# 1992 USAMO Problems/Problem 4

## Problem

Chords $AA'$, $BB'$, and $CC'$ of a sphere meet at an interior point $P$ but are not contained in the same plane. The sphere through $A$, $B$, $C$, and $P$ is tangent to the sphere through $A'$, $B'$, $C'$, and $P$. Prove that $AA'=BB'=CC'$.

## Solution

### Solution 1

Consider the plane through $A,A',B,B'$. This plane, of course, also contains $P$. We can easily find the $\triangle APB$ is isosceles because the base angles are equal. Thus, $AP=BP$. Similarly, $A'P=B'P$. Thus, $AA'=BB'$. By symmetry, $BB'=CC'$ and $CC'=AA'$, and hence $AA'=BB'=CC'$ as desired. $\mathbb{QED.}$

By another person ^v^

The person that came up with the solution did not prove that $\triangle APB$ is isosceles nor the base angles are congruent. I will add on to the solution.

There is a common tangent plane that pass through $P$ for the $2$ spheres that are tangent to each other.

Since any cross section of sphere is a circle. It implies that $A$, $A'$, $B$, $B'$ be on the same circle ( $\omega_1$), $A$, $B$, $P$ be on the same circle ( $\omega_2$), and $A'$, $B'$, $P$ be on the same circle ( $\omega_3$). $m\angle APB= m\angle A'PB'$ because they are vertical angles. By power of point, $(AP)(A'P)=(BP)(B'P)\rightarrow\frac{AP}{B'P}=\frac{BP}{A'P}$

By the SAS triangle simlarity theory, $\triangle APB \sim\triangle B'PA'$. That implies that $\angle ABP\cong\angle B'PA'$.

Let's call the interception of the common tangent plane and the plane containing $A$, $A'$, $B$, $B'$, $P$, line $l$. $l$ must be the common tangent of $\omega_2$ and $\omega_3$.

The acute angles form by $l$ and $\overline{AA'}$ are congruent to each other (vertical angles) and by the tangent-chord theorem, the central angle of chord $\overline{AP}$ and $\overline{A'P}$ are equal.

Similarly the central angle of chord $\overline{BP}$ and $\overline{B'P}$ are equal.

The length of any chord with central angle $2\theta$ and radius $r$ is $2r\sin\left({\theta}\right)$, which can easily been seen if we drop the perpendicular from the center to the chord.

Thus, $\frac{AP}{A'P}=\frac{BP}{B'P}$.

By the SAS triangle simlarity theory, $\triangle APB \sim\triangle A'PB'$. That implies that $\angle ABP\cong\angle B'PA'$.

That implies that $\angle ABP\cong\angle A'PB'\cong\angle B'PA'$. Thus, $\triangle A'PB'$ is an isosceles triangle and since $\triangle APB \sim\triangle A'PB'$, then $\triangle APB$ must be an isosceles triangle too.

### Solution 2

Call the large sphere $O_1$, the one containing $A$ $O_2$, and the one containing $A'$ $O_3$. The centers are $c_1$, $c_2,$ and $c_3$.

Since two spheres always intersect in a circle , we know that A,B, and C must lie on a circle ( $w_1$)completely contained in $O_1$ and $O_2$

Similarly, A', B', and C' must lie on a circle ( $w_2$) completely contained in $O_1$ and $O_3$.

So, we know that 3 lines connecting a point on $w_1$ and P hit a point on $w_2$. This implies that $w_1$ projects through P to $w_2$, which in turn means that $w_1$ is in a plane parallel to that of $w_2$. Then, since $w_1$ and $w_2$ lie on the same sphere, we know that they must have the same central axis, which also must contain P (since the center projects through P to the other center).

So, all line from a point on $w_1$ to P are of the same length, as are all lines from a point on $w_2$ to P. Since AA', BB', and CC' are all composed of one of each type of line, they must all be equal.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 