problem_id int64 1 978 | question stringlengths 86 2.11k | source stringlengths 19 76 | solution stringlengths 94 14.7k | asymptote_code stringlengths 44 17.8k | solution_image_url stringclasses 16
values |
|---|---|---|---|---|---|
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 en... | 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... | // 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);... | [] |
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... | 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 $CD\cdot DE - [AFB] = 4\cdot (2\p... | // Block 1
defaultpen(0.8);
real pi=3.14159265359;
pair A=(0,2), B=(4,0), C=(2*pi+1, 0), D=(2*pi+1,4), E=(0,4), F=(0,0);
draw(A--B--C--D--E--cycle);
draw(circle((A+B)/2,length(B-A)/2));
label("$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);... | [] |
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... | 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 $\overline{DE}$ ... | // 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);... | [] |
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 ... | 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 $SA=... | unitsize(1cm); pair A=(0,1), B=(4,4), C=(4,1), S=(12/9,4/9); dot(A); dot(B); draw( circle(A,1) ); draw( circle(B,4) ); draw( (-1.5,0)--(8.5,0) ); draw( (A+(4,0)) -- A -- (A+(0,-1)) ); draw( A -- B -- (B+(0,-4)) ); label("$A$",A,N); label("$B$",B,N); label("$C$",C,E); label("$S$",S,N); filldraw( circle(S,4/9), lightgra... | [] |
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 $D$?
$... | 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 c... | // 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--cyc... | [] |
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}... | // 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);
dr... | [] |
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 \... | 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)... | [] |
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/... | 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... | // 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... | [] |
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 $HC/JE$.
... | 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... | // 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("$A$"... | [] |
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... | 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 ... | // 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
uni... | [] |
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}{... | // 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); labe... | [] |
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 \... | 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... | [] |
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... | // 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,... | [] |
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 ... | // 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)--... | [] |
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 inte... | // 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*... | [] |
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, $\frac{BC}{AB} = \frac{CD}{AD}$
$\frac{BC}{12} = \frac... | // Block 1
size(12cm, 12cm);
pair A, B, C, D, M, N;
A = (0,0);
B = (88/9, 28*sqrt(5)/9);
C = (16,0);
D = 9/16*C;
M = (B + C)/2;
N = (6,4.27);
draw(A--B--C--cycle);
draw(B--D--M);
draw(D--N--B);
label("$A$", A, SW);
label("$B$", B, N);
label("$C$", C, SE);
label("$D$", D, S);
label("$M$", M, NE);
label("$N$", N, NW);
... | [] |
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 $GD = 2 + 2\sqrt 2$,... | // 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("$F$"... | [] |
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 $... | // Block 1
for(int i=5; i > 0; --i) {
for(int j=0; j < i; ++j) {
draw(circle((j+(5-i)/2,(5-i)*sqrt(3)/2),.2));
};
};
path m1 = brace((2,-.3),(0,-.3),.2);
draw(m1);
label("$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);
... | [] |
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 ... | 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... | // 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... | [] |
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^... | 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("$a$... | ["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)}... | 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 ... | // 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,ES... | [] |
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... | 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\... | 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,sqr... | [] |
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... | 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\... | // 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... | [] |
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\te... | 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 ge... | // 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)... | 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 \... | // 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));
dra... | [] |
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 se... | 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}{... | // 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,... | [] |
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... | // 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(... | [] |
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 ... | // 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);... | [] |
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 s... | // 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);... | [] |
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 tr... | 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("$F$... | [] |
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... | 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, 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 ... | 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("$F$... | [] |
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 s... | 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("$F$... | [] |
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 tr... | 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("$F$... | [] |
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 se... | 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}{... | 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... | [] |
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... | 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 ... | // 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));
dra... | [] |
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\... | 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 i... | // 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... | [] |
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 \mathr... | 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 $CE = FC + x = ... | // Block 1
size(150); defaultpen(fontsize(10)); pair A=(0,0), B=(2,0), C=(2,2), D=(0,2), E=(0,1/2), F=E+(C-E)/abs(C-E)/2; draw(A--B--C--D--cycle);draw(C--E); draw(Arc((1,0),1,0,180));draw((A+B)/2--F); label("$A$",A,(-1,-1)); label("$B$",B,( 1,-1)); label("$C$",C,( 1, 1)); label("$D$",D,(-1, 1)); label("$E$",E,(-1, 0));... | ["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 \... | 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 $... | 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--cyc... | [] |
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 \... | 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... | 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("\(C\)", ... | [] |
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 \mathr... | 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 $CE = FC + x = ... | // Block 1
size(150); defaultpen(fontsize(10)); pair A=(0,0), B=(2,0), C=(2,2), D=(0,2), E=(0,1/2), F=E+(C-E)/abs(C-E)/2; draw(A--B--C--D--cycle);draw(C--E); draw(Arc((1,0),1,0,180));draw((A+B)/2--F); label("$A$",A,(-1,-1)); label("$B$",B,( 1,-1)); label("$C$",C,( 1, 1)); label("$D$",D,(-1, 1)); label("$E$",E,(-1, 0));... | ["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 \... | 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 $... | // 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--... | [] |
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 \... | 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... | // 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);
labe... | [] |
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)}\ ... | 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 gr... | // 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)... | [] |
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} ... | 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-li... | // 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... | [] |
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 \math... | 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 = \b... | // 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); dr... | [] |
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) \ ... | 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 $\triangle AEC... | // Block 1
unitsize(0.5cm);
defaultpen(0.8);
pair C=(0,0), A=(0,12), E=(20,0);
draw(A--C--E--cycle);
pair B=A + 3*(C-A)/length(C-A);
pair D=C + 4*(E-C)/length(E-C);
pair F=E + 5*(A-E)/length(A-E);
draw(B--D--F--cycle);
label("$A$",A,N);
label("$B$",B,W);
label("$C$",C,SW);
label("$D$",D,S);
label("$E$",E,SE);
label("$F... | [] |
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... | 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 $A=(-3,0)$, $B=... | // Block 1
unitsize(1cm);
defaultpen(0.8);
pair A=(-3,0), B=(0,4), C=(15/11,-16/11), D=(1,0), E=(0,-1);
draw(A--B--C--cycle);
draw(A--D);
draw(B--E);
pair T=intersectionpoint(A--D,B--E);
label("$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);
... | [] |
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) \ } ... | 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 th... | // 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);
d... | [] |
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) \ } ... | 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 $2$, usin... | // 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), ... | [] |
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... | 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... | // 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);
d... | [] |
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\sqr... | 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 tria... | // 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("$A$... | [] |
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,... | [] |
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)}\ \fra... | 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 ... | // 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 )... | [] |
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 $\overli... | // 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("\(... | [] |
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) } \f... | 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... | 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 $\triangle ABD$?
... | 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{O... | // 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);dra... | [] |
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 $\triangle ABD$?
... | 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} \longrig... | // 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("$B$",... | [] |
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... | 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 differ... | // 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);
lab... | [] |
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... | 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);
lab... | [] |
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... | 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... | // 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);
lab... | [] |
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 $\triangle ABD$?
... | 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 ... | // 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,SS... | [] |
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 triangl... | // 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)--... | [] |
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} ... | 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... | // 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... | [] |
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*}... | // 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(... | [] |
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); la... | [] |
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 $\sqr... | // Block 1
real Xmin,Xmax,Ymin,Ymax;
real R = 6+6*sqrt(2);
Xmin = -16; Xmax = 16; Ymin = -4; Ymax = 40;
xaxis(Xmin,Xmax,Arrows);
yaxis(Ymin,Ymax,Arrows);
label("$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)); d... | [] |
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... | 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 $\angl... | // 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)); yax... | [] |
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}... | 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 th... | // 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,... | [] |
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?
$\mathr... | 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, ... | 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 ... | // 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,... | [] |
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\q... | 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 alread... | // 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... | [] |
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... | 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... | // 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
dr... | [] |
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}{... | 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 w... | // 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));... | [] |
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}{... | 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 \tri... | // 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));... | [] |
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}\qq... | 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 dif... | // 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-... | [] |
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}\qq... | 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... | // 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-... | [] |
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 = \f... | // 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)--(... | [] |
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... | 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 b... | 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 ... | // 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,... | [] |
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... | 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 b... | // 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,... | [] |
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) \ } \fra... | 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 base... | // 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(... | [] |
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}\qqua... | 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 $DF... | // 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));
labe... | [] |
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 {... | 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 $\sqrt{l^2+w^... | // Block 1
size(7cm);
real l=10,
w=7,
ang=asin(w/sqrt(l*l+w*w))*180/pi;
draw((-l,-w)--(l,-w)--(l,w)--(-l,w)--cycle);
draw(rotate(ang)*ellipse((0,0),2*l+2*w,l*w*2/sqrt(l^2+w^2)));
draw(rotate(ang)*((0,0)--(2l+2w,0)),red);
draw(rotate(ang+90)*((0,0)--(l*w*2/sqrt(l^2+w^2),0)),red);
label("$A$",(-l,w),NW);
label("$B$",... | [] |
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 \math... | 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 ... | // 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.... | [] |
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 \math... | 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... | // 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); ... | [] |
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{(... | 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... | // 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), pla... | [] |
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{(... | 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}... | // 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), pla... | [] |
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 h... | 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 th... | 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)}\ \fr... | 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 $\frac{AH}{CG} = \frac{3}{2}$, ... | 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); d... | [] |
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)}\ \fr... | 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 + trapezoi... | // 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... | [] |
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 figure... | 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,... | [] |
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 h... | 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 th... | // 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("$... | [] |
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 ... | 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 circ... | // 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);... | [] |
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 =... | // 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("$E$",... | [] |
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 sol... | // 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);... | [] |
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... | 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; ... | [] |
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