2025 AIME I Problems/Problem 6

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 isosceles 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, $\left(\frac{r - s}{2}\right)^2 + 6^2 = 12^2.$ $\left(\frac{r-s}{2}\right)^2=108.$ $(r-s)^2=432.$ Noting that $\frac{(r + s)^2 + (r - s)^2}{2} = r^2 + s^2$ we find, $\frac{(24^2+432)}{2}=\boxed{504}$

~mathkiddus

Solution 2 (Trigonometry)

Draw angle bisectors from the bottom left vertex to the center of the circle. Call the angle formed $\theta$ . Drawing a line from the center of the circle to the midway point of the bottom base of the trapezoid makes a right angle, and the other angle has to be $90^{\circ} - \theta$ . Then draw a line segment from the center of the circle to the top left vertex, then you have a right triangle. The smaller angle of this triangle is $180^{\circ} - (180^{\circ} - \theta) = \theta$ . This means $\frac{r}{2} = \frac{3}{\tan \theta} \implies r = \frac{6}{\tan \theta}$ . This also means $\frac{s}{2} = 3 \tan \theta \implies s = 6 \tan \theta$ . Note that $r^2 + s^2 = (r + s)^2 - 2rs.$ $rs = \frac{6}{\tan \theta} \cdot 6 \tan \theta = 36 \implies 2rs = 72$ . The area of the trapezoid is $72 = 6 \cdot \frac{r + s}{2} \implies r + s = 24$ . $(r + s)^2 - 2rs = 576 - 72 = \boxed{504}$ .

$[asy] size(15cm); draw(circle((0,0), 3)); draw((-0.5 * (12 + 6sqrt(3)), -3) -- (0.5 * (12 + 6sqrt(3)), -3) -- (0.5 * (12 - 6sqrt(3)), 3) -- (-0.5 * (12 - 6sqrt(3)), 3) -- cycle); draw((0, 0) -- (-0.5 * (12 + 6*sqrt(3)),-3) -- cycle); draw((0, 0) -- (0.5 * (12 + 6*sqrt(3)),-3) -- cycle); draw((0, 0) -- (-0.5 * (12 - 6*sqrt(3)),3) -- cycle); draw((0, 0) -- (0.5 * (12 - 6*sqrt(3)),3) -- cycle); draw((0, 0) -- (0,3) -- cycle); draw((0, 0) -- (0,-3) -- cycle); draw((-0.5, -3) -- (-0.5,-2.5) -- (0.5, -2.5) -- (0.5, -3) -- cycle); draw((-0.25, 3) -- (-0.25, 2.75) -- (0.25, 2.75) -- (0.25, 3) -- cycle); label("$\theta$", (-0.5 * (12 + 6*sqrt(3)) + 3, -3), NE); label("$\theta$", (0.5 * (12 + 6*sqrt(3)) - 3, -3), NW); label("$\theta$", (-0.5 * (12 + 6*sqrt(3)) + 3, -2), NE); label("$\theta$", (0.5 * (12 + 6*sqrt(3)) - 3, -2), NW); label("$90^{\circ} - \theta$", (0, -0.5), SW); label("$90^{\circ} - \theta$", (0, -0.5), SE); label("$90^{\circ}$", (0 - 0.1, 0), NW); label("$90^{\circ}$", (0 + 0.1, 0), NE); label("$\frac{r}{2}$", (-0.25 * (12 + 6*sqrt(3)), -3), S); label("$\frac{r}{2}$", (0.25 * (12 + 6*sqrt(3)), -3), S); label("$\theta$", (-0.1, 1.75), E); label("$\theta$", (0.1, 1.75), W); label("$\frac{s}{2}$", (-0.25 * (12 - 6*sqrt(3)), 3), N); label("$\frac{s}{2}$", (0.25 * (12 - 6*sqrt(3)), 3), N); [/asy]$

-alwaysgonnagiveyouup

Solution 3 (Fastest formula)

Denote the radius of the inscribed circle as $R$ , and the parallel sides as $r$ and $s$ . By formula, we get $R = 3 = \frac{1}{2} \cdot \sqrt{rs}$ , where $rs = 36$ . Also, by formula, $A = 72 = \frac{1}{2} \cdot \sqrt{rs} \cdot (r + s)$ , where $r + s = 24$ . Therefore, $\begin{aligned} r^{2} + s^{2} = (r + s)^{2} - 2 r s \\ = 24^{2} - 2 \cdot 36 \\ = 504 \end{aligned}$

Solution 4(Double-angle Formula)

Let $∠ O A B = α$ , $\tan (α) = \frac{6}{s}$ . By the double - angle formula for tangent $\tan (2 α) = \frac{2 \tan α}{1 - \tan^{2} α} = \frac{2 \times \frac{6}{s}}{1 - (\frac{6}{s})^{2}} = \frac{\frac{12}{s}}{\frac{s^{2} - 36}{s^{2}}} = \frac{12 s}{s^{2} - 36}$ .

Since $∠ D A B = 2 α$ , $\tan (2 α) = \frac{12}{s - r} = \frac{12 s}{s (s - r)} = \frac{12 s}{s^{2} - s r}$ .

Set $\frac{12 s}{s^{2} - s r} = \frac{12 s}{s^{2} - 36}$ . Since $s \neq 0$ , we can cancel out $12 s$ from both sides of the equation, getting $s^{2} - s r = s^{2} - 36$ . Subtracting $s^{2}$ from both sides, we have $- s r = - 36$ , so $s r = 36$ .

Assume $(r + s)^{2} = 576$ . Using the formula $(r + s)^{2} = r^{2} + 2 r s + s^{2}$ , then $r^{2} + s^{2} = (r + s)^{2} - 2 r s$ .

Substitute $r s = 36$ and $(r + s)^{2} = 576$ into the formula: $r^{2} + s^{2} = 576 - 2 \times 36 = 576 - 72 = 504$ .

So the final answer is $504$ . By Airbus 320-214.

Formula reference to here: https://en.wikipedia.org/wiki/Tangential_trapezoid

~Mitsuihisashi14

~ LaTeX by alwaysgonnagiveyouup

Solution 5

The height of the trapezoid is clearly the diameter of the circle or $6$ . We let the larger base be $r$ , and the smaller one be $s$ . We have by the area of a trapezoid: $\dfrac{r+s}{2} \cdot 6 = 72 \Longrightarrow r+s = 24$ . Now, by Pitot's theorem, and letting the legs of the trapezoid be $x$ , we have that $r+s = 2x = 24 \Longrightarrow x = 12$ . Now, by pythag we have that $\dfrac{r-s}{2} = 6\sqrt3 \Longrightarrow r-s = 12\sqrt3$ . Now, by systems of equations, we find that $r = 12+6\sqrt3$ , and $s = 12-6\sqrt3$ . Now, we have that $r^2 + s^2 = (12+6\sqrt3)^2+(12-6\sqrt3)^2 = \boxed{504}$

-jb2015007

Video Solution(Fast and Easy)

https://youtu.be/zK5C_FkdhlM

~MC

2025 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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