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2021 Fall AMC 10A Problems/Problem 22

Revision as of 21:30, 25 November 2021 by Stevenyiweichen (talk | contribs)

Problem

Inside a right circular cone with base radius $5$ and height $12$ are three congruent spheres with radius $r$. Each sphere is tangent to the other two spheres and also tangent to the base and side of the cone. What is $r$?

$\textbf{(A)}\ \frac{3}{2} \qquad\textbf{(B)}\ \frac{90-40\sqrt{3}}{11} \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ \frac{144-25\sqrt{3}}{44} \qquad\textbf{(E)}\ \frac{5}{2}$

Solution 1 (Coordinates)

We will use coordinates. WLOG, let the coordinates of the center of the base of the cone be the origin. Then, let the center of one of the spheres be $(0, 2r/\sqrt{3}, r)$. Note that the distance between this point and the plane given by $12y+5z=60$ is $r$. Thus, by the point-to-plane distance formula, we have

\[\frac{|12 \cdot 2r/\sqrt{3} + 5r - 60|}{\sqrt{0^2+5^2+12^2}}=r.\]

Solving for $r$ yields $r = \boxed{\textbf{(B) }\dfrac{90 - 40\sqrt3}{11}}$.

~ Leo.Euler

Solution 2 (Cross section & angle bisector)

We can take half of a cross section of the sphere, as such: [asy] unitsize(0.5cm); real r = (90-40*sqrt(3))/11; pair C = (0,0); pair A = (-5,0); pair B = (0,12); pair O = (-((2*sqrt(3))/3) * r, r); draw(A--B--C--cycle); draw(circle(O,r)); pair D = (-(2*sqrt(3))/3 * r - (12/13)*r, (18/13)*r); pair E = (-2.2, 0); draw(O--E); draw(D--O); label("$A$", A, SW); label("$B$", B, N); label("$C$", C, SE); label("$O$", O, N); label("$D$", D, NW); label("$E$", E, SW); dot(D); dot(E); [/asy] Notice that we chose a cross section where one of the spheres was tangent to the lateral surface of the cone at $D$.

To evaluate $r$, we will find $AE$ and $EC$ in terms of $r$; we also know that $AE+EC = 5$, so with this, we can solve $r$. Firstly, to find $EC$, we can take a bird's eye view of the cone: [asy] unitsize(0.8cm); pair C = (0,0); draw(circle(C,5)); label("$C$", C, N); dot(C); real r = (90-40*sqrt(3))/11; real raise = r*(2/3*sqrt(3)); pair E = (-r,raise/-2); pair X = (0,raise); pair Y = (r,raise/-2); label("$E$", E, SW); dot(E); label("$X$", X, NW); dot(X); label("$Y$", Y, SW); dot(Y); draw(circle(X,r),dashed); draw(circle(E,r),dashed); draw(circle(Y,r),dashed); draw(E--X,dashed); draw(X--Y,dashed); draw(E--Y,dashed); [/asy] $C$ is the centroid of equilateral triangle $EXY$. Also, since all of the medians of an equilateral triangle are also altitudes, we want to find two-thirds of the altitude from $E$ to $XY$; this is because medians cut each other into a $2$ to $1$ ratio. This equilateral triangle has a side length of $2r$, therefore it has an altitude of length $r \sqrt{3}$; two thirds of this is $\frac{2r \sqrt{3}}{3}$, so \[EC = \frac{2r \sqrt{3}}{3}.\] [asy] unitsize(0.5cm); real r = (90-40*sqrt(3))/11; pair C = (0,0); pair A = (-5,0); pair B = (0,12); pair O = (-((2*sqrt(3))/3) * r, r); draw(A--B--C--cycle); draw(circle(O,r)); pair D = (-(2*sqrt(3))/3 * r - (12/13)*r, (18/13)*r); pair E = (-2.2, 0); pair F = (-2.2, 6.72); draw(E--F); draw(D--O); draw(A--O, dotted); label("$A$", A, SW); label("$B$", B, N); label("$C$", C, SE); label("$O$", O, NE); label("$D$", D, NW); label("$E$", E, SW); label("$F$", F, NW); dot(D); dot(E); [/asy] To evaluate $AE$ in terms of $r$, we will extend $\overline{OE}$ past point $O$ to $\overline{AB}$ at point $F$.$\triangle AEF$ is similar to $\triangle ACB$. Also, $AO$ is the angle bisector of $\angle EAB$. Therefore, by the angle bisector theorem, $\frac{OE}{OF} = \frac{AE}{AF} = \frac{5}{13}$. Also, $OE = r$, so $\frac{r}{OF} = \frac{5}{13}$, so $OF = \frac{13r}{5}$. This means that\[AE = \frac{5 \cdot EF}{12} = \frac{5 \cdot (OE + OF)}{12} = \frac{5 \cdot (r + \frac{13r}{5})}{12} = \frac{18r}{12} = \frac{3r}{2}.\] We have that $EC = \frac{2r \sqrt{3}}{3}$ and that $AE = \frac{3r}{2}$, so $AC = EC + AE = \frac{2r \sqrt{3}}{3} + \frac{3r}{2} = \frac{4r \sqrt{3} + 9r}{6}$. We also were given that $AC = 5$. Therefore, we have \[\frac{4r \sqrt{3} + 9r}{6} = 5.\] This is a simple linear equation in terms of $r$. We can solve for $r$ to get $r = \boxed{\textbf{(B) } \frac{90 - 40 \sqrt{3}}{11}}.$

Solution 3

Denote by $O_1$, $O_2$, $O_3$ the centers of three spheres.

Because three congruent spheres are tangent to the base of the cone, the plane formed by $O_1$, $O_2$, $O_3$ (denoted as $\alpha$) is parallel to the base, with the distance $r$.

Let $D$ be the point that the sphere with center $O_1$ meets the base of the cone at. Hence, $O_1 D = r$.

Because three congruent spheres are mutually externally tangent to each other, $\triangle O_1 O_2 O_3$ is equilateral, with side length $2 r$.

Let $O$ be the center of the base, $V$ be the vertex of the base. Let line $OV$ and plane $\alpha$ intersect at point $D$. By symmetry, $E$ is the center $\triangle O_1 O_2 O_3$. Hence, $O_1 E = \frac{\sqrt{3}}{3} O_1 O_2 = \frac{2 \sqrt{3}}{3}$.

Let $F$ be the point that the sphere with center $O_1$ meets the side of the cone at. Hence, $O_1 F = r$.

Let line $VF$ and the base intersect at point $A$.

Hence, we only need to analyze the following 2-d geometry problem: In $\triangle VOA$ with $\angle O = 90^\circ$, $VO = 12$, $OA = 5$, there is an interior point $O_1$ whose distances to $OA$, $OV$, $VA$, are $r$, $\frac{2 \sqrt{3}}{3}$, and $r$, respectively. What is $r$?

Now, we solve this problem.

We compute the area of $\triangle VOA$ in two ways.

First, we have \begin{align*} {\rm Area} \ \triangle VOA & = \frac{1}{2} OA \cdot OV = 30 . \end{align*}

Second, we have \begin{align*} {\rm Area} \ \triangle VOA & = {\rm Area} \ \triangle O_1 OA + {\rm Area} \ \triangle O_1 OV + {\rm Area} \ \triangle O_1 VA \\ & = \frac{1}{2} OA \cdot O_1 D + \frac{1}{2} OV \cdot O_1 E + \frac{1}{2} VA \cdot O_1 F \\ & = \frac{1}{2} 5 r + \frac{1}{2} 12 \frac{2 \sqrt{3}}{3} + \frac{1}{2} 13 r \\ & = \left( 9 + 4 \sqrt{3} \right) r . \end{align*}

These two approaches to compute ${\rm Area} \ \triangle VOA$ should give me the same number. Hence, \begin{align*} r & = \frac{30}{9 + 4 \sqrt{3}} \\ & = \frac{90 - 40 \sqrt{3}}{11} . \end{align*}

Therefore, the answer is $\boxed{\textbf{(B) }\frac{90 - 40 \sqrt{3}}{11}}$.

~Steven Chen (www.professorchenedu.com)

~ ihatemath123

See Also

2021 Fall AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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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