Difference between revisions of "2022 AIME I Problems/Problem 10"
(→Solution 1) |
(→Diagrams) |
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Line 41: | Line 41: | ||
label("$x$",midpoint(midpoint(B--OB)--OB),E); | label("$x$",midpoint(midpoint(B--OB)--OB),E); | ||
label("$D$",midpoint(B--OB),E); | label("$D$",midpoint(B--OB),E); | ||
+ | |||
+ | |||
+ | |||
+ | </asy> | ||
+ | |||
+ | <asy> | ||
+ | |||
+ | size(500); | ||
+ | pair A, C, OA, OC; | ||
+ | |||
+ | C = (0,0); | ||
+ | A = (-27.4954541697,0); | ||
+ | OC = (0,-16); | ||
+ | OA = (-27.4954541697,-4); | ||
+ | |||
+ | draw(circle(OC,19)); | ||
+ | draw(circle(OA,11)); | ||
+ | |||
+ | draw((-48,0)--(24,0)); | ||
+ | label("$l$",(-42,1),N); | ||
+ | |||
+ | label("$A$",A,N); | ||
+ | label("$C$",C,N); | ||
+ | label("$O_A$",OA,S); | ||
+ | label("$O_C$",OC,S); | ||
+ | |||
+ | draw(A--OA); | ||
+ | draw(C--OC); | ||
+ | draw(OA--OC); | ||
+ | draw(OA--(0,-4)); | ||
+ | draw(OA--(-33.9112699,0)); | ||
+ | draw(OC--(10.2469508,0)); | ||
+ | label("$30$",midpoint(OA--OC),S); | ||
+ | label("$11$",midpoint(OA--(-33.9112699,0)),S); | ||
+ | label("$19$",midpoint(OC--(10.2469508,0)),S); | ||
+ | label("$r$",midpoint(midpoint(A--C)--A),N); | ||
+ | label("$r$",midpoint(midpoint(A--C)--C),N); | ||
+ | label("$r$",midpoint(A--(-33.9112699,0)),N); | ||
+ | label("$r$",midpoint(C--(10.2469508,0)),N); | ||
+ | label("$E$",midpoint(C--OC),E); | ||
Revision as of 14:58, 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
.
We now look at our second diagram.
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.