from
stringclasses 2
values | value
stringlengths 16
6.77k
|
|---|---|
human
|
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 the length, in feet, of the string?
[asy]
size(150);
draw((0,0)--(0,20)..(1,19.5)..(2,20)--(2,0)..(1,-.5)..(0,0),linewidth(1));
draw((0,20)..(1,20.5)..(2,20),linewidth(1));
draw((1,19.5)--(0,18.5),linewidth(1));
draw((2,.5)--(1,-.5),linewidth(1));
draw((2,16)--(0,14),linewidth(1));
draw((2,11)--(0,9),linewidth(1));
draw((2,6)--(0,4),linewidth(1));
[/asy]
|
gpt
|
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 right triangle has been formed. For each time around the post, the string has length 5. So, the total length of the string will be $4\cdot 5=\boxed{20}$ feet.
|
human
|
A rectangular box has interior dimensions 6-inches by 5-inches by 10-inches. The box is filled with as many solid 3-inch cubes as possible, with all of the cubes entirely inside the rectangular box. What percent of the volume of the box is taken up by the cubes?
|
gpt
|
Three-inch cubes can fill a rectangular box only if the edge lengths of the box are all integer multiples of 3 inches. The largest such box whose dimensions are less than or equal to those of the $6''\times5''\times10''$ box is a $6''\times3''\times9''$ box. The ratio of the volumes of these two boxes is \[
\frac{6\cdot3\cdot9}{6\cdot5\cdot10}=\frac{3\cdot9}{5\cdot10}=\frac{27}{50},
\] which is $\boxed{54}$ percent.
|
human
|
A sphere is inscribed in a right cone with base radius $12$ cm and height $24$ cm, as shown. The radius of the sphere can be expressed as $a\sqrt{c} - a$ cm. What is the value of $a + c$? [asy]
import three; size(120); defaultpen(linewidth(1)); pen dashes = linetype("2 2") + linewidth(1);
currentprojection = orthographic(0,-1,0.16);
void drawticks(triple p1, triple p2, triple tickmarks) {
draw(p1--p2); draw(p1 + tickmarks-- p1 - tickmarks); draw(p2 + tickmarks -- p2 - tickmarks);
}
real r = 6*5^.5-6;
triple O = (0,0,0), A = (0,0,-24);
draw(scale3(12)*unitcircle3); draw((-12,0,0)--A--(12,0,0)); draw(O--(12,0,0),dashes);
draw(O..(-r,0,-r)..(0,0,-2r)..(r,0,-r)..cycle);
draw((-r,0,-r)..(0,-r,-r)..(r,0,-r)); draw((-r,0,-r)..(0,r,-r)..(r,0,-r),dashes);
drawticks((0,0,2.8),(12,0,2.8),(0,0,0.5));
drawticks((-13,0,0),(-13,0,-24),(0.5,0,0));
label("$12$", (6,0,3.5), N); label("$24$",(-14,0,-12), W);
[/asy]
|
gpt
|
Consider a cross-section of the cone that passes through the apex of the cone and the center of the circular base. It looks as follows: [asy] defaultpen(linewidth(1) + fontsize(10)); size(120); pen dashes = linetype("2 2") + linewidth(1); real r = 6*5^.5 - 6;
pair A = (0,-24), O = (0,0), C = (0,-r), P = foot(C,(12,0),A); draw(circle(C,r)); draw((-12,0)--A--(12,0)--cycle); draw(O--A, dashes); dot(C); draw(C--P,dashes); draw(rightanglemark(C,P,A));
label("$A$",A,S); label("$B$",(-12,0),N); label("$C$",(12,0),N); label("$D$",O,N); label("$O$",C,W); label("$P$",P,SE);
[/asy] Let $O$ be the center of the sphere (or the center of the circle in the cross-section), let the triangle be $\triangle ABC$, so that $D$ is the midpoint of $BC$ and $A$ is the apex (as $\triangle ABC$ is isosceles, then $\overline{AD}$ is an altitude). Let $P$ be the point of tangency of the circle with $\overline{AC}$, so that $OP \perp AC$. It follows that $\triangle AOP \sim \triangle ACD$. Let $r$ be the radius of the circle. It follows that $$\frac{OP}{AO} = \frac{CD}{AC} \implies OP \cdot AC = AO \cdot CD.$$We know that $CD = 12$, $AC = \sqrt{12^2 + 24^2} = 12\sqrt{5}$, $OP = r$, and $AO = AD - OP = 24 - r$. Thus, $$12r\sqrt{5} = 12(24-r) = 12^2 \cdot 2 - 12r \implies 12r(1 + \sqrt{5}) = 12^2 \cdot 2.$$Thus, $r = \frac{24}{1+\sqrt{5}}$. Multiplying the numerator and denominator by the conjugate, we find that $$r = \frac{24}{1+\sqrt{5}} \cdot \frac{\sqrt{5} - 1}{\sqrt{5} - 1} = \frac{24(\sqrt{5} - 1)}{5 - 1} = 6\sqrt{5} - 6.$$It follows that $a+c = \boxed{11}$.
|
human
|
Determine an expression for the area of $\triangle QCA$ in terms of $p$. Your answer should be simplified as much as possible. [asy]
size(5cm);defaultpen(fontsize(9));
pair o = (0, 0); pair q = (0, 12); pair b = (12, 0);
pair a = (2, 12); pair t = (2, 0); pair c = (0, 9);
draw((-2, 0)--(15, 0), Arrow);
draw((0, -2)--(0, 15), Arrow);
draw(q--a--b);
//draw(a--t);
draw(a--c--b);
label("$Q(0, 12)$", q, W);
label("$A(2, 12)$", a, NE);
label("$B(12, 0)$", b, S);
label("$O(0, 0)$", o, SW);
label("$x$", (15, 0), E);
label("$y$", (0, 15), N);
//label("$T(2, 0)$", t, S + 0.6 * E);
label("$C(0, p)$", c, W);
[/asy]
|
gpt
|
Since $QA$ is perpendicular to $QC$, we can treat $QC$ as the height of $\triangle QCA$ and $QA$ as the base. The area of $\triangle QCA$ is $$\frac{1}{2}\times QA\times QC=\frac{1}{2}\times(2-0)\times(12-p)=\frac{1}{2}\times2\times (12-p)=\boxed{12-p}.$$
|
human
|
In the diagram, each of the three identical circles touch the other two. The circumference of each circle is 36. What is the perimeter of the shaded region? [asy]
defaultpen(1);
path p = (1, 0){down}..{-dir(30)}dir(-60){dir(30)}..{dir(-30)}((2, 0) + dir(-120)){-dir(-30)}..{up}(1, 0)--cycle;
fill(p, gray(0.75));
draw(unitcircle);
draw(shift(2 * dir(-60)) * unitcircle);
draw(shift(2) * unitcircle);
[/asy]
|
gpt
|
Join the centre of each circle to the centre of the other two. Since each circle touches each of the other two, then these line segments pass through the points where the circles touch, and each is of equal length (that is, is equal to twice the length of the radius of one of the circles). [asy]
import olympiad;
defaultpen(1);
path p = (1, 0){down}..{-dir(30)}dir(-60){dir(30)}..{dir(-30)}((2, 0) + dir(-120)){-dir(-30)}..{up}(1, 0)--cycle;
fill(p, gray(0.75));
draw(unitcircle);
draw(shift(2 * dir(-60)) * unitcircle);
draw(shift(2) * unitcircle);
// Add lines
draw((0, 0)--(2, 0)--(2 * dir(-60))--cycle);
// Add ticks
add(pathticks((0, 0)--(1, 0), s=4)); add(pathticks((1, 0)--(2, 0), s=4));
add(pathticks((0, 0)--dir(-60), s=4)); add(pathticks(dir(-60)--(2 * dir(-60)), s=4));
add(pathticks((2 * dir(-60))--(2 * dir(-60) + dir(60)), s=4)); add(pathticks((2, 0)--(2 * dir(-60) + dir(60)), s=4));
[/asy]
Since each of these line segments have equal length, then the triangle that they form is equilateral, and so each of its angles is equal to $60^\circ$.
Now, the perimeter of the shaded region is equal to the sum of the lengths of the three circular arcs which enclose it. Each of these arcs is the arc of one of the circles between the points where this circle touches the other two circles.
Thus, each arc is a $60^\circ$ arc of one of the circles (since the radii joining either end of each arc to the centre of its circle form an angle of $60^\circ$), so each arc is $\frac{60^\circ}{360^\circ} = \frac{1}{6}$ of the total circumference of the circle, so each arc has length $\frac{1}{6}(36)=6$.
Therefore, the perimeter of the shaded region is $3(6) = \boxed{18}$.
|
human
|
In triangle $ABC$, $AB=AC$ and $D$ is a point on $\overline{AC}$ so that $\overline{BD}$ bisects angle $ABC$. If $BD=BC$, what is the measure, in degrees, of angle $A$?
|
gpt
|
Since $AB=AC$, triangle $ABC$ must be an isosceles triangle and the measures of $\angle ABC$ and $\angle ACB$ must be equal. Continuing, since $\overline{BD}$ bisects angle $ABC$, we have that the measures of $\angle ABD$ and $\angle BDC$ are equal. Finally, since $BD=BC$, triangle $BDC$ must also be an isosceles triangle so the measures of $\angle BDC = \angle BCD$. Now if we consider triangle $BDC$, we know that angles $BDC$ and $BCD$ have equal angle measures and angle $DBC$ has an angle measure that is half that of the other two. Since these three angle measures must add up to $180^\circ$, we have that $\angle DBC$ has measure $36^\circ$ and angles $BDC$ and $BCD$ have measures $72 ^\circ$.
Now, since $\angle ABC \cong \angle ACB$ and $\angle ACB$ has measure $72^\circ$, we know that $\angle A$ must have an angle measure of $180-72-72=\boxed{36}$ degrees.
|
human
|
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),true)));
//markangle(n=1,radius=15,X,C,B,marker(markinterval(stickframe(n=1),true)));
label("$28$",.5*(B+X),SE);
label("$30$",.5*(B+C),N);
label("$21$",.5*(A+C),NW);
[/asy]
|
gpt
|
The Angle Bisector Theorem tells us that \[\frac{AC}{AX}=\frac{BC}{BX}\]so \[AX=\frac{AC\cdot BX}{BC}=\frac{21\cdot28}{30}=\frac{7^2\cdot3\cdot4}{30}=\frac{7^2\cdot2}{5}=\boxed{\frac{98}5}.\]
|
human
|
In triangle $ABC$, we have $\angle A = 90^\circ$ and $\sin B = \frac{4}{7}$. Find $\cos C$.
|
gpt
|
[asy]
pair A,B,C;
A = (0,0);
B = (4,0);
C = (0,sqrt(33));
draw(A--B--C--A);
draw(rightanglemark(B,A,C,10));
label("$A$",A,SW);
label("$B$",B,SE);
label("$C$",C,N);
[/asy]
Since $\triangle ABC$ is a right triangle, we have $\sin B = \frac{AC}{BC}$ and $\cos C = \frac{AC}{BC}$, so $\cos C = \sin B = \boxed{\frac47}$.
|
human
|
Compute $\sin 240^\circ$.
|
gpt
|
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);
O= (0,0);
label("$x$",(1.2,0),SE);
label("$y$",(0,1.2),NE);
P = rotate(240)*A;
D = foot(P,A,-A);
draw(O--P--D);
draw(rightanglemark(O,D,P,2));
draw(Circle(O,1));
label("$O$",O,SE);
label("$P$",P,SW);
//label("$A$",A,SE);
label("$D$",D,N);
[/asy]
Triangle $POD$ is a 30-60-90 triangle, so $DO = \frac{1}{2}$ and $PD = \frac{\sqrt{3}}{2}$. Therefore, the coordinates of $P$ are $\left(-\frac12,-\frac{\sqrt{3}}{2}\right)$, so $\sin240^\circ = \boxed{-\frac{\sqrt{3}}{2}}$.
|
human
|
Point $A$ has coordinates $(x,6)$. When Point $A$ is reflected over the $y$-axis it lands on Point $B$. What is the sum of the four coordinate values of points $A$ and $B$?
|
gpt
|
The coordinates of point $B$ are $(-x,6)$. The sum of all four coordinates is $x+6+(-x)+6=\boxed{12}$.
|
human
|
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?
|
gpt
|
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 the parabola is maximized when the $x$-coordinate is chosen as close to the $x$-coordinate of the vertex as possible. The $x$-coordinate of the vertex is halfway between the zeros at $x=(0+72)/2=36$, so the maximum area is $(36)(36)=1296$ square units. Similarly, to minimize the area we choose the length to be as far from $36$ as possible. The resulting dimensions are $1$ unit and $71$ units, so the minimum area is 71 square units. The difference between 1296 square units and 71 square units is $\boxed{1225}$ square units.
[asy]
import graph; defaultpen(linewidth(0.8));
size(150,IgnoreAspect);
real f(real x)
{
return x*(15-x);
}
xaxis(Arrows(4));
yaxis(ymax=f(7.5),Arrows(4));
draw(graph(f,-3,18),Arrows(4));
label("Area",(0,f(7.5)),N);
label("$l$",(18,0),S);[/asy]
|
human
|
Compute $\cos 210^\circ$.
|
gpt
|
Let $P$ be the point on the unit circle that is $210^\circ$ 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= (0,0);
label("$x$",(1.2,0),SE);
label("$y$",(0,1.2),NE);
P = rotate(210)*A;
D = foot(P,A,-A);
draw(O--P--D);
draw(rightanglemark(O,D,P,2));
draw(Circle(O,1));
label("$O$",O,SE);
label("$P$",P,SW);
//label("$A$",A,SE);
label("$D$",D,N);
[/asy]
Triangle $POD$ is a 30-60-90 triangle, so $DO = \frac{\sqrt{3}}{2}$ and $DP = \frac12$. Therefore, the coordinates of $P$ are $\left(-\frac{\sqrt{3}}{2}, -\frac12\right)$, so $\cos 210^\circ = \boxed{-\frac{\sqrt{3}}{2}}$.
|
human
|
A right cylinder with a height of 5 inches has a radius of 2 inches. What is the area of the curved surface of the cylinder, in square inches? Express your answer in terms of $\pi$.
|
gpt
|
The lateral area of a cylinder is $2\pi rh$. Since $h = 5$ and $r = 2$, our answer is $2\pi\cdot 2\cdot 5 = \boxed{20\pi}$.
|
human
|
From a circular piece of paper with radius $BC$, Jeff removes the unshaded sector shown. Using the larger shaded sector, he joins edge $BC$ to edge $BA$ (without overlap) to form a cone of radius 12 centimeters and of volume $432\pi$ cubic centimeters. What is the number of degrees in the measure of angle $ABC$ of the sector that is not used? [asy]
import graph;
defaultpen(linewidth(0.7));
fill((0,0)--dir(20)..dir(60)..dir(100)..dir(140)..dir(180)..dir(220)..dir(260)..dir(300)--cycle,gray);
draw((0,0)--dir(20)..dir(60)..dir(100)..dir(140)..dir(180)..dir(220)..dir(260)..dir(300)--(0,0));
draw(dir(300)..dir(320)..dir(340)..dir(360)..dir(20),dotted);
label("$C$",dir(20),E);
label("$A$",dir(300),SE);
label("$B$",(0,0),W);[/asy]
|
gpt
|
Solving $\frac{1}{3}\pi(12\text{ cm})^2(h)=432\pi\text{ cm}^3$, we find that the height $h$ of the cone is 9 cm. Since the radius is 12 cm and the height is 9 cm, the slant height of the cone, which is the same as the distance from $B$ to $C$, is $\sqrt{9^2+12^2}=15$ centimeters. The length of major arc $AC$ is equal to the circumference of the cone, which is $2\pi(12\text{ cm})=24\pi$ cm. The distance all the way around the circle is $2\pi(BC)=30\pi$ cm. Therefore, the central angle of major arc $AC$ measures $\left(\frac{24\pi\text{ cm}}{30\pi\text{ cm}}\right)360^\circ=288$ degrees. The measure of angle $ABC$ is $360^\circ-288^\circ=\boxed{72}$ degrees.
|
human
|
Rectangle $ABCD$ is the base of pyramid $PABCD$. If $AB = 8$, $BC = 4$, $\overline{PA}\perp \overline{AB}$, $\overline{PA}\perp\overline{AD}$, and $PA = 6$, then what is the volume of $PABCD$?
|
gpt
|
[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(C--P, dashed);
draw(B--C--D,dashed);
label("$A$",A,S);
label("$B$",B,W);
label("$C$",C,S);
label("$D$",D,E);
label("$P$",P,N);
[/asy]
Since $\overline{PA}$ is perpendicular to both $\overline{AB}$ and $\overline{AD}$, the segment $\overline{PA}$ is the altitude from the apex to the base of the pyramid. The area of the base is $[ABCD] = (AB)(BC) = 32$, and the height from the apex to the base is 6, so the volume of the pyramid is $\frac13(32)(6) = \boxed{64}$ cubic units.
|
human
|
Compute $\tan 225^\circ$.
|
gpt
|
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= (0,0);
label("$x$",(1.2,0),SE);
label("$y$",(0,1.2),NE);
P = rotate(225)*A;
D = foot(P,A,-A);
draw(O--P--D);
draw(rightanglemark(O,D,P,2));
draw(Circle(O,1));
label("$O$",O,NE);
label("$P$",P,SW);
//label("$A$",A,SE);
label("$D$",D,N);
[/asy]
Triangle $POD$ is a 45-45-90 triangle, so $DO = DP = \frac{\sqrt{2}}{2}$. Therefore, the coordinates of $P$ are $\left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right)$, so $\tan 225^\circ = \frac{\sin 225^\circ}{\cos 225^\circ} = \frac{-\sqrt{2}/2}{-\sqrt{2}/2} = \boxed{1}$.
|
human
|
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 $\pi$.
|
gpt
|
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 $(1/3)\pi(5^2)(12)=\boxed{100\pi}$.
|
human
|
A right circular cone is sliced into four pieces by planes parallel to its base, as shown in the figure. All of these pieces have the same height. What is the ratio of the volume of the second-largest piece to the volume of the largest piece? Express your answer as a common fraction.
[asy]
size(150);
pair A, B, C, D, E, F, G, H, I, w, x, y, z;
A=(0,0);
B=(.25,.75);
C=(.5,1.5);
D=(.75,2.25);
E=(1,3);
F=(1.25,2.25);
G=(1.5,1.5);
H=(1.75,.75);
I=(2,0);
w=(A+I)/2;
x=(B+H)/2;
y=(C+G)/2;
z=(D+F)/2;
draw(ellipse(w, 1, .25));
draw(ellipse(x, .75, .1875));
draw(ellipse(y, .5, .125));
draw(ellipse(z, .25, .0625));
draw(A--E--I);
[/asy]
|
gpt
|
Let the height of the smallest cone (the one on top) be $h$ and let the radius of the circular base of that cone be $r$. Consider the 4 cones in the diagram: the smallest one on top (cone A), the top 2 pieces (cone B), the top 3 pieces (cone C), and all 4 pieces together (cone D). Because each piece of the large cone has the same height as the smallest cone and the same angle and vertex at the top, each of the 4 cones is a dilation of the smaller cone at the top. In other words, all four cones are similar. Because cone B has height twice that of cone A, its circular base has twice the radius as that of A. Likewise, cone C has three times the height, and thus 3 times the radius, and cone D has 4 times the height and 4 times the radius. Thus, using the formula for the volume of a cone, we get \begin{align*}
V_B&=\frac{1}{3} \pi (2r)^2 (2h)=\frac{8}{3} \pi r^2 h \\
V_C&=\frac{1}{3} \pi (3r)^2 (3h)=\frac{27}{3} \pi r^2 h \\
V_D&=\frac{1}{3} \pi (4r)^2 (4h)=\frac{64}{3} \pi r^2 h
\end{align*}Looking at the diagram, we can see that the largest piece will be the volume of cone D minus that of cone C: \[V_{1}=\frac{64}{3} \pi r^2 h-\frac{27}{3} \pi r^2 h=\frac{64-27}{3} \pi r^2 h=\frac{37}{3} \pi r^2 h.\]Also notice that the volume of the second largest piece is the volume of cone C minus that of cone B: \[V_{2}=\frac{27}{3} \pi r^2 h-\frac{8}{3} \pi r^2 h=\frac{27-8}{3} \pi r^2 h=\frac{19}{3} \pi r^2 h.\]Thus, the ratio of the volume of the second largest piece to that of the largest piece is \begin{align*}
\frac{V_2}{V_1}=\frac{\frac{19}{3} \pi r^2 h}{\frac{37}{3} \pi r^2 h}
=\frac{\frac{19}{\cancel{3}} \cancel{\pi} \cancel{r^2} \cancel{h}}{\frac{37}{\cancel{3}} \cancel{\pi} \cancel{r^2} \cancel{h}}
=\boxed{\frac{19}{37}}.
\end{align*}
|
human
|
Three circles of radius 1 are externally tangent to each other and internally tangent to a larger circle. What is the radius of the large circle? Express your answer as a common fraction in simplest radical form.
[asy]
draw(Circle((0,-0.58),2.15),linewidth(0.7));
draw(Circle((-1,0),1),linewidth(0.7));
draw(Circle((1,0),1),linewidth(0.7));
draw(Circle((0,-1.73),1),linewidth(0.7));
[/asy]
|
gpt
|
Let $O$ be the center of the large circle, let $C$ be the center of one of the small circles, and let $\overline{OA}$ and $\overline{OB}$ be tangent to the small circle at $A$ and $B$.
[asy]
dot((0.57,1));
label("1",(0.8,1.45),E);
label("1",(0.57,0.5),E);
draw(arc((0,0),2.15,0,90),linewidth(0.7));
//draw((0,2.15)..(-2.15,0)--(2.15,0)..cycle,linewidth(0.7));
//fill((0,2.2)--(0,-0.1)--(-2.2,-0.1)--(-2.2,2.2)--cycle,white);
draw((0,0)--(1.08,1.87),linewidth(0.7));
draw(Circle((0.57,1),1),linewidth(0.7));
draw((0.57,1)--(0.57,0),linewidth(0.7));
draw((-1,1.73)--(0,0)--(2.15,0),linewidth(0.7));
label("$C$",(0.57,1),E);
label("$O$",(0,0),SW);
label("$B$",(-0.29,0.5),W);
label("$A$",(0.57,0),S);
label("$D$",(1.08,1.87),NE);
[/asy]
By symmetry, $\angle AOB =120^{\circ}$ and $\angle AOC = 60^{\circ}$. Thus $\triangle AOC$ is a 30-60-90 degree right triangle, and $AC=1$, so \[
OC= \frac{2}{\sqrt{3}}AC= \frac{2\sqrt{3}}{3}.
\]If $OD$ is a radius of the large circle through $C$, then \[
OD=CD + OC= 1 + \frac{2\sqrt{3}}{3}= \boxed{\frac{3+2\sqrt{3}}{3}}.
\]
|
human
|
Bonnie makes the frame of a cube out of 12 pieces of wire that are each six inches long. Meanwhile Roark uses 1-inch-long pieces of wire to make a collection of unit cube frames that are not connected to each other. The total volume of Roark's cubes is the same as the volume of Bonnie's cube. What is the ratio of the total length of Bonnie's wire to the total length of Roark's wire? Express your answer as a common fraction. [asy]
size(50);
draw((0,0)--(4,0)--(4,4)--(0,4)--cycle);
draw((3,1)--(7,1)--(7,5)--(3,5)--cycle);
draw((0,0)--(3,1));
draw((4,0)--(7,1));
draw((4,4)--(7,5));
draw((0,4)--(3,5));
[/asy]
|
gpt
|
The total length of Bonnie's wire is $12\cdot6=72$ inches, while her total volume is $6^3=216$ cubic inches. Each of Roark's unit cubes has volume $1$ cubic inch, so he needs $216$ cubes.
Since each cube has $12$ edges, each of Roark's cubes has $12\cdot1=12$ inches of wire. So his $216$ cubes have a total of $216\cdot12$ inches of wire.
So the desired fraction is $\dfrac{72}{216\cdot12}=\dfrac{6}{216}=\boxed{\dfrac{1}{36}}$.
|
human
|
Two boards, one four inches wide and the other six inches wide, are nailed together to form an X. The angle at which they cross is 60 degrees. If this structure is painted and the boards are separated what is the area of the unpainted region on the four-inch board? (The holes caused by the nails are negligible.) Express your answer in simplest radical form.
[asy]
draw(6dir(150)--15dir(-30),linewidth(1));
draw((6dir(150)+12/sqrt(3)*dir(30))--(15dir(-30)+12/sqrt(3)*dir(30)),linewidth(1));
draw(6dir(210)--(0,0),linewidth(1));
draw((9dir(210)+8/sqrt(3)*dir(-30))--8/sqrt(3)*dir(-30),linewidth(1));
draw(12/sqrt(3)*dir(30)--(12/sqrt(3)+6)*dir(30),linewidth(1));
draw(12/sqrt(3)*dir(30)+8/sqrt(3)*dir(-30)--(12/sqrt(3)+9)*dir(30)+8/sqrt(3)*dir(-30),linewidth(1));
draw(2dir(150)--2dir(150)+6dir(60),dashed);
draw(2dir(210)--2dir(210)+4dir(-60),dashed);
dot((2,0));
dot((4,-1));
dot((8,1));
dot((6,2));
label("$60^{\circ}$", (11,1), E);
label(rotate(30)*"$4^{\prime\prime}$", .5*(2dir(210)+2dir(210)+4dir(-60))+(0,-.5),W);
label(rotate(-30)*"$6^{\prime\prime}$", .5*(2dir(150)+2dir(150)+6dir(60))+(1,1),W);
[/asy]
|
gpt
|
Note that the unpainted region forms a parallelogram with heights between bases of 4 inches and 6 inches and with one angle 60 degree, as shown.
[asy]
size(150); unitsize(7.5,7.5); import olympiad;
draw(6dir(150)--15dir(-30),dashed);
draw((6dir(150)+12/sqrt(3)*dir(30))--(15dir(-30)+12/sqrt(3)*dir(30)),dashed);
draw(6dir(210)--(0,0),dashed);
draw((9dir(210)+8/sqrt(3)*dir(-30))--8/sqrt(3)*dir(-30),dashed);
draw(12/sqrt(3)*dir(30)--(12/sqrt(3)+6)*dir(30),dashed);
draw(12/sqrt(3)*dir(30)+8/sqrt(3)*dir(-30)--(12/sqrt(3)+9)*dir(30)+8/sqrt(3)*dir(-30),dashed);
label("$60^{\circ}$",+(11,1),+E,fontsize(8pt));
label("$60^{\circ}$",+(9,1),+W,fontsize(8pt));
draw((0,0)--6/sin(pi/3)*dir(30)--(6/sin(pi/3)*dir(30)+4/sin(pi/3)*dir(-30))--4/sin(pi/3)*dir(-30)--cycle, linewidth(1));
draw(4/sin(pi/3)*dir(-30) -- (4/sin(pi/3)*dir(-30) + 6*dir(60)));
draw(rightanglemark(4/sin(pi/3)*dir(-30),4/sin(pi/3)*dir(-30) + 6*dir(60), (6/sin(pi/3)*dir(30)+4/sin(pi/3)*dir(-30))));
label("6",(4/sin(pi/3)*dir(-30) + 4/sin(pi/3)*dir(-30) + 6*dir(60))/2,NW,fontsize(8pt));
[/asy]
The right triangle formed by drawing the height shown is a 30-60-90 triangle, and hence the hypotenuse has length $\frac{6}{\sqrt{3}/2} = 4\sqrt{3}$ inches. Now considering the hypotenuse as the base of the paralleogram, our new height is 4, and thus the area of this parallelogram is $4\cdot 4\sqrt{3} = \boxed{16\sqrt{3}}$.
|
human
|
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 $\pi$?
|
gpt
|
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 $2\pi r = \boxed{6\pi}$.
|
human
|
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)--cycle);
draw((17,3)--(12,3)--(12,0)--(21,0),dashed);
draw((21,3)--(17,3)--(16,0)--(16+3.2,-2.4)--(21,0)--(21,3)--cycle);
draw((17,3)--(21,0));
label("A", (0,3), NW);
label("B", (0,0), SW);
label("C", (9,0), SE);
label("D", (9,3), NE);
label("B", (19.2,-2.4), SE);
label("D", (21,3), NE);
label("E", (16,0), SW);
label("F", (17,3), N);
label("A$\&$C", (21,0), SE);
[/asy]
|
gpt
|
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 $EA$ and call the intersection point $P$. $PFDA$ is a rectangle, so we know that $PA = FD = 4$, so $PE = 5 - 4 = 1$. Furthermore, we know that $FP = DA = 3$. Now, we have right triangle $\triangle{FPE}$ with legs $1$ and $3$, so we can solve for $FE$ by applying the Pythagorean Theorem: $FE = \sqrt{1+9} = \boxed{\sqrt{10}}$.
|
human
|
An equilateral triangle $ABC$ shares a common side $BC$ with a square $BCDE,$ as pictured. What is the number of degrees in $\angle DAE$ (not pictured)? [asy]
pair pA, pB, pC, pD, pE;
pA = (0, 0);
pB = pA + dir(300);
pC = pA + dir(240);
pD = pC + dir(270);
pE = pB + dir(270);
draw(pA--pB--pC--pA);
draw(pB--pC--pD--pE--pB);
label("$A$", pA, N);
label("$B$", pB, E);
label("$C$", pC, W);
label("$D$", pD, SW);
label("$E$", pE, SE);
[/asy]
|
gpt
|
First of all, for our benefit, we should draw in the desired angle: [asy]
pair pA, pB, pC, pD, pE;
pA = (0, 0);
pB = pA + dir(300);
pC = pA + dir(240);
pD = pC + dir(270);
pE = pB + dir(270);
draw(pA--pB--pC--pA);
draw(pB--pC--pD--pE--pB);
draw(pD--pA--pE, red);
label("$A$", pA, N);
label("$B$", pB, E);
label("$C$", pC, W);
label("$D$", pD, SW);
label("$E$", pE, SE);
[/asy] We can see that $AB = BC = EB,$ thus $\triangle ABE$ is an isosceles triangle, where $\angle ABE = 90^\circ + 60^\circ = 150^\circ.$ Since the other two angles are equal and add up to $30^\circ$, we have that $\angle BAE = 15^\circ.$ Likewise, $\angle CAD = 15^\circ.$
Then, $\angle DAE = \angle CAB - \angle CAD - \angle BAE = 60^\circ - 15^\circ - 15^\circ = \boxed{30^\circ.}$
|
human
|
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);
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));
label("$3$",(C+D)/2,WNW+(0,0.3));
label("$5$",(A+D)/2,NW);
label("$2$",(C+E)/2,E);
label("$x$",(B+E)/2,NE);
[/asy]
|
gpt
|
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));
label("$A$", A, S);
label("$B$", B, S);
label("$C$", C, N);
label("$D$", D, NW);
label("$E$", E, NE);
label("$3$",(C+D)/2,WNW+(0,0.3));
label("$5$",(A+D)/2,NW);
label("$2$",(C+E)/2,E);
label("$x$",(B+E)/2,NE);
[/asy] $\triangle ACE$ and $\triangle BCD$ are similar by AA since they share $\angle ACB$ and $\angle AEC$ and $\angle BDC$ are both right angles and hence congruent. So $$\frac{CE}{CD} = \frac{AC}{BC}.$$ Plugging in values, we have $$\frac23 = \frac{8}{x+2}.$$ Solving this gives $x+2 = 12,$ or $x = \boxed{10}.$
|
human
|
In triangle $ABC$, $BC = 4$, $AC = 3 \sqrt{2}$, and $\angle C = 45^\circ$. Altitudes $AD$, $BE$, and $CF$ intersect at the orthocenter $H$. Find $AH:HD$.
|
gpt
|
Since $\angle C = 45^\circ$, triangle $ACD$ is a $45^\circ$-$45^\circ$-$90^\circ$ triangle, which means $AD = CD = AC/\sqrt{2} = 3$. Then $BD = BC - CD = 4 - 3 = 1$.
[asy]
unitsize(1 cm);
pair A, B, C, D, E, F, H;
A = (1,3);
B = (0,0);
C = (4,0);
D = (A + reflect(B,C)*(A))/2;
E = (B + reflect(C,A)*(B))/2;
F = (C + reflect(A,B)*(C))/2;
H = extension(B,E,C,F);
draw(A--B--C--cycle);
draw(A--D);
draw(B--E);
draw(C--F);
label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, S);
label("$E$", E, NE);
label("$F$", F, NW);
label("$H$", H, SE);
[/asy]
Also, $\angle EBC = 90^\circ - \angle BCE = 45^\circ$, so triangle $BHD$ is a $45^\circ$-$45^\circ$-$90^\circ$ triangle. Hence, $HD = BD = 1$. Then $AH = AD - HD = 3 - 1 = 2$, so $AH:HD = \boxed{2}$.
|
human
|
In the diagram, two pairs of identical isosceles triangles are cut off of square $ABCD$, leaving rectangle $PQRS$. The total area cut off is $200 \text{ m}^2$. What is the length of $PR$, in meters? [asy]
size(5cm);
pair a = (0, 1); pair b = (1, 1); pair c = (1, 0); pair d = (0, 0);
pair s = (0, 0.333); pair p = (0.667, 1); pair q = (1, 0.667); pair r = (0.333, 0);
// Thicken pen
defaultpen(linewidth(1));
// Fill triangles
path tri1 = a--p--s--cycle;
path tri2 = p--q--b--cycle;
path tri3 = q--c--r--cycle;
path tri4 = s--r--d--cycle;
fill(tri1, gray(0.75));fill(tri2, gray(0.75));
fill(tri3, gray(0.75));fill(tri4, gray(0.75));
// Draw rectangles
draw(a--b--c--d--cycle); draw(p--q--r--s--cycle);
// Labels
label("$A$", a, NW); label("$B$", b, NE); label("$C$", c, SE); label("$D$", d, SW);
label("$P$", p, N); label("$Q$", q, E); label("$R$", r, S); label("$S$", s, W);
[/asy]
|
gpt
|
Let $AS=x$ and $SD=y$.
Since $\triangle SAP$ and $\triangle SDR$ are isosceles, then $AP=x$ and $DR=y$.
Since there are two pairs of identical triangles, then $BP=BQ=y$ and $CQ=CR=x$. [asy]
size(5cm);
pair a = (0, 1); pair b = (1, 1); pair c = (1, 0); pair d = (0, 0);
pair s = (0, 0.333); pair p = (0.667, 1); pair q = (1, 0.667); pair r = (0.333, 0);
// Thicken pen
defaultpen(linewidth(1));
// Fill triangles
path tri1 = a--p--s--cycle;
path tri2 = p--q--b--cycle;
path tri3 = q--c--r--cycle;
path tri4 = s--r--d--cycle;
fill(tri1, gray(0.75));fill(tri2, gray(0.75));
fill(tri3, gray(0.75));fill(tri4, gray(0.75));
// Draw rectangles
draw(a--b--c--d--cycle); draw(p--q--r--s--cycle);
// Labels
label("$A$", a, NW); label("$B$", b, NE); label("$C$", c, SE); label("$D$", d, SW);
label("$P$", p, N); label("$Q$", q, E); label("$R$", r, S); label("$S$", s, W);
// x and y labels
label("$y$", r / 2, S);
label("$y$", s / 2, W);
label("$y$", p + (b - p) / 2, N);
label("$y$", q + (b - q) / 2, E);
label("$x$", r + (c - r) / 2, S);
label("$x$", s + (a - s) / 2, W);
label("$x$", c + (q - c) / 2, E);
label("$x$", a + (p - a) / 2, N);
[/asy] $\triangle SDR$ is right-angled (since $ABCD$ is a square) and isosceles, so its area (and hence the area of $\triangle BPQ$) is $\frac{1}{2}y^2$.
Similarly, the area of each of $\triangle SAP$ and $\triangle QCR$ is $\frac{1}{2}x^2$.
Therefore, the total area of the four triangles is $2(\frac{1}{2}x^2) + 2(\frac{1}{2}y^2) = x^2 + y^2$, so $x^2 + y^2 = 200$.
Now, by the Pythagorean Theorem, used first in $\triangle PRS$, then in $\triangle SAP$ and $\triangle SDR$, \begin{align*}
PR^2 & = PS^2 + SR^2 \\
& = (SA^2 + AP^2) + (SD^2 + DR^2) \\
& = 2x^2 + 2y^2 \\
& = 2(200) \\
& = 400
\end{align*} so $PR = \boxed{20}$ m.
|
human
|
An isosceles trapezoid has legs of length 30 cm each, two diagonals of length 40 cm each and the longer base is 50 cm. What is the trapezoid's area in sq cm?
|
gpt
|
We can pick a diagonal and a leg of the trapezoid such that, along with the longer base, these lines form a triangle with sides of length 30, 40, and 50. This is a Pythagorean triple, so the triangle is a right triangle. It follows that the altitude to the longer base of the trapezoid is $30\cdot 40/50 = 24$. This altitude is the same length as the height of the trapezoid.
We now look at the right triangle formed by this altitude, the adjacent leg of the trapezoid, and part of the longer base. These three sides form a right triangle, with hypotenuse of 30 and one leg (the altitude) of length 24. It follows that the other leg has length 18.
Because this is an isosceles trapezoid, we can now calculate the shorter base to have length $50 - 2\cdot 18 = 14$. Therefore, the area of the trapezoid is $\dfrac{(50 + 14)(24)}{2} = \boxed{768}$.
|
human
|
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 number of cubic centimeters in the volume of the fitted bowling ball. Express your answer in terms of $\pi$.
|
gpt
|
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 \[\pi\left(\frac{3}{2}\right)^2(8)=9\cdot 2 \pi = 18\pi\] cubic cm. Post hole-removal the fitted ball has volume \[4500\pi - 2\cdot 8\pi - 18\pi = \boxed{4466\pi}\] cubic cm.
|
human
|
The measures of angles $A$ and $B$ are both positive, integer numbers of degrees. The measure of angle $A$ is a multiple of the measure of angle $B$, and angles $A$ and $B$ are complementary angles. How many measures are possible for angle $A$?
|
gpt
|
The given information tells us that $A = 90^\circ -B$ and $A=kB$ for some $k\ge1$. Therefore, we have $kB = 90^\circ - B$. This simplifies to $(k+1)B=90^\circ$. $k+1$ can be any factor of $90$ except one, since $k+1\ge2$. $90=2\cdot3^2\cdot5$ has $2\cdot3\cdot2=12$ factors, so there are 11 possible values of $k$. Each value of $k$ uniquely determines the value of $B$ and therefore the value of $A$, so there are $\boxed{11}$ possible measures for $A$.
|
human
|
Segment $AB$ has midpoint $C$, and segment $BC$ has midpoint $D$. Semi-circles are constructed with diameters $\overline{AB}$ and $\overline{BC}$ to form the entire region shown. Segment $CP$ splits the region into two sections of equal area. What is the degree measure of angle $ACP$? Express your answer as a decimal to the nearest tenth.
[asy]
draw((0,0)--10dir(180),linewidth(2));
draw((0,0)--10dir(67.5),linewidth(2));
draw((0,0)--10dir(0),dashed);
draw(10dir(180)..10dir(90)..10dir(0),linewidth(2));
draw((5,0)+5dir(180)..(5,0)+5dir(-90)..(5,0)+5dir(0),linewidth(2));
dot((0,0));
dot((5,0));
label("A",10dir(180),W);
label("B",10dir(0),E);
label("C",(0,0),SW);
label("D",5dir(0),NE);
label("P",10dir(67.5),NE);
[/asy]
|
gpt
|
The semi-circle with diameter BC has radius $\frac{1}{2}$ that of the semi-circle with diameter AB, and thus, has $\frac{1}{4}$ of the area. (Area of a circle $= \pi \times r^2$ - thus, if $r$ is half as large, that will be squared in the process). Therefore, the sum of their areas represents $\frac{5}{8}$ of a circle with diameter AB, and since the line CP splits this area exactly in half, that area would be $\frac{5}{16}$ of a circle with diameter AB. Therefore, the degree measure of that sector is $360 \times \frac{5}{16} = \boxed{112.5}$
|
human
|
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?
|
gpt
|
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 $\sqrt{25} = 5$ times as long as the side lengths of the smaller triangle. Therefore, the sum of the lengths of the legs of the larger triangle is $5(3+4) = \boxed{35}$.
Proof that the only possibility for the smaller triangle is that it is a 3-4-5 triangle: Let's call the legs of the smaller triangle $a$ and $b$ (with $b$ being the longer leg) and the hypotenuse of the smaller triangle $c$. Similarly, let's call the corresponding legs of the larger triangle $A$ and $B$ and the hypotenuse of the larger triangle $C$. Since the area of the smaller triangle is 6 square inches, we can say $$\frac{1}{2}ab=6.$$ Additionally, we are told that the hypotenuse of the smaller triangle is 5 inches, so $c=5$ and $$a^2+b^2=25.$$ Because $\frac{1}{2}ab=6$, we get $ab=12$ or $a=\frac{12}{b}$. We can now write the equation in terms of $b$. We get \begin{align*}
a^2+b^2&=25\\
\left(\frac{12}{b}\right)^{2}+b^2&=25\\
12^2+b^4&=25b^2\\
b^4-25b^2+144&=0.
\end{align*} Solving for $b$, we get $$b^4-25b^2+144=(b-4)(b+4)(b-3)(b+3)=0.$$ Since we said that $b$ is the longer of the two legs, $b=4$ and $a=3$. Therefore, the triangle must be a 3-4-5 right triangle.
|
human
|
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("C",C,S); label("D",D,S);
draw(rightanglemark(A,B,C,8),linewidth(0.7));
[/asy]
|
gpt
|
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 $\frac{[ABD]}{[ADC]} = \frac{\frac 12 \cdot BD \cdot h}{\frac 12 \cdot DC \cdot h} = \frac{BD}{DC} = \frac{4}{3}$. Thus, $[ADC] = \frac 34 [ABD] = \frac 34 \cdot 24 = \boxed{18}$.
|
human
|
A triangle in a Cartesian coordinate plane has vertices (5, -2), (10, 5) and (5, 5). How many square units are in the area of the triangle? Express your answer as a decimal to the nearest tenth.
|
gpt
|
Plotting the given points in a coordinate plane, we find that the triangle is a right triangle whose legs have length $5-(-2)=7$ and $10-5=5$ units. The area of the triangle is $\frac{1}{2}(\text{base})(\text{height})=\frac{1}{2}(7)(5)=\boxed{17.5}$ square units.
[asy]
defaultpen(linewidth(0.7)+fontsize(8));
dotfactor = 4;
draw((-1,0)--(10,0),Arrows(4));
draw((0,-4)--(0,10),Arrows(4));
pair A=(5,-2), B=(10,5), C=(5,5);
pair[] dots = {A,B,C};
dot(dots);
draw(A--B--C--cycle);
label(rotate(90)*"$5-(-2)$",(0,0.2)+(A+C)/2,W);
label("$10-5$",(B+C)/2,N);
label("$(5,-2)$",A,S);
label("$(10,5)$",B,NE);
label("$(5,5)$",C,NW);
[/asy]
|
human
|
Two congruent cones with radius 12 cm and height 12 cm are enclosed within a cylinder. The base of each cone is a base of the cylinder, and the height of the cylinder is 24 cm. What is the number of cubic centimeters in the volume of the cylinder not occupied by the cones? Express your answer in terms of $\pi$.
|
gpt
|
The cylinder has volume $\pi (12)^2 (24)$ cubic cm. Each cone has volume $(1/3)\pi (12)^2(12)$ cubic cm. Hence the volume of the space in the cylinder not occupied by the cones is \begin{align*}
\pi (12)^2 (24) - (2)(1/3)\pi (12)^2(12) &= 12^3\pi(2-2/3)\\
&=12^3\pi(4/3)\\
&=\boxed{2304\pi} \text{ cubic cm}.
\end{align*}
|
human
|
Compute $\sin 150^\circ$.
|
gpt
|
Let $P$ be the point on the unit circle that is $150^\circ$ 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= (0,0);
label("$x$",(1.2,0),SE);
label("$y$",(0,1.2),NE);
P = rotate(150)*A;
D = foot(P,A,-A);
draw(O--P--D);
draw(rightanglemark(O,D,P,2));
draw(Circle(O,1));
label("$O$",O,SE);
label("$P$",P,NW);
//label("$A$",A,SE);
label("$D$",D,S);
[/asy]
Triangle $POD$ is a 30-60-90 triangle, so $DO = \frac{\sqrt{3}}{2}$ and $DP = \frac12$. Therefore, the coordinates of $P$ are $\left(-\frac{\sqrt{3}}{2}, \frac12\right)$, so $\sin 150^\circ = \boxed{\frac{1}{2}}$.
|
human
|
The lengths of the sides of a non-degenerate triangle are $x$, 13 and 37 units. How many integer values of $x$ are possible?
|
gpt
|
By the triangle inequality, \begin{align*}
x + 13 &> 37, \\
x + 37 &> 13, \\
13 + 37 &> x,
\end{align*} which tell us that $x > 24$, $x > -24$, and $x < 50$. Hence, the possible values of $x$ are $25, 26, \dots, 49$, for a total of $49 - 25 + 1 = \boxed{25}$.
|
human
|
In the diagram below, we have $AB = 24$ and $\angle ADB =90^\circ$. If $\sin A = \frac23$ and $\sin C = \frac13$, then what is $DC$?
[asy]
pair A,B,C,D;
A = (0,0);
B = (8*sqrt(5),16);
D = (8*sqrt(5),0);
C = (8*sqrt(5) + 32*sqrt(2),0);
draw(D--B--A--C--B);
label("$A$",A,SW);
label("$B$",B,N);
label("$C$",C,SE);
label("$D$",D,S);
draw(rightanglemark(B,D,A,63));
[/asy]
|
gpt
|
From right triangle $ABD$, we have $\sin A = \frac{BD}{AB} = \frac{BD}{24}$. Since $\sin A = \frac23$, we have $\frac23 = \frac{BD}{24}$, so $BD = \frac23\cdot 24 = 16$.
From right triangle $BCD$, we have $\sin C = \frac{BD}{BC}=\frac{16}{BC}$. Since $\sin C = \frac13$, we have $\frac{16}{BC} = \frac13$. Therefore, we have $BC = 3\cdot 16=48$. Finally, the Pythagorean Theorem gives us \begin{align*}
CD &= \sqrt{BC^2 - BD^2}\\
&= \sqrt{48^2 - 16^2} \\
&= \sqrt{(3\cdot 16)^2 - 16^2} \\
&= \sqrt{9\cdot 16^2 - 16^2} = \sqrt{8\cdot 16^2} = 2\cdot 16 \sqrt{2} = \boxed{32\sqrt{2}}.\end{align*}
|
human
|
A cone is formed from a 300-degree sector of a circle of radius 18 by aligning the two straight sides. [asy]
size(110);
draw(Arc((0,0),1,0,300));
draw((1,0)--(0,0)--(.5,-.5*sqrt(3)));
label("18",(.5,0),S); label("$300^\circ$",(0,0),NW);
[/asy] What is the result when the volume of the cone is divided by $\pi$?
|
gpt
|
A full circle with radius 18 has circumference $2(\pi)(18)=36\pi$, so a 300-degree sector has arc length (shown in blue below) \[\frac{300^\circ}{360^\circ}\cdot 36\pi = 30\pi.\][asy]
size(110);
draw(Arc((0,0),1,0,300),heavycyan);
draw(Arc((0,0),1,300,360),linetype("2 4"));
draw((1,0)--(0,0)--(.5,-.5*sqrt(3)));
label("18",(.5,0),S); label("$300^\circ$",(0,0),NW);
[/asy]
When we fold the sector into a cone, the arc length of the sector becomes the circumference of the base of the cone, and the radius of the sector becomes the slant height of the cone.
[asy]
size(100);
import geometry;
draw(scale(1,.2)*arc((0,0),3,0,180),heavycyan);
draw(scale(1,.2)*arc((0,0),3,180,360),heavycyan);
draw((3,0.05)--(0,2)); label("18", (3,0.05)--(0,2), NE);
draw((0,2)--(-3,0.05),heavycyan+linetype("2 4"));
draw((0,2)--(0,0)--(3,0)); label("$h$",(0,1),W); label("$r$",(1.5,0),S);
[/asy]
Let the cone that is formed have height $h$ and radius $r$. Thus we have \[2\pi r = 30\pi\]and \[r^2+h^2=18^2\]From the first equation we have $r=15$; from the second equation we have $h=\sqrt{18^2-15^2}=\sqrt{99}=3\sqrt{11}$.
Finally, the desired volume is \[\frac{1}{3}r^2h\pi = \frac{1}{3}(15^2)(3\sqrt{11})\pi = {225\pi\sqrt{11}}.\]So, dividing the volume by $\pi$ gives $\boxed{225\sqrt{11}}$.
|
human
|
Triangle $\triangle ABC$ has a right angle at $C$, $\angle A = 60^\circ$, and $AC=10$. Find the radius of the incircle of $\triangle ABC$.
|
gpt
|
We begin by drawing a diagram:
[asy]
size(100);
pair A,B,C;
real x = sqrt(3);
C=(0,0); A=(10,0); B=(0,10*x);
draw(A--B--C--cycle);
draw(rightanglemark(B,C,A,30));
label("$A$",A,SE); label("$C$",C,SW); label("$B$",B,NW); label("10",(A+C)/2,S);
real r = 5*sqrt(3) - 5;
draw(Circle((r,r),r));
[/asy]
Since $\angle A = 60^\circ$, we have $\angle B = 180^\circ - 90^\circ - 60^\circ = 30^\circ$. Then $\triangle ABC$ is a $30 - 60 - 90$ triangle, so $BC=AC\sqrt{3}=10\sqrt{3}$ and $AB=2AC=20$. We can compute the area of $\triangle ABC$ as \[ [\triangle ABC] = \frac{1}{2}(AC)(BC)=\frac{1}{2}(10)(10\sqrt{3}) = 50\sqrt{3}.\]Let the incircle of $\triangle ABC$ have radius $r$. A triangle with inradius $r$ and semiperimeter $s$ has \[\text{area} = rs,\]so we have \[ [\triangle ABC] = r \left( \frac{10+10\sqrt{3}+20}{2} \right) = r(15+5\sqrt{3}).\]Setting these two area expressions equal gives \[50\sqrt{3}=r(15+5\sqrt{3}).\]Solving for $r$ gives \[r = \frac{10\sqrt{3}}{3+\sqrt{3}} = \frac{10\sqrt{3}(3-\sqrt{3})}{9-3} = \boxed{5(\sqrt{3}-1)}.\]
|
human
|
The set of all solutions of the system $$
\begin{cases}
x+y\leq 3 \\
2x+y\geq 2 \\
x\geq 0 \\
y\geq 0
\end{cases}
$$ is a quadrilateral region. Find the number of units in the length of the longest side. Express your answer in simplest radical form.
|
gpt
|
[asy]
Label f;
f.p=fontsize(6);
xaxis(0,4,Ticks(f, 1.0));
yaxis(0,4,Ticks(f, 1.0));
fill((0,2)--(0,3)--(3,0)--(1,0)--cycle, grey);
draw((-.5,3)--(1.5,-1), dashed, Arrows);
draw((-1,4)--(4,-1), dashed, Arrows);
[/asy] The upper diagonal line is the graph of $x+y=3.$ The lower diagonal line is the graph of $2x+y=2.$ The $y$-axis is the graph of $x=0$ and the $x$-axis is the graph of $y=0.$ The shaded region includes the solutions to the system. The longest side is the upper diagonal side. The length of this side is $\boxed{3\sqrt{2}}.$
|
human
|
If a triangle has two sides of lengths 5 and 7 units, then how many different integer lengths can the third side be?
|
gpt
|
Let $n$ be the length of the third side. Then by the triangle inequality, \begin{align*}
n + 5 &> 7, \\
n + 7 &> 5, \\
5 + 7 &> n,
\end{align*} which tell us that $n > 2$, $n > -2$, and $n < 12$. Hence, the possible values of $n$ are 3, 4, 5, 6, 7, 8, 9, 10, and 11, for a total of $\boxed{9}$.
|
human
|
In $\triangle XYZ$, we have $\angle X = 90^\circ$ and $\tan Y = \frac34$. If $YZ = 30$, then what is $XY$?
|
gpt
|
[asy]
pair X,Y,Z;
X = (0,0);
Y = (16,0);
Z = (0,12);
draw(X--Y--Z--X);
draw(rightanglemark(Y,X,Z,23));
label("$X$",X,SW);
label("$Y$",Y,SE);
label("$Z$",Z,N);
label("$30$",(Y+Z)/2,NE);
label("$3k$",(Z)/2,W);
label("$4k$",Y/2,S);
[/asy]
Since $\triangle XYZ$ is a right triangle with $\angle X = 90^\circ$, we have $\tan Y = \frac{XZ}{XY}$. Since $\tan Y = \frac34$, we have $XZ = 3k$ and $XY = 4k$ for some value of $k$, as shown in the diagram. Therefore, $\triangle XYZ$ is a 3-4-5 triangle. Since the hypotenuse has length $30 = 5\cdot 6$, the legs have lengths $XZ = 3\cdot 6 = 18$ and $XY = 4\cdot 6 = \boxed{24}$.
|
human
|
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 $x$?
|
gpt
|
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);
draw(A--B--C--cycle);
xaxis(-2,10,Arrows(4));
yaxis(-2,20,Arrows(4));
label("$(x,0)$",C,S);
label("$(x,2x)$",B,N);
[/asy]
|
human
|
In triangle $PQR$, we have $\angle P = 90^\circ$, $QR = 20$, and $\tan R = 4\sin R$. What is $PR$?
|
gpt
|
[asy]
pair P,Q,R;
P = (0,0);
Q = (5*sqrt(15),0);
R = (0,5);
draw(P--Q--R--P);
draw(rightanglemark(Q,P,R,18));
label("$P$",P,SW);
label("$Q$",Q,SE);
label("$R$",R,N);
label("$20$",(R+Q)/2,NE);
[/asy]
We have $\tan R = \frac{PQ}{PR}$ and $\sin R = \frac{PQ}{RQ} = \frac{PQ}{20}$, so $\tan R = 4\sin R$ gives us $\frac{PQ}{PR} = 4\cdot \frac{PQ}{20} = \frac{PQ}{5}$. From $\frac{PQ}{PR} = \frac{PQ}{5}$, we have $PR = \boxed{5}$.
|
human
|
How many non-similar triangles have angles whose degree measures are distinct positive integers in arithmetic progression?
|
gpt
|
Let $n-d$, $n$, and $n+d$ be the angles in the triangle. Then \[
180 = n-d+n+n+d= 3n, \quad \text{so} \quad n=60.
\] Because the sum of the degree measures of two angles of a triangle is less than 180, we have $$180 > n + (n+d) = 120 + d,$$ which implies that $0<d<60$.
There are $\boxed{59}$ triangles with this property.
|
human
|
$ABCD$ is a regular tetrahedron (right triangular pyramid). If $M$ is the midpoint of $\overline{CD}$, then what is $\cos \angle AMB$?
|
gpt
|
The tetrahedron is shown below. In order to find $\cos \angle AMB$, we build a right triangle with $\angle AMB$ among its angles. The foot of the altitude from $A$ to face $BCD$ is the centroid, $G$, of triangle $BCD$.
[asy]
import three;
currentprojection = orthographic(1.5,1.1,-1);
triple A = (1,1,1);
triple B = (1,0,0);
triple C = (0,1,0);
triple D = (0,0,1);
draw(A--B--C--A);
draw(A--D,dashed);
draw(C--D--B,dashed);
label("$A$",A,NW);
label("$B$",B,W);
label("$C$",C,S);
label("$D$",D,NW);
triple M = (0,0.5,0.5);
draw(A--M--B,dashed);
label("$M$",M,NE);
triple G = B/3 + 2*M/3;
draw(A--G,dashed);
label("$G$",G,S);
[/asy]
Since $\overline{BM}$ is a median of $\triangle BCD$, point $G$ is on $\overline{BM}$ such that $GM = \frac13BM$. Furthermore, we have $AM = BM$, so \[\cos \angle AMB= \cos \angle AMG = \frac{GM}{AM} = \frac{(BM/3)}{BM}=\boxed{\frac{1}{3}}.\]
|
human
|
A right pyramid with a square base has total surface area 432 square units. The area of each triangular face is half the area of the square face. What is the volume of the pyramid in cubic units?
|
gpt
|
Let $ABCD$ be the base of the pyramid and let $P$ be the pyramid's apex.
[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=(B+C)/2;
draw(P--F--M,dashed);
draw(P--M);
label("$F$",F,S);
label("$M$",M,SW);
[/asy]
Let $F$ be the center of the square base and $M$ be the midpoint of an edge of the square, as shown. There are four triangular faces, each with area half the area of the square face. So, the total surface area of the pyramid is 3 times the area of the square face. Therefore, the area of the square face is $432/3=144$ square units, which means that each side of the square has length 12.
Since the area of the triangle is half the area of the square, we have $(BC)(PM)/2 = 72$, so $(BC)(PM) = 144$, which means $PM = 144/12 = 12$. Since $F$ is the center of the square base, we have $FM = 6$, so $PF = \sqrt{12^2 - 6^2} = 6\sqrt{3}$. Finally, the volume of the pyramid is \[\frac{[ABCD]\cdot PF}{3} = \frac{144\cdot 6\sqrt{3}}{3} = \boxed{288\sqrt{3}}.\]
|
human
|
What is the sum of the squares of the lengths of the $\textbf{medians}$ of a triangle whose side lengths are $10,$ $10,$ and $12$?
|
gpt
|
Let us draw our triangle and medians and label our points of interest: [asy]
pair A, B, C, D, E, F;
A = (0, 8);
B = (-6, 0);
C = (6, 0);
D = (0, 0);
E = (3, 4);
F = (-3, 4);
draw(A--B--C--cycle);
draw(A--D);
draw(B--E);
draw(C--F);
label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, S);
label("$E$", E, NE);
label("$F$", F, NW);
[/asy] We have made $AB = AC = 10$ and $BC = 12.$ We can notice a few useful things. Since $ABC$ is isosceles, it follows that $AD$ is an altitude as well as a median, which is useful for finding lengths, since it means we can use the Pythagorean Theorem. At this point, we can drop additional segments from $E$ and $F$ perpendicular to $BC,$ meeting $BC$ at $G$ and $H,$ respectively: [asy]
pair A, B, C, D, E, F, G, H;
A = (0, 8);
B = (-6, 0);
C = (6, 0);
D = (0, 0);
E = (3, 4);
F = (-3, 4);
G = (3, 0);
H = (-3, 0);
draw(A--B--C--cycle);
draw(A--D);
draw(B--E);
draw(C--F);
draw(E--G, dotted);
draw(F--H, dotted);
draw(D + (-0.4, 0) -- D + (-0.4, 0.4) -- D + (0, 0.4));
draw(G + (-0.4, 0) -- G + (-0.4, 0.4) -- G + (0, 0.4));
draw(H + (-0.4, 0) -- H + (-0.4, 0.4) -- H + (0, 0.4));
label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, S);
label("$E$", E, NE);
label("$F$", F, NW);
label("$G$", G, S);
label("$H$", H, S);
[/asy] Since $DC = 6$ and $AC = 10,$ we have a $3:4:5$ Pythagorean triple and $AD = 8$. Since $\triangle BFH \sim \triangle BAD$ and $BF = \frac{1}{2} \cdot AB$ (since F is the midpoint of AB), we can see that $FH = \frac{1}{2} \cdot AD = 4$ and $BH = \frac{1}{2} \cdot BD = \frac{1}{4} \cdot BC = 3.$ $HC = BC - BH = 12 - 3 = 9.$
To find $CF^2,$ we simply use the Pythagorean Theorem: $CF^2 = FH^2 + HC^2 = 16 + 81 = 97.$ By symmetry, we can see that $BE^2 = 97.$ From before, we have that $AD^2 = 8^2 = 64.$ Our answer is therefore $AD^2 + BE^2 + CF^2 = 64 + 97 + 97 = \boxed{258}.$
|
human
|
How many non-congruent triangles with only integer side lengths have a perimeter of 15 units?
|
gpt
|
In a triangle, the lengths of any two sides must add up to a value larger than the third length's side. This is known as the Triangle Inequality. Keeping this in mind, we list out cases based on the length of the shortest side.
Case 1: shortest side has length $1$. Then the other two sides must have lengths $7$ and $7$. This leads to the set $\{1,7,7\}$.
Case 2: shortest side has length $2$. Then the other two sides must have lengths $6$ and $7$. This leads to the set $\{2,6,7\}$.
Case 3: shortest side has length $3$. Then the other two sides can have lengths $6$ and $6$ or $5$ and $7$. This leads to the sets $\{3,6,6\}$ and $\{3,5,7\}$.
Case 4: shortest side has length $4$. Then the other two sides can have lengths $5$ and $6$ or $4$ and $7$. This leads to the sets $\{4,5,6\}$ and $\{4,4,7\}$.
Case 5: shortest side has length $5$. Then the other two sides must have lengths $5$ and $5$. This leads to the set $\{5,5,5\}$.
Hence there are $\boxed{7}$ sets of non-congruent triangles with a perimeter of $15$ units.
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.