = (5,0); C = (5,3); D = (0,3); F = (1,3); G = (3,3); E = extension(A,F,B,G); draw(A--B--C--D--cycle); draw(A--E--B); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, NE); label("$D$", D, NW); label("$E$", E, N); label("$F$", F, SE); label("$G$", G, SW); label("$2/5$", (D + F)/2, N); label("$4/5$", (G + C)/2, N); label("$6/5$", (B + C)/2, dir(0)); label("$6/5$", (A + D)/2, W); label("$2$", (A + B)/2, S); | [] |
599 | A regular octagon $ABCDEFGH$ has an area of one square unit. What is the area of the rectangle $ABEF$?
$\textbf{(A)}\ 1-\frac{\sqrt2}{2}\qquad\textbf{(B)}\ \frac{\sqrt2}{4}\qquad\textbf{(C)}\ \sqrt2-1\qquad\textbf{(D)}\ \frac{1}2\qquad\textbf{(E)}\ \frac{1+\sqrt2}{4}$ | 2003 AMC 12B Problem 15 | Here is a less complicated way than that of the user below. If you draw a line segment from each vertex to the center of the octagon and draw the rectangle ABEF (in red), you can see that $2$ of the triangles (in blue) share the same base and height with $\dfrac{1}{2}$ the rectangle. Therefore, the rectangle's area is the same as $2\cdot2$ of the $8$ triangles, and is $\boxed{\textbf{(D)}\ \frac{1}{2}}$ the area of the octagon. | unitsize(1cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); pair C=dir(22.5), B=dir(67.5), A=dir(112.5), H=dir(157.5), G=dir(202.5), F=dir(247.5), E=dir(292.5), D=dir(337.5); draw(A--B--C--D--E--F--G--H--cycle); label("$A$",A,NNW); label("$B$",B,NNE); label("$C$",C,ENE); label("$D$",D,ESE); label("$E$",E,SSE); label("$F$",F,SSW); label("$G$",G,WSW); label("$H$",H,WNW); draw(A--E--F--B--C--G--H--D); draw(A--E--F--B--A,blue); draw(A--F--E--B--A,red); | [] |
599 | A regular octagon $ABCDEFGH$ has an area of one square unit. What is the area of the rectangle $ABEF$?
$\textbf{(A)}\ 1-\frac{\sqrt2}{2}\qquad\textbf{(B)}\ \frac{\sqrt2}{4}\qquad\textbf{(C)}\ \sqrt2-1\qquad\textbf{(D)}\ \frac{1}2\qquad\textbf{(E)}\ \frac{1+\sqrt2}{4}$ | 2003 AMC 12B Problem 15 | Drawing lines $AD$, $BG$, $CF$, and $EH$, we can see that the octagon is comprised of $1$ square, $4$ rectangles, and $4$ triangles. The triangles each are $45-45-90$ triangles, and since their diagonal is length $x$, each of their sides is $\frac{\sqrt{2}}{2}x$. The area of the entire figure is, likewise, $x^2$ (the square)$+4x^2\frac{\sqrt{2}}{2}$ (the 4 rectangles)$+2\cdot(\frac{\sqrt{2}}{2}x)^2$ (the triangles), which simplifies to $2x^2 + 2\sqrt{2}x^2$. The area of $ABEF$ is just $x(x+\frac{2\sqrt{2}}{2}x)$, or $x^2$ + $x^2\sqrt{2}$, which we can see is the area of $\frac{ABCDEFGH}{2} = \boxed{\textbf{(D)}\ \frac{1}{2}}$ the area of the octagon. | unitsize(1cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); pair C=dir(22.5), B=dir(67.5), A=dir(112.5), H=dir(157.5), G=dir(202.5), F=dir(247.5), E=dir(292.5), D=dir(337.5); draw(A--B--C--D--E--F--G--H--cycle); label("$A$",A,NNW); label("$B$",B,NNE); label("$C$",C,ENE); label("$D$",D,ESE); label("$E$",E,SSE); label("$F$",F,SSW); label("$G$",G,WSW); label("$H$",H,WNW); draw(A--D--E--H--G--B--C--F); | [] |
599 | A regular octagon $ABCDEFGH$ has an area of one square unit. What is the area of the rectangle $ABEF$?
$\textbf{(A)}\ 1-\frac{\sqrt2}{2}\qquad\textbf{(B)}\ \frac{\sqrt2}{4}\qquad\textbf{(C)}\ \sqrt2-1\qquad\textbf{(D)}\ \frac{1}2\qquad\textbf{(E)}\ \frac{1+\sqrt2}{4}$ | 2003 AMC 12B Problem 15 | First, we are going to divide the diagram. Now we need to find the ratio of the area of the rectangle to the area of the trapezoid.
The area of a trapezoid is $\frac{b_1 + b_2}{2}h$
Note that the trapezoid is made out of 2 45-45-90 triangles and a rectangle, and that $AH=FG=1$. By realizing that, the area of the trapezoid is $(\frac{2+\sqrt{2}}{2})(\frac{\sqrt{2}}{2}$). To make this product easier, note there is two trapezoids, so the new product is now this, $(\frac{2+\sqrt{2}}{2})(\sqrt{2}) = \sqrt{2} + 1$
Notice how the rectangle has side lengths $\sqrt{2}+1$ and $1$, so it's area is also $\sqrt{2}+1$.
The ratio of the area of rectangle $ABEF$ to the two trapezoids is $1:1$, meaning they share half the area of the octagon. Since the area of the octagon is 1, $\therefore$ the area of the rectangle is $\boxed{\textbf{(D)}\ \frac{1}{2}}$.
~ghfhgvghj10 | unitsize(1cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); pair C=dir(22.5), B=dir(67.5), A=dir(112.5), H=dir(157.5), G=dir(202.5), F=dir(247.5), E=dir(292.5), D=dir(337.5); draw(A--B--C--D--E--F--G--H--cycle); label("$A$",A,NNW); label("$B$",B,NNE); label("$C$",C,ENE); label("$D$",D,ESE); label("$E$",E,SSE); label("$F$",F,SSW); label("$G$",G,WSW); label("$H$",H,WNW);draw(A--F);draw(B--E);draw(D--G);draw(C--H); | [] |
600 | Three semicircles of radius $1$ are constructed on diameter $\overline{AB}$ of a semicircle of radius $2$. The centers of the small semicircles divide $\overline{AB}$ into four line segments of equal length, as shown. What is the area of the shaded region that lies within the large semicircle but outside the smaller semicircles?
$\textbf{(A) } \pi - \sqrt{3} \qquad\textbf{(B) } \pi - \sqrt{2} \qquad\textbf{(C) } \frac{\pi + \sqrt{2}}{2} \qquad\textbf{(D) } \frac{\pi +\sqrt{3}}{2} \qquad\textbf{(E) } \frac{7}{6}\pi - \frac{\sqrt{3}}{2}$ | 2003 AMC 12B Problem 16 | By drawing four lines from the intersect of the semicircles to their centers, we have split the white region into $\frac{5}{6}$ of a circle with radius $1$ and two equilateral triangles with side length $1$.
This gives the area of the white region as $\frac{5}{6}\pi+\frac{2\cdot\sqrt3}{4}=\frac{5}{6}\pi+\frac{\sqrt3}{2}$.
The area of the shaded region is the area of the white region subtracted from the area of the large semicircle. This is equivalent to $2\pi-\left(\frac{5}{6}\pi+\frac{\sqrt3}{2}\right)=\frac{7}{6}\pi-\frac{\sqrt3}{2}$.
Thus the answer is $\boxed{\textbf{(E)}\ \frac{7}{6}\pi-\frac{\sqrt3}{2}}$.
Note
The reason why it is $\frac{5}{6}$ of a circle and why the triangles are equilateral are because, first, the radii are the same and they make up the equilateral triangles.
Secondly, the reason it is $\frac{5}{6}$ of a circle is because the middle sector has a degree of $180-2 \cdot 60 = 60$ and thus $\frac{60}{360}=\frac{1}{6}$ of a circle.
The other two have areas of $\frac{180-60}{360}=\frac{1}{3}$ of a triangle each.
Therefore, the total fraction of the circle(since they have the same radii) is $\frac{1}{6} + 2 \cdot \frac{1}{3} = \frac{1}{6} + \frac{4}{6} = \frac{5}{6}.$
~mathboy282 | import graph; unitsize(14mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); dashed=linetype("4 4"); dotfactor=3; pair A=(-2,0), B=(2,0); fill(Arc((0,0),2,0,180)--cycle,mediumgray); fill(Arc((-1,0),1,0,180)--cycle,white); fill(Arc((0,0),1,0,180)--cycle,white); fill(Arc((1,0),1,0,180)--cycle,white); draw(Arc((-1,0),1,60,180)); draw(Arc((0,0),1,0,60),dashed); draw(Arc((0,0),1,60,120)); draw(Arc((0,0),1,120,180),dashed); draw(Arc((1,0),1,0,120)); draw(Arc((0,0),2,0,180)--cycle); dot((0,0)); dot((-1,0)); dot((1,0)); draw((-2,-0.1)--(-2,-0.3),gray); draw((-1,-0.1)--(-1,-0.3),gray); draw((1,-0.1)--(1,-0.3),gray); draw((2,-0.1)--(2,-0.3),gray); label("$A$",A,W); label("$B$",B,E); label("1",(-1.5,-0.1),S); label("2",(0,-0.1),S); label("1",(1.5,-0.1),S); draw((1,0)--(0.5,0.866)); draw((0,0)--(0.5,0.866)); draw((-1,0)--(-0.5,0.866)); draw((0,0)--(-0.5,0.866)); | [] |
601 | Positive integers $a,b,$ and $c$ are chosen so that $a<b<c$, and the system of equations
$2x + y = 2003 \quad$ and $\quad y = |x-a| + |x-b| + |x-c|$
has exactly one solution. What is the minimum value of $c$?
$\mathrm{(A)}\ 668 \qquad\mathrm{(B)}\ 669 \qquad\mathrm{(C)}\ 1002 \qquad\mathrm{(D)}\ 2003 \qquad\mathrm{(E)}\ 2004$ | 2003 AMC 12B Problem 24 | Consider the graph of $f(x)=|x-a|+|x-b|+|x-c|$.
When $x<a$, the slope is $-3$.
When $a<x<b$, the slope is $-1$.
When $b<x<c$, the slope is $1$.
When $c<x$, the slope is $3$.
Setting $x=b$ gives $y=|b-a|+|b-b|+|b-c|=c-a$, so $(b,c-a)$ is a point on $f(x)$. In fact, it is the minimum of $f(x)$ considering the slope of lines to the left and right of $(b,c-a)$. Thus, graphing this will produce a figure that looks like a cup:
From the graph, it is clear that $f(x)$ and $2x+y=2003$ have one intersection point if and only if they intersect at $x=a$. Since the line where $a<x<b$ has slope $-1$, the positive difference in $y$-coordinates from $x=a$ to $x=b$ must be $b-a$. Together with the fact that $(b,c-a)$ is on $f(x)$, we see that $P=(a,c-a+b-a)$. Since this point is on $x=a$, the only intersection point with $2x+y=2003$, we have $2 \cdot a+(b+c-2a)=2003 \implies b+c=2003$. As $c>b$, the smallest possible value of $c$ occurs when $b=1001$ and $c=1002$. This is indeed a solution as $a=1000$ puts $P$ on $y=2003-2x$, and thus the answer is $\boxed{\mathrm{(C)}\ 1002}$.
This indeed works for the two right segments of slope $1$ and $3$. We already know that the minimum is achieved between slopes $-3$ and $-1$ with $b+c=2003$:
\[2003-2x=-a-b+c+x\longrightarrow 3x\ne a+b-c+2003 \{b<x<c\}\rightarrow (3b,3c)\ne a+2b\rightarrow b>a\text{ (true)}\]
\[2003-2x=-a-b-c+3x\longrightarrow 5x\ne a+b+c+2003 \{x>c\}\rightarrow (5c,+5\infty)\ne a+2b+2c\rightarrow 3c>a+2b\text{ (true)}\]
Indeed, within the restricted domain of $x$ in each segment, these inequalities prove to be unequal everywhere. So $y=2003-2x$ is strictly below $y=|x-a|+|x-b|+|x-c|$ at these domains. | // Block 1
import graph;
size(100);
draw((0,6)--(3,0));
xaxis(0,8.5);
yaxis(0,10);
real f(real x)
{
return -3(x-2)+5;
}
real f2(real x)
{
return -1(x-2)+5;
}
real f3(real x)
{
return 1(x-4)+3;
}
real f4(real x)
{
return 3(x-7)+6;
}
draw(graph(f,0,2));
draw(graph(f2,2,4));
draw(graph(f3,4,7));
draw(graph(f4,7,8.5));
draw((2,-0.25)--(2,0.25));
label("a",(2,0),N);
draw((4,-0.25)--(4,0.25));
label("b",(4,0),N);
draw((7,-0.25)--(7,0.25));
label("c",(7,0),N);
dot((2,5));
label("P",(1.9,5.2),E);
// Block 2
import graph; size(100); draw((0,6)--(3,0)); xaxis(0,8.5); yaxis(0,10); real f(real x) { return -3(x-2)+5; } real f2(real x) { return -1(x-2)+5; } real f3(real x) { return 1(x-4)+3; } real f4(real x) { return 3(x-7)+6; } draw(graph(f,0,2)); draw(graph(f2,2,4)); draw(graph(f3,4,7)); draw(graph(f4,7,8.5)); draw((2,-0.25)--(2,0.25)); label("a",(2,0),N); draw((4,-0.25)--(4,0.25)); label("b",(4,0),N); draw((7,-0.25)--(7,0.25)); label("c",(7,0),N); dot((2,5)); label("P",(1.9,5.2),E); | [] |
602 | Three points are chosen randomly and independently on a circle. What is the probability that all three pairwise distances between the points are less than the radius of the circle?
$\mathrm{(A)}\ \dfrac{1}{36} \qquad\mathrm{(B)}\ \dfrac{1}{24} \qquad\mathrm{(C)}\ \dfrac{1}{18} \qquad\mathrm{(D)}\ \dfrac{1}{12} \qquad\mathrm{(E)}\ \dfrac{1}{9}$ | 2003 AMC 12B Problem 25 | We will use geometric probability.
The first point can be anywhere. Each point must be $\frac{\pi}{3}$ or less away from each other.
Define $x$ be the amount of radians away the second point is from the first. We limit $x$ to be in the interval $[-\pi, \pi]$. Define $y$ be the amount of radians away the third point is from the first. We limit $y$ to be in the interval $[-\pi, \pi]$. Now, we can deduct that: \[|x| \le \frac{\pi}{3},\] \[|y| \le \frac{\pi}{3},\] and \[|x-y| \le \frac{\pi}{3}.\] We now begin plotting these on the coordinate grid.
First note that the area of the points that $x$ and $y$ can be (ignoring the conditions) is $(2\pi)^2$ (remember what we restricted $x$ and $y$ to).
Now, we can graph the equations we deduced on the coordinate grid. That should look like this:
The area of the shaded region can be calculated in many ways. Eventually, you will find that the area is $\frac{\pi^2}{3}$.
Thus, the probability is $\frac{\frac{\pi^2}{3}}{(2\pi)^2} = \frac{1}{12}$, or $\boxed{\text{(D)}}$.
~superagh | // Block 1
fill((40, 0)--(40, 40)--(0, 40)--(-40, 0)--(-40, -40)--(0, -40)--cycle, green);
draw((-50, 0)--(50, 0));
draw((0, -50)--(0, 50));
draw((-50, -10)--(10, 50),red);
draw((-10, -50)--(50, 10),red);
draw((-40, 40)--(40, 40),red);
draw((-40, -40)--(40, -40),red);
draw((40, 40)--(40, -40),red);
draw((-40, 40)--(-40, -40),red);
label("$\frac{\pi}{3}$", (40, 0), SE);
label("$\frac{\pi}{3}$", (-40, 0), SW);
label("$\frac{\pi}{3}$", (0, 40), NE);
label("$\frac{\pi}{3}$", (0, -40), NW);
// Block 2
fill((40, 0)--(40, 40)--(0, 40)--(-40, 0)--(-40, -40)--(0, -40)--cycle, green); draw((-50, 0)--(50, 0)); draw((0, -50)--(0, 50)); draw((-50, -10)--(10, 50),red); draw((-10, -50)--(50, 10),red); draw((-40, 40)--(40, 40),red); draw((-40, -40)--(40, -40),red); draw((40, 40)--(40, -40),red); draw((-40, 40)--(-40, -40),red); label("$\frac{\pi}{3}$", (40, 0), SE); label("$\frac{\pi}{3}$", (-40, 0), SW); label("$\frac{\pi}{3}$", (0, 40), NE); label("$\frac{\pi}{3}$", (0, -40), NW); | [] |
603 | Square $ABCD$ has side length $2$. A semicircle with diameter $\overline{AB}$ is constructed inside the square, and the tangent to the semicircle from $C$ intersects side $\overline{AD}$ at $E$. What is the length of $\overline{CE}$?
$\mathrm{(A) \ } \frac{2+\sqrt{5}}{2} \qquad \mathrm{(B) \ } \sqrt{5} \qquad \mathrm{(C) \ } \sqrt{6} \qquad \mathrm{(D) \ } \frac{5}{2} \qquad \mathrm{(E) \ } 5-\sqrt{5}$ | 2004 AMC 10A Problem 22 | Solution 1
Let the point of tangency be $F$. By the Two Tangent Theorem $BC = FC = 2$ and $AE = EF = x$. Thus $DE = 2-x$. The Pythagorean Theorem on $\triangle CDE$ yields
\begin{align*} DE^2 + CD^2 &= CE^2\\ (2-x)^2 + 2^2 &= (2+x)^2\\ x^2 - 4x + 8 &= x^2 + 4x + 4\\ x &= \frac{1}{2}\end{align*}
Hence $CE = FC + x = \frac{5}{2} \Rightarrow\boxed{\mathrm{(D)}\ \frac{5}{2}}$.
Solution 2
Call the point of tangency point $F$ and the midpoint of $AB$ as $G$. $CF=2$ by Tangent Theorem. Notice that $\angle EGF=\frac{180-2\cdot\angle CGF}{2}=90-\angle CGF$. Thus, $\angle EGF=\angle FCG$ and $\tan EGF=\tan FCG=\frac{1}{2}$. Solving $EF=\frac{1}{2}$. Adding, the answer is $\frac{5}{2}$.
Solution 3
[Image: images/amc/2004_AMC_10A_Problem_22_0.png]
Clearly, $EA = EF = BG$. Thus, the sides of right triangle $CDE$ are in arithmetic progression. Thus it is similar to the triangle $3 - 4 - 5$ and since $DC = 2$, $CE = \frac{5}{2} \Rightarrow\boxed{\mathrm{(D)}\ \frac{5}{2}}$.
Solution 4
Let us call the midpoint of side $AB$, point $G$. Since the semicircle has radius 1, we can do the Pythagorean theorem on sides $GB, BC, GC$. We get $GC=\sqrt{5}$. We then know that $CF=2$ by Pythagorean theorem. Then by connecting $EG$, we get similar triangles $EFG$ and $GFC$. Solving the ratios, we get $x=\frac{1}{2}$, so the answer is $\frac{5}{2} \Rightarrow\boxed{\mathrm{(D)}\ \frac{5}{2}}$.
Alternatively, we could apply the Pythagorean theorem on triangle $CDE$ to get the equation \[(2-x)^2+2^2=(x+2)^2\] which would give us that $x=\frac{1}{2}.$ Adding up $CF$ and $EF,$ we get $2+\frac{1}{2}=\boxed{\mathrm{(D)}\ \frac{5}{2}}$ again. Note that we know $DE=2-x$ because $\triangle EFG\cong \triangle EAG$ by $\text{HL.}$
~Alternate solution by Tinsel
Solution 5
Using the diagram as drawn in Solution 4, let the total area of square $ABCD$ be divided into the triangles $DCE$, $EAG$, $CGB$, and $EGC$. Let x be the length of AE. Thus, the area of each triangle can be determined as follows:
\[DCE = \frac{DC\cdot{DE}}{2} = \frac{2\cdot(2-x)}{2} = 2-x\]
\[EAG= \frac{AE\cdot{AG}}{2} = \frac{1\cdot{x}}{2} = \frac{x}{2}\]
\[CGB = \frac{GB\cdot{CB}}{2} = \frac{1\cdot(2)}{2} = 1\]
\[EGC= \frac{EG\cdot{GC}}{2} = \frac{\sqrt{5x^2 + 5}}{2}\] (the length of CE is calculated with the Pythagorean Theorem, lines GE and
CE are perpendicular by definition of tangent)
Adding up the areas and equating to the area of the total square \[(2 \cdot 2=4)\], we get
\[x = \frac{1}{2}\]
So, \[CE = 2 + \frac{1}{2} = 5/2\].
~Typo Fix by doulai1
Video Solution
https://youtu.be/pM0zICtH6Lg
Education, the Study of Everything | // Block 1
size(150); defaultpen(fontsize(10)); pair A=(0,0), B=(2,0), C=(2,2), D=(0,2), E=(0,1/2), F=E+(C-E)/abs(C-E)/2; draw(A--B--C--D--cycle);draw(C--E); draw(Arc((1,0),1,0,180));draw((A+B)/2--F); label("$A$",A,(-1,-1)); label("$B$",B,( 1,-1)); label("$C$",C,( 1, 1)); label("$D$",D,(-1, 1)); label("$E$",E,(-1, 0)); label("$F$",F,( 0, 1)); label("$x$",(A+E)/2,(-1, 0)); label("$x$",(E+F)/2,( 0, 1)); label("$2$",(F+C)/2,( 0, 1)); label("$2$",(D+C)/2,( 0, 1)); label("$2$",(B+C)/2,( 1, 0)); label("$2-x$",(D+E)/2,(-1, 0));
// Block 2
size(150); defaultpen(fontsize(10)); pair A=(0,0), B=(2,0), C=(2,2), D=(0,2), E=(0,1/2), F=E+(C-E)/abs(C-E)/2, G=(1,0); draw(A--B--C--D--cycle);draw(C--E); draw(Arc((1,0),1,0,180));draw((A+B)/2--F); label("$A$",A,(-1,-1)); label("$B$",B,( 1,-1)); label("$C$",C,( 1, 1)); label("$D$",D,(-1, 1)); label("$E$",E,(-1, 0)); label("$F$",F,( 0, 1)); label("$x$",(A+E)/2,(-1, 0)); label("$x$",(E+F)/2,( 0, 1)); label("$2$",(F+C)/2,( 0, 1)); label("$2$",(D+C)/2,( 0, 1)); label("$2$",(B+C)/2,( 1, 0)); label("$2-x$",(D+E)/2,(-1, 0)); label("$G$",G,(0,-1)); dot(G); draw(G--C); label("$\sqrt{5}$",(G+C)/2,(-1,0)); | ["https://artofproblemsolving.com/wiki/images/e/ed/2004_AMC12A-18.png"] |
604 | Circles $A, B$ and $C$ are externally tangent to each other, and internally tangent to circle $D$. Circles $B$ and $C$ are congruent. Circle $A$ has radius $1$ and passes through the center of $D$. What is the radius of circle $B$?
$\text{(A) } \frac23 \qquad \text{(B) } \frac {\sqrt3}{2} \qquad \text{(C) } \frac78 \qquad \text{(D) } \frac89 \qquad \text{(E) } \frac {1 + \sqrt3}{3}$ | 2004 AMC 10A Problem 23 | Let $O_{i}$ be the center of circle $i$ for all $i \in \{A,B,C,D\}$ and let $E$ be the tangent point of $B,C$. Since the radius of $D$ is the diameter of $A$, the radius of $D$ is $2$.
Let the radius of $B,C$ be $r$ and let $O_{D}E = x$. If we connect $O_{A},O_{B},O_{C}$, we get an isosceles triangle with lengths $1 + r, 2r$.
Then right triangle $O_{D}O_{B}O_{E}$ has legs $r, x$ and hypotenuse $2-r$. Solving for $x$, we get $x^2 = (2-r)^2 - r^2 \Longrightarrow x = \sqrt{4-4r}$.
Also, right triangle $O_{A}O_{B}O_{E}$ has legs $r, 1+x$, and hypotenuse $1+r$. Solving,
\begin{eqnarray*} r^2 + (1+\sqrt{4-4r})^2 &=& (1+r)^2\\ 1+4-4r+2\sqrt{4-4r}&=& 2r + 1\\ 1-r &=& \left(\frac{6r-4}{4}\right)^2\\ \frac{9}{4}r^2-2r&=& 0\\ r &=& \frac 89 \end{eqnarray*}
So the answer is $\boxed{\mathrm{(D)}\ \frac{8}{9}}$. | import graph; size(400); defaultpen(fontsize(10)); pair OA=(-1,0),OB=(2/3,8/9),OC=(2/3,-8/9),OD=(0,0),E=(2/3,0); real t = 2.5; pair OA1=(-2+2*t,0),OB1=(4/3+2*t,16/9),OC1=(4/3+2*t,-16/9),OD1=(0+2*t,0),E1=(4/3+2*t,0); draw(Circle(OD,2)); draw(Circle(OA,1)); draw(Circle(OB,8/9)); draw(Circle(OC,8/9)); draw(OA--OB--OC--cycle); draw(OD--OB--OB+(OB-OD)*4/5); draw(OA--E); label("$O_{A}$",OA,(-1,0)); label("$O_{B}$",OB,(-1,1)); label("$O_{C}$",OC,(-1,-1)); label("$O_{D}$",OD,(-1,-1)); label("$E$",E,(0.5,-1)); label("$r$",OB+(OB-OD)*2/5,(-0.5,1)); label("$r$",(1*OA+3*OB)/4,(-0.5,1)); dot(OA^^OB^^OC^^OD^^E); draw(OA1--OB1--OC1--cycle); draw(OD1--OB1); draw(OA1--E1); label("$O_{A}$",OA1,(-1,0)); label("$O_{B}$",OB1,(1,1)); label("$O_{C}$",OC1,(1,-1)); label("$O_{D}$",OD1,(0,-1)); label("$E$",E1,(1,0)); label("$1+r$",(OA1+OB1)/2,(-0.5,1)); label("$r$",(E1+OB1)/2,(1,0)); label("$r$",(E1+OC1)/2,(1,0)); label("$2-r$",(OB1+OD1)/2,(-1,0)); label("$1$",(OA1+OD1)/2,(0,-1)); label("$x$",(E1+OD1)/2,(0,-1)); dot(OA1^^OB1^^OC1^^OD1^^E1); | [] |
604 | Circles $A, B$ and $C$ are externally tangent to each other, and internally tangent to circle $D$. Circles $B$ and $C$ are congruent. Circle $A$ has radius $1$ and passes through the center of $D$. What is the radius of circle $B$?
$\text{(A) } \frac23 \qquad \text{(B) } \frac {\sqrt3}{2} \qquad \text{(C) } \frac78 \qquad \text{(D) } \frac89 \qquad \text{(E) } \frac {1 + \sqrt3}{3}$ | 2004 AMC 10A Problem 23 | Note that $BD= 2-r$ since $D$ is the center of the larger circle of radius $2$. Using the Pythagorean Theorem on $\triangle BDE$,
\begin{align*} r^2 + h^2 &= (2-r)^2 \\ r^2 + h^2 &= 4 - 4r + r^2 \\ h^2 &= 4 - 4r \\ h &= 2\sqrt{1-r} \end{align*}
Now using the Pythagorean Theorem on $\triangle BAE$,
\begin{align*} r^2 + (h+1)^2 &= (r+1)^2 \\ r^2 + h^2 + 2h + 1 &= r^2 + 2r + 1 \\ h^2 + 2h &= 2r \end{align*}
Substituting $h$,
\begin{align*} (4-4r) + 4\sqrt{1-r} &= 2r \\ 4\sqrt{1-r} &= 6r - 4 \\ 16-16r &= 36r^2 - 48r + 16 \\ 0 &= 36r^2 - 32r \\ r &= \frac{32}{36} = \frac{8}{9} \Longrightarrow \qquad \textbf{(D)} \end{align*} | unitsize(15mm); pair A=(0,1),B=(-8/9,-2/3),C=(8/9,-2/3),D=(0,0), E=(0,-2/3); draw(Circle(D,2)); draw(Circle(A,1)); draw(Circle(B,8/9)); draw(Circle(C,8/9)); draw(A--B--C--A); draw(B--D--C); draw(A--E); dot(A);dot(B);dot(C);dot(D);dot(E); label("\(D\)", D,NW); label("\(A\)", A,N); label("\(B\)", B,W); label("\(C\)", C,E); label("\(E\)", E,SE); label("\(1\)",(-.4,.7)); label("\(1\)",(0,0.5),W); label("\(r\)", (-.8,-.1)); label("\(r\)", (-4/9,-2/3),S); label("\(h\)", (0,-1/3), W); | [] |
605 | Square $ABCD$ has side length $2$. A semicircle with diameter $\overline{AB}$ is constructed inside the square, and the tangent to the semicircle from $C$ intersects side $\overline{AD}$ at $E$. What is the length of $\overline{CE}$?
$\mathrm{(A) \ } \frac{2+\sqrt{5}}{2} \qquad \mathrm{(B) \ } \sqrt{5} \qquad \mathrm{(C) \ } \sqrt{6} \qquad \mathrm{(D) \ } \frac{5}{2} \qquad \mathrm{(E) \ } 5-\sqrt{5}$ | 2004 AMC 12A Problem 18 | Solution 1
Let the point of tangency be $F$. By the Two Tangent Theorem $BC = FC = 2$ and $AE = EF = x$. Thus $DE = 2-x$. The Pythagorean Theorem on $\triangle CDE$ yields
\begin{align*} DE^2 + CD^2 &= CE^2\\ (2-x)^2 + 2^2 &= (2+x)^2\\ x^2 - 4x + 8 &= x^2 + 4x + 4\\ x &= \frac{1}{2}\end{align*}
Hence $CE = FC + x = \frac{5}{2} \Rightarrow\boxed{\mathrm{(D)}\ \frac{5}{2}}$.
Solution 2
Call the point of tangency point $F$ and the midpoint of $AB$ as $G$. $CF=2$ by Tangent Theorem. Notice that $\angle EGF=\frac{180-2\cdot\angle CGF}{2}=90-\angle CGF$. Thus, $\angle EGF=\angle FCG$ and $\tan EGF=\tan FCG=\frac{1}{2}$. Solving $EF=\frac{1}{2}$. Adding, the answer is $\frac{5}{2}$.
Solution 3
[Image: images/amc/2004_AMC_12A_Problem_18_0.png]
Clearly, $EA = EF = BG$. Thus, the sides of right triangle $CDE$ are in arithmetic progression. Thus it is similar to the triangle $3 - 4 - 5$ and since $DC = 2$, $CE = \frac{5}{2} \Rightarrow\boxed{\mathrm{(D)}\ \frac{5}{2}}$.
Solution 4
Let us call the midpoint of side $AB$, point $G$. Since the semicircle has radius 1, we can do the Pythagorean theorem on sides $GB, BC, GC$. We get $GC=\sqrt{5}$. We then know that $CF=2$ by Pythagorean theorem. Then by connecting $EG$, we get similar triangles $EFG$ and $GFC$. Solving the ratios, we get $x=\frac{1}{2}$, so the answer is $\frac{5}{2} \Rightarrow\boxed{\mathrm{(D)}\ \frac{5}{2}}$.
Alternatively, we could apply the Pythagorean theorem on triangle $CDE$ to get the equation \[(2-x)^2+2^2=(x+2)^2\] which would give us that $x=\frac{1}{2}.$ Adding up $CF$ and $EF,$ we get $2+\frac{1}{2}=\boxed{\mathrm{(D)}\ \frac{5}{2}}$ again. Note that we know $DE=2-x$ because $\triangle EFG\cong \triangle EAG$ by $\text{HL.}$
~Alternate solution by Tinsel
Solution 5
Using the diagram as drawn in Solution 4, let the total area of square $ABCD$ be divided into the triangles $DCE$, $EAG$, $CGB$, and $EGC$. Let x be the length of AE. Thus, the area of each triangle can be determined as follows:
\[DCE = \frac{DC\cdot{DE}}{2} = \frac{2\cdot(2-x)}{2} = 2-x\]
\[EAG= \frac{AE\cdot{AG}}{2} = \frac{1\cdot{x}}{2} = \frac{x}{2}\]
\[CGB = \frac{GB\cdot{CB}}{2} = \frac{1\cdot(2)}{2} = 1\]
\[EGC= \frac{EG\cdot{GC}}{2} = \frac{\sqrt{5x^2 + 5}}{2}\] (the length of CE is calculated with the Pythagorean Theorem, lines GE and
CE are perpendicular by definition of tangent)
Adding up the areas and equating to the area of the total square \[(2 \cdot 2=4)\], we get
\[x = \frac{1}{2}\]
So, \[CE = 2 + \frac{1}{2} = 5/2\].
~Typo Fix by doulai1
Video Solution
https://youtu.be/pM0zICtH6Lg
Education, the Study of Everything | // Block 1
size(150); defaultpen(fontsize(10)); pair A=(0,0), B=(2,0), C=(2,2), D=(0,2), E=(0,1/2), F=E+(C-E)/abs(C-E)/2; draw(A--B--C--D--cycle);draw(C--E); draw(Arc((1,0),1,0,180));draw((A+B)/2--F); label("$A$",A,(-1,-1)); label("$B$",B,( 1,-1)); label("$C$",C,( 1, 1)); label("$D$",D,(-1, 1)); label("$E$",E,(-1, 0)); label("$F$",F,( 0, 1)); label("$x$",(A+E)/2,(-1, 0)); label("$x$",(E+F)/2,( 0, 1)); label("$2$",(F+C)/2,( 0, 1)); label("$2$",(D+C)/2,( 0, 1)); label("$2$",(B+C)/2,( 1, 0)); label("$2-x$",(D+E)/2,(-1, 0));
// Block 2
size(150); defaultpen(fontsize(10)); pair A=(0,0), B=(2,0), C=(2,2), D=(0,2), E=(0,1/2), F=E+(C-E)/abs(C-E)/2, G=(1,0); draw(A--B--C--D--cycle);draw(C--E); draw(Arc((1,0),1,0,180));draw((A+B)/2--F); label("$A$",A,(-1,-1)); label("$B$",B,( 1,-1)); label("$C$",C,( 1, 1)); label("$D$",D,(-1, 1)); label("$E$",E,(-1, 0)); label("$F$",F,( 0, 1)); label("$x$",(A+E)/2,(-1, 0)); label("$x$",(E+F)/2,( 0, 1)); label("$2$",(F+C)/2,( 0, 1)); label("$2$",(D+C)/2,( 0, 1)); label("$2$",(B+C)/2,( 1, 0)); label("$2-x$",(D+E)/2,(-1, 0)); label("$G$",G,(0,-1)); dot(G); draw(G--C); label("$\sqrt{5}$",(G+C)/2,(-1,0)); | ["https://artofproblemsolving.com/wiki/images/e/ed/2004_AMC12A-18.png"] |
606 | Circles $A, B$ and $C$ are externally tangent to each other, and internally tangent to circle $D$. Circles $B$ and $C$ are congruent. Circle $A$ has radius $1$ and passes through the center of $D$. What is the radius of circle $B$?
$\text{(A) } \frac23 \qquad \text{(B) } \frac {\sqrt3}{2} \qquad \text{(C) } \frac78 \qquad \text{(D) } \frac89 \qquad \text{(E) } \frac {1 + \sqrt3}{3}$ | 2004 AMC 12A Problem 19 | Let $O_{i}$ be the center of circle $i$ for all $i \in \{A,B,C,D\}$ and let $E$ be the tangent point of $B,C$. Since the radius of $D$ is the diameter of $A$, the radius of $D$ is $2$.
Let the radius of $B,C$ be $r$ and let $O_{D}E = x$. If we connect $O_{A},O_{B},O_{C}$, we get an isosceles triangle with lengths $1 + r, 2r$.
Then right triangle $O_{D}O_{B}O_{E}$ has legs $r, x$ and hypotenuse $2-r$. Solving for $x$, we get $x^2 = (2-r)^2 - r^2 \Longrightarrow x = \sqrt{4-4r}$.
Also, right triangle $O_{A}O_{B}O_{E}$ has legs $r, 1+x$, and hypotenuse $1+r$. Solving,
\begin{eqnarray*} r^2 + (1+\sqrt{4-4r})^2 &=& (1+r)^2\\ 1+4-4r+2\sqrt{4-4r}&=& 2r + 1\\ 1-r &=& \left(\frac{6r-4}{4}\right)^2\\ \frac{9}{4}r^2-2r&=& 0\\ r &=& \frac 89 \end{eqnarray*}
So the answer is $\boxed{\mathrm{(D)}\ \frac{8}{9}}$. | // Block 1
import graph;
size(400);
defaultpen(fontsize(10));
pair OA=(-1,0),OB=(2/3,8/9),OC=(2/3,-8/9),OD=(0,0),E=(2/3,0);
real t = 2.5;
pair OA1=(-2+2*t,0),OB1=(4/3+2*t,16/9),OC1=(4/3+2*t,-16/9),OD1=(0+2*t,0),E1=(4/3+2*t,0);
draw(Circle(OD,2));
draw(Circle(OA,1));
draw(Circle(OB,8/9));
draw(Circle(OC,8/9));
draw(OA--OB--OC--cycle);
draw(OD--OB--OB+(OB-OD)*4/5);
draw(OA--E);
label("$O_{A}$",OA,(-1,0));
label("$O_{B}$",OB,(-1,1));
label("$O_{C}$",OC,(-1,-1));
label("$O_{D}$",OD,(-1,-1));
label("$E$",E,(0.5,-1));
label("$r$",OB+(OB-OD)*2/5,(-0.5,1));
label("$r$",(1*OA+3*OB)/4,(-0.5,1));
dot(OA^^OB^^OC^^OD^^E);
draw(OA1--OB1--OC1--cycle);
draw(OD1--OB1);
draw(OA1--E1);
label("$O_{A}$",OA1,(-1,0));
label("$O_{B}$",OB1,(1,1));
label("$O_{C}$",OC1,(1,-1));
label("$O_{D}$",OD1,(0,-1));
label("$E$",E1,(1,0));
label("$1+r$",(OA1+OB1)/2,(-0.5,1));
label("$r$",(E1+OB1)/2,(1,0));
label("$r$",(E1+OC1)/2,(1,0));
label("$2-r$",(OB1+OD1)/2,(-1,0));
label("$1$",(OA1+OD1)/2,(0,-1));
label("$x$",(E1+OD1)/2,(0,-1));
dot(OA1^^OB1^^OC1^^OD1^^E1);
// Block 2
import graph; size(400); defaultpen(fontsize(10)); pair OA=(-1,0),OB=(2/3,8/9),OC=(2/3,-8/9),OD=(0,0),E=(2/3,0); real t = 2.5; pair OA1=(-2+2*t,0),OB1=(4/3+2*t,16/9),OC1=(4/3+2*t,-16/9),OD1=(0+2*t,0),E1=(4/3+2*t,0); draw(Circle(OD,2)); draw(Circle(OA,1)); draw(Circle(OB,8/9)); draw(Circle(OC,8/9)); draw(OA--OB--OC--cycle); draw(OD--OB--OB+(OB-OD)*4/5); draw(OA--E); label("$O_{A}$",OA,(-1,0)); label("$O_{B}$",OB,(-1,1)); label("$O_{C}$",OC,(-1,-1)); label("$O_{D}$",OD,(-1,-1)); label("$E$",E,(0.5,-1)); label("$r$",OB+(OB-OD)*2/5,(-0.5,1)); label("$r$",(1*OA+3*OB)/4,(-0.5,1)); dot(OA^^OB^^OC^^OD^^E); draw(OA1--OB1--OC1--cycle); draw(OD1--OB1); draw(OA1--E1); label("$O_{A}$",OA1,(-1,0)); label("$O_{B}$",OB1,(1,1)); label("$O_{C}$",OC1,(1,-1)); label("$O_{D}$",OD1,(0,-1)); label("$E$",E1,(1,0)); label("$1+r$",(OA1+OB1)/2,(-0.5,1)); label("$r$",(E1+OB1)/2,(1,0)); label("$r$",(E1+OC1)/2,(1,0)); label("$2-r$",(OB1+OD1)/2,(-1,0)); label("$1$",(OA1+OD1)/2,(0,-1)); label("$x$",(E1+OD1)/2,(0,-1)); dot(OA1^^OB1^^OC1^^OD1^^E1); | [] |
606 | Circles $A, B$ and $C$ are externally tangent to each other, and internally tangent to circle $D$. Circles $B$ and $C$ are congruent. Circle $A$ has radius $1$ and passes through the center of $D$. What is the radius of circle $B$?
$\text{(A) } \frac23 \qquad \text{(B) } \frac {\sqrt3}{2} \qquad \text{(C) } \frac78 \qquad \text{(D) } \frac89 \qquad \text{(E) } \frac {1 + \sqrt3}{3}$ | 2004 AMC 12A Problem 19 | Note that $BD= 2-r$ since $D$ is the center of the larger circle of radius $2$. Using the Pythagorean Theorem on $\triangle BDE$,
\begin{align*} r^2 + h^2 &= (2-r)^2 \\ r^2 + h^2 &= 4 - 4r + r^2 \\ h^2 &= 4 - 4r \\ h &= 2\sqrt{1-r} \end{align*}
Now using the Pythagorean Theorem on $\triangle BAE$,
\begin{align*} r^2 + (h+1)^2 &= (r+1)^2 \\ r^2 + h^2 + 2h + 1 &= r^2 + 2r + 1 \\ h^2 + 2h &= 2r \end{align*}
Substituting $h$,
\begin{align*} (4-4r) + 4\sqrt{1-r} &= 2r \\ 4\sqrt{1-r} &= 6r - 4 \\ 16-16r &= 36r^2 - 48r + 16 \\ 0 &= 36r^2 - 32r \\ r &= \frac{32}{36} = \frac{8}{9} \Longrightarrow \qquad \textbf{(D)} \end{align*} | // Block 1
unitsize(15mm);
pair A=(0,1),B=(-8/9,-2/3),C=(8/9,-2/3),D=(0,0), E=(0,-2/3);
draw(Circle(D,2));
draw(Circle(A,1));
draw(Circle(B,8/9));
draw(Circle(C,8/9));
draw(A--B--C--A);
draw(B--D--C);
draw(A--E);
dot(A);dot(B);dot(C);dot(D);dot(E);
label("\(D\)", D,NW);
label("\(A\)", A,N);
label("\(B\)", B,W);
label("\(C\)", C,E);
label("\(E\)", E,SE);
label("\(1\)",(-.4,.7));
label("\(1\)",(0,0.5),W);
label("\(r\)", (-.8,-.1));
label("\(r\)", (-4/9,-2/3),S);
label("\(h\)", (0,-1/3), W);
// Block 2
unitsize(15mm); pair A=(0,1),B=(-8/9,-2/3),C=(8/9,-2/3),D=(0,0), E=(0,-2/3); draw(Circle(D,2)); draw(Circle(A,1)); draw(Circle(B,8/9)); draw(Circle(C,8/9)); draw(A--B--C--A); draw(B--D--C); draw(A--E); dot(A);dot(B);dot(C);dot(D);dot(E); label("\(D\)", D,NW); label("\(A\)", A,N); label("\(B\)", B,W); label("\(C\)", C,E); label("\(E\)", E,SE); label("\(1\)",(-.4,.7)); label("\(1\)",(0,0.5),W); label("\(r\)", (-.8,-.1)); label("\(r\)", (-4/9,-2/3),S); label("\(h\)", (0,-1/3), W); | [] |
607 | Select numbers $a$ and $b$ between $0$ and $1$ independently and at random, and let $c$ be their sum. Let $A, B$ and $C$ be the results when $a, b$ and $c$, respectively, are rounded to the nearest integer. What is the probability that $A + B = C$?
$\text {(A)}\ \frac14 \qquad \text {(B)}\ \frac13 \qquad \text {(C)}\ \frac12 \qquad \text {(D)}\ \frac23 \qquad \text {(E)}\ \frac34$ | 2004 AMC 12A Problem 20 | The conditions under which $A+B=C$ are as follows.
(i) If $a+b< 1/2$, then $A=B=C=0$.
(ii) If $a\geq 1/2$ and $b<1/2$, then $B=0$ and $A=C=1$.
(iii) If $a<1/2$ and $b\geq 1/2$, then $A=0$ and $B=C=1$.
(iv) If $a+b\geq 3/2$, then $A=B=1$ and $C=2$.
These conditions correspond to the shaded regions of the graph shown. The combined area of those regions is 3/4, and the area of the entire square is 1, so the requested probability is $\boxed{3/4}$. | // Block 1
unitsize(2cm);
draw((1.1,0)--(0,0)--(0,1.1),linewidth(1));
fill((0,0)--(1,0)--(1,1)--(0,1)--cycle,gray(0.7));
fill((0.5,0)--(0.5,0.5)--(0,0.5)--cycle,white);
fill((0.5,0.5)--(1,0.5)--(0.5,1)--cycle,white);
label("$a$",(1.1,0),E);
label("$b$",(0,1.1),N);
label("1",(1,0),S);
label("1",(0,1),W);
label("0",(0,0),SW);
// Block 2
unitsize(2cm); draw((1.1,0)--(0,0)--(0,1.1),linewidth(1)); fill((0,0)--(1,0)--(1,1)--(0,1)--cycle,gray(0.7)); fill((0.5,0)--(0.5,0.5)--(0,0.5)--cycle,white); fill((0.5,0.5)--(1,0.5)--(0.5,1)--cycle,white); label("$a$",(1.1,0),E); label("$b$",(0,1.1),N); label("1",(1,0),S); label("1",(0,1),W); label("0",(0,0),SW); | [] |
608 | A plane contains points $A$ and $B$ with $AB = 1$. Let $S$ be the union of all disks of radius $1$ in the plane that cover $\overline{AB}$. What is the area of $S$?
$\textbf {(A) } 2\pi + \sqrt3 \qquad \textbf {(B) } \frac {8\pi}{3} \qquad \textbf {(C) } 3\pi - \frac {\sqrt3}{2} \qquad \textbf {(D) } \frac {10\pi}{3} - \sqrt3 \qquad \textbf {(E) }4\pi - 2\sqrt3$ | 2004 AMC 12A Problem 24 | As the red circles move about segment $AB$, they cover the area we are looking for.
On the left side, the circle must move around pivoted on $B$.
On the right side, the circle must move pivoted on $A$
However, at the top and bottom, the circle must lie on both A and B, giving us our upper and lower bounds.
This egg-like shape is $S$.
The area of the region can be found by dividing it into several sectors, namely
\begin{align*} A &= 2(\mathrm{Blue\ Sector}) + 2(\mathrm{Red\ Sector}) - 2(\mathrm{Equilateral\ Triangle}) \\ A &= 2\left(\frac{120^\circ}{360^\circ} \cdot \pi (2)^2\right) + 2\left(\frac{60^\circ}{360^\circ} \cdot \pi (1)^2\right) - 2\left(\frac{(1)^2\sqrt{3}}{4}\right) \\ A &= \frac{8\pi}{3} + \frac{\pi}{3} - \frac{\sqrt{3}}{2} \\ A &= 3\pi - \frac{\sqrt{3}}{2} \Longrightarrow \textbf {(C)}\end{align*} | // Block 1
pair A=(-.5,0), B=(.5,0), C=(0,3**(.5)/2), D=(0,-3**(.5)/2);
draw(arc(A,2,-60,60),blue);
draw(arc(B,2,120,240),blue);
draw(circle(C,1),red);
draw(A--(.5,3^.5));
draw(B--(-.5,3^.5));
draw(A--(.5,-3^.5));
draw(B--(-.5,-3^.5));
draw(A--B);
dot(A);dot(B);dot(C);dot(D);
label("\(1\)",(0,0),N);
label("\(1\)",A/2+D/2,W);
label("\(1\)",A/2+C/2,W);
label("\(1\)",B/2+D/2,E);
label("\(1\)",B/2+C/2,E);
label("\(1\)",A/2+3D/2,W);
label("\(1\)",A/2+3C/2,W);
label("\(1\)",B/2+3D/2,E);
label("\(1\)",B/2+3C/2,E);
label("\(A\)",A,W);
label("\(B\)",B,E);
label("\(C\)",C,W);
label("\(D\)",D,E);
// Block 2
pair A=(-.5,0), B=(.5,0), C=(0,3**(.5)/2), D=(0,-3**(.5)/2);
draw(arc(A,2,-60,60),blue);
draw(arc(B,2,120,240),blue);
draw(arc(C,1,60,120),red);
draw(arc(D,1,-120,-60),red);
draw(A--(.5,3^.5));
draw(B--(-.5,3^.5));
draw(A--(.5,-3^.5));
draw(B--(-.5,-3^.5));
draw(A--B);
dot(A);dot(B);dot(C);dot(D);
label("\(A\)",A,W);
label("\(B\)",B,E);
label("\(C\)",C,W);
label("\(D\)",D,E);
label("\(1\)",(0,0),N);
label("\(1\)",A/2+D/2,W);
label("\(1\)",A/2+C/2,W);
label("\(1\)",B/2+D/2,E);
label("\(1\)",B/2+C/2,E);
label("\(1\)",A/2+3D/2,W);
label("\(1\)",A/2+3C/2,W);
label("\(1\)",B/2+3D/2,E);
label("\(1\)",B/2+3C/2,E);
// Block 3
pair A=(-.5,0), B=(.5,0), C=(0,3**(.5)/2), D=(0,-3**(.5)/2); draw(arc(A,2,-60,60),blue); draw(arc(B,2,120,240),blue); draw(circle(C,1),red); draw(A--(.5,3^.5)); draw(B--(-.5,3^.5)); draw(A--(.5,-3^.5)); draw(B--(-.5,-3^.5)); draw(A--B); dot(A);dot(B);dot(C);dot(D); label("\(1\)",(0,0),N); label("\(1\)",A/2+D/2,W); label("\(1\)",A/2+C/2,W); label("\(1\)",B/2+D/2,E); label("\(1\)",B/2+C/2,E); label("\(1\)",A/2+3D/2,W); label("\(1\)",A/2+3C/2,W); label("\(1\)",B/2+3D/2,E); label("\(1\)",B/2+3C/2,E); label("\(A\)",A,W); label("\(B\)",B,E); label("\(C\)",C,W); label("\(D\)",D,E);
// Block 4
pair A=(-.5,0), B=(.5,0), C=(0,3**(.5)/2), D=(0,-3**(.5)/2); draw(arc(A,2,-60,60),blue); draw(arc(B,2,120,240),blue); draw(arc(C,1,60,120),red); draw(arc(D,1,-120,-60),red); draw(A--(.5,3^.5)); draw(B--(-.5,3^.5)); draw(A--(.5,-3^.5)); draw(B--(-.5,-3^.5)); draw(A--B); dot(A);dot(B);dot(C);dot(D); label("\(A\)",A,W); label("\(B\)",B,E); label("\(C\)",C,W); label("\(D\)",D,E); label("\(1\)",(0,0),N); label("\(1\)",A/2+D/2,W); label("\(1\)",A/2+C/2,W); label("\(1\)",B/2+D/2,E); label("\(1\)",B/2+C/2,E); label("\(1\)",A/2+3D/2,W); label("\(1\)",A/2+3C/2,W); label("\(1\)",B/2+3D/2,E); label("\(1\)",B/2+3C/2,E); | [] |
609 | A square has sides of length $10$, and a circle centered at one of its vertices has radius $10$. What is the area of the union of the regions enclosed by the square and the circle?
$\mathrm{(A)\ }200+25\pi\quad\mathrm{(B)\ }100+75\pi\quad\mathrm{(C)\ }75+100\pi\quad\mathrm{(D)\ }100+100\pi\quad\mathrm{(E)\ }100+125\pi$ | 2004 AMC 10B Problem 9 | The area of the circle is $S_{\bigcirc}=100\pi$; the area of the square is $S_{\square}=100$.
Exactly $\frac{1}{4}$ of the circle lies inside the square. Thus the total area is $\dfrac34 S_{\bigcirc}+S_{\square}=\boxed{\mathrm{(B)\ }100+75\pi}$. | Draw(Circle((0,0),10)); Draw((0,0)--(10,0)--(10,10)--(0,10)--(0,0)); label("$10$",(5,0),S); label("$10$",(0,5),W); dot((0,0)); | [] |
610 | Three circles of radius $1$ are externally tangent to each other and internally tangent to a larger circle. What is the radius of the large circle?
$\mathrm{(A) \ } \frac{2 + \sqrt{6}}{3} \qquad \mathrm{(B) \ } 2 \qquad \mathrm{(C) \ } \frac{2 + 3\sqrt{2}}{2} \qquad \mathrm{(D) \ } \frac{3 + 2\sqrt{3}}{3} \qquad \mathrm{(E) \ } \frac{3 + \sqrt{3}}{2}$ | 2004 AMC 10B Problem 16 | The situation is shown in the picture below. The radius we seek is $SD = AD + AS$. Clearly $AD=1$. The point $S$ is the center of the equilateral triangle $ABC$, thus $AS$ is $2/3$ of the altitude of this triangle. We get that $AS = \frac23 \cdot \sqrt 3$. Therefore the radius we seek is
$1 + \frac23 \cdot \sqrt 3 = \boxed{\mathrm{(D)\ }\frac{3+2\sqrt{3}}3}$. | // Block 1
unitsize(2cm);
pair A=(0,0), B=dir(0)*2, C=dir(60)*2;
draw(circle(A,1));
draw(circle(B,1));
draw(circle(C,1));
dot(A); dot(B); dot(C);
draw(A--B--C--cycle);
pair D=A+dir(210), E=B+dir(-30), F=C+dir(90);
draw(circumcircle(D,E,F));
dot(D); dot(E); dot(F);
pair S=(A+B+C)/3;
dot(S);
draw(S--D); draw(S--E); draw(S--F);
label("$S$",S,S);
label("$A$",A,SE);
label("$D$",D,SW);
label("$B$",B,NE);
label("$C$",C,ENE);
// Block 2
unitsize(2cm); pair A=(0,0), B=dir(0)*2, C=dir(60)*2; draw(circle(A,1)); draw(circle(B,1)); draw(circle(C,1)); dot(A); dot(B); dot(C); draw(A--B--C--cycle); pair D=A+dir(210), E=B+dir(-30), F=C+dir(90); draw(circumcircle(D,E,F)); dot(D); dot(E); dot(F); pair S=(A+B+C)/3; dot(S); draw(S--D); draw(S--E); draw(S--F); label("$S$",S,S); label("$A$",A,SE); label("$D$",D,SW); label("$B$",B,NE); label("$C$",C,ENE); | [] |
611 | In the right triangle $\triangle ACE$, we have $AC=12$, $CE=16$, and $EA=20$. Points $B$, $D$, and $F$ are located on $AC$, $CE$, and $EA$, respectively, so that $AB=3$, $CD=4$, and $EF=5$. What is the ratio of the area of $\triangle DBF$ to that of $\triangle ACE$?
$\mathrm{(A) \ } \frac{1}{4} \qquad \mathrm{(B) \ } \frac{9}{25} \qquad \mathrm{(C) \ } \frac{3}{8} \qquad \mathrm{(D) \ } \frac{11}{25} \qquad \mathrm{(E) \ } \frac{7}{16}$ | 2004 AMC 10B Problem 18 | First of all, note that $\frac{AB}{AC} = \frac{CD}{CE} = \frac{EF}{EA} = \frac 14$, and therefore
$\frac{BC}{AC} = \frac{DE}{CE} = \frac{FA}{EA} = \frac 34$.
Draw the height from $F$ onto $AB$ as in the picture below:
Now consider the area of $\triangle ABF$. Clearly the triangles $\triangle AFG$ and $\triangle AEC$ are similar, as they have all angles equal. Their ratio is $\frac {AF}{AE} = \frac 34$, hence $FG = \frac 34 \cdot CE$.
Now the area $S_{ABF}$ of $\triangle ABF$ can be computed as $S_{ABF} = \frac 12 \cdot AB \cdot FG$ = $\frac 12 \cdot \left( \frac 14 \cdot AC \right) \cdot \left( \frac 34 \cdot EC \right) = \frac 14 \cdot \frac 34 \cdot S_{ACE}$.
Similarly we can find that $S_{BCD} = S_{DEF} = \frac 3{16}\cdot S_{ACE}$ as well.
Hence $S_{BDF} = S_{ACE} - 3\cdot\left( \frac 3{16} \cdot S_{ACE} \right) = \frac 7{16} \cdot S_{ACE}$, and the answer is $\frac{S_{BDF}}{S_{ACE}} = \boxed{\frac 7{16}}$. | // Block 1
unitsize(0.5cm);
defaultpen(0.8);
pair C=(0,0), A=(0,12), E=(20,0);
draw(A--C--E--cycle);
pair B=A + 3*(C-A)/length(C-A);
pair D=C + 4*(E-C)/length(E-C);
pair F=E + 5*(A-E)/length(A-E);
draw(B--D--F--cycle);
label("$A$",A,N);
label("$B$",B,W);
label("$C$",C,SW);
label("$D$",D,S);
label("$E$",E,SE);
label("$F$",F,NE);
label("$3$",A--B,W);
label("$9$",0.5*C + 0.5*B,4*W);
label("$4$",C--D,S);
label("$12$",D--E,S);
label("$5$",E--F,NE);
label("$15$",F--A,NE);
pair G = intersectionpoint(F -- (F-(100,0)), A--C);
draw(F--G, dashed);
label("$G$",G,W);
// Block 2
unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,12), E=(20,0); draw(A--C--E--cycle); pair B=A + 3*(C-A)/length(C-A); pair D=C + 4*(E-C)/length(E-C); pair F=E + 5*(A-E)/length(A-E); draw(B--D--F--cycle); label("$A$",A,N); label("$B$",B,W); label("$C$",C,SW); label("$D$",D,S); label("$E$",E,SE); label("$F$",F,NE); label("$3$",A--B,W); label("$9$",0.5*C + 0.5*B,4*W); label("$4$",C--D,S); label("$12$",D--E,S); label("$5$",E--F,NE); label("$15$",F--A,NE); pair G = intersectionpoint(F -- (F-(100,0)), A--C); draw(F--G, dashed); label("$G$",G,W); | [] |
612 | In $\triangle ABC$ points $D$ and $E$ lie on $BC$ and $AC$, respectively. If $AD$ and $BE$ intersect at $T$ so that $\frac{AT}{DT}=3$ and $\frac{BT}{ET}=4$, what is $\frac{CD}{BD}$?
$\mathrm{(A) \ } \frac{1}{8} \qquad \mathrm{(B) \ } \frac{2}{9} \qquad \mathrm{(C) \ } \frac{3}{10} \qquad \mathrm{(D) \ } \frac{4}{11} \qquad \mathrm{(E) \ } \frac{5}{12}$ | 2004 AMC 10B Problem 20 | Affine transformations preserve ratios of distances, and for any pair of triangles, there is an affine transformation that maps the first one onto the second one. This is why the answer is the same for any $\triangle ABC$, and we just need to compute it for any single triangle.
We can choose the points $A=(-3,0)$, $B=(0,4)$, and $D=(1,0)$. This way we will have $T=(0,0)$, and $E=(0,-1)$. The situation is
shown in the picture below:
The point $C$ is the intersection of the lines $BD$ and $AE$. The points on the first line have the form $(t,4-4t)$, the points on the second line have the form $(t,-1-t/3)$. Solving for $t$ we get $t=15/11$, hence $C=(15/11,-16/11)$.
The ratio $CD/BD$ can now be computed simply by observing the $x$ coordinates of $B$, $C$, and $D$:
\[\frac{CD}{BD} = \frac{15/11 - 1}{1 - 0} = \boxed{\textbf{(D)}\frac 4{11}}\] | // Block 1
unitsize(1cm);
defaultpen(0.8);
pair A=(-3,0), B=(0,4), C=(15/11,-16/11), D=(1,0), E=(0,-1);
draw(A--B--C--cycle);
draw(A--D);
draw(B--E);
pair T=intersectionpoint(A--D,B--E);
label("$A$",A,SW);
label("$B$",B,N);
label("$C$",C,SE);
label("$D$",D,NE);
label("$E$",E,S);
label("$T$",T,NW);
label("$3$",A--T,N);
label("$4$",B--T,W);
label("$1$",D--T,N);
label("$1$",E--T,W);
// Block 2
unitsize(1cm); defaultpen(0.8); pair A=(-3,0), B=(0,4), C=(15/11,-16/11), D=(1,0), E=(0,-1); draw(A--B--C--cycle); draw(A--D); draw(B--E); pair T=intersectionpoint(A--D,B--E); label("$A$",A,SW); label("$B$",B,N); label("$C$",C,SE); label("$D$",D,NE); label("$E$",E,S); label("$T$",T,NW); label("$3$",A--T,N); label("$4$",B--T,W); label("$1$",D--T,N); label("$1$",E--T,W); | [] |
613 | A triangle with sides of 5, 12, and 13 has both an inscribed and a circumscribed circle. What is the distance between the centers of those circles?
$\mathrm{(A) \ } \frac{3\sqrt{5}}{2} \qquad \mathrm{(B) \ } \frac{7}{2} \qquad \mathrm{(C) \ } \sqrt{15} \qquad \mathrm{(D) \ } \frac{\sqrt{65}}{2} \qquad \mathrm{(E) \ } \frac{9}{2}$ | 2004 AMC 10B Problem 22 | This is a right triangle. Pick a coordinate system so that the right angle is at $(0,0)$ and the other two vertices are at $(12,0)$ and $(0,5)$.
As this is a right triangle, the center of the circumcircle is in the middle of the hypotenuse, at $(6,2.5)$.
The radius $r$ of the inscribed circle can be computed using the well-known identity $\frac{rP}2=S$, where $S$ is the area of the triangle and $P$ its perimeter. In our case, $S=\frac{5\cdot 12}{2}=30$ and $P=5+12+13=30$. Thus, $r=2$. As the inscribed circle touches both legs, its center must be at $(r,r)=(2,2)$.
The distance of these two points is then $\sqrt{ (6-2)^2 + (2.5-2)^2 } = \sqrt{16.25} = \sqrt{\frac{65}4} = \boxed{\frac{\sqrt{65}}2}$. | // Block 1
import geometry;
unitsize(0.6 cm);
pair A, B, C, D, E, F, I, O;
A = (5^2/13,5*12/13);
B = (0,0);
C = (13,0);
I = incenter(A,B,C);
D = (I + reflect(B,C)*(I))/2;
E = (I + reflect(C,A)*(I))/2;
F = (I + reflect(A,B)*(I))/2;
O = (B + C)/2;
draw(A--B--C--cycle);
draw(incircle(A,B,C));
draw(I--D);
draw(I--E);
draw(I--F);
draw(I--O);
label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);
dot("$D$", D, S);
dot("$E$", E, NE);
dot("$F$", F, NW);
dot("$I$", I, N);
dot("$O$", O, S);
// Block 2
import geometry; unitsize(0.6 cm); pair A, B, C, D, E, F, I, O; A = (5^2/13,5*12/13); B = (0,0); C = (13,0); I = incenter(A,B,C); D = (I + reflect(B,C)*(I))/2; E = (I + reflect(C,A)*(I))/2; F = (I + reflect(A,B)*(I))/2; O = (B + C)/2; draw(A--B--C--cycle); draw(incircle(A,B,C)); draw(I--D); draw(I--E); draw(I--F); draw(I--O); label("$A$", A, N); label("$B$", B, SW); label("$C$", C, SE); dot("$D$", D, S); dot("$E$", E, NE); dot("$F$", F, NW); dot("$I$", I, N); dot("$O$", O, S); | [] |
613 | A triangle with sides of 5, 12, and 13 has both an inscribed and a circumscribed circle. What is the distance between the centers of those circles?
$\mathrm{(A) \ } \frac{3\sqrt{5}}{2} \qquad \mathrm{(B) \ } \frac{7}{2} \qquad \mathrm{(C) \ } \sqrt{15} \qquad \mathrm{(D) \ } \frac{\sqrt{65}}{2} \qquad \mathrm{(E) \ } \frac{9}{2}$ | 2004 AMC 10B Problem 22 | Construct $\triangle{ABC}$ such that $AB=5$, $AC=12$, and $BC=13$. Since this is a pythagorean triple, $\angle{A}=90$. By a property of circumcircles and right triangles, the circumcenter, $G$, lies on the midpoint of $\overline{BC}$, so $BG=\frac{13}{2}$. Turning to the incircle, we find that the inradius is $2$, using the formula $A=rs$, where $A$ is the area of the triangle, $r$ is the inradius, and $s$ is the semiperimeter. We then denote the incenter $I$, along with the points of tangency $D$, $E$, and $F$. Because $\angle{IDA}=\angle{IEA}=90$ by a property of tangency, $\angle{EID}=90$, and so $IDAE$ is a square. Then, since $IE=2$, $AD=2$. As $AB=5$, $BD=3$, and because $\triangle{BID}\cong\triangle{BIF}$ by HL, $BD=BF=3$. Therefore, $FG=\frac{7}{2}$. Because $IF=2$, pythagorean theorem gives $IG=\boxed{\frac{\sqrt{65}}{2}}$ | // Block 1
size(15cm);
draw((0,0)--(0,5), linewidth(2));
draw((0,0)--(12,0), linewidth(2));
draw((12,0)--(0,5), linewidth(2));
draw((2,0)--(2,2), linewidth(2));
draw((2,2)--(2.770565628817799,3.8455976546592505), linewidth(2));
draw((2,2)--(6.023716614191289,2.4901180774202962), linewidth(2));
draw((2,2)--(0,5), linewidth(2));
draw((2,2)--(12,0), linewidth(2));
draw((0,2)--(2,2), linewidth(2));
label("$A$", (0.14164244785738467,0.25966489738837517), NE);
label("$B$", (0.14164244785738467,5.311734560831129), NE);
label("$C$", (12.120449134133493,0.3232129434694161), NE);
label("$D$", (0.14164244785738467,2.324976395022205), NE);
label("$E$", (2.111631876369636,0.3232129434694161), NE);
label("$F$", (2.9059824523826405,4.167869731372392), NE);
label("$G$", (6.146932802515699,2.801586740630012), NE);
label("$I$", (2.1, 2.1), NE);
// Block 2
size(15cm); draw((0,0)--(0,5), linewidth(2)); draw((0,0)--(12,0), linewidth(2)); draw((12,0)--(0,5), linewidth(2)); draw((2,0)--(2,2), linewidth(2)); draw((2,2)--(2.770565628817799,3.8455976546592505), linewidth(2)); draw((2,2)--(6.023716614191289,2.4901180774202962), linewidth(2)); draw((2,2)--(0,5), linewidth(2)); draw((2,2)--(12,0), linewidth(2)); draw((0,2)--(2,2), linewidth(2)); label("$A$", (0.14164244785738467,0.25966489738837517), NE); label("$B$", (0.14164244785738467,5.311734560831129), NE); label("$C$", (12.120449134133493,0.3232129434694161), NE); label("$D$", (0.14164244785738467,2.324976395022205), NE); label("$E$", (2.111631876369636,0.3232129434694161), NE); label("$F$", (2.9059824523826405,4.167869731372392), NE); label("$G$", (6.146932802515699,2.801586740630012), NE); label("$I$", (2.1, 2.1), NE); | [] |
614 | In triangle $ABC$ we have $AB=7$, $AC=8$, $BC=9$. Point $D$ is on the circumscribed circle of the triangle so that $AD$ bisects angle $BAC$. What is the value of $\frac{AD}{CD}$?
$\text{(A) } \dfrac{9}{8} \qquad \text{(B) } \dfrac{5}{3} \qquad \text{(C) } 2 \qquad \text{(D) } \dfrac{17}{7} \qquad \text{(E) } \dfrac{5}{2}$ | 2004 AMC 10B Problem 24 | Let $E = \overline{BC}\cap \overline{AD}$. Observe that $\angle ABC \cong \angle ADC$ because they both subtend arc $\overarc{AC}.$
Furthermore, $\angle BAE \cong \angle EAC$ because $\overline{AE}$ is an angle bisector, so $\triangle ABE \sim \triangle ADC$ by $\text{AA}$ similarity. Then $\dfrac{AD}{AB} = \dfrac{CD}{BE}$. By the Angle Bisector Theorem, $\dfrac{7}{BE} = \dfrac{8}{CE}$, so $\dfrac{7}{BE} = \dfrac{8}{9-BE}$. This in turn gives $BE = \frac{21}{5}$. Plugging this into the similarity proportion gives:
$\dfrac{AD}{7} = \dfrac{CD}{\tfrac{21}{5}} \implies \dfrac{AD}{CD} = {\dfrac{5}{3}} = \boxed{\text{(B)}}$. | // Block 1
import graph;
import geometry;
import markers;
unitsize(0.5 cm);
pair A, B, C, D, E, I;
A = (11/3,8*sqrt(5)/3);
B = (0,0);
C = (9,0);
I = incenter(A,B,C);
D = intersectionpoint(I--(I + 2*(I - A)), circumcircle(A,B,C));
E = extension(A,D,B,C);
draw(A--B--C--cycle);
draw(circumcircle(A,B,C));
draw(D--A);
draw(D--B);
draw(D--C);
label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, S);
label("$E$", E, NE);
markangle(radius = 20,B, A, C, marker(markinterval(2,stickframe(1,2mm),true)));
markangle(radius = 20,B, C, D, marker(markinterval(1,stickframe(1,2mm),true)));
markangle(radius = 20,D, B, C, marker(markinterval(1,stickframe(1,2mm),true)));
markangle(radius = 20,C, B, A, marker(markinterval(1,stickframe(2,2mm),true)));
markangle(radius = 20,C, D, A, marker(markinterval(1,stickframe(2,2mm),true)));
// Block 2
import graph; import geometry; import markers; unitsize(0.5 cm); pair A, B, C, D, E, I; A = (11/3,8*sqrt(5)/3); B = (0,0); C = (9,0); I = incenter(A,B,C); D = intersectionpoint(I--(I + 2*(I - A)), circumcircle(A,B,C)); E = extension(A,D,B,C); draw(A--B--C--cycle); draw(circumcircle(A,B,C)); draw(D--A); draw(D--B); draw(D--C); label("$A$", A, N); label("$B$", B, SW); label("$C$", C, SE); label("$D$", D, S); label("$E$", E, NE); markangle(radius = 20,B, A, C, marker(markinterval(2,stickframe(1,2mm),true))); markangle(radius = 20,B, C, D, marker(markinterval(1,stickframe(1,2mm),true))); markangle(radius = 20,D, B, C, marker(markinterval(1,stickframe(1,2mm),true))); markangle(radius = 20,C, B, A, marker(markinterval(1,stickframe(2,2mm),true))); markangle(radius = 20,C, D, A, marker(markinterval(1,stickframe(2,2mm),true))); | [] |
615 | A circle of radius $1$ is internally tangent to two circles of radius $2$ at points $A$ and $B$, where $AB$ is a diameter of the smaller circle. What is the area of the region, shaded in the picture, that is outside the smaller circle and inside each of the two larger circles?
$\mathrm{(A) \ } \frac{5}{3} \pi - 3\sqrt 2 \qquad \mathrm{(B) \ } \frac{5}{3} \pi - 2\sqrt 3 \qquad \mathrm{(C) \ } \frac{8}{3} \pi - 3\sqrt 3 \qquad \mathrm{(D) \ } \frac{8}{3} \pi - 3\sqrt 2 \qquad \mathrm{(E) \ } \frac{8}{3} \pi - 2\sqrt 3$ | 2004 AMC 10B Problem 25 | The area of the small circle is $\pi$. We can add it to the shaded region, compute the area of the new region, and then subtract the area of the small circle from the result.
Let $C$ and $D$ be the intersections of the two large circles. Connect them to $A$ and $B$ to get the picture below:
We can see that the triangles $\triangle ABC$ and $\triangle ABD$ are both equilateral with side $2$.
Take a look at the lower circle. The angle $ABC$ is $60^\circ$, thus sector $ABC$ is $1/6$ of the circle. The same is true for sector $ABD$ of the lower circle, and sectors $CAB$ and $BAD$ of the upper circle.
If we now sum the areas of these four sectors, we will almost get the area of the new shaded region - except that each of the two equilateral triangles will be counted twice. The area of an equilateral triangle given its side, $s,$ is $\frac{\sqrt3}4s^2.$
Therefore, the area of the new shaded region is $4\cdot \left( \frac 16 \cdot \pi\cdot 2^2 \right) - 2 \cdot \left( 4 \cdot \frac{\sqrt3}4 \right) = \frac 83 \pi - 2\sqrt 3$. Lastly, we must subtract the area of the circle that we added earlier, $\pi$, and we get
$\left( \frac 83 \pi - 2\sqrt 3 \right) - \pi = \boxed{\mathrm{(B)\ } \frac 53 \pi - 2\sqrt 3 }$. | // Block 1
unitsize(1.5cm);
defaultpen(0.8);
pair O=(0,0), A=(0,1), B=(0,-1);
path bigc1 = Circle(A,2), bigc2 = Circle(B,2), smallc = Circle(O,1);
pair[] P = intersectionpoints(bigc1, bigc2);
filldraw( arc(A,P[0],P[1])--arc(B,P[1],P[0])--cycle, lightgray, black );
draw(bigc1);
draw(bigc2);
dot(A); dot(B); label("$A$",A,N); label("$B$",B,S);
/*
dot(O);
unfill(smallc);
draw(smallc);
draw( O--dir(30) );
draw( A--(A+2*dir(30)) );
draw( B--(B+2*dir(210)) );
label("$1$", O--dir(30), N );
label("$2$", A--(A+2*dir(30)), N );
label("$2$", B--(B+2*dir(210)), S );
*/
label("$C$",P[0],W);
label("$D$",P[1],E);
draw( P[0]--A--P[1] );
draw( P[0]--B--P[1] );
draw( A--B );
// Block 2
unitsize(1.5cm); defaultpen(0.8); pair O=(0,0), A=(0,1), B=(0,-1); path bigc1 = Circle(A,2), bigc2 = Circle(B,2), smallc = Circle(O,1); pair[] P = intersectionpoints(bigc1, bigc2); filldraw( arc(A,P[0],P[1])--arc(B,P[1],P[0])--cycle, lightgray, black ); draw(bigc1); draw(bigc2); dot(A); dot(B); label("$A$",A,N); label("$B$",B,S); /* dot(O); unfill(smallc); draw(smallc); draw( O--dir(30) ); draw( A--(A+2*dir(30)) ); draw( B--(B+2*dir(210)) ); label("$1$", O--dir(30), N ); label("$2$", A--(A+2*dir(30)), N ); label("$2$", B--(B+2*dir(210)), S ); */ label("$C$",P[0],W); label("$D$",P[1],E); draw( P[0]--A--P[1] ); draw( P[0]--B--P[1] ); draw( A--B ); | [] |
616 | A square has sides of length $10$, and a circle centered at one of its vertices has radius $10$. What is the area of the union of the regions enclosed by the square and the circle?
$\mathrm{(A)\ }200+25\pi\quad\mathrm{(B)\ }100+75\pi\quad\mathrm{(C)\ }75+100\pi\quad\mathrm{(D)\ }100+100\pi\quad\mathrm{(E)\ }100+125\pi$ | 2004 AMC 12B Problem 7 | The area of the circle is $S_{\bigcirc}=100\pi$; the area of the square is $S_{\square}=100$.
Exactly $\frac{1}{4}$ of the circle lies inside the square. Thus the total area is $\dfrac34 S_{\bigcirc}+S_{\square}=\boxed{\mathrm{(B)\ }100+75\pi}$. | // Block 1
Draw(Circle((0,0),10));
Draw((0,0)--(10,0)--(10,10)--(0,10)--(0,0));
label("$10$",(5,0),S);
label("$10$",(0,5),W);
dot((0,0));
// Block 2
Draw(Circle((0,0),10)); Draw((0,0)--(10,0)--(10,10)--(0,10)--(0,0)); label("$10$",(5,0),S); label("$10$",(0,5),W); dot((0,0)); | [] |
617 | The point $(-3,2)$ is rotated $90^\circ$ clockwise around the origin to point $B$. Point $B$ is then reflected over the line $x=y$ to point $C$. What are the coordinates of $C$?
$\mathrm{(A)}\ (-3,-2) \qquad \mathrm{(B)}\ (-2,-3) \qquad \mathrm{(C)}\ (2,-3) \qquad \mathrm{(D)}\ (2,3) \qquad \mathrm{(E)}\ (3,2)$ | 2004 AMC 12B Problem 9 | The entire situation is in the picture below. The correct answer is $\boxed{\mathrm{(E)}\ (3,2)}$. | // Block 1
unitsize(1cm);
defaultpen(0.8);
pair A=(-3,2), B=rotate(-90)*A, C=(3,2);
dot(A); dot(B); dot(C);
draw( A -- (0,0) -- B -- C, Dotted );
draw( (-3,-3) -- (4,4), dashed );
label("$A=(-3,2)$", A, NW );
label("$B=(2,3)$", B, N );
label("$C=(3,2)$", C, E );
label("$x=y$",(4,4),NE);
dot((0,0));
label("$(0,0)$", (0,0), SE);
// Block 2
unitsize(1cm); defaultpen(0.8); pair A=(-3,2), B=rotate(-90)*A, C=(3,2); dot(A); dot(B); dot(C); draw( A -- (0,0) -- B -- C, Dotted ); draw( (-3,-3) -- (4,4), dashed ); label("$A=(-3,2)$", A, NW ); label("$B=(2,3)$", B, N ); label("$C=(3,2)$", C, E ); label("$x=y$",(4,4),NE); dot((0,0)); label("$(0,0)$", (0,0), SE); | [] |
618 | In $\triangle ABC$, $AB=13$, $AC=5$, and $BC=12$. Points $M$ and $N$ lie on $AC$ and $BC$, respectively, with $CM=CN=4$. Points $J$ and $K$ are on $AB$ so that $MJ$ and $NK$ are perpendicular to $AB$. What is the area of pentagon $CMJKN$?
$\mathrm{(A)}\ 15 \qquad \mathrm{(B)}\ \frac{81}{5} \qquad \mathrm{(C)}\ \frac{205}{12} \qquad \mathrm{(D)}\ \frac{240}{13} \qquad \mathrm{(E)}\ 20$ | 2004 AMC 12B Problem 14 | Solution 1
The triangle $ABC$ is clearly a right triangle, its area is $\frac{5\cdot 12}2 = 30$. If we knew the areas of triangles $AMJ$ and $BNK$, we could subtract them to get the area of the pentagon.
Draw the height $CL$ from $C$ onto $AB$. As $AB=13$ and the area is $30$, we get $CL=\frac{60}{13}$. The situation is shown in the picture below:
Now note that the triangles $ABC$, $AMJ$, $ACL$, $CBL$ and $NBK$ all have the same angles and therefore they are similar. We already know some of their sides, and we will use this information to compute their areas. Note that if two polygons are similar with ratio $k$, their areas have ratio $k^2$. We will use this fact repeatedly.
Below we will use $[XYZ]$ to denote the area of the triangle $XYZ$.
We have $\frac{CL}{BC} = \frac{60/13}{12} = \frac 5{13}$, hence $[ACL] = \frac{ 25[ABC] }{169} = \frac{750}{169}$.
Also, $\frac{CL}{AC} = \frac{60/13}5 = \frac{12}{13}$, hence $[CBL] = \frac{ 144[ABC] }{169} = \frac{4320}{169}$.
Now for the smaller triangles:
We know that $\frac{AM}{AC}=\frac 15$, hence $[AMJ] = \frac{[ACL]}{25} = \frac{30}{169}$.
Similarly, $\frac{BN}{BC}=\frac 8{12} = \frac 23$, hence $[NBK] = \frac{4[CBL]}9 = \frac{1920}{169}$.
Finally, the area of the pentagon is $30 - \frac{30}{169} - \frac{1920}{169} = \boxed{\frac{240}{13}}$.
Solution 1a
Observe that all of the triangles in the problem are right triangles and similar with a ratio of 5-12-13. The largest triangle $\Delta ABC$ has area $\dfrac{5\cdot12}2=30$. $\Delta AMJ$ is linearly scaled down from $\Delta ABC$ by a factor of $\dfrac{5-4}{13}=\dfrac1{13}$ (as we can see from comparing the two hypotenuses), while $\Delta NBK$ is scaled by a factor of $\dfrac{12-4}{13}=\dfrac8{13}$. The area we desire is the combined area of $\Delta AMJ$ and $\Delta NBK$ subtracted from the area of $\Delta ABC$, which is
\[30-30\left(\dfrac1{13}\right)^2-30\left(\dfrac8{13}\right)^2=30\left(1-\left(\dfrac1{13}\right)^2-\left(\dfrac8{13}\right)^2\right)=30\left(1-\dfrac{65}{169}\right)=30\left(\dfrac8{13}\right)=\boxed{\textbf{(D) }\dfrac{240}{13}}.\]
~Technodoggo
Solution 2
Split the pentagon along a different diagonal as follows:
The area of the pentagon is then the sum of the areas of the resulting right triangle and trapezoid. As before, triangles $ABC$, $AMJ$, and $NBK$ are all similar.
Since $BN=12-4=8$, $NK=\frac{5}{13}(8)=\frac{40}{13}$ and $BK=\frac{12}{13}(8)=\frac{96}{13}$. Since $AM=5-4=1$, $JM=\frac{12}{13}$ and $AJ=\frac{5}{13}$.
The trapezoid's height is therefore $13-\frac{5}{13}-\frac{96}{13}=\frac{68}{13}$, and its area is $\frac{1}{2}\left(\frac{68}{13}\right)\left(\frac{12}{13}+\frac{40}{13}\right)=\frac{34}{13}(4)=\frac{136}{13}$.
Triangle $MCN$ has area $\frac{1}{2}(4)(4)=8$, and the total area is $\frac{104+136}{13}=\boxed{\frac{240}{13}}$.
Solution 3
Right triangle $ABC$ has area $1/2 \cdot AC \cdot BC = 1/2 \cdot 5 \cdot 12 = 30$.
We are given that $AC = 5$ and $CM = 4$, so $AM = AC - CM = 1$. Right triangles $AMJ$ and $ABC$ are similar, with hypotenuses 1 and 13, respectively, so triangle $AMJ$ has area
\[\left( \frac{1}{13} \right)^2 \cdot 30.\]
Also, we are given that $BC = 12$ and $CN = 4$, so $BN = BC - CN = 8$. Right triangles $NBK$ and $ABC$ are similar, with hypotenuses 8 and 13, respectively, so triangle $NBK$ has area
\[\left( \frac{8}{13} \right)^2 \cdot 30.\]
Therefore, the area of pentagon $CMJKN$ is
\[30 - \left( \frac{1}{13} \right)^2 \cdot 30 - \left( \frac{8}{13} \right)^2 \cdot 30 = \boxed{\frac{240}{13}}.\]
~math31415926535
Solution 4
Because triangle ABC, triangle NBK, and triangle AMJ are similar right triangles whose hypotenuses are in the ratio 13 : 8 : 1, their areas are in
the ratio 169 : 64 : 1. The area of triangle ABC is 1/2 (12)(5) = 30, so the areas of triangle NBK and triangle AMJ are (64/169) (30) and (1/169)(30), respectively. Thus the area of pentagon CMJKN is (1 − 64/169 - 1/169)(30)= $\boxed{\mathbf{(D)}240/13}$
Credit to http://billingswest.billings.k12.mt.us/math/AMC%201012/AMC%2012%20work%20sheets/2004%20AMC%2012B%20ws-15.pdf for Solution 3.
Solution 5
We fake-solve. Draw a perpendicular line to side $CB$ that goes through $N$. Similarly, draw a perpendicular line to side $AC$ that goes through $M$. Let them intersect at $O$. It follows that quadrilateral $MONC$ is a square with side $4$ and thus area $16$. Now, draw a perpendicular from $J$ to $MO$ and let them intersect at $Z$. There is a obvious pair of congruent triangles. Fill in the gap. Hence, we see that the area of the pentagon is definitely greater than $16$. How much greater? Well, if we let the intersection of $AB$ and $NO$ be $X$, we can approximate that $\triangle NXK$ has an area greater than $2$. So we place bets on $\boxed{\mathbf{(D)}240/13}$ which is correct!
This would be much easier to show and more convincing with a diagram but idk how to upload GeoGebra to AoPSwiki.
Solution by franzliszt | // Block 1
unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,5), B=(12,0), M=(0,4), N=(4,0); pair J=intersectionpoint(A--B, M--(M+rotate(90)*(B-A)) ); pair K=intersectionpoint(A--B, N--(N+rotate(90)*(B-A)) ); pair L=intersectionpoint(A--B, C--(C+rotate(90)*(B-A)) ); draw( A--B--C--cycle ); draw( M--J ); draw( N--K ); draw( C--L, dashed ); label("$A$",A,NW); label("$B$",B,SE); label("$C$",C,SW); label("$M$",M,SW); label("$N$",N,S); label("$J$",J,NE); label("$K$",K,NE); label("$L$",L,NE);
// Block 2
unitsize(0.5cm); defaultpen(0.8); pair C=(0,0), A=(0,5), B=(12,0), M=(0,4), N=(4,0); pair J=intersectionpoint(A--B, M--(M+rotate(90)*(B-A)) ); pair K=intersectionpoint(A--B, N--(N+rotate(90)*(B-A)) ); pair L=intersectionpoint(A--B, C--(C+rotate(90)*(B-A)) ); draw( A--B--C--cycle ); draw( M--J ); draw( N--K ); draw( M--N, dashed ); label("$A$",A,NW); label("$B$",B,SE); label("$C$",C,SW); label("$M$",M,SW); label("$N$",N,S); label("$J$",J,NE); label("$K$",K,NE); label("$L$",L,NE); | [] |
619 | A truncated cone has horizontal bases with radii $18$ and $2$. A sphere is tangent to the top, bottom, and lateral surface of the truncated cone. What is the radius of the sphere?
$\mathrm{(A)}\ 6 \qquad\mathrm{(B)}\ 4\sqrt{5} \qquad\mathrm{(C)}\ 9 \qquad\mathrm{(D)}\ 10 \qquad\mathrm{(E)}\ 6\sqrt{3}$ | 2004 AMC 12B Problem 19 | Consider a trapezoid (label it $ABCD$ as follows) cross-section of the truncate cone along a diameter of the bases:
Above, $E,F,$ and $G$ are points of tangency. By the Two Tangent Theorem, $BF = BE = 18$ and $CF = CG = 2$, so $BC = 20$. We draw $H$ such that it is the foot of the altitude $\overline{HD}$ to $\overline{AB}$:
By the Pythagorean Theorem,
\[r = \frac{DH}2 = \frac{\sqrt{20^2 - 16^2}}2 = 12\]
Therefore, the answer is $\boxed{{A (6)}}.$ | // Block 1
import olympiad;
size(220);
defaultpen(0.7);
pair A = (0,0), B = (36,0), C = (20,12), D = (16,12), E=(A+B)/2, F=(20+1.6,12-1.2), G = (C+D)/2;
draw(A--B--C--D--cycle);
draw(circumcircle(E,F,G));
dot(E);
dot(F);
dot(G);
label("\(A\)",A,S);
label("\(B\)",B,S);
label("\(C\)",C,NE);
label("\(D\)",D,NW);
label("\(E\)",E,S);
label("\(F\)",F,NE);
label("\(G\)",G,N);
// Block 2
import olympiad;
size(220);
defaultpen(0.7);
pair A = (0,0), B = (36,0), C = (20,12), D = (16,12), E=(A+B)/2, F=(20+1.6,12-1.2), G = (C+D)/2, H=(16,0);
pair P=(D+G)/2, Q=(D+H)/2, R=(B+E)/2, T=(A+H)/2, O=(E+G)/2;
draw(A--B--C--D--cycle);
draw(G--E--H--D);
draw(circumcircle(E,F,G));
dot(E);dot(F);dot(G);dot(H);dot(O);
label("\(A\)",A,S);
label("\(B\)",B,S);
label("\(C\)",C,NE);
label("\(D\)",D,NW);
label("\(E\)",E,S);
label("\(F\)",F,NE);
label("\(G\)",G,N);
label("\(H\)",H,S);
label("\(O\)",O,NE);
label("\(2\)",P,N);
label("\(12\)",Q,W);
label("\(18\)",R,S);
label("\(16\)",T,S);
label("\(20\)",(A+D)/2,NW);
label("\(r\)",(O+E)/2,SE);
// Block 3
import olympiad; size(220); defaultpen(0.7); pair A = (0,0), B = (36,0), C = (20,12), D = (16,12), E=(A+B)/2, F=(20+1.6,12-1.2), G = (C+D)/2; draw(A--B--C--D--cycle); draw(circumcircle(E,F,G)); dot(E); dot(F); dot(G); label("\(A\)",A,S); label("\(B\)",B,S); label("\(C\)",C,NE); label("\(D\)",D,NW); label("\(E\)",E,S); label("\(F\)",F,NE); label("\(G\)",G,N);
// Block 4
import olympiad; size(220); defaultpen(0.7); pair A = (0,0), B = (36,0), C = (20,12), D = (16,12), E=(A+B)/2, F=(20+1.6,12-1.2), G = (C+D)/2, H=(16,0); pair P=(D+G)/2, Q=(D+H)/2, R=(B+E)/2, T=(A+H)/2, O=(E+G)/2; draw(A--B--C--D--cycle); draw(G--E--H--D); draw(circumcircle(E,F,G)); dot(E);dot(F);dot(G);dot(H);dot(O); label("\(A\)",A,S); label("\(B\)",B,S); label("\(C\)",C,NE); label("\(D\)",D,NW); label("\(E\)",E,S); label("\(F\)",F,NE); label("\(G\)",G,N); label("\(H\)",H,S); label("\(O\)",O,NE); label("\(2\)",P,N); label("\(12\)",Q,W); label("\(18\)",R,S); label("\(16\)",T,S); label("\(20\)",(A+D)/2,NW); label("\(r\)",(O+E)/2,SE); | [] |
620 | An equiangular octagon has four sides of length $1$ and four sides of length $\sqrt{2}/2$, arranged so that no two consecutive sides have the same length. What is the area of the octagon?
$\textbf{(A) } \frac{7}{2}\qquad \textbf{(B) } \frac{7\sqrt{2}}{2}\qquad \textbf{(C) } \frac{5+4\sqrt{2}}{2}\qquad \textbf{(D) } \frac{4+5\sqrt{2}}{2}\qquad \textbf{(E) } 7$ | 2005 AMC 10A Problem 20 | The sum of the octagon's angles is $180\cdot(8-2)^{\circ} = 1080^{\circ}$, so since it is equiangular, each angle is $\frac{1080^{\circ}}{8} = 135^{\circ}$, i.e. the same as in a regular octagon. This means that this octagon, just like with a regular octagon, can be divided into squares (whose diagonals form $45^{\circ}$ angles) and right triangles.
In particular, as shown in the diagram below, the area of this octagon can be divided up into $5$ squares of side length $\frac{\sqrt{2}}{2}$ and $4$ right triangles, each of which has half the area of each of the squares (since each square has diagonal $\frac{\sqrt{2}}{2} \cdot \sqrt{2} = 1$, so each right triangle is congruent by side-side-side to one of the triangles formed by slicing a square along one of its diagonals).
Therefore, the area of the octagon is equal to the area of $5 + 4 \cdot \frac{1}{2} = 7$ of the squares, and the area of each square is simply $\left(\frac{\sqrt{2}}{2}\right)^2 = \frac{1}{2}$, so the answer is is $7 \cdot \frac{1}{2} = \boxed{\textbf{(A) } \frac{7}{2}}$. | pair A=(0.5, 0), B=(0, 0.5), C=(0, 1.5), D=(0.5, 2), E=(1.5, 2), F=(2, 1.5), G=(2, 0.5), H=(1.5, 0); draw(A--B); draw(B--C); draw(C--D); draw(D--E);draw(E--F);draw(F--G); draw(G--H); draw(H--A);draw(A--F, blue);draw(E--B,blue);draw(C--H, blue); draw(D--G,blue);dot(A);dot(B);dot(C);dot(D);dot(E);dot(F);dot(G);dot(H); | [] |
621 | Let $\overline{AB}$ be a diameter of a circle and $C$ be a point on $\overline{AB}$ with $2 \cdot AC = BC$. Let $D$ and $E$ be points on the circle such that $\overline{DC} \perp \overline{AB}$ and $\overline{DE}$ is a second diameter. What is the ratio of the area of $\triangle DCE$ to the area of $\triangle ABD$?
$\textbf{(A) } \frac{1}{6}\qquad \textbf{(B) } \frac{1}{4}\qquad \textbf{(C) } \frac{1}{3}\qquad \textbf{(D) } \frac{1}{2}\qquad \textbf{(E) } \frac{2}{3}$ | 2005 AMC 10A Problem 23 | Since $\triangle DCE$ and $\triangle ABD$ share a base, the ratio of their areas is the ratio of their altitudes. Draw the altitude from $C$ to $DE$.
$OD=r, OC=\frac{1}{3}r$.
Since $m\angle DCO=m\angle DFC=90^\circ$, then $\triangle DCO\sim \triangle DFC$. So the ratio of the two altitudes is $\frac{CF}{DC}=\frac{OC}{DO}=\boxed{\textbf{(C) }\frac{1}{3}}$ | // Block 1
import graph;
import olympiad;
pair O,A,B,C,D,E,F;
O=(0,0);A=(15,0);B=(-15,0);C=(5,0);D=(5,14.142135623730950488016887242097);E=(-5,-14.142135623730950488016887242097);F=(0.5555555555555555,1.5713484026367722764463208046774);
draw(Circle((0,0),15));
draw(A--B);draw(D--E);draw(C--D);draw(C--E);draw(C--F);draw(A--D);draw(D--B);
label("A",A,NE);label("B",B,W);label("C",C,SE);label("D",D,NE);label("E",E,SW);label("O",O,SW);label("F",F,NW);
markscalefactor=0.2;
draw(anglemark(C,F,D),blue);draw(anglemark(D,C,B),blue);
// Block 2
import graph; import olympiad; pair O,A,B,C,D,E,F; O=(0,0);A=(15,0);B=(-15,0);C=(5,0);D=(5,14.142135623730950488016887242097);E=(-5,-14.142135623730950488016887242097);F=(0.5555555555555555,1.5713484026367722764463208046774); draw(Circle((0,0),15)); draw(A--B);draw(D--E);draw(C--D);draw(C--E);draw(C--F);draw(A--D);draw(D--B); label("A",A,NE);label("B",B,W);label("C",C,SE);label("D",D,NE);label("E",E,SW);label("O",O,SW);label("F",F,NW); markscalefactor=0.2; draw(anglemark(C,F,D),blue);draw(anglemark(D,C,B),blue); | [] |
621 | Let $\overline{AB}$ be a diameter of a circle and $C$ be a point on $\overline{AB}$ with $2 \cdot AC = BC$. Let $D$ and $E$ be points on the circle such that $\overline{DC} \perp \overline{AB}$ and $\overline{DE}$ is a second diameter. What is the ratio of the area of $\triangle DCE$ to the area of $\triangle ABD$?
$\textbf{(A) } \frac{1}{6}\qquad \textbf{(B) } \frac{1}{4}\qquad \textbf{(C) } \frac{1}{3}\qquad \textbf{(D) } \frac{1}{2}\qquad \textbf{(E) } \frac{2}{3}$ | 2005 AMC 10A Problem 23 | Let the point G be the reflection of point $D$ across $\overline{AB}$. (Point G is on the circle).
Let $AC=x$, then $BC=2x$. The diameter is $3x$. To find $DC$, there are two ways (presented here):
1. Since $\overline{AB}$ is the diameter, $CD=CG$. Using power of points,
\[AC\cdot BC=x\cdot2x=2x^{2}=CD^{2} \longrightarrow CD=x\sqrt{2}\]
2. Use the geometric mean theorem,
\[AC\cdot BC=x\cdot2x=2x^{2}=CD^{2} \longrightarrow CD=x\sqrt{2}\]
(These are the same equations but obtained through different formulae)
Therefore $DG=2x\sqrt{2}$. Since $\overline{DE}$ is a diameter, $\triangle DGE$ is right. By the Pythagorean theorem,
\[DE^{2}=GD^{2}+GE^{2} \longrightarrow \left(3x\right)^{2}=\left(2x\sqrt{2}\right)^{2}+GE^{2}\]
\[9x^{2}=8x^{2}+GE^{2} \longrightarrow GE^{2}=x^{2} \longrightarrow GE=x\]
As established before, $\angle DGE$ is right (if you are unsure, look up "inscribed angle theorem", this is a special case of the theorem where the central angle measures $180^{\circ}$) so $GE=x$ is the altitude of $\triangle DCE$, and $DC=x\sqrt{2}$ is the base. Therefore
\[\left[DCE\right]=\frac{1}{2}\cdot DC\cdot GE=\frac{1}{2}\cdot x\sqrt{2}\cdot x=\frac{x^{2}\sqrt{2}}{2}\]
$AB=3x$ is the base of $\triangle ABD$ and $CD=x\sqrt{2}$ is the height.
\[\left[ABD\right]=\frac{1}{2}\cdot3x\cdot x\sqrt{2}=\frac{3x^{2}\sqrt{2}}{2}\]
The required ratio is
\[\frac{\left[DCE\right]}{\left[ABD\right]}=\frac{\frac{x^{2}\sqrt{2}}{2}}{\frac{3x^{2}\sqrt{2}}{2}}=\frac{x^{2}\sqrt{2}}{2}\cdot\frac{2}{3x^{2}\sqrt{2}}=\frac{x^{2}\sqrt{2}}{3x^{2}\sqrt{2}}=\frac{1}{3}\]
The answer is $\boxed{\textbf{(C) } \frac{1}{3}}$.
~JH. L | // Block 1
unitsize(2.5cm);
defaultpen(fontsize(10pt)+linewidth(.8pt));
dotfactor=3;
pair O=(0,0), C=(-1/3.0), B=(1,0), A=(-1,0);
pair D=dir(aCos(C.x)), E=(-D.x,-D.y);
draw(A--B--D--cycle);
draw(D--E--C);
draw(unitcircle,white);
drawline(D,C);
dot(O);
clip(unitcircle);
draw(unitcircle);
label("$E$",E,SSE);
label("$B$",B,E);
label("$A$",A,W);
label("$D$",D,NNW);
label("$C$",C,SW);
draw(rightanglemark(D,C,B,2));
// Block 2
unitsize(2.5cm); defaultpen(fontsize(10pt)+linewidth(.8pt)); dotfactor=3; pair O=(0,0), C=(-1/3.0), B=(1,0), A=(-1,0); pair D=dir(aCos(C.x)), E=(-D.x,-D.y); draw(A--B--D--cycle); draw(D--E--C); draw(unitcircle,white); drawline(D,C); dot(O); clip(unitcircle); draw(unitcircle); label("$E$",E,SSE); label("$B$",B,E); label("$A$",A,W); label("$D$",D,NNW); label("$C$",C,SW); draw(rightanglemark(D,C,B,2)); | [] |
622 | In $\triangle ABC$ we have $AB = 25$, $BC = 39$, and $AC = 42$. Points $D$ and $E$ are on $\overline{AB}$ and $\overline{AC}$ respectively, with $AD = 19$ and $AE = 14$. What is the ratio of the area of triangle $ADE$ to the area of the quadrilateral $BCED$?
$\textbf{(A) } \frac{266}{1521}\qquad \textbf{(B) } \frac{19}{75}\qquad \textbf{(C) } \frac{1}{3}\qquad \textbf{(D) } \frac{19}{56}\qquad \textbf{(E) } 1$ | 2005 AMC 10A Problem 25 | We have
\[\frac{[ADE]}{[ABC]} = \frac{AD}{AB} \cdot \frac{AE}{AC} = \frac{19}{25} \cdot \frac{14}{42} = \frac{19}{75}.\]
(Area of a triangle is base times height, so the area ratio of triangles, that have a common vertex (height) and bases on a common line, is the base length ratio. This is applied twice, using different pairs of bases, and corresponding altitudes for height.).
$[BCED] = [ABC] - [ADE]$, so
\begin{align*} \frac{[ADE]}{[BCED]} &= \frac{[ADE]}{[ABC] - [ADE]} \\ &= \frac{1}{\frac{ABC}{ADE} - 1} \\ &= \frac{1}{\frac{75}{19} - 1} \\ &= \boxed{\textbf{(D) }\frac{19}{56}}. \end{align*}
Note: If it is hard to understand why \[\frac{[ADE]}{[ABC]} = \frac{AD}{AB} \cdot \frac{AE}{AC}\], you can use the fact that the area of a triangle equals $\frac{1}{2} \cdot ab \cdot \sin(C)$. If angle $DAE = Z$, we have that \[\frac{[ADE]}{[ABC]} = \frac{\frac{1}{2} \cdot 19 \cdot 14 \cdot \sin(Z)}{\frac{1}{2} \cdot 25 \cdot 42 \cdot \sin(Z)} = \frac{19 \cdot 14}{25 \cdot 42} = \frac{ab}{cd}\]. | // Block 1
unitsize(0.15 cm);
pair A, B, C, D, E;
A = (191/39,28*sqrt(1166)/39);
B = (0,0);
C = (39,0);
D = (6*A + 19*B)/25;
E = (28*A + 14*C)/42;
draw(A--B--C--cycle);
draw(D--E);
label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, W);
label("$E$", E, NE);
label("$19$", (A + D)/2, W);
label("$6$", (B + D)/2, W);
label("$14$", (A + E)/2, NE);
label("$28$", (C + E)/2, NE);
// Block 2
unitsize(0.15 cm); pair A, B, C, D, E; A = (191/39,28*sqrt(1166)/39); B = (0,0); C = (39,0); D = (6*A + 19*B)/25; E = (28*A + 14*C)/42; draw(A--B--C--cycle); draw(D--E); label("$A$", A, N); label("$B$", B, SW); label("$C$", C, SE); label("$D$", D, W); label("$E$", E, NE); label("$19$", (A + D)/2, W); label("$6$", (B + D)/2, W); label("$14$", (A + E)/2, NE); label("$28$", (C + E)/2, NE); | [] |
622 | In $\triangle ABC$ we have $AB = 25$, $BC = 39$, and $AC = 42$. Points $D$ and $E$ are on $\overline{AB}$ and $\overline{AC}$ respectively, with $AD = 19$ and $AE = 14$. What is the ratio of the area of triangle $ADE$ to the area of the quadrilateral $BCED$?
$\textbf{(A) } \frac{266}{1521}\qquad \textbf{(B) } \frac{19}{75}\qquad \textbf{(C) } \frac{1}{3}\qquad \textbf{(D) } \frac{19}{56}\qquad \textbf{(E) } 1$ | 2005 AMC 10A Problem 25 | We can let $[ADE]=x$.
Since $EC=2 \cdot EA$, $[DEC]=2x$.
So, $[ADC]=3x$.
This means that $[BDC]=\frac{6}{19}\cdot3x=\frac{18x}{19}$.
Thus, $\frac{[ADE]}{[BCED]} = \frac{x}{\frac{18x}{19}+2x}= \boxed{\textbf{(D) }\frac{19}{56}}.$
-Conantwiz2023 | // Block 1
unitsize(0.15 cm);
pair A, B, C, D, E;
A = (191/39,28*sqrt(1166)/39);
B = (0,0);
C = (39,0);
D = (6*A + 19*B)/25;
E = (28*A + 14*C)/42;
draw(A--B--C--cycle);
draw(D--E);
label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, W);
label("$E$", E, NE);
label("$19$", (A + D)/2, W);
label("$6$", (B + D)/2, W);
label("$14$", (A + E)/2, NE);
label("$28$", (C + E)/2, NE);
// Block 2
unitsize(0.15 cm); pair A, B, C, D, E; A = (191/39,28*sqrt(1166)/39); B = (0,0); C = (39,0); D = (6*A + 19*B)/25; E = (28*A + 14*C)/42; draw(A--B--C--cycle); draw(D--E); label("$A$", A, N); label("$B$", B, SW); label("$C$", C, SE); label("$D$", D, W); label("$E$", E, NE); label("$19$", (A + D)/2, W); label("$6$", (B + D)/2, W); label("$14$", (A + E)/2, NE); label("$28$", (C + E)/2, NE); | [] |
622 | In $\triangle ABC$ we have $AB = 25$, $BC = 39$, and $AC = 42$. Points $D$ and $E$ are on $\overline{AB}$ and $\overline{AC}$ respectively, with $AD = 19$ and $AE = 14$. What is the ratio of the area of triangle $ADE$ to the area of the quadrilateral $BCED$?
$\textbf{(A) } \frac{266}{1521}\qquad \textbf{(B) } \frac{19}{75}\qquad \textbf{(C) } \frac{1}{3}\qquad \textbf{(D) } \frac{19}{56}\qquad \textbf{(E) } 1$ | 2005 AMC 10A Problem 25 | Let the area of triangle ABC be denoted by [ABC] and the area of quadrilateral ABCD be denoted by [ABCD].
\[\text{Diagram:}\]
Let the area of $\triangle ABC$ be $x$. $\triangle ABE$ and $\triangle BEC$ share a height, and the ratio of their bases are $1:2$, so the area of $\triangle ABE$ is $\frac{x}{3}$.
Similarly, $\triangle AED$ and $\triangle DEB$ share a height, and the ratio of their bases is $19:6$, so the ratio of $\frac{[AED]}{[AEB]}=\frac{19}{25}$. Therefore,
\[[AED]=\frac{19}{25}\cdot\left[AEB\right]=\frac{19}{25}\cdot\frac{1}{3}\cdot\left[ABC\right]=\frac{19}{25}\cdot\frac{1}{3}\cdot x=\frac{19}{75}x\]
\[[DECB]=[ABC]-[AED]=x-\frac{19}{75}x=\frac{56}{75}x\]
The ratio $\frac{[AED]}{[DECB]}=\frac{\frac{19}{75}}{\frac{56}{75}}=\frac{19}{56}$ which is answer choice $\boxed{\textbf{(D) } \frac{19}{56}}$.
~JH. L | // Block 1
unitsize(0.15 cm);
pair A, B, C, D, E;
A = (191/39,28*sqrt(1166)/39);
B = (0,0);
C = (39,0);
D = (6*A + 19*B)/25;
E = (28*A + 14*C)/42;
draw(A--B--C--cycle);
draw(D--E);
label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, W);
label("$E$", E, NE);
label("$19$", (A + D)/2, W);
label("$6$", (B + D)/2, W);
label("$14$", (A + E)/2, NE);
label("$28$", (C + E)/2, NE);
// Block 2
unitsize(0.15 cm); pair A, B, C, D, E; A = (191/39,28*sqrt(1166)/39); B = (0,0); C = (39,0); D = (6*A + 19*B)/25; E = (28*A + 14*C)/42; draw(A--B--C--cycle); draw(D--E); label("$A$", A, N); label("$B$", B, SW); label("$C$", C, SE); label("$D$", D, W); label("$E$", E, NE); label("$19$", (A + D)/2, W); label("$6$", (B + D)/2, W); label("$14$", (A + E)/2, NE); label("$28$", (C + E)/2, NE); | [] |
623 | Let $\overline{AB}$ be a diameter of a circle and $C$ be a point on $\overline{AB}$ with $2 \cdot AC = BC$. Let $D$ and $E$ be points on the circle such that $\overline{DC} \perp \overline{AB}$ and $\overline{DE}$ is a second diameter. What is the ratio of the area of $\triangle DCE$ to the area of $\triangle ABD$?
$(\text {A}) \ \frac {1}{6} \qquad (\text {B}) \ \frac {1}{4} \qquad (\text {C})\ \frac {1}{3} \qquad (\text {D}) \ \frac {1}{2} \qquad (\text {E})\ \frac {2}{3}$ | 2005 AMC 12A Problem 15 | Solution 1
Notice that the bases of both triangles are diameters of the circle. Hence the ratio of the areas is just the ratio of the heights of the triangles, or $\frac{CF}{CD}$ ($F$ is the foot of the perpendicular from $C$ to $DE$).
Call the radius $r$. Then $AC = \frac 13(2r) = \frac 23r$, $CO = \frac 13r$. Using the Pythagorean Theorem in $\triangle OCD$, we get $\frac{1}{9}r^2 + CD^2 = r^2 \Longrightarrow CD = \frac{2\sqrt{2}}3r$.
Now we have to find $CF$. Notice $\triangle OCD \sim \triangle OFC$, so we can write the proportion:
$\frac{OF}{OC} = \frac{OC}{OD}$
$\frac{OF}{\frac{1}{3}r} = \frac{\frac{1}{3}r}{r}$
$OF = \frac 19r$
By the Pythagorean Theorem in $\triangle OFC$, we have $\left(\frac{1}{9}r\right)^2 + CF^2 = \left(\frac{1}{3}r\right)^2 \Longrightarrow CF = \sqrt{\frac{8}{81}r^2} = \frac{2\sqrt{2}}{9}r$.
Our answer is $\frac{CF}{CD} = \frac{\frac{2\sqrt{2}}{9}r}{\frac{2\sqrt{2}}{3}r} = \frac 13 \Longrightarrow \mathrm{(C)}$.
Solution 2
Let the center of the circle be $O$.
Note that $2 \cdot AC = BC \Rightarrow 3 \cdot AC = AB$.
$O$ is midpoint of $AB \Rightarrow \frac{3}{2}AC = AO \Rightarrow CO = \frac{1}{3}AO \Rightarrow CO = \frac{1}{6} AB$.
$O$ is midpoint of $DE \Rightarrow$ Area of $\triangle DCE = 2 \cdot$ Area of $\triangle DCO = 2 \cdot (\frac{1}{6} \cdot$ Area of $\triangle ABD) = \frac{1}{3} \cdot$ Area of $\triangle ABD \Longrightarrow \mathrm{(C)}$.
Solution 3
Let $r$ be the radius of the circle. Note that $AC+BC = 2r$ so $AC = \frac{2}{3}r$.
By Power of a Point Theorem, $CD^2= AC \cdot BC = 2\cdot AC^2$, and thus $CD = \sqrt{2} \cdot AC = \frac{2\sqrt{2}}{3}r$
Then the area of $\triangle ABD$ is $\frac{1}{2} AB \cdot CD = \frac{2\sqrt{2}}{3}r^2$. Similarly, the area of $\triangle DCE$ is $\frac{1}{2}(r-AC) \cdot 2 \cdot CD = \frac{2\sqrt{2}}{9}r^2$, so the desired ratio is $\frac{\frac{2\sqrt{2}}{9}r^2}{\frac{2\sqrt{2}}{3}r^2} = \frac{1}{3} \Longrightarrow \mathrm{(C)}$
Solution 4
Let the center of the circle be $O$.
Without loss of generality, let the radius of the circle be equal to $3$. Thus, $AO=3$ and $OB=3$. As a consequence of $2(AC)=BC$, $AC=2$ and $CO=1$. Also, we know that $DO$ and $OE$ are both equal to $3$ due to the fact that they are both radii. Thus from the Pythagorean Theorem, we have DC being equal to $\sqrt{3^2-1^2}$ or $2\sqrt{2}$. Now we know that the area of $[ABD]$ is equal to $\frac{(3+2+1)(2\sqrt{2})}{2}$ or $6\sqrt{2}$. Know we need to find the area of $[DCE]$. By simple inspection $[COD]$ $\cong$ $[HOE]$ due to angles being equal and CPCTC. Thus $HE=2\sqrt{2}$ and $OH=1$. Know we know the area of $[CHE]=\frac{(1+1)(2\sqrt{2})}{2}$ or $2\sqrt{2}$. We also know that the area of $[OHE]=\frac{(1)(2\sqrt{2})}{2}$ or $\sqrt{2}$. Thus the area of $[COE]=2\sqrt{2}-\sqrt{2}$ or $\sqrt{2}$. We also can calculate the area of $[DOC]$ to be $\frac{(1)(2\sqrt{2})}{2}$ or $\sqrt{2}$. Thus $[DCE]$ is equal to $[COE]$ + $[DOC]$ or $\sqrt{2}+\sqrt{2}$ or $2\sqrt{2}$. The ratio between $[DCE]$ and $[ABD]$ is equal to $\frac{2\sqrt{2}}{6\sqrt{2}}$ or $\frac{1}{3}$ $\Longrightarrow \mathrm{(C)}$.
Solution 5
We will use the shoelace formula. Our origin is the center of the circle. Denote the ordered pair for $D (x,y)$, and notice how $E$ is a 180 degree rotation of $D$, using the rotation matrix formula we get $E = (-x,-y)$. WLOG say that this circle has radius $3$. We can now find points $A$, $B$, and $C$ which are $(-3,0)$, $(3,0)$, and $(-1,0)$ respectively.
By shoelace the area of $CED$ is $Y$, and the area of $ADB$ is $3Y$. Using division we get that the answer is $\mathrm{(C)}$.
Solution 6 (Mass Points)
We set point $A$ as a mass of 2. This means that point $B$ has a mass of $1$ since $2\times{AC} = 1\times{BC}$. This implies that point $C$ has a mass of $2+1 = 3$ and the center of the circle has a mass of $3+1 = 4$. After this, we notice that points $D$ and $E$ both must have a mass of $2$ since $2+2 = 4$ and they are both radii of the circle.
To find the ratio of the areas, we do the reciprocal of the multiplication of the mass points of DCE and the multiplication of ABD divided by each other. Which is simply $\frac{3\times2\times2}{2\times2\times1}$ which is $\boxed{\frac{1}{3}}$ (the reciprocal of 3)
-Brudder
Solution 7 (Slight Trigonometry)
Let the center of the circle be $O$.
The area of $ADB$ is $\frac{1}{2} \cdot 6x \cdot 2x\sqrt{2}$ = $6x^2\sqrt{2}$. The area of $DCE$ is $\frac{1}{2} \cdot DC \cdot DE \cdot$ sin $CDE$. We find sin $CDE$ is $\frac{OC}{OD}$ = $\frac{1}{3}$. Substituting $DC = 2x\sqrt{2}$ and $DE = 6x$, we get $[DCE]$ = $\frac{1}{2} \cdot 2x\sqrt{2} \cdot 6x \cdot \frac{1}{3}$ = $2x^2\sqrt{2}$. Hence, the ratio between the areas of $DCE$ and $ADB$ is equal to $\frac{2x^2\sqrt{2}}{6x^2\sqrt{2}}$ or $\frac{1}{3}$ = $\boxed{(C)}$.
~Math_Genius_164
Solution 8
In my opinion, the solution below is the easiest and quickest.
Since both $DE$ and $AB$ are diameters, they intersect at the center of the circle. Call this center $O$. WLOG, let $AC=2, CO=1,OB=3$. Call the point where the extension of $DC$ hits the circle $F$. Notice that $\angle DCB = 90^\circ$. This implies that $DC=CF$. WOLG, let $DC=CF=1$. Then, $[DCE]=\frac 12 bh = \frac 12 * DF*CO = \frac 12 *2*1=1$ and $[ABD]=\frac 12 bh = \frac 12 *AB*CD = \frac 12 *6*1 = 3$. Thus, the answer is $\frac{1}{3}$ = $\boxed{(C)}$.
Solution by franzliszt
Solution 9 (Slick construction)
Let $X$ be the reflection of $C$ over the center $O.$ Since $\triangle DXO\cong\triangle ECO$ by SAS, it follows that the area of $\triangle DEC$ is equal to the area of $\triangle DCX.$ However, we know that
\[BC = 2AC\implies CO=\frac{1}{2}AC\implies CX=AC,\]
so the ratio of the area of $\triangle DEC$ to the area of $\triangle ABD$ is $\frac{\frac{AC\cdot DC}{2}}{\frac{3AC\cdot DC}{2}}=\frac{1}{3}\implies \boxed{(C)}$ | // Block 1
unitsize(2.5cm); defaultpen(fontsize(10pt)+linewidth(.8pt)); dotfactor=3; pair O=(0,0), C=(-1/3.0), B=(1,0), A=(-1,0); pair D=dir(aCos(C.x)), e=(-D.x,-D.y); pair H=(e.x,0); draw(A--B--D--cycle); draw(D--e--C); draw(unitcircle,white); drawline(D,C); dot(O); clip(unitcircle); draw(unitcircle); label("$E$",e,SSE); label("$B$",B,NE); label("$A$",A,W); label("$D$",D,NNW); label("$C$",C,SW); label("$H$",H,SE); draw(e--H,dashed); label("O",(0,0),NE); label("1",(C--O),N); label("1",(H--O),N); label("2",(A--C),N); label("2",(H--B),N); label("3",(O--D),NE); label("3",(O--e),NE); label("$2\sqrt{2}$",(D--C),W); label("$2\sqrt{2}$",(H--e),E); draw(rightanglemark(e,(e.x,0),A,2)); draw(rightanglemark(D,C,B,2));
// Block 2
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(7cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.24313994860289, xmax = 6.360350402147026, ymin = -8.17642986522568, ymax = 4.1323989018072735; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw(circle((0.9223776980185863,-1.084225871478444), 3.171161249925393), linewidth(2) + wrwrwr); draw((-0.5623104956355617,1.717909856970905)--(-0.39884850438732933,-3.9670413347199647), linewidth(2) + wrwrwr); draw((-2.2487343499798245,-1.1018908670262892)--(4.0935388680341624,-1.0849377800575102), linewidth(2) + wrwrwr); draw((-0.4813673187299407,-1.0971666418223938)--(2.1729847859273126,-3.904641555130568), linewidth(2) + wrwrwr); draw((-0.5623104956355617,1.717909856970905)--(2.1561002471522333,-3.887757016355488), linewidth(2) + wrwrwr); draw((-2.2487343499798245,-1.1018908670262892)--(-0.5623104956355617,1.717909856970905), linewidth(2) + wrwrwr); draw((-0.5623104956355617,1.717909856970905)--(4.0935388680341624,-1.0849377800575102), linewidth(2) + wrwrwr); draw(circle((-2.858607769046372,-1.1524617347926092), 0.45463011998128017), linewidth(2) + wrwrwr); draw(circle((4.790088296064635,-1.144019465405069), 0.45463011998128006), linewidth(2) + wrwrwr); draw(circle((-0.9168858099122313,-1.6336710898823752), 0.45463011998127983), linewidth(2) + wrwrwr); draw(circle((1.4976032349241348,-1.3128648531558642), 0.4546301199812797), linewidth(2) + wrwrwr); draw(circle((-0.815578577261755,2.334195522261307), 0.4546301199812809), linewidth(2) + wrwrwr); draw(circle((2.6119827940793803,-4.5546962979711285), 0.45463011998128033), linewidth(2) + wrwrwr); /* dots and labels */ dot((0.9223776980185863,-1.084225871478444),dotstyle); dot((-0.5623104956355617,1.717909856970905),dotstyle); label("$D$", (-0.49477234053524466,1.8867552447217006), NE * labelscalefactor); dot((-0.39884850438732933,-3.9670413347199647),dotstyle); dot((-2.2487343499798245,-1.1018908670262892),dotstyle); label("A", (-2.183226218043193,-0.9329627307165757), NE * labelscalefactor); dot((4.0935388680341624,-1.0849377800575102),dotstyle); label("B", (4.165360361386693,-0.9160781919414961), NE * labelscalefactor); dot((-0.4813673187299407,-1.0971666418223938),linewidth(4pt) + dotstyle); label("C", (-0.41034964665984724,-0.9667318082667347), NE * labelscalefactor); dot((2.1729847859273126,-3.904641555130568),dotstyle); dot((2.1561002471522333,-3.887757016355488),dotstyle); label("E", (2.2236384022525524,-3.7189116286046926), NE * labelscalefactor); label("2", (-2.9261459241466903,-1.2031153511178476), NE * labelscalefactor,wrwrwr); label("1", (4.722550140964316,-1.186230812342768), NE * labelscalefactor,wrwrwr); label("3", (-0.9844239650125497,-1.6758824368200735), NE * labelscalefactor,wrwrwr); label("4", (1.4300650798238166,-1.355076200093563), NE * labelscalefactor,wrwrwr); label("2", (-0.8831167323620728,2.2919841753236083), NE * labelscalefactor,wrwrwr); label("2", (2.5444446389790625,-4.5969076449088275), NE * labelscalefactor,wrwrwr); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */
// Block 3
unitsize(3.5cm); defaultpen(fontsize(12pt)+linewidth(.8pt)); dotfactor=3; pair O=(0,0), C=(-1/3.0), B=(1,0), A=(-1,0); pair D=dir(aCos(C.x)), E=(-D.x,-D.y); draw(A--B--D--cycle); draw(D--E--C); draw(unitcircle,white); drawline(D,C); dot(O); clip(unitcircle); draw(unitcircle); label("$E$",E,SSE); label("$B$",B,NE); label("$A$",A,W); label("$D$",D,NNW); label("$C$",C,SW); label("O",(0,0),NE); label("x",(C--O),N); label("3x",(B--O),N); label("2x",(A--C),N); label("3x",(O--D),NE); label("3x",(O--E),NE); label("$2x\sqrt{2}$",(D--C),W); draw(rightanglemark(D,C,B,2));
// Block 4
unitsize(2.5cm); defaultpen(fontsize(10pt)+linewidth(.8pt)); dotfactor=3; pair O=(0,0), C=(-1/3.0), B=(1,0), A=(-1,0), X = (1/3.0); pair D=dir(aCos(C.x)), E=(-D.x,-D.y); draw(A--B--D--cycle); draw(D--E--C); draw(D--X); draw(unitcircle,white); drawline(D,C); dot(O); clip(unitcircle); draw(unitcircle); label("$E$",E,SSE); label("$B$",B,E); label("$A$",A,W); label("$D$",D,NNW); label("$C$",C,SW); label("$X$",X,NE); draw(rightanglemark(D,C,B,2)); | [] |
624 | Let $S$ be the set of all points with coordinates $(x,y,z)$, where $x$, $y$, and $z$ are each chosen from the set $\{0,1,2\}$. How many equilateral triangles all have their vertices in $S$?
$(\mathrm {A}) \ 72\qquad (\mathrm {B}) \ 76 \qquad (\mathrm {C})\ 80 \qquad (\mathrm {D}) \ 84 \qquad (\mathrm {E})\ 88$ | 2005 AMC 12A Problem 25 | For this solution, we will just find as many equilateral triangles as possible, until it becomes intuitive that there are no more size of triangles left.
First, we observe that we can form an equilateral triangle with vertices in $S$ by taking any point in $S$ and connecting it to the $2$ adjacent points. This triangle will have a side length of $\sqrt{2}$; a quick further examination of this cube will show us that this is the only possible side length (the red triangle in the diagram below). Each of these triangles is determined by one vertex of the cube, so in one cube we have $8$ equilateral triangles. We have $8$ unit cubes, as well as the entire $2 \times 2 \times 2$ cube (giving the green triangle in the diagram), for a total of $8+1 = 9$ cubes, and thus $9 \cdot 8 = 72$ equilateral triangles.
(Note that connecting the centers of the faces will actually give an octahedron, not a cube, because it only has $6$ vertices.)
Now we look for any further equilateral triangles. Connecting the midpoints of $3$ non-adjacent, non-parallel edges indeed gives us more equilateral triangles (e.g. the blue triangle in the diagram below). Notice that picking these $3$ edges leaves $2$ vertices alone (labelled A and B in the diagram), and that picking any $2$ opposite vertices determines $2$ equilateral triangles. Hence there are $\frac{8 \cdot 2}{2} = 8$ of these equilateral triangles, so adding these to the triangles already found above gives a total of $72+8 = \boxed{\textbf{(C) }80}$. | // Block 1
import three;
unitsize(1cm);
size(200);
currentprojection=perspective(1/3,-1,1/2);
draw((0,0,0)--(2,0,0)--(2,2,0)--(0,2,0)--cycle);
draw((0,0,0)--(0,0,2));
draw((0,2,0)--(0,2,2));
draw((2,2,0)--(2,2,2));
draw((2,0,0)--(2,0,2));
draw((0,0,2)--(2,0,2)--(2,2,2)--(0,2,2)--cycle);
draw((2,0,0)--(0,2,0)--(0,0,2)--cycle,green);
draw((1,0,0)--(0,1,0)--(0,0,1)--cycle,red);
label("$x=2$",(1,0,0),S);
label("$z=2$",(2,2,1),E);
label("$y=2$",(2,1,0),SE);
// Block 2
import three;
unitsize(1cm);
size(200);
currentprojection=perspective(1/3,-1,1/2);
draw((0,0,0)--(2,0,0)--(2,2,0)--(0,2,0)--cycle);
draw((0,0,0)--(0,0,2));
draw((0,2,0)--(0,2,2));
draw((2,2,0)--(2,2,2));
draw((2,0,0)--(2,0,2));
draw((0,0,2)--(2,0,2)--(2,2,2)--(0,2,2)--cycle);
draw((1,0,0)--(2,2,1)--(0,1,2)--cycle,blue);
label("$x=2$",(1,0,0),S);
label("$z=2$",(2,2,1),E);
label("$y=2$",(2,1,0),SE);
label("$A$",(0,2,0), NW);
label("$B$",(2,0,2), NW);
// Block 3
import three; unitsize(1cm); size(200); currentprojection=perspective(1/3,-1,1/2); draw((0,0,0)--(2,0,0)--(2,2,0)--(0,2,0)--cycle); draw((0,0,0)--(0,0,2)); draw((0,2,0)--(0,2,2)); draw((2,2,0)--(2,2,2)); draw((2,0,0)--(2,0,2)); draw((0,0,2)--(2,0,2)--(2,2,2)--(0,2,2)--cycle); draw((2,0,0)--(0,2,0)--(0,0,2)--cycle,green); draw((1,0,0)--(0,1,0)--(0,0,1)--cycle,red); label("$x=2$",(1,0,0),S); label("$z=2$",(2,2,1),E); label("$y=2$",(2,1,0),SE);
// Block 4
import three; unitsize(1cm); size(200); currentprojection=perspective(1/3,-1,1/2); draw((0,0,0)--(2,0,0)--(2,2,0)--(0,2,0)--cycle); draw((0,0,0)--(0,0,2)); draw((0,2,0)--(0,2,2)); draw((2,2,0)--(2,2,2)); draw((2,0,0)--(2,0,2)); draw((0,0,2)--(2,0,2)--(2,2,2)--(0,2,2)--cycle); draw((1,0,0)--(2,2,1)--(0,1,2)--cycle,blue); label("$x=2$",(1,0,0),S); label("$z=2$",(2,2,1),E); label("$y=2$",(2,1,0),SE); label("$A$",(0,2,0), NW); label("$B$",(2,0,2), NW); | [] |
625 | A circle is inscribed in a square, then a square is inscribed in this circle, and finally, a circle is inscribed in this square. What is the ratio of the area of the smallest circle to the area of the largest square?
$\textbf{(A) } \frac{\pi}{16} \qquad \textbf{(B) } \frac{\pi}{8} \qquad \textbf{(C) } \frac{3\pi}{16} \qquad \textbf{(D) } \frac{\pi}{4} \qquad \textbf{(E) } \frac{\pi}{2}$ | 2005 AMC 10B Problem 7 | Let the radius of the smallest circle be $r$. Then the side length of the smaller square is $2r$. The radius of the larger circle is half the length of the diagonal of the smaller square, so it is $\sqrt{2}r$. Hence the largest square has sides of length $2\sqrt{2}r$. The ratio of the area of the smallest circle to the area of the largest square is therefore $\frac{\pi r^2}{\left(2\sqrt{2}r\right)^2} =\boxed{\textbf{(B) }\frac{\pi}{8}}.$
Tip: When facing a geometry problem, it is very helpful to draw a diagram. | // Block 1
draw(Circle((0,0),10),linewidth(0.7));
draw(Circle((0,0),14.1),linewidth(0.7));
draw((0,14.1)--(14.1,0)--(0,-14.1)--(-14.1,0)--cycle,linewidth(0.7));
draw((-14.1,14.1)--(14.1,14.1)--(14.1,-14.1)--(-14.1,-14.1)--cycle,linewidth(0.7));
draw((0,0)--(-14.1,0),linewidth(0.7));
draw((-7.1,7.1)--(0,0),linewidth(0.7));
label("$\sqrt{2}r$",(-6,0),S);
label("$r$",(-3.5,3.5),NE);
label("$2r$",(-7.1,7.1),W);
label("$2\sqrt{2}r$",(0,14.1),N);
// Block 2
draw(Circle((0,0),10),linewidth(0.7)); draw(Circle((0,0),14.1),linewidth(0.7)); draw((0,14.1)--(14.1,0)--(0,-14.1)--(-14.1,0)--cycle,linewidth(0.7)); draw((-14.1,14.1)--(14.1,14.1)--(14.1,-14.1)--(-14.1,-14.1)--cycle,linewidth(0.7)); draw((0,0)--(-14.1,0),linewidth(0.7)); draw((-7.1,7.1)--(0,0),linewidth(0.7)); label("$\sqrt{2}r$",(-6,0),S); label("$r$",(-3.5,3.5),NE); label("$2r$",(-7.1,7.1),W); label("$2\sqrt{2}r$",(0,14.1),N); | [] |
626 | What is the area enclosed by the graph of $|3x|+|4y|=12$?
$\mathrm{(A)}\ 6 \qquad \mathrm{(B)}\ 12 \qquad \mathrm{(C)}\ 16 \qquad \mathrm{(D)}\ 24 \qquad \mathrm{(E)}\ 25$ | 2005 AMC 12B Problem 7 | If we get rid of the absolute values, we are left with the following 4 equations (using the logic that if $|a|=b$, then $a$ is either $b$ or $-b$):
\begin{align*} 3x+4y=12 \\ -3x+4y=12 \\ 3x-4y=12 \\ -3x-4y=12 \end{align*}
We can then put these equations in slope-intercept form in order to graph them.
\begin{align*} 3x+4y=12 \,\implies\, y=-\dfrac{3}{4}x+3\\ -3x+4y=12\,\implies\, y=\dfrac{3}{4}x+3\\ 3x-4y=12\,\implies\, y=\dfrac{3}{4}x-3\\ -3x-4y=12\,\implies\, y=-\dfrac{3}{4}x-3\end{align*}
Now you can graph the lines to find the shape of the graph:
We can easily see that it is a rhombus with diagonals of $6$ and $8$. The area is $\dfrac{1}{2}\times 6\times8$, or $\boxed{\mathrm{(D)}\ 24}$ | // Block 1
Label f;
f.p=fontsize(6);
xaxis(-8,8,Ticks(f, 4.0));
yaxis(-6,6,Ticks(f, 3.0));
fill((0,-3)--(4,0)--(0,3)--(-4,0)--cycle,grey);
draw((-4,-6)--(8,3), Arrows(4));
draw((4,-6)--(-8,3), Arrows(4));
draw((-4,6)--(8,-3), Arrows(4));
draw((4,6)--(-8,-3), Arrows(4));
// Block 2
Label f; f.p=fontsize(6); xaxis(-8,8,Ticks(f, 4.0)); yaxis(-6,6,Ticks(f, 3.0)); fill((0,-3)--(4,0)--(0,3)--(-4,0)--cycle,grey); draw((-4,-6)--(8,3), Arrows(4)); draw((4,-6)--(-8,3), Arrows(4)); draw((-4,6)--(8,-3), Arrows(4)); draw((4,6)--(-8,-3), Arrows(4)); | [] |
626 | What is the area enclosed by the graph of $|3x|+|4y|=12$?
$\mathrm{(A)}\ 6 \qquad \mathrm{(B)}\ 12 \qquad \mathrm{(C)}\ 16 \qquad \mathrm{(D)}\ 24 \qquad \mathrm{(E)}\ 25$ | 2005 AMC 12B Problem 7 | The graph is symmetric with respect to both coordinate axes, and in the first quadrant it coincides with the graph of the line $3x + 4y = 12.$ Therefore the region is a rhombus, and the area is
\[\text{Area} = 4\left(\frac{1}{2}(4\cdot 3)\right) = 24 \rightarrow \boxed{D}\]
~Alcumus | // Block 1
draw((-5,0)--(5,0),Arrow);
draw((0,-4)--(0,4),Arrow);
label("$x$",(5,0),S);
label("$y$",(0,4),E);
label("4",(4,0),S);
label("-4",(-4,0),S);
label("3",(0,3),NW);
label("-3",(0,-3),SW);
draw((4,0)--(0,3)--(-4,0)--(0,-3)--cycle,linewidth(0.7));
// Block 2
draw((-5,0)--(5,0),Arrow); draw((0,-4)--(0,4),Arrow); label("$x$",(5,0),S); label("$y$",(0,4),E); label("4",(4,0),S); label("-4",(-4,0),S); label("3",(0,3),NW); label("-3",(0,-3),SW); draw((4,0)--(0,3)--(-4,0)--(0,-3)--cycle,linewidth(0.7)); | [] |
627 | A circle having center $(0,k)$, with $k>6$, is tangent to the lines $y=x$, $y=-x$ and $y=6$. What is the radius of this circle?
$\mathrm{(A)}\ 6\sqrt{2}-6 \qquad \mathrm{(B)}\ 6 \qquad \mathrm{(C)}\ 6\sqrt{2} \qquad \mathrm{(D)}\ 12 \qquad \mathrm{(E)}\ 6+6\sqrt{2}$ | 2005 AMC 12B Problem 14 | Let $R$ be the radius of the circle. Draw the two radii that meet the points of tangency to the lines $y = \pm x$. We can see that a square is formed by the origin, two tangency points, and the center of the circle. The side lengths of this square are $R$ and the diagonal is $k = R+6$. The diagonal of a square is $\sqrt{2}$ times the side length. Therefore, $R+6 = R\sqrt{2} \Rightarrow R = \dfrac{6}{\sqrt{2}-1} = 6+6\sqrt{2} \Rightarrow \boxed{\mathrm{E}}$. | // Block 1
real Xmin,Xmax,Ymin,Ymax;
real R = 6+6*sqrt(2);
Xmin = -16; Xmax = 16; Ymin = -4; Ymax = 40;
xaxis(Xmin,Xmax,Arrows);
yaxis(Ymin,Ymax,Arrows);
label("$x$",(Xmax+0.25,0),S);
label("$y$",(0,Ymax+0.25),E);
draw((Xmin,-Xmin)--(-Ymin,Ymin));
draw((Xmax,Xmax)--(Ymin,Ymin));
draw((Xmin,6)--(Xmax,6));
dot((0,0)); dot((R/sqrt(2),R/sqrt(2))); dot((-R/sqrt(2),R/sqrt(2))); dot((0,R*sqrt(2)));
draw((0,0)--(R/sqrt(2),R/sqrt(2))--(0,R*sqrt(2))--(-R/sqrt(2),R/sqrt(2))--(0,0));
draw(Circle((0,6+R),R));
label("$R$",(0,6+R/2),(0,0));
label("$R$",(-R/(2*sqrt(2)),3*R/(2*sqrt(2))),0.5*NW);
label("$R$",(R/(2*sqrt(2)),3*R/(2*sqrt(2))),0.5*NE);
label("$R$",(-R/(2*sqrt(2)),R/(2*sqrt(2))),0.5*SW);
label("$R$",(R/(2*sqrt(2)),R/(2*sqrt(2))),0.5*SE);
label("$6$",(0,3),(0,0));
// Block 2
real Xmin,Xmax,Ymin,Ymax; real R = 6+6*sqrt(2); Xmin = -16; Xmax = 16; Ymin = -4; Ymax = 40; xaxis(Xmin,Xmax,Arrows); yaxis(Ymin,Ymax,Arrows); label("$x$",(Xmax+0.25,0),S); label("$y$",(0,Ymax+0.25),E); draw((Xmin,-Xmin)--(-Ymin,Ymin)); draw((Xmax,Xmax)--(Ymin,Ymin)); draw((Xmin,6)--(Xmax,6)); dot((0,0)); dot((R/sqrt(2),R/sqrt(2))); dot((-R/sqrt(2),R/sqrt(2))); dot((0,R*sqrt(2))); draw((0,0)--(R/sqrt(2),R/sqrt(2))--(0,R*sqrt(2))--(-R/sqrt(2),R/sqrt(2))--(0,0)); draw(Circle((0,6+R),R)); label("$R$",(0,6+R/2),(0,0)); label("$R$",(-R/(2*sqrt(2)),3*R/(2*sqrt(2))),0.5*NW); label("$R$",(R/(2*sqrt(2)),3*R/(2*sqrt(2))),0.5*NE); label("$R$",(-R/(2*sqrt(2)),R/(2*sqrt(2))),0.5*SW); label("$R$",(R/(2*sqrt(2)),R/(2*sqrt(2))),0.5*SE); label("$6$",(0,3),(0,0)); | [] |
628 | Let $A(2,2)$ and $B(7,7)$ be points in the plane. Define $R$ as the region in the first quadrant consisting of those points $C$ such that $\triangle ABC$ is an acute triangle. What is the closest integer to the area of the region $R$?
$\mathrm{(A)}\ 25 \qquad \mathrm{(B)}\ 39 \qquad \mathrm{(C)}\ 51 \qquad \mathrm{(D)}\ 60 \qquad \mathrm{(E)}\ 80$ | 2005 AMC 12B Problem 18 | For angle $A$ and $B$ to be acute, $C$ must be between the two lines that are perpendicular to $\overline{AB}$ and contain points $A$ and $B$. For angle $C$ to be acute, first draw a $45-45-90$ triangle with $\overline{AB}$ as the hypotenuse. Note $C$ cannot be inside this triangle's circumscribed circle or else $\angle C > 90^\circ$. Hence, the area of $R$ is the area of the large triangle minus the area of the small triangle minus the area of the circle, which is $\frac{14^2}{2}-\frac{4^2}{2}-(\frac{5\sqrt{2}}{2})^2\pi=98-8-\frac{25\pi}{2}$, which is approximately $51$. The answer is $\boxed{\mathrm{C}}$.
Note: When doing this problem, it is important to double check that the circle with diameter $AB$ lies fully in the first quadrant, which luckily it does. | // Block 1
Label f;
f.p=fontsize(6);
xaxis(-1,15,Ticks(f, 2.0));
yaxis(-1,15,Ticks(f, 2.0));
pair A = MP("A",(2,2),SW), B = MP("B",(7,7),NE);
D(A--B);
filldraw((0,4)--(4,0)--(14,0)--(0,14)--cycle,gray);
filldraw(CP(0.5(A+B),A),white);
D(A);
D(B);
// Block 2
Label f; f.p=fontsize(6); xaxis(-1,15,Ticks(f, 2.0)); yaxis(-1,15,Ticks(f, 2.0)); pair A = MP("A",(2,2),SW), B = MP("B",(7,7),NE); D(A--B); filldraw((0,4)--(4,0)--(14,0)--(0,14)--cycle,gray); filldraw(CP(0.5(A+B),A),white); D(A); D(B); | [] |
629 | All three vertices of an equilateral triangle are on the parabola $y = x^2$, and one of its sides has a slope of $2$. The $x$-coordinates of the three vertices have a sum of $m/n$, where $m$ and $n$ are relatively prime positive integers. What is the value of $m + n$?
$\mathrm{(A)}\ {{{14}}}\qquad\mathrm{(B)}\ {{{15}}}\qquad\mathrm{(C)}\ {{{16}}}\qquad\mathrm{(D)}\ {{{17}}}\qquad\mathrm{(E)}\ {{{18}}}$ | 2005 AMC 12B Problem 24 | Using the slope formula and differences of squares, we find:
$a+b$ = the slope of $AB$,
$b+c$ = the slope of $BC$,
$a+c$ = the slope of $AC$.
So the value that we need to find is the sum of the slopes of the three sides of the triangle divided by $2$. Without loss of generality, let $AB$ be the side that has the smallest angle with the positive $x$-axis. Let $J$ be an arbitrary point with the coordinates $(1, 0)$. Translate the triangle so $A$ is at the origin. Then $\tan(BOJ) = 2$. Since the slope of a line is equal to the tangent of the angle formed by the line and the positive x- axis, the answer is $\dfrac{\tan(BOJ) + \tan(BOJ+60) + \tan(BOJ-60)}{2}$.
Using $\tan(BOJ) = 2$, and the tangent addition formula, this simplifies to $\dfrac{3}{11}$, so the answer is $3 + 11 = \boxed{\mathrm{(A)}\ 14}$ | // Block 1
import graph;
real f(real x) {return x^2;}
unitsize(1 cm);
pair A, B, C;
real a, b, c;
a = (-5*sqrt(3) + 11)/11;
b = (5*sqrt(3) + 11)/11;
c = -19/11;
A = (a,f(a));
B = (b,f(b));
C = (c,f(c));
draw(graph(f,-2,2));
draw((-2,0)--(2,0),Arrows);
draw((0,-0.5)--(0,4),Arrows);
draw(A--B--C--cycle);
label("$x$", (2,0), NE);
label("$y$", (0,4), NE);
dot("$A(a,a^2)$", A, S);
dot("$B(b,b^2)$", B, E);
dot("$C(c,c^2)$", C, W);
// Block 2
import graph; real f(real x) {return x^2;} unitsize(1 cm); pair A, B, C; real a, b, c; a = (-5*sqrt(3) + 11)/11; b = (5*sqrt(3) + 11)/11; c = -19/11; A = (a,f(a)); B = (b,f(b)); C = (c,f(c)); draw(graph(f,-2,2)); draw((-2,0)--(2,0),Arrows); draw((0,-0.5)--(0,4),Arrows); draw(A--B--C--cycle); label("$x$", (2,0), NE); label("$y$", (0,4), NE); dot("$A(a,a^2)$", A, S); dot("$B(b,b^2)$", B, E); dot("$C(c,c^2)$", C, W); | [] |
630 | Six ants simultaneously stand on the six vertices of a regular octahedron, with each ant at a different vertex. Simultaneously and independently, each ant moves from its vertex to one of the four adjacent vertices, each with equal probability. What is the probability that no two ants arrive at the same vertex?
$\mathrm{(A)}\ \frac {5}{256} \qquad\mathrm{(B)}\ \frac {21}{1024} \qquad\mathrm{(C)}\ \frac {11}{512} \qquad\mathrm{(D)}\ \frac {23}{1024} \qquad\mathrm{(E)}\ \frac {3}{128}$ | 2005 AMC 12B Problem 25 | Solution 1
We approach this problem by counting the number of ways ants can do their desired migration, and then multiply this number by the probability that each case occurs.
Let the octahedron be $ABCDEF$, with points $B,C,D,E$ coplanar. Then the ant from $A$ and the ant from $F$ must move to plane $BCDE$. Suppose, without loss of generality, that the ant from $A$ moved to point $B$. Then, we must consider three cases.
Case 1: Ant from point $F$ moved to point $C$
On the plane, points $B$ and $C$ are taken. The ant that moves to $D$ can come from either $E$ or $C$. The ant that moves to $E$ can come from either $B$ or $D$. Once these two ants are fixed, the other two ants must migrate to the "poles" of the octahedron, points $A$ and $F$. Thus, there are two degrees of freedom in deciding which ant moves to $D$, two degrees of freedom in deciding which ant moves to $E$, and two degrees of freedom in deciding which ant moves to $A$. Hence, there are $2 \times 2 \times 2=8$ ways the ants can move to different points.
Case 2: Ant from point $F$ moved to point $D$
On the plane, points $B$ and $D$ are taken. The ant that moves to $C$ must be from $B$ or $D$, but the ant that moves to $E$ must also be from $B$ or $D$. The other two ants, originating from points $C$ and $E$, must move to the poles. Therefore, there are two degrees of freedom in deciding which ant moves to $C$ and two degrees of freedom in choosing which ant moves to $A$. Hence, there are $2 \times 2=4$ ways the ants can move to different points.
Case 3: Ant from point $F$ moved to point $E$
By symmetry to Case 1, there are $8$ ways the ants can move to different points.
Given a point $B$, there is a total of $8+4+8=20$ ways the ants can move to different points. We oriented the square so that point $B$ was defined as the point to which the ant from point $A$ moved. Since the ant from point $A$ can actually move to four different points, there is a total of $4 \times 20=80$ ways the ants can move to different points.
Each ant acts independently, having four different points to choose from. Hence, each ant has probability $1/4$ of moving to the desired location. Since there are six ants, the probability of each case occuring is $\frac{1}{4^6} = \frac{1}{4096}$. Thus, the desired answer is $\frac{80}{4096}= \boxed{\frac{5}{256}} \Rightarrow \mathrm{(A)}$.
Solution 2
Let $f(n)$ be the number of cycles of length $n$ the can be walked among the vertices of an octahedron. For example, $f(3)$ would represent the number of ways in which an ant could navigate $2$ vertices and then return back to the original spot. Since an ant cannot stay still, $f(1) = 0$. We also easily see that $f(2) = 1, f(3) = 2$.
Now consider any four vertices of the octahedron. All four vertices will be connected by edges except for one pair. Let’s think of this as a square with one diagonal (from top left to bottom right). EDIT: This part is wrong as if you choose the 4 vertices that have a cross section as a square, there exists no connecting diagonal.
Suppose an ant moved across this diagonal; then the ant at the other end can only move across the diagonal (which creates 2-cycle, bad) or it can move to another vertex, but then the ant at that vertex must move to the spot of the original ant (which creates 3-cycle, bad). Thus none of the ants can navigate the diagonal and can either shift clockwise or counterclockwise, and so $f(4) = 2$.
For $f(6)$, consider an ant at the top of the octahedron. It has four choices. Afterwards, it can either travel directly to the bottom, and then it has $2$ ways back up, or it can travel along the sides and then go to the bottom, of which simple counting gives us $6$ ways back up. Hence, this totals $4 \times (2+6) = 32$.
Now, the number of possible ways is given by the sum of all possible cycles,
\[a \cdot f(2) \cdot f(2) \cdot f(2) + b \cdot f(2) \cdot f(4) + c \cdot f(3) \cdot f(3) + d \cdot f(6)\]
where the coefficients represent the number of ways we can configure these cycles. To find $a$, fix any face, there are $4$ adjacent faces to select from to complete the cycle. From the four remaining faces there are only $2$ ways to create cycles, hence $a = 8$.
To find $b$, each cycle of $2$ faces is distinguished by their common edge, and there are $12$ edges, so $b = 12$.
To find $c$, each three-cycle is distinguished by the vertex, and there are $8$ edges. However, since the two three-cycles are indistinguishable, $c = 8/2 = 4$.
Clearly $d = 1$. Finally,
\[8(1)(1)(1) + 12(1)(2) + 4(2)(2) + (32) = 80\]
Each bug has $4$ possibilities to choose from, so the probability is $\frac{80}{4^6} = \frac{5}{256}$. | size(30); defaultpen(0.6); pair A = (0,0), B=(5,0), C=(5,5), D=(0,5); draw(A--B--C--D--cycle); draw(B--D); | [] |
631 | The $8\times18$ rectangle $ABCD$ is cut into two congruent hexagons, as shown, in such a way that the two hexagons can be repositioned without overlap to form a square. What is $y$?
$\mathrm{(A)}\ 6\qquad\mathrm{(B)}\ 7\qquad\mathrm{(C)}\ 8\qquad\mathrm{(D)}\ 9\qquad\mathrm{(E)}\ 10$ | 2006 AMC 10A Problem 7 | Since the two hexagons are going to be repositioned to form a square without overlap, the area will remain the same. The rectangle's area is $18\cdot8=144$. This means the square will have four sides of length 12. The only way to do this is shown below.
As you can see from the diagram, the line segment denoted as $y$ is half the length of the side of the square, which leads to $y = \frac{12}{2} = \boxed{\textbf{(A) }6}$.
Solution 2 (Shortcut)
As solution 1 says, the two hexagons are going to be repositioned to form a square without overlap. Thus we create this square out of the original rectangle.
As you can see from the diagram, the length $y$ fits into the previously blank side, so we know that it is equal to $y$.
From there we can say $3y = 18$ so $y = \frac{18}{3} = \boxed{\textbf{(A) }6}$.
~Ezraft | // Block 1
size(175); pair A,B,C,D,E,F,G,H; A=(0,8); B=(12,12); C=(12,4); D=(0,0); E=(0,12); F=(12,0); G=(6,4); H=(6,8); draw(A--E--B--C--G--H--A--D--F--C); label("$A$",A,W); label("$B$",B,NE); label("$C$",(12.6,4)); label("$D$",D,SW); label("$12$",E--B,N); label("$12$",D--F,S); label("$4$",E--A,W); label("$4$",(12.4,-1.75),E); label("$8$",A--D,W); label("$8$",(12.4,4),E); label("$y$",A--H,S); label("$y$",G--C,N);
// Block 2
size(175); pair A,B,C,D,E,F,G,H; A=(0,8); B=(12,12); C=(12,4); D=(0,0); E=(0,12); F=(12,0); G=(6,4); H=(6,8); draw(A--E--B--C--G--H--A--D--F--C); label("$y$",A--H,S); label("$y$",G--C,N);
// Block 3
unitsize(3mm); defaultpen(fontsize(10pt)+linewidth(.8pt)); dotfactor=4; draw((0,4)--(18,4)--(18,-4)--(0,-4)--cycle); draw((6,4)--(6,0)--(12,0)--(12,-4)); label("$y$",(9,-2),NW); label("$A$",(0,4),NW); label("$B$",(18,4),NE); label("$C$",(18,-4),SE); label("$D$",(0,-4),SW); label("$y$",(3,4),S); label("$y$",(15,-4),N); label("$18$",(9,4),N); label("$18$",(9,-4),S); label("$8$",(0,0),W); label("$8$",(18,0),E); dot((0,4)); dot((18,4)); dot((18,-4)); dot((0,-4)); | [] |
632 | Rolly wishes to secure his dog with an 8-foot rope to a square shed that is 16 feet on each side. His preliminary drawings are shown.
Which of these arrangements give the dog the greater area to roam, and by how many square feet?
$\textbf{(A) } I,\,\textrm{ by }\,8\pi\qquad \textbf{(B) } I,\,\textrm{ by }\,6\pi\qquad \textbf{(C) } II,\,\textrm{ by }\,4\pi\qquad \textbf{(D) } II,\,\textrm{ by }\,8\pi\qquad \textbf{(E) } II,\,\textrm{ by }\,10\pi$ | 2006 AMC 10A Problem 12 | Let us first examine the area of both possible arrangements. The rope outlines a circular boundary that the dog may dwell in. Arrangement $I$ allows the dog
$\frac12\cdot(\pi\cdot8^2) = 32\pi$ square feet of area. Arrangement $II$ allows $32\pi$ square feet plus a little more on the top part of the fence. So we already know that Arrangement $II$ allows more freedom - only thing left is to find out how much. The extra area can be represented by a quarter of a circle with radius 4. So the extra area is
$\frac14\cdot(\pi\cdot4^2) = 4\pi$. Thus the answer is $\boxed{\textbf{(C) } II,\,\textrm{ by }\,4\pi}$. | // Block 1
size(150); pathpen = linewidth(0.7); defaultpen(linewidth(0.7)+fontsize(10));
filldraw(arc((16,-8),8,-90,90)--cycle, rgb(0.9,0.9,0.6));
filldraw(arc((16,-26),8,-90,90)--cycle, rgb(0.9,0.9,0.6));
filldraw(arc((16,-22),4,90,180)--(16,-22)--cycle, rgb(0.9,0.9,0.6));
D((0,0)--(16,0)--(16,-16)--(0,-16)--cycle);
D((16,-8)--(24,-8));
label('Dog', (24, -8), SE);
MP('I', (8,-8), (0,0));
MP('8', (16,-4), W);
MP('8', (16,-12),W);
MP('8', (20,-8), N);
label('Rope', (20,-8),S);
pair sD = (0,-2);
D(shift(sD)*((0,-20)--(16,-20)--(16,-36)--(0,-36)--cycle));
D(shift(sD)*((16,-24)--(24,-24)));
MP("II", (8,-28)+sD, (0,0));
MP('4', (16,-22)+sD, W);
MP('8', (20,-24)+sD, N);
label("Dog",(24,-24)+sD,SE);
label("Rope", (20,-24)+sD, S);
// Block 2
size(150); pathpen = linewidth(0.7); defaultpen(linewidth(0.7)+fontsize(10)); filldraw(arc((16,-8),8,-90,90)--cycle, rgb(0.9,0.9,0.6)); filldraw(arc((16,-26),8,-90,90)--cycle, rgb(0.9,0.9,0.6)); filldraw(arc((16,-22),4,90,180)--(16,-22)--cycle, rgb(0.9,0.9,0.6)); D((0,0)--(16,0)--(16,-16)--(0,-16)--cycle); D((16,-8)--(24,-8)); label('Dog', (24, -8), SE); MP('I', (8,-8), (0,0)); MP('8', (16,-4), W); MP('8', (16,-12),W); MP('8', (20,-8), N); label('Rope', (20,-8),S); pair sD = (0,-2); D(shift(sD)*((0,-20)--(16,-20)--(16,-36)--(0,-36)--cycle)); D(shift(sD)*((16,-24)--(24,-24))); MP("II", (8,-28)+sD, (0,0)); MP('4', (16,-22)+sD, W); MP('8', (20,-24)+sD, N); label("Dog",(24,-24)+sD,SE); label("Rope", (20,-24)+sD, S); | [] |
633 | Odell and Kershaw run for $30$ minutes on a circular track. Odell runs clockwise at $\text{250 m/min}$ and uses the inner lane with a radius of $50$ meters. Kershaw runs counterclockwise at $\text{300 m/min}$ and uses the outer lane with a radius of $60$ meters, starting on the same radial line as Odell. How many times after the start do they pass each other?
$\textbf{(A) } 29\qquad\textbf{(B) } 42\qquad\textbf{(C) } 45\qquad\textbf{(D) } 47\qquad\textbf{(E) } 50\qquad$ | 2006 AMC 10A Problem 15 | Since $d = rt$, we note that Odell runs one lap in $\frac{2 \cdot 50\pi}{250} = \frac{2\pi}{5}$ minutes, while Kershaw also runs one lap in $\frac{2 \cdot 60\pi}{300} = \frac{2\pi}{5}$ minutes. They take the same amount of time to run a lap, and since they are running in opposite directions they will meet exactly twice per lap (once at the starting point, the other at the half-way point). Thus, there are $\frac{30}{\frac{2\pi}{5}} \approx 23.8$ laps run by both, or $\lfloor 2\cdot 23.8\rfloor = 23 \cdot 2 + 1 =\boxed{\textbf{(D) } 47}$ meeting points. | // Block 1
draw((5,0){up}..{left}(0,5),red);
draw((-5,0){up}..{right}(0,5),red);
draw((5,0){down}..{left}(0,-5),red);
draw((-5,0){down}..{right}(0,-5),red);
draw((6,0){up}..{left}(0,6),blue);
draw((-6,0){up}..{right}(0,6),blue);
draw((6,0){down}..{left}(0,-6),blue);
draw((-6,0){down}..{right}(0,-6),blue);
// Block 2
draw((5,0){up}..{left}(0,5),red); draw((-5,0){up}..{right}(0,5),red); draw((5,0){down}..{left}(0,-5),red); draw((-5,0){down}..{right}(0,-5),red); draw((6,0){up}..{left}(0,6),blue); draw((-6,0){up}..{right}(0,6),blue); draw((6,0){down}..{left}(0,-6),blue); draw((-6,0){down}..{right}(0,-6),blue); | [] |
634 | A circle of radius $1$ is tangent to a circle of radius $2$. The sides of $\triangle ABC$ are tangent to the circles as shown, and the sides $\overline{AB}$ and $\overline{AC}$ are congruent. What is the area of $\triangle ABC$?
$\textbf{(A) } \frac{35}{2}\qquad\textbf{(B) } 15\sqrt{2}\qquad\textbf{(C) } \frac{64}{3}\qquad\textbf{(D) } 16\sqrt{2}\qquad\textbf{(E) } 24\qquad$ | 2006 AMC 10A Problem 16 | Let the centers of the smaller and larger circles be $O_1$ and $O_2$ , respectively.
Let their tangent points to $\triangle ABC$ be $D$ and $E$, respectively.
We can then draw the following diagram:
We see that $\triangle ADO_1 \sim \triangle AEO_2 \sim \triangle AFC$. Using the first pair of similar triangles, we write the proportion:
$\frac{AO_1}{AO_2} = \frac{DO_1}{EO_2} \Longrightarrow \frac{AO_1}{AO_1 + 3} = \frac{1}{2} \Longrightarrow AO_1 = 3$
By the Pythagorean Theorem, we have $AD = \sqrt{3^2-1^2} = \sqrt{8}$.
Now using $\triangle ADO_1 \sim \triangle AFC$,
$\frac{AD}{AF} = \frac{DO_1}{FC} \Longrightarrow \frac{2\sqrt{2}}{8} = \frac{1}{FC} \Longrightarrow FC = 2\sqrt{2}$
Hence, the area of the triangle is \[\frac{1}{2}\cdot AF \cdot BC = \frac{1}{2}\cdot AF \cdot (2\cdot CF) = AF \cdot CF = 8\left(2\sqrt{2}\right) = \boxed{\textbf{(D) } 16\sqrt{2}}\] | // Block 1
size(200); pathpen = linewidth(0.7); pointpen = black; pointfontpen = fontsize(10);
real t=2^0.5;
D((0,0)--(4*t,0)--(2*t,8)--cycle);
D(CR(D((2*t,2)),2));
D(CR(D((2*t,5)),1));
D('B', (0,0),SW); D('C',(4*t,0), SE); D('A', (2*t, 8), N);
pair A = foot((2*t,2),(2*t,8),(4*t,0)), B = foot((2*t,5),(2*t,8),(4*t,0));
D((2*t,0)--(2*t,8)); D((2*t,2)--D(A)); D((2*t,5)--D(B)); D(rightanglemark((2*t,2),A,(4*t,0))); D(rightanglemark((2*t,5),B,(4*t,0)));
MP('2', (2*t,3), W); MP('1',(2*t, 5.5), W); MP("F",(2*t,0)); MP("O_1",(2*t,5),W); MP("O_2",(2*t,2),W); MP("D",B,NE); MP("E",A,NE);
// Block 2
size(200); pathpen = linewidth(0.7); pointpen = black; pointfontpen = fontsize(10); real t=2^0.5; D((0,0)--(4*t,0)--(2*t,8)--cycle); D(CR(D((2*t,2)),2)); D(CR(D((2*t,5)),1)); D('B', (0,0),SW); D('C',(4*t,0), SE); D('A', (2*t, 8), N); pair A = foot((2*t,2),(2*t,8),(4*t,0)), B = foot((2*t,5),(2*t,8),(4*t,0)); D((2*t,0)--(2*t,8)); D((2*t,2)--D(A)); D((2*t,5)--D(B)); D(rightanglemark((2*t,2),A,(4*t,0))); D(rightanglemark((2*t,5),B,(4*t,0))); MP('2', (2*t,3), W); MP('1',(2*t, 5.5), W); MP("F",(2*t,0)); MP("O_1",(2*t,5),W); MP("O_2",(2*t,2),W); MP("D",B,NE); MP("E",A,NE); | [] |
634 | A circle of radius $1$ is tangent to a circle of radius $2$. The sides of $\triangle ABC$ are tangent to the circles as shown, and the sides $\overline{AB}$ and $\overline{AC}$ are congruent. What is the area of $\triangle ABC$?
$\textbf{(A) } \frac{35}{2}\qquad\textbf{(B) } 15\sqrt{2}\qquad\textbf{(C) } \frac{64}{3}\qquad\textbf{(D) } 16\sqrt{2}\qquad\textbf{(E) } 24\qquad$ | 2006 AMC 10A Problem 16 | Since $\triangle{A O_1 D} \sim \triangle{A O_2 E},$ we have that $\frac{A O_1}{A O_2} = \frac{O_1 D}{O_2 E} = \frac{1}{2}$.
Since we know that $O_1 O_2 = 1 + 2 = 3,$ the total length of $A O_2 = 2 \cdot 3 = 6.$
We also know that $O_2 F = 2$, so $A F = A O_2 + O_2 F = 6 + 2 = 8.$
Also, since $\triangle{ABF} \sim \triangle{A E O_2},$ we have that $\frac{AC}{A O_2} = \frac{FC}{O_2 E}.$
Since we know that $A O_2 = 6$ and $O_2 E = 2,$ we have that $\frac{AC}{6} = \frac{FC}{2}.$
This equation simplified gets us $AC = 3 \cdot FC.$
Let $FC = a$
By the Pythagorean Theorem on $\triangle{AFC},$ we have that $AF^2 + FC^2 = AC^2.$
We know that $AF = 8$, $FC = a$ and $AC = 3a$ so we have $8^2 + a^2 = (3a)^2.$
Simplifying, we have that $64 = 8a^2 \Rightarrow a^2 = 8 \Rightarrow a = \sqrt{8} = 2 \sqrt{2}.$
Recall that $FC=a$.
Therefore, $BC = 2 \cdot FC = 2 \cdot 2 \sqrt{2} = 4 \sqrt{2}.$
Since the height is $AF = 8,$ we have the area equal to $\frac{4 \sqrt{2} \cdot 8}{2}=16\sqrt{2}.$
Thus our answer is $\boxed{\textbf{(D) }16 \sqrt{2}}$.
~mathboy282 | // Block 1
size(200); pathpen = linewidth(0.7); pointpen = black; pointfontpen = fontsize(10);
real t=2^0.5;
D((0,0)--(4*t,0)--(2*t,8)--cycle);
D(CR(D((2*t,2)),2));
D(CR(D((2*t,5)),1));
D('B', (0,0),SW); D('C',(4*t,0), SE); D('A', (2*t, 8), N);
pair A = foot((2*t,2),(2*t,8),(4*t,0)), B = foot((2*t,5),(2*t,8),(4*t,0));
D((2*t,0)--(2*t,8)); D((2*t,2)--D(A)); D((2*t,5)--D(B)); D(rightanglemark((2*t,2),A,(4*t,0))); D(rightanglemark((2*t,5),B,(4*t,0)));
MP('2', (2*t,3), W); MP('1',(2*t, 5.5), W); MP("F",(2*t,0)); MP("O_1",(2*t,5),W); MP("O_2",(2*t,2),W); MP("D",B,NE); MP("E",A,NE);
// Block 2
size(200); pathpen = linewidth(0.7); pointpen = black; pointfontpen = fontsize(10); real t=2^0.5; D((0,0)--(4*t,0)--(2*t,8)--cycle); D(CR(D((2*t,2)),2)); D(CR(D((2*t,5)),1)); D('B', (0,0),SW); D('C',(4*t,0), SE); D('A', (2*t, 8), N); pair A = foot((2*t,2),(2*t,8),(4*t,0)), B = foot((2*t,5),(2*t,8),(4*t,0)); D((2*t,0)--(2*t,8)); D((2*t,2)--D(A)); D((2*t,5)--D(B)); D(rightanglemark((2*t,2),A,(4*t,0))); D(rightanglemark((2*t,5),B,(4*t,0))); MP('2', (2*t,3), W); MP('1',(2*t, 5.5), W); MP("F",(2*t,0)); MP("O_1",(2*t,5),W); MP("O_2",(2*t,2),W); MP("D",B,NE); MP("E",A,NE); | [] |
635 | In rectangle $ADEH$, points $B$ and $C$ trisect $\overline{AD}$, and points $G$ and $F$ trisect $\overline{HE}$. In addition, $AH=AC=2$, and $AD=3$. What is the area of quadrilateral $WXYZ$ shown in the figure?
$\textbf{(A) } \frac{1}{2}\qquad\textbf{(B) } \frac{\sqrt{2}}{2}\qquad\textbf{(C) } \frac{\sqrt{3}}{2}\qquad\textbf{(D) } \sqrt{2} \qquad\textbf{(E) } \frac{2\sqrt{3}}{3}\qquad$ | 2006 AMC 10A Problem 17 | By symmetry, $WXYZ$ is a square.
Draw $\overline{BZ}$. $BZ = \frac 12AH = 1$, so $\triangle BWZ$ is a $45-45-90 \triangle$. Hence $WZ = \frac{1}{\sqrt{2}}$, and $[WXYZ] = \left(\frac{1}{\sqrt{2}}\right)^2 =\boxed{\textbf{(A) }\frac 12}$.
There are many different similar ways to come to the same conclusion using different 45-45-90 triangles.
Solution 2
Drawing lines as shown above and piecing together the triangles, we see that $ABCD$ is made up of $12$ squares congruent to $WXYZ$. Hence $[WXYZ] = \frac{2\cdot 3}{12} =\boxed{\textbf{(A) }\frac 12}$.
Solution 3
We see that if we draw a line to $BZ$ it is half the width of the rectangle so that length would be $1$, and the resulting triangle is a $45-45-90$ so using the Pythagorean Theorem we can get that each side is $\sqrt{\frac{1^2}{2}}$ so the area of the middle square would be $(\sqrt{\frac{1^2}{2}})^2=(\sqrt{\frac{1}{2}})^2=\boxed{\textbf{(A) }\frac{1}{2}}$ which is our answer.
Solution 4
Since $B$ and $C$ are trisection points and $AC = 2$, we see that $AD = 3$. Also, $AC = AH$, so triangle $ACH$ is a right isosceles triangle, i.e. $\angle ACH = \angle AHC = 45^\circ$. By symmetry, triangles $AFH$, $DEG$, and $BED$ are also right isosceles triangles. Therefore, $\angle WAD = \angle WDA = 45^\circ$, which means triangle $AWD$ is also a right isosceles triangle. Also, triangle $AXC$ is a right isosceles triangle.
Then $AW = AD/\sqrt{2} = 3/\sqrt{2}$, and $AX = AC/\sqrt{2} = 2/\sqrt{2}$. Hence, $XW = AW - AX = 3/\sqrt{2} - 2/\sqrt{2} = 1/\sqrt{2}$.
By symmetry, quadrilateral $WXYZ$ is a square, so its area is
\[XW^2 = \left( \frac{1}{\sqrt{2}} \right)^2 = \boxed{\textbf{(A) }\frac{1}{2}}.\]
~made by AoPS (somewhere) -put here by qkddud~ | // Block 1
size(7cm); pathpen = linewidth(0.7); pointpen = black; pointfontpen = fontsize(10);
pair A,B,C,D,E,F,G,H,W,X,Y,Z;
A=(0,2); B=(1,2); C=(2,2); D=(3,2);
H=(0,0); G=(1,0); F=(2,0); E=(3,0);
D('A',A, N); D('B',B,N); D('C',C,N); D('D',D,N); D('E',E,SE); D('F',F,SE); D('G',G,SW); D('H',H,SW);
D(A--F); D(B--E); D(D--G); D(C--H);
Z=IP(A--F, C--H); Y=IP(A--F, D--G); X=IP(B--E,D--G); W=IP(B--E,C--H);
D('W',W,1.6*N); D('X',X,1.6*plain.E); D('Y',Y,1.6*S); D('Z',Z,1.6*plain.W);
MP("1",(A+B)/2,2*N); MP("2",(A+H)/2,plain.W); D(B--Z); MP("1",(B+Z)/2,plain.W); MP("\frac{\sqrt{2}}{2}",(W+Z)/2,plain.SE);
D(A--D--E--H--cycle);
// Block 2
size(7cm); pathpen = linewidth(0.7); pointpen = black; pointfontpen = fontsize(10); pair A,B,C,D,E,F,G,H,W,X,Y,Z; A=(0,2); B=(1,2); C=(2,2); D=(3,2); H=(0,0); G=(1,0); F=(2,0); E=(3,0); D('A',A, N); D('B',B,N); D('C',C,N); D('D',D,N); D('E',E,SE); D('F',F,SE); D('G',G,SW); D('H',H,SW); D(A--F); D(B--E); D(D--G); D(C--H); Z=IP(A--F, C--H); Y=IP(A--F, D--G); X=IP(B--E,D--G); W=IP(B--E,C--H); D('W',W,1.6*N); D('X',X,1.6*plain.E); D('Y',Y,1.6*S); D('Z',Z,1.6*plain.W); MP("1",(A+B)/2,2*N); MP("2",(A+H)/2,plain.W); D(B--Z); MP("1",(B+Z)/2,plain.W); MP("\frac{\sqrt{2}}{2}",(W+Z)/2,plain.SE); D(A--D--E--H--cycle);
// Block 3
size(7cm); pathpen = linewidth(0.7); pointpen = black; pointfontpen = fontsize(10); pair A,B,C,D,E,F,G,H,W,X,Y,Z; A=(0,2); B=(1,2); C=(2,2); D=(3,2); H=(0,0); G=(1,0); F=(2,0); E=(3,0); D('A',A, N); D('B',B,N); D('C',C,N); D('D',D,N); D('E',E,SE); D('F',F,SE); D('G',G,SW); D('H',H,SW); D(A--F); D(B--E); D(D--G); D(C--H); Z=IP(A--F, C--H); Y=IP(A--F, D--G); X=IP(B--E,D--G); W=IP(B--E,C--H); D('W',W,1.6*N); D('X',X,1.6*plain.E); D('Y',Y,1.6*S); D('Z',Z,1.6*plain.W); D(B--D((A+H)/2)--G);D(C--D((E+D)/2)--F); D(A--D--E--H--cycle); | [] |
635 | In rectangle $ADEH$, points $B$ and $C$ trisect $\overline{AD}$, and points $G$ and $F$ trisect $\overline{HE}$. In addition, $AH=AC=2$, and $AD=3$. What is the area of quadrilateral $WXYZ$ shown in the figure?
$\textbf{(A) } \frac{1}{2}\qquad\textbf{(B) } \frac{\sqrt{2}}{2}\qquad\textbf{(C) } \frac{\sqrt{3}}{2}\qquad\textbf{(D) } \sqrt{2} \qquad\textbf{(E) } \frac{2\sqrt{3}}{3}\qquad$ | 2006 AMC 10A Problem 17 | By symmetry, quadrilateral $WXYZ$ is a square.
First step, proving that $\triangle BXD \sim \triangle BWC$.
We can tell quadrilateral $CDGH$ is a parallelogram because $CD \parallel GH$ and $CD \cong GH$.
By knowing that, we can say that $CW \parallel DX$.
Finally, we can now prove $\triangle BXD \sim \triangle BWC$ by AA, with a ratio of 2:1.
Since $BD = DE = 2$ and $\angle BDE = 90$. Then $\triangle BDE$ is a 45-45-90 triangle.
This will make $\angle DBE = 45$ making $\triangle BXD$ and $\triangle BWC$ a 45-45-90 triangle.
This will make, $BD = 2, BC=1, DX=BX=\sqrt 2, BW=WX=\frac{\sqrt 2}{2}$. Since $WX$ is the length of the square, our answer will be $(\frac{\sqrt 2}{2})^2 = \frac{2}{4} = \boxed{\textbf{(A) }\frac{1}{2}}.$
~ghfhgvghj10 | // Block 1
size(7cm); pathpen = linewidth(0.7); pointpen = black; pointfontpen = fontsize(10);
pair A,B,C,D,E,F,G,H,W,X,Y,Z;
A=(0,2); B=(1,2); C=(2,2); D=(3,2);
H=(0,0); G=(1,0); F=(2,0); E=(3,0);
D('A',A, N); D('B',B,N); D('C',C,N); D('D',D,N); D('E',E,SE); D('F',F,SE); D('G',G,SW); D('H',H,SW);
D(A--F); D(B--E); D(D--G); D(C--H);
Z=IP(A--F, C--H); Y=IP(A--F, D--G); X=IP(B--E,D--G); W=IP(B--E,C--H);
D('W',W,1.6*N); D('X',X,1.6*plain.E); D('Y',Y,1.6*S); D('Z',Z,1.6*plain.W);
D(A--D--E--H--cycle);
// Block 2
size(7cm); pathpen = linewidth(0.7); pointpen = black; pointfontpen = fontsize(10); pair A,B,C,D,E,F,G,H,W,X,Y,Z; A=(0,2); B=(1,2); C=(2,2); D=(3,2); H=(0,0); G=(1,0); F=(2,0); E=(3,0); D('A',A, N); D('B',B,N); D('C',C,N); D('D',D,N); D('E',E,SE); D('F',F,SE); D('G',G,SW); D('H',H,SW); D(A--F); D(B--E); D(D--G); D(C--H); Z=IP(A--F, C--H); Y=IP(A--F, D--G); X=IP(B--E,D--G); W=IP(B--E,C--H); D('W',W,1.6*N); D('X',X,1.6*plain.E); D('Y',Y,1.6*S); D('Z',Z,1.6*plain.W); D(A--D--E--H--cycle); | [] |
636 | Centers of adjacent faces of a unit cube are joined to form a regular octahedron. What is the volume of this octahedron?
$\textbf{(A) } \frac{1}{8}\qquad\textbf{(B) } \frac{1}{6}\qquad\textbf{(C) } \frac{1}{4}\qquad\textbf{(D) } \frac{1}{3}\qquad\textbf{(E) } \frac{1}{2}\qquad$ | 2006 AMC 10A Problem 24 | We can break the octahedron into two square pyramids by cutting it along a plane perpendicular to one of its internal diagonals.
The cube has edges of length 1 so all edges of the regular octahedron have length $\frac{\sqrt{2}}{2}$. Then the square base of the pyramid has area $\left(\frac{1}{2}\sqrt{2}\right)^2 = \frac{1}{2}$.
We also know that the height of the pyramid is half the height of the cube, so it is $\frac{1}{2}$. The volume of a pyramid with base area $B$ and height $h$ is $A=\frac{1}{3}Bh$ so each of the pyramids has volume $\frac{1}{3}\left(\frac{1}{2}\right)\left(\frac{1}{2}\right) = \frac{1}{12}$. The whole octahedron is twice this volume, so $\frac{1}{12} \cdot 2 = \boxed{\textbf{(B) }\frac{1}{6}}$. | // Block 1
import three;
real r = 1/2;
triple A = (-0.5,1.5,0);
size(400);
currentprojection=orthographic(1,1/4,1/2);
draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--(0,0,0)^^(0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--(0,0,1)^^(0,0,0)--(0,0,1)^^(1,0,0)--(1,0,1)^^(0,1,0)--(0,1,1)^^(1,1,0)--(1,1,1),gray(0.8));
draw((0,r,r)--(r,1,r)--(1,r,r)--(r,0,r)--cycle^^(r,r,0)--(0,r,r)--(r,r,1)--(r,1,r)--(r,r,0)--(1,r,r)--(r,r,1)--(r,0,r)--(r,r,0));
draw((0,r,r)+A--(r,1,r)+A--(1,r,r)+A--(r,0,r)+A--cycle^^(0,r,r)+A--(r,r,1)+A--(1,r,r)+A^^(r,1,r)+A--(r,r,1)+A--(r,0,r)+A);
// Block 2
import three; real r = 1/2; triple A = (-0.5,1.5,0); size(400); currentprojection=orthographic(1,1/4,1/2); draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--(0,0,0)^^(0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--(0,0,1)^^(0,0,0)--(0,0,1)^^(1,0,0)--(1,0,1)^^(0,1,0)--(0,1,1)^^(1,1,0)--(1,1,1),gray(0.8)); draw((0,r,r)--(r,1,r)--(1,r,r)--(r,0,r)--cycle^^(r,r,0)--(0,r,r)--(r,r,1)--(r,1,r)--(r,r,0)--(1,r,r)--(r,r,1)--(r,0,r)--(r,r,0)); draw((0,r,r)+A--(r,1,r)+A--(1,r,r)+A--(r,0,r)+A--cycle^^(0,r,r)+A--(r,r,1)+A--(1,r,r)+A^^(r,1,r)+A--(r,r,1)+A--(r,0,r)+A); | [] |
637 | A bug starts at one vertex of a cube and moves along the edges of the cube according to the following rule. At each vertex the bug will choose to travel along one of the three edges emanating from that vertex. Each edge has equal probability of being chosen, and all choices are independent. What is the probability that after seven moves the bug will have visited every vertex exactly once?
$\textbf{(A) } \frac{1}{2187}\qquad \textbf{(B) } \frac{1}{729}\qquad \textbf{(C) } \frac{2}{243}\qquad \textbf{(D) } \frac{1}{81} \qquad \textbf{(E) } \frac{5}{243}$ | 2006 AMC 10A Problem 25 | Let us count the good paths. The bug starts at an arbitrary vertex, moves to a neighboring vertex ($3$ ways), and then to a new neighbor ($2$ more ways). So, without loss of generality, let the cube have vertices $ABCDEFGH$ such that $ABCD$ and $EFGH$ are two opposite faces with $A$ above $E$ and $B$ above $F$. The bug starts at $A$ and moves first to $B$, then to $C$.
From this point, there are two cases.
Case $1$: the bug moves to $D$. From $D$, there is only one good move available, to $H$. From $H$, there are two ways to finish the trip, either by going $H \to G \to F \to E$ or $H \to E \to F \to G$. So there are $2$ good paths in this case.
Case $2$: the bug moves to $G$.
Case $2a$: the bug moves $G \to H$. In this case, there are $0$ good paths because it will not be possible to visit both $D$ and $E$ without double-visiting some vertex.
Case $2b$: the bug moves $G \to F$. There is $1$ good path in this case, $F \to E \to H \to D$.
Thus, there are a total of $3\cdot 3 \cdot 2 = 18$ good paths. There were $3^7 = 2187$ possible paths the bug could have taken, so the probability of a random path is good is $\frac{18}{2187} = \boxed{\textbf{(C) }\frac2{243}}$. | import three; unitsize(1cm); size(50); currentprojection=orthographic(1/2,-1,1/2); /* three - currentprojection, orthographic */ draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--cycle); draw((0,0,0)--(0,0,1)); draw((0,1,0)--(0,1,1)); draw((1,1,0)--(1,1,1)); draw((1,0,0)--(1,0,1)); draw((0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--cycle); | [] |
638 | The $8\times18$ rectangle $ABCD$ is cut into two congruent hexagons, as shown, in such a way that the two hexagons can be repositioned without overlap to form a square. What is $y$?
$\mathrm{(A)}\ 6\qquad\mathrm{(B)}\ 7\qquad\mathrm{(C)}\ 8\qquad\mathrm{(D)}\ 9\qquad\mathrm{(E)}\ 10$ | 2006 AMC 12A Problem 6 | Since the two hexagons are going to be repositioned to form a square without overlap, the area will remain the same. The rectangle's area is $18\cdot8=144$. This means the square will have four sides of length 12. The only way to do this is shown below.
As you can see from the diagram, the line segment denoted as $y$ is half the length of the side of the square, which leads to $y = \frac{12}{2} = \boxed{\textbf{(A) }6}$.
Solution 2 (Shortcut)
As solution 1 says, the two hexagons are going to be repositioned to form a square without overlap. Thus we create this square out of the original rectangle.
As you can see from the diagram, the length $y$ fits into the previously blank side, so we know that it is equal to $y$.
From there we can say $3y = 18$ so $y = \frac{18}{3} = \boxed{\textbf{(A) }6}$.
~Ezraft | // Block 1
size(175);
pair A,B,C,D,E,F,G,H;
A=(0,8);
B=(12,12);
C=(12,4);
D=(0,0);
E=(0,12);
F=(12,0);
G=(6,4);
H=(6,8);
draw(A--E--B--C--G--H--A--D--F--C);
label("$A$",A,W); label("$B$",B,NE); label("$C$",(12.6,4)); label("$D$",D,SW);
label("$12$",E--B,N); label("$12$",D--F,S);
label("$4$",E--A,W); label("$4$",(12.4,-1.75),E);
label("$8$",A--D,W); label("$8$",(12.4,4),E);
label("$y$",A--H,S); label("$y$",G--C,N);
// Block 2
size(175); pair A,B,C,D,E,F,G,H; A=(0,8); B=(12,12); C=(12,4); D=(0,0); E=(0,12); F=(12,0); G=(6,4); H=(6,8); draw(A--E--B--C--G--H--A--D--F--C); label("$A$",A,W); label("$B$",B,NE); label("$C$",(12.6,4)); label("$D$",D,SW); label("$12$",E--B,N); label("$12$",D--F,S); label("$4$",E--A,W); label("$4$",(12.4,-1.75),E); label("$8$",A--D,W); label("$8$",(12.4,4),E); label("$y$",A--H,S); label("$y$",G--C,N);
// Block 3
size(175); pair A,B,C,D,E,F,G,H; A=(0,8); B=(12,12); C=(12,4); D=(0,0); E=(0,12); F=(12,0); G=(6,4); H=(6,8); draw(A--E--B--C--G--H--A--D--F--C); label("$y$",A--H,S); label("$y$",G--C,N);
// Block 4
unitsize(3mm); defaultpen(fontsize(10pt)+linewidth(.8pt)); dotfactor=4; draw((0,4)--(18,4)--(18,-4)--(0,-4)--cycle); draw((6,4)--(6,0)--(12,0)--(12,-4)); label("$y$",(9,-2),NW); label("$A$",(0,4),NW); label("$B$",(18,4),NE); label("$C$",(18,-4),SE); label("$D$",(0,-4),SW); label("$y$",(3,4),S); label("$y$",(15,-4),N); label("$18$",(9,4),N); label("$18$",(9,-4),S); label("$8$",(0,0),W); label("$8$",(18,0),E); dot((0,4)); dot((18,4)); dot((18,-4)); dot((0,-4)); | [] |
639 | Which of the following describes the graph of the equation $(x+y)^2=x^2+y^2$?
$\mathrm{(A)}\ \text{the empty set}\qquad\mathrm{(B)}\ \text{one point}\qquad\mathrm{(C)}\ \text{two lines}\qquad\mathrm{(D)}\ \text{a circle}\qquad\mathrm{(E)}\ \text{the entire plane}$ | 2006 AMC 12A Problem 11 | \begin{align*}(x+y)^2&=x^2+y^2\\ x^2 + 2xy + y^2 &= x^2 + y^2\\ 2xy &= 0\end{align*}
Either $x = 0$ or $y = 0$. The union of them is 2 lines, and thus the answer is $\mathrm{(C)}$. | draw((0,-50)--(0,50));draw((-50,0)--(50,0)); | [] |
640 | A bug starts at one vertex of a cube and moves along the edges of the cube according to the following rule. At each vertex the bug will choose to travel along one of the three edges emanating from that vertex. Each edge has equal probability of being chosen, and all choices are independent. What is the probability that after seven moves the bug will have visited every vertex exactly once?
$\textbf{(A) } \frac{1}{2187}\qquad \textbf{(B) } \frac{1}{729}\qquad \textbf{(C) } \frac{2}{243}\qquad \textbf{(D) } \frac{1}{81} \qquad \textbf{(E) } \frac{5}{243}$ | 2006 AMC 12A Problem 20 | Let us count the good paths. The bug starts at an arbitrary vertex, moves to a neighboring vertex ($3$ ways), and then to a new neighbor ($2$ more ways). So, without loss of generality, let the cube have vertices $ABCDEFGH$ such that $ABCD$ and $EFGH$ are two opposite faces with $A$ above $E$ and $B$ above $F$. The bug starts at $A$ and moves first to $B$, then to $C$.
From this point, there are two cases.
Case $1$: the bug moves to $D$. From $D$, there is only one good move available, to $H$. From $H$, there are two ways to finish the trip, either by going $H \to G \to F \to E$ or $H \to E \to F \to G$. So there are $2$ good paths in this case.
Case $2$: the bug moves to $G$.
Case $2a$: the bug moves $G \to H$. In this case, there are $0$ good paths because it will not be possible to visit both $D$ and $E$ without double-visiting some vertex.
Case $2b$: the bug moves $G \to F$. There is $1$ good path in this case, $F \to E \to H \to D$.
Thus, there are a total of $3\cdot 3 \cdot 2 = 18$ good paths. There were $3^7 = 2187$ possible paths the bug could have taken, so the probability of a random path is good is $\frac{18}{2187} = \boxed{\textbf{(C) }\frac2{243}}$. | // Block 1
import three;
unitsize(1cm);
size(50);
currentprojection=orthographic(1/2,-1,1/2);
/* three - currentprojection, orthographic */ draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--cycle);
draw((0,0,0)--(0,0,1));
draw((0,1,0)--(0,1,1));
draw((1,1,0)--(1,1,1));
draw((1,0,0)--(1,0,1));
draw((0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--cycle);
// Block 2
import three; unitsize(1cm); size(50); currentprojection=orthographic(1/2,-1,1/2); /* three - currentprojection, orthographic */ draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--cycle); draw((0,0,0)--(0,0,1)); draw((0,1,0)--(0,1,1)); draw((1,1,0)--(1,1,1)); draw((1,0,0)--(1,0,1)); draw((0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--cycle); | [] |
641 | A circle of radius $2$ is centered at $O$. Square $OABC$ has side length $1$. Sides $AB$ and $CB$ are extended past $B$ to meet the circle at $D$ and $E$, respectively. What is the area of the shaded region in the figure, which is bounded by $BD$, $BE$, and the minor arc connecting $D$ and $E$?
$\mathrm{(A) \ } \frac{\pi}{3}+1-\sqrt{3}\qquad \mathrm{(B) \ } \frac{\pi}{2}(2-\sqrt{3}) \qquad \mathrm{(C) \ } \pi(2-\sqrt{3})\qquad \mathrm{(D) \ } \frac{\pi}{6}+\frac{\sqrt{3}+1}{2}\qquad \mathrm{(E) \ } \frac{\pi}{3}-1+\sqrt{3}$ | 2006 AMC 10B Problem 19 | From the Pythagorean Theorem, we can see that $DA$ is $\sqrt{3}$. Then, $DB = DA - BA = \sqrt{3} - 1$. The area of the shaded element is the area of sector $DOE$ minus the areas of triangle $DBO$ and triangle $EBO$ combined. Below is an image to help.
Using the Base Altitude formula, where $DB$ and $BE$ are the bases and $OA$ and $CO$ are the altitudes, respectively, $[DBO] = [EBO] = \frac{\sqrt{3}-1}{2}$. The area of sector $DOE$ is $\frac{1}{12}$ of circle $O$. The area of circle $O$ is $4\pi$, and therefore we have the area of sector $DBE$ to be $\boxed{\textbf{(A) }\frac{\pi}{3} + 1 - \sqrt{3}}$. | // Block 1
defaultpen(linewidth(0.8));
pair O=origin, A=(1,0), C=(0,1), B=(1,1), D=(1, sqrt(3)), E=(sqrt(3), 1), point=B;
fill(Arc(O, 2, 0, 90)--O--cycle, mediumgray);
clip(B--Arc(O, 2, 30, 60)--cycle);
draw(Circle(origin, 2));
draw((-2,0)--(2,0)^^(0,-2)--(0,2));
draw(A--D^^C--E^^D--O^^E--O^^B--O);
label("$A$", A, dir(point--A));
label("$C$", C, dir(point--C));
label("$O$", O, dir(point--O));
label("$D$", D, dir(point--D));
label("$E$", E, dir(point--E));
label("$B$", (1.33,1.04), SW);
// Block 2
defaultpen(linewidth(0.8)); pair O=origin, A=(1,0), C=(0,1), B=(1,1), D=(1, sqrt(3)), E=(sqrt(3), 1), point=B; fill(Arc(O, 2, 0, 90)--O--cycle, mediumgray); clip(B--Arc(O, 2, 30, 60)--cycle); draw(Circle(origin, 2)); draw((-2,0)--(2,0)^^(0,-2)--(0,2)); draw(A--D^^C--E^^D--O^^E--O^^B--O); label("$A$", A, dir(point--A)); label("$C$", C, dir(point--C)); label("$O$", O, dir(point--O)); label("$D$", D, dir(point--D)); label("$E$", E, dir(point--E)); label("$B$", (1.33,1.04), SW); | [] |
642 | A triangle is partitioned into three triangles and a quadrilateral by drawing two lines from vertices to their opposite sides. The areas of the three triangles are 3, 7, and 7, as shown. What is the area of the shaded quadrilateral?
$\mathrm{(A) \ } 15\qquad \mathrm{(B) \ } 17\qquad \mathrm{(C) \ } \frac{35}{2}\qquad \mathrm{(D) \ } 18\qquad \mathrm{(E) \ } \frac{55}{3}$ | 2006 AMC 10B Problem 23 | Label the points in the figure as shown below, and draw the segment $CF$. This segment divides the quadrilateral into two triangles, let their areas be $x$ and $y$.
Since triangles $AFB$ and $DFB$ share an altitude from $B$ and have equal area, their bases must be equal, hence $AF=DF$.
Since triangles $AFC$ and $DFC$ share an altitude from $C$ and their respective bases are equal, their areas must be equal, hence $x+3=y$.
Since triangles $EFA$ and $BFA$ share an altitude from $A$ and their respective areas are in the ratio $3:7$, their bases must be in the same ratio, hence $EF:FB = 3:7$.
Since triangles $EFC$ and $BFC$ share an altitude from $C$ and their respective bases are in the ratio $3:7$, their areas must be in the same ratio, hence $x:(y+7) = 3:7$, which gives us $7x = 3(y+7)$.
Substituting $y=x+3$ into the second equation we get $7x = 3(x+10)$, which solves to $x=\frac{15}{2}$. Then $y=x+3 = \frac{15}{2}+3 = \frac{21}{2}$, and the total area of the quadrilateral is $x+y = \frac{15}{2}+\frac{21}{2} = \boxed{\textbf{(D) }18}$. | // Block 1
unitsize(2cm);
defaultpen(.8);
pair A = (0,0), B = (3,0), C = (1.4, 2), D = B + 0.4*(C-B), Ep = A + 0.3*(C-A);
pair F = intersectionpoint( A--D, B--Ep );
draw( A -- B -- C -- cycle );
draw( A -- D );
draw( B -- Ep );
filldraw( D -- F -- Ep -- C -- cycle, mediumgray, black );
label("$7$",(1.45,0.15));
label("$7$",(2.2,0.45));
label("$3$",(0.45,0.35));
draw( C -- F, dashed );
label("$A$",A,SW);
label("$B$",B,SE);
label("$C$",C,N);
label("$D$",D,NE);
label("$E$",Ep,NW);
label("$F$",F,S);
label("$x$",(1,1));
label("$y$",(1.6,1));
// Block 2
unitsize(2cm); defaultpen(.8); pair A = (0,0), B = (3,0), C = (1.4, 2), D = B + 0.4*(C-B), Ep = A + 0.3*(C-A); pair F = intersectionpoint( A--D, B--Ep ); draw( A -- B -- C -- cycle ); draw( A -- D ); draw( B -- Ep ); filldraw( D -- F -- Ep -- C -- cycle, mediumgray, black ); label("$7$",(1.45,0.15)); label("$7$",(2.2,0.45)); label("$3$",(0.45,0.35)); draw( C -- F, dashed ); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N); label("$D$",D,NE); label("$E$",Ep,NW); label("$F$",F,S); label("$x$",(1,1)); label("$y$",(1.6,1)); | [] |
643 | Rectangle $ABCD$ has area $2006$. An ellipse with area $2006\pi$ passes through $A$ and $C$ and has foci at $B$ and $D$. What is the perimeter of the rectangle? (The area of an ellipse is $ab\pi$ where $2a$ and $2b$ are the lengths of the axes.)
$\mathrm{(A)}\ \frac {16\sqrt {2006}}{\pi} \qquad \mathrm{(B)}\ \frac {1003}4 \qquad \mathrm{(C)}\ 8\sqrt {1003} \qquad \mathrm{(D)}\ 6\sqrt {2006} \qquad \mathrm{(E)}\ \frac {32\sqrt {1003}}\pi$ | 2006 AMC 12B Problem 21 | Let the rectangle have side lengths $l$ and $w$. Let the axis of the ellipse on which the foci lie have length $2a$, and let the other axis have length $2b$. We have
\[lw=ab=2006\]
From the definition of an ellipse, $l+w=2a\Longrightarrow \frac{l+w}{2}=a$. Also, the diagonal of the rectangle has length $\sqrt{l^2+w^2}$. Comparing the lengths of the axes and the distance from the foci to the center, we have
\[a^2=\frac{l^2+w^2}{4}+b^2\Longrightarrow \frac{l^2+2lw+w^2}{4}=\frac{l^2+w^2}{4}+b^2\Longrightarrow \frac{lw}{2}=b^2\Longrightarrow b=\sqrt{1003}\]
Since $ab=2006$, we now know $a\sqrt{1003}=2006\Longrightarrow a=2\sqrt{1003}$ and because $a=\frac{l+w}{2}$, or one-fourth of the rectangle's perimeter, we multiply by four to get an answer of $\boxed{8\sqrt{1003}}$. | // Block 1
size(7cm);
real l=10,
w=7,
ang=asin(w/sqrt(l*l+w*w))*180/pi;
draw((-l,-w)--(l,-w)--(l,w)--(-l,w)--cycle);
draw(rotate(ang)*ellipse((0,0),2*l+2*w,l*w*2/sqrt(l^2+w^2)));
draw(rotate(ang)*((0,0)--(2l+2w,0)),red);
draw(rotate(ang+90)*((0,0)--(l*w*2/sqrt(l^2+w^2),0)),red);
label("$A$",(-l,w),NW);
label("$B$",(-l,-w),SW);
label("$C$",(l,-w),SE);
label("$D$",(l,w),SE);
// Made by chezbgone2
// Block 2
size(7cm); real l=10, w=7, ang=asin(w/sqrt(l*l+w*w))*180/pi; draw((-l,-w)--(l,-w)--(l,w)--(-l,w)--cycle); draw(rotate(ang)*ellipse((0,0),2*l+2*w,l*w*2/sqrt(l^2+w^2))); draw(rotate(ang)*((0,0)--(2l+2w,0)),red); draw(rotate(ang+90)*((0,0)--(l*w*2/sqrt(l^2+w^2),0)),red); label("$A$",(-l,w),NW); label("$B$",(-l,-w),SW); label("$C$",(l,-w),SE); label("$D$",(l,w),SE); // Made by chezbgone2 | [] |
644 | Isosceles $\triangle ABC$ has a right angle at $C$. Point $P$ is inside $\triangle ABC$, such that $PA=11$, $PB=7$, and $PC=6$. Legs $\overline{AC}$ and $\overline{BC}$ have length $s=\sqrt{a+b\sqrt{2}}$, where $a$ and $b$ are positive integers. What is $a+b$?
$\mathrm{(A)}\ 85 \qquad \mathrm{(B)}\ 91 \qquad \mathrm{(C)}\ 108 \qquad \mathrm{(D)}\ 121 \qquad \mathrm{(E)}\ 127$ | 2006 AMC 12B Problem 23 | Using the Law of Cosines on $\triangle PBC$, we have:
\begin{align*} PB^2&=BC^2+PC^2-2\cdot BC\cdot PC\cdot \cos(\alpha) \Rightarrow 49 = 36 + s^2 - 12s\cos(\alpha) \Rightarrow \cos(\alpha) = \dfrac{s^2-13}{12s}. \end{align*}
Using the Law of Cosines on $\triangle PAC$, we have:
\begin{align*} PA^2&=AC^2+PC^2-2\cdot AC\cdot PC\cdot \cos(90^\circ-\alpha) \Rightarrow 121 = 36 + s^2 - 12s\sin(\alpha) \Rightarrow \sin(\alpha) = \dfrac{s^2-85}{12s}. \end{align*}
Now we use $\sin^2(\alpha) + \cos^2(\alpha) = 1$.
\begin{align*} \sin^2(\alpha)+\cos^2(\alpha) = 1 &\Rightarrow \frac{s^4-26s^2+169}{144s^2} + \frac{s^4-170s^2+7225}{144s^2} = 1 \\ &\Rightarrow s^4-170s^2+3697 = 0 \\ &\Rightarrow s^2 = \dfrac{170 \pm 84\sqrt{2}}{2} = 85 \pm 42\sqrt2 \end{align*}
Note that we know that we want the solution with $s^2 > 85$ since we know that $\sin(\alpha) > 0$. Thus, $a+b=85+42=\boxed{127}$. | // Block 1
pathpen = linewidth(0.7);
pen f = fontsize(10);
size(5cm);
pair B = (0,sqrt(85+42*sqrt(2)));
pair A = (B.y,0);
pair C = (0,0);
pair P = IP(arc(B,7,180,360),arc(C,6,0,90));
D(A--B--C--cycle);
D(P--A);
D(P--B);
D(P--C);
MP("A",D(A),plain.E,f);
MP("B",D(B),plain.N,f);
MP("C",D(C),plain.SW,f);
MP("P",D(P),plain.NE,f);
MP("\alpha",C,5*dir(80),f);
MP("90^\circ-\alpha",C,3*dir(30),f);
MP("s",(A+C)/2,plain.S,f);
MP("s",(B+C)/2,plain.W,f);
// Block 2
pathpen = linewidth(0.7); pen f = fontsize(10); size(5cm); pair B = (0,sqrt(85+42*sqrt(2))); pair A = (B.y,0); pair C = (0,0); pair P = IP(arc(B,7,180,360),arc(C,6,0,90)); D(A--B--C--cycle); D(P--A); D(P--B); D(P--C); MP("A",D(A),plain.E,f); MP("B",D(B),plain.N,f); MP("C",D(C),plain.SW,f); MP("P",D(P),plain.NE,f); MP("\alpha",C,5*dir(80),f); MP("90^\circ-\alpha",C,3*dir(30),f); MP("s",(A+C)/2,plain.S,f); MP("s",(B+C)/2,plain.W,f); | [] |
644 | Isosceles $\triangle ABC$ has a right angle at $C$. Point $P$ is inside $\triangle ABC$, such that $PA=11$, $PB=7$, and $PC=6$. Legs $\overline{AC}$ and $\overline{BC}$ have length $s=\sqrt{a+b\sqrt{2}}$, where $a$ and $b$ are positive integers. What is $a+b$?
$\mathrm{(A)}\ 85 \qquad \mathrm{(B)}\ 91 \qquad \mathrm{(C)}\ 108 \qquad \mathrm{(D)}\ 121 \qquad \mathrm{(E)}\ 127$ | 2006 AMC 12B Problem 23 | Rotate triangle $PAC$ 90 degrees counterclockwise about $C$ so that the image of $A$ rests on $B$. Now let the image of $P$ be $\widetilde{P}$.
Note that $\widetilde{P}C=6$, meaning triangle $PC\widetilde{P}$ is right isosceles, and $\angle P\widetilde{P}C=45^\circ$. Then $P\widetilde{P}=6\sqrt{2}$. Now because $PB=7$ and $\widetilde{P}B=11$, we observe that $\angle \widetilde{P}PB=90^\circ$, by the Pythagorean Theorem on $\widetilde{P}PB$. Now we have that $\angle APC=\angle B\widetilde{P}C=\angle B\widetilde{P}P + \angle P\widetilde{P}C$. So we take the cosine of the second equality, using that fact that $\angle P\widetilde{P}C=45^\circ$, to get $\cos(B\widetilde{P}C)=\frac{6\sqrt{2}-7}{11\sqrt{2}}$. Finally, we use the fact that $\cos(B\widetilde{P}C)=\cos(CPA)$ and use the Law of Cosines on triangle $CPA$ to arrive at the value of $s^2$.
Or notice that since $\angle \widetilde{P}PB=90^\circ$ and $\angle P\widetilde{P}C=45^\circ$, we have $\angle BPC=135^\circ$, and Law of Cosines on triangle $BPC$ gives the value of $s^2$. | // Block 1
pathpen = linewidth(0.7);
pen f = fontsize(10); size(10cm);
pair B = (0,sqrt(85+42*sqrt(2)));
pair A = (B.y,0);
pair C = (0,0);
pair P = IP(arc(B,7,180,360),arc(C,6,0,90));
draw(A--B--C--cycle, black+0.8); D(P--A); D(P--B); D(P--C);
MP("A",D(A),E,f); MP("B, \widetilde{A}",D(B),N,f); MP("C",D(C),S,f); MP("P",D(P),NE,f);
pair Bp = dir(90)*(B-origin);
pair Pp = dir(90)*(P-origin);
D(B--Bp--C--Pp--B);
MP("\widetilde{P}",D(Pp),SW,f); MP("\widetilde{B}",D(Bp),W,f);
// Block 2
pathpen = linewidth(0.7); pen f = fontsize(10); size(10cm); pair B = (0,sqrt(85+42*sqrt(2))); pair A = (B.y,0); pair C = (0,0); pair P = IP(arc(B,7,180,360),arc(C,6,0,90)); draw(A--B--C--cycle, black+0.8); D(P--A); D(P--B); D(P--C); MP("A",D(A),E,f); MP("B, \widetilde{A}",D(B),N,f); MP("C",D(C),S,f); MP("P",D(P),NE,f); pair Bp = dir(90)*(B-origin); pair Pp = dir(90)*(P-origin); D(B--Bp--C--Pp--B); MP("\widetilde{P}",D(Pp),SW,f); MP("\widetilde{B}",D(Bp),W,f); | [] |
645 | Let $S$ be the set of all point $(x,y)$ in the coordinate plane such that $0 \le x \le \frac{\pi}{2}$ and $0 \le y \le \frac{\pi}{2}$. What is the area of the subset of $S$ for which \[\sin^2x-\sin x \sin y + \sin^2y \le \frac34?\]
$\mathrm{(A)}\ \dfrac{\pi^2}{9} \qquad \mathrm{(B)}\ \dfrac{\pi^2}{8} \qquad \mathrm{(C)}\ \dfrac{\pi^2}{6} \qquad \mathrm{(D)}\ \dfrac{3\pi^2}{16} \qquad \mathrm{(E)}\ \dfrac{2\pi^2}{9}$ | 2006 AMC 12B Problem 24 | We start out by solving the equality first.
\begin{align*} \sin^2x - \sin x \sin y + \sin^2y &= \frac34 \\ \sin x &= \frac{\sin y \pm \sqrt{\sin^2 y - 4 ( \sin^2y - \frac34 ) }}{2} \\ &= \frac{\sin y \pm \sqrt{3 - 3 \sin^2 y }}{2} \\ &= \frac{\sin y \pm \sqrt{3 \cos^2 y }}{2} \\ &= \frac12 \sin y \pm \frac{\sqrt3}{2} \cos y \\ \sin x &= \sin (y \pm \frac{\pi}{3}) \end{align*}
We end up with three lines that matter: $x = y + \frac\pi3$, $x = y - \frac\pi3$, and $x = \pi - (y + \frac\pi3) = \frac{2\pi}{3} - y$. We plot these lines below.
Note that by testing the point $(\pi/6,\pi/6)$, we can see that we want the area of the pentagon. We can calculate that by calculating the area of the square and then subtracting the area of the 3 triangles. (Note we could also do this by adding the areas of the isosceles triangle in the bottom left corner and the rectangle with the previous triangle's hypotenuse as the longer side.)
\begin{align*} A &= \left(\frac{\pi}{2}\right)^2 - 2 \cdot \frac12 \cdot \left(\frac{\pi}{6}\right)^2 - \frac12 \cdot \left(\frac{\pi}{3}\right)^2 \\ &= \pi^2 \left ( \frac14 - \frac1{36} - \frac1{18}\right ) \\ &= \pi^2 \left ( \frac{9-1-2}{36} \right ) = \boxed{\text{(C)}\ \frac{\pi^2}{6}} \end{align*} | // Block 1
size(5cm);
D((0,0)--(3,0)--(3,3)--(0,3)--cycle);
D((1,-0.1)--(1,0.1));
D((2,-0.1)--(2,0.1));
D((-0.1,1)--(0.1,1));
D((-0.1,2)--(0.1,2));
D((2,0)--(3,1)--(1,3)--(0,2));
MP("\frac{\pi}{6}", (1,0), plain.S);
MP("\frac{\pi}{3}", (2,0), plain.S);
MP("\frac{\pi}{2}", (3,0), plain.S);
MP("\frac{\pi}{6}", (0,1), plain.W);
MP("\frac{\pi}{3}", (0,2), plain.W);
MP("\frac{\pi}{2}", (0,3), plain.W);
// Block 2
size(5cm); D((0,0)--(3,0)--(3,3)--(0,3)--cycle); D((1,-0.1)--(1,0.1)); D((2,-0.1)--(2,0.1)); D((-0.1,1)--(0.1,1)); D((-0.1,2)--(0.1,2)); D((2,0)--(3,1)--(1,3)--(0,2)); MP("\frac{\pi}{6}", (1,0), plain.S); MP("\frac{\pi}{3}", (2,0), plain.S); MP("\frac{\pi}{2}", (3,0), plain.S); MP("\frac{\pi}{6}", (0,1), plain.W); MP("\frac{\pi}{3}", (0,2), plain.W); MP("\frac{\pi}{2}", (0,3), plain.W); | [] |
645 | Let $S$ be the set of all point $(x,y)$ in the coordinate plane such that $0 \le x \le \frac{\pi}{2}$ and $0 \le y \le \frac{\pi}{2}$. What is the area of the subset of $S$ for which \[\sin^2x-\sin x \sin y + \sin^2y \le \frac34?\]
$\mathrm{(A)}\ \dfrac{\pi^2}{9} \qquad \mathrm{(B)}\ \dfrac{\pi^2}{8} \qquad \mathrm{(C)}\ \dfrac{\pi^2}{6} \qquad \mathrm{(D)}\ \dfrac{3\pi^2}{16} \qquad \mathrm{(E)}\ \dfrac{2\pi^2}{9}$ | 2006 AMC 12B Problem 24 | We can write the given equation as \[\sin^3x + \sin^3y \le \frac{3}{4}(\sin x + \sin y).\] Note that when $x = 0$, we have $\sin y \le \frac{\sqrt{3}}{2}$ which implies $y \le \frac{\pi}{3}$. Similary we have $x \le \frac{\pi}{3}$ when $y = 0$. Then we see what happens at $x = \frac{\pi}{2}$. Clearly at $x = \frac{\pi}{2}$, we see that $y \le \frac{\pi}{6}$. By symmetry we have $x \le \frac{\pi}{6}$ when $y = \frac{\pi}{2}$. So we get a graph like
Thus the area of what we are interested in is \[\frac{\pi^2}{4} - (\frac{\pi^2}{18} + \frac{\pi^2}{36}) = \frac{\pi^2}{6}.\]
~coolmath_2018 | // Block 1
size(5cm);
D((0,0)--(3,0)--(3,3)--(0,3)--cycle);
D((1,-0.1)--(1,0.1));
D((2,-0.1)--(2,0.1));
D((-0.1,1)--(0.1,1));
D((-0.1,2)--(0.1,2));
D((2,0)--(3,1)--(1,3)--(0,2));
MP("\frac{\pi}{6}", (1,0), plain.S);
MP("\frac{\pi}{3}", (2,0), plain.S);
MP("\frac{\pi}{2}", (3,0), plain.S);
MP("\frac{\pi}{6}", (0,1), plain.W);
MP("\frac{\pi}{3}", (0,2), plain.W);
MP("\frac{\pi}{2}", (0,3), plain.W);
// Block 2
size(5cm); D((0,0)--(3,0)--(3,3)--(0,3)--cycle); D((1,-0.1)--(1,0.1)); D((2,-0.1)--(2,0.1)); D((-0.1,1)--(0.1,1)); D((-0.1,2)--(0.1,2)); D((2,0)--(3,1)--(1,3)--(0,2)); MP("\frac{\pi}{6}", (1,0), plain.S); MP("\frac{\pi}{3}", (2,0), plain.S); MP("\frac{\pi}{2}", (3,0), plain.S); MP("\frac{\pi}{6}", (0,1), plain.W); MP("\frac{\pi}{3}", (0,2), plain.W); MP("\frac{\pi}{2}", (0,3), plain.W); | [] |
646 | The following problem is from the 2007 AMC 12A #9 and 2007 AMC 10A #13, so those problems redirect to this page.
The trip from Carville to Nikpath requires $4\frac 12$ hours when traveling at an average speed of 70 miles per hour. How many hours does the trip require when traveling at an average speed of 60 miles per hour? Express your answer as a decimal to the nearest hundredth.
Contents
1 Simple Solution
2 Solution 1
3 Solution 2
4 Solution 3
5 Solution 4
6 Video Solution
7 See also | 2007 AMC 10A Problem 13 | Let $H$ represent Yan's home, $S$ represent the stadium, and $Y$ represent Yan's current position. If Yan walks directly to the stadium, he will reach Point $P$ the same time he will reach $H$ if he is walking home.
Since he bikes $7$ times as fast as he walks and the time is the same, the distance from his home to the stadium must be $7$ times the distance from $P$ to the stadium. If $PS=x$, then $HS=7x$ and $HP=6x$. Since Y is the midpoint of $\overline{HP}$, $HY=YP=3x$. Therefore, the ratio of Yan's distance from his home to his distance from the stadium is $\frac{YH}{YS}=\frac{3x}{4x}=\boxed{\mathrm{(B)}\ \frac{3}{4}}$ | draw((0,0)--(7,0)); dot((0,0)); dot((3,0)); dot((6,0)); dot((7,0)); label("$H$",(0,0),S); label("$Y$",(3,0),S); label("$P$",(6,0),S); label("$S$",(7,0),S); | [] |
647 | Consider the $12$-sided polygon $ABCDEFGHIJKL$, as shown. Each of its sides has length $4$, and each two consecutive sides form a right angle. Suppose that $\overline{AG}$ and $\overline{CH}$ meet at $M$. What is the area of quadrilateral $ABCM$?
$\text{(A)}\ \frac {44}{3}\qquad \text{(B)}\ 16 \qquad \text{(C)}\ \frac {88}{5}\qquad \text{(D)}\ 20 \qquad \text{(E)}\ \frac {62}{3}$ | 2007 AMC 10A Problem 18 | Solution 1
We can obtain the solution by calculating the area of rectangle $ABGH$ minus the combined area of triangles $\triangle AHG$ and $\triangle CGM$.
We know that triangles $\triangle AMH$ and $\triangle GMC$ are similar because $\overline{AH} \parallel \overline{CG}$. Also, since $\frac{AH}{CG} = \frac{3}{2}$, the ratio of the distance from $M$ to $\overline{AH}$ to the distance from $M$ to $\overline{CG}$ is also $\frac{3}{2}$. Solving with the fact that the distance from $\overline{AH}$ to $\overline{CG}$ is 4, we see that the distance from $M$ to $\overline{CG}$ is $\frac{8}{5}$.
The area of $\triangle AHG$ is simply $\frac{1}{2} \cdot 4 \cdot 12 = 24$, the area of $\triangle CGM$ is $\frac{1}{2} \cdot \frac{8}{5} \cdot 8 = \frac{32}{5}$, and the area of rectangle $ABGH$ is $4 \cdot 12 = 48$.
Taking the area of rectangle $ABGH$ and subtracting the combined area of $\triangle AHG$ and $\triangle CGM$ yields $48 - \left(24 + \frac{32}{5}\right) = \boxed{\frac{88}{5}}\ \text{(C)}$.
Solution 2
Extend $AB$ and $CH$ and call their intersection $N$.
The triangles $CBN$ and $CGH$ are clearly similar with ratio $1:2$, hence $BN=2$ and thus $AN=6$. The area of the triangle $BCN$ is $\frac{2\cdot 4}2 = 4$.
The triangles $MAN$ and $MGH$ are similar as well, and we now know that the ratio of their dimensions is $AN:GH = 6:4 = 3:2$.
Draw altitudes from $M$ onto $AN$ and $GH$, let their feet be $M_1$ and $M_2$. We get that $MM_1 : MM_2 = 3:2$. Hence $MM_1 = \frac 35 \cdot 12 = \frac {36}5$. (An alternate way is by seeing that the set-up AHGCM is similar to the 2 pole problem (http://www.artofproblemsolving.com/wiki/index.php/1951_AHSME_Problems/Problem_30). Therefore, $MM_2$ must be $\frac{1}{\frac{1}{8}+\frac{1}{12}} = \frac{24}{5}$, by the harmonic mean. Thus, $MM_1$ must be $\frac{36}{5}$.)
Then the area of $AMN$ is $\frac 12 \cdot AN \cdot MM_1 = \frac{108}5$, and the area of $ABCM$ can be obtained by subtracting the area of $BCN$, which is $4$. Hence the answer is $\frac{108}5 - 4 = \boxed{\frac{88}5}$. | unitsize(2cm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(1,3), B=(2,3), C=(2,2), D=(3,2), Ep=(3,1), F=(2,1), G=(2,0), H=(1,0), I=(1,1), J=(0,1), K=(0,2), L=(1,2); pair M=intersectionpoints(A--G,H--C)[0]; pair Z=(2.5,3); draw(A--B--C--D--Ep--F--G--H--I--J--K--L--cycle); draw(A--G); draw(H--C); draw(B--Z--C); draw(C--F); dot(M); label("$A$",A,NW); label("$B$",B,N); label("$C$",C,SE); label("$D$",D,NE); label("$E$",Ep,SE); label("$F$",F,SE); label("$G$",G,SE); label("$H$",H,SW); label("$I$",I,SW); label("$J$",J,SW); label("$K$",K,NW); label("$L$",L,NW); label("$M$",M,W); label("$N$",Z,NE); | [] |
647 | Consider the $12$-sided polygon $ABCDEFGHIJKL$, as shown. Each of its sides has length $4$, and each two consecutive sides form a right angle. Suppose that $\overline{AG}$ and $\overline{CH}$ meet at $M$. What is the area of quadrilateral $ABCM$?
$\text{(A)}\ \frac {44}{3}\qquad \text{(B)}\ 16 \qquad \text{(C)}\ \frac {88}{5}\qquad \text{(D)}\ 20 \qquad \text{(E)}\ \frac {62}{3}$ | 2007 AMC 10A Problem 18 | We can use coordinates to solve this. Let $H=(0,0).$ Thus, we have $A=(0,12), C=(4,8), G=(4,0).$ Therefore, $AG$ has equation $-3x+12=y$ and $HC$ has equation $2x=y.$ Solving, we have $M=(\frac {12}{5},\frac {24}{5}).$ Using the Shoelace Theorem (or you could connect $LC$ and solve for the resulting triangle + trapezoid areas), we find $[ABCM]=\boxed{\mathrm{(C) \ }\dfrac{88}{5}}.$
~(minor edits by Arcticturn) | // Block 1
unitsize(13mm);
defaultpen(linewidth(.8pt)+fontsize(10pt));
dotfactor=4;
pair A=(1,3), B=(2,3), C=(2,2), D=(3,2), Ep=(3,1), F=(2,1), G=(2,0), H=(1,0), I=(1,1), J=(0,1), K=(0,2), L=(1,2);
pair M=intersectionpoints(A--G,H--C)[0];
draw(A--B--C--D--Ep--F--G--H--I--J--K--L--cycle);
draw(A--G);
draw(H--C);
dot(M);
label("$A$",A,NW);
label("$B$",B,NE);
label("$C$",C,NE);
label("$D$",D,NE);
label("$E$",Ep,SE);
label("$F$",F,SE);
label("$G$",G,SE);
label("$H$",H,SW);
label("$I$",I,SW);
label("$J$",J,SW);
label("$K$",K,NW);
label("$L$",L,NW);
label("$M$",M,W);
// Block 2
unitsize(13mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(1,3), B=(2,3), C=(2,2), D=(3,2), Ep=(3,1), F=(2,1), G=(2,0), H=(1,0), I=(1,1), J=(0,1), K=(0,2), L=(1,2); pair M=intersectionpoints(A--G,H--C)[0]; draw(A--B--C--D--Ep--F--G--H--I--J--K--L--cycle); draw(A--G); draw(H--C); dot(M); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,NE); label("$D$",D,NE); label("$E$",Ep,SE); label("$F$",F,SE); label("$G$",G,SE); label("$H$",H,SW); label("$I$",I,SW); label("$J$",J,SW); label("$K$",K,NW); label("$L$",L,NW); label("$M$",M,W); | [] |
648 | A sphere is inscribed in a cube that has a surface area of $24$ square meters. A second cube is then inscribed within the sphere. What is the surface area in square meters of the inner cube?
$\text{(A)}\ 3 \qquad \text{(B)}\ 6 \qquad \text{(C)}\ 8 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 12$ | 2007 AMC 10A Problem 21 | Solution 1
We rotate the smaller cube around the sphere such that two opposite vertices of the cube are on opposite faces of the larger cube. Thus the main diagonal of the smaller cube is the side length of the outer square.
Let $S$ be the surface area of the inner square. The ratio of the areas of two similar figures is equal to the square of the ratio of their sides. As the diagonal of a cube has length $s\sqrt{3}$ where $s$ is a side of the cube, the ratio of a side of the inner square to a side of the outer square is $\frac{1}{\sqrt{3}}$ (since the side of the outer square = the diagonal of the inner square). So we have $\frac{S}{24} = \left(\frac{1}{\sqrt{3}}\right)^2$. Thus $S = 8\Rightarrow \mathrm{\boxed{(C)}}$.
Solution 2 (computation)
The area of each face of the outer cube is $\frac {24}{6} = 4$, and the edge length of the outer cube is $2$. This is also the diameter of the sphere, and thus the length of a long diagonal of the inner cube.
A long diagonal of a cube is the hypotenuse of a right triangle with a side of the cube and a face diagonal of the cube as legs. If a side of the cube is $x$, we see that $2 = \sqrt {x^{2} + (\sqrt {2}x)^{2}}\Rightarrow x = \frac {2}{\sqrt {3}}$.
Thus the surface area of the inner cube is $6x^{2} = 6\left(\frac {2}{\sqrt {3}}\right )^{2} = 8$.
Solution 3, from AoPS
Since the surface area of the original cube is 24 square meters, each face of the cube has a surface area of $24/6 = 4$ square meters, and the side length of this cube is 2 meters. The sphere inscribed within the cube has diameter 2 meters, which is also the length of the diagonal of the cube inscribed in the sphere. Let $l$ represent the side length of the inscribed cube. Applying the Pythagorean Theorem twice gives\[l^2 + l^2 + l^2 = 2^2 = 4.\]Hence each face has surface area\[l^2 = \frac{4}{3} \ \text{square meters}.\]So the surface area of the inscribed cube is $6\cdot \frac43 = \boxed{8}$ square meters.
Solution 4
First of all, it is pretty easy to see the length of each edge of the bigger cube is $2$ so the radius of the sphere is $1$. We know that when a cube is inscribed in a sphere, assuming the edge length of the square is $x$ the radius can be presented as $\frac {\sqrt3x}{2} = 1$, we can solve that $x=\frac {2\sqrt3}{3}$ and now we only need to apply the basic formula to find the surface and we got our final answer as $\frac {2\sqrt3}{3}*\frac {2\sqrt3}{3}*6=8$ and the final answer is $\boxed{C}$ | import three; draw(((0,0,0)--(0,1,0)--(1,1,0)--(1,0,0)--(0,0,0))^^((0,0,1)--(0,1,1)--(1,1,1)--(1,0,1)--(0,0,1))^^((0,0,0)--(0,0,1))^^((0,1,0)--(0,1,1))^^((1,1,0)--(1,1,1))^^((1,0,0)--(1,0,1))); draw(shift((0.5,0.5,0.5))*scale3(1/sqrt(3))*shift((-0.5,-0.5,-0.5))*rotate(aTan(sqrt(2)),(0,0,0.5),(1,1,0.5))*(((0,0,0)--(0,1,0)--(1,1,0)--(1,0,0)--(0,0,0))^^((0,0,1)--(0,1,1)--(1,1,1)--(1,0,1)--(0,0,1))^^((0,0,0)--(0,0,1))^^((0,1,0)--(0,1,1))^^((1,1,0)--(1,1,1))^^((1,0,0)--(1,0,1)))); dot((0.5,0.5,1)^^(0.5,0.5,0)); | [] |
649 | The following problem is from the 2007 AMC 12A #9 and 2007 AMC 10A #13, so those problems redirect to this page.
The trip from Carville to Nikpath requires $4\frac 12$ hours when traveling at an average speed of 70 miles per hour. How many hours does the trip require when traveling at an average speed of 60 miles per hour? Express your answer as a decimal to the nearest hundredth.
Contents
1 Simple Solution
2 Solution 1
3 Solution 2
4 Solution 3
5 Solution 4
6 Video Solution
7 See also | 2007 AMC 12A Problem 9 | Let $H$ represent Yan's home, $S$ represent the stadium, and $Y$ represent Yan's current position. If Yan walks directly to the stadium, he will reach Point $P$ the same time he will reach $H$ if he is walking home.
Since he bikes $7$ times as fast as he walks and the time is the same, the distance from his home to the stadium must be $7$ times the distance from $P$ to the stadium. If $PS=x$, then $HS=7x$ and $HP=6x$. Since Y is the midpoint of $\overline{HP}$, $HY=YP=3x$. Therefore, the ratio of Yan's distance from his home to his distance from the stadium is $\frac{YH}{YS}=\frac{3x}{4x}=\boxed{\mathrm{(B)}\ \frac{3}{4}}$ | // Block 1
draw((0,0)--(7,0));
dot((0,0)); dot((3,0)); dot((6,0)); dot((7,0));
label("$H$",(0,0),S);
label("$Y$",(3,0),S);
label("$P$",(6,0),S);
label("$S$",(7,0),S);
// Block 2
draw((0,0)--(7,0)); dot((0,0)); dot((3,0)); dot((6,0)); dot((7,0)); label("$H$",(0,0),S); label("$Y$",(3,0),S); label("$P$",(6,0),S); label("$S$",(7,0),S); | [] |
650 | In a certain land, all Arogs are Brafs, all Crups are Brafs, all Dramps are Arogs, and all Crups are Dramps. Which of the following statements is implied by these facts?
$\textbf{(A) } \text{All Dramps are Brafs and are Crups.}\\ \textbf{(B) } \text{All Brafs are Crups and are Dramps.}\\ \textbf{(C) } \text{All Arogs are Crups and are Dramps.}\\ \textbf{(D) } \text{All Crups are Arogs and are Brafs.}\\ \textbf{(E) } \text{All Arogs are Dramps and some Arogs may not be Crups.}$ | 2007 AMC 10B Problem 5 | It may be easier to visualize this by drawing some sort of diagram. From the first statement, you can draw an Arog circle inside of the Braf circle, since all Arogs are Brafs, but not all Brafs are Arogs. Ignore the second statement for now, and draw a Dramp circle in the Arog circle and a Crup circle in the Dramp circle. You can see the second statement is already true because all Crups are Arogs. As you can see, the only statement that is true is $\boxed{\mathrm{(D)}}$ | // Block 1
unitsize(6mm);
defaultpen(linewidth(.8pt)+fontsize(9pt));
dotfactor=4;
real r1=1, r2=2, r3=3, r4=4;
pair O1=(0,0), O2=(0,-0.5), O3=(0,-1), O4=(0,-1.5);
path circleA=Circle(O1,r1); draw(circleA);
path circleB=Circle(O2,r2); draw(circleB);
path circleC=Circle(O3,r3); draw(circleC);
path circleD=Circle(O4,r4); draw(circleD);
label("$Crups$",(0,-.5));
label("$Dramps$",(0,-2));
label("$Arogs$",(0,-3.5));
label("$Brafs$",(0,-5));
// Block 2
unitsize(6mm); defaultpen(linewidth(.8pt)+fontsize(9pt)); dotfactor=4; real r1=1, r2=2, r3=3, r4=4; pair O1=(0,0), O2=(0,-0.5), O3=(0,-1), O4=(0,-1.5); path circleA=Circle(O1,r1); draw(circleA); path circleB=Circle(O2,r2); draw(circleB); path circleC=Circle(O3,r3); draw(circleC); path circleD=Circle(O4,r4); draw(circleD); label("$Crups$",(0,-.5)); label("$Dramps$",(0,-2)); label("$Arogs$",(0,-3.5)); label("$Brafs$",(0,-5)); | [] |
651 | All sides of the convex pentagon $ABCDE$ are of equal length, and $\angle A= \angle B = 90^\circ.$ What is the degree measure of $\angle E?$
$\textbf{(A) } 90 \qquad\textbf{(B) } 108 \qquad\textbf{(C) } 120 \qquad\textbf{(D) } 144 \qquad\textbf{(E) } 150$ | 2007 AMC 10B Problem 7 | $AB = EC$ because they are opposite sides of a square. Also, $ED = DC = AB$ because all sides of the convex pentagon are of equal length. Since $ABCE$ is a square and $\triangle CED$ is an equilateral triangle, $\angle AEC = 90$ and $\angle CED = 60.$ Use angle addition:
\[\angle E = \angle AEC + \angle CED = 90 + 60 = \boxed{\textbf{(E)} 150}\] | // Block 1
unitsize(1.5cm);
defaultpen(linewidth(.8pt)+fontsize(10pt));
pair A=(0,2), B=(0,0), C=(2,0), D=(2+sqrt(3),1), E=(2,2);
draw(A--B--C--D--E--cycle);
draw(E--C,gray);
draw(rightanglemark(B,A,E));
draw(rightanglemark(A,B,C));
label("$A$",A,NW);
label("$B$",B,SW);
label("$C$",C,SE);
label("$D$",D,E);
label("$E$",E,NE);
// Block 2
unitsize(1.5cm); defaultpen(linewidth(.8pt)+fontsize(10pt)); pair A=(0,2), B=(0,0), C=(2,0), D=(2+sqrt(3),1), E=(2,2); draw(A--B--C--D--E--cycle); draw(E--C,gray); draw(rightanglemark(B,A,E)); draw(rightanglemark(A,B,C)); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,E); label("$E$",E,NE); | [] |
652 | A circle passes through the three vertices of an isosceles triangle that has two sides of length $3$ and a base of length $2$. What is the area of this circle?
$\textbf{(A) } 2\pi \qquad\textbf{(B) } \frac{5}{2}\pi \qquad\textbf{(C) } \frac{81}{32}\pi \qquad\textbf{(D) } 3\pi \qquad\textbf{(E) } \frac{7}{2}\pi$ | 2007 AMC 10B Problem 11 | Solution 1
Let $\triangle ABC$ have vertex $A$ and center $O$, with foot of altitude from $A$ intersecting $BC$ at $D$.
Then by the Pythagorean Theorem (with radius $r$ and height $OD = h$) on $\triangle OBD$ and $\triangle ABD$
\begin{align*} h^2 + 1 & = r^2 \\ (h + r)^2 + 1 & = 9 \end{align*}
Substituting and solving gives $r = \frac {9}{4\sqrt {2}}$. Then the area of the circle is $r^2 \pi = \left(\frac {9}{4\sqrt {2}}\right)^2 \pi = \boxed{\mathrm{(C) \ } \frac {81}{32} \pi}$.
Solution 2
By $A = \frac {1}{2}Bh = \frac {abc}{4R}$ (or we could use $s = 4$ and Heron's formula),
\[R = \frac {abc}{2Bh} = \frac {3 \cdot 3 \cdot 2}{2(2)(2\sqrt {2})} = \frac {9}{4\sqrt {2}}\]
and the answer is $R^2 \pi = \boxed{\mathrm{(C) \ } \frac {81}{32} \pi}$
Alternatively, by the Extended Law of Sines,
\[2R = \frac {AC}{\sin \angle ABC} = \frac {3}{\frac {2\sqrt {2}}{3}} \Longrightarrow R = \frac {9}{4\sqrt {2}}\]
Answer follows as above.
Solution 3
Extend segment $AD$ to $R$ on Circle $O$.
By the Pythagorean Theorem \[AD^2 = 3^2 - 1^2\] \[AD = 2\sqrt{2}\]
$\triangle ADC$ is similar to $\triangle ACR$, so \[\frac {2\sqrt{2}}{3} = \frac {3}{2r}\] which gives us \[2r = \frac {9}{2\sqrt{2}} = \frac {9\sqrt{2}}{4}\] therefore \[r = \frac {9\sqrt{2}}{8}\]
The area of the circle is therefore $\pi r^2 = \left(\frac {9\sqrt{2}}{8}\right)^2 \pi = \boxed{\mathrm{(C) \ } \frac {81}{32} \pi}$
Solution 4
First, we extend $AD$ to intersect the circle at $E.$
By the Pythagorean Theorem, we know that
\[1^2+AD^2=3^2\implies AD=2\sqrt{2}.\]
We also know that, from the Power of a Point Theorem, \[AD\cdot DE=BD\cdot DC.\]
We can substitute the values we know to get
\[2\sqrt{2}\cdot DE=1\]
We can simplify this to get that
\[DE=\dfrac{2\sqrt{2}}{8}.\]
We add $AD$ and $DE$ together to get the length of the diameter, and then we can find the area.
\[AE=AD+DE=2\sqrt{2}+\dfrac{2\sqrt{2}}{8}=\dfrac{9\sqrt{2}}{4}.\]
Therefore, the radius is $\dfrac{9\sqrt{2}}{8}$, so the area is
\[\left(\dfrac{9\sqrt{2}}{8}\right)^2\pi=\boxed{(\text{C})\dfrac{81}{32}\pi.}\]
Solution 5
Another possible solution is to plot the circle and triangle on a graph with the circle having center $(0,0)$.
Let the radius of the circle = $r$.
Let the distance between the origin and the base of triangle = $a$.
\[1 + a^2 = r^2\]
\[r + a = 2\sqrt{2}\]
\[a = 2\sqrt{2} - r\]
\[9 - 4r\sqrt{2} = 0\]
\[r = \frac{9\sqrt{2}}{8}\]
\[\pi r^2 = \boxed{(\text{C})\frac{81}{32}\pi}\]
Solution 6
We can also use LOC to solve this problem. If O is the center of the circle, angle $OBC$ is double angle $BAC$. From law of cosines, we have that $\cos(\angle BAC) = \frac{7}{9}$. We now apply law of cosines on triangle $OBC$ to get that $4 = r^2+r^2-2r^2\cos(2 \angle BAC)$ but we know that $2 \cos (\angle BAC) = 2 \cdot (\frac{7}{9})^2 - 1 = \frac{17}{81}$. Plugging this in and simplifying, we get $r = \frac{9}{8} \cdot \sqrt{2}$ and thus the area of the circle is equal to $\frac{81}{32} \pi$. | // Block 1
import olympiad; pair B=(0,0), C=(2,0), A=(1,3), D=(1,0); pair O=circumcenter(A,B,C); draw(A--B--C--A--D); draw(B--O--C); draw(circumcircle(A,B,C)); dot(O); label("\(A\)",A,N); label("\(B\)",B,S); label("\(C\)",C,S); label("\(D\)",D,S); label("\(O\)",O,W); label("\(r\)",(O+A)/2,SE); label("\(r\)",(O+B)/2,N); label("\(h\)",(O+D)/2,SE); label("\(3\)",(A+B)/2,NW); label("\(1\)",(B+D)/2,N);
// Block 2
import olympiad; pair B=(0,0), C=(2,0), A=(1,3), D=(1,0), R=(1,-0.35); pair O=circumcenter(A,B,C); draw(A--B--C--A--D--R--C); draw(B--O--C); draw(circumcircle(A,B,C)); dot(O); label("\(A\)",A,N); label("\(B\)",B,S); label("\(C\)",C,S); label("\(D\)",D,S); label("\(O\)",O,W); label("\(R\)",R,S); label("\(r\)",(O+A)/2,SE); label("\(r\)",(O+R)/2,SE); label("\(3\)",(A+C)/2,NE); label("\(1\)",(C+D)/2,N);
// Block 3
import olympiad; pair B=(0,0), C=(2,0), A=(1,3), D=(1,0), E=(1,-(8^0.5)/8); pair O=circumcenter(A,B,C); draw(A--B--C--A--E); draw(circumcircle(A,B,C)); dot(O); dot(D); dot(B); dot(C); dot(A); dot(E); label("$A$",A,N); label("$B$",B,S); label("\(C\)",C,S); label("$D$",D,NE); label("$O$",O,W); label("$E$",E,S); label("$3$",(A+B)/2,NW); label("$1$",(B+D)/2,N); | [] |
653 | Rhombus $ABCD$, with side length $6$, is rolled to form a cylinder of volume $6$ by taping $\overline{AB}$ to $\overline{DC}$. What is $\sin(\angle ABC)$?
$\mathrm{(A)}\ \frac{\pi}{9} \qquad \mathrm{(B)}\ \frac{1}{2} \qquad \mathrm{(C)}\ \frac{\pi}{6} \qquad \mathrm{(D)}\ \frac{\pi}{4} \qquad \mathrm{(E)}\ \frac{\sqrt{3}}{2}$ | 2007 AMC 12B Problem 19 | $V_{\mathrm{Cylinder}} = \pi r^2 h$
Where $C = 2\pi r = 6$ and $h=6\sin\theta$
$r = \frac{3}{\pi}$
$V = \pi \left(\frac{3}{\pi}\right)^2\cdot 6\sin\theta$
$6 = \frac{9}{\pi} \cdot 6\sin\theta$
$\sin\theta = \frac{\pi}{9} \Rightarrow \mathrm{(A)}$ | // Block 1
pair B=(0,0), A=(6*dir(60)), C=(6,0);
pair D=A+C;
draw(A--B--C--D--A);
draw(A--(3,0));
label("\(A\)",A,NW);label("\(B\)",B,SW);label("\(C\)",C,SE);label("\(D\)",D,NE);
label("\(6\)",A/2,NW);
label("\(\theta\)",(.8,.5));
label("\(h\)",(3,2.6),E);
// Block 2
pair B=(0,0), A=(6*dir(60)), C=(6,0); pair D=A+C; draw(A--B--C--D--A); draw(A--(3,0)); label("\(A\)",A,NW);label("\(B\)",B,SW);label("\(C\)",C,SE);label("\(D\)",D,NE); label("\(6\)",A/2,NW); label("\(\theta\)",(.8,.5)); label("\(h\)",(3,2.6),E); | [] |
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