Difference between revisions of "2022 AIME I Problems/Problem 10"
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==Solution 2== | ==Solution 2== | ||
− | 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 | + | 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</math>, we have <math>a=4</math>, <math>b=8</math>, and <math>c=16</math>. |
− | The desired value is <math>(11+19)^2-(16-4)^2=\boxed{756}</math> | + | The desired value is <math>(11+19)^2-(16-4)^2=\boxed{756}</math>. |
~bluesoul | ~bluesoul |
Revision as of 16:46, 12 January 2023
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
.
Diagram
~MRENTHUSIASM
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, as shown below:
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
.
We also notice that since we had
means that
and since we know that
,
.
We take a cross-section that contains and
, which contains these two spheres but not the third, as shown below:
We have
. Since
, we have
. Using Pythagorean theorem,
. Therefore,
.
~KingRavi
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
, we have
,
, and
.
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.