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2014 AMC 10B Problems/Problem 23

Revision as of 12:13, 28 December 2019 by Gr8song (talk | contribs) (→‎Solution 1)

Problem

A sphere is inscribed in a truncated right circular cone as shown. The volume of the truncated cone is twice that of the sphere. What is the ratio of the radius of the bottom base of the truncated cone to the radius of the top base of the truncated cone?

$\text{(A) } \dfrac32 \quad \text{(B) } \dfrac{1+\sqrt5}2 \quad \text{(C) } \sqrt3 \quad \text{(D) } 2 \quad \text{(E) } \dfrac{3+\sqrt5}2$

[asy] real r=(3+sqrt(5))/2; real s=sqrt(r); real Brad=r; real brad=1; real Fht = 2*s; import graph3; import solids; currentprojection=orthographic(1,0,.2); currentlight=(10,10,5); revolution sph=sphere((0,0,Fht/2),Fht/2); draw(surface(sph),green+white+opacity(0.5)); //triple f(pair t) {return (t.x*cos(t.y),t.x*sin(t.y),t.x^(1/n)*sin(t.y/n));} triple f(pair t) { triple v0 = Brad*(cos(t.x),sin(t.x),0); triple v1 = brad*(cos(t.x),sin(t.x),0)+(0,0,Fht); return (v0 + t.y*(v1-v0)); } triple g(pair t) { return (t.y*cos(t.x),t.y*sin(t.x),0); } surface sback=surface(f,(3pi/4,0),(7pi/4,1),80,2); surface sfront=surface(f,(7pi/4,0),(11pi/4,1),80,2); surface base = surface(g,(0,0),(2pi,Brad),80,2); draw(sback,gray(0.9)); draw(sfront,gray(0.5)); draw(base,gray(0.9)); draw(surface(sph),gray(0.4));[/asy]

Solution 1

First, we draw the vertical cross-section passing through the middle of the frustum. Let the top base have a diameter of $2$, and the bottom base have a diameter of $2r$. [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("$r-1$",(r/2+1/2,0),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] (If you don't get why the blue hypotenuse is $1+r$, reference this (https://www.ck12.org/geometry/Tangent-Lines/lesson/Tangent-Lines-GEOM/)


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 end up with \[s=\sqrt{r}.\] Next, we can find the volume of the frustum (truncated cone) and of the sphere. Since we know $V_{\text{frustum}}=2V_{\text{sphere}}$, we can solve for $s$ using $V_{\text{frustum}}=\frac{\pi h}{3}(R^2+r^2+Rr)$ we get: \[V_{\text{frustum}}=\frac{\pi\cdot2\sqrt{r}}{3}(r^2+r+1)\] Using $V_{\text{sphere}}=\dfrac{4r^{3}\pi}{3}$, we get \[V_{\text{sphere}}=\dfrac{4(\sqrt{r})^{3}\pi}{3}\] so we have: \[\frac{\pi\cdot2\sqrt{r}}{3}(r^2+r+1)=2\cdot\dfrac{4(\sqrt{r})^{3}\pi}{3}.\] Dividing by $\frac{2\pi\sqrt{r}}{3}$, we get \[r^2+r+1=4r\] which is equivalent to \[r^2-3r+1=0\] by the Quadratic Formula, $r=\dfrac{3\pm\sqrt{(-3)^2-4\cdot1\cdot1}}{2\cdot1}$ , so \[r=\dfrac{3+\sqrt{5}}{2} \longrightarrow \boxed{\textbf{(E)}}\]

Solution 2

Similar to above, draw a smaller cone top with the base of the smaller circle with radius $r_1$ and height $h$. The smaller right triangle is similar to the blue highlighted one in Solution 1. Then $\frac{r_1}{r_2-r_1}=\frac{h}{2R}$ where $R$ is the radius of the sphere. Then $h=\frac{2Rr_1}{r_2-r_1}$.

From the Pythagorean theorem on the blue triangle in Solution 1, we get similarly that $R=\sqrt{r_2r_1}$.

From the volume requirements, we get that $\frac{8}{3}\pi R^3=\frac{\pi r_2^2(h+2R)-\pi r_1^2h}{3}$ which yields $8R^3=r_2^2(h+2R)-r_1^2h$.

The small right triangle on top is similar to the big right triangle of the entire big cone. So $\frac{r_1}{r_2}=\frac{h}{h+2R}\implies h+2R=\frac{r_2h}{r_1}$.

Substituting yields $8R^3=\frac{h(r_2^3-r_1^3)}{r_1}$.

Substituting $R=\sqrt{r_2r_1}$ and $h=\frac{2Rr_1}{r_2-r_1}$ yields $8(r_2r_1)^{\frac{3}{2}}=2(r_2r_1)^{\frac{1}{2}}(r_2^2+r_2r_1+r_1^2)$ which yields $r_2^2-3r_2r_1+r_1^2=0$.

Solving in $r_2$ yields $r_2=\frac{3r_1\pm r_1\sqrt{5}}{2}$ so $\frac{r_2}{r_1}=\frac{3+\sqrt{5}}{2}$.

Solution 3: Another Way of Simplifying the Equation in Solution 2

Similar to the previous solutions, we let $R$, $r$, and $s$ be the radii of the larger base, smaller base, and the sphere, respectively. By the Pythagorean Theorem, we have $s=\sqrt{Rr}$. Let $H$, $h$ be the heights of the large and small cones, respectively, then we have $H=h+2s$ and \[h=2s(\frac{r}{R-r})==\frac{2sr}{R-r}=\frac{2(\sqrt{Rr})r}{R-r}\] Thus we have the equation \[V_{large cone}-V_{small cone}=2V_{sphere}\] \[\frac{\pi R^2H}{3}-\frac{\pi r^2h}{3}=2(\frac{4\pi s^3}{3})\] \[\pi R^2H-\pi r^2h=8\pi s^3\] \[R^2H-r^2h=8s^3\] Substituting the expressions for $H$, $h$, $s$ in order into the equation yields \[R^2(h+2s)+r^2h=8s^3\] \[R^2((\frac{2sr}{R-r})+2s)+r^2(\frac{2sr}{R-r})=8s^3\] \[R^2(\frac{2sR}{R-r})+r^2(\frac{2sr}{R-r})=8s^3\] \[\frac{2sR^3-2sr^3}{R-r}=8s^3\] \[\frac{R^3-r^3}{R-r}=4s^2\] The numerator $R^3-r^3$ can be factored to yield \[\frac{(R-r)(R^2+Rr+r^2)}{R-r}=4s^2\] \[R^2+Rr+r^2{R-r}=4s^2\] \[R^2+Rr+r^2=4(\sqrt{Rr})^2\] \[R^2+Rr+r^2=4Rr\] \[R^2-3Rr+r^2=0\] Solving the quadratic to obtain the ratio $\frac{R}{r}=\frac{3+\sqrt{5}}{2}$ $\boxed{\textbf{(E)}}$

~ Nafer

See Also

2014 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions
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