Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2025 AIME I Problems/Problem 6"
m
Line 32: Line 32:
 
==Solution 1==
 
==Solution 1==
 
To begin with, because of tangents from the circle to the bases, the height is <math>2\cdot3=6.</math> The formula for the area of a trapezoid is <math>\frac{h(b_1+b_2)}{2}.</math> Plugging in our known values we have <cmath>\frac{6(r+s)}{2}=72.</cmath> <cmath>r+s=24.</cmath>
 
To begin with, because of tangents from the circle to the bases, the height is <math>2\cdot3=6.</math> The formula for the area of a trapezoid is <math>\frac{h(b_1+b_2)}{2}.</math> Plugging in our known values we have <cmath>\frac{6(r+s)}{2}=72.</cmath> <cmath>r+s=24.</cmath>
−
Next, we use Pitot's Theorem which states for tangential quadrilaterals <math>AB+CD=AD+BC.</math> Since we are given <math>ABCD</math> is an isosceles trapezoid we have <math>AD=BC=x.</math> Using Pitot's we find, <cmath>AB+CD=r+s=2x=24.</cmath> <cmath>x=12.</cmath> Finally we can use the Pythagorean Theorem by dropping an altitude from D, <cmath>(\frac{r - s}{2})^2 + 6^2 = 12^2.</cmath> <cmath>(\frac{r-s}{2})^2=108.</cmath> <cmath>(r-s)^2=324.</cmath> Noting that <math>\frac{(r + s)^2 + (r - s)^2}{2} = r^2 + s^2</math> we find, <cmath>\frac{(24^2+324)}{2}=\boxed{504}</cmath>
+
Next, we use Pitot's Theorem which states for tangential quadrilaterals <math>AB+CD=AD+BC.</math> Since we are given <math>ABCD</math> is an isosceles trapezoid we have <math>AD=BC=x.</math> Using Pitot's we find, <cmath>AB+CD=r+s=2x=24.</cmath> <cmath>x=12.</cmath> Finally we can use the Pythagorean Theorem by dropping an altitude from D, <cmath>\left(\frac{r - s}{2}\right)^2 + 6^2 = 12^2.</cmath> <cmath>\left(\frac{r-s}{2}\right)^2=108.</cmath> <cmath>(r-s)^2=324.</cmath> Noting that <math>\frac{(r + s)^2 + (r - s)^2}{2} = r^2 + s^2</math> we find, <cmath>\frac{(24^2+324)}{2}=\boxed{504}</cmath>
  
 
~[[User:Mathkiddus|mathkiddus]]
 
~[[User:Mathkiddus|mathkiddus]]

Revision as of 20:53, 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

See also

2025 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2025_AIME_I_Problems/Problem_6&oldid=242340"