AIME I 2025 (Problem 6)An isosceles trapezoid has an inscribed circle tangent to each of its four sides. The radius of the circle is 333, and the area of the trapezoid is 727272. Let the parallel sides of the trapezoid have lengths rrr and sss, with r≠sr\ne sr=s. Find r2+s2r^2+s^2r2+s2.Related TopicsToolkit 67 — When We Have a Trapezoid in a Problem, Draw the AltitudesToolkit 68 — Circles, Tangents, and Radical AxisHints (6)Hint 1Hint 2Hint 3[ABCD]=6(r+s)2=72⇒r+s=24[ABCD]=\frac{6(r+s)}{2}=72\Rightarrow r+s=24[ABCD]=26(r+s)=72⇒r+s=24Hint 4By the Pythagorean Theorem in △AED\triangle AED△AED:AE2+DE2=AD2AE^2+DE^2=AD^2AE2+DE2=AD262+(s−r2)2=(s+r2)26^2+\left(\frac{s-r}{2}\right)^2=\left(\frac{s+r}{2}\right)^262+(2s−r)2=(2s+r)2Hint 5By Hints 3 and 4:(s−r2)2=(s+r2)2−62=122−62=144−36=108\left(\frac{s-r}{2}\right)^2=\left(\frac{s+r}{2}\right)^2-6^2=12^2-6^2=144-36=108(2s−r)2=(2s+r)2−62=122−62=144−36=108(s−r)2=4⋅108=432(s-r)^2=4\cdot108=432(s−r)2=4⋅108=432Hint 6r2+s2=12((r+s)2+(s−r)2)=12(242+432)r^2+s^2=\frac12\left((r+s)^2+(s-r)^2\right)=\frac12\left(24^2+432\right)r2+s2=21((r+s)2+(s−r)2)=21(242+432)=12(576+432)=12(1008)=504=\frac12(576+432)=\frac12(1008)=504=21(576+432)=21(1008)=504Final Answer504504504
