# 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

$[asy] size(10cm); pair A, B, C, D, O, P, H, L, X, Y; A = (-1, 0); B = (1, 0); H = (0, 0); C = (5, 0); D = (9, 0); L = (7, 0); O = (0, sqrt(160/13 - 1)); P = (7, sqrt(160/13 - 4)); X = (0, sqrt(160/13 - 4)); Y = (O + P) / 2; draw(A -- O -- B -- cycle); draw(C -- P -- D -- cycle); draw(B -- C); draw(O -- P); draw(P -- X, dashed); draw(O -- H, dashed); draw(P -- L, dashed); draw(circle(O, sqrt(160/13))); draw(circle(P, sqrt(160/13))); path b = brace(L, H); draw(b); label("R", O -- Y, N); label("R", Y -- P, N); label("R", O -- A, NW); label("R", P -- D, NE); label("1", A -- H, N); label("2", L -- D, N); label("7", b, S); [/asy]$

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 2-D, 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: $$(\sqrt{r^2-1} - \sqrt{r^2-4})^2 + 7^2 = (2r)^2.$$ Simplifying a few times, $$r^2 - 1 - 2\left(\sqrt{(r^2-1)(r^2-4)}\right) + r^2 - 4 + 49 = 4r^2$$ $$2r^2-44= -2\left(\sqrt{(r^2-1)(r^2-4)}\right)$$ $$22-r^2=\left(\sqrt{r^4 - 5r^2 + 4}\right)$$ $$r^4 -44r^2 + 484 = r^4 - 5r^2 + 4$$ $$39r^2=480$$ $$r^2=\frac{480}{39} = \frac{160}{13}.$$ Therefore, our answer is $\boxed{173}$.

-molocyxu

## Solution 2 (Official MAA)

Consider a cross section of the board and spheres with a plane that passes through the centers of the holes and centers of the spheres as shown.

$[asy] unitsize(1.5 cm); pair A, B, C, D, E, F, G, P, Q; C = dir(175); D = dir(175 + 180); P = (-2,-0.8); Q = (2,-0.8); A = (C + reflect(P,Q)*(C))/2; B = (D + reflect(P,Q)*(D))/2; E = intersectionpoint(P--A, Circle(C,1)); F = intersectionpoint(B--Q, Circle(D,1)); G = (D + reflect(A,C)*(D))/2; draw(Circle(C,1)); draw(Circle(D,1)); draw(P--Q); draw(A--(C + (0,1))); draw(B--(D + (0,1))); draw(E--C--D--F); draw(D--G); dot("A", A, NE); dot("B", B, NW); dot("C", C, NW); dot("D", D, NE); dot("E", E, SW); dot("F", F, SE); dot("G", G, SE); [/asy]$

Let $A$, $C$, and $E$ be, respectively, the center of the hole with radius $1,$ the center of the sphere resting in that hole, and a point on the edge of that hole. Let $B$, $D$, and $F$ be the corresponding points for the hole with radius $2.$ Let $G$ be the point on $\overline{AC}$ such that $\overline{AC} \perp \overline{GD}$. Let the radius of the spheres be $r = CE = DF$. Because $r^2 = AE^2 + AC^2 = 1 + AC^2$ and $r^2 = BF^2 + BD^2 = 4 + BD^2$, it follows that$$CG = AC - AG = AC - BD = \sqrt{r^2 - 1} - \sqrt{r^2-4}.$$Because $DG = 7$, $CD = 2r$, and $CD^2 = CG^2 +GD^2$, it follows that$$4r^2 = \left(\sqrt{r^2 - 1} - \sqrt{r^2-4}\right)^{\!2} + 7^2,$$which simplifies to $r^2 = \frac{160}{13}$. The requested sum is $160+13 = 173$. The value of $r$ is approximately $3.5082.$