Difference between revisions of "2022 AIME I Problems/Problem 10"

Revision as of 12:28, 21 February 2022

Problem

Three spheres with radii $11$ , $13$ , and $19$ are mutually externally tangent. A plane intersects the spheres in three congruent circles centered at $A$ , $B$ , and $C$ , respectively, and the centers of the spheres all lie on the same side of this plane. Suppose that $AB^2 = 560$ . Find $AC^2$ .

Diagrams

$[asy] size(500); pair A, B, OA, OB; B = (0,0); A = (-23.6643191,0); OB = (0,-8); OA = (-23.6643191,-4); draw(circle(OB,13)); draw(circle(OA,11)); draw((-48,0)--(24,0)); label("$l$",(-42,1),N); label("$A$",A,N); label("$B$",B,N); label("$O_A$",OA,S); label("$O_B$",OB,S); draw(A--OA); draw(B--OB); draw(OA--OB); draw(OA--(0,-4)); draw(OA--(-33.9112699,0)); draw(OB--(10.2469508,0)); label("$24$",midpoint(OA,OB),S); label("$\sqrt{560}$",midpoint(A,B),N); label("$11$",midpoint(OA,(-33.9112699,0)),S); label("$13$",midpoint(OB,(10.2469508,0)),S); label("$r$",midpoint(midpoint(A,B),A),N); label("$r$",midpoint(midpoint(A,B),B),N); [/asy]$

Solution 1

Let the distance between the center of the sphere to the center of those circular intersections as $a,b,c$ separately. $a-11,b-13,c-19$ . 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 $11^2-a^2=13^2-b^2; a=4,b=8,c=16$

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

~bluesoul

Solution 2

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, $O_A O_B = 11 + 13 = 24$ , $O_A O_C = 11 + 19 = 32$ .

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

Recall that $\begin{align*} O_B B^2 - O_A A^2 & = \left( 13^2 - r^2 \right) - \left( 11^2 - r^2 \right) \\ & = 48 . \hspace{1cm} (2) \end{align*}$

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, $\begin{align*} AC^2 & = O_A O_C^2 - \left( O_C C - O_A A \right)^2 \\ & = \boxed{\textbf{(756) }} . \end{align*}$

$\textbf{FINAL NOTE:}$ In our solution, we do not use the conditio that spheres $A$ and $B$ are externally tangent. This condition is redundant in solving this problem.

~Steven Chen (www.professorcheneeu.com)

Video Solution

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

~Steven Chen (www.professorcheneeu.com)

Video Solution 2 (Mathematical Dexterity)

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

2022 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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