Difference between revisions of "2021 Fall AMC 10A Problems/Problem 22"
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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) == | ||
+ | |||
+ | 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 19:49, 22 November 2021
Problem
Inside a right circular cone with base radius and height are three congruent spheres with radius . Each sphere is tangent to the other two spheres and also tangent to the base and side of the cone. What is ?
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 . Note that the distance between this point and the plane given by is . Thus, by the point-to-plane distance formula, we have
Solving for yields .
~ Leo.Euler