AIME 2025 I · 第 6 题

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 $72 = 6 \cdot \frac{r + s}{2} \implies r + s = 24$. $(r + s)^2 - 2rs = 576 - 72 = \boxed{504}$.

-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,

$$\begin{aligned} &r^2 + s^2 = (r + s)^2 - 2rs \\ &= 24^2 - 2 \cdot 36 \\ &= \boxed{504} \end{aligned}$$

Solution 4(Double-angle Formula)

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

Since $\angle DAB = 2\alpha$, $\tan(2\alpha)=\frac{12}{s - r}=\frac{12s}{s(s - r)}=\frac{12s}{s^{2}-sr}$.

Set $\frac{12s}{s^{2}-sr}=\frac{12s}{s^{2}-36}$. Since $s\neq0$, we can cancel out $12s$ from both sides of the equation, getting $s^{2}-sr=s^{2}-36$. Subtracting $s^{2}$ from both sides, we have $-sr=-36$, so $sr = 36$.

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

Substitute $rs = 36$ and $(r + s)^{2}=576$ into the formula: $r^{2}+s^{2}=576-2\times36=576 - 72=504$.

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

Solution 6 (Brahmagupta's)

The area of the trapezoid is $\frac{r + s}{2} \cdot 6 = 72 \implies r + s = 24.$ Note that external tangents of a circle are equal. Let $a$ and $s - a$ be the two external tangents making up the bottom base and $b$ and $r - b$ be the two external tangents making up the top base. Since the trapezoid is isosceles, $a + b = r + s - (a + b) \implies a + b = 12.$ So the legs of the trapezoid have length $12.$ From here, we can apply Brahmagupta's Formula to the trapezoid since isosceles trapezoids are cyclic: $\sqrt{\frac{r + s}{2} \cdot \frac{r + s}{2} \cdot \frac{24 - r + s}{2} \cdot \frac{24 + r - s}{2}} = 72.$ Substituting $r + s = 24$ and solving, we get $(r - s)^2 = 432 \implies s - r = 12\sqrt{3}.$ Since we know $s + r = 24,$ we can just solve this system to get $s = 12 + 6\sqrt{3}$ and $r = 12 - 6\sqrt{3}.$ The answer is $(12 + 6\sqrt{3})^2 + (12 - 6\sqrt{3})^2 = \boxed{504}.$

~grogg007

Video Solution - Trapezoid and Inscribed Circle by HungryCalculator

https://www.youtube.com/watch?v=4xaFu0RF1j4&pp=0gcJCYcKAYcqIYzv

~HungryCalculator

Video Solution(Fast and Easy)

https://youtu.be/zK5C_FkdhlM

~MC