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Difference between revisions of "2021 Fall AMC 10A Problems/Problem 22"

(Created page with "== Problem == Inside a right circular cone with base radius <math>5</math> and height <math>12</math> are three congruent spheres with radius <math>r</math>. Each sphere is ta...")
 
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<math>\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}</math>
 
<math>\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}</math>
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== Solution 1 (Coordinates) ==
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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 <math>(0, 2r/\sqrt{3}, r)</math>. Note that the distance between this point and the plane given by <math>12y+5z=60</math> is <math>r</math>. Thus, by the point-to-plane distance formula, we have
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<cmath> \frac{|12 \cdot 2r/\sqrt{3} + 5r - 60|}{\sqrt{0^2+5^2+12^2}}=r. </cmath>
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Solving for <math>r</math> yields <math>r = \boxed{\textbf{(B) }\dfrac{90 - 40\sqrt3}{11}}</math>.
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~ Leo.Euler

Revision as of 20:49, 22 November 2021

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

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