Question

An isosceles trapezoid has an inscribed circle tangent to each of its four sides. The radius of the circle is $3,$ and the area of the trapezoid is $72.$ Let the parallel sides of the trapezoid have lengths $r$ and $s,$ with $r \ne s.$ Find $r^2 + s^2.$

Answer

Solution:

The circle is tangent to both parallel sides, so the height of the trapezoid is $2 \cdot 3 = 6.$ From the area, $\frac{r + s}{2} \cdot 6 = 72,$ so $r + s = 24.$ By the Pitot theorem the legs together also sum to $24,$ and since the trapezoid is isosceles each leg is $12.$

Dropping a perpendicular from an endpoint of the shorter base, the leg is the hypotenuse of a right triangle with legs $6$ and $\frac{|r - s|}{2}:$ $144 = 36 + \left(\frac{r - s}{2}\right)^2,$ so $(r - s)^2 = 432.$ Therefore $r^2 + s^2 = \frac{(r+s)^2 + (r-s)^2}{2} = \frac{576 + 432}{2} = 504.$

2025 AIME I Problem 6

Solution:

Problem 6 in Other Years