2020 AIME I Problems/Problem 6

Problem

A flat board has a circular hole with radius $1$ and a circular hole with radius $2$ such that the distance between the centers of the two holes is $7.$ Two spheres with equal radii sit in the two holes such that the spheres are tangent to each other. The square of the radius of the spheres is $\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution

$[asy] size(10cm); pair A, B, C, D, O, P, H, L, X, Y; A = (-1, 0); B = (1, 0); H = (0, 0); C = (5, 0); D = (9, 0); L = (7, 0); O = (0, sqrt(160/13 - 1)); P = (7, sqrt(160/13 - 4)); X = (0, sqrt(160/13 - 4)); Y = (O + P) / 2; draw(A -- O -- B -- cycle); draw(C -- P -- D -- cycle); draw(B -- C); draw(O -- P); draw(P -- X, dashed); draw(O -- H, dashed); draw(P -- L, dashed); draw(circle(O, sqrt(160/13))); draw(circle(P, sqrt(160/13))); path b = brace(L-(0,1), H-(0,1),0.5); draw(b); label("$r$", O -- Y, N); label("$r$", Y -- P, N); label("$r$", O -- A, NW); label("$r$", P -- D, NE); label("$1$", A -- H, N); label("$2$", L -- D, N); label("$7$", b, S); dot(A^^B^^C^^D^^O^^P^^H^^L^^X^^Y,linewidth(4)); [/asy]$ Set the common radius to $r$ . First, take the cross section of the sphere sitting in the hole of radius $1$ . If we draw the perpendicular bisector of the chord (the hole) through the circle, this line goes through the center. Connect the center also to where the chord hits the circle, for a right triangle with hypotenuse $r$ and base $1$ . Therefore, the height of this circle outside of the hole is $\sqrt{r^2-1}$ .

The other circle follows similarly for a height (outside the hole) of $\sqrt{r^2-4}$ . Now, if we take the cross section of the entire board, essentially making it two-dimensional, we can connect the centers of the two spheres, then form another right triangle with base $7$ , as given by the problem. The height of this triangle is the difference between the heights of the parts of the two spheres outside the holes, which is $\sqrt{r^2-1} - \sqrt{r^2-4}$ . Now we can set up an equation in terms of $r$ with the Pythagorean theorem: $\left(\sqrt{r^2-1} - \sqrt{r^2-4}\right)^2 + 7^2 = (2r)^2.$ Simplifying a few times, $\begin{align*} r^2 - 1 - 2\left(\sqrt{(r^2-1)(r^2-4)}\right) + r^2 - 4 + 49 &= 4r^2 \\ 2r^2-44 &= -2\left(\sqrt{(r^2-1)(r^2-4)}\right) \\ 22-r^2 &= \left(\sqrt{r^4 - 5r^2 + 4}\right) \\ r^4 -44r^2 + 484 &= r^4 - 5r^2 + 4 \\ 39r^2&=480 \\ r^2&=\frac{480}{39} = \frac{160}{13}. \end{align*}$ Therefore, our answer is $\boxed{173}$ .

~molocyxu

Solution 2 (Tangential distance)

Let $A$ and $B$ be the centers of the holes, $C$ is point of crossing $AB$ and radical axes of the circles. So $C$ has equal tangential distance to any point of both spheres. In particular to the circles (https://en.wikipedia.org/wiki/Radical_axis.)

$CA = \frac {AB} {2} – \frac {r_A^2 – r_B^2}{2 AB} = \frac{23}{7}, CB = AB - AC =\frac{26}{7},$ $CA' = CB'= BC^2 – r_B^2 = \frac {4}{7} \sqrt{30}.$

Let $D$ be the point of tangency of the spheres common radius $R$ centered at $O$ and $O'.$ Let $\alpha$ be the angle between $OO'$ and flat board. In the plane, perpendicular to board $DC \perp OO', DC = \frac {4}{7} \sqrt{30}.$

$\sin \alpha = \frac {AC – BC}{DC} = \frac {3}{8\sqrt 30}.$ $\cos \alpha = \sqrt {1 – \frac {9}{64 \cdot 30}} = \sqrt{ \frac {637}{640}} = \frac {7}{2} \sqrt {\frac{13}{160}}.$ $2R \cos \alpha = AB = 7 \implies R = \frac {\frac{7}{2} } {\frac{7}{2}\sqrt \frac{13}{160}} = \sqrt {\frac{160}{13}} .$ ~vvsss, www.deoma-cmd.ru

Video solution (With Animation)

https://youtu.be/cOf9uTJ9J40

Video Solution

https://www.youtube.com/watch?v=qCTq8KhZfYQ

2020 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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