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2025 AIME I Problems/Problem 6

Revision as of 17:19, 13 February 2025 by Mathkiddus (talk | contribs) (→‎Solution 1)

Problem

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 $r \neq s$. Find $r^2+s^2$

Diagram

[asy] unitsize(0.5 cm); real r = 12 + 6*sqrt(3); real s = 12 - 6*sqrt(3); real h = 6; pair A = (-r/2, 0); pair B = ( r/2, 0); pair C = ( s/2, h); pair D = (-s/2, h); draw(A--B--C--D--cycle); pair O = (0, h/2); draw(circle(O, 3)); dot(A); label("$A$", A, SW); dot(B); label("$B$", B, SE); dot(C); label("$C$", C, NE); dot(D); label("$D$", D, NW); dot(O); label("$O$", (0,h/2), E); label("$r$", midpoint(A--B), S); label("$s$", midpoint(C--D), N); [/asy]

Solution 1

To begin with, because of tangents from the circle to the bases, the height is $2\cdot3=6.$ The formula for the area of a trapezoid is $\frac{h(b_1+b_2)}{2}.$ Plugging in our known values we have \[\frac{6(r+s)}{2}=72.\] \[r+s=24.\] Next, we use Pitot's Theorem which states for tangential quadrilaterals $AB+CD=AD+BC.$ Since we are given ABCD is an isocelese trapezoid we have $AD=BC=x.$ Using Pitot's we find, \[AB+CD=r+s=2x=24.\] \[x=12.\] Finally we can use the Pythagorean Theorem by dropping an altitude from D, \[(\frac{r - s}{2})^2 + 6^2 = 12^2.\] \[(\frac{r-s}{2})^2=108.\] \[(r-s)^2=324.\] Noting that $\frac{(r + s)^2 + (r - s)^2}{2} = r^2 + s^2$ we find, \[\frac{(24^2+324)}{2}=\boxed{504}\]

-mathkiddus

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