problem stringlengths 16 4.31k | level stringclasses 6
values | type stringclasses 7
values | solution stringlengths 29 6.77k | index int64 0 25.9k |
|---|---|---|---|---|
The length of the median to the hypotenuse of an isosceles, right triangle is $10$ units. What is the length of a leg of the triangle, in units? Express your answer in simplest radical form. | Level 4 | Geometry | The length of the median to the hypotenuse is half the length of the hypotenuse, so the hypotenuse is $10\cdot2=20$ units long. Since the right triangle is isosceles, the length of a leg is $20/\sqrt{2}=\boxed{10\sqrt{2}}$ units. | 599 |
In $\Delta ABC$, $AC = BC$, $m\angle DCB = 40^{\circ}$, and $CD \parallel AB$. What is the number of degrees in $m\angle ECD$?
[asy] pair A,B,C,D,E; B = dir(-40); A = dir(-140); D = (.5,0); E = .4 * dir(40);
draw(C--B--A--E,EndArrow); draw(C--D,EndArrow);
label("$A$",A,W); label("$C$",C,NW);label("$B$",B,E);label("$D$... | Level 2 | Geometry | Angles $\angle DCB$ and $\angle B$ are alternate interior angles, so they are congruent. Therefore, $m\angle B=40^\circ$.
Since $AC=BC$, triangle $\triangle ABC$ is isosceles with equal angles at $A$ and $B$. Therefore, $m\angle A=40^\circ$.
Finally, $\angle A$ and $\angle ECD$ are corresponding angles, so $m\angle E... | 647 |
Two vertices of an obtuse triangle are $(6,4)$ and $(0,0)$. The third vertex is located on the negative branch of the $x$-axis. What are the coordinates of the third vertex if the area of the triangle is 30 square units? | Level 4 | Geometry | We know that, for a triangle, area = 1/2(base)(height), which equals 30 in this problem. We also know that the height of the triangle is 4 if we use the horizontal leg on the x-axis as the base. Now we can plug this information into the equation to find the length of the base that runs along the x-axis. The equation is... | 1,131 |
Suppose that we are given 40 points equally spaced around the perimeter of a square, so that four of them are located at the vertices and the remaining points divide each side into ten congruent segments. If $P$, $Q$, and $R$ are chosen to be any three of these points which are not collinear, then how many different p... | Level 5 | Geometry | Without loss of generality, assume that our square has vertices at $(0,0)$, $(10,0)$, $(10,10)$, and $(0,10)$ in the coordinate plane, so that the 40 equally spaced points are exactly those points along the perimeter of this square with integral coordinates. We first note that if $P$, $Q$, and $R$ are three of these p... | 324 |
In right triangle $DEF$, we have $\angle D = 25^\circ$, $\angle E = 90^\circ$, and $EF = 9$. Find $DE$ to the nearest tenth. You may use a calculator for this problem. | Level 4 | Geometry | We start with a diagram:
[asy]
pair D,EE,F;
EE = (0,0);
F = (8,0);
D = (0,8*Tan(65));
draw(D--EE--F--D);
draw(rightanglemark(F,EE,D,18));
label("$E$",EE,SW);
label("$F$",F,SE);
label("$D$",D,N);
label("$9$",F/2,S);
[/asy]
We seek $DE$, and we have $EF$ and $\angle D$. We can relate these three with the tangent func... | 1,135 |
[asy] draw(circle((0,6sqrt(2)),2sqrt(2)),black+linewidth(.75)); draw(circle((0,3sqrt(2)),sqrt(2)),black+linewidth(.75)); draw((-8/3,16sqrt(2)/3)--(-4/3,8sqrt(2)/3)--(0,0)--(4/3,8sqrt(2)/3)--(8/3,16sqrt(2)/3),dot); MP("B",(-8/3,16*sqrt(2)/3),W);MP("B'",(8/3,16*sqrt(2)/3),E); MP("A",(-4/3,8*sqrt(2)/3),W);MP("A'",(4/3,8*s... | Level 5 | Geometry | Using the tangent-tangent theorem, $PA=AB=PA'=A'B'=4$. We can then drop perpendiculars from the centers of the circles to the points of tangency and use similar triangles. Let us let the center of the smaller circle be point $S$ and the center of the larger circle be point $L$. If we let the radius of the larger circle... | 6,195 |
How many cubic feet are in the volume of a round swimming pool which is 16 feet in diameter and 4 feet deep throughout? Express your answer in terms of $\pi$. | Level 2 | Geometry | The radius of this pool is $16/2=8$ feet; the volume of this pool is thus $\pi(8^2)(4)=\boxed{256\pi}$ cubic feet. | 427 |
The number of inches in the perimeter of an equilateral triangle equals the number of square inches in the area of its circumscribed circle. What is the radius, in inches, of the circle? Express your answer in terms of pi and in simplest radical form. | Level 5 | Geometry | Let the triangle have vertices $A$, $B$, and $C$, let $O$ be the center of the circle, and let $D$ be the midpoint of $
\overline{BC}$. Triangle $COD$ is a $30 - 60 - 90$ degree triangle. If $r$ is the radius of the circle, then the sides of $\triangle COD$ are $r$, $\frac{r}{2}$, and $\frac{r\sqrt{3}}{2}$. The perimet... | 807 |
The shaded region shown consists of 11 unit squares and rests along the $x$-axis and the $y$-axis. The shaded region is rotated about the $x$-axis to form a solid. In cubic units, what is the volume of the resulting solid? Express your answer in simplest form in terms of $\pi$.
[asy]
defaultpen(linewidth(0.7));
fill((... | Level 4 | Geometry | The resulting solid is the union of two cylinders: one whose radius is 5 units and whose height is 1 unit (the squares shown in light gray produce this cylinder), and the other whose radius is 2 units and whose height is 3 units (shown in dark gray). The sum of these volumes is $\pi(5)^2(1)+\pi(2)^2(3)=\boxed{37\pi}$ ... | 1,001 |
A square piece of paper has sides of length $100$. From each corner a wedge is cut in the following manner: at each corner, the two cuts for the wedge each start at a distance $\sqrt{17}$ from the corner, and they meet on the diagonal at an angle of $60^{\circ}$ (see the figure below). The paper is then folded up along... | Level 5 | Geometry | [asy] import three; import math; import cse5; size(500); pathpen=blue; real r = (51^0.5-17^0.5)/200, h=867^0.25/100; triple A=(0,0,0),B=(1,0,0),C=(1,1,0),D=(0,1,0); triple F=B+(r,-r,h),G=(1,-r,h),H=(1+r,0,h),I=B+(0,0,h); draw(B--F--H--cycle); draw(B--F--G--cycle); draw(G--I--H); draw(B--I); draw(A--B--C--D--cycle); tri... | 6,108 |
Segment $s_1$ has endpoints at $(4,1)$ and $(-8,5)$. Segment $s_2$ is obtained by translating $s_1$ by $2$ units to the right and $3$ units up. Find the midpoint of segment $s_2$. Express your answer as $(a,b)$ with $a$ and $b$ integers. | Level 2 | Geometry | The midpoint of segment $s_1$ can be found using the midpoint formula: $\left(\frac{4-8}2,\frac{1+5}2\right)=(-2,3).$ The midpoint of $s_2$ is the translation of the midpoint of $s_1$ be $2$ units to the right and $3$ units up. Thus its coordinates are $(-2+2,3+3)=\boxed{(0,6)}.$ | 332 |
$ABCDE$ is a regular pentagon. $AP$, $AQ$ and $AR$ are the perpendiculars dropped from $A$ onto $CD$, $CB$ extended and $DE$ extended, respectively. Let $O$ be the center of the pentagon. If $OP = 1$, then find $AO + AQ + AR$.
[asy]
unitsize(2 cm);
pair A, B, C, D, E, O, P, Q, R;
A = dir(90);
B = dir(90 - 360/5... | Level 5 | Geometry | To solve the problem, we compute the area of regular pentagon $ABCDE$ in two different ways. First, we can divide regular pentagon $ABCDE$ into five congruent triangles.
[asy]
unitsize(2 cm);
pair A, B, C, D, E, O, P, Q, R;
A = dir(90);
B = dir(90 - 360/5);
C = dir(90 - 2*360/5);
D = dir(90 - 3*360/5);
E = dir... | 1,079 |
A solid right prism $ABCDEF$ has a height of $16,$ as shown. Also, its bases are equilateral triangles with side length $12.$ Points $X,$ $Y,$ and $Z$ are the midpoints of edges $AC,$ $BC,$ and $DC,$ respectively. A part of the prism above is sliced off with a straight cut through points $X,$ $Y,$ and $Z.$ Determine th... | Level 5 | Geometry | To determine the surface area of solid $CXYZ,$ we determine the area of each of the four triangular faces and sum them.
Areas of $\triangle CZX$ and $\triangle CZY:$
Each of these triangles is right-angled and has legs of lengths 6 and 8; therefore, the area of each is $\frac{1}{2}(6)(8)=24$.
Area of $\triangle CXY:... | 86 |
In triangle $ABC$, $\angle ABC = 90^\circ$ and $AD$ is an angle bisector. If $AB = 90,$ $BC = x$, and $AC = 2x - 6,$ then find the area of $\triangle ADC$. Round your answer to the nearest integer. | Level 5 | Geometry | First, we shall sketch! [asy]
pair A, B, C, D;
A = (0,90);
B = (0,0);
C = (56,0);
D = (56*90/(90+106),0);
draw(A--B--C--cycle);
draw(A--D);
label("$A$", A, NW);
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, NE);
label("$90$", (A + B)/2, W);
label("$x$", (B + C)/2, S);
label("$2x-6$", (A + C)/2, NE);
draw(ri... | 255 |
A cylindrical glass is half full of lemonade. The ratio of lemon juice to water in the lemonade is 1:11. If the glass is 6 inches tall and has a diameter of 2 inches, what is the volume of lemon juice in the glass? Express your answer as a decimal to the nearest hundredth. | Level 4 | Geometry | We can begin by calculating the volume of the liquid in the glass. Since the glass is half full, the portion filled with liquid has height 3 inches. The volume will be $\pi r^2 h=\pi\cdot 1^2 \cdot 3 = 3\pi$. Now, since the ratio of lemon juice to water is 1:11, the ratio of lemon juice to the liquid will be 1:(1+11... | 93 |
A square flag has a red cross of uniform width with a blue square in the center on a white background as shown. (The cross is symmetric with respect to each of the diagonals of the square.) If the entire cross (both the red arms and the blue center) takes up 36% of the area of the flag, what percent of the area of the ... | Level 5 | Geometry | The diagram can be quartered as shown:[asy] draw((0,0)--(0,5)--(5,5)--(5,0)--(0,0)); draw((0,1)--(4,5)); draw((1,0)--(5,4)); draw((0,4)--(4,0)); draw((1,5)--(5,1)); draw((0,0)--(5,5),dotted); draw((0,5)--(5,0),dotted); [/asy]and reassembled into two smaller squares of side $k$, each of which looks like this:[asy] draw(... | 6,200 |
Two cross sections of a right hexagonal pyramid are obtained by cutting the pyramid with planes parallel to the hexagonal base. The areas of the cross sections are $216\sqrt{3}$ square feet and $486\sqrt{3}$ square feet. The two planes are $8$ feet apart. How far from the apex of the pyramid is the larger cross section... | Level 5 | Geometry | The ratio of the areas of the cross sections is equal to $\frac{216\sqrt{3}}{486\sqrt{3}} = \frac 49$. Since the ratio of the area of two similar figures is the square of the ratio of their corresponding sides, it follows that the ratio of the corresponding sides of the cross-sections is equal to $\sqrt{\frac 49} = \fr... | 25,569 |
Equilateral triangle $T$ is inscribed in circle $A$, which has radius $10$. Circle $B$ with radius $3$ is internally tangent to circle $A$ at one vertex of $T$. Circles $C$ and $D$, both with radius $2$, are internally tangent to circle $A$ at the other two vertices of $T$. Circles $B$, $C$, and $D$ are all externally ... | Level 5 | Geometry | Let $X$ be the intersection of the circles with centers $B$ and $E$, and $Y$ be the intersection of the circles with centers $C$ and $E$. Since the radius of $B$ is $3$, $AX =4$. Assume $AE$ = $p$. Then $EX$ and $EY$ are radii of circle $E$ and have length $4+p$. $AC = 8$, and angle $CAE = 60$ degrees because we are gi... | 6,119 |
Compute $\sin 225^\circ$. | Level 3 | Geometry | Let $P$ be the point on the unit circle that is $225^\circ$ counterclockwise from $(1,0)$, and let $D$ be the foot of the altitude from $P$ to the $x$-axis, as shown below.
[asy]
pair A,C,P,O,D;
draw((0,-1.2)--(0,1.2),p=black+1.2bp,Arrows(0.15cm));
draw((-1.2,0)--(1.2,0),p=black+1.2bp,Arrows(0.15cm));
A = (1,0);
... | 146 |
In triangle $ABC$, $BC = 20 \sqrt{3}$ and $\angle C = 30^\circ$. Let the perpendicular bisector of $BC$ intersect $BC$ and $AC$ at $D$ and $E$, respectively. Find the length of $DE$. | Level 4 | Geometry | We have that $D$ is the midpoint of $BC$, and that $CD = BC/2 = 20 \sqrt{3}/2 = 10 \sqrt{3}$.
[asy]
unitsize(3 cm);
pair A, B, C, D, E;
A = dir(133);
B = dir(193);
C = dir(-13);
D = (B + C)/2;
E = extension(A, C, D, D + rotate(90)*(B - C));
draw(A--B--C--cycle);
draw(D--E);
label("$A$", A, N);
label("$B$", B, SW);... | 311 |
What is the ratio of the area of a square inscribed in a semicircle with radius $r$ to the area of a square inscribed in a circle with radius $r$? Express your answer as a common fraction. | Level 5 | Geometry | Let $s_1$ be the side length of the square inscribed in the semicircle of radius $r$. Applying the Pythagorean theorem to the right triangle shown in the diagram, we have $(s_1/2)^2+s_1^2=r^2$, which implies $s_1^2=\frac{4}{5}r^2$. Let $s_2$ be the side length of the square inscribed in the circle of radius $r$. App... | 234 |
In the figure with circle $Q$, angle $KAT$ measures 42 degrees. What is the measure of minor arc $AK$ in degrees? [asy]
import olympiad; size(150); defaultpen(linewidth(0.8)); dotfactor=4;
draw(unitcircle);
draw(dir(84)--(-1,0)--(1,0));
dot("$A$",(-1,0),W); dot("$K$",dir(84),NNE); dot("$T$",(1,0),E); dot("$Q$",(0,0),S)... | Level 3 | Geometry | Since $\angle A$ is inscribed in arc $KT$, the measure of arc $KT$ is $2\angle A = 84^\circ$. Since arc $AKT$ is a semicircle, arc $KA$ has measure $180 - 84 = \boxed{96}$ degrees. | 957 |
Stuart has drawn a pair of concentric circles, as shown. He draws chords $\overline{AB}$, $\overline{BC}, \ldots$ of the large circle, each tangent to the small one. If $m\angle ABC=75^\circ$, then how many segments will he draw before returning to his starting point at $A$? [asy]
size(100); defaultpen(linewidth(0.8)... | Level 5 | Geometry | We look at $\angle ABC$. $\angle ABC$ cuts off minor arc $\widehat{AC}$, which has measure $2\cdot m\angle ABC = 150^\circ$, so minor arcs $\widehat{AB}$ and $\widehat{BC}$ each have measure $\frac{360^\circ-150^\circ}{2}=105^\circ$.
Stuart cuts off one $105^\circ$ minor arc with each segment he draws. By the time St... | 589 |
Find the area of rhombus $ABCD$ given that the radii of the circles circumscribed around triangles $ABD$ and $ACD$ are $12.5$ and $25$, respectively.
| Level 5 | Geometry | The diagonals of the rhombus perpendicularly bisect each other. Call half of diagonal BD $a$ and half of diagonal AC $b$. The length of the four sides of the rhombus is $\sqrt{a^2+b^2}$.
The area of any triangle can be expressed as $\frac{a\cdot b\cdot c}{4R}$, where $a$, $b$, and $c$ are the sides and $R$ is the circu... | 6,082 |
Jimmy owns a cube-shaped container that measures $10$ inches on each side. He fills this container with water until it is half full. Then he throws ten giant ice cubes that measure $2$ inches on each side into the container. In inches cubed, how much of the container is unoccupied by ice or water? | Level 3 | Geometry | First let's compute the volume of the container. The container measures $10\times10\times10$ so its volume is $$10\cdot10\cdot10=10^3=1000$$inches cubed. Since the container is only half full, there are $$\frac{1}{2}\cdot10^3=500$$inches cubed of water in it. Additionally, there are ten ice cubes each with a volume of... | 225 |
$ABC$ is an equilateral triangle with sides equal to 2cm. $\overline{BC}$ is extended its own length to $D$, and $E$ is the midpoint of $\overline{AB}$. Suppose $\overline{ED}$ meets $\overline{AC}$ at $F$. Find the area of the quadrilateral $BEFC$ in square centimeters.
[asy]
size(200);
draw( (0,0) -- (1/2, .866) --(... | Level 5 | Geometry | [asy]
size(200);
draw( (0,0) -- (1/2, .866) --(1,0)--cycle); label("$B$", (0,0), W); label("$C$", (1,0), S); label( "$A$", (1/2, .866), N);
draw( (1/4 , .866/2)--(2,0)); label("$E$", (1/4, .866/2), NW); label("$D$", (2, 0), E); draw((0,0)-- (2,0));
pair t = intersectionpoint( (1/4 , .866/2)--(2,0), (1/2, .866) --(1,0))... | 251 |
What is the number of square units in the area of the hexagon below?
[asy]
unitsize(0.5cm);
defaultpen(linewidth(0.7)+fontsize(10));
dotfactor = 4;
int i,j;
for(i=0;i<=4;++i)
{
for(j=-3;j<=3;++j)
{
dot((i,j));
}
}
for(i=1;i<=4;++i)
{
draw((i,-1/3)--(i,1/3));
}
for(j=1;j<=3;++j)
{
draw((-1/3,j)--(1/3,j));
... | Level 2 | Geometry | Each of the four shaded triangles in the diagram below has area $\frac{1}{2}(1)(3)=\frac{3}{2}$ square units, and the shaded triangles along with the hexagon form a rectangular region whose area is $6\cdot4=24$ square units. Therefore, the area of the hexagon is $24-4\cdot \frac{3}{2}=\boxed{18}$ square units.
[asy]
... | 133 |
The rectangle $ABCD$ below has dimensions $AB = 12 \sqrt{3}$ and $BC = 13 \sqrt{3}$. Diagonals $\overline{AC}$ and $\overline{BD}$ intersect at $P$. If triangle $ABP$ is cut out and removed, edges $\overline{AP}$ and $\overline{BP}$ are joined, and the figure is then creased along segments $\overline{CP}$ and $\overlin... | Level 5 | Geometry | Let $\triangle{ABC}$ (or the triangle with sides $12\sqrt {3}$, $13\sqrt {3}$, $13\sqrt {3}$) be the base of our tetrahedron. We set points $C$ and $D$ as $(6\sqrt {3}, 0, 0)$ and $( - 6\sqrt {3}, 0, 0)$, respectively. Using Pythagoras, we find $A$ as $(0, \sqrt {399}, 0)$. We know that the vertex of the tetrahedron ($... | 6,013 |
A pyramid with volume 40 cubic inches has a rectangular base. If the length of the base is doubled, the width tripled and the height increased by $50\%$, what is the volume of the new pyramid, in cubic inches? | Level 4 | Geometry | Since the volume of a pyramid is linear in each of length, width, and height (in particular, $V = \frac{1}{3} lwh$), multiplying any of these dimensions by a scalar multiplies the volume by the same scalar. So the new volume is $2\cdot 3\cdot 1.50 = 9$ times the old one, or $\boxed{360}$ cubic inches. | 480 |
Kadin makes a snowman by stacking snowballs of radius 2 inches, 3 inches, and 5 inches. Assuming all his snowballs are spherical, what is the total volume of snow he uses, in cubic inches? Express your answer in terms of $\pi$. | Level 3 | Geometry | A sphere with radius $r$ has volume $\frac{4}{3}\pi r^3$. Thus, the snowballs with radius 2, 3, and 5 inches have volumes $\frac{4}{3}\pi(2^3)$, $\frac{4}{3}\pi(3^3)$, and $\frac{4}{3}\pi(5^3)$ cubic inches respectively. The total volume of snow used is thus \begin{align*}
\frac{4}{3}\pi(2^3)+\frac{4}{3}\pi(3^3)+\fr... | 22 |
Points $P$ and $Q$ are midpoints of two sides of the square. What fraction of the interior of the square is shaded? Express your answer as a common fraction.
[asy]
filldraw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)--cycle,gray,linewidth(1));
filldraw((0,1)--(1,2)--(2,2)--(0,1)--cycle,white,linewidth(1));
label("P",(0,1),W);
l... | Level 3 | Geometry | Let the side length of the square be $x$. The triangle has $\frac{1}{2} x$ as both its base and height. Therefore, its area is $\frac{1}{8} x^2$, and since the area of the square is $x^2$, the shaded area is $\boxed{\frac{7}{8}}$ of the total. | 919 |
In triangle $ABC,\,$ angle $C$ is a right angle and the altitude from $C\,$ meets $\overline{AB}\,$ at $D.\,$ The lengths of the sides of $\triangle ABC\,$ are integers, $BD=29^3,\,$ and $\cos B=m/n\,$, where $m\,$ and $n\,$ are relatively prime positive integers. Find $m+n.\,$
| Level 5 | Geometry | Since $\triangle ABC \sim \triangle CBD$, we have $\frac{BC}{AB} = \frac{29^3}{BC} \Longrightarrow BC^2 = 29^3 AB$. It follows that $29^2 | BC$ and $29 | AB$, so $BC$ and $AB$ are in the form $29^2 x$ and $29 x^2$, respectively, where x is an integer.
By the Pythagorean Theorem, we find that $AC^2 + BC^2 = AB^2 \Longri... | 6,028 |
In the figure, $\angle EAB$ and $\angle ABC$ are right angles, $AB=4,$ $BC=6,$ $AE=8,$ and $\overline{AC}$ and $\overline{BE}$ intersect at $D.$ What is the difference between the areas of $\triangle ADE$ and $\triangle BDC?$ [asy]
pair A,B,C,D,I;
I=(0,8);
A=(0,0);
B=(4,0);
C=(4,6);
D=(2.5,4);
draw(A--B--I--cycle,linew... | Level 3 | Geometry | Let $x,$ $y,$ and $z$ be the areas of $\triangle ADE,$ $\triangle BDC,$ and $\triangle ABD,$ respectively. The area of $\triangle ABE$ is \[\frac 1 2\cdot 4\cdot 8= 16= x+z,\]and the area of $\triangle BAC$ is \[\frac 1 2\cdot 4\cdot 6 = 12= y+z.\]Subtracting these equations gives $$(x+z) - (y+z) = 16-12\implies x - y ... | 672 |
An $\textit{annulus}$ is the region between two concentric circles. The concentric circles in the figure have radii $b$ and $c$, with $b>c$. Let $\overline{OX}$ be a radius of the larger circle, let $\overline{XZ}$ be tangent to the smaller circle at $Z$, and let $\overline{OY}$ be the radius of the larger circle that ... | Level 5 | Geometry | The area of the annulus is the difference between the areas of the two circles, which is $\pi b^2 -\pi c^2$. Because the tangent $\overline{XZ}$ is perpendicular to the radius $\overline{OZ}$, $b^2 -
c^2 = a^2$, so the area is $\boxed{\pi a^2}$. | 325 |
The base of a triangular piece of paper $ABC$ is $12\text{ cm}$ long. The paper is folded down over the base, with the crease $DE$ parallel to the base of the paper. The area of the triangle that projects below the base is $16\%$ that of the area of the triangle $ABC.$ What is the length of $DE,$ in cm?
[asy]
draw((0,... | Level 5 | Geometry | Let $X$ and $Y$ be the points where the folded portion of the triangle crosses $AB,$ and $Z$ be the location of the original vertex $C$ after folding.
[asy]
draw((0,0)--(12,0)--(9.36,3.3)--(1.32,3.3)--cycle,black+linewidth(1));
draw((1.32,3.3)--(4,-3.4)--(9.36,3.3),black+linewidth(1));
draw((1.32,3.3)--(4,10)--(9.36,3... | 611 |
The sides of triangle $CAB$ are in the ratio of $2:3:4$. Segment $BD$ is the angle bisector drawn to the shortest side, dividing it into segments $AD$ and $DC$. What is the length, in inches, of the longer subsegment of side $AC$ if the length of side $AC$ is $10$ inches? Express your answer as a common fraction. | Level 4 | Geometry | Without loss of generality, suppose that $BA < BC$. Since $BD$ is the angle bisector of $\angle B$, by the Angle Bisector Theorem, it follows that $$\frac{AD}{CD} = \frac{BA}{BC} = \frac 34.$$ Thus, $AD < CD$, so $CD$ is the longer subsegment of $AC$. Solving for $AD$, it follows that $AD = \frac{3CD}{4}$. Also, we kno... | 849 |
The convex pentagon $ABCDE$ has $\angle A = \angle B = 120^\circ$, $EA = AB = BC = 2$ and $CD = DE = 4$. What is the area of $ABCDE$?
[asy]
unitsize(1 cm);
pair A, B, C, D, E;
A = (0,0);
B = (1,0);
C = B + dir(60);
D = C + 2*dir(120);
E = dir(120);
draw(A--B--C--D--E--cycle);
label("$A$", A, SW);
label("$B$... | Level 4 | Geometry | We can divide the pentagon into 7 equilateral triangles of side length 2.
[asy]
unitsize(1 cm);
pair A, B, C, D, E;
A = (0,0);
B = (1,0);
C = B + dir(60);
D = C + 2*dir(120);
E = dir(120);
draw(A--B--C--D--E--cycle);
draw(C--E);
draw((C + D)/2--(D + E)/2);
draw(A--(C + D)/2);
draw(B--(D + E)/2);
label("$A$", A, SW... | 674 |
In rectangle $ABCD$, $P$ is a point on $BC$ so that $\angle APD=90^{\circ}$. $TS$ is perpendicular to $BC$ with $BP=PT$, as shown. $PD$ intersects $TS$ at $Q$. Point $R$ is on $CD$ such that $RA$ passes through $Q$. In $\triangle PQA$, $PA=20$, $AQ=25$ and $QP=15$. [asy]
size(7cm);defaultpen(fontsize(9));
real sd = ... | Level 2 | Geometry | Since $\angle ABP=90^{\circ}$, $\triangle ABP$ is a right-angled triangle. By the Pythagorean Theorem, $$BP^2=AP^2-AB^2=20^2-16^2=144$$ and so $BP=12$, since $BP>0$.
Since $\angle QTP=90^{\circ}$, $\triangle QTP$ is a right-angled triangle with $PT=12$. Since $PT=BP=12$, then by the Pythagorean Theorem, $$QT^2=QP^2-PT... | 290 |
Find the number of cubic centimeters in the volume of the cylinder formed by rotating a square with side length 14 centimeters about its vertical line of symmetry. Express your answer in terms of $\pi$. | Level 5 | Geometry | [asy]
size(100);
draw((-5,-.2)--(-3,-.2)--(-3,1.8)--(-5,1.8)--cycle); label("14",((-3,1.8)--(-5,1.8)),N); label("14",((-5,-.2)--(-5,1.8)),W);
draw((-2.5,.9)--(-1.5,.9),EndArrow);
import solids; import three; currentprojection = orthographic(5,0,2);
revolution c = cylinder((0,0,0), 1, 2);
draw((0,-1,2)--(0,1,2)); label(... | 850 |
In a 5 by 5 grid, each of the 25 small squares measures 2 cm by 2 cm and is shaded. Five unshaded circles are then placed on top of the grid as shown. The area of the visible shaded region can be written in the form $A-B\pi$ square cm. What is the value $A+B$?
[asy]
for(int i = 0; i < 5; ++i)
{
for(int j = 0; j < 5; ... | Level 3 | Geometry | The area of the visible shaded region is equal to the area of the grid minus the area of the five circles. The diameter of the four smaller circles is equal to a side of a small square, or 2 cm, so the radius of each of the smaller circles is 1 cm. The area of all four circles is then $4\cdot\pi \cdot1^2=4\pi$. The dia... | 732 |
An isosceles, obtuse triangle has one angle with a degree measure that is 50$\%$ larger than the measure of a right angle. What is the measure, in degrees, of one of the two smallest angles in the triangle? Express your answer as a decimal to the nearest tenth. | Level 1 | Geometry | An angle with measure $50\%$ larger than the measure of a right angle has measure $\frac{3}{2}\cdot 90^{\circ}=135^{\circ}$.
Thus the other two angles have a combined measure of $45^{\circ}$. Each one has a measure of
$$\frac{45^{\circ}}{2}=\boxed{22.5^{\circ}}.$$ | 501 |
In triangle $ABC$, angle $ACB$ is 50 degrees, and angle $CBA$ is 70 degrees. Let $D$ be the foot of the perpendicular from $A$ to $BC$, $O$ the center of the circle circumscribed about triangle $ABC$, and $E$ the other end of the diameter which goes through $A$. Find the angle $DAE$, in degrees.
[asy]
unitsize(1.5 ... | Level 3 | Geometry | Since triangle $ACD$ is right, $\angle CAD = 90^\circ - \angle ACD = 90^\circ - 50^\circ = 40^\circ$.
[asy]
unitsize(2 cm);
pair A, B, C, D, E, O;
A = dir(90);
B = dir(90 + 100);
C = dir(90 - 140);
D = (A + reflect(B,C)*(A))/2;
E = -A;
O = (0,0);
draw(Circle(O,1));
draw(A--B--C--cycle);
draw(A--D);
draw(A--E);
draw... | 622 |
A right cone has a base with a circumference of $16\pi$ inches and a height of 30 inches. The height of this cone is reduced while the circumference stays the same. The volume of the shorter cone is $192\pi$ cubic inches. What is the ratio of the shorter height to the original height? Express your answer as a common fr... | Level 4 | Geometry | Let the cone have radius $r$ inches; we have $2\pi r = 16\pi$, so $r = 8$. Let the new height of the cone be $h$ inches. We have $192\pi = (1/3)\pi(8^2)(h)$; solving yields $h = 9$. Thus the ratio of the new height to the original height is $9/30 = \boxed{\frac{3}{10}}$. | 978 |
Brad has a cylindrical barrel with radius 10 inches and height 15 inches. He wants to fill it from a well, using a bucket in the shape of a hemisphere with a radius of 10 inches. How many trips must Brad make to the well in order to fill the barrel? | Level 3 | Geometry | We start by finding the volume of the bucket and the volume of the barrel. Let $r$ be ten inches. The bucket is half of a sphere of radius $r$, so the volume it can hold is \[ \frac{1}{2} \cdot \frac{4}{3} \pi r^3 = \frac{2}{3} \pi r^3 . \]On the other hand, the barrel is a cylinder of radius $r$ and height $15 \tex... | 148 |
A bowling ball is a solid ball with a spherical surface and diameter 30 cm. To custom fit a bowling ball for each bowler, three holes are drilled in the ball. Bowler Kris has holes drilled that are 8 cm deep and have diameters of 2 cm, 2 cm, and 3 cm. Assuming the three holes are right circular cylinders, find the numb... | Level 4 | Geometry | The untampered bowling ball has radius $30/2=15$ cm and volume \[\frac{4}{3}\pi(15^3)=4\cdot 15^2\cdot 5\pi=225\cdot 20\pi = 4500\pi\] cubic cm. The 2 cm cylindrical holes each have radius $2/2=1$ cm and volume \[\pi (1^2)(8)=8\pi\] cubic cm; the 3 cm cylindrical hole has radius $3/2$ cm and volume \[\pi\left(\frac{3}... | 543 |
Twelve congruent disks are placed on a circle $C$ of radius 1 in such a way that the twelve disks cover $C$, no two of the disks overlap, and so that each of the twelve disks is tangent to its two neighbors. The resulting arrangement of disks is shown in the figure below. The sum of the areas of the twelve disks can be... | Level 5 | Geometry | We wish to find the radius of one circle, so that we can find the total area.
Notice that for them to contain the entire circle, each pair of circles must be tangent on the larger circle. Now consider two adjacent smaller circles. This means that the line connecting the radii is a segment of length $2r$ that is tangent... | 6,016 |
What is the volume in cubic inches of a right, rectangular prism with side, front and bottom faces having an area 15 square inches, 10 square inches and 6 square inches, respectively? | Level 2 | Geometry | Let the sides of the prism have lengths $x$, $y$, and $z$. We have the equations $xy=15$, $yz=10$ and $xz=6$. Multiplying these equations together, we have $xy\cdot yz \cdot xz = 15\cdot10\cdot6 \Rightarrow x^2y^2z^2=900$. Since the volume of the prism is equal to $xyz$, we take the square root of both sides to get the... | 385 |
Find $AX$ in the diagram if $CX$ bisects $\angle ACB$.
[asy]
import markers;
real t=.56;
pair A=(0,0);
pair B=(3,2);
pair C=(.5,1.5);
pair X=t*A+(1-t)*B;
draw(C--A--B--C--X);
label("$A$",A,SW);
label("$B$",B,E);
label("$C$",C,N);
label("$X$",X,SE);
//markangle(n=1,radius=15,A,C,X,marker(markinterval(stickframe(n=1... | Level 3 | Geometry | The Angle Bisector Theorem tells us that \[\frac{AC}{AX}=\frac{BC}{BX}\]so \[AX=\frac{AC\cdot BX}{BC}=\frac{21\cdot28}{30}=\frac{7^2\cdot3\cdot4}{30}=\frac{7^2\cdot2}{5}=\boxed{\frac{98}5}.\] | 513 |
Two rectangles have integer dimensions, and both have a perimeter of 144 cm. What is the greatest possible difference between the areas of two such rectangles? | Level 3 | Geometry | Let the dimensions of the rectangle be $l$ and $w$. We are given $2l+2w=144$, which implies $l+w=72$. Solving for $w$, we have $w=72-l$. The area of the rectangle is $lw=l(72-l)$. As a function of $l$, this expression is a parabola whose zeros are at $l=0$ and $l=72$ (see graph). The $y$-coordinate of a point on t... | 520 |
Circle $C$ has radius 6 cm. How many square centimeters are in the area of the largest possible inscribed triangle having one side as a diameter of circle $C$? | Level 3 | Geometry | We may consider the diameter of circle $C$ as the base of the inscribed triangle; its length is $12\text{ cm}$. Then the corresponding height extends from some point on the diameter to some point on the circle $C$. The greatest possible height is a radius of $C$, achieved when the triangle is right isosceles: [asy]
uni... | 220 |
A machine-shop cutting tool has the shape of a notched circle, as shown. The radius of the circle is $\sqrt{50}$ cm, the length of $AB$ is $6$ cm and that of $BC$ is $2$ cm. The angle $ABC$ is a right angle. Find the square of the distance (in centimeters) from $B$ to the center of the circle.
[asy] size(150); default... | Level 5 | Geometry | We use coordinates. Let the circle have center $(0,0)$ and radius $\sqrt{50}$; this circle has equation $x^2 + y^2 = 50$. Let the coordinates of $B$ be $(a,b)$. We want to find $a^2 + b^2$. $A$ and $C$ with coordinates $(a,b+6)$ and $(a+2,b)$, respectively, both lie on the circle. From this we obtain the system of equa... | 6,003 |
The diameter, in inches, of a sphere with twice the volume of a sphere of radius 9 inches can be expressed in the form $a\sqrt[3]{b}$ where $a$ and $b$ are positive integers and $b$ contains no perfect cube factors. Compute $a+b$. | Level 5 | Geometry | A sphere with radius 9 inches has volume $\frac{4}{3}\pi(9^3)=4\cdot 9^2 \cdot 3\pi$ cubic inches; twice this is $8\cdot 9^2\cdot 3 \pi$ cubic inches. Let the radius of the larger sphere be $r$, so we have \[\frac{4}{3}\pi r^3= 8\cdot 9^2\cdot 3\pi .\] Solving for $r$ yields \[r^3 =2\cdot 9^3 \Rightarrow r = 9\sqrt[3]... | 863 |
Points $(1,-5)$ and $(11,7)$ are the opposite vertices of a parallelogram. What are the coordinates of the point where the diagonals of the parallelogram intersect? | Level 3 | Geometry | The diagonals of a parallelogram intersect at the midpoint of each diagonal. So, we simply find the midpoint of $(1,-5)$ and $(11,7)$, which is $\left(\frac{1+11}{2}, \frac{-5+7}{2}\right)=\boxed{(6,1)}$. | 683 |
In a regular tetrahedron the centers of the four faces are the vertices of a smaller tetrahedron. The ratio of the volume of the smaller tetrahedron to that of the larger is $m/n$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
| Level 5 | Geometry | Embed the tetrahedron in 4-space to make calculations easier. Its vertices are $(1,0,0,0)$, $(0,1,0,0)$, $(0,0,1,0)$, $(0,0,0,1)$.
To get the center of any face, we take the average of the three coordinates of that face. The vertices of the center of the faces are: $(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}, 0)$,$(\frac{1... | 6,079 |
When each edge of a cube is increased by $50\%$, by what percent is the surface area of the cube increased? | Level 4 | Geometry | Let $s$ represent the length of the edge of a cube. The surface area of the cube is 6 times the area of each face (since there are 6 faces), or $6s^2$. Increasing $s$ by $50\%$ gives us $1.5s$. The new surface area is $6(1.5s)^2=6s^2(2.25)$. Increasing the surface area by $x\%$ is $6s^2\left(1+\frac{x}{100}\right)$, so... | 858 |
A frustum of a right circular cone is formed by cutting a small cone off of the top of a larger cone. If a particular frustum has an altitude of $24$ centimeters, the area of its lower base is $225\pi$ sq cm and the area of its upper base is $25\pi$ sq cm, what is the altitude of the small cone that was cut off? [asy]s... | Level 5 | Geometry | The two bases are circles, and the area of a circle is $\pi r^2$. If the area of the upper base (which is also the base of the small cone) is $25\pi$ sq cm, then its radius is $5$ cm, and the radius of the lower base is $15$ cm. The upper base, therefore, has a radius that is $\frac{1}{3}$ the size of the radius of t... | 498 |
A rectangle measures 6 meters by 10 meters. Drawn on each side of the rectangle is a semicircle that has the endpoints of its diameter on the vertices of the rectangle. What percent larger is the area of the large semicircles than the area of the small semicircles? Express your answer to the nearest whole number. | Level 5 | Geometry | The two large semicircles together make a circle of radius 5, which has area $25\pi$. The two small circles together make a circle with radius 3, which has area $9\pi$. Therefore, the ratio of the large semicircles' area to the small semicircles' area is $\frac{25\pi}{9\pi} = \frac{25}{9} \approx 2.78$. Since the ... | 583 |
Amy and Belinda each roll a sheet of 6-inch by 8-inch paper to form a cylindrical tube. Amy tapes the two 8-inch sides together without overlap. Belinda tapes the two 6-inch sides together without overlap. What is $\pi$ times the positive difference of the volumes of the two tubes? | Level 5 | Geometry | Amy's cylinder has a height of 8 and a base circumference of 6. Let her cylinder have volume $V_A$ and radius $r_A$; we have $2\pi r_A = 6$ so $r_A = 3/\pi$ and $V_A = \pi r_A ^2 h = \pi (3/\pi)^2 (8) = 72/\pi$.
Belinda's cylinder has a height of 6 and a base circumference of 8. Similarly, let her cylinder have volu... | 729 |
Either increasing the radius or the height of a cylinder by six inches will result in the same volume. The original height of the cylinder is two inches. What is the original radius in inches? | Level 4 | Geometry | Let the original radius be $r$. The volume of the cylinder with the increased radius is $\pi \cdot (r+6)^2 \cdot 2$. The volume of the cylinder with the increased height is $\pi \cdot r^2 \cdot 8$. Since we are told these two volumes are the same, we have the equation $\pi \cdot (r+6)^2 \cdot 2 = \pi \cdot r^2 \cdot 8$... | 294 |
In the triangle shown, $n$ is a positive integer, and $\angle A > \angle B > \angle C$. How many possible values of $n$ are there? [asy]
draw((0,0)--(1,0)--(.4,.5)--cycle);
label("$A$",(.4,.5),N); label("$B$",(1,0),SE); label("$C$",(0,0),SW);
label("$2n + 12$",(.5,0),S); label("$3n - 3$",(.7,.25),NE); label("$2n + 7$"... | Level 5 | Geometry | The sides of the triangle must satisfy the triangle inequality, so $AB + AC > BC$, $AB + BC > AC$, and $AC + BC > AB$. Substituting the side lengths, these inequalities turn into \begin{align*}
(3n - 3) + (2n + 7) &> 2n + 12, \\
(3n - 3) + (2n + 12) &> 2n + 7, \\
(2n + 7) + (2n + 12) &> 3n - 3,
\end{align*} which give... | 720 |
The image of the point with coordinates $(-3,-1)$ under the reflection across the line $y=mx+b$ is the point with coordinates $(5,3)$. Find $m+b$. | Level 4 | Geometry | The line of reflection is the perpendicular bisector of the segment connecting the point with its image under the reflection. The slope of the segment is $\frac{3-(-1)}{5-(-3)}=\frac{1}{2}$. Since the line of reflection is perpendicular, its slope, $m$, equals $-2$. By the midpoint formula, the coordinates of the mi... | 447 |
Two different points, $C$ and $D$, lie on the same side of line $AB$ so that $\triangle ABC$ and $\triangle BAD$ are congruent with $AB = 9$, $BC=AD=10$, and $CA=DB=17$. The intersection of these two triangular regions has area $\tfrac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
| Level 5 | Geometry | [asy] unitsize(10); pair A = (0,0); pair B = (9,0); pair C = (15,8); pair D = (-6,8); pair E = (-6,0); draw(A--B--C--cycle); draw(B--D--A); label("$A$",A,dir(-120)); label("$B$",B,dir(-60)); label("$C$",C,dir(60)); label("$D$",D,dir(120)); label("$E$",E,dir(-135)); label("$9$",(A+B)/2,dir(-90)); label("$10$",(D+A)/2,di... | 6,154 |
In triangle $ABC$, the angle bisectors are $AD$, $BE$, and $CF$, which intersect at the incenter $I$. If $\angle ACB = 38^\circ$, then find the measure of $\angle AIE$, in degrees. | Level 5 | Geometry | Since $AD$ is an angle bisector, $\angle BAI = \angle BAC/2$. Since $BE$ is an angle bisector, $\angle ABI = \angle ABC/2$. As an angle that is external to triangle $ABI$, $\angle AIE = \angle BAI + \angle ABI = \angle BAC/2 + \angle ABC/2$.
[asy]
import geometry;
unitsize(0.3 cm);
pair A, B, C, D, E, F, I;
A = (... | 767 |
Triangle $ABC$ lies in the cartesian plane and has an area of $70$. The coordinates of $B$ and $C$ are $(12,19)$ and $(23,20),$ respectively, and the coordinates of $A$ are $(p,q).$ The line containing the median to side $BC$ has slope $-5.$ Find the largest possible value of $p+q.$
[asy]defaultpen(fontsize(8)); size(1... | Level 5 | Geometry | The midpoint $M$ of line segment $\overline{BC}$ is $\left(\frac{35}{2}, \frac{39}{2}\right)$. The equation of the median can be found by $-5 = \frac{q - \frac{39}{2}}{p - \frac{35}{2}}$. Cross multiply and simplify to yield that $-5p + \frac{35 \cdot 5}{2} = q - \frac{39}{2}$, so $q = -5p + 107$.
Use determinants to f... | 6,091 |
A right pyramid has a square base with side length 10 cm. Its peak is 12 cm above the center of its base. What is the total surface area of the pyramid, in square centimeters? | Level 3 | Geometry | [asy]
import three;
triple A = (0,0,0);
triple B = (1,0,0);
triple C = (1,1,0);
triple D = (0,1,0);
triple P = (0.5,0.5,1);
draw(B--C--D--P--B);
draw(P--C);
draw(B--A--D,dashed);
draw(P--A,dashed);
label("$A$",A,NW);
label("$B$",B,W);
label("$C$",C,S);
label("$D$",D,E);
label("$P$",P,N);
triple F= (0.5,0.5,0);
triple M... | 776 |
Point $F$ is taken on the extension of side $AD$ of parallelogram $ABCD$. $BF$ intersects diagonal $AC$ at $E$ and side $DC$ at $G$. If $EF = 32$ and $GF = 24$, then $BE$ equals:
[asy] size(7cm); pair A = (0, 0), B = (7, 0), C = (10, 5), D = (3, 5), F = (5.7, 9.5); pair G = intersectionpoints(B--F, D--C)[0]; pair E = i... | Level 5 | Geometry | Let $BE = x$ and $BC = y$. Since $AF \parallel BC$, by AA Similarity, $\triangle AFE \sim \triangle CBE$. That means $\frac{AF}{CB} = \frac{FE}{BE}$. Substituting in values results in\[\frac{AF}{y} = \frac{32}{x}\]Thus, $AF = \frac{32y}{x}$, so $FD = \frac{32y - xy}{x}$.
In addition, $DC \parallel AB$, so by AA Similar... | 6,223 |
What is the smallest possible area, in square units, of a right triangle with two sides measuring $4$ units and $5$ units? | Level 1 | Geometry | Since $5>4$, $4$ cannot be the length of the hypotenuse. Thus either $4$ and $5$ are the lengths of the two smaller sides, or $5$ is the hypotenuse, meaning the two smaller sides are $4$ and $3$. In this latter case, the area will be smaller, so the area is $\frac{(3)(4)}{2} = \boxed{6}$. | 806 |
Consider the set of all triangles $OPQ$ where $O$ is the origin and $P$ and $Q$ are distinct points in the plane with nonnegative integer coordinates $(x,y)$ such that $41x + y = 2009$. Find the number of such distinct triangles whose area is a positive integer.
| Level 5 | Geometry | Let the two points $P$ and $Q$ be defined with coordinates; $P=(x_1,y_1)$ and $Q=(x_2,y_2)$
We can calculate the area of the parallelogram with the determinant of the matrix of the coordinates of the two points(shoelace theorem).
$\det \left(\begin{array}{c} P \\ Q\end{array}\right)=\det \left(\begin{array}{cc}x_1 &y_1... | 6,112 |
$\triangle DEF$ is inscribed inside $\triangle ABC$ such that $D,E,F$ lie on $BC, AC, AB$, respectively. The circumcircles of $\triangle DEC, \triangle BFD, \triangle AFE$ have centers $O_1,O_2,O_3$, respectively. Also, $AB = 23, BC = 25, AC=24$, and $\stackrel{\frown}{BF} = \stackrel{\frown}{EC},\ \stackrel{\frown}{AF... | Level 5 | Geometry | [asy] size(150); defaultpen(linewidth(0.8)); import markers; pair B = (0,0), C = (25,0), A = (578/50,19.8838); draw(A--B--C--cycle); label("$B$",B,SW); label("$C$",C,SE); label("$A$",A,N); pair D = (13,0), E = (11*A + 13*C)/24, F = (12*B + 11*A)/23; draw(D--E--F--cycle); label("$D$",D,dir(-90)); label("$E$",E,dir(0)); ... | 6,232 |
Circle $\Gamma$ is the incircle of $\triangle ABC$ and is also the circumcircle of $\triangle XYZ$. The point $X$ is on $\overline{BC}$, the point $Y$ is on $\overline{AB}$, and the point $Z$ is on $\overline{AC}$. If $\angle A=40^\circ$, $\angle B=60^\circ$, and $\angle C=80^\circ$, what is the measure of $\angle YZ... | Level 4 | Geometry | A diagram will probably help.
[asy]
size(200);
pair X=(1,0);
pair Y=dir(120)*(1,0);
pair Z=dir(-100)*(1,0);
real t =60;
pair B=dir(t)*(2.0,0);
pair A=dir(t+130)*(2.86,0);
pair C=dir(t+250)*(1.6,0);
draw(unitcircle);
draw(A--B--C--A);
draw(X--Y--Z--X);
label("$A$",A,W);
label("$B$",B,NE);
label("$C$",C,SE);
label("... | 205 |
The sides of triangle $PQR$ are tangent to a circle with center $C$ as shown. Given that $\angle PQR = 65^\circ$ and $\angle QRC = 30^\circ$, find $\angle QPR$, in degrees.
[asy]
unitsize(1.0 cm);
pair Q, P, R, C;
Q = (2.43,3.46);
P = (0,0);
R = (4.43,0);
C = incenter(Q,P,R);
draw(Q--P--R--cycle);
draw(incircle(Q,P... | Level 2 | Geometry | The circle with center $C$ is the incircle of $\triangle PQR$. So, any segment from a vertex of the triangle to $C$ is an angle bisector.
From $\angle QRC = 30^\circ$, we find that $\angle QRP = 60^\circ$ because $RC$ is an angle bisector.
The sum of the measures of the internal angles of a triangle is $180^\circ$, s... | 298 |
A right circular cone has a volume of $12\pi$ cubic centimeters. The height of the cone is 4 cm. How many centimeters is the circumference of the base of the cone, in terms of $\pi$? | Level 3 | Geometry | The volume of a cone is $\frac{1}{3}\pi r^2 h$. We're given that the volume is $12\pi$ and the height is $4$. Thus, $\frac{1}{3}\pi r^2 \cdot 4 = 12\pi$. Solving for $r$, we find $r = 3$. Therefore, the circumference of the base is $2\pi r = \boxed{6\pi}$. | 534 |
In right triangle $ABC$ with $\angle A = 90^\circ$, we have $AB = 6$ and $BC = 10$. Find $\cos C$. | Level 2 | Geometry | The triangle is shown below:
[asy]
pair A,B,C;
A = (0,0);
B = (6,0);
C = (0,8);
draw(A--B--C--A);
draw(rightanglemark(B,A,C,10));
label("$A$",A,SW);
label("$B$",B,SE);
label("$C$",C,N);
label("$10$",(B+C)/2,NE);
label("$6$",B/2,S);
[/asy]
The Pythagorean Theorem gives us $AC = \sqrt{BC^2 - AB^2} = \sqrt{1... | 831 |
Points $A$, $B$, $C$, and $T$ are in space such that each of $\overline{TA}$, $\overline{TB}$, and $\overline{TC}$ is perpendicular to the other two. If $TA = TB = 12$ and $TC = 6$, then what is the distance from $T$ to face $ABC$? | Level 5 | Geometry | [asy]
import three;
triple A = (4,8,0);
triple B= (4,0,0);
triple C = (0,0,0);
triple D = (0,8,0);
triple P = (4,8,6);
draw(B--P--D--A--B);
draw(A--P);
draw(B--D,dashed);
label("$T$",A,S);
label("$B$",B,W);
label("$C$",D,E);
label("$A$",P,N);
label("$M$",(P+B)/2,NW);
draw(D--((P+B)/2),dashed);
[/asy]
We can think of $... | 494 |
The sides of triangle $PQR$ are tangent to a circle with center $C$ as shown. Given that $\angle PQR = 63^\circ$ and $\angle QPR = 59^\circ$, find $\angle QRC$, in degrees.
[asy]
unitsize(1.0 cm);
pair Q, P, R, C;
Q = (2.43,3.46);
P = (0,0);
R = (4.43,0);
C = incenter(Q,P,R);
draw(Q--P--R--cycle);
draw(incircl... | Level 2 | Geometry | The circle with center $C$ is the incircle of $\triangle PQR$. So, any segment from a vertex of the triangle to $C$ is an angle bisector.
The sum of the measures of the internal angles of a triangle is $180^\circ$, so
\begin{align*}
\angle QRP &= 180^\circ - \angle PQR - \angle QPR \\
&= 180^\circ - 63^\circ - 59^\ci... | 130 |
Let $A_0=(0,0)$. Distinct points $A_1,A_2,\dots$ lie on the $x$-axis, and distinct points $B_1,B_2,\dots$ lie on the graph of $y=\sqrt{x}$. For every positive integer $n,\ A_{n-1}B_nA_n$ is an equilateral triangle. What is the least $n$ for which the length $A_0A_n\geq100$?
$\textbf{(A)}\ 13\qquad \textbf{(B)}\ 15\qqua... | Level 5 | Geometry | Let $a_n=|A_{n-1}A_n|$. We need to rewrite the recursion into something manageable. The two strange conditions, $B$'s lie on the graph of $y=\sqrt{x}$ and $A_{n-1}B_nA_n$ is an equilateral triangle, can be compacted as follows:\[\left(a_n\frac{\sqrt{3}}{2}\right)^2=\frac{a_n}{2}+a_{n-1}+a_{n-2}+\cdots+a_1\]which uses $... | 6,180 |
Compute $\sin 0^\circ$. | Level 1 | Geometry | Rotating the point $(1,0)$ by $0^\circ$ counterclockwise about the origin gives us the point $(1,0)$, so $\sin 0^\circ = \boxed{0}$. | 78 |
Sector $OAB$ is a quarter of a circle of radius 3 cm. A circle is drawn inside this sector, tangent at three points as shown. What is the number of centimeters in the radius of the inscribed circle? Express your answer in simplest radical form. [asy]
import olympiad; import geometry; size(100); defaultpen(linewidth(0.8... | Level 5 | Geometry | Call the center of the inscribed circle $C$, and let $D$ be the point shared by arc $AB$ and the inscribed circle. Let $E$ and $F$ be the points where the inscribed circle is tangent to $OA$ and $OB$ respectively. Since angles $CEO$, $CFO$, and $EOF$ are all right angles, angle $FCE$ is a right angle as well. Theref... | 264 |
The six edges of a tetrahedron $ABCD$ measure $7, 13, 18, 27, 36$ and $41$ units. If the length of edge $AB$ is $41$, then the length of edge $CD$ is
$\textbf{(A)}\ 7\qquad \textbf{(B)}\ 13\qquad \textbf{(C)}\ 18\qquad \textbf{(D)}\ 27\qquad \textbf{(E)}\ 36$
| Level 5 | Geometry | By the triangle inequality in $\triangle ABC$, we find that $BC$ and $CA$ must sum to greater than $41$, so they must be (in some order) $7$ and $36$, $13$ and $36$, $18$ and $27$, $18$ and $36$, or $27$ and $36$. We try $7$ and $36$, and now by the triangle inequality in $\triangle ABD$, we must use the remaining numb... | 6,204 |
A right square pyramid with base edges of length $8\sqrt{2}$ units each and slant edges of length 10 units each is cut by a plane that is parallel to its base and 3 units above its base. What is the volume, in cubic units, of the new pyramid that is cut off by this plane? [asy]
import three;
size(2.5inch);
currentproje... | Level 5 | Geometry | Define the points $A$, $B$, $C$ , and $D$, $E$, and $F$ as shown so that $AC$ is perpendicular to the base of the pyramid. Segment $DC$ is a leg of the isosceles right triangle $CDF$ whose hypotenuse is $8\sqrt{2}$. Therefore, $CD=8\sqrt{2}/\sqrt{2}=8$. Applying the Pythagorean theorem to triangle $ACD$ gives $AC=6$... | 44 |
In acute triangle $ABC$ points $P$ and $Q$ are the feet of the perpendiculars from $C$ to $\overline{AB}$ and from $B$ to $\overline{AC}$, respectively. Line $PQ$ intersects the circumcircle of $\triangle ABC$ in two distinct points, $X$ and $Y$. Suppose $XP=10$, $PQ=25$, and $QY=15$. The value of $AB\cdot AC$ can be w... | Level 5 | Geometry | Let $AP=a, AQ=b, \cos\angle A = k$
Therefore $AB= \frac{b}{k} , AC= \frac{a}{k}$
By power of point, we have $AP\cdot BP=XP\cdot YP , AQ\cdot CQ=YQ\cdot XQ$ Which are simplified to
$400= \frac{ab}{k} - a^2$
$525= \frac{ab}{k} - b^2$
Or
$a^2= \frac{ab}{k} - 400$
$b^2= \frac{ab}{k} - 525$
(1)
Or
$k= \frac{ab}{a^2+400} = \... | 6,157 |
Compute $\cos 240^\circ$. | Level 3 | Geometry | Let $P$ be the point on the unit circle that is $240^\circ$ counterclockwise from $(1,0)$, and let $D$ be the foot of the altitude from $P$ to the $x$-axis, as shown below.
[asy]
pair A,C,P,O,D;
draw((0,-1.2)--(0,1.2),p=black+1.2bp,Arrows(0.15cm));
draw((-1.2,0)--(1.2,0),p=black+1.2bp,Arrows(0.15cm));
A = (1,0);
... | 925 |
Regular hexagon $ABCDEF$ is the base of right pyramid $\allowbreak PABCDEF$. If $PAD$ is an equilateral triangle with side length 8, then what is the volume of the pyramid? | Level 5 | Geometry | [asy]
import three;
triple A = (1,0,0);
triple B = (0.5,sqrt(3)/2,0);
triple C = (-0.5,sqrt(3)/2,0);
triple D = (-1,0,0);
triple EE = (-0.5,-sqrt(3)/2,0);
triple F = (0.5,-sqrt(3)/2,0);
triple P = (0,0,1);
draw(F--A--B--C);
draw(C--D--EE--F,dashed);
draw(A--P--C);
draw(EE--P--D,dashed);
draw(B--P--F);
label("$A$",A,S... | 723 |
In triangle $ABC$, $\angle C=90^\circ$, $AC=6$ and $BC=8$. Points $D$ and $E$ are on $\overline{AB}$ and $\overline{BC}$, respectively, and $\angle BED=90^\circ$. If $DE=4$, then what is the length of $BD$? [asy]
import olympiad; import geometry; size(150); defaultpen(linewidth(0.8));
draw(origin--(6,0)--(6,8)--cycle);... | Level 3 | Geometry | Applying the Pythagorean Theorem to triangle $ABC$ gives $BA=10$. Since $\triangle DBE\sim\triangle ABC$, $$\frac{BD}{BA}=\frac{DE}{AC}.\qquad{\rm So}\qquad
BD=\frac{DE}{AC}(BA)=\frac 46(10)=\boxed{\frac{20}{3}}.$$ | 794 |
In triangle $PQR$, we have $\angle P = 90^\circ$, $QR = 15$, and $\tan R = 5\cos Q$. What is $PQ$? | Level 4 | Geometry | [asy]
pair P,Q,R;
P = (0,0);
Q = (3*sqrt(24),0);
R = (0,3);
draw(P--Q--R--P);
draw(rightanglemark(Q,P,R,18));
label("$P$",P,SW);
label("$Q$",Q,SE);
label("$R$",R,N);
label("$15$",(R+Q)/2,NE);
[/asy]
We have $\tan R = \frac{PQ}{PR}$ and $\cos Q = \frac{PQ}{QR} = \frac{PQ}{15}$, so $\tan R = 5\cos Q$ gives us $\frac{PQ}... | 759 |
The image of the point with coordinates $(1,1)$ under the reflection across the line $y=mx+b$ is the point with coordinates $(9,5)$. Find $m+b$. | Level 5 | Geometry | The line of reflection is the perpendicular bisector of the segment connecting the point with its image under the reflection. The slope of the segment is $\frac{5-1}{9-1}=\frac{1}{2}$. Since the line of reflection is perpendicular, its slope, $m$, equals $-2$. By the midpoint formula, the coordinates of the midpoint... | 597 |
Which of the following could NOT be the lengths of the external diagonals of a right regular prism [a "box"]? (An $\textit{external diagonal}$ is a diagonal of one of the rectangular faces of the box.)
$\text{(A) }\{4,5,6\} \quad \text{(B) } \{4,5,7\} \quad \text{(C) } \{4,6,7\} \quad \text{(D) } \{5,6,7\} \quad \text... | Level 5 | Geometry | Let $a,$ $b,$ and $c$ be the side lengths of the rectangular prism. By Pythagoras, the lengths of the external diagonals are $\sqrt{a^2 + b^2},$ $\sqrt{b^2 + c^2},$ and $\sqrt{a^2 + c^2}.$ If we square each of these to obtain $a^2 + b^2,$ $b^2 + c^2,$ and $a^2 + c^2,$ we observe that since each of $a,$ $b,$ and $c$ are... | 6,188 |
Let $\triangle{PQR}$ be a right triangle with $PQ = 90$, $PR = 120$, and $QR = 150$. Let $C_{1}$ be the inscribed circle. Construct $\overline{ST}$ with $S$ on $\overline{PR}$ and $T$ on $\overline{QR}$, such that $\overline{ST}$ is perpendicular to $\overline{PR}$ and tangent to $C_{1}$. Construct $\overline{UV}$ with... | Level 5 | Geometry | [asy] pointpen = black; pathpen = black + linewidth(0.7); pair P = (0,0), Q = (90, 0), R = (0, 120), S=(0, 60), T=(45, 60), U = (60,0), V=(60, 40), O1 = (30,30), O2 = (15, 75), O3 = (70, 10); D(MP("P",P)--MP("Q",Q)--MP("R",R,W)--cycle); D(MP("S",S,W) -- MP("T",T,NE)); D(MP("U",U) -- MP("V",V,NE)); D(O2 -- O3, rgb(0.2,... | 6,067 |
The graphs of the equations
$y=k, \qquad y=\sqrt{3}x+2k, \qquad y=-\sqrt{3}x+2k,$
are drawn in the coordinate plane for $k=-10,-9,-8,\ldots,9,10.\,$ These 63 lines cut part of the plane into equilateral triangles of side $2/\sqrt{3}.\,$ How many such triangles are formed?
| Level 5 | Geometry | We note that the lines partition the hexagon of the six extremal lines into disjoint unit regular triangles, and forms a series of unit regular triangles along the edge of the hexagon.
[asy] size(200); picture pica, picb, picc; int i; for(i=-10;i<=10;++i){ if((i%10) == 0){draw(pica,(-20/sqrt(3)-abs((0,i))/sqrt(3),i)--(... | 6,031 |
In rectangle $ABCD$, angle $C$ is trisected by $\overline{CF}$ and $\overline{CE}$, where $E$ is on $\overline{AB}$, $F$ is on $\overline{AD}$, $BE=6$, and $AF=2$. Find the area of $ABCD$.
[asy]
import olympiad; import geometry; size(150); defaultpen(linewidth(0.8)); dotfactor=4;
real length = 2 * (6*sqrt(3) - 2), wid... | Level 5 | Geometry | From $30^\circ$-$60^\circ$-$90^\circ$ triangle $CEB$, we have $BC=6\sqrt{3}$. Therefore, $FD=AD-AF=6\sqrt{3}-2$. In the $30^\circ$-$60^\circ$-$90^\circ$ triangle $CFD$, $CD=FD\sqrt{3}=18-2\sqrt{3}$. The area of rectangle $ABCD$ is $$(BC)(CD)=\left(6\sqrt{3}\right)\left(18-2\sqrt{3}\right)=
\boxed{108\sqrt{3}-36}.$$ | 94 |
A hexagon inscribed in a circle has three consecutive sides, each of length 3, and three consecutive sides, each of length 5. The chord of the circle that divides the hexagon into two trapezoids, one with three sides, each of length 3, and the other with three sides, each of length 5, has length equal to $m/n$, where $... | Level 5 | Geometry | In hexagon $ABCDEF$, let $AB=BC=CD=3$ and let $DE=EF=FA=5$. Since arc $BAF$ is one third of the circumference of the circle, it follows that $\angle BCF = \angle BEF=60^{\circ}$. Similarly, $\angle CBE =\angle CFE=60^{\circ}$. Let $P$ be the intersection of $\overline{BE}$ and $\overline{CF}$, $Q$ that of $\overline{BE... | 1,070 |
In an isosceles triangle, one of the angles opposite an equal side is $40^{\circ}$. How many degrees are in the measure of the triangle's largest angle? [asy] draw((0,0)--(6,0)--(3,2)--(0,0)); label("$\backslash$",(1.5,1)); label("{/}",(4.5,1));
label("$40^{\circ}$",(.5,0),dir(45));
[/asy] | Level 1 | Geometry | The two angles opposite the equal sides of an isosceles triangle are congruent, so in this case, both are $40^\circ$. Since the three angles of a triangle add up to $180^\circ$, the third angle in this triangle is $(180-40-40)^\circ = \boxed{100}^\circ$. | 6 |
An isosceles triangle has side lengths 8 cm, 8 cm and 10 cm. The longest side of a similar triangle is 25 cm. What is the perimeter of the larger triangle, in centimeters? | Level 1 | Geometry | The ratio of the length of the longest sides of the small triangle to the large triangle is $10/25 = 2/5$, which must hold constant for all sides of the two triangles since they are similar. Thus the perimeters of the two triangles are also in the ratio of $2/5$. The small triangle has perimeter $8+8+10=26$, so the l... | 272 |
The point $O$ is the center of the circle circumscribed about $\triangle ABC$, with $\angle BOC = 120^{\circ}$ and $\angle AOB =
140^{\circ}$, as shown. What is the degree measure of $\angle
ABC$?
[asy]
pair A,B,C;
draw(Circle((0,0),20),linewidth(0.7));
label("$O$",(0,0),S);
A=(-16,-12);
C=(16,-12);
B=(3,19.7);
draw(A... | Level 2 | Geometry | Since $OA=OB=OC$, triangles $AOB$, $BOC$, and $COA$ are all isosceles. Hence \[
\angle ABC = \angle ABO + \angle OBC =
\frac{180^{\circ}-140^{\circ}}{2}+
\frac{180^{\circ}-120^{\circ}}{2}=\boxed{50^{\circ}}.
\]OR
Since \[
\angle AOC = 360^{\circ}-140^{\circ}-120^{\circ}=100^{\circ},
\]the Central Angle Theorem implie... | 314 |
Triangle $ABC$ has positive integer side lengths with $AB=AC$. Let $I$ be the intersection of the bisectors of $\angle B$ and $\angle C$. Suppose $BI=8$. Find the smallest possible perimeter of $\triangle ABC$.
| Level 5 | Geometry | Let $D$ be the midpoint of $\overline{BC}$. Then by SAS Congruence, $\triangle ABD \cong \triangle ACD$, so $\angle ADB = \angle ADC = 90^o$.
Now let $BD=y$, $AB=x$, and $\angle IBD = \dfrac{\angle ABD}{2} = \theta$.
Then $\mathrm{cos}{(\theta)} = \dfrac{y}{8}$
and $\mathrm{cos}{(2\theta)} = \dfrac{y}{x} = 2\mathrm{cos... | 6,141 |
Compute $\tan 225^\circ$. | Level 1 | Geometry | Let $P$ be the point on the unit circle that is $225^\circ$ counterclockwise from $(1,0)$, and let $D$ be the foot of the altitude from $P$ to the $x$-axis, as shown below.
[asy]
pair A,C,P,O,D;
draw((0,-1.2)--(0,1.2),p=black+1.2bp,Arrows(0.15cm));
draw((-1.2,0)--(1.2,0),p=black+1.2bp,Arrows(0.15cm));
A = (1,0);
O= (0... | 527 |
Let $ABC$ be a triangle with $\angle BAC = 90^\circ$. A circle is tangent to the sides $AB$ and $AC$ at $X$ and $Y$ respectively, such that the points on the circle diametrically opposite $X$ and $Y$ both lie on the side $BC$. Given that $AB = 6$, find the area of the portion of the circle that lies outside the tria... | Level 4 | Geometry | Let $O$ be the center of the circle, and $r$ its radius, and let $X'$ and $Y'$ be the points diametrically opposite $X$ and $Y$, respectively. We have $OX' = OY' = r$, and $\angle X'OY' = 90^\circ$. Since triangles $X'OY'$ and $BAC$ are similar, we see that $AB = AC$. Let $X''$ be the foot of the altitude from $Y'$ to ... | 304 |
If an arc of $45^{\circ}$ on circle $A$ has the same length as an arc of $30^{\circ}$ on circle $B$, then what is the ratio of the area of circle $A$ to the area of circle $B$? Express your answer as a common fraction. | Level 4 | Geometry | Let $C_A= 2\pi R_A$ be the circumference of circle $A$, let $C_B= 2\pi R_B$ be the circumference of circle $B$, and let $L$ the common length of the two arcs. Then $$
\frac{45}{360}C_A = L = \frac{30}{360}C_B.
$$Therefore $$
\frac{C_A}{C_B} = \frac{2}{3}\quad\text{so}\quad
\frac{2}{3}=\frac{2\pi R_A}{2\pi R_B} =\frac... | 851 |
Circles $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$ intersect at two points, one of which is $(9,6)$, and the product of the radii is $68$. The x-axis and the line $y = mx$, where $m > 0$, are tangent to both circles. It is given that $m$ can be written in the form $a\sqrt {b}/c$, where $a$, $b$, and $c$ are positive integ... | Level 5 | Geometry | Let the smaller angle between the $x$-axis and the line $y=mx$ be $\theta$. Note that the centers of the two circles lie on the angle bisector of the angle between the $x$-axis and the line $y=mx$. Also note that if $(x,y)$ is on said angle bisector, we have that $\frac{y}{x}=\tan{\frac{\theta}{2}}$. Let $\tan{\frac{\t... | 6,071 |
The slant height of a cone is 13 cm, and the height from the vertex to the center of the base is 12 cm. What is the number of cubic centimeters in the volume of the cone? Express your answer in terms of $\pi$. | Level 3 | Geometry | We create a right triangle with the slant height as the hypotenuse, the height from the vertex to the center of the base as one of the legs, and a radius as the other leg. By Pythagorean theorem, the radius measures $\sqrt{13^2-12^2}=5$ cm. It follows that the volume of the cone is $(1/3)\pi(5^2)(12)=\boxed{100\pi}$. | 529 |
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