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.

See Also

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

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