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.

$\mathbb{QED.}$

Add-on

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.

1992 USAMO (Problems • Resources)
Preceded by Problem 3	Followed by Problem 5
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions

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