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 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 (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 $3.5082.$

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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