Summer Problem Solving Marathon Solution #20

Let A be the vertex opposite the side of length 39, B be the vertex opposite the side of length 40, and C be the vertex opposite the side of length 25. Drop a perpendicular from A to BC. Let D be the point at which the perpendicular intersects BC.

Then BDA and CDA are both right triangles. Let BD be of length x. Then CD is of length 39 - x. Let y be the length of AD. Then, using the Pythagorean Theorem on the two right triangles, we get the following two equations:

x² + y² = 25²
(39 - x)² + y² = 40²

Subtracting the first equation from the second, we have:

(39 - x)² - x² = 40² - 25²

We factor each side using the difference of squares:

(39 - x + x)(39 - x - x) = (40 - 25)(40 + 25)
39(39 - 2x) = 15(65)
1521 - 78x = 975
78x = 546
x = 7

We then determine y:

7² + y² = 25²
49 + y² = 625
y² = 576
y = 24

Now let O be the center of the circle. Draw the two radii OA and OC. Angle CBA and angle COA both cut off the same arc of the circle. CBA is an inscribed angle and COA is a central angle, so COA is twice the size of CBA. Now drop a perpendicular from O to AC, intersecting AC at E. This perpendicular bisects AC, so AEO and CEO are congruent triangles, and the two angles AEO and CEO are both congruent to angle CBA.

Angle CBA is also an angle in the right triangle DBA. Its sine is thus AD/AB = 24/25. So the sine of angle AEO is also 24/25. Now side AE is of length 20, so OA is 20/(24/25) = 500/24 = 125/6.

This gives us the length of a radius of the circle. The length of the diameter is then twice that, or 125/3.