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

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== Problem ==
 
== Problem ==
 
Three spheres with radii <math>11</math>, <math>13</math>, and <math>19</math> are mutually externally tangent. A plane intersects the spheres in three congruent circles centered at <math>A</math>, <math>B</math>, and <math>C</math>, respectively, and the centers of the spheres all lie on the same side of this plane. Suppose that <math>AB^2 = 560</math>. Find <math>AC^2</math>.
 
Three spheres with radii <math>11</math>, <math>13</math>, and <math>19</math> are mutually externally tangent. A plane intersects the spheres in three congruent circles centered at <math>A</math>, <math>B</math>, and <math>C</math>, respectively, and the centers of the spheres all lie on the same side of this plane. Suppose that <math>AB^2 = 560</math>. Find <math>AC^2</math>.
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==solution 1==
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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>
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The desired value is <math>(11+19)^2-(16-4)^2=\boxed{756}</math>
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~bluesoul
  
 
==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 23:02, 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

See Also

2022 AIME I (ProblemsAnswer KeyResources)
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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