sentence1 stringlengths 16 4.31k | sentence2 stringlengths 29 6.77k | label stringclasses 1
value |
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The area of rectangle $ABCD$ with vertices $A$(0, 0), $B$(0, 4), $C$($x$, 4) and $D$($x$, 0) is 28 square units. If $x > 0$, what is the value of $x$? | Plotting the points, we see that the dimensions of the rectangle are $x$ and $4$. The area of the rectangle is $(\text{length})(\text{width})=4x$, so $4x=28$ and $x=\boxed{7}$. [asy]
size(5cm);
import graph;
defaultpen(linewidth(0.7)+fontsize(12));
real x = 7;
pair A=(0,0), B=(0,4), C=(x,4), D=(x,0);
pair[] dots = {A,... | entailment |
Two cylindrical cans have the same volume. The height of one can is triple the height of the other. If the radius of the narrower can is 12 units, how many units are in the length of the radius of the wider can? Express your answer in simplest radical form. | Let the height of the wide can be $h$ and the height of the narrow can be $3h$. Let the wide can have radius $x$ units. Since the two volumes are equal, we have \[\pi (12^2) (3h) = \pi (x^2) (h).\] Solving yields $x = 12\sqrt{3}$ so the wide can has radius $\boxed{12\sqrt{3}}$ units. | entailment |
In the following diagram, $AB=50$. Find $AX$.
[asy]
import markers;
real t=.67;
pair A=(0,0);
pair B=(3,-2);
pair C=(1.5,1.5);
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,SW);
markangle(n=1,radius=15,A,C,X,marker(markinterval(stickframe(n=1),true))... | The Angle Bisector Theorem tells us that \[\frac{AX}{AC}=\frac{BX}{BC}\] so cross multiplying and substituting tells us \[56AX=28BX\] or $BX=2AX$.
We want to find $AX$, so we write \[50=AB=AX+XB=AX+2AX=3AX.\] Solving gives us $AX=\boxed{\frac{50}3}$. | entailment |
When plotted in the standard rectangular coordinate system, trapezoid $ABCD$ has vertices $A(1, -2)$, $B(1, 1)$, $C(5, 7)$ and $D(5, 1)$. What is the area of trapezoid $ABCD$? | The two bases of the trapezoids are the segments $AB$ and $CD$, and the height is the perpendicular distance between the bases, which in this case is the difference of the $x$-coordinates: $5 - 1 = 4$. Similarly, the lengths of the bases are the differences of the $y$-coordinates of their two endpoints. Using the formu... | entailment |
Compute $\sin 60^\circ$. | Let $P$ be the point on the unit circle that is $60^\circ$ 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... | entailment |
What is the total area, in square units, of the four triangular faces of a right, square-based pyramid that has base edges measuring 6 units and lateral edges measuring 5 units? | The triangular faces are isosceles triangles. We drop an altitude from the apex to the base, and, since the triangle is isosceles, it will also be a median. So it forms a right triangle with hypotenuse $5$ and one leg $3$, and thus the other leg, the altitude, is $4$. The area of the triangle is then $\frac{4(6)}{2}... | entailment |
Two identical rectangular crates are packed with cylindrical pipes, using different methods. Each pipe has diameter $10\text{ cm}.$ A side view of the first four rows of each of the two different methods of packing is shown below.
[asy]
draw(circle((1,1),1),black+linewidth(1));
draw(circle((3,1),1),black+linewidth(1))... | Join the centres $A,$ $B,$ and $C$ of the three circles. The lines $AB,$ $BC,$ and $CA$ will pass through the points where the circles touch, so will each have length $10\text{ cm}$ (that is, twice the radius of one of the circles).
We can break the height of the pile into three pieces: the distance from the bottom of... | entailment |
A unit cube is cut twice to form three triangular prisms, two of which are congruent, as shown in Figure 1. The cube is then cut in the same manner along the dashed lines shown in Figure 2. This creates nine pieces. What is the volume of the piece that contains vertex $W$?
[asy]
path a=(0,0)--(10,0)--(10,10)--(0,10)--... | The piece that contains $W$ is shown. It is a pyramid with vertices $V, W, X,Y$, and $Z$. Its base $WXYZ$ is a square with sides of length $1/2$ and its altitude $VW$ is 1. Hence the volume of this pyramid is \[
\frac{1}{3}\left(\frac{1}{2}\right)^2(1)=\boxed{\frac{1}{12}}.
\][asy]
unitsize(0.3cm);
draw((0,0)--(10,0)-... | entailment |
In the diagram, $PQRS$ is a trapezoid with an area of $12.$ $RS$ is twice the length of $PQ.$ What is the area of $\triangle PQS?$
[asy]
draw((0,0)--(1,4)--(7,4)--(12,0)--cycle);
draw((7,4)--(0,0));
label("$S$",(0,0),W);
label("$P$",(1,4),NW);
label("$Q$",(7,4),NE);
label("$R$",(12,0),E);
[/asy] | Since $PQ$ is parallel to $SR,$ the height of $\triangle PQS$ (considering $PQ$ as the base) and the height of $\triangle SRQ$ (considering $SR$ as the base) are the same (that is, the vertical distance between $PQ$ and $SR$).
Since $SR$ is twice the length of $PQ$ and the heights are the same, the area of $\triangle ... | entailment |
[asy] draw((0,0)--(2,2)--(5/2,1/2)--(2,0)--cycle,dot); MP("A",(0,0),W);MP("B",(2,2),N);MP("C",(5/2,1/2),SE);MP("D",(2,0),S); MP("a",(1,0),N);MP("b",(17/8,1/8),N); [/asy]
In the accompanying figure, segments $AB$ and $CD$ are parallel, the measure of angle $D$ is twice that of angle $B$, and the measures of segments $AD... | With reference to the diagram above, let $E$ be the point on $AB$ such that $DE||BC$. Let $\angle ABC=\alpha$. We then have $\alpha =\angle AED = \angle EDC$ since $AB||CD$, so $\angle ADE=\angle ADC-\angle BDC=2\alpha-\alpha = \alpha$, which means $\triangle AED$ is isosceles.
Therefore, $AB=AE+EB=\boxed{a+b}$. | entailment |
A circular cylindrical post with a circumference of 4 feet has a string wrapped around it, spiraling from the bottom of the post to the top of the post. The string evenly loops around the post exactly four full times, starting at the bottom edge and finishing at the top edge. The height of the post is 12 feet. What is ... | Each time the string spirals around the post, it travels 3 feet up and 4 feet around the post. If we were to unroll this path, it would look like: [asy]
size(150);
draw((0,0)--(0,3)--(4,3)--(4,0)--cycle, linewidth(.7));
draw((0,0)--(4,3),linewidth(.7));
label("3",(0,1.5),W);
label("4",(2,3),N);
[/asy] Clearly, a 3-4-5... | entailment |
In the figure, $m\angle A = 28^{\circ}$, $m\angle B = 74^\circ$ and $m\angle C = 26^{\circ}$. If $x$ and $y$ are the measures of the angles in which they are shown, what is the value of $x + y$? [asy]
size(150);
draw((0,5)--(0,0)--(15,0)--(15,5),linewidth(1));
draw((0,5)--(2,2)--(5,5)--(12,-2)--(15,5),linewidth(.7));
l... | Starting from the right triangle that contains angle $C$, we can see the third angle in this triangle is $90-26=64$ degrees. By vertical angles, this makes the rightmost angle in the triangle containing angle $y$ also equal to 64 degrees. Thus, the third angle in that triangle has measure $180-(y+64)=116-y$ degrees. ... | entailment |
The diagonals of a regular hexagon have two possible lengths. What is the ratio of the shorter length to the longer length? Express your answer as a common fraction in simplest radical form. | The two possible diagonal lengths are $AB$ and $AC$. The measure of an interior angle of a regular hexagon is $180(6-2)/6=120$ degrees. Therefore, angle $BCA$ measures $120/2=60$ degrees. Also, the base angles of the isosceles triangle with the marked 120-degree angle each measure $(180-120)/2=30$ degrees. This impli... | entailment |
Line $l_1$ has equation $3x - 2y = 1$ and goes through $A = (-1, -2)$. Line $l_2$ has equation $y = 1$ and meets line $l_1$ at point $B$. Line $l_3$ has positive slope, goes through point $A$, and meets $l_2$ at point $C$. The area of $\triangle ABC$ is $3$. What is the slope of $l_3$? | We find the coordinates of point $B$ by solving $3x-2y = 1$ and $y = 1$ simultaneously. With $y=1,$ we get $3x-2=1,$ and so $x=1.$ Thus, $B=(1,1).$ The distance from $A$ to line $l_2$ is $1 - (-2) = 3,$ so we have \[\tfrac{1}{2} \cdot BC \cdot 3 = [\triangle ABC] = 3,\]and thus $BC = 2.$ Therefore, either $C = (3, 1)$ ... | entailment |
Altitudes $\overline{AP}$ and $\overline{BQ}$ of an acute triangle $\triangle ABC$ intersect at point $H$. If $HP=5$ while $HQ=2$, then calculate $(BP)(PC)-(AQ)(QC)$. [asy]
size(150); defaultpen(linewidth(0.8));
pair B = (0,0), C = (3,0), A = (2,2), P = foot(A,B,C), Q = foot(B,A,C),H = intersectionpoint(B--Q,A--P);
dr... | We use similar triangles: $\triangle BPH \sim \triangle APC$ since they are both right triangles and the angles at $A$ and $B$ are each complementary to $\angle C$, and thus congruent. Similarly, $\triangle AQH \sim \triangle BQC$. We know that $HP=5$ and $HQ=2$, so we have the ratios \[ \frac{BP}{5} = \frac{AH+5}{PC}... | entailment |
Square $ABCD$ has area $200$. Point $E$ lies on side $\overline{BC}$. Points $F$ and $G$ are the midpoints of $\overline{AE}$ and $\overline{DE}$, respectively. Given that quadrilateral $BEGF$ has area $34$, what is the area of triangle $GCD$? | We begin by drawing a diagram: [asy]
size(5cm);
pair a=(0,1); pair b=(1,1); pair c=(1,0); pair d=(0,0); pair e=(1,.82); pair f=(a+e)/2; pair g=(d+e)/2;
fill(b--e--g--f--cycle,gray);
fill(g--c--d--cycle,pink);
dot(a);dot(b);dot(c);dot(d);dot(e);dot(f);dot(g);
draw(a--b--c--d--a);
draw(a--e--d);
draw(e--g--f--b);
draw(g-... | entailment |
In convex quadrilateral $ABCD$, $AB=8$, $BC=4$, $CD=DA=10$, and $\angle CDA=60^\circ$. If the area of $ABCD$ can be written in the form $\sqrt{a}+b\sqrt{c}$ where $a$ and $c$ have no perfect square factors (greater than 1), what is $a+b+c$? | We begin by drawing a diagram: [asy]
pair A,B,C,D;
A=(0,5*sqrt(3));
B=(10-13/5,5*sqrt(3)+(1/5)*sqrt(231));
C=(10,5*sqrt(3));
D=(5,0);
draw(A--B--C--D--cycle);
label("$A$",A,W); label("$B$",B,N); label("$C$",C,E); label("$D$",D,S);
draw(A--C);
label("$60^\circ$",(5,1.8));
label("$8$",(A--B),NW); label("$4$",(B--C),NE); ... | entailment |
With all angles measured in degrees, the product $\prod_{k=1}^{45} \csc^2(2k-1)^\circ=m^n$, where $m$ and $n$ are integers greater than 1. Find $m+n$.
| Let $x = \cos 1^\circ + i \sin 1^\circ$. Then from the identity\[\sin 1 = \frac{x - \frac{1}{x}}{2i} = \frac{x^2 - 1}{2 i x},\]we deduce that (taking absolute values and noticing $|x| = 1$)\[|2\sin 1| = |x^2 - 1|.\]But because $\csc$ is the reciprocal of $\sin$ and because $\sin z = \sin (180^\circ - z)$, if we let our... | entailment |
The cube below has sides of length 4 feet. If a cylindrical section of radius 2 feet is removed from the solid, what is the total remaining volume of the cube? Express your answer in cubic feet in terms of $\pi$.
[asy]
import solids; size(150); import three; defaultpen(linewidth(0.8)); currentprojection = orthographic... | The cube has volume $4^3=64$ cubic feet. The cylinder has radius 2, height 4, and volume $\pi(2^2)(4)=16\pi$ cubic feet. It follows that when the cylindrical section is removed from the solid, the remaining volume is $\boxed{64-16\pi}$ cubic feet. | entailment |
In triangle $ABC$, $AB = 3$, $BC = 4$, $AC = 5$, and $BD$ is the angle bisector from vertex $B$. If $BD = k \sqrt{2}$, then find $k$. | By Pythagoras, $\angle ABC = 90^\circ$. Let $P$ and $Q$ be the projections of $D$ onto $BC$ and $AB$, respectively.
[asy]
unitsize(1 cm);
pair A, B, C, D, P, Q;
A = (0,3);
B = (0,0);
C = (4,0);
D = (12/7,12/7);
P = (12/7,0);
Q = (0,12/7);
draw(A--B--C--cycle);
draw(B--D);
draw(P--D--Q);
label("$A$", A, NW);
label... | entailment |
An equilateral triangle has sides 8 units long. An equilateral triangle with sides 4 units long is cut off at the top, leaving an isosceles trapezoid. What is the ratio of the area of the smaller triangle to the area of the trapezoid? Express your answer as a common fraction. | Connect the midpoints of the sides of the equilateral triangle as shown. The triangle is divided into four congruent equilateral triangles, and the isosceles trapezoid is made up of 3 of these 4 triangles. Therefore, the ratio of the area of one of the triangles to the area of the trapezoid is $\boxed{\frac{1}{3}}$.
... | entailment |
Compute $\cos 0^\circ$. | Rotating the point $(1,0)$ about the origin by $0^\circ$ counterclockwise gives us the point $(1,0)$, so $\cos 0^\circ = \boxed{1}$. | entailment |
In triangle $\triangle JKL$ shown, $\tan K = \frac{3}{2}$. What is $KL$?
[asy]
pair J,K,L;
L = (0,0);
J = (0,3);
K = (2,3);
draw(L--J--K--L);
draw(rightanglemark(L,J,K,7));
label("$L$",L,SW);
label("$J$",J,NW);
label("$K$",K,NE);
label("$2$",(J+K)/2,N);
[/asy] | Because $\triangle JKL$ is a right triangle, $\tan K = \frac{JL}{JK}$. So $\tan K = \frac{3}{2} = \frac{JL}{2}$. Then $JL = 3$.
By the Pythagorean Theorem, $KL = \sqrt{JL^2 + JK^2} = \sqrt{3^2 + 2^2} = \boxed{\sqrt{13}}$. | entailment |
In triangle $ABC$, $A'$, $B'$, and $C'$ are on the sides $BC$, $AC$, and $AB$, respectively. Given that $AA'$, $BB'$, and $CC'$ are concurrent at the point $O$, and that $\frac{AO}{OA'}+\frac{BO}{OB'}+\frac{CO}{OC'}=92$, find $\frac{AO}{OA'}\cdot \frac{BO}{OB'}\cdot \frac{CO}{OC'}$.
| Let $K_A=[BOC], K_B=[COA],$ and $K_C=[AOB].$ Due to triangles $BOC$ and $ABC$ having the same base,\[\frac{AO}{OA'}+1=\frac{AA'}{OA'}=\frac{[ABC]}{[BOC]}=\frac{K_A+K_B+K_C}{K_A}.\]Therefore, we have\[\frac{AO}{OA'}=\frac{K_B+K_C}{K_A}\]\[\frac{BO}{OB'}=\frac{K_A+K_C}{K_B}\]\[\frac{CO}{OC'}=\frac{K_A+K_B}{K_C}.\]Thus, w... | entailment |
A circle centered at $A$ with a radius of $1$ and a circle centered at $B$ with a radius of $4$ are externally tangent. A third circle is tangent to the first two and to one of their common external tangents as shown. What is the radius of the third circle? [asy]
draw((-3,0)--(7.5,0));
draw(Circle((-1,1),1),linewidth(0... | Let $C$ be the intersection of the horizontal line through $A$ and the vertical line through $B.$ In right triangle $ABC,$ we have $BC=3$ and $AB=5,$ so $AC=4.$ Let $x$ be the radius of the third circle, and $D$ be the center. Let $E$ and $F$ be the points of intersection of the horizontal line through $D$ with the ver... | entailment |
A park is in the shape of a regular hexagon $2$ km on a side. Starting at a corner, Alice walks along the perimeter of the park for a distance of $5$ km. How many kilometers is she from her starting point?
$\textbf{(A)}\ \sqrt{13}\qquad \textbf{(B)}\ \sqrt{14}\qquad \textbf{(C)}\ \sqrt{15}\qquad \textbf{(D)}\ \sqrt{16}... | We imagine this problem on a coordinate plane and let Alice's starting position be the origin. We see that she will travel along two edges and then go halfway along a third. Therefore, her new $x$-coordinate will be $1 + 2 + \frac{1}{2} = \frac{7}{2}$ because she travels along a distance of $2 \cdot \frac{1}{2} = 1$ km... | entailment |
What is the area of the triangle bounded by the lines $y=x,$ $y=-x,$ and $y=6$? | [asy]
size(200); defaultpen(linewidth(0.8));
xlimits(-8,8); ylimits(-1,10);
xaxis(Label("$x$"),-8,8,EndArrow(size=5));
yaxis(Label("$y$"),-2,8,EndArrow(size=5));
dot("$A(6,6)$",(6,6)); dot("$B(-6,6)$",(-6,6),W); dot("$O$",(0,0),SW);
draw((0,0) -- (6,6));
draw((0,0) -- (-6,6));
draw((-6,6)--(6,6));
[/asy]
Let $O = (0,0)... | entailment |
In trapezoid $ABCD$ with $\overline{BC}\parallel\overline{AD}$, let $BC = 1000$ and $AD = 2008$. Let $\angle A = 37^\circ$, $\angle D = 53^\circ$, and $M$ and $N$ be the midpoints of $\overline{BC}$ and $\overline{AD}$, respectively. Find the length $MN$.
| Extend $\overline{AB}$ and $\overline{CD}$ to meet at a point $E$. Then $\angle AED = 180 - 53 - 37 = 90^{\circ}$.
[asy] size(220); defaultpen(0.7+fontsize(10)); real f=100, r=1004/f; pair A=(0,0), D=(2*r, 0), N=(r,0), E=N+r*expi(74*pi/180); pair B=(126*A+125*E)/251, C=(126*D + 125*E)/251; pair[] M = intersectionpoints... | entailment |
The isosceles trapezoid shown has side lengths as labeled. How long is segment AC? [asy]
unitsize(1.5mm);
defaultpen(linewidth(.7pt)+fontsize(10pt));
dotfactor=3;
pair A=(0,0), B=(21,0), C=(15,8), D=(6,8);
pair[] dots={A,B,C,D};
draw(A--B--C--D--cycle);
dot(dots);
label("A",A,SW);
label("B",B,SE);
label("C",C,NE);
la... | Define $E$ and $F$ to be the feet of the perpendiculars drawn to $AB$ from $C$ and $D$ respectively. Since $EF=CD=9$, we find $AF=(21-9)/2=6$ and $AE=AF+FE=15$. Also, from the Pythagorean theorem, $CE=DF=\sqrt{10^2-6^2}=8$. Again using the Pythagorean theorem, $AC=\sqrt{CE^2+AE^2}=\sqrt{8^2+15^2}=\boxed{17}$ units.
... | entailment |
Let $\triangle ABC$ be a right triangle such that $B$ is a right angle. A circle with diameter of $BC$ meets side $AC$ at $D.$ If the area of $\triangle ABC$ is $150$ and $AC = 25,$ then what is $BD$? | We might try sketching a diagram: [asy]
pair pA, pB, pC, pO, pD;
pA = (-15, 0);
pB = (0, 0);
pC = (0, 20);
pO = (0, 10);
pD = (-9.6, 7.2);
draw(pA--pB--pC--pA);
draw(pD--pB);
draw(circle(pO, 10));
label("$A$", pA, SW);
label("$B$", pB, S);
label("$C$", pC, N);
label("$D$", pD, W);
[/asy] Since $BC$ is a diameter of the... | entailment |
A right circular cone is inscribed in a right prism as shown. What is the ratio of the volume of the cone to the volume of the prism? Express your answer as a common fraction in terms of $\pi$. [asy]
import three; import graph3;
defaultpen(linewidth(0.8));
size(200);
draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--cycle);
dra... | Since the cone is tangent to all sides of the base of the prism, the base of the prism is a square. Furthermore, if the radius of the base of the cone is $r$, then the side length of the square is $2r$.
Let $h$ be the common height of the cone and the prism. Then the volume of the cone is \[\frac{1}{3} \pi r^2 h,\] ... | entailment |
The vertices of a square are the centers of four circles as shown below. Given each side of the square is 6cm and the radius of each circle is $2\sqrt{3}$cm, find the area in square centimeters of the shaded region. [asy]
fill( (-1,-1)-- (1,-1) -- (1,1) -- (-1,1)--cycle, gray);
fill( Circle((1,1), 1.2), white);
fill( C... | [asy]
fill( (-1,-1)-- (1,-1) -- (1,1) -- (-1,1)--cycle, gray);
fill( Circle((1,1), 1.2), white);
fill( Circle((-1,-1), 1.2), white);
fill( Circle((1,-1),1.2), white);
fill( Circle((-1,1), 1.2), white);
draw( Arc((1,1),1.2 ,180,270));
draw( Arc((1,-1),1.2,90,180));
draw( Arc((-1,-1),1.2,0,90));
draw( Arc((-1,1),1.2,0,-9... | entailment |
The radius of a sphere is $p$ units and the radius of a hemisphere is $2p$ units. What is the ratio of the volume of the sphere to the volume of the hemisphere? | The volume of the sphere is \[\frac{4}{3}\pi p^3\] and the volume of the hemisphere is \[\frac{1}{2}\cdot \frac{4}{3}\pi (2p)^3 = \frac{4}{3}\pi p^3 \cdot 4.\] Thus the ratio of the volume of the sphere to the volume of the hemisphere is $\boxed{\frac{1}{4}}$. | entailment |
[asy] draw((0,0)--(1,sqrt(3)),black+linewidth(.75),EndArrow); draw((0,0)--(1,-sqrt(3)),black+linewidth(.75),EndArrow); draw((0,0)--(1,0),dashed+black+linewidth(.75)); dot((1,0)); MP("P",(1,0),E); [/asy]
Let $S$ be the set of points on the rays forming the sides of a $120^{\circ}$ angle, and let $P$ be a fixed point ins... | We have all the angles we need, but most obviously, we see that right angle in triangle $ABD$.
Note also that angle $BAD$ is 6 degrees, so length $AB = cos(6)$ because the diameter, $AD$, is 1.
Now, we can concentrate on triangle $ABX$ (after all, now we can decipher all angles easily and use Law of Sines).
We get:
$\f... | entailment |
Three circles, each of radius $3$, are drawn with centers at $(14, 92)$, $(17, 76)$, and $(19, 84)$. A line passing through $(17,76)$ is such that the total area of the parts of the three circles to one side of the line is equal to the total area of the parts of the three circles to the other side of it. What is the ab... | First of all, we can translate everything downwards by $76$ and to the left by $14$. Then, note that a line passing through a given point intersecting a circle with a center as that given point will always cut the circle in half, so we can re-phrase the problem:
Two circles, each of radius $3$, are drawn with centers a... | entailment |
In the figure, $AB \perp BC, BC \perp CD$, and $BC$ is tangent to the circle with center $O$ and diameter $AD$. In which one of the following cases is the area of $ABCD$ an integer?
[asy] pair O=origin, A=(-1/sqrt(2),1/sqrt(2)), B=(-1/sqrt(2),-1), C=(1/sqrt(2),-1), D=(1/sqrt(2),-1/sqrt(2)); draw(unitcircle); dot(O); dr... | Let $E$ and $F$ be the intersections of lines $AB$ and $BC$ with the circle. One can prove that $BCDE$ is a rectangle, so $BE=CD$.
In order for the area of trapezoid $ABCD$ to be an integer, the expression $\frac{(AB+CD)BC}2=(AB+CD)BF$ must be an integer, so $BF$ must be rational.
By Power of a Point, $AB\cdot BE=BF^2\... | entailment |
The radius of the inscribed circle is 6 cm. What is the number of centimeters in the length of $\overline{AB}$? Express your answer in simplest radical form. [asy]
import olympiad; import geometry; size(150); defaultpen(linewidth(0.8));
draw((sqrt(3),0)--origin--(0,1)--cycle);
real r1 = (sqrt(3) - 1)/2;
draw(Circle((r1... | Define points $C$, $D$, $E$, $F$ and $O$ as shown in the figure. Triangles $BCO$ and $BFO$ are right triangles that share a hypotenuse, and $CO=6\text{ cm}=OF$. By the hypotenuse-leg congruency theorem, triangles $BCO$ and $BFO$ are congruent. Therefore, angles $CBO$ and $FBO$ each measure 30 degrees, so angle $BOC$... | entailment |
Two right triangles share a side as follows: [asy]
pair pA, pB, pC, pD, pE;
pA = (0, 0);
pB = pA + 6 * dir(0);
pC = pA + 10 * dir(90);
pD = pB + 6 * dir(90);
pE = (6 * pA + 10 * pD) / 16;
draw(pA--pB--pC--pA);
draw(pA--pB--pD--pA);
label("$A$", pA, SW);
label("$B$", pB, SE);
label("$C$", pC, NW);
label("$D$", pD, NE);
... | Since $AB = BD,$ we see that $\triangle ABD$ is an isosceles right triangle, therefore $\angle DAB = 45^\circ.$ That means that $AD$, and consequently $AE,$ bisects $\angle CAB.$
Relating our areas to side lengths and applying the Angle Bisector Theorem, we have that: \begin{align*}
\frac{[\triangle ABE]}{[\triangle A... | entailment |
Two of the altitudes of the scalene triangle $ABC$ have length $4$ and $12$. If the length of the third altitude is also an integer, what is the biggest it can be?
$\textbf{(A)}\ 4\qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ \text{none of these}$
| Assume we have a scalene triangle $ABC$. Arbitrarily, let $12$ be the height to base $AB$ and $4$ be the height to base $AC$. Due to area equivalences, the base $AC$ must be three times the length of $AB$.
Let the base $AB$ be $x$, thus making $AC = 3x$. Thus, setting the final height to base $BC$ to $h$, we note that ... | entailment |
In quadrilateral $ABCD$, $\angle{BAD}\cong\angle{ADC}$ and $\angle{ABD}\cong\angle{BCD}$, $AB = 8$, $BD = 10$, and $BC = 6$. The length $CD$ may be written in the form $\frac {m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.
| Extend $\overline{AD}$ and $\overline{BC}$ to meet at $E$. Then, since $\angle BAD = \angle ADC$ and $\angle ABD = \angle DCE$, we know that $\triangle ABD \sim \triangle DCE$. Hence $\angle ADB = \angle DEC$, and $\triangle BDE$ is isosceles. Then $BD = BE = 10$.
[asy] /* We arbitrarily set AD = x */ real x = 60^.5, a... | entailment |
A cylinder has a radius of 3 cm and a height of 8 cm. What is the longest segment, in centimeters, that would fit inside the cylinder? | The longest segment stretches from the bottom to the top of the cylinder and across a diameter, and is thus the hypotenuse of a right triangle where one leg is the height $8$, and the other is a diameter of length $2(3)=6$. Thus its length is
$$\sqrt{6^2+8^2}=\boxed{10}$$ | entailment |
The dimensions of a triangle are tripled to form a new triangle. If the area of the new triangle is 54 square feet, how many square feet were in the area of the original triangle? | If two similar triangles have side ratios of $r : 1,$ the ratio of their areas must be $r^2 : 1.$ That means that when a triangle is tripled to form a new triangle, the new triangle has 9 times the area of the original. That means the original triangle must have an area of $\dfrac{54\text{ ft}^2}{9} = \boxed{6}\text{ f... | entailment |
Two of the altitudes of an acute triangle divide the sides into segments of lengths $5,3,2$ and $x$ units, as shown. What is the value of $x$? [asy]
defaultpen(linewidth(0.7)); size(75);
pair A = (0,0);
pair B = (1,0);
pair C = (74/136,119/136);
pair D = foot(B, A, C);
pair E = /*foot(A,B,C)*/ (52*B+(119-52)*C)/(119);
... | Let us label this diagram. [asy]
defaultpen(linewidth(0.7)); size(120);
pair A = (0,0);
pair B = (1,0);
pair C = (74/136,119/136);
pair D = foot(B, A, C);
pair E = /*foot(A, B, C)*/ (52*B+(119-52)*C)/(119);
draw(A--B--C--cycle);
draw(B--D);
draw(A--E);
draw(rightanglemark(A,D,B,1.2));
draw(rightanglemark(A,E,B,1.2));
l... | entailment |
Square $ABCD$ is inscribed in a circle. Square $EFGH$ has vertices $E$ and $F$ on $\overline{CD}$ and vertices $G$ and $H$ on the circle. If the area of square $ABCD$ is $1$, then the area of square $EFGH$ can be expressed as $\frac {m}{n}$ where $m$ and $n$ are relatively prime positive integers and $m < n$. Find $10n... | Let $O$ be the center of the circle, and $2a$ be the side length of $ABCD$, $2b$ be the side length of $EFGH$. By the Pythagorean Theorem, the radius of $\odot O = OC = a\sqrt{2}$.
[asy] size(150); pointpen = black; pathpen = black+linewidth(0.7); pen d = linetype("4 4") + blue + linewidth(0.7); pair C=(1,1), D=(1,-1),... | entailment |
Two similar right triangles have areas of 6 square inches and 150 square inches. The length of the hypotenuse of the smaller triangle is 5 inches. What is the sum of the lengths of the legs of the larger triangle? | Since the smaller triangle has hypotenuse 5, we guess that it is a 3-4-5 triangle. Sure enough, the area of a right triangle with legs of lengths 3 and 4 is $(3)(4)/2 = 6$, so this works. The area of the larger triangle is $150/6=25$ times the area of the smaller triangle, so its side lengths are $\sqrt{25} = 5$ time... | entailment |
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... | 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 ($... | entailment |
The midpoints of the sides of a triangle with area $T$ are joined to form a triangle with area $M$. What is the ratio of $M$ to $T$? Express your answer as a common fraction. | When you connect the midpoints of two sides of a triangle, you get a segment which is half as long as the third side of the triangle. Therefore, every side in the smaller triangle is $\frac{1}{2}$ the side length of the original triangle. Therefore, the area of the smaller triangle is $\left(\frac{1}{2}\right)^2 = \box... | entailment |
$A$, $B$, $C$, and $D$ are points on a circle, and segments $\overline{AC}$ and $\overline{BD}$ intersect at $P$, such that $AP=8$, $PC=1$, and $BD=6$. Find $BP$, given that $BP < DP.$
[asy]
unitsize(0.6 inch);
draw(circle((0,0),1));
draw((-0.3,0.94)--(0.3,-0.94));
draw((-0.7,-0.7)--(0.7,-0.7));
label("$A$",(-0.3,0... | Writing $BP=x$ and $PD=6-x$, we have that $BP < 3$. Power of a point at $P$ gives $AP \cdot PC = BP \cdot PD$ or $8=x(6-x)$. This can be solved for $x=2$ and $x=4$, and we discard the latter, leaving $BP = \boxed{2}$. | entailment |
A hexagon is inscribed in a circle. Five of the sides have length $81$ and the sixth, denoted by $\overline{AB}$, has length $31$. Find the sum of the lengths of the three diagonals that can be drawn from $A$.
| [asy]defaultpen(fontsize(9)); pair A=expi(-pi/2-acos(475/486)), B=expi(-pi/2+acos(475/486)), C=expi(-pi/2+acos(475/486)+acos(7/18)), D=expi(-pi/2+acos(475/486)+2*acos(7/18)), E=expi(-pi/2+acos(475/486)+3*acos(7/18)), F=expi(-pi/2-acos(475/486)-acos(7/18)); draw(unitcircle);draw(A--B--C--D--E--F--A);draw(A--C..A--D..A--... | entailment |
Compute $\tan 3825^\circ$. | Rotating $360^\circ$ is the same as doing nothing, so rotating $3825^\circ$ is the same as rotating $3825^\circ - 10\cdot 360^\circ = 225^\circ$. Therefore, we have $\tan 3825^\circ = \tan (3825^\circ - 10\cdot 360^\circ) = \tan 225^\circ$.
Let $P$ be the point on the unit circle that is $225^\circ$ counterclockwise ... | entailment |
In triangle $ABC$, $AB = 3$, $AC = 5$, and $BC = 4$. The medians $AD$, $BE$, and $CF$ of triangle $ABC$ intersect at the centroid $G$. Let the projections of $G$ onto $BC$, $AC$, and $AB$ be $P$, $Q$, and $R$, respectively. Find $GP + GQ + GR$.
[asy]
import geometry;
unitsize(1 cm);
pair A, B, C, D, E, F, G, P, Q... | By Pythagoras, triangle $ABC$ is right with $\angle B = 90^\circ$. Then the area of triangle $ABC$ is $1/2 \cdot AB \cdot BC = 1/2 \cdot 3 \cdot 4 = 6$.
Since $G$ is the centroid of triangle $ABC$, the areas of triangles $BCG$, $CAG$, and $ABG$ are all one-third the area of triangle $ABC$, namely $6/3 = 2$.
We can v... | entailment |
In the middle of a vast prairie, a firetruck is stationed at the intersection of two perpendicular straight highways. The truck travels at $50$ miles per hour along the highways and at $14$ miles per hour across the prairie. Consider the set of points that can be reached by the firetruck within six minutes. The area of... | Let the intersection of the highways be at the origin $O$, and let the highways be the x and y axes. We consider the case where the truck moves in the positive x direction.
After going $x$ miles, $t=\frac{d}{r}=\frac{x}{50}$ hours has passed. If the truck leaves the highway it can travel for at most $t=\frac{1}{10}-\fr... | entailment |
Which of the cones below can be formed from a $252^{\circ}$ sector of a circle of radius 10 by aligning the two straight sides?
[asy]
draw((5.8,8.1)..(-10,0)--(0,0)--(3.1,-9.5)..cycle);
label("10",(-5,0),S);
label("$252^{\circ}$",(0,0),NE);
[/asy]
A. base radius = 6, slant =10
B. base radius = 6, height =10
C. base... | The slant height of the cone is equal to the radius of the sector, or $10$. The circumference of the base of the cone is equal to the length of the sector's arc, or $\frac{252^\circ}{360^\circ}(20\pi) = 14\pi$. The radius of a circle with circumference $14\pi$ is $7$. Hence the answer is $\boxed{C}$. | entailment |
In right $\Delta ABC$, $\angle CAB$ is a right angle. Point $M$ is the midpoint of $\overline{BC}$. What is the number of centimeters in the length of median $\overline{AM}$? Express your answer as a decimal to the nearest tenth. [asy] pair A,B,C,M;
A = (0,0); B = (4,0); C = (0,3); M = (B+C)/2;
draw(M--A--B--C--A);
lab... | The length of the median to the hypotenuse of a right triangle is half the hypotenuse. The hypotenuse of $\triangle ABC$ is $\sqrt{3^2+4^2} = 5$, so $AM = BC/2 = \boxed{2.5}$. | entailment |
The perimeter of triangle $APM$ is $152$, and the angle $PAM$ is a right angle. A circle of radius $19$ with center $O$ on $\overline{AP}$ is drawn so that it is tangent to $\overline{AM}$ and $\overline{PM}$. Given that $OP=m/n$ where $m$ and $n$ are relatively prime positive integers, find $m+n$.
| Let the circle intersect $\overline{PM}$ at $B$. Then note $\triangle OPB$ and $\triangle MPA$ are similar. Also note that $AM = BM$ by power of a point. Using the fact that the ratio of corresponding sides in similar triangles is equal to the ratio of their perimeters, we have\[\frac{19}{AM} = \frac{152-2AM-19+19}{152... | entailment |
$\triangle ABC$ is similar to $\triangle DEF$ . What is the number of centimeters in the length of $\overline{EF}$ ? Express your answer as a decimal to the nearest tenth.
[asy]
draw((0,0)--(8,-2)--(5,4)--cycle);
label("8cm",(2.5,2),NW);
label("5cm",(6.1,1),NE);
draw((12,0)--(18,-1.5)--(15.7,2.5)--cycle);
label("$A$",... | Because $\triangle ABC \sim \triangle DEF,$ we know that: \begin{align*}
\frac{EF}{ED} &= \frac{BC}{BA} \\
\frac{EF}{3\text{ cm}} &= \frac{8\text{ cm}}{5\text{ cm}} \\
EF &= \frac{8\text{ cm}\cdot3\text{ cm}}{5\text{ cm}} = \boxed{4.8}\text{ cm}.
\end{align*} | entailment |
In the figure shown, the ratio of $BD$ to $DC$ is $4$ to $3$. The area of $\triangle ABD$ is $24$ square centimeters. What is the area of $\triangle ADC$? [asy] size(85); defaultpen(linewidth(1)+fontsize(10));
pair A = (0,5.5), B=(0,0), D = (2,0), C = (3,0);
draw(A--B--C--A--D); label("A",A,N); label("B",B,S); label(... | The area of a triangle is given by the formula $\frac 12 bh$. Both $\triangle ABD$ and $\triangle ADC$ share the same height $AB$. Let $[ABD]$ be the area of $\triangle ABD$ and $[ADC]$ be the area of $\triangle ADC$. It follows that $\frac{[ABD]}{[ADC]} = \frac{\frac 12 \cdot BD \cdot h}{\frac 12 \cdot DC \cdot h} = \... | entailment |
In the figure, triangles $ABC$ and $BCD$ are equilateral triangles. What is the value of $AD \div BC$ when expressed in simplest radical form?
[asy]
draw((0,0)--(5,8.7)--(10,0)--cycle);
draw((10,0)--(15,8.7)--(5,8.7));
label("$A$",(0,0),SW);
label("$B$",(5,8.7),N);
label("$C$",(10,0),SE);
label("$D$",(15,8.7),NE);
[/a... | Let $BC = s$. We can see that $AD$ consists of the altitudes from $A$ and $D$ to $BC$, each of which has length $s\sqrt{3}/2$. Thus, $AD = s\sqrt{3}$. Therefore, $AD\div BC = s\sqrt{3}/s = \boxed{\sqrt{3}}$. | entailment |
Find the ratio of the area of $\triangle BCX$ to the area of $\triangle ACX$ in the diagram if $CX$ bisects $\angle ACB$. Express your answer as a common fraction. [asy]
import markers;
real t=27/(27+30);
pair A=(-15.57,0);
pair B=(8.43,0);
pair C=(0,25.65);
pair X=t*A+(1-t)*B;
draw(C--A--B--C--X);
label("$A$",A,SW)... | The Angle Bisector Theorem tells us that \[\frac{BX}{AX}=\frac{BC}{AC}=\frac{27}{30}=\frac{9}{10}.\]Since $\triangle BCX$ and $\triangle ACX$ share the same height, the ratio of their areas is simply the ratio of their bases, so our answer is \[\frac{BX}{AX}=\boxed{\frac{9}{10}}.\] | entailment |
Quadrilateral $ABCD$ is a square. A circle with center $D$ has arc $AEC$. A circle with center $B$ has arc $AFC$. If $AB = 2$ cm, what is the total number of square centimeters in the football-shaped area of regions II and III combined? Express your answer as a decimal to the nearest tenth.
[asy]
path a=(7,13)..(0,0)-... | Regions I, II, and III combine to form a sector of a circle whose central angle measures 90 degrees. Therefore, the area of this sector is $\frac{90}{360}\pi(\text{radius})^2=\frac{1}{4}\pi(2)^2=\pi$ square centimeters. Also, regions I and II combine to form an isosceles right triangle whose area is $\frac{1}{2}(\tex... | entailment |
$ABCD$ is a trapezoid with the measure of base $\overline{AB}$ twice the measure of the base $\overline{CD}$. Point $E$ is the point of intersection of the diagonals. The measure of diagonal $\overline{AC}$ is 11. Find the length of segment $\overline{EC}$. Express your answer as a common fraction.
[asy]
size(200);
p... | Since the bases of the trapezoid are $\overline{AB}$ and $\overline{CD}$, these two line segments must be parallel. Now, since $\overline{AC}$ intersects these two parallel lines, $\angle DCE$ and $\angle BAE$ are alternate interior angles and therefore must be congruent. Similarly, $\overline{DB}$ intersects the bases... | entailment |
Two circles are centered at the origin, as shown. The point $P(8,6)$ is on the larger circle and the point $S(0,k)$ is on the smaller circle. If $QR=3$, what is the value of $k$?
[asy]
unitsize(0.2 cm);
defaultpen(linewidth(.7pt)+fontsize(10pt));
dotfactor=4;
draw(Circle((0,0),7)); draw(Circle((0,0),10));
dot((0,0)... | We can determine the distance from $O$ to $P$ by dropping a perpendicular from $P$ to $T$ on the $x$-axis. [asy]
unitsize(0.2 cm);
defaultpen(linewidth(.7pt)+fontsize(10pt));
dotfactor=4;
draw(Circle((0,0),7)); draw(Circle((0,0),10));
dot((0,0)); dot((7,0)); dot((10,0)); dot((0,7)); dot((8,6));
draw((0,0)--(8,6)--(8,0)... | entailment |
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... | 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... | entailment |
A parallelogram has adjacent sides of lengths $s$ units and $2s$ units forming a 45-degree angle. The area of the parallelogram is $8\sqrt 2$ square units. What is the value of $s$? Express your answer in simplest radical form. | If we let the side of length $2s$ be the base of the parallelogram, we can use our 45-degree angle to find the height of the parallelogram. The height and sides of the parallelogram form a 45-45-90 triangle, with the side of length $s$ as the hypotenuse. Thus, the height of the parallelogram is $s/\sqrt{2}$.
It follow... | entailment |
In a triangle with integer side lengths, one side is three times as long as a second side, and the length of the third side is 15. What is the greatest possible perimeter of the triangle? | Let the sides of the triangle have lengths $x$, $3x$, and 15. The Triangle Inequality implies that $3x<x+15$, so $x<7.5$. Because $x$ is an integer, the greatest possible perimeter is $7+21+15=\boxed{43}$. | entailment |
In rectangle $ABCD$, $AB = 3$ and $BC = 9$. The rectangle is folded so that points $A$ and $C$ coincide, forming the pentagon $ABEFD$. What is the length of segment $EF$? Express your answer in simplest radical form.
[asy]
size(200);
defaultpen(linewidth(.8pt)+fontsize(10pt));
draw((0,0)--(9,0)--(9,3)--(0,3)--(0,0)--c... | To begin with, let $DF = x$ and $FA = 9 - x$. $\triangle{DFA}$ is a right triangle, so we can solve for $x$ by applying the Pythagorean Theorem: $x^2 + 9 = 81 - 18x + x^2$, so $18x = 72$, or $x = 4$. By applying the same argument to $\triangle{EAB}$, we can see that $FA = EA = 5$. Drop a perpendicular from $F$ to $EA$ ... | entailment |
Triangle $ABC$ is isosceles with angle $B$ congruent to angle $C$. The measure of angle $C$ is four times the measure of angle $A$. What is the number of degrees in the measure of angle $B$? | Let $x$ be the number of degrees in $\angle A$. Then $\angle C=4x^\circ$, and $\angle B$ is also $4x^\circ$ (since $\angle B$ is congruent to $\angle C$).
Since the sum of angles in a triangle is $180^\circ$, we have $$x + 4x + 4x = 180,$$ which we can solve for $x=20$. Therefore, $\angle B = 4\cdot 20 = \boxed{80}$ d... | entailment |
In right triangle $ABC$ with $\angle A = 90^\circ$, we have $AB = 6$ and $BC = 10$. Find $\cos C$. | 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... | entailment |
A circle with center $O$ has radius $8$ units and circle $P$ has radius $2$ units. The circles are externally tangent to each other at point $Q$. Segment $TS$ is the common external tangent to circle $O$ and circle $P$ at points $T$ and $S$, respectively. What is the length of segment $OS$? Express your answer in s... | We create a diagram with the given information from the problem: [asy]
draw(Circle((0,0),8));
draw(Circle((10,0),2));
dot((0,0));dot((10,0));
label("$O$",(0,0),SW); label("$P$",(10,0),SW);
dot((8,0)); label("$Q$",(8,0),SW);
label("$T$",(4.6,6.6),NE); label("$S$",(11,1.7),NE);
draw((4.6,6.6)--(11,1.7));
[/asy]
We dra... | entailment |
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? | 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... | entailment |
Square $BCFE$ is inscribed in right triangle $AGD$, as shown below. If $AB = 28$ units and $CD = 58$ units, what is the area of square $BCFE$?
[asy]
draw((9,15.6)--(10.7,14.6)--(11.7,16.3));
draw((0,0)--(10,17.3)--(40,0)--cycle);
draw((7,0)--(19,0)--(19,12.1)--(7,12.1)--cycle);
label("$A$",(0,0),SW);
label("$B$",(7,0)... | $\angle CDF = \angle AEB$ and $\angle BAE = \angle CFD$, so we know that $\bigtriangleup AEB \sim \bigtriangleup FDC$. Thus, denoting the side length of $BEFC$ as $x$, we can create the ratios: $\frac{28}{x} = \frac{x}{58} \Rightarrow x^2 = 1624$, and since $x^2$ is the area of square $BCFE$, $\boxed{1624}$ is the ans... | entailment |
Compute $\cos 135^\circ$. | Let $P$ be the point on the unit circle that is $135^\circ$ 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... | entailment |
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... | 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:... | entailment |
Suppose $\cos Q = 0.4$ in the diagram below. What is $QR$?
[asy]
pair P,Q,R;
P = (0,0);
Q = (6,0);
R = (0,6*tan(acos(0.4)));
draw(P--Q--R--P);
draw(rightanglemark(Q,P,R,18));
label("$P$",P,SW);
label("$Q$",Q,SE);
label("$R$",R,N);
label("$12$",Q/2,S);
[/asy] | Since $\cos Q = 0.4$ and $\cos Q = \frac{QP}{QR}=\frac{12}{QR}$, we have $\frac{12}{QR} = 0.4$, so $QR = \frac{12}{0.4} = \boxed{30}$. | entailment |
The diagonals of rectangle $PQRS$ intersect at point $X$. If $PS = 10$ and $RS=24$, then what is $\cos \angle PXS$? | [asy]
pair P,Q,R,SS,X,F;
SS = (0,0);
P = (0,5);
R = (12,0);
Q= R+P;
X = Q/2;
F = foot(SS,P,R);
draw(F--SS--R--Q--P--SS--Q);
draw(P--R);
label("$P$",P,NW);
label("$Q$",Q,NE);
label("$R$",R,SE);
label("$S$",SS,SW);
label("$X$",X,S);
label("$F$",F,SW);
draw(rightanglemark(S,F,X,12));
[/asy]
To find $\cos \angle PXS$, we ... | entailment |
An isosceles right triangle is removed from each corner of a square piece of paper, as shown, to create a rectangle. If $AB = 12$ units, what is the combined area of the four removed triangles, in square units? [asy]
unitsize(5mm);
defaultpen(linewidth(.7pt)+fontsize(8pt));
pair A=(1,4), Ap=(0,3), B=(3,0), Bp=(4,1);
... | Each of the sides of the square is divided into two segments by a vertex of the rectangle. Call the lengths of these two segments $r$ and $s$. Also, let $C$ be the foot of the perpendicular dropped from $A$ to the side containing the point $B$. Since $AC=r+s$ and $BC=|r-s|$, \[
(r+s)^2+(r-s)^2=12^2,
\] from the Pyth... | entailment |
The sum of the lengths of the twelve edges of a rectangular box is $140$, and the distance from one corner of the box to the farthest corner is $21$. What is the total surface area of the box? | Let $a$, $b$, and $c$ be the dimensions of the box. It is given that \[140=4a+4b+4c{\qquad \rm and
\qquad}21=\sqrt{a^2+b^2+c^2}\] hence \[35=a+b+c{\qquad (1)\qquad
\rm and\qquad}441=a^2+b^2+c^2{\qquad (2)}.\]
Square both sides of $(1)$ and combine with $(2)$ to obtain \begin{align*}
1225 & = (a+b+c)^2 \\
&= a^2+b^2+c^... | entailment |
Find the area of a triangle with side lengths 8, 9, and 9. | We draw the longest altitude of this triangle, which breaks the triangle into two right triangles. [asy]
size(100);
pair A,B,C,D;
A=(0,0); B=(8,0); C=(4,sqrt(65)); D=(4,0);
draw(A--B--C--cycle); draw(C--D);
draw(rightanglemark(C,D,A,18));
label("8",D,S); label("9",(A+C)/2,W); label("9",(B+C)/2,E);
[/asy] The right tr... | entailment |
Pyramid $OABCD$ has square base $ABCD,$ congruent edges $\overline{OA}, \overline{OB}, \overline{OC},$ and $\overline{OD},$ and $\angle AOB=45^\circ.$ Let $\theta$ be the measure of the dihedral angle formed by faces $OAB$ and $OBC.$ Given that $\cos \theta=m+\sqrt{n},$ where $m$ and $n$ are integers, find $m+n.$
| [asy] import three; // calculate intersection of line and plane // p = point on line // d = direction of line // q = point in plane // n = normal to plane triple lineintersectplan(triple p, triple d, triple q, triple n) { return (p + dot(n,q - p)/dot(n,d)*d); } // projection of point A onto line BC triple projection... | entailment |
A jar of peanut butter which is 3 inches in diameter and 4 inches high sells for $\$$0.60. At the same rate, what would be the price for a jar that is 6 inches in diameter and 6 inches high? | The first jar has a volume of $V=\pi r^2h=\pi(\frac{3}{2})^24=9\pi$. The second jar has a volume of $V=\pi r^2h=\pi(\frac{6}{2})^26=54\pi$. Note that the volume of the second jar is 6 times greater than that of the first jar. Because peanut butter is sold by volume, the second jar will be six times more expensive than ... | entailment |
The lengths of the three sides of a triangle are $7$, $x+4$ and $2x+1$. The perimeter of the triangle is 36. What is the length of the longest side of the triangle? | Since the perimeter of the triangle is 36, then $7+(x+4)+(2x+1)=36$ or $3x+12=36$ or $3x=24$ or $x=8$.
Thus, the lengths of the three sides of the triangle are $7$, $8+4=12$ and $2(8)+1=17$, of which the longest is $\boxed{17}.$ | entailment |
Regular hexagon $ABCDEF$ has vertices $A$ and $C$ at $(0,0)$ and $(7,1)$, respectively. What is its area? | Diagonals $\overline{AC}$, $\overline{CE}$, $\overline{EA}$, $\overline{AD}$, $\overline{CF}$, and $\overline{EB}$ divide the hexagon into twelve congruent 30-60-90 triangles, six of which make up equilateral $\triangle ACE$.
[asy]
unitsize(0.5 cm);
pair A, B, C, D, E, F, G;
A = (0,0);
C = (7,1);
E = rotate(60)*(C);... | entailment |
Two triangles are similar. The ratio of their areas is 1:4. If the height of the smaller triangle is 3 cm, how long is the corresponding height of the larger triangle, in centimeters? | If any linear dimension (such as radius, side length, height, etc.) of a closed, two-dimensional figure is multiplied by $k$ while the shape of the figure remains the same, the area of the figure is multiplied by $k^2$. Since the area is multiplied by 4 in going from the smaller triangle to the larger triangle, we hav... | entailment |
A bridge is built by suspending a plank of wood between two triangular wedges with equal heights, as in the following diagram: [asy]
import olympiad;
import math;
// Draw triangles
pair A = (0, 1);
pair B = (-cos(1.3962), 0);
pair C = (cos(1.3962), 0);
pair D = (2, 1);
pair E = (2-cos(1.3089), 0);
pair F = (2+cos(1.30... | There are several ways to proceed, and here is one. Since $\triangle ABC$ and $\triangle DEF$ are both isosceles, it should be easy to find that $\angle B = \angle C = 80^\circ$ and $\angle E = \angle F = 75^\circ.$ Now, connect $C$ and $E$:
[asy]
import olympiad;
import math;
// Draw triangles
pair A = (0, 1);
pair ... | entailment |
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. | 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
$$\frac{45^{\circ}}{2}=\boxed{22.5^{\circ}}.$$ | entailment |
Let $\triangle ABC$ be an acute scalene triangle with circumcircle $\omega$. The tangents to $\omega$ at $B$ and $C$ intersect at $T$. Let $X$ and $Y$ be the projections of $T$ onto lines $AB$ and $AC$, respectively. Suppose $BT = CT = 16$, $BC = 22$, and $TX^2 + TY^2 + XY^2 = 1143$. Find $XY^2$.
| Assume $O$ to be the center of triangle $ABC$, $OT$ cross $BC$ at $M$, link $XM$, $YM$. Let $P$ be the middle point of $BT$ and $Q$ be the middle point of $CT$, so we have $MT=3\sqrt{15}$. Since $\angle A=\angle CBT=\angle BCT$, we have $\cos A=\frac{11}{16}$. Notice that $\angle XTY=180^{\circ}-A$, so $\cos XYT=-\cos ... | entailment |
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 $ABC$? | [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 $... | entailment |
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. | 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 = (... | entailment |
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? | 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 $\pi \cdot (r+6)^2 \cdot 2 = \pi \cdot r^2 \cdot 8$... | entailment |
In $\triangle ABC, AB = 8, BC = 7, CA = 6$ and side $BC$ is extended, as shown in the figure, to a point $P$ so that $\triangle PAB$ is similar to $\triangle PCA$. The length of $PC$ is
[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=origin, P=(1.5,5), B=(8,0), C=P+2.5*dir(P--B); draw(A--P--C--A--B--C); label("A"... | Since we are given that $\triangle{PAB}\sim\triangle{PCA}$, we have $\frac{PC}{PA}=\frac{6}{8}=\frac{PA}{PC+7}$.
Solving for $PA$ in $\frac{PC}{PA}=\frac{6}{8}=\frac{3}{4}$ gives us $PA=\frac{4PC}{3}$.
We also have $\frac{PA}{PC+7}=\frac{3}{4}$. Substituting $PA$ in for our expression yields $\frac{\frac{4PC}{3}}{PC+7}... | entailment |
An isosceles trapezoid is circumscribed around a circle. The longer base of the trapezoid is $16$, and one of the base angles is $\arcsin(.8)$. Find the area of the trapezoid.
$\textbf{(A)}\ 72\qquad \textbf{(B)}\ 75\qquad \textbf{(C)}\ 80\qquad \textbf{(D)}\ 90\qquad \textbf{(E)}\ \text{not uniquely determined}$
| Let the trapezium have diagonal legs of length $x$ and a shorter base of length $y$. Drop altitudes from the endpoints of the shorter base to the longer base to form two right-angled triangles, which are congruent since the trapezium is isosceles. Thus using the base angle of $\arcsin(0.8)$ gives the vertical side of t... | entailment |
Let $\triangle ABC$ have side lengths $AB=13$, $AC=14$, and $BC=15$. There are two circles located inside $\angle BAC$ which are tangent to rays $\overline{AB}$, $\overline{AC}$, and segment $\overline{BC}$. Compute the distance between the centers of these two circles. | The two circles described in the problem are shown in the diagram. The circle located inside $\triangle ABC$ is called the incircle; following convention we will label its center $I$. The other circle is known as an excircle, and we label its center $E$. To begin, we may compute the area of triangle $ABC$ using Hero... | entailment |
A cube has side length $6$. Its vertices are alternately colored black and purple, as shown below. What is the volume of the tetrahedron whose corners are the purple vertices of the cube? (A tetrahedron is a pyramid with a triangular base.)
[asy]
import three;
real t=-0.05;
triple A,B,C,D,EE,F,G,H;
A = (0,0,0);
B = (c... | The volume of any pyramid is $\frac 13$ the product of the base area and the height. However, determining the height of the purple tetrahedron is somewhat tricky! Instead of doing that, we observe that the total volume of the cube consists of the purple tetrahedron and four other "clear" tetrahedra. Each clear tetrahed... | entailment |
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... | 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 $$(BC)(CD)=\left(6\sqrt{3}\right)\left(18-2\sqrt{3}\right)=
\boxed{108\sqrt{3}-36}.$$ | entailment |
For $x > 0$, the area of the triangle with vertices $(0, 0), (x, 2x)$, and $(x, 0)$ is 64 square units. What is the value of $x$? | Plotting the given points, we find that the triangle is a right triangle whose legs measure $x$ and $2x$ units. Therefore, $\frac{1}{2}(x)(2x)=64$, which we solve to find $x=\boxed{8}$ units. [asy]
import graph;
defaultpen(linewidth(0.7));
real x=8;
pair A=(0,0), B=(x,2*x), C=(x,0);
pair[] dots = {A,B,C};
dot(dots);
d... | entailment |
Five identical rectangles are arranged to form a larger rectangle $PQRS$, as shown. The area of $PQRS$ is $4000$. What is the length, $x$, rounded off to the nearest integer? [asy]
real x = 1; real w = 2/3;
// Draw outer square and labels
pair s = (0, 0); pair r = (2 * x, 0); pair q = (3 * w, x + w); pair p = (0, x ... | Let $w$ be the width of each of the identical rectangles. Since $PQ=3w$, $RS=2x$ and $PQ=RS$ (because $PQRS$ is a rectangle), then $2x = 3w$, or $$w=\frac{2}{3}x.$$ Therefore, the area of each of the five identical rectangles is $$x\left(\frac{2}{3}x\right)=\frac{2}{3}x^2.$$ Since the area of $PQRS$ is 4000 and it is m... | entailment |
A triangle with side lengths in the ratio 3:4:5 is inscribed in a circle of radius 3. What is the area of the triangle? Provide your answer as a decimal rounded to the nearest hundredth. | Let the sides of the triangle have lengths $3x$, $4x$, and $5x$. The triangle is a right triangle, so its hypotenuse is a diameter of the circle. Thus $5x=2\cdot 3=6$, so $x=6/5$. The area of the triangle is \[
\frac{1}{2}\cdot 3x\cdot 4x =\frac{1}{2}\cdot \frac{18}{5}\cdot \frac{24}{5}
=\frac{216}{25}=\boxed{8.64}.... | entailment |
A triangle is made of wood sticks of lengths 8, 15 and 17 inches joined end-to-end. Pieces of the same integral length are cut from each of the sticks so that the three remaining pieces can no longer form a triangle. How many inches are in the length of the smallest piece that can be cut from each of the three sticks t... | Our current triangle lengths are 8, 15, and 17. Let us say that $x$ is the length of the piece that we cut from each of the three sticks. Then, our lengths will be $8 - x,$ $15 - x,$ and $17 - x.$ These lengths will no longer form a triangle when the two shorter lengths added together is shorter than or equal to the lo... | entailment |
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... | 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. | entailment |
What is the radius of the circle inscribed in triangle $ABC$ if $AB = 5, AC=6, BC=7$? Express your answer in simplest radical form. | Let $r$ be the radius of the inscribed circle. Let $s$ be the semiperimeter of the triangle, that is, $s=\frac{AB+AC+BC}{2}=9$. Let $K$ denote the area of $\triangle ABC$.
Heron's formula tells us that \begin{align*}
K &= \sqrt{s(s-AB)(s-AC)(s-BC)} \\
&= \sqrt{9\cdot 4\cdot 3\cdot 2} \\
&= \sqrt{3^3\cdot 2^3} \\
&= 6\... | entailment |
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