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Difference between revisions of "2022 AIME I Problems/Problem 10"
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==solution 1==
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==Solution 1==
 
Let the distance between the center of the sphere to the center of those circular intersections as <math>a,b,c</math> separately. <math>a-11,b-13,c-19</math>. According to the problem, we have <math>a^2-11^2=b^2-13^2=c^2-19^2;(11+13)^2-(b-a)^2=560</math>. After solving we have <math>b-a=4</math>, plug this back to <math>11^2-a^2=13^2-b^2; a=4,b=8,c=16</math>
 
Let the distance between the center of the sphere to the center of those circular intersections as <math>a,b,c</math> separately. <math>a-11,b-13,c-19</math>. According to the problem, we have <math>a^2-11^2=b^2-13^2=c^2-19^2;(11+13)^2-(b-a)^2=560</math>. After solving we have <math>b-a=4</math>, plug this back to <math>11^2-a^2=13^2-b^2; a=4,b=8,c=16</math>
  
Line 9: Line 9:
  
 
~bluesoul
 
~bluesoul
 +
 +
==Solution 2==
 +
 +
Denote by <math>r</math> the radius of three congruent circles formed by the cutting plane.
 +
Denote by <math>O_A</math>, <math>O_B</math>, <math>O_C</math> the centers of three spheres that intersect the plane to get circles centered at <math>A</math>, <math>B</math>, <math>C</math>, respectively.
 +
 +
Because three spheres are mutually tangent, <math>O_A O_B = 11 + 13 = 24</math>, <math>O_A O_C = 11 + 19 = 32</math>.
 +
 +
We have <math>O_A A^2 = 11^2 - r^2</math>, <math>O_B B^2 = 13^2 - r^2</math>, <math>O_C C^2 = 19^2 - r^2</math>.
 +
 +
Because <math>O_A A</math> and <math>O_B B</math> are perpendicular to the plane, <math>O_A AB O_B</math> is a right trapezoid, with <math>\angle O_A A B = \angle O_B BA = 90^\circ</math>.
 +
 +
Hence,
 +
<cmath>
 +
\begin{align*}
 +
O_B B - O_A A & = \sqrt{O_A O_B^2 - AB^2} \\
 +
& = 4 . \hspace{1cm} (1)
 +
\end{align*}
 +
</cmath>
 +
 +
Recall that
 +
<cmath>
 +
\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*}
 +
</cmath>
 +
 +
Hence, taking <math>\frac{(2)}{(1)}</math>, we get
 +
<cmath>
 +
\[
 +
O_B B + O_A A = 12 . \hspace{1cm} (3)
 +
\]
 +
</cmath>
 +
 +
Solving (1) and (3), we get <math>O_B B = 8</math> and <math>O_A A = 4</math>.
 +
 +
Thus, <math>r^2 = 11^2 - O_A A^2 = 105</math>.
 +
 +
Thus, <math>O_C C = \sqrt{19^2 - r^2} = 16</math>.
 +
 +
Because <math>O_A A</math> and <math>O_C C</math> are perpendicular to the plane, <math>O_A AC O_C</math> is a right trapezoid, with <math>\angle O_A A C = \angle O_C CA = 90^\circ</math>.
 +
 +
Therefore,
 +
<cmath>
 +
\begin{align*}
 +
AC^2 & = O_A O_C^2 - \left( O_C C - O_A A \right)^2 \\
 +
& = \boxed{\textbf{(756) }} .
 +
\end{align*}
 +
</cmath>
 +
 +
<math>\textbf{FINAL NOTE:}</math> In our solution, we do not use the conditio that spheres <math>A</math> and <math>B</math> 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)
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2022|n=I|num-b=9|num-a=11}}
 
{{AIME box|year=2022|n=I|num-b=9|num-a=11}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 22:43, 17 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 $AC^2$.


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)

See Also

2022 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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