1992 USAMO Problems/Problem 4
Contents
Problem
Chords , , and of a sphere meet at an interior point but are not contained in the same plane. The sphere through , , , and is tangent to the sphere through , , , and . Prove that .
Solution
Consider the plane through . This plane, of course, also contains . We can easily find the is isosceles because the base angles are equal. Thus, . Similarly, . Thus, . By symmetry, and , and hence as desired.
Add-on
By another person ^v^
The person that came up with the solution did not prove that is isosceles nor the base angles are congruent. I will add on to the solution.
There is a common tangent plane that pass through for the spheres that are tangent to each other.
Since any cross section of sphere is a circle. It implies that , , , be on the same circle (), , , be on the same circle (), and , , be on the same circle ().
$m\angle APB= m\angleA'PB'$ (Error compiling LaTeX. ! Undefined control sequence.) because they are vertical angles. By power of point,
By the SAS triangle simlarity theory, . That implies that $\angle ABP\cong\angleB'PA'$ (Error compiling LaTeX. ! Undefined control sequence.).
Let's call the interception of the common tangent plane and the plane containing , , , , , line .
must be the common tangent of and .
The acute angles form by and $\overbar{AA'}$ (Error compiling LaTeX. ! Undefined control sequence.) are congruent to each other (vertical angles) and by the tangent-chord theorem, the central angle of chord $\overbar{AP}$ (Error compiling LaTeX. ! Undefined control sequence.) and $\overbar{A'P}$ (Error compiling LaTeX. ! Undefined control sequence.) are equal.
Similarly the central angle of chord $\overbar{BP}$ (Error compiling LaTeX. ! Undefined control sequence.) and $\overbar{B'P}$ (Error compiling LaTeX. ! Undefined control sequence.) are equal.
The length of any chord with central angle and radius is , which can easily been seen if we drop the perpendicular from the center to the chord.
Thus, .
By the SAS triangle simlarity theory, . That implies that $\angle ABP\cong\angleB'PA'$ (Error compiling LaTeX. ! Undefined control sequence.).
That implies that $\angle ABP\cong\angleA'PB'\cong\angleB'PA'$ (Error compiling LaTeX. ! Undefined control sequence.). Thus, is an isosceles triangle and since , is an isosceles triangle too.
Resources
1992 USAMO (Problems • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |