2014 AMC 12B Problems/Problem 19

Solution

First, we draw the vertical cross-section passing through the middle of the frustum. let the top base equal 2 and the bottom base to be equal to 2r $[asy] size(7cm); pair A,B,C,D; real r = (3+sqrt(5))/2; real s = sqrt(r); A = (-r,0); B = (r,0); C = (1,2*s); D = (-1,2*s); draw(A--B--C--D--cycle); pair O = (0,s); draw(shift(O)*scale(s)*unitcircle); dot(O); pair X,Y; X = (0,0); Y = (0,2*s); draw(X--Y); label("$r-1$",(X+B)/2,S); label("$1$",(Y+C)/2,N); label("$s$",(O+Y)/2,W); label("$s$",(O+X)/2,W); draw(B--C--(1,0)--cycle,blue+1bp); pair P = 0.73*C+0.27*B; draw(O--P); dot(P); label("$1$",(C+P)/2,NE); label("$r$",(B+P)/2,NE); [/asy]$ (diagram by DivideBy0?)

then using the Pythagorean theorem we have: $(r+1)^2=(2s)^2+(r-1)^2$ which is equivalent to: $r^2+2r+1=4s^2+r^2-2r+1$ subtracting $r^2-2r+1$ from both sides $4r=4s^2$ solving for s we get: $s=\sqrt{r}$ next we can find the area of the frustum and of the sphere and we know $V_{frustum}=2V_{sphere}$ so we can solve for $s$ using $V_{frustum}=\frac{\pi*h}{3}(R^2+r^2+Rr)$ we get: $V_{frustum}=\frac{\pi*2\sqrt{r}}{3}(r^2+r+1)$ using $V_{sphere}=\dfrac{4r^{3}\pi}{3}$ we get $V_{sphere}=\dfrac{4(\sqrt{r})^{3}\pi}{3}$ so we have: $\frac{\pi*2\sqrt{r}}{3}(r^2+r+1)=2*\dfrac{4(\sqrt{r})^{3}\pi}{3}$ dividing by $\frac{2\pi*\sqrt{r}}{3}$ we get $r^2+r+1=4r$ which is equivalent to $r^2-3r+1=0$ $r=\dfrac{3\pm\sqrt{(-3)^2-4*1*1}}{2*1}$ so $r=\dfrac{3+\sqrt{5}}{2}\longrightarrow E$

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2014 AMC 12B Problems/Problem 19

Solution