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Difference between revisions of "2025 AIME I Problems/Problem 6"

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-alwaysgonnagiveyouup
 
-alwaysgonnagiveyouup
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==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 = 1/2 * \sqrt{rs}, where rs = 36
 +
Also, by formula, A = 72 = 1/2 * \sqrt{rs} * (r + s), where r + s = 24
 +
Therefore, r^2 + s^2 = (r + s)^2 - 2 * r * s
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= 24^2 - 2 * 36
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= 504
 +
 +
Formula reference to here: https://en.wikipedia.org/wiki/Tangential_trapezoid
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 +
(Could anyone please help me fix the latex, really appreciate it)
 +
~Mitsuihisashi14
  
 
==See also==
 
==See also==

Revision as of 21:00, 13 February 2025

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, \[\left(\frac{r - s}{2}\right)^2 + 6^2 = 12^2.\] \[\left(\frac{r-s}{2}\right)^2=108.\] \[(r-s)^2=324.\] Noting that $\frac{(r + s)^2 + (r - s)^2}{2} = r^2 + s^2$ we find, \[\frac{(24^2+324)}{2}=\boxed{504}\]

~mathkiddus

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 = 1/2 * \sqrt{rs}, where rs = 36 Also, by formula, A = 72 = 1/2 * \sqrt{rs} * (r + s), where r + s = 24 Therefore, r^2 + s^2 = (r + s)^2 - 2 * r * s = 24^2 - 2 * 36 = 504

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

(Could anyone please help me fix the latex, really appreciate it) ~Mitsuihisashi14

See also

2025 AIME I (Problems • Answer Key • Resources)
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Problem 5 		Followed by
Problem 7
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All AIME Problems and Solutions

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