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human
|
In triangle $VWX$ shown, $VX = \sqrt{13}$ and $VW = 3$. What is $\tan V$?
[asy]
pair V,W,X;
W = (0,0);
V = (0,3);
X = (2,0);
draw(V--W--X--V);
draw(rightanglemark(V,W,X,7));
label("$W$",W,SW);
label("$V$",V,NW);
label("$X$",X,SE);
label("$3$",(V+W)/2,NW);
label("$\sqrt{13}$",(V+X)/2,NE);
[/asy]
|
gpt
|
Because $\triangle VWX$ is a right triangle, $\tan V = \frac{WX}{VW}$.
By the Pythagorean Theorem, $$WX = \sqrt{VX^2 - WX^2} = \sqrt{13 - 9} = \sqrt{4} = 2.$$Then $\tan V = \boxed{\frac{2}{3}}$.
|
human
|
An isosceles triangle has side lengths 8 cm, 8 cm and 10 cm. The longest side of a similar triangle is 25 cm. What is the perimeter of the larger triangle, in centimeters?
|
gpt
|
The ratio of the length of the longest sides of the small triangle to the large triangle is $10/25 = 2/5$, which must hold constant for all sides of the two triangles since they are similar. Thus the perimeters of the two triangles are also in the ratio of $2/5$. The small triangle has perimeter $8+8+10=26$, so the large triangle has perimeter $\frac{5}{2}\cdot 26 = \boxed{65}$.
|
human
|
Compute $\sin 315^\circ$.
|
gpt
|
Let $P$ be the point on the unit circle that is $315^\circ$ 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(315)*A;
D = foot(P,A,-A);
draw(O--P--D);
draw(rightanglemark(O,D,P,2));
draw(Circle(O,1));
label("$O$",O,NW);
label("$P$",P,SE);
//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 $\sin 315^\circ = \boxed{-\frac{\sqrt{2}}{2}}$.
|
human
|
Find $AX$ in the diagram.
[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("$24$",.5*(B+X),SE);
label("$28$",.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 \[AX=\frac{AC\cdot BX}{BC}=\frac{21\cdot24}{28}=\frac{7\cdot3\cdot6\cdot 4}{7\cdot4}=\boxed{18}.\]
|
human
|
In the diagram below, points $A$, $B$, $C$, and $P$ are situated so that $PA=2$, $PB=3$, $PC=4$, and $BC=5$. What is the maximum possible area of $\triangle ABC$? [asy]
defaultpen(linewidth(0.8)); size(150);
pair B = (0,0), C = (5,0), A = (2,3), P = (2.2,2);
draw(A--B--C--cycle^^B--P^^C--P^^A--P);
label("$A$",A,N); label("$B$",B,S); label("$C$",C,S); label("$P$",P,S);
[/asy]
|
gpt
|
We first observe that by the Pythagorean theorem $\triangle PBC$ must be a right triangle with right angle at $P$, since $PB=3$, $PC=4$, and $BC=5$.
$[\triangle PBC]=\frac{1}{2}(3)(4) = 6=\frac{1}{2}(PH)(5)$. Hence, the altitude $\overline{PH}$ from $P$ to $\overline{BC}$ has length $\frac{12}{5}$. Let $h$ be the length of the altitude from $A$ to $\overline{BC}$. Then $[\triangle ABC] = \frac{1}{2}(h)(5)$, so the area is maximized when $A$ is most high above $\overline {BC}$. Since $AP=2$, maximization occurs when $A$ is directly over $P$, leading to a height of $h=\frac{12}{5}+2=\frac{22}{5}$. In this case, \[[\triangle ABC] = \frac{1}{2} \left( \frac{22}{5} \right)(5)=\boxed{11}.\]
|
human
|
A right triangle with legs $10$ and $8$ is similar to another right triangle with corresponding legs $x$ and $5$, respectively. What is the value of $x$? Express your answer as a decimal to the nearest hundredth.
|
gpt
|
By similarity, we know that $\frac{10}{x} = \frac{8}{5}$, so therefore $x = \frac{50}{8} = \boxed{6.25}$.
|
human
|
Pentagon ABCDE has a vertical line of symmetry. What is the $y$-coordinate of vertex C so that the area of the pentagon is 40 square units? [asy]
unitsize(2mm);
defaultpen(linewidth(.7pt)+fontsize(8pt));
dotfactor=4;
pair A=(0,0), B=(0,4), C=(2,16), D=(4,4), E=(4,0);
pair[] dots={A,B,C,D,E};
draw(B--C--D--E);
dot(dots);
axes(Arrows(4));
label("A(0,0)",A,SW);
label("E(4,0)",E,SE);
label("D(4,4)",D,NE);
label("C",C,NE);
label("B(0,4)",B,NW);
[/asy]
|
gpt
|
Write the area of pentagon $ABCDE$ as sum the areas of square $ABDE$ and triangle $BCD$. Since square $ABDE$ has area $4^2=16$ square units, triangle $BCD$ has area $40-16=24$ square units. If $h$ is the $y$-coordinate of point $C$, the height of triangle $BCD$ is $h-4$ units and its base is $4$ units. Solving $\frac{1}{2}(4)(h-4)=24$, we find $h=\boxed{16}$.
|
human
|
Altitudes $\overline{AD}$ and $\overline{BE}$ of $\triangle ABC$ intersect at $H$. If $\angle BAC = 46^\circ$ and $\angle ABC = 71^\circ$, then what is $\angle AHB$?
[asy]
size(150); defaultpen(linewidth(0.8));
pair B = (0,0), C = (3,0), A = (1.8,2), P = foot(A,B,C), Q = foot(B,A,C),H = intersectionpoint(B--Q,A--P);
draw(A--B--C--cycle);
draw(A--P^^B--Q);
label("$A$",A,N); label("$B$",B,W); label("$C$",C,E); label("$D$",P,S); label("$E$",Q,E); label("$H$",H,NW);
draw(rightanglemark(C,P,H,3.5));
draw(rightanglemark(H,Q,C,3.5));
[/asy]
|
gpt
|
First, we build a diagram:
[asy]
size(150); defaultpen(linewidth(0.8));
pair B = (0,0), C = (3,0), A = (1.8,2), P = foot(A,B,C), Q = foot(B,A,C),H = intersectionpoint(B--Q,A--P);
draw(A--B--C--cycle);
draw(A--P^^B--Q);
label("$A$",A,N); label("$B$",B,W); label("$C$",C,E); label("$D$",P,S); label("$E$",Q,E); label("$H$",H,NW);
draw(rightanglemark(C,P,H,3.5));
draw(rightanglemark(H,Q,C,3.5));
[/asy]
We have $\angle AHB = \angle DHE$, and from quadrilateral $CDHE$, we have \begin{align*}
\angle DHE &= 360^\circ - \angle HEC - \angle ECD - \angle CDH \\
&= 360^\circ - 90^\circ - \angle ACB - 90^\circ\\
&= 180^\circ - \angle ACB.
\end{align*} From triangle $ABC$, we have $180^\circ - \angle ACB = \angle BAC + \angle ABC = 46^\circ + 71^\circ = \boxed{117^\circ}$
|
human
|
What is the total volume in cubic feet of three boxes if each box is a cube with edge length 4 feet?
|
gpt
|
Each box has volume $4^3=64$ cubic feet. Thus, three boxes have volume $64\cdot3=\boxed{192}$ cubic feet.
|
human
|
The endpoints of a diameter of circle $M$ are $(-1,-4)$ and $(-7,6)$. What are the coordinates of the center of circle $M$? Express your answer as an ordered pair.
|
gpt
|
The midpoint of a diameter of a circle is its center. Thus, $M$ is the midpoint of the segment from $(-1,-4)$ to $(-7,6),$ so $M$ has coordinates \[\left(\frac{-1+(-7)}{2},\frac{-4+6}{2} \right)=\boxed{(-4,1)}.\]
|
human
|
Compute $\sin 45^\circ$.
|
gpt
|
Let $P$ be the point on the unit circle that is $45^\circ$ 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(45)*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,NE);
//label("$A$",A,SE);
label("$D$",D,S);
[/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 $\sin 45^\circ = \boxed{\frac{\sqrt{2}}{2}}$.
|
human
|
Compute $\cos 270^\circ$.
|
gpt
|
Rotating the point $(1,0)$ about the origin by $270^\circ$ counterclockwise gives us the point $(0,-1)$, so $\cos 270^\circ = \boxed{0}$.
|
human
|
The rectangle with vertices $(-1, y), (7, y), (-1, 3)$, and $(7, 3)$ has an area of 72 square units, and $y$ is positive. What is the value of $y$?
|
gpt
|
First we write the rectangle's side lengths in terms of the coordinates provided. The length is $7-(-1)=8$ and the height is $y-3.$ It follows that $8(y-3)=72,$ and $y=\boxed{12}.$ [asy]
import graph;
size(4cm);
defaultpen(linewidth(0.7)+fontsize(10));
dotfactor=4;
xaxis(Arrows(4));
yaxis(ymin=-2,Arrows(4));
pair A=(-1,12), B=(7,12), C=(-1,3), D=(7,3);
pair[] dots = {A,B,C,D};
dot(dots);
draw(A--B--D--C--cycle);
label("$8$",(A+B)/2,N);
label("$y-3$",(B+D)/2,E);
[/asy]
|
human
|
In rectangle $ABCD$, $P$ is a point on $BC$ so that $\angle APD=90^{\circ}$. $TS$ is perpendicular to $BC$ with $BP=PT$, as shown. $PD$ intersects $TS$ at $Q$. Point $R$ is on $CD$ such that $RA$ passes through $Q$. In $\triangle PQA$, $PA=20$, $AQ=25$ and $QP=15$. [asy]
size(7cm);defaultpen(fontsize(9));
real sd = 7/9 * 12;
path extend(pair a, pair b) {return a--(10 * (b - a));}
// Rectangle
pair a = (0, 0); pair b = (0, 16); pair d = (24 + sd, 0); pair c = (d.x, b.y);
draw(a--b--c--d--cycle);
label("$A$", a, SW);label("$B$", b, NW);label("$C$", c, NE);label("$D$", d, SE);
// Extra points and lines
pair q = (24, 7); pair s = (q.x, 0); pair t = (q.x, b.y);
pair r = IP(c--d, extend(a, q));
pair p = (12, b.y);
draw(q--a--p--d--r--cycle);draw(t--s);
label("$R$", r, E); label("$P$", p, N);label("$Q$", q, 1.2 * NE + 0.2 * N);label("$S$", s, S); label("$T$", t, N);
// Right angles and tick marks
markscalefactor = 0.1;
draw(rightanglemark(a, b, p)); draw(rightanglemark(p, t, s)); draw(rightanglemark(q, s, d));draw(rightanglemark(a, p, q));
add(pathticks(b--p, 2, spacing=3.4, s=10));add(pathticks(p--t, 2, spacing=3.5, s=10));
// Number labels
label("$16$", midpoint(a--b), W); label("$20$", midpoint(a--p), NW); label("$15$", midpoint(p--q), NE);
label("$25$", midpoint(a--q), 0.8 * S + E);
[/asy] Find the lengths of $BP$ and $QT$.
When writing your answer, first write the length of $BP$, then a comma, and then the length of $QT$. For example, if you find that these lengths are $5$ and $3/4$, respectively, your final answer should be written "5,3/4" (without the quotes).
|
gpt
|
Since $\angle ABP=90^{\circ}$, $\triangle ABP$ is a right-angled triangle. By the Pythagorean Theorem, $$BP^2=AP^2-AB^2=20^2-16^2=144$$ and so $BP=12$, since $BP>0$.
Since $\angle QTP=90^{\circ}$, $\triangle QTP$ is a right-angled triangle with $PT=12$. Since $PT=BP=12$, then by the Pythagorean Theorem, $$QT^2=QP^2-PT^2=15^2-12^2 = 81$$ and so $QT=9$, since $QT>0$.
Our final answer is then $\boxed{12,9}$.
|
human
|
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);
label("D",D,NW);
label("9",midpoint(C--D),N);
label("10",midpoint(D--A),NW);
label("21",midpoint(A--B),S);
label("10",midpoint(B--C),NE);
[/asy]
|
gpt
|
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.
[asy]
unitsize(1.5mm);
defaultpen(linewidth(.7pt)+fontsize(10pt));
dotfactor=3;
pair A=(0,0), B=(21,0), C=(15,8), D=(6,8), E=(15,0), F=(6,0);
pair[] dots={A,B,C,D,E,F};
draw(A--B--C--D--cycle);
draw(C--E);
dot(dots);
label("A",A,SW);
label("B",B,SE);
label("C",C,NE);
label("D",D,NW);
label("E",E,S);
label("F",F,S);
label("9",midpoint(C--D),N);
label("10",midpoint(D--A),NW);
label("21",midpoint(A--B)+(0,-2),S);
label("10",midpoint(B--C),NE);[/asy]
|
human
|
What is the smallest possible perimeter, in units, of a triangle whose side-length measures are consecutive integer values?
|
gpt
|
The smallest such triangle has lengths 1, 2, and 3. However, this triangle doesn't work since the sum of any two side lengths must be greater than the third side length (by the Triangle Inequality). The next smallest triangle has lengths 2, 3, and 4, which works. Thus, the smallest possible perimeter is $2+3+4=\boxed{9}$ units.
|
human
|
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?
|
gpt
|
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 $\pi \cdot (r+6)^2 \cdot 2 = \pi \cdot r^2 \cdot 8$. Simplifying, we have $(r+6)^2=4r^2 \Rightarrow r^2+12r+36=4r^2 \Rightarrow r^2-4r-12=0$. Factoring, we have $(r-6)(r+2)=0$, so $r=\boxed{6}$ (it cannot be negative because it is a physical length).
|
human
|
Two sectors of a circle of radius $12$ overlap as shown, with $P$ and $R$ as the centers of the respective circles. Determine the area of the shaded region.
[asy]
draw((0,0)--(10.3923,-6)--(20.7846,0)--(10.3923,6)--cycle,black+linewidth(1));
filldraw((10.3923,6)..(12,0)..(10.3923,-6)--cycle,gray,black+linewidth(1));
filldraw((10.3923,6)..(8.7846,0)..(10.3923,-6)--cycle,gray,black+linewidth(1));
label("$P$",(0,0),W);
label("$Q$",(10.3923,6),N);
label("$R$",(20.7846,0),E);
label("$S$",(10.3923,-6),S);
label("$60^\circ$",(0,0),2E);
label("$60^\circ$",(20.7846,0),2W);
[/asy]
|
gpt
|
By symmetry, the areas of the two parts of the shaded region are equal. Consider the right part of the shaded region and the left triangle.
[asy]
draw((0,0)--(10.3923,-6)--(10.3923,6)--cycle,black+linewidth(1));
filldraw((10.3923,6)..(12,0)..(10.3923,-6)--cycle,gray,black+linewidth(1));
draw((0,0)--(10.3923,0),black+linewidth(1));
draw((10.3923,0)--(9.3923,0)--(9.3923,1)--(10.3923,1),black+linewidth(1));
label("$P$",(0,0),W);
label("$Q$",(10.3923,6),N);
label("$S$",(10.3923,-6),S);
label("$Z$",(10.3923,0),SW);
[/asy]
The shaded area is equal to the area of sector $PQS$ minus the area of triangle $PQS.$
Since $\angle PQS = 60^\circ$ and $PQ = 12,$ the area of sector $PQS$ is
\[\frac{1}{6} \cdot 12^2 \cdot \pi = 24 \pi.\]Also, triangle $PQS$ is equilateral with side length 12, so its area is
\[\frac{\sqrt{3}}{4} \cdot 12^2 = 36 \sqrt{3}.\]Thus, the area of the right part of the shaded region is $24\pi - 36\sqrt{3},$ so the area of the entire shaded region is $$2(24\pi-36\sqrt{3})=\boxed{48\pi-72\sqrt{3}}.$$
|
human
|
The sides of triangle $PQR$ are tangent to a circle with center $C$ as shown. Given that $\angle PQR = 65^\circ$ and $\angle QRC = 30^\circ$, find $\angle QPR$, in degrees.
[asy]
unitsize(1.0 cm);
pair Q, P, R, C;
Q = (2.43,3.46);
P = (0,0);
R = (4.43,0);
C = incenter(Q,P,R);
draw(Q--P--R--cycle);
draw(incircle(Q,P,R));
draw(R--C);
label("$Q$", Q, N);
label("$P$", P, SW);
label("$R$", R, SE);
label("$C$", C, N);
[/asy]
|
gpt
|
The circle with center $C$ is the incircle of $\triangle PQR$. So, any segment from a vertex of the triangle to $C$ is an angle bisector.
From $\angle QRC = 30^\circ$, we find that $\angle QRP = 60^\circ$ because $RC$ is an angle bisector.
The sum of the measures of the internal angles of a triangle is $180^\circ$, so
\begin{align*}
\angle QPR &= 180^\circ - \angle PQR - \angle QRP \\
&= 180^\circ - 65^\circ - 60^\circ.
\end{align*}This yields $\angle QPR = \boxed{55^\circ}$.
|
human
|
The figure shows two concentric circles. If the length of chord AB is 80 units and chord AB is tangent to the smaller circle, what is the area of the shaded region? Express your answer in terms of $\pi$.
[asy]
defaultpen(linewidth(.8pt));
dotfactor=4;
filldraw(circle((0,0),50),gray);
filldraw(circle((0,0),30),white);
dot((0,0));
draw((-40,30)--(40,30));
label("$A$",(-40,30),W);
label("$B$",(40,30),E);
[/asy]
|
gpt
|
Call the point of tangency between the two circles $P$ and the center $O$. [asy]
defaultpen(linewidth(.8pt));
dotfactor=4;
filldraw(circle((0,0),50),gray);
filldraw(circle((0,0),30),white);
draw((-40,30)--(40,30));
draw((0,30)--(0,0)--(-40,30));
label("$P$",(0,30),N);
label("$O$",(0,0),S);
label("$A$",(-40,30),W);
label("$B$",(40,30),E);
[/asy] $\overline{OP}\perp\overline{AB}$, so $\overline{OP}$ bisects $\overline{AB}$. This means $AP=40$. By the Pythagorean Theorem, $AP^2=1600=AO^2-OP^2$. The area of the shaded region is \[
AO^2\pi-OP^2\pi=\pi\left(AO^2-OP^2\right)=\boxed{1600\pi}\text{ square units.}
\]
|
human
|
$\Delta ABC$ is isosceles with $AC = BC$. If $m\angle C = 40^{\circ}$, what is the number of degrees in $m\angle CBD$? [asy] pair A,B,C,D,E;
C = dir(65); B = C + dir(-65); D = (1.5,0); E = (2,0);
draw(B--C--A--E); dot(D);
label("$A$",A,S); label("$B$",B,S); label("$D$",D,S); label("$C$",C,N);
[/asy]
|
gpt
|
Let $x$ be the number of degrees in $\angle ABC$. Since $\triangle ABC$ is isosceles with $AC=BC$, we have $\angle BAC=\angle ABC$.
So, the three interior angles of $\triangle ABC$ measure $x^\circ$, $x^\circ$, and $40^\circ$. The sum of the angles in a triangle is $180^\circ$, so we have $$x+x+40 = 180,$$which we can solve to obtain $x=70$. Finally, $\angle CBD$ is supplementary to angle $\angle ABC$, so \begin{align*}
m\angle CBD &= 180^\circ - m\angle ABC \\
&= 180^\circ - 70^\circ \\
&= \boxed{110}^\circ.
\end{align*}
|
human
|
Find the number of units in the length of diagonal $DA$ of the regular hexagon shown. Express your answer in simplest radical form. [asy]
size(120);
draw((1,0)--(3,0)--(4,1.732)--(3,3.464)--(1,3.464)--(0,1.732)--cycle);
draw((1,0)--(1,3.464));
label("10",(3.5,2.598),NE);
label("$A$",(1,0),SW);
label("$D$",(1,3.464),NW);
[/asy]
|
gpt
|
Label point $X$ as shown below, and let $Y$ be the foot of the perpendicular from $X$ to $AD$. [asy]
size(120);
pair A,B,C,D,E,F;
A = dir(0); B = dir(60); C = dir(120); D = dir(180); E = dir(240); F = dir(300); label("$10$",(A+B)/2,NE);
pair H=(E+C)/2; draw(D--H); draw(E--C); label("$D$",C,NW);label("$X$",D,W);label("$A$",E,SW);label("$Y$",H,E);
draw(A--B--C--D--E--F--A);
[/asy] Since the hexagon is regular, $\angle DXA = 120^\circ$ and $\angle AXY = \angle DXY = 120^\circ / 2 = 60^\circ$. Thus, $\triangle AXY$ and $\triangle DXY$ are congruent $30^\circ - 60^\circ - 90^\circ$ triangles. These triangles are each half an equilateral triangle, so their short leg is half as long as their hypotenuse.
Since the side length of the hexagon is 10, we have $AX=XD=10$. It follows that $XY = AX/2 = 5$ and $AY = DY = \sqrt{10^2-5^2} = \sqrt{75} = 5\sqrt{3}$. (Notice that this value is $\sqrt{3}$ times the length of $XY$, the short leg. In general, the ratio of the sides in a $30^\circ - 60^\circ - 90^\circ$ is $1:\sqrt{3}:2$, which can be shown by the Pythagorean Theorem.) Then, $DA = 2\cdot 5\sqrt{3} = \boxed{10\sqrt{3}}$.
|
human
|
The shaded region consists of 16 congruent squares. If $PQ = 6$ cm, what is the area of the entire shaded region?
[asy]
for(int i = 0; i < 5; ++i)
{
for(int j = 0; j < 2; ++j)
{
filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,gray,linewidth(2));
}
}
for(int i = 0; i < 2; ++i)
{
for(int j = 0; j < 5; ++j)
{
filldraw((i,j)--(i+1,j)--(i+1,j+1)--(i,j+1)--(i,j)--cycle,gray,linewidth(2));
}
}
draw((2,2)--(5,2)--(2,5)--(2,2)--cycle,linewidth(2));
label("P",(2,5),N);
label("Q",(5,2),E);
[/asy]
|
gpt
|
Imagine the square whose diagonal would be PQ. Clearly, that square would be formed of 9 of the shaded squares. The formula for the area of a square from its diagonal is $A = \frac{d^2}{2}$, therefore, the area of that imaginary square is 18. Thus, each smaller shaded square has area 2, making for a total of $\boxed{32\text{ square cm}}$ for the entire shaded area.
|
human
|
A quadrilateral in the plane has vertices $(1, 3)$, $(1, 1)$, $(2, 1)$ and $(2006, 2007)$. How many square units is the area of the quadrilateral?
|
gpt
|
The quadrilateral is shown below: [asy]
size(100);
defaultpen(linewidth(.8));
draw((1,1)--(2,1)--(10,10.1)--(1,3)--cycle);
draw((1,1)--(10,10.1),dashed);
label("$A$", (1,1), S);
label("$B$", (2,1), E);
label("$C$", (10,10.1), N);
label("$D$", (1,3), W);
[/asy] Divide the quadrilateral into two triangles with the dashed line. We will find the area of these two triangles separately. Since $AB$ is horizontal, the area of triangle $ABC$ is half the product of the length $AB$ multiplied by the length of the vertical altitude from $C$ to line $AB$, or $\frac{1\cdot2006}{2}=1003$. Since $AD$ is vertical, the area of triangle $ACD$ is half the product of the length $AD$ multiplied by the length of the horizontal altitude from $C$ to line $AD$, or $\frac{2\cdot2005}{2}=2005$. The area of the entire quadrilateral is $1003+2005=\boxed{3008}$ square units.
|
human
|
Let $ABC$ be a triangle with $\angle BAC = 90^\circ$. A circle is tangent to the sides $AB$ and $AC$ at $X$ and $Y$ respectively, such that the points on the circle diametrically opposite $X$ and $Y$ both lie on the side $BC$. Given that $AB = 6$, find the area of the portion of the circle that lies outside the triangle.
[asy]
import olympiad;
import math;
import graph;
unitsize(4cm);
pair A = (0,0);
pair B = A + right;
pair C = A + up;
pair O = (1/3, 1/3);
pair Xprime = (1/3,2/3);
pair Yprime = (2/3,1/3);
fill(Arc(O,1/3,0,90)--Xprime--Yprime--cycle,0.7*white);
draw(A--B--C--cycle);
draw(Circle(O, 1/3));
draw((0,1/3)--(2/3,1/3));
draw((1/3,0)--(1/3,2/3));
draw((1/16,0)--(1/16,1/16)--(0,1/16));
label("$A$",A, SW);
label("$B$",B, down);
label("$C$",C, left);
label("$X$",(1/3,0), down);
label("$Y$",(0,1/3), left);
[/asy]
|
gpt
|
Let $O$ be the center of the circle, and $r$ its radius, and let $X'$ and $Y'$ be the points diametrically opposite $X$ and $Y$, respectively. We have $OX' = OY' = r$, and $\angle X'OY' = 90^\circ$. Since triangles $X'OY'$ and $BAC$ are similar, we see that $AB = AC$. Let $X''$ be the foot of the altitude from $Y'$ to $\overline{AB}$. Since $X''BY'$ is similar to $ABC$, and $X''Y' = r$, we have $X''B = r$. It follows that $AB = 3r$, so $r = 2$.
[asy]
import olympiad;
import math;
import graph;
unitsize(4cm);
pair A = (0,0);
pair B = A + right;
pair C = A + up;
pair O = (1/3, 1/3);
pair Xprime = (1/3,2/3);
pair Yprime = (2/3,1/3);
fill(Arc(O,1/3,0,90)--Xprime--Yprime--cycle,0.7*white);
draw(A--B--C--cycle);
draw(Circle(O, 1/3));
draw((0,1/3)--(2/3,1/3));
draw((1/3,0)--(1/3,2/3));
draw((2/3, 0)--(2/3, 1/3));
draw((1/16,0)--(1/16,1/16)--(0,1/16));
label("$A$",A, SW);
label("$B$",B, down);
label("$C$",C, left);
label("$X$",(1/3,0), down);
label("$Y$",(0,1/3), left);
label("$X'$", (1/3, 2/3), NE);
label("$Y'$", (2/3, 1/3), NE);
label("$X''$", (2/3, 0), down);
label("$O$", O, NE);
[/asy]
Then, the desired area is the area of the quarter circle minus that of the triangle $X'OY'$. And the answer is $\frac 1 4 \pi r^2 - \frac 1 2 r^2 = \boxed{\pi - 2}$.
|
human
|
A sphere is inscribed in a cube. What is the ratio of the volume of the inscribed sphere to the volume of the cube? Express your answer as a common fraction in terms of $\pi$.
|
gpt
|
[asy]
size(60);
draw(Circle((6,6),4.5));
draw((10.5,6)..(6,6.9)..(1.5,6),linetype("2 4"));
draw((10.5,6)..(6,5.1)..(1.5,6));
draw((0,0)--(9,0)--(9,9)--(0,9)--cycle);
draw((0,9)--(3,12)--(12,12)--(9,9));
draw((12,12)--(12,3)--(9,0));
draw((0,0)--(3,3)--(12,3),dashed); draw((3,3)--(3,12),dashed);
[/asy]
Let the side length of the cube be $s$. The side length of the cube is equal to diameter of the inscribed sphere, so the radius of the sphere has length $\frac{s}{2}$. Thus, the volume of the sphere is equal to $\frac{4}{3}\pi \left(\frac{s}{2}\right)^3 = \frac{\pi s^3}{6}$ and the volume of the cube is equal to $s^3$. Hence the ratio of the sphere's volume to the cube's volume is $\boxed{\frac{\pi}{6}}$.
|
human
|
The angles of quadrilateral $ABCD$ satisfy $\angle A = 2\angle B =
3\angle C = 4\angle D$. What is the degree measure of $\angle A$, rounded to the nearest whole number?
|
gpt
|
Let $x$ be the degree measure of $\angle A$. Then the degree measures of angles $B$, $C$, and $D$ are $x/2$, $x/3$, and $x/4$, respectively. The degree measures of the four angles have a sum of 360, so \[
360 = x+\frac{x}{2}+\frac{x}{3}+\frac{x}{4} =
\frac{25x}{12}.
\]Thus $x=(12\cdot 360)/25 = 172.8\approx \boxed{173}$.
|
human
|
A frustum of a right circular cone is formed by cutting a small cone off of the top of a larger cone. If a particular frustum has a lower base radius of 6 inches, an upper base radius of 3 inches, and a height of 4 inches, what is its lateral surface area? (The lateral surface area of a cone or frustum is the curved surface excluding the base(s).)
[asy]size(200);
import three; defaultpen(linewidth(.8)); currentprojection = orthographic(0,-3,0.5); pen dots = linetype("0 3") + linewidth(1);
real h = 2.3, ratio = (91-24)/(171-24);
picture p1, p2; /* p1 is left-hand picture */
triple A = (0,0,0), B = (0,0,h); draw(p1,(-1,0,0)..(0,-1,0)..(1,0,0)); draw(p1,(-1,0,0)..(0,1,0)..(1,0,0),dots); draw(p1,(-1,0,0)--B--(1,0,0));
add(p1);
triple vlift = (0,0,0.5);
path3 toparc1 = shift((0,0,h*(1-ratio)))*scale3(ratio)*((-1,0,0)..(0,1,0)..(1,0,0)), toparc2 = shift((0,0,h*(1-ratio)))*scale3(ratio)*((1,0,0)..(0,-1,0)..(-1,0,0));
draw(p2,(-1,0,0)..(0,-1,0)..(1,0,0)); draw(p2,(-1,0,0)..(0,1,0)..(1,0,0),dots);
draw(p2,(-1,0,0)--ratio*(-1,0,0)+(1-ratio)*B^^ratio*(1,0,0)+(1-ratio)*B--(1,0,0));
draw(p2,shift(vlift)*(ratio*(-1,0,0)+(1-ratio)*B--B--ratio*(1,0,0)+(1-ratio)*B));
draw(p2,toparc1--toparc2); draw(p2,shift(vlift)*toparc1,dots); draw(p2,shift(vlift)*toparc2);
draw(p2,shift(vlift)*((1-ratio)*B--B),linewidth(0.7)); dot(p2,shift(vlift)*((1-ratio)*B),linewidth(1.5));
label(p2,"frustum",(0,0,h/4));
add(shift((3.4,0,0))*p2);
[/asy]
|
gpt
|
We start by drawing the frustum. Let the top and bottom circles have centers $O_1$ and $O_2$ respectively, and label points $A$ and $B$ on the circumferences as shown such that $O_1$, $O_2$, $A$, and $B$ lie in the same plane.
[asy]
unitsize(0.5 cm);
import geometry; defaultpen(linewidth(.8)+fontsize(10));
label("$O_1$",(0,4),W); label("$O_2$",(0,0),SW); label("$B$",(6,0),SE); label("$A$",(3,4),NE);
draw((3,4)--(0,4)--(0,0)--(6,0));
draw(scale(1,.2)*arc((0,0),6,0,180),linetype("2 4"));
draw(scale(1,.2)*arc((0,0),6,180,360));
draw(scale(1,.2)*arc((0,20),3,0,180));
draw(scale(1,.2)*arc((0,20),3,180,360));
draw((6,0)--(3,4)); draw((-6,0)--(-3,4));
label("6",(3,0),S); label("4",(0,2),W); label("3",(1.5,4),N);
[/asy]
Because the frustum was cut from a right circular cone, $\angle AO_1O_2$ and $\angle BO_2O_1$ are both right angles. We drop a perpendicular from $A$ to $\overline{O_2B}$ and let the intersection point be $X$. Then $O_1AXO_2$ is a rectangle and \[XB=O_2B-O_1A=6-3=3.\]Pythagorean theorem on right $\triangle AXB$ gives \[AB=\sqrt{AX^2 + BX^2}=\sqrt{4^2+3^2}=5.\]Thus the slant height of the frustum is 5.
Extend $\overline{O_1O_2}$ and $\overline{AB}$ above the frustum, and let them intersect at point $C$. $C$ is the tip of the full cone that the frustum was cut from. To compute the lateral surface area of the frustum, we compute the lateral surface area of the full cone and subtract off the lateral surface area of the smaller cone that was removed.
[asy]
unitsize(0.5 cm);
import geometry; defaultpen(linewidth(.8)+fontsize(10));
label("$O_1$",(0,4),W); label("$O_2$",(0,0),SW); label("$B$",(6,0),SE); label("$A$",(3,4),NE);
draw((3,4)--(0,4)--(0,0)--(6,0)); draw((3,4)--(0,8)--(-3,4)); draw((0,4)--(0,8)); label("$C$",(0,8),NE);
draw(scale(1,.2)*arc((0,0),6,0,180),linetype("2 4"));
draw(scale(1,.2)*arc((0,0),6,180,360));
draw(scale(1,.2)*arc((0,20),3,0,180),linetype("2 4"));
draw(scale(1,.2)*arc((0,20),3,180,360));
draw((6,0)--(3,4)); draw((-6,0)--(-3,4));
label("6",(3,0),S); label("4",(0,2),W); label("3",(1.5,4),N); label("5",(4.5,2),NE); [/asy]
To find the height of the whole cone, we take a vertical cross-section of the cone that includes $O_1$, $O_2$, $A$, and $B$. This cross-section is an isosceles triangle.
[asy]
unitsize(0.5 cm);
defaultpen(linewidth(.8)+fontsize(10));
draw((0,0)--(12,0)--(6,8)--cycle); draw((6,0)--(6,8)); draw((6,4)--(9,4));
label("$B$",(12,0),E); label("$C$",(6,8),NE); label("$O_1$",(6,4),W); label("$O_2$",(6,0),SW); label("$A$",(9,4),E);
label("6",(9,0),S); label("3",(7.5,4),S); label("4",(6,2),W); label("5",(10.5,2),NE);
[/asy]
$\triangle CO_1A$ and $\triangle CO_2B$ are similar, so \[\frac{CO_1}{CO_2} = \frac{CA}{CB}=\frac{O_1A}{O_2B}=\frac{3}{6}.\]Thus $CO_1=4$ and $CA=5$ (and we see the small removed cone has half the height of the full cone). Also, $CB=10$.
Now we unroll the lateral surface area of the full cone. (The desired frustum lateral area is shown in blue.)
[asy]
unitsize(0.2 cm);
import graph;
defaultpen(linewidth(.8)+fontsize(10));
fill(Arc((0,0),10,0,240)--cycle,heavycyan); fill(Arc((0,0),5,0,240)--cycle,white); fill((5,0)--(10,0)--(-5,-5*sqrt(3))--(-2.5,-2.5*sqrt(3))--cycle,white);
draw(Arc((0,0),10,0,240)); draw(Arc((0,0),5,0,240));
draw(Arc((0,0),10,240,360),linetype("2 4")); draw(Arc((0,0),5,240,360),linetype("2 4"));
draw((10,0)--(0,0)--(-5,-5*sqrt(3)));
label("$C$",(0,0),SE); label("$A$",(5,0),SE); label("$B$",(10,0),SE); label("10",(-2.5,-2.5*sqrt(3)),SE);
[/asy]
When unrolled, the full cone's lateral surface area is a sector whose arc length is the cone's base perimeter and whose radius is the cone's slant height. So, the sector has arc length $2\cdot \pi \cdot 6 = 12\pi$ and radius $10$. A full circle with radius 10 has arc length $2\cdot \pi \cdot 10 = 20\pi$, so the sector has $\frac{12\pi}{20\pi}=\frac{3}{5}$ of the circle's arc length and thus has 3/5 of the circle's area. Thus, the full cone has lateral surface area \[\frac{3}{5}\pi (10^2) = 60\pi.\]Similarly, the small removed cone's lateral surface area is a sector with radius 5 and arc length $2\cdot \pi \cdot 3 = 6\pi$ (which is $3/5$ of the arc length of a full circle with radius 5), so its lateral surface area is \[\frac{3}{5}\pi (5^2)=15\pi.\]The lateral surface area of the frustum, in blue, is the full cone's lateral surface area minus the small removed cone's lateral surface area, which is \[60\pi-15\pi=\boxed{45\pi}.\]
|
human
|
In regular pentagon $ABCDE$, diagonal $AC$ is drawn, as shown. Given that each interior angle of a regular pentagon measures 108 degrees, what is the measure of angle $CAB$?
[asy]
size(4cm,4cm);
defaultpen(linewidth(1pt)+fontsize(10pt));
pair A,B,C,D,E;
A = (0,0);
B = dir(108);
C = B+dir(39);
D = C+dir(-39);
E = (1,0);
draw(A--B--C--D--E--cycle,linewidth(1));
draw(A--C,linewidth(1)+linetype("0 4"));
label("A",A,S);
label("B",B,W);
label("C",C,N);
label("D",D,E);
label("E",E,S);
label("$108^\circ$",B,E);;
[/asy]
|
gpt
|
Since $ABCDE$ is a regular pentagon, we know by symmetry that the measures of $\angle CAB$ and $\angle BCA$ are equal. We also know that the sum of the measures of the angles of $\triangle ABC$ equals $180$ degrees. Thus, if we let $x = $ the measure of $\angle CAB$ = the measure of $\angle BCA$, we have that $180 = 108 + x + x \Rightarrow 2x = 72 \Rightarrow x = 36$. The measure of angle $CAB$ is $\boxed{36}$ degrees.
|
human
|
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?
|
gpt
|
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 $\boxed{17}.$
|
human
|
In triangle $ABC$, $BC = 20 \sqrt{3}$ and $\angle C = 30^\circ$. Let the perpendicular bisector of $BC$ intersect $BC$ and $AC$ at $D$ and $E$, respectively. Find the length of $DE$.
|
gpt
|
We have that $D$ is the midpoint of $BC$, and that $CD = BC/2 = 20 \sqrt{3}/2 = 10 \sqrt{3}$.
[asy]
unitsize(3 cm);
pair A, B, C, D, E;
A = dir(133);
B = dir(193);
C = dir(-13);
D = (B + C)/2;
E = extension(A, C, D, D + rotate(90)*(B - C));
draw(A--B--C--cycle);
draw(D--E);
label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, S);
label("$E$", E, NE);
[/asy]
Also, triangle $CED$ is a $30^\circ$-$60^\circ$-$90^\circ$ triangle, so $DE = CD/\sqrt{3} = 10 \sqrt{3}/\sqrt{3} = \boxed{10}$.
|
human
|
What is the volume, in cubic units, of a cube whose surface area is 600 square units?
|
gpt
|
The surface area of a cube equals 6 times the area of each face (since there are 6 faces). If the cube has a sidelength of $s$, then the surface area equals $6s^2$. We set that equal to 600 and solve for $s$, which must be positive. $$600=6s^2\qquad\Rightarrow 100=s^2\qquad\Rightarrow 10=s$$ The volume of the cube is $s^3=10^3=\boxed{1000}$ cubic units.
|
human
|
In the diagram, the smaller circles touch the larger circle and touch each other at the center of the larger circle. The radius of the larger circle is $6.$ What is the area of the shaded region?
[asy]
size(100);
import graph;
filldraw(Circle((0,0),2),mediumgray);
filldraw(Circle((-1,0),1),white);
filldraw(Circle((1,0),1),white);
[/asy]
|
gpt
|
Label the center of the larger circle $O$ and the points of contact between the larger circle and the smaller circles $A$ and $B.$ Draw the radius $OA$ of the larger circle.
[asy]
size(120);
import graph;
filldraw(Circle((0,0),2),mediumgray);
filldraw(Circle((-1,0),1),white);
filldraw(Circle((1,0),1),white);
draw((-2,0)--(0,0));
label("$A$",(-2,0),W); label("$O$",(0,0),E); label("$B$",(2,0),E);
[/asy]
Since the smaller circle and the larger circle touch at $A,$ the diameter through $A$ of the smaller circle lies along the diameter through $A$ of the larger circle. (This is because each diameter is perpendicular to the common tangent at the point of contact.)
Since $AO$ is a radius of the larger circle, it is a diameter of the smaller circle.
Since the radius of the larger circle is $6,$ the diameter of the smaller circle is $6,$ so the radius of the smaller circle on the left is $3.$
Similarly, we can draw a radius through $O$ and $B$ and deduce that the radius of the smaller circle on the right is also $3.$ The area of the shaded region equals the area of the larger circle minus the combined area of the two smaller circles. Thus, the area of the shaded region is $$6^2\pi - 3^2\pi - 3^2\pi = 36\pi - 9\pi - 9\pi = \boxed{18\pi}.$$
|
human
|
The point $O$ is the center of the circle circumscribed about $\triangle ABC$, with $\angle BOC = 120^{\circ}$ and $\angle AOB =
140^{\circ}$, as shown. What is the degree measure of $\angle
ABC$?
[asy]
pair A,B,C;
draw(Circle((0,0),20),linewidth(0.7));
label("$O$",(0,0),S);
A=(-16,-12);
C=(16,-12);
B=(3,19.7);
draw(A--B--C--cycle,linewidth(0.7));
label("$140^{\circ}$",(0,0),W);
label("$120^{\circ}$",(0,0.3),E);
draw(C--(0,0)--B);
draw(A--(0,0));
label("$A$",A,SW);
label("$B$",B,NE);
label("$C$",C,SE);
[/asy]
|
gpt
|
Since $OA=OB=OC$, triangles $AOB$, $BOC$, and $COA$ are all isosceles. Hence \[
\angle ABC = \angle ABO + \angle OBC =
\frac{180^{\circ}-140^{\circ}}{2}+
\frac{180^{\circ}-120^{\circ}}{2}=\boxed{50^{\circ}}.
\]OR
Since \[
\angle AOC = 360^{\circ}-140^{\circ}-120^{\circ}=100^{\circ},
\]the Central Angle Theorem implies that \[
\angle ABC = \frac{1}{2}\angle AOC = \boxed{50^{\circ}}.
\]
|
human
|
In the diagram, $AOB$ is a sector of a circle with $\angle AOB=60^\circ.$ $OY$ is drawn perpendicular to $AB$ and intersects $AB$ at $X.$ What is the length of $XY
?$ [asy]
draw((0,0)--(12,0),black+linewidth(1));
draw((0,0)--(10.3923,-6)..(12,0)..(10.3923,6)--(0,0),black+linewidth(1));
draw((10.3923,-6)--(10.3923,6),black+linewidth(1));
label("$O$",(0,0),W);
label("$A$",(10.3923,6),N);
label("$B$",(10.3923,-6),S);
label("$X$",(10.3923,0),NW);
label("$Y$",(12,0),E);
label("12",(0,0)--(10.3923,6),NW);
label("12",(0,0)--(10.3923,-6),SW);
[/asy]
|
gpt
|
Since $OY$ is a radius of the circle with centre $O,$ we have $OY=12.$ To find the length of $XY,$ we must find the length of $OX.$
Since $OA=OB,$ we know that $\triangle OAB$ is isosceles.
Since $\angle AOB = 60^\circ,$ we have $$\angle OAB=\frac{1}{2}(180^\circ-60^\circ)=60^\circ.$$ Therefore, $$
\angle AOX = 180^\circ-60^\circ-90^\circ
=30^\circ,
$$ so $\triangle OAX$ is a $30^\circ$-$60^\circ$-$90^\circ$ triangle.
Since $OA=12,$ we have $AX = \frac{1}{2}OA=6$ and $OX = \sqrt{3}AX=6\sqrt{3}.$ Thus, $$XY=OY-OX = \boxed{12 - 6\sqrt{3}} \approx 1.61.$$
|
human
|
In triangle $ABC$, medians $AD$ and $CE$ intersect at $P$, $PE=1.5$, $PD=2$, and $DE=2.5$. What is the area of $AEDC$?
|
gpt
|
Note that $1.5^2 + 2^2 = 2.5^2,$ so $\triangle PED$ has a right angle at $P.$ (Alternatively, you could note that $(1.5, 2, 2.5)$ is half of the Pythagorean triple $(3,4,5).$) [asy]size(6cm);pair P=(0,0),D=(0,-2),E=(-1.5,0),C=(3,0),A=(0,4),B=extension(A,E,D,C);draw(A--B--C--cycle^^C--E^^A--D);draw(rightanglemark(E,P,D));draw(E--D);dot("$A$",A,N);dot("$B$",B,SW);dot("$C$",C,dir(0));dot("$D$",D,SSE);dot("$E$",E,NW);dot("$P$",P,NE);[/asy] Since the centroid $P$ divides medians $AD$ and $CE$ in the ratio $2 : 1,$ we have $CP = 2 \cdot EP = 2 \cdot 1.5 = 3$ and $AP = 2 \cdot DP = 2 \cdot 2 = 4.$ Then quadrilateral $AEDC$ consists of four right triangles; we can then compute its area as \[[AEDC] = \tfrac12 (4 \cdot 1.5 + 2 \cdot 1.5 + 3 \cdot 2 + 4 \cdot 3) = \boxed{13.5}.\]
|
human
|
In the diagram, $PQ$ and $RS$ are diameters of a circle with radius 4. If $PQ$ and $RS$ are perpendicular, what is the area of the shaded region?
[asy]
size(120);
import graph;
fill((-1,-1)--(-1,1)--(1,1)--(1,-1)--cycle,mediumgray);
fill(Arc((0,0),sqrt(2),45,135)--cycle,mediumgray);fill(Arc((0,0),sqrt(2),225,315)--cycle,mediumgray);
draw(Circle((0,0),sqrt(2)));
draw((-1,-1)--(1,1)--(1,-1)--(-1,1)--cycle);
label("$P$",(-1,1),NW); label("$R$",(1,1),NE); label("$S$",(-1,-1),SW); label("$Q$",(1,-1),SE);
[/asy]
|
gpt
|
Diameters $PQ$ and $RS$ cross at the center of the circle, which we call $O$.
The area of the shaded region is the sum of the areas of $\triangle POS$ and $\triangle ROQ$ plus the sum of the areas of sectors $POR$ and $SOQ$.
Each of $\triangle POS$ and $\triangle ROQ$ is right-angled and has its two perpendicular sides of length 4 (the radius of the circle).
Therefore, the area of each of these triangles is $\frac{1}{2}(4)(4)=8$.
Each of sector $POR$ and sector $SOQ$ has area $\frac{1}{4}$ of the total area of the circle, as each has central angle $90^\circ$ (that is, $\angle POR = \angle SOQ = 90^\circ$) and $90^\circ$ is one-quarter of the total central angle.
Therefore, each sector has area $\frac{1}{4}(\pi(4^2))=\frac{1}{4}(16\pi)=4\pi$.
Thus, the total shaded area is $2(8)+2(4\pi)=\boxed{16+8\pi}$.
|
human
|
We have a triangle $\triangle ABC$ such that $AB = AC = 8$ and $BC = 10.$ What is the length of the median $AM$?
|
gpt
|
Let's draw a sketch first. Since $\triangle ABC$ is isosceles, we know that $AM$ must form a right angle with $BC.$ [asy]
pair A, B, C, M;
A = (0, 6.24);
B = (-5, 0);
C = (5, 0);
M = 0.5 * B + 0.5 * C;
draw(A--B--C--cycle);
draw(A--M);
label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);
label("$M$", M, S);
draw(rightanglemark(A,M,B,10));
[/asy] We know that $BM = MC = \frac{BC}{2} = 5.$ Now we just simply apply the Pythagorean Theorem on the right triangle $\triangle ABM.$ \begin{align*}
AM^2 &= AB^2 - BM^2\\
AM^2 &= 8^2 - 5^2 = 39\\
AM &= \boxed{\sqrt{39}}
\end{align*}
|
human
|
The radius of a cylinder is doubled and its height is tripled. If its original volume was 10 cubic feet, what is its volume now, in cubic feet?
|
gpt
|
Let the cylinder's original radius and height be $r$ and $h$, respectively. The new cylinder has volume \[
\pi (2r)^2(3h)=12\pi r^2 h,
\] which is 12 times larger than the original volume. Since the original volume was 10 cubic feet, the new volume is $\boxed{120}$ cubic feet.
|
human
|
Find the ratio of the volume of the cone to the volume of the cylinder. Express your answer as a common fraction.
[asy]
import solids; size(150); import three; defaultpen(linewidth(0.8)); currentprojection = orthographic(5,0,3);
revolution c = cylinder((0,0,0), 1, 3);
revolution c2 = cone((0,0,0), 1,1.5);
draw(c,black);
draw(c2,black);
draw((1.5,0)--(1.5,2.5),Arrows(TeXHead));
label("12",(1.5,0)--(1.5,2.5),E);
draw((-1.5,0)--(-1.5,1.25),Arrows(TeXHead));
label("6",(-1.5,0)--(-1.5,1.25),W);
draw((0,0)--(1,0),Arrows(TeXHead));
label("4",(0,0)--(1,0),S);
[/asy]
|
gpt
|
The volume of a cone with radius $r$ and height $h$ is $(1/3) \pi r^2 h$; the volume of a cylinder with similar dimensions is $\pi r^2 h$. The cone has the same radius as the cylinder and half the height, so it has $1/3$ the volume of half the cylinder and thus has $1/2\cdot 1/3 = 1/6$ the volume of the whole cylinder. Hence the desired ratio is $\boxed{\frac{1}{6}}$.
|
human
|
Suppose that we are given 40 points equally spaced around the perimeter of a square, so that four of them are located at the vertices and the remaining points divide each side into ten congruent segments. If $P$, $Q$, and $R$ are chosen to be any three of these points which are not collinear, then how many different possible positions are there for the centroid of $\triangle PQR$?
|
gpt
|
Without loss of generality, assume that our square has vertices at $(0,0)$, $(10,0)$, $(10,10)$, and $(0,10)$ in the coordinate plane, so that the 40 equally spaced points are exactly those points along the perimeter of this square with integral coordinates. We first note that if $P$, $Q$, and $R$ are three of these points which are not collinear, then the centroid of $\triangle PQR$ must lie in the interior of the square, not along one of its sides. And secondly, we recall that the coordinates of the centroid are found by averaging the coordinates of $P$, $Q$, and $R$. Therefore the coordinates of the centroid must be of the form $\left(\frac{m}{3}, \frac{n}{3}\right)$ where $m$ and $n$ are integers with $1\le m,n\le 29$.
To show that every point of the form $\left( \frac{m}{3}, \frac{n}{3} \right)$ can be a centroid, we divide into cases.
If $1 \le m \le 10$ and $1 \le n \le 10$, then we can take the points as $(0,0)$, $(m,0)$, and $(0,n)$.
If $10 \le m \le 19$ and $1 \le n \le 10$, then we can take the points as $(m - 10,0)$, $(10,0)$, and $(0,n)$.
If $20 \le m \le 29$ and $1 \le n \le 10$, then we can take the points as $(m - 20,0)$, $(10,0)$, and $(10,n)$.
If $1 \le m \le 10$ and $11 \le n \le 19$, then we can take the points as $(m,0)$, $(0,n - 10)$, and $(0,10)$.
If $10 \le m \le 19$ and $11 \le n \le 19$, then we can take the points as $(10,0)$, $(0,n - 10)$, and $(m - 10,10)$.
If $20 \le m \le 29$ and $11 \le n \le 19$, then we can take the points as $(m - 20,0)$, $(10,n - 10)$, and $(10,10)$.
If $1 \le m \le 10$ and $20 \le n \le 29$, then we can take the points as $(0,n - 20)$, $(0,10)$, and $(m,10)$.
If $10 \le m \le 19$ and $20 \le n \le 29$, then we can take the points as $(0,n - 20)$, $(m - 10,10)$, and $(10,10)$.
If $20 \le m \le 29$ and $20 \le n \le 29$, then we can take the points as $(m - 20,10)$, $(10,n - 20)$, and $(10,10)$.
Thus, every point of the form $\left( \frac{m}{3}, \frac{n}{3} \right)$ can be a centroid. This means that there are $29^2=\boxed{841}$ positions for the centroid.
|
human
|
An $\textit{annulus}$ is the region between two concentric circles. The concentric circles in the figure have radii $b$ and $c$, with $b>c$. Let $\overline{OX}$ be a radius of the larger circle, let $\overline{XZ}$ be tangent to the smaller circle at $Z$, and let $\overline{OY}$ be the radius of the larger circle that contains $Z$. Let $a=XZ$, $d=YZ$, and $e=XY$. What is the area of the annulus? Express your answer in terms of $\pi$ and at most one of the variables $a,b,c,d,e$.
[asy]
pair O,X,Y,Z;
O=(0,0);
X=(16,12);
Y=(0,20);
Z=(0,12);
fill(Circle(0,20),gray(0.7));
fill(Circle(0,12),white);
draw(Circle(O,20),linewidth(0.7));
draw(Circle(O,12),linewidth(0.7));
dot(O);
dot(X);
dot(Y);
dot(Z);
draw(O--X--Y--cycle,linewidth(0.7));
draw(Z--X,linewidth(0.7));
label("$b$",(8,6),SE);
label("$a$",(8,12),S);
label("$e$",(8,16),SW);
label("$c$",(0,6),W);
label("$d$",(0,16),W);
label("$O$",O,S);
label("$X$",X,NE);
label("$Y$",Y,N);
label("$Z$",Z,SW);
[/asy]
|
gpt
|
The area of the annulus is the difference between the areas of the two circles, which is $\pi b^2 -\pi c^2$. Because the tangent $\overline{XZ}$ is perpendicular to the radius $\overline{OZ}$, $b^2 -
c^2 = a^2$, so the area is $\boxed{\pi a^2}$.
|
human
|
The square with vertices $(-a, -a), (a, -a), (-a, a), (a, a)$ is cut by the line $y = x/2$ into congruent quadrilaterals. The perimeter of one of these congruent quadrilaterals divided by $a$ equals what? Express your answer in simplified radical form.
|
gpt
|
The line $y=\frac x2$ will intersect the two vertical sides of the square, as shown below:
[asy]
real f(real x)
{
return x/2;
}
import graph;
size(6cm);
real a = 8;
pair A=(-a,a), B=(a,a), C=(a,-a), D=(-a,-a);
draw(A--B--C--D--cycle);
draw(graph(f,-11,11),Arrows);
axes(Arrows(4));
dot("$(-a,a)$",A,N);
dot("$(a,a)$",B,N);
dot("$(a,-a)$",C,S);
dot("$(-a,-a)$",D,S);
real eps=0.2;
dot((8,4)^^(-8,-4));
draw(shift((10,0))*"$2a$",(-a+eps,-a/2-.5)--(a-eps,-a/2-.5),Arrows);
draw(shift((0,10))*"$a$",(a+2*eps,-a/2)--(a+2*eps,a/2),Arrows);[/asy]
The equation of the right side of the square is $x=a,$ so we have $y= \frac x2 = \frac a2,$ which means that the intersection point with the right side of the square is $\left(a, \frac a2 \right).$ Similarly, the equation of the left side of the square is $x=-a,$ so we have $y= \frac x2 = -\frac a2,$ which means that the intersection point with the left side of the square is $\left(-a, -\frac a2 \right).$ It follows that the sides of each quadrilateral have lengths $\frac a2,$ $2a,$ $\frac{3a}2,$ and $\sqrt{a^2 + (2a)^2} = a\sqrt{5},$ by the Pythagorean theorem. Hence, the perimeter of the quadrilateral is \[\frac a2 + 2a + \frac{3a}2 + a\sqrt{5} = \left(4+\sqrt5\right)a,\]and when this is divided by $a,$ we get $\boxed{4+\sqrt{5}}.$
|
human
|
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 $ABCD$?
|
gpt
|
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 formula $A = \frac{1}{2}(b_1+ b_2)h$, the area is $\frac{1}{2}(3+6)(4) = \boxed{18}$ square units.
|
human
|
Polygon $ABCDEF$ is a regular hexagon. What is the measure in degrees of angle $ABF$?
|
gpt
|
In triangle $ABF$, the two acute angles are equal since $AB=AF$. Also, the measure of $\angle A$ is $180^\circ(6-2)/6=120^\circ$. Letting $x$ be the measure of $\angle ABF$, we have \[
120^\circ+x+x=180^\circ \implies x=\boxed{30}\text{ degrees}.
\] [asy]
size(5cm);
defaultpen(linewidth(0.7));
int i;
pair A=dir(0), B=dir(60), C=dir(120), D=dir(180), Ep=dir(240), F=dir(300);
pair[] dots = {A,B,C,D,Ep,F};
string[] alphabet={"$A$","$B$","$C$","$D$","$E$","$F$"};
dot(dots);
for(i=0;i<6;++i)
{
draw(dir(60*i)--dir(60*i+60));
label(alphabet[i],dots[i],dots[i]);
}
draw(A--B--F);
draw(anglemark(F,B,A));
draw(anglemark(A,F,B));
[/asy]
|
human
|
In right triangle $JKL$, angle $J$ measures 60 degrees and angle $K$ measures 30 degrees. When drawn, the angle bisectors of angles $J$ and $K$ intersect at a point $M$. What is the measure of obtuse angle $JMK$?
[asy]
import geometry;
import olympiad;
unitsize(0.8inch);
dotfactor = 3;
defaultpen(linewidth(1pt)+fontsize(10pt));
pair J,K,L,M,U,V;
J = (0,0);
K = (1,2);
L = (1,0);
draw(J--K--L--cycle);
draw(rightanglemark(J,L,K,5));
label("$J$",J,W);
label("$K$",K,N);
label("$L$",L,E);
U = (1,2/3);
V = (2/(2+sqrt(3)),0);
draw(J--U);
draw(K--V);
M = intersectionpoint(J--U,K--V);
dot("M",M,NW);
[/asy]
|
gpt
|
Since $JM$ bisects $\angle J$, we know that the measure of $\angle KJM$ is $60/2 = 30$ degrees. Similarly, since $MK$ bisects $\angle K$, we know that the measure of $\angle JKM$ is $30/2 = 15$ degrees. Finally, since the sum of the measures of the angles of a triangle always equals $180$ degrees, we know that the sum of the measures of $\angle JKM$, $\angle KJM$, and $\angle JMK$ equals $180$ degrees. Thus, the measure of $\angle JMK = 180 - 30 - 15 = \boxed{135}$ degrees.
|
human
|
Segment $s_1$ has endpoints at $(4,1)$ and $(-8,5)$. Segment $s_2$ is obtained by translating $s_1$ by $2$ units to the right and $3$ units up. Find the midpoint of segment $s_2$. Express your answer as $(a,b)$ with $a$ and $b$ integers.
|
gpt
|
The midpoint of segment $s_1$ can be found using the midpoint formula: $\left(\frac{4-8}2,\frac{1+5}2\right)=(-2,3).$ The midpoint of $s_2$ is the translation of the midpoint of $s_1$ be $2$ units to the right and $3$ units up. Thus its coordinates are $(-2+2,3+3)=\boxed{(0,6)}.$
|
human
|
Circle $T$ has a circumference of $12\pi$ inches, and segment $XY$ is a diameter. If the measure of angle $TXZ$ is $60^{\circ}$, what is the length, in inches, of segment $XZ$?
[asy]
size(150);
draw(Circle((0,0),13),linewidth(1));
draw((-12,-5)--(-5,-12)--(12,5)--cycle,linewidth(1));
dot((0,0));
label("T",(0,0),N);
label("X",(-12,-5),W);
label("Z",(-5,-12),S);
label("Y",(12,5),E);
[/asy]
|
gpt
|
We can begin by using the circumference to solve for the radius of the circle. If the circumference is $12\pi$, then $2\pi r=12\pi$ which implies $r=6$. Now, we can draw in the radius $TZ$ as shown: [asy]
size(150);
draw(Circle((0,0),13),linewidth(1));
draw((-12,-5)--(-5,-12)--(12,5)--cycle,linewidth(1));
draw((0,0)--(-5,-12),linewidth(1)+linetype("0 4"));
dot((0,0));
label("T",(0,0),N);
label("X",(-12,-5),W);
label("Z",(-5,-12),S);
label("Y",(12,5),E);
[/asy]
We know that $TX=TZ$, since both are radii of length 6. We are given $\angle TXZ=60^{\circ}$, so $\angle TZX=60^{\circ}$, and triangle $TXZ$ is equilateral. Thus, $TX=TZ=XZ=\boxed{6}$.
|
human
|
The lengths of two sides of a triangle are 33 units and 42 units. The third side also has an integral length. What is the least possible number of units in the perimeter of the triangle?
|
gpt
|
The sum of the smaller two sides must exceed the greatest side, so if $x$ is the missing side then $x+33>42\implies x>9$. The smallest integer greater than 9 is 10, so the least perimeter is $10+33+42=\boxed{85}$ units.
|
human
|
A cube has eight vertices (corners) and twelve edges. A segment, such as $x$, which joins two vertices not joined by an edge is called a diagonal. Segment $y$ is also a diagonal. How many diagonals does a cube have? [asy]
/* AMC8 1998 #17 Problem */
pair A=(0,48), B=(0,0), C=(48,0), D=(48,48);
pair E=(24,72), F=(24,24), G=(72,24), H=(72,72);
pen d = linetype("8 8");
draw(A--D--C--B--cycle);
draw(D--H--G--C);
draw(A--E--H);
draw(B--F--E);
draw(F--G);
draw(H--A--G, d);
label("$x$", (40, 61), N);
label("$y$", (40, 35), N);
[/asy]
|
gpt
|
There are two diagonals, such as $x$, in each of the six faces for a total of twelve face diagonals. There are also four space diagonals, such as $y$, which are within the cube. This makes a total of $\boxed{16}$.
|
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