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 |
|---|---|---|---|---|---|
654 | The parallelogram bounded by the lines $y=ax+c$, $y=ax+d$, $y=bx+c$, and $y=bx+d$ has area $18$. The parallelogram bounded by the lines $y=ax+c$, $y=ax-d$, $y=bx+c$, and $y=bx-d$ has area $72$. Given that $a$, $b$, $c$, and $d$ are positive integers, what is the smallest possible value of $a+b+c+d$?
$\mathrm {(A)} 13\... | 2007 AMC 12B Problem 20 | Plotting the parallelogram on the coordinate plane, the 4 corners are at $(0,c),(0,d),\left(\frac{d-c}{a-b},\frac{ad-bc}{a-b}\right),\left(\frac{c-d}{a-b},\frac{bc-ad}{a-b}\right)$. Because $72= 4\cdot 18$, we have that $4(c-d)\left(\frac{c-d}{a-b}\right) = (c+d)\left(\frac{c+d}{a-b}\right)$ or that $2(c-d)=c+d$, which... | pathpen = linewidth(0.7);
real a = 3, b = 1, c = 9, d = 3;
D((0,c) -- ((d-c)/(a-b),(a*d-b*c)/(a-b)) -- (0,d) -- ((c-d)/(a-b),(b*c-a*d)/(a-b)) -- cycle);
D((0,c) -- ((-d-c)/(a-b),(-a*d-b*c)/(a-b)) -- (0,-d) -- ((c+d)/(a-b),-(-a*d-b*c)/(a-b)) -- cycle); | [] |
655 | Points $A$ and $B$ lie on a circle centered at $O$, and $\angle AOB = 60^\circ$. A second circle is internally tangent to the first and tangent to both $\overline{OA}$ and $\overline{OB}$. What is the ratio of the area of the smaller circle to that of the larger circle?
$\mathrm{(A)}\ \frac{1}{16}\qquad\mathrm{(B)}\ \... | 2008 AMC 10A Problem 16 | Let $P$ be the center of the small circle with radius $r$, and let $Q$ be the point where the small circle is tangent to $OA$. Also, let $C$ be the point where the small circle is tangent to the big circle with radius $R$.
Then $PQO$ is a right triangle. Angle $POQ$ is $30$ degrees because line $OP$ bisects angle $AOB... | size(200); defaultpen(fontsize(10)); pair O=(0,0), A=(3,0), B=(3/2,3/2*3^.5), C=(3^.5,1), D=(3^.5,0), F=(1.5*3^.5,1.5), G=(2*3^.5,2); picture p = new picture; draw(p,Circle(O,0.2)); clip(p,O--C--A--cycle); add(p); draw(Circle(O,3)); dot(A); dot(B); dot(C); dot(O); draw(A--O--B); draw(O--C--D); draw(C--F); draw(D-(0.2,... | [] |
656 | An equilateral triangle has side length $6$. What is the area of the region containing all points that are outside the triangle but not more than $3$ units from a point of the triangle?
$\mathrm{(A)}\ 36+24\sqrt{3}\qquad\mathrm{(B)}\ 54+9\pi\qquad\mathrm{(C)}\ 54+18\sqrt{3}+6\pi\qquad\mathrm{(D)}\ \left(2\sqrt{3}+3\ri... | 2008 AMC 10A Problem 17 | The region described contains three rectangles of dimensions $3 \times 6$, and three $120^{\circ}$ degree arcs of circles of radius $3$. Thus the answer is \[3(3 \times 6) + 3 \left( \frac{120^{\circ}}{360^{\circ}} \times 3^2 \pi\right) = 54 + 9\pi \Longrightarrow \mathrm{(B)}.\] | // Block 1
pointpen = black; pathpen = black+linewidth(0.7); pen d = linetype("6 6")+linewidth(0.7);
pair A=(0,0),B=(6,0),C=6*expi(-pi/3);
D(arc(A,3,90,210)); D(arc(B,3,-30,90)); D(arc(C,3,210,330));
D(arc(A,-3,90,210),d); D(arc(B,-3,-30,90),d); D(arc(C,-3,210,330),d);
D(D(A)--D(B)--D(C)--cycle,linewidth(1));
D(A--(0,3... | [] |
657 | Rectangle $PQRS$ lies in a plane with $PQ=RS=2$ and $QR=SP=6$. The rectangle is rotated $90^\circ$ clockwise about $R$, then rotated $90^\circ$ clockwise about the point $S$ moved to after the first rotation. What is the length of the path traveled by point $P$?
$\mathrm{(A)}\ \left(2\sqrt{3}+\sqrt{5}\right)\pi\qquad\... | 2008 AMC 10A Problem 19 | We let $P'Q'R'S'$ be the rectangle after the first rotation, and $P''Q''R''S''$ be the rectangle after the second rotation. Point $P$ pivots about $R$ in an arc of a circle of radius $\sqrt{2^2+6^2} = 2\sqrt{10}$, and since $\angle PRS,\, \angle P'RQ'$ are complementary, it follows that the arc has a degree measure of ... | // Block 1
size(220);pathpen=black+linewidth(0.65);pointpen=black;
/* draw in rectangles */
D(MP("R",(0,0))--MP("Q",(-6,0))--MP("P",(-6,2),N)--MP("S",(0,2),NW)--cycle);
D((0,0)--MP("Q'",(0,6),SW)--MP("P'",(2,6),SE)--MP("S'",(2,0))--cycle);
D(MP("R''",(2,2),NE)--MP("Q''",(8,2),N)--MP("P''",(8,0))--(2,0)--cycle);
D(arc((... | [] |
658 | Trapezoid $ABCD$ has bases $\overline{AB}$ and $\overline{CD}$ and diagonals intersecting at $K.$ Suppose that $AB = 9$, $DC = 12$, and the area of $\triangle AKD$ is $24.$ What is the area of trapezoid $ABCD$?
$\mathrm{(A)}\ 92\qquad\mathrm{(B)}\ 94\qquad\mathrm{(C)}\ 96\qquad\mathrm{(D)}\ 98 \qquad\mathrm{(E)}\ 100$ | 2008 AMC 10A Problem 20 | Since $\overline{AB} \parallel \overline{DC}$ it follows that $\triangle ABK \sim \triangle CDK$. Thus $\frac{KA}{KC} = \frac{KB}{KD} = \frac{AB}{DC} = \frac{3}{4}$.
We now introduce the concept of area ratios: given two triangles that share the same height, the ratio of the areas is equal to the ratio of their bases... | // Block 1
pointpen = black; pathpen = black + linewidth(0.62); /* cse5 */
pen sm = fontsize(10); /* small font pen */
pair D=(0,0),C=(6,0), K=(3.5,8/3); /* note that K.x is arbitrary, as generator for A,B */
pair A=7*K/4-3*C/4, B=7*K/4-3*D/4;
D(MP("A",A,N)--MP("B",B,N)--MP("C",C)--MP("D",D)--A--C);D(B--D)... | [] |
659 | A cube with side length $1$ is sliced by a plane that passes through two diagonally opposite vertices $A$ and $C$ and the midpoints $B$ and $D$ of two opposite edges not containing $A$ or $C$, as shown. What is the area of quadrilateral $ABCD$?
$\mathrm{(A)}\ \frac{\sqrt{6}}{2}\qquad\mathrm{(B)}\ \frac{5}{4}\qquad\m... | 2008 AMC 10A Problem 21 | Since $AB = AD = CB = CD = \sqrt{\left(\frac{1}{2}\right)^2+1^2}$, it follows that $ABCD$ is a rhombus. The area of the rhombus can be computed by the formula $A = \frac 12 d_1d_2$, where $d_1,\,d_2$ are the diagonals of the rhombus (or of a kite in general). $BD$ has the same length as a face diagonal, or $\sqrt{1^2 +... | // Block 1
import three;
unitsize(3cm);
defaultpen(fontsize(8)+linewidth(0.7));
currentprojection=obliqueX;
triple A=(0.5,0,0),C=(0,1,1),D=(0.5,1,0.5),B=(0,0,0.5);
draw((0.5,0,0)--(0,0,0)--(0,0,1)--(0,0,0)--(0,1,0),linetype("4 4"));
draw((0.5,0,1)--(0,0,1)--(0,1,1)--(0.5,1,1)--(0.5,0,1)--(0.5,0,0)--(0.5,1,0)--(0.5,1,1... | [] |
660 | A round table has radius $4$. Six rectangular place mats are placed on the table. Each place mat has width $1$ and length $x$ as shown. They are positioned so that each mat has two corners on the edge of the table, these two corners being end points of the same side of length $x$. Further, the mats are positioned so th... | 2008 AMC 10A Problem 25 | Let one of the mats be $ABCD$, and the center be $O$ as shown:
Since there are $6$ mats, $\Delta BOC$ is equilateral (the hexagon with side length $x$ is regular). So, $BC=CO=x$. Also, $\angle OCD = \angle OCB + \angle BCD = 60^\circ+90^\circ=150^\circ$.
By the Law of Cosines: $4^2=1^2+x^2-2\cdot1\cdot x \cdot \c... | unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; draw(mat); draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); label("\(x\)",(-1.55,2.1),E); la... | [] |
660 | A round table has radius $4$. Six rectangular place mats are placed on the table. Each place mat has width $1$ and length $x$ as shown. They are positioned so that each mat has two corners on the edge of the table, these two corners being end points of the same side of length $x$. Further, the mats are positioned so th... | 2008 AMC 10A Problem 25 | Draw $OD$ and $OC$ as in the diagram. Draw the altitude from $O$ to $DC$ and call the intersection $E$.
As proved in the first solution, $\angle OCD = 150^\circ$.
That makes $\Delta OCE$ a $30-60-90$ triangle, so $OE = \frac{x}{2}$ and $CE= \frac{x\sqrt 3}{2}$
Since $\Delta ODE$ is a right triangle,
$\left({\fr... | unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=((-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle); draw(mat); draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); pair D = rotate(300)*(-3.687,1... | [] |
660 | A round table has radius $4$. Six rectangular place mats are placed on the table. Each place mat has width $1$ and length $x$ as shown. They are positioned so that each mat has two corners on the edge of the table, these two corners being end points of the same side of length $x$. Further, the mats are positioned so th... | 2008 AMC 10A Problem 25 | By symmetry, $E$ is the midpoint of $DF$ and $OE$ is an extension of $OC$. Thus $\angle OED = 90^\circ$. Since $OD = 4$ and $DE = \frac{1}{2}$, $OE = \sqrt{16-\frac{1}{4}} = \frac{\sqrt{63}}{2} = \frac{3\sqrt{7}}{2}$. Since $\triangle CED$ is $30-60-90$, $CE = \frac{\sqrt{3}}{2}$ (this can also be deduced from Pythagor... | unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; draw(mat); draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); label("\(x\)",(-1.55,2.1),E); la... | [] |
660 | A round table has radius $4$. Six rectangular place mats are placed on the table. Each place mat has width $1$ and length $x$ as shown. They are positioned so that each mat has two corners on the edge of the table, these two corners being end points of the same side of length $x$. Further, the mats are positioned so th... | 2008 AMC 10A Problem 25 | Looking at the diagram above, we know that $BE$ is a diameter of circle $O$ due to symmetry. Due to Thales' theorem, triangle $ABE$ is a right triangle with $A = 90 ^\circ$. $AE$ lies on $AD$ and $GE$ because $BAD$ is also a right angle. To find the length of $DG$, notice that if we draw a line from $F$ to $M$, the mid... | unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; draw(mat); draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); label("\(x\)",(-1.95,3),E); labe... | [] |
660 | A round table has radius $4$. Six rectangular place mats are placed on the table. Each place mat has width $1$ and length $x$ as shown. They are positioned so that each mat has two corners on the edge of the table, these two corners being end points of the same side of length $x$. Further, the mats are positioned so th... | 2008 AMC 10A Problem 25 | We will let $O(0,0)$ be the origin. This way the coordinates of $C$ will be $(0,y)$. By $30-60-90$, the coordinates of $D$ will be $\left(-\frac{1}{2}, y + \frac{\sqrt{3}}{2}\right)$. The distance any point with coordinates $(x, y)$ is from the origin is $\sqrt{x^2 + y^2}$. Therefore, the distance $D$ is from the origi... | unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; draw(mat); draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); label("\(x\)",(-1.55,2.1),E); la... | [] |
660 | A round table has radius $4$. Six rectangular place mats are placed on the table. Each place mat has width $1$ and length $x$ as shown. They are positioned so that each mat has two corners on the edge of the table, these two corners being end points of the same side of length $x$. Further, the mats are positioned so th... | 2008 AMC 10A Problem 25 | Notice that $\overarc{AE}$ is $\frac16$ the circumference of the circle. Therefore, $\overline{AE}$ is the side length of an inscribed hexagon with side length $4$. $\triangle AFE$ is a right triangle with $\overline{AF}=\frac12$. The length of $\overline{EF}=x+\frac{\sqrt{3}}{2}$. Using the Pythagorean Theorem, we get... | unitsize(8mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; draw(mat); draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); label("\(x\)",(-1.55,2.1),E); la... | [] |
661 | Points $A$ and $B$ lie on a circle centered at $O$, and $\angle AOB = 60^\circ$. A second circle is internally tangent to the first and tangent to both $\overline{OA}$ and $\overline{OB}$. What is the ratio of the area of the smaller circle to that of the larger circle?
$\mathrm{(A)}\ \frac{1}{16}\qquad\mathrm{(B)}\ \... | 2008 AMC 12A Problem 13 | Let $P$ be the center of the small circle with radius $r$, and let $Q$ be the point where the small circle is tangent to $OA$. Also, let $C$ be the point where the small circle is tangent to the big circle with radius $R$.
Then $PQO$ is a right triangle. Angle $POQ$ is $30$ degrees because line $OP$ bisects angle $AOB... | // Block 1
size(200);
defaultpen(fontsize(10));
pair O=(0,0), A=(3,0), B=(3/2,3/2*3^.5), C=(3^.5,1), D=(3^.5,0), F=(1.5*3^.5,1.5), G=(2*3^.5,2);
picture p = new picture;
draw(p,Circle(O,0.2));
clip(p,O--C--A--cycle);
add(p);
draw(Circle(O,3));
dot(A); dot(B); dot(C); dot(O);
draw(A--O--B);
draw(O--C--D);
draw(C--F);
d... | [] |
662 | Triangle $ABC$, with sides of length $5$, $6$, and $7$, has one vertex on the positive $x$-axis, one on the positive $y$-axis, and one on the positive $z$-axis. Let $O$ be the origin. What is the volume of tetrahedron $OABC$?
$\mathrm{(A)}\ \sqrt{85}\qquad\mathrm{(B)}\ \sqrt{90}\qquad\mathrm{(C)}\ \sqrt{95}\qquad\math... | 2008 AMC 12A Problem 18 | Without loss of generality, let $A$ be on the $x$ axis, $B$ be on the $y$ axis, and $C$ be on the $z$ axis, and let $AB, BC, CA$ have respective lengths of 5, 6, and 7. Let $a,b,c$ denote the lengths of segments $OA,OB,OC,$ respectively. Then by the Pythagorean Theorem,
\begin{align*} a^2+b^2 &=5^2 , \\ b^2+c^2&=6^2... | // Block 1
defaultpen(fontsize(8));
draw((0,10)--(0,0)--(8,0));draw((-3,-4)--(0,0));draw((0,10)--(-3,-4)--(8,0)--cycle);
label("A",(8,0),(1,0));label("B",(0,10),(0,1));label("C",(-3,-4),(-1,-1));label("O",(0,0),(1,1));
label("$a$",(4,0),(0,1));label("$b$",(0,5),(1,0));label("$c$",(-1.5,-2),(1,0));
label("$5$",(4,5),(1,... | [] |
663 | Triangle $ABC$ has $AC=3$, $BC=4$, and $AB=5$. Point $D$ is on $\overline{AB}$, and $\overline{CD}$ bisects the right angle. The inscribed circles of $\triangle ADC$ and $\triangle BCD$ have radii $r_a$ and $r_b$, respectively. What is $r_a/r_b$?
$\mathrm{(A)}\ \frac{1}{28}\left(10-\sqrt{2}\right)\qquad\mathrm{(B)}\ \... | 2008 AMC 12A Problem 20 | By the Angle Bisector Theorem,
\[\frac{BD}{4} = \frac{5-BD}{3} \Longrightarrow BD = \frac{20}7\]
By Law of Sines on $\triangle BCD$,
\[\frac{BD}{\sin 45^{\circ}} = \frac{CD}{\sin \angle B} \Longrightarrow \frac{20/7}{\sqrt{2}/2} = \frac{CD}{3/5} \Longrightarrow CD=\frac{12\sqrt{2}}{7}\]
Since the area of a triangle s... | // Block 1
import olympiad;
size(300);
defaultpen(0.8);
pair C=(0,0),A=(0,3),B=(4,0),D=(4-2.28571,1.71429);
pair O=incenter(A,C,D), P=incenter(B,C,D);
picture p = new picture;
draw(p,Circle(C,0.2)); draw(p,Circle(B,0.2));
clip(p,B--C--D--cycle);
add(p);
draw(A--B--C--D--C--cycle);
draw(incircle(A,C,D));
draw(incircle(... | [] |
663 | Triangle $ABC$ has $AC=3$, $BC=4$, and $AB=5$. Point $D$ is on $\overline{AB}$, and $\overline{CD}$ bisects the right angle. The inscribed circles of $\triangle ADC$ and $\triangle BCD$ have radii $r_a$ and $r_b$, respectively. What is $r_a/r_b$?
$\mathrm{(A)}\ \frac{1}{28}\left(10-\sqrt{2}\right)\qquad\mathrm{(B)}\ \... | 2008 AMC 12A Problem 20 | We start by finding the length of $AD$ and $BD$ as in solution 1. Using the angle bisector theorem, we see that $AD = \frac{15}{7}$ and $BD = \frac{20}{7}$. Using Stewart's Theorem gives us the equation $5d^2 + \frac{1500}{49} = \frac{240}{7} + \frac{180}{7}$, where $d$ is the length of $CD$. Solving gives us $d = \fra... | // Block 1
import olympiad;
import geometry;
size(300);
defaultpen(0.8);
pair C=(0,0),A=(0,3),B=(4,0),D=(4-2.28571,1.71429);
pair O=incenter(A,C,D), P=incenter(B,C,D);
line cd = line(C, D);
picture p = new picture;
picture q = new picture;
picture r = new picture;
picture s = new picture;
draw(p,Circle(C,0.2));
cli... | [] |
664 | A round table has radius $4$. Six rectangular place mats are placed on the table. Each place mat has width $1$ and length $x$ as shown. They are positioned so that each mat has two corners on the edge of the table, these two corners being end points of the same side of length $x$. Further, the mats are positioned so th... | 2008 AMC 12A Problem 22 | Let one of the mats be $ABCD$, and the center be $O$ as shown:
Since there are $6$ mats, $\Delta BOC$ is equilateral (the hexagon with side length $x$ is regular). So, $BC=CO=x$. Also, $\angle OCD = \angle OCB + \angle BCD = 60^\circ+90^\circ=150^\circ$.
By the Law of Cosines: $4^2=1^2+x^2-2\cdot1\cdot x \cdot \c... | // Block 1
unitsize(8mm);
defaultpen(linewidth(.8)+fontsize(8));
draw(Circle((0,0),4));
path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle;
draw(mat);
draw(rotate(60)*mat);
draw(rotate(120)*mat);
draw(rotate(180)*mat);
draw(rotate(240)*mat);
draw(rotate(300)*mat);
label("\(x\)",(-1.55,... | [] |
664 | A round table has radius $4$. Six rectangular place mats are placed on the table. Each place mat has width $1$ and length $x$ as shown. They are positioned so that each mat has two corners on the edge of the table, these two corners being end points of the same side of length $x$. Further, the mats are positioned so th... | 2008 AMC 12A Problem 22 | Draw $OD$ and $OC$ as in the diagram. Draw the altitude from $O$ to $DC$ and call the intersection $E$.
As proved in the first solution, $\angle OCD = 150^\circ$.
That makes $\Delta OCE$ a $30-60-90$ triangle, so $OE = \frac{x}{2}$ and $CE= \frac{x\sqrt 3}{2}$
Since $\Delta ODE$ is a right triangle,
$\left({\fr... | // Block 1
unitsize(8mm);
defaultpen(linewidth(.8)+fontsize(8));
draw(Circle((0,0),4));
path mat=((-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle);
draw(mat);
draw(rotate(60)*mat);
draw(rotate(120)*mat);
draw(rotate(180)*mat);
draw(rotate(240)*mat);
draw(rotate(300)*mat);
pair D = rotate(300... | [] |
664 | A round table has radius $4$. Six rectangular place mats are placed on the table. Each place mat has width $1$ and length $x$ as shown. They are positioned so that each mat has two corners on the edge of the table, these two corners being end points of the same side of length $x$. Further, the mats are positioned so th... | 2008 AMC 12A Problem 22 | By symmetry, $E$ is the midpoint of $DF$ and $OE$ is an extension of $OC$. Thus $\angle OED = 90^\circ$. Since $OD = 4$ and $DE = \frac{1}{2}$, $OE = \sqrt{16-\frac{1}{4}} = \frac{\sqrt{63}}{2} = \frac{3\sqrt{7}}{2}$. Since $\triangle CED$ is $30-60-90$, $CE = \frac{\sqrt{3}}{2}$ (this can also be deduced from Pythagor... | // Block 1
unitsize(8mm);
defaultpen(linewidth(.8)+fontsize(8));
draw(Circle((0,0),4));
path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle;
draw(mat);
draw(rotate(60)*mat);
draw(rotate(120)*mat);
draw(rotate(180)*mat);
draw(rotate(240)*mat);
draw(rotate(300)*mat);
label("\(x\)",(-1.55,... | [] |
664 | A round table has radius $4$. Six rectangular place mats are placed on the table. Each place mat has width $1$ and length $x$ as shown. They are positioned so that each mat has two corners on the edge of the table, these two corners being end points of the same side of length $x$. Further, the mats are positioned so th... | 2008 AMC 12A Problem 22 | Looking at the diagram above, we know that $BE$ is a diameter of circle $O$ due to symmetry. Due to Thales' theorem, triangle $ABE$ is a right triangle with $A = 90 ^\circ$. $AE$ lies on $AD$ and $GE$ because $BAD$ is also a right angle. To find the length of $DG$, notice that if we draw a line from $F$ to $M$, the mid... | // Block 1
unitsize(8mm);
defaultpen(linewidth(.8)+fontsize(8));
draw(Circle((0,0),4));
path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle;
draw(mat);
draw(rotate(60)*mat);
draw(rotate(120)*mat);
draw(rotate(180)*mat);
draw(rotate(240)*mat);
draw(rotate(300)*mat);
label("\(x\)",(-1.95,... | [] |
664 | A round table has radius $4$. Six rectangular place mats are placed on the table. Each place mat has width $1$ and length $x$ as shown. They are positioned so that each mat has two corners on the edge of the table, these two corners being end points of the same side of length $x$. Further, the mats are positioned so th... | 2008 AMC 12A Problem 22 | We will let $O(0,0)$ be the origin. This way the coordinates of $C$ will be $(0,y)$. By $30-60-90$, the coordinates of $D$ will be $\left(-\frac{1}{2}, y + \frac{\sqrt{3}}{2}\right)$. The distance any point with coordinates $(x, y)$ is from the origin is $\sqrt{x^2 + y^2}$. Therefore, the distance $D$ is from the origi... | // Block 1
unitsize(8mm);
defaultpen(linewidth(.8)+fontsize(8));
draw(Circle((0,0),4));
path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle;
draw(mat);
draw(rotate(60)*mat);
draw(rotate(120)*mat);
draw(rotate(180)*mat);
draw(rotate(240)*mat);
draw(rotate(300)*mat);
label("\(x\)",(-1.55,... | [] |
664 | A round table has radius $4$. Six rectangular place mats are placed on the table. Each place mat has width $1$ and length $x$ as shown. They are positioned so that each mat has two corners on the edge of the table, these two corners being end points of the same side of length $x$. Further, the mats are positioned so th... | 2008 AMC 12A Problem 22 | Notice that $\overarc{AE}$ is $\frac16$ the circumference of the circle. Therefore, $\overline{AE}$ is the side length of an inscribed hexagon with side length $4$. $\triangle AFE$ is a right triangle with $\overline{AF}=\frac12$. The length of $\overline{EF}=x+\frac{\sqrt{3}}{2}$. Using the Pythagorean Theorem, we get... | // Block 1
unitsize(8mm);
defaultpen(linewidth(.8)+fontsize(8));
draw(Circle((0,0),4));
path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle;
draw(mat);
draw(rotate(60)*mat);
draw(rotate(120)*mat);
draw(rotate(180)*mat);
draw(rotate(240)*mat);
draw(rotate(300)*mat);
label("\(x\)",(-1.55,... | [] |
665 | Triangle $ABC$ has $\angle C = 60^{\circ}$ and $BC = 4$. Point $D$ is the midpoint of $BC$. What is the largest possible value of $\tan{\angle BAD}$?
$\mathrm{(A)}\ \frac{\sqrt{3}}{6}\qquad\mathrm{(B)}\ \frac{\sqrt{3}}{3}\qquad\mathrm{(C)}\ \frac{\sqrt{3}}{2\sqrt{2}}\qquad\mathrm{(D)}\ \frac{\sqrt{3}}{4\sqrt{2}-3}\q... | 2008 AMC 12A Problem 24 | Let $x = CA$. Then $\tan\theta = \tan(\angle BAF - \angle DAE)$, and since $\tan\angle BAF = \frac{2\sqrt{3}}{x-2}$ and $\tan\angle DAE = \frac{\sqrt{3}}{x-1}$, we have
\[\tan\theta = \frac{\frac{2\sqrt{3}}{x-2} - \frac{\sqrt{3}}{x-1}}{1 + \frac{2\sqrt{3}}{x-2}\cdot\frac{\sqrt{3}}{x-1}}= \frac{x\sqrt{3}}{x^2-3x+8}\]
... | // Block 1
unitsize(12mm);
pair C=(0,0), B=(4 * dir(60)), A = (8,0), D=(2 * dir(60));
pair E=(1,0), F=(2,0);
draw(C--B--A--C);
draw(A--D);draw(D--E);draw(B--F);
dot(A);dot(B);dot(C);dot(D);dot(E);dot(F);
label("\(C\)",C,SW);
label("\(B\)",B,N);
label("\(A\)",A,SE);
label("\(D\)",D,NW);
label("\(E\)",E,S);
label("\(F\)"... | [] |
666 | Points $B$ and $C$ lie on $\overline{AD}$. The length of $\overline{AB}$ is $4$ times the length of $\overline{BD}$, and the length of $\overline{AC}$ is $9$ times the length of $\overline{CD}$. The length of $\overline{BC}$ is what fraction of the length of $\overline{AD}$?
$\textbf{(A)}\ \frac{1}{36}\qquad\textbf{(B... | 2008 AMC 10B Problem 6 | Let $x = BD$ and $y = CD$. Therefore, $AB = 4x$ and $AC = 9y$, as shown in the diagram(the labels on the bottom are for that line segment while the labels on the top are from one point to the left to one point to the right).
From this, we can see that $AD = 10y = 5x$, and since $BC = BD - CD = x-y$. Now, our ratio ... | // Block 1
dot((0,0));
label("A", (0,0), S);
dot((5,0));
label("B", (5,0), S);
dot((10,0));
label("C", (10,0), S);
dot((15,0));
label("D", (15,0), S);
draw((0,0)--(5,0));
draw((5,0)--(10,0));
draw((10,0)--(15,0));
draw((0,0)--(10,0));
draw((10,0)--(15,0));
label("$4x$", (0,0)--(5,0), S);
label("$9y$", (0,0)--(10... | [] |
667 | An equilateral triangle of side length $10$ is completely filled in by non-overlapping equilateral triangles of side length $1$. How many small triangles are required?
$\mathrm{(A)}\ 10\qquad\mathrm{(B)}\ 25\qquad\mathrm{(C)}\ 100\qquad\mathrm{(D)}\ 250\qquad\mathrm{(E)}\ 1000$ | 2008 AMC 10B Problem 7 | The number of triangles is $1+3+\dots+19 = \boxed{100}$.
Also, another way to do it is to notice that as you go row by row (from the bottom), the number of triangles decrease by 2 from 19, so we have:
$19+17+15...+3+1 = \frac{19+1}{2}\cdot 10 = \boxed{100}$
A fourth solution is to notice that the small triangles are ... | // Block 1
unitsize(0.5cm);
defaultpen(0.8);
for (int i=0; i<10; ++i) { draw( (i*dir(60)) -- ( (10,0) + (i*dir(120)) ) ); }
for (int i=0; i<10; ++i) { draw( (i*dir(0)) -- ( 10*dir(60) + (i*dir(-60)) ) ); }
for (int i=0; i<10; ++i) { draw( ((10-i)*dir(60)) -- ((10-i)*dir(0)) ); }
// Block 2
unitsize(0.5cm); defaultpen(0... | [] |
668 | Points $A$ and $B$ are on a circle of radius $5$ and $AB=6$. Point $C$ is the midpoint of the minor arc $AB$. What is the length of the line segment $AC$?
$\mathrm{(A)}\ \sqrt{10}\qquad\mathrm{(B)}\ \frac{7}{2}\qquad\mathrm{(C)}\ \sqrt{14}\qquad\mathrm{(D)}\ \sqrt{15}\qquad\mathrm{(E)}\ 4$ | 2008 AMC 10B Problem 10 | Let the center of the circle be $O$, and let $D$ be the intersection of $\overline{AB}$ and $\overline{OC}$ (then $D$ is the midpoint of $\overline{AB}$). $OA=OB=5$, since they are both radii of the circle.
By the Pythagorean Theorem, $OD = \sqrt{OA^2 - DA^2} = 4$, and by subtraction, $CD=OC - OD = 1$.
Using the Pyth... | // Block 1
pen d = linewidth(0.8); pathpen = d; pointpen = black; pen f = fontsize(9);
path p = CR((0,0),5);
pair O = (0,0), A=(5,0), B = IP(p,CR(A,6)), C = IP(p,CR(A,3)), D=IP(A--B,O--C);
D(p); D(MP("A",A,E)--D(MP("O",O))--MP("B",B,NE)--cycle); D(A--MP("C",C,ENE),dashed+d); D(O--C,dashed+d); D(rightanglemark(O,D(MP("D... | [] |
669 | A cylindrical tank with radius $4$ feet and height $9$ feet is lying on its side. The tank is filled with water to a depth of $2$ feet. What is the volume of water, in cubic feet?
$\mathrm{(A)}\ 24\pi - 36 \sqrt {2} \qquad \mathrm{(B)}\ 24\pi - 24 \sqrt {3} \qquad \mathrm{(C)}\ 36\pi - 36 \sqrt {3} \qquad \mathrm{(D)}... | 2008 AMC 10B Problem 19 | Any vertical cross-section of the tank parallel with its base looks as follows:
The volume of water can be computed as the height of the tank times the area of the shaded part.
Let $\theta$ be the size of the smaller angle $DAC$. We then have $\cos\theta = \frac{AD}{AC}=\frac 12$, hence $\theta=60^\circ$.
The figur... | // Block 1
unitsize(0.8cm);
defaultpen(0.8);
pair s=(0,0), bottom=(0,-4), mid=(0,-2);
pair x[]=intersectionpoints( (-10,-2)--(10,-2), circle(s,4) );
fill( arc(s,x[0],x[1]) -- cycle, lightgray );
draw( circle(s,4) );
dot(s);
draw( s -- bottom );
label( "$2$", (mid+bottom)/2, E );
draw ( s -- x[0] -- x[1] -- s );
label( ... | [] |
670 | Quadrilateral $ABCD$ has $AB = BC = CD$, $m\angle ABC = 70^\circ$ and $m\angle BCD = 170^\circ$. What is the degree measure of $\angle BAD$?
$\mathrm{(A)}\ 75\qquad\mathrm{(B)}\ 80\qquad\mathrm{(C)}\ 85\qquad\mathrm{(D)}\ 90\qquad\mathrm{(E)}\ 95$ | 2008 AMC 10B Problem 24 | First, connect the diagonal $DB$, then, draw line $DE$ such that it is congruent to $DC$ and is parallel to $AB$. Because triangle $DCB$ is isosceles and angle $DCB$ is $170^\circ$, the angles $CDB$ and $CBD$ are both $\frac{180-170}{2} = 5^\circ$. Because angle $ABC$ is $70^\circ$, we get angle $ABD$ is $65^\circ$. Ne... | // Block 1
unitsize(1cm);
defaultpen(.8);
real a=4;
pair A=(0,0), B=a*dir(0), C=B+a*dir(110), D=C+a*dir(120), E=D+a*dir(0);
draw(A--B--C--D--cycle);
draw(E--C);
draw(B--D);
draw(B--E);
draw(D--E);
label("$A$",A,SW);
label("$B$",B,SE);
label("$C$",C,SE);
label("$D$",D,N);
label("$E$",E,NE);
label("$60^\circ$",C + .75*di... | [] |
670 | Quadrilateral $ABCD$ has $AB = BC = CD$, $m\angle ABC = 70^\circ$ and $m\angle BCD = 170^\circ$. What is the degree measure of $\angle BAD$?
$\mathrm{(A)}\ 75\qquad\mathrm{(B)}\ 80\qquad\mathrm{(C)}\ 85\qquad\mathrm{(D)}\ 90\qquad\mathrm{(E)}\ 95$ | 2008 AMC 10B Problem 24 | Let the unknown $\angle BAD$ be $x$.
First, we draw diagonal $BD$ and $AC$.
$I$ is the intersection of the two diagonals. The diagonals each form two isosceles triangles, $\triangle BCD$ and $\triangle ABC$.
Using this, we find: $\angle DBC = \angle CDB = 5^\circ$ and $\angle BAC = \angle BCA = 55^\circ$. Expanding ... | // Block 1
unitsize(3 cm);
pair A, B, C, D;
A = (0,0);
B = dir(85);
C = B + dir(-25);
D = C + dir(-35);
draw(A--B--C--D--cycle);
draw(A--C);
draw(B--D);
draw(((A + B)/2 + scale(0.02)*rotate(90)*(B - A))--((A + B)/2 + scale(0.02)*rotate(90)*(A - B)));
draw(((B + C)/2 + scale(0.02)*rotate(90)*(C - B))--((B + C)/2 + sca... | [] |
671 | Michael walks at the rate of $5$ feet per second on a long straight path. Trash pails are located every $200$ feet along the path. A garbage truck traveling at $10$ feet per second in the same direction as Michael stops for $30$ seconds at each pail. As Michael passes a pail, he notices the truck ahead of him just leav... | 2008 AMC 10B Problem 25 | Pick a coordinate system where Michael's starting pail is $0$ and the one where the truck starts is $200$.
Let $M(t)$ and $T(t)$ be the coordinates of Michael and the truck after $t$ seconds.
Let $D(t)=T(t)-M(t)$ be their (signed) distance after $t$ seconds.
Meetings occur whenever $D(t)=0$.
We have $D(0)=200$.
The t... | import graph; size(400,300,IgnoreAspect); real[] xt = new real[21]; real[] yt = new real[21]; for (int i=0; i<11; ++i) xt[2*i]=50*i; for (int i=0; i<10; ++i) xt[2*i+1]=50*i+20; for (int i=0; i<11; ++i) yt[2*i]=200*(i+1); for (int i=0; i<10; ++i) yt[2*i+1]=200*(i+2); real[] xm={0,500}; real[] ym={0,2500}; draw(graph... | [] |
672 | Points $A$ and $B$ are on a circle of radius $5$ and $AB = 6$. Point $C$ is the midpoint of the minor arc $AB$. What is the length of the line segment $AC$?
$\textbf{(A)}\ \sqrt {10} \qquad \textbf{(B)}\ \frac {7}{2} \qquad \textbf{(C)}\ \sqrt {14} \qquad \textbf{(D)}\ \sqrt {15} \qquad \textbf{(E)}\ 4$ | 2008 AMC 12B Problem 9 | Solution 1
Let $\alpha$ be the angle that subtends the arc $AB$. By the law of cosines,
$6^2=5^2+5^2-2\cdot 5\cdot 5\cos(\alpha)$ implies $\cos(\alpha) = 7/25$.
The half-angle formula says that
$\cos(\alpha/2) = \frac{\sqrt{1+\cos(\alpha)}}{2} = \sqrt{\frac{32/25}{2}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$. The law of... | pen d = linewidth(0.7); pathpen = d; pointpen = black; pen f = fontsize(9); path p = CR((0,0),5); pair O = (0,0), A=(5,0), B = IP(p,CR(A,6)), C = IP(p,CR(A,3)), D=IP(A--B,O--C); D(p); D(MP("A",A,E)--D(MP("O",O))--MP("B",B,NE)--cycle); D(A--MP("C",C,ENE),dashed+d); D(O--C,dashed+d); D(rightanglemark(O,D(MP("D",D,W)),A))... | [] |
673 | Vertex $E$ of equilateral $\triangle{ABE}$ is in the interior of unit square $ABCD$. Let $R$ be the region consisting of all points inside $ABCD$ and outside $\triangle{ABE}$ whose distance from $AD$ is between $\frac{1}{3}$ and $\frac{2}{3}$. What is the area of $R$?
$\textbf{(A)}\ \frac{12-5\sqrt3}{72} \qquad \text... | 2008 AMC 12B Problem 13 | The region is the shaded area:
We can find the area of the shaded region by subtracting the pentagon from the middle third of the square. The area of the middle third of the square is $\left(\frac13\right)(1)=\frac13$. The pentagon can be split into a rectangle and an equilateral triangle.
The base of the equilat... | // Block 1
pair A,B,C,D,E;
A=(0,1);
B=(1,1);
C=(1,0);
D=(0,0);
E=(1/2,1-sqrt(3)/2);
draw(A--B--C--D--cycle);
label("A",A,NW);
dot(A);
label("B",B,NE);
dot(B);
label("C",C,SE);
dot(C);
label("D",D,SW);
dot(D);
draw(A--E--B--cycle);
label("E",E,S);
dot(E);
draw((1/3,0)--(1/3,1));
draw((2/3,0)--(2/3,1));
fill((1/3,0)--(1/... | [] |
674 | On each side of a unit square, an equilateral triangle of side length 1 is constructed. On each new side of each equilateral triangle, another equilateral triangle of side length 1 is constructed. The interiors of the square and the 12 triangles have no points in common. Let $R$ be the region formed by the union of the... | 2008 AMC 12B Problem 15 | The equilateral triangles form trapezoids with side lengths $1, 1, 1, 2$ (half a unit hexagon) on each face of the unit square. The four triangles "in between" these trapezoids are isosceles triangles with two side lengths $1$ and an angle of $30^{\circ}$ in between them, so the total area of these triangles (which is... | // Block 1
real a = 1/2, b = sqrt(3)/2;
draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);
draw((0,0)--(a,-b)--(1,0)--(1+b,a)--(1,1)--(a,1+b)--(0,1)--(-b,a)--(0,0));
draw((0,0)--(-1+a,-b)--(1+a,-b)--(1,0)--(1+b,-1+a)--(1+b,1+a)--(1,1)--(1+a,1+b)--(-1+a,1+b)--(0,1)--(-b,1+a)--(-b,-1+a)--(0,0));
filldraw((1+a,-b)--(1,0)--(1+b,-1+a... | [] |
674 | On each side of a unit square, an equilateral triangle of side length 1 is constructed. On each new side of each equilateral triangle, another equilateral triangle of side length 1 is constructed. The interiors of the square and the 12 triangles have no points in common. Let $R$ be the region formed by the union of the... | 2008 AMC 12B Problem 15 | The area of the largest square is $(1+\sqrt{3})^2=4+2\sqrt{3}$. The area of region $R$ is $1+12\frac{1}{2}\frac{\sqrt{3}}{2}=1+3\sqrt{3}$. The total area of four small 45-45-90 triangles at corner is $4*\frac{1}{2}(\frac{\sqrt{3}+1-2}{2})^2=2-\sqrt{3}$. $S=4+2\sqrt{3}-(2-\sqrt{3})=2+3\sqrt{3}$, $S-R=1$.
~Bran Qin | // Block 1
real a = 1/2, b = sqrt(3)/2;
draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);
draw((a+b+a,a+b+a)--(a+b+a,-b)--(-b,-b)--(-b,a+a+b)--cycle,dashed);
draw((0,0)--(a,-b)--(1,0)--(1+b,a)--(1,1)--(a,1+b)--(0,1)--(-b,a)--(0,0));
draw((0,0)--(-1+a,-b)--(1+a,-b)--(1,0)--(1+b,-1+a)--(1+b,1+a)--(1,1)--(1+a,1+b)--(-1+a,1+b)--(0,... | [] |
675 | Michael walks at the rate of $5$ feet per second on a long straight path. Trash pails are located every $200$ feet along the path. A garbage truck traveling at $10$ feet per second in the same direction as Michael stops for $30$ seconds at each pail. As Michael passes a pail, he notices the truck ahead of him just leav... | 2008 AMC 12B Problem 20 | Pick a coordinate system where Michael's starting pail is $0$ and the one where the truck starts is $200$.
Let $M(t)$ and $T(t)$ be the coordinates of Michael and the truck after $t$ seconds.
Let $D(t)=T(t)-M(t)$ be their (signed) distance after $t$ seconds.
Meetings occur whenever $D(t)=0$.
We have $D(0)=200$.
The t... | // Block 1
import graph;
size(400,300,IgnoreAspect);
real[] xt = new real[21];
real[] yt = new real[21];
for (int i=0; i<11; ++i) xt[2*i]=50*i;
for (int i=0; i<10; ++i) xt[2*i+1]=50*i+20;
for (int i=0; i<11; ++i) yt[2*i]=200*(i+1);
for (int i=0; i<10; ++i) yt[2*i+1]=200*(i+2);
real[] xm={0,500};
real[] ym={0,2500};
... | [] |
676 | Let $ABCD$ be a trapezoid with $AB||CD, AB=11, BC=5, CD=19,$ and $DA=7$. Bisectors of $\angle A$ and $\angle D$ meet at $P$, and bisectors of $\angle B$ and $\angle C$ meet at $Q$. What is the area of hexagon $ABQCDP$?
$\textbf{(A)}\ 28\sqrt{3}\qquad \textbf{(B)}\ 30\sqrt{3}\qquad \textbf{(C)}\ 32\sqrt{3}\qquad \textb... | 2008 AMC 12B Problem 25 | $P$ is the intersection of the angle bisectors of $\angle A$ and $\angle D$. By definition, angle bisectors are always equidistant from the sides of the angle, so $P$ is equidistant from $\overline{AB}$, $\overline{AD}$, and $\overline{CD}$. Likewise, point $Q$ is equidistant from $\overline{AB}$, $\overline{BC}$, and... | // Block 1
unitsize(0.6cm);
import olympiad;
pair A,B,C,D,P,Q,M,N,W,X,Y,Z;
A=(11/2,5sqrt(3)/2);
B=(33/2,5sqrt(3)/2);
C=(19,0);
D=(0,0);
P=incenter(A,D,(99999,5sqrt(3)/4));
Q=incenter(B,C,(-99999,5sqrt(3)/4));
W=P+(0,5sqrt(3)/4);
X=P-(0,5sqrt(3)/4);
Y=Q+(0,5sqrt(3)/4);
Z=Q-(0,5sqrt(3)/4);
M=reflect(A,P)*W;
N=reflect(B,Q... | [] |
676 | Let $ABCD$ be a trapezoid with $AB||CD, AB=11, BC=5, CD=19,$ and $DA=7$. Bisectors of $\angle A$ and $\angle D$ meet at $P$, and bisectors of $\angle B$ and $\angle C$ meet at $Q$. What is the area of hexagon $ABQCDP$?
$\textbf{(A)}\ 28\sqrt{3}\qquad \textbf{(B)}\ 30\sqrt{3}\qquad \textbf{(C)}\ 32\sqrt{3}\qquad \textb... | 2008 AMC 12B Problem 25 | Let $W, X, Y, Z = \overline{AP} \cap \overline{CD}, \overline{BQ} \cap \overline{CD}, \overline{CQ} \cap \overline{AB}, \overline{DP} \cap \overline{AB}$ respectively. Since $\angle{DCQ} = \angle{BCQ}, \angle{CBQ} = \angle{ABQ}$ we have $\angle{QCB} + \angle{CBQ} = 90 \iff \overline{BX} \perp \overline{CY};$ similarly ... | // Block 1
import olympiad;
unitsize(0.5cm);
pair A, B, C, D;
A = 5*(Cos(120), Sin(120));
B = A + (-11, 0);
C = origin + (-19, 0);
D = origin;
label("$A$", A, dir(30));
label("$B$", B, dir(150));
label("$C$", C, dir(150));
label("$D$", D, dir(30));
pair E, F, G, H;
E = bisectorpoint(B, A, D);
F = bisectorpoint(A, B,... | [] |
677 | Triangle $ABC$ has a right angle at $B$. Point $D$ is the foot of the altitude from $B$, $AD=3$, and $DC=4$. What is the area of $\triangle ABC$?
$\mathrm{(A)}\ 4\sqrt3 \qquad \mathrm{(B)}\ 7\sqrt3 \qquad \mathrm{(C)}\ 21 \qquad \mathrm{(D)}\ 14\sqrt3 \qquad \mathrm{(E)}\ 42$ | 2009 AMC 10A Problem 10 | Draw the circumcircle $\omega$ of the $\Delta ABC$. Because $\Delta ABC$ is a right angle triangle, AC is the diameter of the circumcircle. And we know that AC is the diameter because of Perpendicular Chord Bisector Theorem. By applying Power of a Point Theorem, we can have $BD=DE$ and $AD\cdot CD=BD^2$ $\Rightarrow BD... | // Block 1
unitsize(6mm);
defaultpen(linewidth(.8pt)+fontsize(8pt));
dotfactor=4;
pair B=(0,0), C=(sqrt(28),0), A=(0,sqrt(21)), E=(6*sqrt(28)/7,8*sqrt(21)/7);
pair D=foot(B,A,C);
pair[] ps={B,C,A,D};
filldraw(Circle((sqrt(28)/2,sqrt(21)/2),sqrt(49)/2),white,black);
draw(A--B--C--cycle);
draw(B--D);
draw(E--D);
draw(... | [] |
678 | The figures $F_1$, $F_2$, $F_3$, and $F_4$ shown are the first in a sequence of figures. For $n\ge3$, $F_n$ is constructed from $F_{n - 1}$ by surrounding it with a square and placing one more diamond on each side of the new square than $F_{n - 1}$ had on each side of its outside square. For example, figure $F_3$ has $... | 2009 AMC 10A Problem 15 | Color the diamond layers alternately blue and red, starting from the outside. You'll get the following pattern:
In the figure $F_n$, the blue diamonds form a $n\times n$ square, and the red diamonds form a $(n-1)\times(n-1)$ square.
Hence the total number of diamonds in $F_{20}$ is $20^2 + 19^2 = \boxed{761}$. | unitsize(3mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); path d=(1/2,0)--(0,sqrt(3)/2)--(-1/2,0)--(0,-sqrt(3)/2)--cycle; marker mred=marker(scale(5)*d,red,Fill); marker mblue=marker(scale(5)*d,blue,Fill); path f1=(0,0); path f2=(-1,1)--(1,1)--(1,-1)--(-1,-1)--cycle; path f3=(-2,-2)--(-2,0)--(-2,2)--(0,2)--(2,2)--(2,0... | [] |
679 | Rectangle $ABCD$ has $AB=4$ and $BC=3$. Segment $EF$ is constructed through $B$ so that $EF$ is perpendicular to $DB$, and $A$ and $C$ lie on $DE$ and $DF$, respectively. What is $EF$?
$\mathrm{(A)}\ 9 \qquad \mathrm{(B)}\ 10 \qquad \mathrm{(C)}\ \frac {125}{12} \qquad \mathrm{(D)}\ \frac {103}{9} \qquad \mathrm{(E)}\... | 2009 AMC 10A Problem 17 | Solution 1
The situation is shown in the picture below.
From the Pythagorean theorem we have $BD=5$.
Triangle $EAB$ is similar to $BAD$, as they have the same angles. Segment $BA$ is perpendicular to $DA$, meaning that angle $DAB$ and $BAE$ are right angles and congruent. Also, angle $DBE$ is a right angle. Because... | unitsize(0.6cm); defaultpen(0.8); pair A=(0,0), B=(4,0), C=(4,3), D=(0,3); pair EF=rotate(90)*(D-B); pair E=intersectionpoint( (0,-100)--(0,100), (B-100*EF)--(B+100*EF) ); pair F=intersectionpoint( (-100,3)--(100,3), (B-100*EF)--(B+100*EF) ); draw(A--B--C--D--cycle); draw(B--D, dashed); draw(E--F); draw(A--E, dashed); ... | [] |
680 | Many Gothic cathedrals have windows with portions containing a ring of congruent circles that are circumscribed by a larger circle, In the figure shown, the number of smaller circles is four. What is the ratio of the sum of the areas of the four smaller circles to the area of the larger circle?
$\mathrm{(A)}\ 3-2\s... | 2009 AMC 10A Problem 21 | Draw some of the radii of the small circles as in the picture below.
Out of symmetry, the quadrilateral in the center must be a square. Its side is $2r$, and therefore its diagonal is $2r\sqrt{2}$. We can now compute the length of the vertical diameter of the large circle as $2r + 2r\sqrt{2}$. Hence $2R=2r + 2r\sqrt... | // Block 1
unitsize(12mm);
defaultpen(linewidth(.8pt));
draw(Circle((0,0),1+sqrt(2)));
draw(Circle((sqrt(2),0),1));
draw(Circle((0,sqrt(2)),1));
draw(Circle((-sqrt(2),0),1));
draw(Circle((0,-sqrt(2)),1));
draw( (sqrt(2),0) -- (0,sqrt(2)) -- (-sqrt(2),0) -- (0,-sqrt(2)) -- cycle );
draw( (0,sqrt(2)) -- (0,1+sqrt(2)) );... | [] |
681 | Convex quadrilateral $ABCD$ has $AB = 9$ and $CD = 12$. Diagonals $AC$ and $BD$ intersect at $E$, $AC = 14$, and $\triangle AED$ and $\triangle BEC$ have equal areas. What is $AE$?
$\textbf{(A)}\ \frac {9}{2}\qquad \textbf{(B)}\ \frac {50}{11}\qquad \textbf{(C)}\ \frac {21}{4}\qquad \textbf{(D)}\ \frac {17}{3}\qquad... | 2009 AMC 10A Problem 23 | Let $[ABC]$ denote the area of triangle $ABC$. $[AED] = [BEC]$, so $[ABD] = [AED] + [AEB] = [BEC] + [AEB] = [ABC]$. Since triangles $ABD$ and $ABC$ share a base, they also have the same height and thus $\overline{AB}||\overline{CD}$ and $\triangle{AEB}\sim\triangle{CED}$ with a ratio of $3: 4$. $AE = \frac {3}{7}\times... | pathpen = linewidth(0.7);pointpen = black; pair D=MP("D",(0,0)),C=MP("C",(12,0)),A=MP("A",C+14*expi(145*pi/180),N),B=MP("B",A+(9,0),N),E=IP(A--C,B--D);MP("9",(A+B)/2,N);MP("12",(C+D)/2); fill(A--D--E--cycle,rgb(0.8,0.8,0.8));fill(B--C--E--cycle,rgb(0.8,0.8,0.8));D(A--B--C--D--cycle);D(A--C);D(B--D);D(E); | [] |
682 | The figures $F_1$, $F_2$, $F_3$, and $F_4$ shown are the first in a sequence of figures. For $n\ge3$, $F_n$ is constructed from $F_{n - 1}$ by surrounding it with a square and placing one more diamond on each side of the new square than $F_{n - 1}$ had on each side of its outside square. For example, figure $F_3$ has $... | 2009 AMC 12A Problem 11 | Color the diamond layers alternately blue and red, starting from the outside. You'll get the following pattern:
In the figure $F_n$, the blue diamonds form a $n\times n$ square, and the red diamonds form a $(n-1)\times(n-1)$ square.
Hence the total number of diamonds in $F_{20}$ is $20^2 + 19^2 = \boxed{761}$. | // Block 1
unitsize(3mm);
defaultpen(linewidth(.8pt)+fontsize(8pt));
path d=(1/2,0)--(0,sqrt(3)/2)--(-1/2,0)--(0,-sqrt(3)/2)--cycle;
marker mred=marker(scale(5)*d,red,Fill);
marker mblue=marker(scale(5)*d,blue,Fill);
path f1=(0,0);
path f2=(-1,1)--(1,1)--(1,-1)--(-1,-1)--cycle;
path f3=(-2,-2)--(-2,0)--(-2,2)--(0,2)--... | [] |
683 | A ship sails $10$ miles in a straight line from $A$ to $B$, turns through an angle between $45^{\circ}$ and $60^{\circ}$, and then sails another $20$ miles to $C$. Let $AC$ be measured in miles. Which of the following intervals contains $AC^2$?
$\textbf{(A)}\ [400,500] \qquad \textbf{(B)}\ [500,600] \qquad \textbf{(C... | 2009 AMC 12A Problem 13 | Answering the question
To answer the question we are asked, it is enough to compute $AC^2$ for two different angles, preferably for both extremes ($45$ and $60$ degrees). You can use the law of cosines to do so.
Alternately, it is enough to compute $AC^2$ for one of the extreme angles. In case it falls inside one of t... | // Block 1
unitsize(3mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair B=(0,0), A=(-10,0), C1=20*dir(45), C2=20*dir(60); draw(A--B--C1); draw(A--C1,linetype("4 4")); draw(A--B--C2); draw(A--C2,linetype("4 4")); draw( arc(A, length(C1-A), 0, 55 ), dotted ); draw( arc(A, length(C2-A), 0, 55 ), dotted ... | [] |
684 | A triangle has vertices $(0,0)$, $(1,1)$, and $(6m,0)$, and the line $y = mx$ divides the triangle into two triangles of equal area. What is the sum of all possible values of $m$?
$\textbf{A} - \!\frac {1}{3} \qquad \textbf{(B)} - \!\frac {1}{6} \qquad \textbf{(C)}\ \frac {1}{6} \qquad \textbf{(D)}\ \frac {1}{3} \qqua... | 2009 AMC 12A Problem 14 | Let's label the three points as $A=(0,0)$, $B=(1,1)$, and $C=(6m,0)$.
Clearly, whenever the line $y=mx$ intersects the inside of the triangle, it will intersect the side $BC$. Let $D$ be the point of intersection.
The triangles $ABD$ and $ACD$ have the same height, which is the distance between the point $A$ and the... | // Block 1
unitsize(1cm);
defaultpen(0.8);
real m=1.5;
draw( (0,0)--(1,1)--(6*m,0)--cycle );
draw( ((-0.5)*(1,m)) -- ((1.5)*(1,m)), dashed );
label("$A$",(0,0),NW);
label("$B$",(1,1),NE);
label("$C$",(6*m,0),E);
// Block 2
unitsize(2cm);
defaultpen(0.8);
real m=0.5;
draw( (0,0)--(1,1)--(6*m,0)--cycle );
draw( ((-1)*(1,... | [] |
685 | A circle with center $C$ is tangent to the positive $x$ and $y$-axes and externally tangent to the circle centered at $(3,0)$ with radius $1$. What is the sum of all possible radii of the circle with center $C$?
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ ... | 2009 AMC 12A Problem 16 | Let $r$ be the radius of our circle. For it to be tangent to the positive $x$ and $y$ axes, we must have $C=(r,r)$. For the circle to be externally tangent to the circle centered at $(3,0)$ with radius $1$, the distance between $C$ and $(3,0)$ must be exactly $r+1$.
By the Pythagorean theorem the distance between $(r,... | // Block 1
unitsize(0.5cm);
defaultpen(0.8);
filldraw( Circle( (3,0), 1 ), lightgray, black );
draw( (0,0) -- (15,0), Arrow );
draw( (0,0) -- (0,15), Arrow );
draw( (0,0) -- (15,15), dashed );
real r1 = 4 - 2*sqrt(2), r2 = 4 + 2*sqrt(2);
pair S1=(r1,r1), S2=(r2,r2);
dot(S1); dot(S2); dot((3,0));
draw( Circle(S1,r1) );
... | [] |
686 | Convex quadrilateral $ABCD$ has $AB = 9$ and $CD = 12$. Diagonals $AC$ and $BD$ intersect at $E$, $AC = 14$, and $\triangle AED$ and $\triangle BEC$ have equal areas. What is $AE$?
$\textbf{(A)}\ \frac {9}{2}\qquad \textbf{(B)}\ \frac {50}{11}\qquad \textbf{(C)}\ \frac {21}{4}\qquad \textbf{(D)}\ \frac {17}{3}\qquad... | 2009 AMC 12A Problem 20 | Let $[ABC]$ denote the area of triangle $ABC$. $[AED] = [BEC]$, so $[ABD] = [AED] + [AEB] = [BEC] + [AEB] = [ABC]$. Since triangles $ABD$ and $ABC$ share a base, they also have the same height and thus $\overline{AB}||\overline{CD}$ and $\triangle{AEB}\sim\triangle{CED}$ with a ratio of $3: 4$. $AE = \frac {3}{7}\times... | // Block 1
pathpen = linewidth(0.7);pointpen = black;
pair D=MP("D",(0,0)),C=MP("C",(12,0)),A=MP("A",C+14*expi(145*pi/180),N),B=MP("B",A+(9,0),N),E=IP(A--C,B--D);MP("9",(A+B)/2,N);MP("12",(C+D)/2);
fill(A--D--E--cycle,rgb(0.8,0.8,0.8));fill(B--C--E--cycle,rgb(0.8,0.8,0.8));D(A--B--C--D--cycle);D(A--C);D(B--D);D(E);
// ... | [] |
687 | Functions $f$ and $g$ are quadratic, $g(x) = - f(100 - x)$, and the graph of $g$ contains the vertex of the graph of $f$. The four $x$-intercepts on the two graphs have $x$-coordinates $x_1$, $x_2$, $x_3$, and $x_4$, in increasing order, and $x_3 - x_2 = 150$. Then $x_4 - x_1 = m + n\sqrt p$, where $m$, $n$, and $p$ ar... | 2009 AMC 12A Problem 23 | The two quadratics are $180^{\circ}$ rotations of each other about $(50,0)$. Since we are only dealing with differences of roots, we can translate them to be symmetric about $(0,0)$. Now $x_3 = - x_2 = 75$ and $x_4 = - x_1$. Say our translated versions of $f$ and $g$ are $p$ and $q$, respectively, so that $p(x) = - q( ... | // Block 1
import graph; size(250);Label k; k.p=fontsize(6); int ymax = 400, ymin = -400; real rt = 175+150*2^.5;
real f(real x){return 1/400*(x-125)*(x+rt);} real g(real x){return -f(100-x);}
xaxis(-600,600,Ticks(k, 5),Arrows(6));yaxis(ymin,ymax,Ticks(k, 5),Arrows(6));
draw(graph(f,-450,300),red+linewidth(0.8),Arrows... | [] |
688 | Points $A$ and $C$ lie on a circle centered at $O$, each of $\overline{BA}$ and $\overline{BC}$ are tangent to the circle, and $\triangle ABC$ is equilateral. The circle intersects $\overline{BO}$ at $D$. What is $\frac{BD}{BO}$?
$\text{(A) } \frac {\sqrt2}{3} \qquad \text{(B) } \frac {1}{2} \qquad \text{(C) } \frac {... | 2009 AMC 10B Problem 16 | Solution 1
As $\triangle ABC$ is equilateral, we have $\angle BAC = \angle BCA = 60^\circ$, hence $\angle OAC = \angle OCA = 30^\circ$. Then $\angle AOC = 120^\circ$, and from symmetry we have $\angle AOB = \angle COB = 60^\circ$. Thus, this gives us $\angle ABO = \angle CBO = 30^\circ$.
We know that $DO = AO$, as $... | // Block 1
unitsize(1.5cm); defaultpen(0.8); pair B=(0,0), A=(3,0), C=3*dir(60), O=intersectionpoint( C -- (C+3*dir(-30)), A -- (A+3*dir(90)) ); pair D=intersectionpoint(B--O, circle(O,length(A-O))); draw(circle(O,length(A-O))); draw(A--B--C--O--A); draw(B--O); draw(rightanglemark(B,A,O)); draw(rightanglemark(B,C,O)); ... | [] |
689 | Rectangle $ABCD$ has $AB=8$ and $BC=6$. Point $M$ is the midpoint of diagonal $\overline{AC}$, and $E$ is on $AB$ with $\overline{ME}\perp\overline{AC}$. What is the area of $\triangle AME$?
$\text{(A) } \frac{65}{8} \qquad \text{(B) } \frac{25}{3} \qquad \text{(C) } 9 \qquad \text{(D) } \frac{75}{8} \qquad \text{(E)... | 2009 AMC 10B Problem 18 | By the Pythagorean theorem we have $AC=10$, hence $AM=5$.
The triangles $AME$ and $ABC$ have the same angle at $A$ and a right angle, thus all their angles are equal, and therefore these two triangles are similar.
The ratio of their sides is $\frac{AM}{AB} = \frac 58$, hence the ratio of their areas is $\left( \frac... | // Block 1
unitsize(0.75cm);
defaultpen(0.8);
pair A=(0,0), B=(8,0), C=(8,6), D=(0,6), M=(A+C)/2;
path ortho = shift(M)*rotate(-90)*(A--C);
pair Ep = intersectionpoint(ortho, A--B);
draw( A--B--C--D--cycle );
draw( A--C );
draw( M--Ep );
filldraw( A--M--Ep--cycle, lightgray, black );
draw( rightanglemark(A,M,Ep) );
dra... | [] |
689 | Rectangle $ABCD$ has $AB=8$ and $BC=6$. Point $M$ is the midpoint of diagonal $\overline{AC}$, and $E$ is on $AB$ with $\overline{ME}\perp\overline{AC}$. What is the area of $\triangle AME$?
$\text{(A) } \frac{65}{8} \qquad \text{(B) } \frac{25}{3} \qquad \text{(C) } 9 \qquad \text{(D) } \frac{75}{8} \qquad \text{(E)... | 2009 AMC 10B Problem 18 | Draw $EC$ as shown from the diagram. Since $AC$ is of length $10$, we have that $AM$ is of length $5$, because of the midpoint $M$. Through the Pythagorean theorem, we know that $AE^2 = AM^2 + ME^2 \implies 25 + ME^2$, which means $AE = \sqrt{25 + ME^2}$. Define $ME$ to be $x$ for the sake of clarity. We know that $EB ... | // Block 1
unitsize(0.75cm);
defaultpen(0.8);
pair A=(0,0), B=(8,0), C=(8,6), D=(0,6), M=(A+C)/2;
path ortho = shift(M)*rotate(-90)*(A--C);
pair Ep = intersectionpoint(ortho, A--B);
draw( A--B--C--D--cycle );
draw( A--C );
draw( M--Ep );
filldraw( A--M--Ep--cycle, lightgray, black );
draw( rightanglemark(A,M,Ep) );
dra... | [] |
689 | Rectangle $ABCD$ has $AB=8$ and $BC=6$. Point $M$ is the midpoint of diagonal $\overline{AC}$, and $E$ is on $AB$ with $\overline{ME}\perp\overline{AC}$. What is the area of $\triangle AME$?
$\text{(A) } \frac{65}{8} \qquad \text{(B) } \frac{25}{3} \qquad \text{(C) } 9 \qquad \text{(D) } \frac{75}{8} \qquad \text{(E)... | 2009 AMC 10B Problem 18 | By the Pythagorean Theorem, we claim that $AC = 10$. It then follows that $AM \cong MC = 5.$
Because we have $AM \cong MC, \angle AME \cong \angle CME,$ and reflexive side $EM$, it follows that $\triangle AME \cong \triangle CME.$ By CPCTC, we have $AE \cong EC.$ For the sake of simplicity, we'll call those side lengt... | // Block 1
unitsize(0.75cm);
defaultpen(0.8);
pair A=(0,0), B=(8,0), C=(8,6), D=(0,6), M=(A+C)/2;
path ortho = shift(M)*rotate(-90)*(A--C);
pair Ep = intersectionpoint(ortho, A--B);
draw( A--B--C--D--cycle );
draw( A--C );
draw( M--Ep );
filldraw( A--M--Ep--cycle, lightgray, black );
draw( rightanglemark(A,M,Ep) );
dra... | [] |
690 | A cubical cake with edge length $2$ inches is iced on the sides and the top. It is cut vertically into three pieces as shown in this top view, where $M$ is the midpoint of a top edge. The piece whose top is triangle $B$ contains $c$ cubic inches of cake and $s$ square inches of icing. What is $c+s$?
$\text{(A) } \f... | 2009 AMC 10B Problem 22 | Let's label the points as in the picture above. Let $[RNQ]$ be the area of $\triangle RNQ$. Then the volume of the corresponding piece is $c=2[RNQ]$. This cake piece has icing on the top and on the vertical side that contains the edge $QR$. Hence the total area with icing is $[RNQ]+2^2 = [RNQ]+4$. Thus the answer to ou... | // Block 1
unitsize(2cm);
defaultpen(linewidth(.8pt)+fontsize(8pt));
draw((-1,-1)--(1,-1)--(1,1)--(-1,1)--cycle);
draw((1,1)--(-1,0));
pair P=foot((1,-1),(1,1),(-1,0));
draw((1,-1)--P);
draw(rightanglemark((-1,0),P,(1,-1),4));
label("$M$",(-1,0),W);
label("$C$",(-0.1,-0.3));
label("$A$",(-0.4,0.7));
label("$B$",(0.7,... | [] |
691 | The keystone arch is an ancient architectural feature. It is composed of congruent isosceles trapezoids fitted together along the non-parallel sides, as shown. The bottom sides of the two end trapezoids are horizontal. In an arch made with $9$ trapezoids, let $x$ be the angle measure in degrees of the larger interior a... | 2009 AMC 10B Problem 24 | Extend all the legs of the trapezoids. They will all intersect in the middle of the bottom side of the picture, forming the situation shown below.
Each of the angles at $X$ is $\frac{180^\circ}9 = 20^\circ$. From $\triangle XYZ$, the degree measure of the smaller interior angle of the trapezoid is $\frac{180^\circ -... | // Block 1
unitsize(6mm);
defaultpen(linewidth(.8pt));
int i;
real r=5, R=6;
path t=r*dir(0)--r*dir(20)--R*dir(20)--R*dir(0);
for(i=0; i<9; ++i)
{
draw(rotate(20*i)*t);
}
draw((-r,0)--(R+1,0));
draw((-R,0)--(-R-1,0));
for (int i=1; i<9; ++i) draw( (0,0) -- (rotate(20*i)*(r,0)), dotted );
label("$X$",(0,0),S);
label(... | [] |
692 | Triangle $ABC$ has vertices $A = (3,0)$, $B = (0,3)$, and $C$, where $C$ is on the line $x + y = 7$. What is the area of $\triangle ABC$?
$\mathrm{(A)}\ 6\qquad \mathrm{(B)}\ 8\qquad \mathrm{(C)}\ 10\qquad \mathrm{(D)}\ 12\qquad \mathrm{(E)}\ 14$ | 2009 AMC 12B Problem 9 | Solution 1
Because the line $x + y = 7$ is parallel to $\overline {AB}$, the area of $\triangle ABC$ is independent of the location of $C$ on that line. Therefore it may be assumed that $C$ is $(7,0)$. In that case the triangle has base $AC = 4$ and altitude $3$, so its area is $\frac 12 \cdot 4 \cdot 3 = \boxed {6}$... | unitsize(0.75cm); defaultpen(0.8); pair A=(3,0), B=(0,3); draw ( (-1,0) -- (9,0), dashed ); draw ( (0,-1) -- (0,9), dashed ); dot(A); dot(B); draw(A--B); draw ( (-1,8) -- (8,-1) ); label( "$A$", A, S ); label( "$B$", B, W ); label( "$3$", A--(0,0), S ); label( "$3$", B--(0,0), W ); label( "$x+y=7$", (8,-1), SE ); pair ... | [] |
693 | A convex polyhedron $Q$ has vertices $V_1,V_2,\ldots,V_n$, and $100$ edges. The polyhedron is cut by planes $P_1,P_2,\ldots,P_n$ in such a way that plane $P_k$ cuts only those edges that meet at vertex $V_k$. In addition, no two planes intersect inside or on $Q$. The cuts produce $n$ pyramids and a new polyhedron $R$. ... | 2009 AMC 12B Problem 20 | Solution 1
Each edge of $Q$ is cut by two planes, so $R$ has $200$ vertices. Three edges of $R$ meet at each vertex, so $R$ has a total of $\frac 12 \cdot 3 \cdot 200 = \boxed {300}$ edges.
Solution 2
At each vertex, as many new edges are created by this process as there are original edges meeting at that vertex. Thu... | // Block 1
pair A,B,C,D,E,F; A=(0,50); B=(0,-50); C=(86.602540378443864676372317075294,0); D=(300,0); E=(370.7,70.7); F=(370.7,-70.7); draw(A--C--B); draw(C--D); draw(E--D--F);
// Block 2
pair A,B,C,D,E,F,G,H,I,J,K; A=(0,50); B=(0,-50); C=(86.602540378443864676372317075294,0); D=(300,0); E=(370.7,70.7); F=(370.7,-70.7)... | [] |
694 | The set $G$ is defined by the points $(x,y)$ with integer coordinates, $3\le|x|\le7$, $3\le|y|\le7$. How many squares of side at least $6$ have their four vertices in $G$?
$\textbf{(A)}\ 125\qquad \textbf{(B)}\ 150\qquad \textbf{(C)}\ 175\qquad \textbf{(D)}\ 200\qquad \textbf{(E)}\ 225$ | 2009 AMC 12B Problem 25 | We need to find a reasonably easy way to count the squares.
First, obviously the maximum distance between two points in the same quadrant is $4\sqrt 2 < 6$, hence each square has exactly one vertex in each quadrant.
Given any square, we can circumscribe another axes-parallel square around it. In the picture below, t... | // Block 1
defaultpen(black+0.75bp+fontsize(8pt));
size(7.5cm);
path p = scale(.15)*unitcircle;
draw((-8,0)--(8.5,0),Arrow(HookHead,1mm));
draw((0,-8)--(0,8.5),Arrow(HookHead,1mm));
int i,j;
for (i=-7;i<8;++i) {
for (j=-7;j<8;++j) {
if (((-7 <= i) && (i <= -3)) || ((3 <= i) && (i<= 7))) { if (((-7 <= j) && (j <= -3)) ... | [] |
694 | The set $G$ is defined by the points $(x,y)$ with integer coordinates, $3\le|x|\le7$, $3\le|y|\le7$. How many squares of side at least $6$ have their four vertices in $G$?
$\textbf{(A)}\ 125\qquad \textbf{(B)}\ 150\qquad \textbf{(C)}\ 175\qquad \textbf{(D)}\ 200\qquad \textbf{(E)}\ 225$ | 2009 AMC 12B Problem 25 | Consider any square that meets the requirements described in the problem. Then, take the vertices of the square and translate them to the first quadrant (This is the "mapping" described in Solution 2). For example, consider a square with vertices $(7, 7), (-7, 7), (-7, -7),$ and $(7, -7)$:
After following the ma... | // Block 1
defaultpen(black+0.75bp+fontsize(8pt));
size(5cm);
path p = scale(.15)*unitcircle;
draw((-8,0)--(8.5,0),Arrow(HookHead,1mm));
draw((0,-8)--(0,8.5),Arrow(HookHead,1mm));
int i,j;
for (i=-7;i<8;++i) {
for (j=-7;j<8;++j) {
if (((-7 <= i) && (i <= -3)) || ((3 <= i) && (i<= 7))) { if (((-7 <= j) && (j <= -3)) ||... | [] |
695 | Crystal has a running course marked out for her daily run. She starts this run by heading due north for one mile. She then runs northeast for one mile, then southeast for one mile. The last portion of her run takes her on a straight line back to where she started. How far, in miles is this last portion of her run?
$\m... | 2010 AMC 10A Problem 7 | Crystal first runs north for one mile. Changing directions, she runs northeast for another mile. The angle difference between north and northeast is 45 degrees. She then switches directions to southeast, meaning a 90 degree angle change. The distance now from traveling north for one mile, and her current destination is... | // Block 1
import olympiad;
draw((0,0)--(0,1));
draw((0,1)--(0,1.7071067811865476), dotted);
draw((0,1)--(0.7071067811865476, 1.7071067811865476));
draw((0.7071067811865476, 1.7071067811865476)--(1.4142135623730951,1));
draw(anglemark((0.7071067811865476, 1.7071067811865476),(0,1),(0,1.7071067811865476)));
label("$45^{... | [] |
696 | Equiangular hexagon $ABCDEF$ has side lengths $AB=CD=EF=1$ and $BC=DE=FA=r$. The area of $\triangle ACE$ is $70\%$ of the area of the hexagon. What is the sum of all possible values of $r$?
$\textbf{(A)}\ \frac{4\sqrt{3}}{3} \qquad \textbf{(B)} \frac{10}{3} \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ \frac{17}{4} \qqu... | 2010 AMC 10A Problem 19 | Drawing not to scale.*
To find r, we'll form an equation by finding the area of hexagon $ABCDEF$ and $\triangle{ACE}$ separately. First, connect a directly opposite diagonal such as $CF$. Because the hexagon is equiangular, edges $AF$ and $BC$ protrude from points $A$ and $B$ at the same angle but in opposite direction... | import graph; size(8cm); pen dps = fontsize(10); defaultpen(dps); real r = 0.7; // Define hexagon vertices clockwise with AB on top pair A = (0, 0); pair B = (1, 0); pair C = B + r * dir(-60); pair D = C + dir(-120); pair E = D + r * dir(180); pair F = E + dir(120); // Draw the hexagon draw(A--B--C--D--E--F--cycle);... | [] |
697 | A fly trapped inside a cubical box with side length $1$ meter decides to relieve its boredom by visiting each corner of the box. It will begin and end in the same corner and visit each of the other corners exactly once. To get from a corner to any other corner, it will either fly or crawl in a straight line. What is th... | 2010 AMC 10A Problem 20 | The path of the fly consists of eight line segments, where each line segment goes from one corner to another corner. The distance of each such line segment is $1$, $\sqrt{2}$, or $\sqrt{3}$.
The only way to obtain a line segment of length $\sqrt{3}$ is to go from one corner of the cube to the opposite corner. Since th... | // Block 1
import graph;
unitsize(2 cm);
draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);
draw((1,0)--(1.3,0.3));
draw((1,1)--(1.3,1.3));
draw((0,1)--(0.3,1.3));
draw((1.3,0.3)--(1.3,1.3)--(0.3,1.3));
draw((0,0)--(0.3,0.3),dashed);
draw((0.3,0.3)--(1.3,0.3),dashed);
draw((0.3,0.3)--(0.3,1.3),dashed);
label("$A$", (0,0), SW)... | [] |
698 | Rectangle $ABCD$, pictured below, shares $50\%$ of its area with square $EFGH$. Square $EFGH$ shares $20\%$ of its area with rectangle $ABCD$. What is $\frac{AB}{AD}$?
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 5 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 10$ | 2010 AMC 12A Problem 3 | If we shift $A$ to coincide with $E$, and add new horizontal lines to divide $EFGH$ into five equal parts:
This helps us to see that $AD=a/5$ and $AB=2a$, where $a=EF$.
Hence $\dfrac{AB}{AD}=\dfrac{2a}{a/5}=10$. | // Block 1
unitsize(1mm);
defaultpen(linewidth(.8pt)+fontsize(8pt));
draw((0,0)--(0,25)--(25,25)--(25,0)--cycle);
fill((0,25)--(0,20)--(25,20)--(25,25)--cycle,gray);
draw((25,20)--(25,25)--(50,25)--(50,20)--cycle);
draw((0,5)--(25,5));
draw((0,10)--(25,10));
draw((0,15)--(25,15));
label("$A=E$",(0,25),W);
label("$B$"... | [] |
699 | For how many integer values of $k$ do the graphs of $x^2+y^2=k^2$ and $xy = k$ not intersect?
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 8$ | 2010 AMC 12A Problem 13 | The image below shows the two curves for $k=4$. The blue curve is $x^2+y^2=k^2$, which is clearly a circle with radius $k$, and the red curve is a part of the curve $xy=k$.
In the special case $k=0$ the blue curve is just the point $(0,0)$, and as $0\cdot 0=0$, this point is on the red curve as well, hence they inte... | // Block 1
import graph;
size(200);
real f(real x) {return 4/x;};
real g1(real x) {return sqrt(4*4-x*x);};
real g2(real x) {return -sqrt(4*4-x*x);};
draw(graph(f,-20./3,-0.6),red);
draw(graph(f,0.6,20./3),red);
draw(graph(g1,-4,4),blue);
draw(graph(g2,-4,4),blue);
axes("$x$","$y$");
// Block 2
import graph; size(200);... | [] |
700 | Equiangular hexagon $ABCDEF$ has side lengths $AB=CD=EF=1$ and $BC=DE=FA=r$. The area of $\triangle ACE$ is $70\%$ of the area of the hexagon. What is the sum of all possible values of $r$?
$\textbf{(A)}\ \frac{4\sqrt{3}}{3} \qquad \textbf{(B)} \frac{10}{3} \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ \frac{17}{4} \qqu... | 2010 AMC 12A Problem 17 | Drawing not to scale.*
To find r, we'll form an equation by finding the area of hexagon $ABCDEF$ and $\triangle{ACE}$ separately. First, connect a directly opposite diagonal such as $CF$. Because the hexagon is equiangular, edges $AF$ and $BC$ protrude from points $A$ and $B$ at the same angle but in opposite direction... | // Block 1
import graph;
size(8cm);
pen dps = fontsize(10); defaultpen(dps);
real r = 0.7;
// Define hexagon vertices clockwise with AB on top
pair A = (0, 0);
pair B = (1, 0);
pair C = B + r * dir(-60);
pair D = C + dir(-120);
pair E = D + r * dir(180);
pair F = E + dir(120);
// Draw the hexagon
draw(A--B--C--D--E-... | [] |
701 | A 16-step path is to go from $(-4,-4)$ to $(4,4)$ with each step increasing either the $x$-coordinate or the $y$-coordinate by 1. How many such paths stay outside or on the boundary of the square $-2 \le x \le 2$, $-2 \le y \le 2$ at each step?
$\textbf{(A)}\ 92 \qquad \textbf{(B)}\ 144 \qquad \textbf{(C)}\ 1568 \qqua... | 2010 AMC 12A Problem 18 | We can draw an $8$ by $8$ square with a hollow $4$ by $4$ center. The ways to get a to a point or corner of a coordinate point is equal to the ways of getting to the point to the left of the desired point and the bottom of the desired point, since we can only move right and up. Using basic addition and symmetry (along ... | // Block 1
// Set up the Asymptote units
unitsize(1cm);
// Draw the 8x8 outer grid
for (int i = 0; i <= 8; ++i) {
// Vertical lines of the grid
draw((i, 0)--(i, 8));
// Horizontal lines of the grid
draw((0, i)--(8, i));
}
// Draw the hollow 4x4 center (white rectangle over the center part)
filldraw(bo... | [] |
701 | A 16-step path is to go from $(-4,-4)$ to $(4,4)$ with each step increasing either the $x$-coordinate or the $y$-coordinate by 1. How many such paths stay outside or on the boundary of the square $-2 \le x \le 2$, $-2 \le y \le 2$ at each step?
$\textbf{(A)}\ 92 \qquad \textbf{(B)}\ 144 \qquad \textbf{(C)}\ 1568 \qqua... | 2010 AMC 12A Problem 18 | Let $N(A,C,B)$ denote the number of paths from $A$ to $C$ then to $B$, $N(A,D,B)$ denote the number of paths from $A$ to $D$ then to $B$, $N(A,E,B)$ denote the number of paths from $A$ to $E$ then to $B$, $N(A,C,D, B)$ denote the number of paths from $A$ to $C$ to $D$ then to $B$, $N(A,C,D,E,B)$ denote the number of p... | // Block 1
unitsize(1cm);
for (int i = 0; i <= 8; ++i) {
draw((i, 0)--(i, 8));
draw((0, i)--(8, i));
}
filldraw(box((2, 2), (6, 6)), white, white);
dot((0,0));
dot((2,6));
dot((2,7));
dot((2,8));
dot((8,8));
label("A",(0,0), align=W);
label("B",(8,8), align=E);
label("C",(2,6), align=NE);
label("D",(2,7),... | [] |
702 | A circle is centered at $O$, $\overline{AB}$ is a diameter and $C$ is a point on the circle with $\angle COB = 50^\circ$. What is the degree measure of $\angle CAB$?
$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 50 \qquad \textbf{(E)}\ 65$ | 2010 AMC 10B Problem 6 | Note that $\angle AOC = 180^\circ - 50^\circ = 130^\circ$. Because triangle $AOC$ is isosceles, $\angle CAB = (180^\circ - 130^\circ)/2 = \boxed{25^\circ}$. | // Block 1
import graph;
unitsize(2 cm);
pair O, A, B, C;
O = (0,0);
A = (-1,0);
B = (1,0);
C = dir(50);
draw(Circle(O,1));
draw(B--A--C--O);
label("$A$", A, W);
label("$B$", B, E);
label("$C$", C, NE);
label("$O$", O, S);
// Block 2
import graph; unitsize(2 cm); pair O, A, B, C; O = (0,0); A = (-1,0); B = (1,0... | [] |
703 | A square of side length $1$ and a circle of radius $\dfrac{\sqrt{3}}{3}$ share the same center. What is the area inside the circle, but outside the square?
$\textbf{(A)}\ \dfrac{\pi}{3}-1 \qquad \textbf{(B)}\ \dfrac{2\pi}{9}-\dfrac{\sqrt{3}}{3} \qquad \textbf{(C)}\ \dfrac{\pi}{18} \qquad \textbf{(D)}\ \dfrac{1}{4} \qq... | 2010 AMC 10B Problem 16 | The radius of the circle is $\frac{\sqrt{3}}{3} = \sqrt{\frac{1}{3}}$. Half the diagonal of the square is $\frac{\sqrt{1^2+1^2}}{2} = \frac{\sqrt{2}}{2} = \sqrt{\frac12}$. We can see that the circle passes outside the square, but the square is NOT completely contained in the circle. Therefore the picture will look some... | // Block 1
unitsize(5cm);
defaultpen(linewidth(.8pt)+fontsize(10pt));
dotfactor=3;
real r=sqrt(1/3);
pair O=(0,0);
pair W=(0.5,0.5), X=(0.5,-0.5), Y=(-0.5,-0.5), Z=(-0.5,0.5);
pair A=(-sqrt(1/12),0.5), B=(sqrt(1/12),0.5);
pair V=(0,0.5);
path outer=Circle(O,r);
draw(outer);
draw(W--X--Y--Z--cycle);
draw(O--A);
draw(O-... | [] |
704 | A circle with center $O$ has area $156\pi$. Triangle $ABC$ is equilateral, $\overline{BC}$ is a chord on the circle, $OA = 4\sqrt{3}$, and point $O$ is outside $\triangle ABC$. What is the side length of $\triangle ABC$?
$\textbf{(A)}\ 2\sqrt{3} \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 4\sqrt{3} \qquad \textbf{(D)}... | 2010 AMC 10B Problem 19 | The formula for the area of a circle is $\pi r^2$ so the radius of this circle is $\sqrt{156}.$
Because $OA=4\sqrt{3}=\sqrt{48} < \sqrt{156}, A$ must be in the interior of circle $O.$
Let $s$ be the side length of the triangle, the unknown value, and let $X$ be the point on $BC$ where $OX \perp BC.$ Since $\triangle... | // Block 1
unitsize(3mm);
defaultpen(linewidth(.8pt)+fontsize(11pt));
dotfactor=3;
real r=sqrt(156);
pair A=(0,sqrt(48)), B=(-3,sqrt(147)), C=(3,sqrt(147));
pair O=(0,0);
pair X=(0,7sqrt(3));
path outer=Circle(O,r);
draw(outer);
draw(A--B--C--cycle);
draw(O--X); draw(O--B);
pair[] ps={A,B,C,O,X};
dot(ps);
label("$A$... | [] |
705 | Circles $A, B,$ and $C$ each has radius $1$. Circles $A$ and $B$ share one point of tangency. Circle $C$ has a point of tangency with the midpoint of $\overline{AB}.$ What is the area inside circle $C$ but outside circle $A$ and circle $B?$
$\textbf{(A)}\ 3 - \frac{\pi}{2} \qquad \textbf{(B)}\ \frac{\pi}{2} \qquad... | 2011 AMC 10A Problem 18 | The requested area is the area of $C$ minus the area shared between circles $A$, $B$ and $C$.
Let $M$ be the midpoint of $\overline{AB}$ and $D$ be the other intersection of circles $C$ and $B$.
The area shared between $C$, $A$ and $B$ is $4$ of the regions between arc $\widehat {MD}$ and line $\overline{MD}$, which ... | unitsize(1.1cm); defaultpen(linewidth(.8pt)); dotfactor=4; pair A=(0,0), B=(2,0), C=(1,-1); pair M=(1,0); pair D=(2,-1); dot (A); dot (B); dot (C); dot (D); dot (M); draw(Circle(A,1)); draw(Circle(B,1)); draw(Circle(C,1)); draw(A--B); draw(M--D); draw(D--B); label("$A$",A,W); label("$B$",B,E); label("$C$",C,W); lab... | [] |
705 | Circles $A, B,$ and $C$ each has radius $1$. Circles $A$ and $B$ share one point of tangency. Circle $C$ has a point of tangency with the midpoint of $\overline{AB}.$ What is the area inside circle $C$ but outside circle $A$ and circle $B?$
$\textbf{(A)}\ 3 - \frac{\pi}{2} \qquad \textbf{(B)}\ \frac{\pi}{2} \qquad... | 2011 AMC 10A Problem 18 | We can move the area above the part of the circle above the segment $EF$ down, and similarly for the other side. Then, we have a square, whose diagonal is $2$, so the area is then just $\left(\frac{2}{\sqrt{2}}\right)^2 = \boxed{\textbf{2 = C}}$.
~ Minor Edits, Challengees24 | unitsize(1.1cm); defaultpen(linewidth(.8pt)); dotfactor=4; pair A=(0,0), B=(2,0), C=(1,-1); pair M=(1,0); pair D=(2,-1); dot (A); dot (B); dot (C); dot (D); dot (M); draw(Circle(A,1)); draw(Circle(B,1)); draw(Circle(C,1)); draw(A--B); draw(M--D); draw(D--B); label("$A$",A,W); label("$B$",B,E); label("$C$",C,W); lab... | [] |
706 | Circles $A, B,$ and $C$ each has radius $1$. Circles $A$ and $B$ share one point of tangency. Circle $C$ has a point of tangency with the midpoint of $\overline{AB}.$ What is the area inside circle $C$ but outside circle $A$ and circle $B?$
$\textbf{(A)}\ 3 - \frac{\pi}{2} \qquad \textbf{(B)}\ \frac{\pi}{2} \qquad... | 2011 AMC 12A Problem 11 | The requested area is the area of $C$ minus the area shared between circles $A$, $B$ and $C$.
Let $M$ be the midpoint of $\overline{AB}$ and $D$ be the other intersection of circles $C$ and $B$.
The area shared between $C$, $A$ and $B$ is $4$ of the regions between arc $\widehat {MD}$ and line $\overline{MD}$, which ... | // Block 1
unitsize(1.1cm);
defaultpen(linewidth(.8pt));
dotfactor=4;
pair A=(0,0), B=(2,0), C=(1,-1);
pair M=(1,0);
pair D=(2,-1);
dot (A);
dot (B);
dot (C);
dot (D);
dot (M);
draw(Circle(A,1));
draw(Circle(B,1));
draw(Circle(C,1));
draw(A--B);
draw(M--D);
draw(D--B);
label("$A$",A,W);
label("$B$",B,E);
label("$C$... | [] |
706 | Circles $A, B,$ and $C$ each has radius $1$. Circles $A$ and $B$ share one point of tangency. Circle $C$ has a point of tangency with the midpoint of $\overline{AB}.$ What is the area inside circle $C$ but outside circle $A$ and circle $B?$
$\textbf{(A)}\ 3 - \frac{\pi}{2} \qquad \textbf{(B)}\ \frac{\pi}{2} \qquad... | 2011 AMC 12A Problem 11 | We can move the area above the part of the circle above the segment $EF$ down, and similarly for the other side. Then, we have a square, whose diagonal is $2$, so the area is then just $\left(\frac{2}{\sqrt{2}}\right)^2 = \boxed{\textbf{2 = C}}$.
~ Minor Edits, Challengees24 | // Block 1
unitsize(1.1cm);
defaultpen(linewidth(.8pt));
dotfactor=4;
pair A=(0,0), B=(2,0), C=(1,-1);
pair M=(1,0);
pair D=(2,-1);
dot (A);
dot (B);
dot (C);
dot (D);
dot (M);
draw(Circle(A,1));
draw(Circle(B,1));
draw(Circle(C,1));
draw(A--B);
draw(M--D);
draw(D--B);
label("$A$",A,W);
label("$B$",B,E);
label("$C$... | [] |
707 | A power boat and a raft both left dock $A$ on a river and headed downstream. The raft drifted at the speed of the river current. The power boat maintained a constant speed with respect to the river. The power boat reached dock $B$ downriver, then immediately turned and traveled back upriver. It eventually met the raft ... | 2011 AMC 12A Problem 12 | Solution 1
WLOG, let the speed of the river be 0. This is allowed because the problem never states that the speed of the current has to have a magnitude greater than 0. In this case, when the powerboat travels from $A$ to $B$, the raft remains at $A$. Thus the trip from $A$ to $B$ takes the same time as the trip from $... | size(8cm,8cm); pair A, B; A=(-3,4); B=(3,4); draw(A--B); label("$A$",A,S); label("$B$",B,S); arrow((-2.5,6.4),dir(180),blue); arrow((-2.5,6.2),dir(180),red); pair A, B; A=(-3,5); B=(3,5); draw(A--B); label("$A$",A,S); label("$B$",B,S); arrow((2.6,5.4),dir(360),blue); arrow((-0.5,5.2),dir(180),red); pair A, B; A=(-3,6);... | [] |
708 | Circles with radii $1$, $2$, and $3$ are mutually externally tangent. What is the area of the triangle determined by the points of tangency?
$\textbf{(A)}\ \frac{3}{5} \qquad \textbf{(B)}\ \frac{4}{5} \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ \frac{6}{5} \qquad \textbf{(E)}\ \frac{4}{3}$ | 2011 AMC 12A Problem 17 | The centers of these circles form a 3-4-5 triangle, which has an area equal to 6.
The areas of the three triangles determined by the center and the two points of tangency of each circle are, using Triangle Area by Sine,
$\frac{1}{2} \cdot 1 \cdot 1 \cdot 1 = \frac{1}{2}$
$\frac{1}{2} \cdot 2 \cdot 2 \cdot \frac{4}{5... | // Block 1
unitsize(.5cm);
defaultpen(linewidth(.8pt));
dotfactor=4;
pair A=(0,0), B=(3,0), C=(0,4);
dot (A);
dot (B);
dot (C);
draw(A--B);
draw(A--C);
draw(B--C);
draw(Circle(A,1));
draw(Circle(B,2));
draw(Circle(C,3));
// Block 2
unitsize(.5cm); defaultpen(linewidth(.8pt)); dotfactor=4; pair A=(0,0), B=(3,0), C=... | [] |
709 | Triangle $ABC$ has $\angle BAC = 60^{\circ}$, $\angle CBA \leq 90^{\circ}$, $BC=1$, and $AC \geq AB$. Let $H$, $I$, and $O$ be the orthocenter, incenter, and circumcenter of $\triangle ABC$, respectively. Assume that the area of pentagon $BCOIH$ is the maximum possible. What is $\angle CBA$?
$\textbf{(A)}\ 60^{\circ} ... | 2011 AMC 12A Problem 25 | By the Inscribed Angle Theorem, \[\angle BOC = 2\angle BAC = 120^\circ .\]Let $D$ and $E$ be the feet of the altitudes of $\triangle ABC$ from $B$ and $C$, respectively. In $\triangle ACE$ we get $\angle ACE = 30^\circ$, and as exterior angle \[\angle BHC = 90^\circ + \angle ACE = 120^\circ .\]Because the lines $BI$ an... | // Block 1
import geometry;
size(200);
defaultpen(fontsize(12)+0.8);
pair O,A,B,C,D,E,I,H;
real h=2*sqrt(3);
O=(0,1/h); B=(-0.5,0); C=(0.5,0);
path c1=CR(O,length(O-B));
// A=IP(c1,O--O+5*dir(120));
A=IP(c1,B--B+5*dir(80));
I=incenter(A,B,C);
H=orthocenter(A,B,C);
D=extension(A,C,B,H); E=extension(A,B,C,H);
path c2=c... | [] |
710 | A dart board is a regular octagon divided into regions as shown. Suppose that a dart thrown at the board is equally likely to land anywhere on the board. What is the probability that the dart lands within the center square?
$\textbf{(A)}\ \frac{\sqrt{2} - 1}{2} \qquad\textbf{(B)}\ \frac{1}{4} \qquad\textbf{(C)}\ \fr... | 2011 AMC 10B Problem 16 | If the side lengths of the dart board and the side lengths of the center square are all $\sqrt{2},$ then the side length of the legs of the triangles are $1$.
\begin{align*} \text{area of center square} &: \sqrt{2} \times \sqrt{2} = 2\\ \text{total area} &: (\sqrt{2})^2 + 4(1 \times \sqrt{2}) + 4(1 \times 1 \times \fr... | // Block 1
unitsize(10mm);
defaultpen(linewidth(.8pt)+fontsize(10pt));
dotfactor=1;
pair A=(0,1), B=(1,0), C=(1+sqrt(2),0), D=(2+sqrt(2),1), E=(2+sqrt(2),1+sqrt(2)), F=(1+sqrt(2),2+sqrt(2)), G=(1,2+sqrt(2)), H=(0,1+sqrt(2));
pair I=(1,1), J=(1+sqrt(2),1), K=(1+sqrt(2),1+sqrt(2)), L=(1,1+sqrt(2));
draw(A--B--C--D--E--... | [] |
711 | Rectangle $ABCD$ has $AB = 6$ and $BC = 3$. Point $M$ is chosen on side $AB$ so that $\angle AMD = \angle CMD$. What is the degree measure of $\angle AMD$?
$\textbf{(A)}\ 15 \qquad\textbf{(B)}\ 30 \qquad\textbf{(C)}\ 45 \qquad\textbf{(D)}\ 60 \qquad\textbf{(E)}\ 75$ | 2011 AMC 10B Problem 18 | It is given that $\angle AMD \sim \angle CMD$. Since $\angle AMD$ and $\angle CDM$ are alternate interior angles and $\overline{AB} \parallel \overline{DC}$, $\angle AMD \cong \angle CDM \longrightarrow \angle CMD \cong \angle CDM$. Use the Base Angle Theorem to show $\overline{DC} \cong \overline{MC}$. We know that $A... | // Block 1
unitsize(10mm);
defaultpen(linewidth(.5pt)+fontsize(10pt));
dotfactor=3;
pair A=(0,3), B=(6,3), C=(6,0), D=(0,0);
pair M=(0.80385,3);
draw(A--B--C--D--cycle);
draw(M--C);
draw(M--D);
draw(anglemark(A,M,D));
draw(anglemark(D,M,C));
pair[] ps={A,B,C,D,M};
dot(ps);
label("$A$",A,NW);
label("$B$",B,NE);
labe... | [] |
712 | Rhombus $ABCD$ has side length $2$ and $\angle B = 120$°. Region $R$ consists of all points inside the rhombus that are closer to vertex $B$ than any of the other three vertices. What is the area of $R$?
$\textbf{(A)}\ \frac{\sqrt{3}}{3} \qquad\textbf{(B)}\ \frac{\sqrt{3}}{2} \qquad\textbf{(C)}\ \frac{2\sqrt{3}}{3} \q... | 2011 AMC 10B Problem 20 | Suppose that $P$ is a point in the rhombus $ABCD$ and let $\ell_{BC}$ be the perpendicular bisector of $\overline{BC}$. Then $PB < PC$ if and only if $P$ is on the same side of $\ell_{BC}$ as $B$. The line $\ell_{BC}$ divides the plane into two half-planes; let $S_{BC}$ be the half-plane containing $B$. Let us define s... | // Block 1
unitsize(8mm);
defaultpen(linewidth(0.8pt)+fontsize(10pt));
dotfactor=4;
pair A=(4,0), B=(2,2sqrt(3)), C=(-2,2sqrt(3)), D=(0,0), E=(B+C)/2, F=(B+C+D)/3, G=(A+C)/2, H=(A+B+D)/3, I=(A+B)/2;
fill((0,2sqrt(3))--B--(3,sqrt(3))--(2,(2sqrt(3))/3)--(0,(4sqrt(3))/3)--cycle,gray);
draw(A--B--C--D--cycle);
draw(D--(0,... | [] |
712 | Rhombus $ABCD$ has side length $2$ and $\angle B = 120$°. Region $R$ consists of all points inside the rhombus that are closer to vertex $B$ than any of the other three vertices. What is the area of $R$?
$\textbf{(A)}\ \frac{\sqrt{3}}{3} \qquad\textbf{(B)}\ \frac{\sqrt{3}}{2} \qquad\textbf{(C)}\ \frac{2\sqrt{3}}{3} \q... | 2011 AMC 10B Problem 20 | Since $H$ and $E$ are halfway between $AB$ and $CB$, respectively, we know that $\overline{BH}=\overline{BE}=1$. By symmetry, $\Delta BFG$ is equilateral, so $\angle FBG=60^\circ\implies\angle EBF=\angle HBG=30^\circ$ and therefore $\Delta EBF$ and $\Delta HBG$ are 30-60-90 right triangles.
Thus, $[\Delta EBF]=[\Delta... | // Block 1
unitsize(8mm);
defaultpen(linewidth(0.8pt)+fontsize(10pt));
dotfactor=4;
pair A=(4,0), B=(2,2sqrt(3)), C=(-2,2sqrt(3)), D=(0,0), E=(B+C)/2, F=(B+C+D)/3, G=(A+B+D)/3, H=(A+B)/2;
fill((0,2sqrt(3))--B--(3,sqrt(3))--(2,(2sqrt(3))/3)--(0,(4sqrt(3))/3)--cycle,gray);
draw(A--B--C--D--cycle);
draw(D--(0,2sqrt(3)));... | [] |
712 | Rhombus $ABCD$ has side length $2$ and $\angle B = 120$°. Region $R$ consists of all points inside the rhombus that are closer to vertex $B$ than any of the other three vertices. What is the area of $R$?
$\textbf{(A)}\ \frac{\sqrt{3}}{3} \qquad\textbf{(B)}\ \frac{\sqrt{3}}{2} \qquad\textbf{(C)}\ \frac{2\sqrt{3}}{3} \q... | 2011 AMC 10B Problem 20 | To keep it simple, break rhombus $ABCD$ into two triangles, $ABD$ and $BCD$. To see the area closest to the point $B$, notice that a third of each triangle, which contains all the points nearest to $B$ in each triangle, is easily visualizable. Thus, a third of rhombus $ABCD$ must be found.
We find the total area of rh... | // Block 1
unitsize(8mm);
defaultpen(linewidth(0.8pt)+fontsize(10pt));
dotfactor=4;
pair A=(4,0), B=(2,2sqrt(3)), C=(-2,2sqrt(3)), D=(0,0), E=(B+C)/2, F=(B+C+D)/3, G=(A+B+D)/3, H=(A+B)/2;
fill((0,2sqrt(3))--B--(3,sqrt(3))--(2,(2sqrt(3))/3)--(0,(4sqrt(3))/3)--cycle,gray);
draw(A--B--C--D--cycle);
draw(D--F); draw(D--G)... | [] |
713 | Brian writes down four integers $w > x > y > z$ whose sum is $44$. The pairwise positive differences of these numbers are $1, 3, 4, 5, 6,$ and $9$. What is the sum of the possible values for $w$?
$\textbf{(A)}\ 16 \qquad\textbf{(B)}\ 31 \qquad\textbf{(C)}\ 48 \qquad\textbf{(D)}\ 62 \qquad\textbf{(E)}\ 93$ | 2011 AMC 10B Problem 21 | The largest difference, $9,$ must be between $w$ and $z.$
The smallest difference, $1,$ must be directly between two integers. This also means the differences directly between the other two should add up to $8.$ The only remaining differences that would make this possible are $3$ and $5.$ However, those two difference... | // Block 1
unitsize(14mm);
defaultpen(linewidth(.8pt)+fontsize(10pt));
dotfactor=4;
pair Z1=(0,1), Y1=(1,1), X1=(2,1), W1=(3,1);
pair Z4=(4,1), Y4=(5,1), X4=(6,1), W4=(7,1);
draw(Z1--W1); draw(Z4--W4);
pair[] ps={W1,W4,X1,X4,Y1,Y4,Z1,Z4};
dot(ps);
label("$z$",Z1,N); label("$y$",Y1,N); label("$x$",X1,N); label("$w$",... | [] |
714 | A pyramid has a square base with sides of length $1$ and has lateral faces that are equilateral triangles. A cube is placed within the pyramid so that one face is on the base of the pyramid and its opposite face has all its edges on the lateral faces of the pyramid. What is the volume of this cube?
$\textbf{(A)}\ 5\sq... | 2011 AMC 10B Problem 22 | It is often easier to first draw a diagram for such a problem.
[Image: images/amc/2011_AMC_10B_Problem_22_0.png]
Sometimes, it may also be easier to think of the problem in 2D. Take a cross section of the pyramid through the apex and two points from the base that are opposite to each other. Place it in two dimensions.... | // Block 1
unitsize(35mm);
defaultpen(linewidth(2pt)+fontsize(10pt));
pair A=(0,0), B=(sqrt(2),0), C=(0.5sqrt(2),0.5sqrt(2));
pair W=(sqrt(2)-1,0), X=(1,0), Y=(1,sqrt(2)-1), Z=(sqrt(2)-1,sqrt(2)-1);
draw(A--B--C--cycle);
draw(W--X--Y--Z--cycle,red);
// Block 2
unitsize(35mm);
defaultpen(linewidth(2pt)+fontsize(12pt));
... | ["https://artofproblemsolving.com/wiki/images/c/ce/2011AMC10B22.png"] |
715 | Rhombus $ABCD$ has side length $2$ and $\angle B = 120$°. Region $R$ consists of all points inside the rhombus that are closer to vertex $B$ than any of the other three vertices. What is the area of $R$?
$\textbf{(A)}\ \frac{\sqrt{3}}{3} \qquad\textbf{(B)}\ \frac{\sqrt{3}}{2} \qquad\textbf{(C)}\ \frac{2\sqrt{3}}{3} \q... | 2011 AMC 12B Problem 16 | Suppose that $P$ is a point in the rhombus $ABCD$ and let $\ell_{BC}$ be the perpendicular bisector of $\overline{BC}$. Then $PB < PC$ if and only if $P$ is on the same side of $\ell_{BC}$ as $B$. The line $\ell_{BC}$ divides the plane into two half-planes; let $S_{BC}$ be the half-plane containing $B$. Let us define s... | unitsize(8mm); defaultpen(linewidth(0.8pt)+fontsize(10pt)); dotfactor=4; pair A=(4,0), B=(2,2sqrt(3)), C=(-2,2sqrt(3)), D=(0,0), E=(B+C)/2, F=(B+C+D)/3, G=(A+C)/2, H=(A+B+D)/3, I=(A+B)/2; fill((0,2sqrt(3))--B--(3,sqrt(3))--(2,(2sqrt(3))/3)--(0,(4sqrt(3))/3)--cycle,gray); draw(A--B--C--D--cycle); draw(D--(0,2sqrt(3)));... | [] |
715 | Rhombus $ABCD$ has side length $2$ and $\angle B = 120$°. Region $R$ consists of all points inside the rhombus that are closer to vertex $B$ than any of the other three vertices. What is the area of $R$?
$\textbf{(A)}\ \frac{\sqrt{3}}{3} \qquad\textbf{(B)}\ \frac{\sqrt{3}}{2} \qquad\textbf{(C)}\ \frac{2\sqrt{3}}{3} \q... | 2011 AMC 12B Problem 16 | Since $H$ and $E$ are halfway between $AB$ and $CB$, respectively, we know that $\overline{BH}=\overline{BE}=1$. By symmetry, $\Delta BFG$ is equilateral, so $\angle FBG=60^\circ\implies\angle EBF=\angle HBG=30^\circ$ and therefore $\Delta EBF$ and $\Delta HBG$ are 30-60-90 right triangles.
Thus, $[\Delta EBF]=[\Delta... | unitsize(8mm); defaultpen(linewidth(0.8pt)+fontsize(10pt)); dotfactor=4; pair A=(4,0), B=(2,2sqrt(3)), C=(-2,2sqrt(3)), D=(0,0), E=(B+C)/2, F=(B+C+D)/3, G=(A+B+D)/3, H=(A+B)/2; fill((0,2sqrt(3))--B--(3,sqrt(3))--(2,(2sqrt(3))/3)--(0,(4sqrt(3))/3)--cycle,gray); draw(A--B--C--D--cycle); draw(D--(0,2sqrt(3))); draw(D--(3... | [] |
715 | Rhombus $ABCD$ has side length $2$ and $\angle B = 120$°. Region $R$ consists of all points inside the rhombus that are closer to vertex $B$ than any of the other three vertices. What is the area of $R$?
$\textbf{(A)}\ \frac{\sqrt{3}}{3} \qquad\textbf{(B)}\ \frac{\sqrt{3}}{2} \qquad\textbf{(C)}\ \frac{2\sqrt{3}}{3} \q... | 2011 AMC 12B Problem 16 | To keep it simple, break rhombus $ABCD$ into two triangles, $ABD$ and $BCD$. To see the area closest to the point $B$, notice that a third of each triangle, which contains all the points nearest to $B$ in each triangle, is easily visualizable. Thus, a third of rhombus $ABCD$ must be found.
We find the total area of rh... | unitsize(8mm); defaultpen(linewidth(0.8pt)+fontsize(10pt)); dotfactor=4; pair A=(4,0), B=(2,2sqrt(3)), C=(-2,2sqrt(3)), D=(0,0), E=(B+C)/2, F=(B+C+D)/3, G=(A+B+D)/3, H=(A+B)/2; fill((0,2sqrt(3))--B--(3,sqrt(3))--(2,(2sqrt(3))/3)--(0,(4sqrt(3))/3)--cycle,gray); draw(A--B--C--D--cycle); draw(D--F); draw(D--G); draw(F--C... | [] |
716 | A bug travels in the coordinate plane, moving only along the lines that are parallel to the $x$-axis or $y$-axis. Let $A = (-3, 2)$ and $B = (3, -2)$. Consider all possible paths of the bug from $A$ to $B$ of length at most $20$. How many points with integer coordinates lie on at least one of these paths?
$\textbf{(A)... | 2011 AMC 12B Problem 23 | (Diagram for Solution 2 by Technodoggo also don't ask about the patterns, I just figured out how to use the patterns module)
Notice that the bug is basically moving from A to B (length 10) but optionally going on a detour anywhere along its route.
Specifically, the detour would be of total length 10, 5 to some point... | // Block 1
import graph;
import patterns;
unitsize(1cm);
add("tile",tile());
add("checker",checker());
filldraw((-3,2)--(3,2)--(3,-2)--(-3,-2)--cycle,pattern("checker"));
filldraw((-3,2)--(-3,7)--(3,7)--(3,2)--cycle,pattern("tile"));
filldraw((-3,-2)--(-3,-7)--(3,-7)--(3,-2)--cycle,pattern("tile"));
filldraw((-8,2)-... | [] |
717 | For every $m$ and $k$ integers with $k$ odd, denote by $\left[\frac{m}{k}\right]$ the integer closest to $\frac{m}{k}$. For every odd integer $k$, let $P(k)$ be the probability that
\[\left[\frac{n}{k}\right] + \left[\frac{100 - n}{k}\right] = \left[\frac{100}{k}\right]\]
for an integer $n$ randomly chosen from the i... | 2011 AMC 12B Problem 25 | Solution 1
Answer: $(D) \frac{34}{67}$
First of all, you have to realize that
if $\left[\frac{n}{k}\right] + \left[\frac{100 - n}{k}\right] = \left[\frac{100}{k}\right]$
then $\left[\frac{n - k}{k}\right] + \left[\frac{100 - (n - k)}{k}\right] = \left[\frac{100}{k}\right]$
So, we can consider what happen in $1\... | // Block 1
size(10cm); for (int i = 0; i < 67; ++i) { if (i<=33) dot((i,0)); else dot((i,1)); } label("(0,0)",(0,0),SW); label("(66,1)",(66,1),NE);
// Block 2
size(10cm); for (int i = 0; i < 67; ++i) { dot((i,0)); } label("(0,0)",(0,0),NW); label("(66,0)",(66,0),SE); | [] |
718 | A bug crawls along a number line, starting at $-2$. It crawls to $-6$, then turns around and crawls to $5$. How many units does the bug crawl altogether?
$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 15$ | 2012 AMC 10A Problem 3 | Crawling from $-2$ to $-6$ takes it a distance of $4$ units. Crawling from $-6$ to $5$ takes it a distance of $11$ units. Add $4$ and $11$ to get $\boxed{\textbf{(E)}\ 15}$ | // Block 1
draw((-2,1)--(-6,1),red+dashed,EndArrow);
draw((-6,2)--(5,2),blue+dashed,EndArrow);
dot((-2,0));
dot((-6,0));
dot((5,0));
label("$-2$",(-2,0),dir(270));
label("$-6$",(-6,0),dir(270));
label("$5$",(5,0),dir(270));
label("$4$",(-4,0.9),dir(270));
label("$11$",(-1.5,2.5),dir(90));
// Block 2
draw((-2,1)--(-6,1)... | [] |
719 | Externally tangent circles with centers at points $A$ and $B$ have radii of lengths $5$ and $3$, respectively. A line externally tangent to both circles intersects ray $AB$ at point $C$. What is $BC$?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 4.8\qquad\textbf{(C)}\ 10.2\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 14.4$ | 2012 AMC 10A Problem 11 | Let $D$ and $E$ be the points of tangency on circles $A$ and $B$ with line $CD$. $AB=8$. Also, let $BC=x$. As $\angle ADC$ and $\angle BEC$ are right angles (a radius is perpendicular to a tangent line at the point of tangency) and both triangles share $\angle ACD$, $\triangle ADC \sim \triangle BEC$. From this we can ... | // Block 1
unitsize(3.5mm);
defaultpen(linewidth(.8pt)+fontsize(10pt));
dotfactor=4;
pair A=(0,0), B=(8,0); pair C=(20,0); pair D=(1.25,-0.25sqrt(375)); pair E=(8.75,-0.15sqrt(375));
path a=Circle(A,5);
path b=Circle(B,3);
draw(a); draw(b);
draw(C--D);
draw(A--C);
draw(A--D);
draw(B--E);
pair[] ps={A,B,C,D,E};
dot(ps... | [] |
720 | Three unit squares and two line segments connecting two pairs of vertices are shown. What is the area of $\triangle ABC$?
$\textbf{(A)}\ \frac16 \qquad\textbf{(B)}\ \frac15 \qquad\textbf{(C)}\ \frac29 \qquad\textbf{(D)}\ \frac13 \qquad\textbf{(E)}\ \frac{\sqrt{2}}{4}$ | 2012 AMC 10A Problem 15 | $AC$ intersects $BC$ at a right angle, (this can be proved by noticing that the slopes of the two lines are negative reciprocals of each other) so $\triangle ABC \sim \triangle BED$. The hypotenuse of right triangle $BED$ is $\sqrt{1^2+2^2}=\sqrt{5}$.
\[\frac{AC}{BC}=\frac{BD}{ED} \Rightarrow \frac{AC}{BC} = \frac21 \... | // Block 1
unitsize(2cm);
defaultpen(linewidth(.8pt)+fontsize(10pt));
dotfactor=4;
pair A=(0,0), B=(1,0); pair C=(0.8,-0.4);
pair D=(1,-2), E=(0,-2);
draw(A--(2,0)); draw((0,-1)--(2,-1)); draw(E--D);
draw(A--E); draw(B--D); draw((2,0)--(2,-1));
draw(A--(2,-1)); draw(B--E);
pair[] ps={A,B,C,D,E};
dot(ps);
label("$A$",... | [] |
720 | Three unit squares and two line segments connecting two pairs of vertices are shown. What is the area of $\triangle ABC$?
$\textbf{(A)}\ \frac16 \qquad\textbf{(B)}\ \frac15 \qquad\textbf{(C)}\ \frac29 \qquad\textbf{(D)}\ \frac13 \qquad\textbf{(E)}\ \frac{\sqrt{2}}{4}$ | 2012 AMC 10A Problem 15 | Let $\text{E}$ be the origin. Then,
$\text{D}=(1, 0)$
$\text{A}=(0, 2)$
$\text{B}=(1, 2)$
$\text{F}=(2, 1)$
${EB}$ can be represented by the line $y=2x$
Also, ${AF}$ can be represented by the line $y=-\frac{1}{2}x+2$
Subtracting the second equation from the first gives us $\frac{5}{2}x-2=0$.
Thus, $x=\frac{4}{5}$.
P... | // Block 1
unitsize(2cm);
defaultpen(linewidth(.8pt)+fontsize(10pt));
dotfactor=4;
pair A=(0,0), B=(1,0); pair C=(0.8,-0.4);
pair D=(1,-2), E=(0,-2); pair F=(2,-1); pair G=(0.8,0);
draw(A--(2,0)); draw((0,-1)--F); draw(E--D);
draw(A--E); draw(B--D); draw((2,0)--F);
draw(A--F); draw(B--E); draw(C--G);
pair[] ps={A,B,C... | [] |
720 | Three unit squares and two line segments connecting two pairs of vertices are shown. What is the area of $\triangle ABC$?
$\textbf{(A)}\ \frac16 \qquad\textbf{(B)}\ \frac15 \qquad\textbf{(C)}\ \frac29 \qquad\textbf{(D)}\ \frac13 \qquad\textbf{(E)}\ \frac{\sqrt{2}}{4}$ | 2012 AMC 10A Problem 15 | Triangle $EAB$ is similar to triangle $EHI$; line $HI = 1/2$
Triangle $ACB$ is similar to triangle $FCI$ and the ratio of line $AB$ to line $IF = 1 : \frac{3}{2} = 2: 3$.
Based on similarity the length of the height of $GC$ is thus $\frac{2}{5}\cdot1 = \frac{2}{5}$.
Thus, $[ABC]=\frac{1}{2} \cdot {AB} \cdot {CG}=\fr... | // Block 1
unitsize(2cm);
defaultpen(linewidth(.8pt)+fontsize(10pt));
dotfactor=4;
pair A=(0,0), B=(1,0); pair C=(0.8,-0.4);
pair D=(1,-2), E=(0,-2); pair F=(2,-1); pair G=(0.8,0);
pair H=(0,-1), I=(0.5,-1);
draw(A--(2,0)); draw((0,-1)--F); draw(E--D);
draw(A--E); draw(B--D); draw((2,0)--F);
draw(A--F); draw(B--E); d... | [] |
721 | The closed curve in the figure is made up of 9 congruent circular arcs each of length $\frac{2\pi}{3}$, where each of the centers of the corresponding circles is among the vertices of a regular hexagon of side 2. What is the area enclosed by the curve?
$\textbf{(A)}\ 2\pi+6\qquad\textbf{(B)}\ 2\pi+4\sqrt{3}\qquad\te... | 2012 AMC 10A Problem 18 | We can draw the hexagon between the centers of the circles, and compute its area. The hexagon is made of $6$ equilateral triangles each with length $2$, so the area is:
\[\frac{\sqrt{3}}{4} \cdot 2^2 \cdot 6=6 \sqrt{3}.\]
Then, we add the areas of the three sectors outside the hexagon:
\[\frac 23 \pi \cdot 3=2\pi.\]
We... | // Block 1
unitsize(2cm);
defaultpen(fontsize(6pt));
dotfactor=4;
label("$\circ$",(0,1));
label("$\circ$",(0.865,0.5));
label("$\circ$",(-0.865,0.5));
label("$\circ$",(0.865,-0.5));
label("$\circ$",(-0.865,-0.5));
label("$\circ$",(0,-1));
dot((0,1.5));
dot((-0.4325,0.75));
dot((0.4325,0.75));
dot((-0.4325,-0.75));
dot(... | [] |
721 | The closed curve in the figure is made up of 9 congruent circular arcs each of length $\frac{2\pi}{3}$, where each of the centers of the corresponding circles is among the vertices of a regular hexagon of side 2. What is the area enclosed by the curve?
$\textbf{(A)}\ 2\pi+6\qquad\textbf{(B)}\ 2\pi+4\sqrt{3}\qquad\te... | 2012 AMC 10A Problem 18 | As you can see, this diagram looks like a fidget spinner ;). Fidget spinners aside, we need to add stuff to our diagram to make it look easier. In the directions, they were talking about the centers of each arc create a hexagon, so let's add that to our diagram.
The side length of the hexagon is 2 and if we plug it... | // Block 1
defaultpen(fontsize(6pt));
dotfactor=4;
label("$\circ$",(0,1));
label("$\circ$",(0.865,0.5));
label("$\circ$",(-0.865,0.5));
label("$\circ$",(0.865,-0.5));
label("$\circ$",(-0.865,-0.5));
label("$\circ$",(0,-1));
dot((0,1.5));
dot((-0.432,0.75));
dot((0.4325,0.75));
dot((-0.4325,-0.75));
dot((0.4325,-0.75));... | [] |
722 | Let points $A$ = $(0 ,0 ,0)$, $B$ = $(1, 0, 0)$, $C$ = $(0, 2, 0)$, and $D$ = $(0, 0, 3)$. Points $E$, $F$, $G$, and $H$ are midpoints of line segments $\overline{BD},\text{ } \overline{AB}, \text{ } \overline {AC},$ and $\overline{DC}$ respectively. What is the area of $EFGH$?
$\textbf{(A)}\ \sqrt{2}\qquad\textbf{(B... | 2012 AMC 10A Problem 21 | Consider a tetrahedron with vertices at $A,B,C,D$ in the $xyz$-space. The length of $EF$ is just one-half of $AD$ because it is the midsegment of $\triangle ABD.$ The same concept applies to the other side lengths. $AD=3$ and $BC=\sqrt{1^2+2^2}=\sqrt{5}$. Then $EF=HG=\frac32$ and $EH=FG=\frac{\sqrt{5}}{2}$. The line se... | // Block 1
import three;
draw((0,0,0)--(1,0,0)--(0,0,3)--cycle);
draw((0,0,0)--(0,2,0));
draw((0,2,0)--(0,0,3));
//EFGH
draw((0.5,0,1.5)--(0.5,0,0)--(0,1,0)--(0,1,1.5)--(0.5,0,1.5),red);
//Points
label("$E$",(0.5,0,1.5),NW);
label("$F$",(0.5,0,0),S);
label("$G$",(0,1,0),S);
label("$H$",(0,1,1.5),NE);
label("$A$",(0,0,0... | [] |
723 | Real numbers $x$, $y$, and $z$ are chosen independently and at random from the interval $[0,n]$ for some positive integer $n$. The probability that no two of $x$, $y$, and $z$ are within 1 unit of each other is greater than $\frac {1}{2}$. What is the smallest possible value of $n$?
$\textbf{(A)}\ 7\qquad\textbf{(B)}\... | 2012 AMC 10A Problem 25 | Since $x,y,z$ are all reals located in $[0, n]$, the number of choices for each one is continuous so we use geometric probability.
WLOG(Without loss of generality), assume that $n\geq x \geq y \geq z \geq 0$. Then the set of points $(x,y,z)$ is a tetrahedron, or a triangular pyramid. The point $(x,y,z)$ distributes un... | // Block 1
import three;
unitsize(10cm);
size(150);
currentprojection=orthographic(1/2,-1,2/3);
// three - currentprojection, orthographic
draw((1,1,0)--(0,1,0)--(0,0,0),dashed);
draw((0,0,0)--(0,0,1));
draw((0,1,0)--(0,1,1),dashed);
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,... | [] |
724 | A bug crawls along a number line, starting at $-2$. It crawls to $-6$, then turns around and crawls to $5$. How many units does the bug crawl altogether?
$\textbf{(A)}\ 9\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 13\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 15$ | 2012 AMC 12A Problem 1 | Crawling from $-2$ to $-6$ takes it a distance of $4$ units. Crawling from $-6$ to $5$ takes it a distance of $11$ units. Add $4$ and $11$ to get $\boxed{\textbf{(E)}\ 15}$ | draw((-2,1)--(-6,1),red+dashed,EndArrow); draw((-6,2)--(5,2),blue+dashed,EndArrow); dot((-2,0)); dot((-6,0)); dot((5,0)); label("$-2$",(-2,0),dir(270)); label("$-6$",(-6,0),dir(270)); label("$5$",(5,0),dir(270)); label("$4$",(-4,0.9),dir(270)); label("$11$",(-1.5,2.5),dir(90)); | [] |
725 | A triangle has area $30$, one side of length $10$, and the median to that side of length $9$. Let $\theta$ be the acute angle formed by that side and the median. What is $\sin{\theta}$?
$\textbf{(A)}\ \frac{3}{10}\qquad\textbf{(B)}\ \frac{1}{3}\qquad\textbf{(C)}\ \frac{9}{20}\qquad\textbf{(D)}\ \frac{2}{3}\qquad\tex... | 2012 AMC 12A Problem 10 | $AB$ is the side of length $10$, and $CD$ is the median of length $9$. The altitude of $C$ to $AB$ is $6$ because the 0.5(altitude)(base)=Area of the triangle. $\theta$ is $\angle CDE$. To find $\sin{\theta}$, just use opposite over hypotenuse with the right triangle $\triangle DCE$. This is equal to $\frac69=\boxed{\t... | // Block 1
unitsize(5mm);
defaultpen(linewidth(.8pt)+fontsize(10pt));
dotfactor=4;
pair A=(-5,0), B=(5,0), C=(sqrt(45),6), D=(0,0), E=(sqrt(45),0);
draw(A--B--C--cycle);
draw(D--C); draw(E--C);
pair[] ps={A,B,C,D,E};
dot(ps);
label("$A$",A,SW);
label("$B$",B,SE);
label("$C$",C,N);
label("$D$",D,SW);
label("$E$",E,S)... | [] |
726 | A square region $ABCD$ is externally tangent to the circle with equation $x^2+y^2=1$ at the point $(0,1)$ on the side $CD$. Vertices $A$ and $B$ are on the circle with equation $x^2+y^2=4$. What is the side length of this square?
$\textbf{(A)}\ \frac{\sqrt{10}+5}{10}\qquad\textbf{(B)}\ \frac{2\sqrt{5}}{5}\qquad\text... | 2012 AMC 12A Problem 12 | The circles have radii of $1$ and $2$. Draw the triangle shown in the figure above and write expressions in terms of $s$ (length of the side of the square) for the sides of the triangle. Because $AO$ is the radius of the larger circle, which is equal to $2$, we can write the Pythagorean Theorem.
\begin{align*} \left( ... | // Block 1
unitsize(15mm);
defaultpen(linewidth(.8pt)+fontsize(10pt));
dotfactor=4;
real a=1; real b=2;
pair O=(0,0);
pair A=(-(sqrt(19)-2)/5,1);
pair B=((sqrt(19)-2)/5,1);
pair C=((sqrt(19)-2)/5,1+2(sqrt(19)-2)/5);
pair D=(-(sqrt(19)-2)/5,1+2(sqrt(19)-2)/5);
pair E=(-(sqrt(19)-2)/5,0);
path inner=Circle(O,a);
path ou... | [] |
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