2025 AIME I Problem 6

Attempt Problem 6 of the 2025 AIME I below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2025 AIME I solutions, or check the answer key.

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6.

An isosceles trapezoid has an inscribed circle tangent to each of its four sides. The radius of the circle is 3,3, and the area of the trapezoid is 72.72. Let the parallel sides of the trapezoid have lengths rr and s,s, with rs.r \ne s. Find r2+s2.r^2 + s^2.

Answer: 504
Concepts:trapezoidincircle, incenter, and inradiusPythagorean Theorem
Difficulty rating: 2230
Solution:

The circle is tangent to both parallel sides, so the height of the trapezoid is 23=6.2 \cdot 3 = 6. From the area, r+s26=72,\frac{r + s}{2} \cdot 6 = 72, so r+s=24.r + s = 24. By the Pitot theorem the legs together also sum to 24,24, and since the trapezoid is isosceles each leg is 12.12.

Dropping a perpendicular from an endpoint of the shorter base, the leg is the hypotenuse of a right triangle with legs 66 and rs2:\frac{|r - s|}{2}: 144=36+(rs2)2,144 = 36 + \left(\frac{r - s}{2}\right)^2, so (rs)2=432.(r - s)^2 = 432. Therefore r2+s2=(r+s)2+(rs)22r^2 + s^2 = \frac{(r+s)^2 + (r-s)^2}{2} =576+4322=504.= \frac{576 + 432}{2} = 504.

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