Difference between revisions of "2022 AIME I Problems/Problem 10"
(→Solution 1) |
(→Solution 1) |
||
Line 48: | Line 48: | ||
==Solution 1== | ==Solution 1== | ||
− | We let <math>l</math> be the plane that passes through the spheres and <math>O_A</math> and <math>O_B</math> be the centers of the spheres with radii <math>11</math> and <math>13</math>. We take a cross-section that contains <math>A</math> and <math>B</math>, which contains these two spheres but not the third. Because the plane cuts out congruent circles, they have the same radius and from the given information, <math>AB = \sqrt{560}</math>. Since <math>ABO_BO_A</math> is a trapezoid, we can drop an altitude from <math>O_A</math> to <math>BO_B</math> to create a rectangle and triangle to use Pythagorean theorem. We know that the length of the altitude is <math>\sqrt{560}</math> and let the distance from <math>O_B</math> to <math>D</math> be <math>x</math>. Then we have <math> | + | We let <math>l</math> be the plane that passes through the spheres and <math>O_A</math> and <math>O_B</math> be the centers of the spheres with radii <math>11</math> and <math>13</math>. We take a cross-section that contains <math>A</math> and <math>B</math>, which contains these two spheres but not the third. Because the plane cuts out congruent circles, they have the same radius and from the given information, <math>AB = \sqrt{560}</math>. Since <math>ABO_BO_A</math> is a trapezoid, we can drop an altitude from <math>O_A</math> to <math>BO_B</math> to create a rectangle and triangle to use Pythagorean theorem. We know that the length of the altitude is <math>\sqrt{560}</math> and let the distance from <math>O_B</math> to <math>D</math> be <math>x</math>. Then we have <math>x^2 = 576-560 \implies x = 4</math>. |
− | We have <math>AO_A = BD</math> because of the rectangle, so <math>\sqrt{ | + | We have <math>AO_A = BD</math> because of the rectangle, so <math>\sqrt{11^2-r^2} = \sqrt{13^2-r^2}-4</math>. |
− | Squaring, we have <math>121- | + | Squaring, we have <math>121-r^2 = 169-r^2 + 16 - 8 \cdot \sqrt{169-r^2}</math>. |
− | Subtracting, we get <math>8 \cdot \sqrt{169- | + | Subtracting, we get <math>8 \cdot \sqrt{169-r^2} = 64 \implies \sqrt{169-r^2} = 8 \implies 169-r^2 = 64 \implies r^2 = 105</math>. |
==Solution 2== | ==Solution 2== |
Revision as of 14:44, 21 February 2022
Contents
Problem
Three spheres with radii ,
, and
are mutually externally tangent. A plane intersects the spheres in three congruent circles centered at
,
, and
, respectively, and the centers of the spheres all lie on the same side of this plane. Suppose that
. Find
.
Diagrams
Solution 1
We let be the plane that passes through the spheres and
and
be the centers of the spheres with radii
and
. We take a cross-section that contains
and
, which contains these two spheres but not the third. Because the plane cuts out congruent circles, they have the same radius and from the given information,
. Since
is a trapezoid, we can drop an altitude from
to
to create a rectangle and triangle to use Pythagorean theorem. We know that the length of the altitude is
and let the distance from
to
be
. Then we have
.
We have because of the rectangle, so
.
Squaring, we have
.
Subtracting, we get
.
Solution 2
Let the distance between the center of the sphere to the center of those circular intersections as separately.
. According to the problem, we have
. After solving we have
, plug this back to
The desired value is
~bluesoul
Solution 3
Denote by the radius of three congruent circles formed by the cutting plane.
Denote by
,
,
the centers of three spheres that intersect the plane to get circles centered at
,
,
, respectively.
Because three spheres are mutually tangent, ,
.
We have ,
,
.
Because and
are perpendicular to the plane,
is a right trapezoid, with
.
Hence,
Recall that
Hence, taking , we get
Solving (1) and (3), we get and
.
Thus, .
Thus, .
Because and
are perpendicular to the plane,
is a right trapezoid, with
.
Therefore,
In our solution, we do not use the conditio that spheres
and
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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.