AIME 2022 I · 第 10 题

Solution 1

This solution refers to the Diagram section.

We let $\ell$ be the plane that passes through the spheres and $O_A$ and $O_B$ be the centers of the spheres with radii $11$ and $13$. We take a cross-section that contains $A$ and $B$, which contains these two spheres but not the third, as shown below:

Because the plane cuts out congruent circles, they have the same radius and from the given information, $AB = \sqrt{560}$. Since $ABO_BO_A$ is a trapezoid, we can drop an altitude from $O_A$ to $BO_B$ to create a rectangle and triangle to use Pythagorean theorem. We know that the length of the altitude is $\sqrt{560}$ and let the distance from $O_B$ to $D$ be $x$. Then we have $x^2 = 576-560 \implies x = 4$.

We have $AO_A = BD$ because of the rectangle, so $\sqrt{11^2-r^2} = \sqrt{13^2-r^2}-4$. Squaring, we have $121-r^2 = 169-r^2 + 16 - 8 \cdot \sqrt{169-r^2}$. Subtracting, we get $8 \cdot \sqrt{169-r^2} = 64 \implies \sqrt{169-r^2} = 8 \implies 169-r^2 = 64 \implies r^2 = 105$. We also notice that since we had $\sqrt{169-r^2} = 8$ means that $BO_B = 8$ and since we know that $x = 4$, $AO_A = 4$.

We take a cross-section that contains $A$ and $C$, which contains these two spheres but not the third, as shown below:

We have $CO_C = \sqrt{19^2-r^2} = \sqrt{361 - 105} = \sqrt{256} = 16$. Since $AO_A = 4$, we have $EO_C = 16-4 = 12$. Using Pythagorean theorem, $O_AE = \sqrt{30^2 - 12^2} = \sqrt{900-144} = \sqrt{756}$. Therefore, $O_AE^2 = AC^2 = \boxed{756}$.

~KingRavi

Solution 2

Let the distance between the center of the sphere to the center of those circular intersections as $a,b,c$ separately.

According to the problem, we have $a^2-11^2=b^2-13^2=c^2-19^2; (11+13)^2-(b-a)^2=560.$ After solving we have $b-a=4,$ plug this back to $11^2-a^2=13^2-b^2,$ we have $a=4, b=8,$ and $c=16.$

The desired value is $(11+19)^2-(16-4)^2=\boxed{756}.$

~bluesoul

Solution 3

Denote by $r$ the radius of three congruent circles formed by the cutting plane. Denote by $O_A$, $O_B$, $O_C$ the centers of three spheres that intersect the plane to get circles centered at $A$, $B$, $C$, respectively.

Because three spheres are mutually tangent, $O_A O_B = 11 + 13 = 24$, $O_A O_C = 11 + 19 = 30$.

We have $O_A A^2 = 11^2 - r^2$, $O_B B^2 = 13^2 - r^2$, $O_C C^2 = 19^2 - r^2$.

Because $O_A A$ and $O_B B$ are perpendicular to the plane, $O_A AB O_B$ is a right trapezoid, with $\angle O_A A B = \angle O_B BA = 90^\circ$.

Hence,

$$\begin{aligned} O_B B - O_A A & = \sqrt{O_A O_B^2 - AB^2} \\ & = 4 . \hspace{1cm} (1) \end{aligned}$$ Recall that

$$\begin{aligned} O_B B^2 - O_A A^2 & = \left( 13^2 - r^2 \right) - \left( 11^2 - r^2 \right) \\ & = 48 . \hspace{1cm} (2) \end{aligned}$$ Hence, taking $\frac{(2)}{(1)}$, we get

$$O_B B + O_A A = 12 . \hspace{1cm} (3)$$ Solving (1) and (3), we get $O_B B = 8$ and $O_A A = 4$.

Thus, $r^2 = 11^2 - O_A A^2 = 105$.

Thus, $O_C C = \sqrt{19^2 - r^2} = 16$.

Because $O_A A$ and $O_C C$ are perpendicular to the plane, $O_A AC O_C$ is a right trapezoid, with $\angle O_A A C = \angle O_C CA = 90^\circ$.

Therefore,

$$\begin{aligned} AC^2 & = O_A O_C^2 - \left( O_C C - O_A A \right)^2 \\ & = \boxed{756}. \end{aligned}$$ $\textbf{FINAL NOTE:}$ In our solution, we do not use the condition that spheres $A$ and $B$ are externally tangent. This condition is redundant in solving this problem.

$\textbf{MORE FINAL NOTE:}$ the above note is incorrect because that condition was used at the start when claiming $O_AO_B=24$. Perhaps the note is referring to spheres $B$ and $C$.

~Steven Chen (www.professorcheneeu.com)

~anonymous (minor edits)

Video Solution 1 by Challenge 25

https://www.youtube.com/watch?v=yeuJDQ1LTlY

Video Solution 2

https://www.youtube.com/watch?v=SqLiV2pbCpY&t=15s

~Steven Chen (www.professorcheneeu.com)

Video Solution 3 Mathematical Dexterity

https://www.youtube.com/watch?v=HbBU13YiopU

Video Solution 4 by StressedPineapple

https://youtube.com/watch?v=x66z0IMj0as&t=303s