Difference between revisions of "2020 AIME I Problems/Problem 6"
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Let <math>A</math>, <math>C</math>, and <math>E</math> be, respectively, the center of the hole with radius <math>1,</math> the center of the sphere resting in that hole, and a point on the edge of that hole. Let <math>B</math>, <math>D</math>, and <math>F</math> be the corresponding points for the hole with radius <math>2.</math> Let <math>G</math> be the point on <math>\overline{AC}</math> such that <math>\overline{AC} \perp \overline{GD}</math>. Let the radius of the spheres be <math>r = CE = DF</math>. Because <math>r^2 = AE^2 + AC^2 = 1 + AC^2</math> and <math>r^2 = BF^2 + BD^2 = 4 + BD^2</math>, it follows that<cmath>CG = AC - AG = AC - BD = \sqrt{r^2 - 1} - \sqrt{r^2-4}.</cmath>Because <math>DG = 7</math>, <math>CD = 2r</math>, and <math>CD^2 = CG^2 +GD^2</math>, it follows that<cmath>4r^2 = \left(\sqrt{r^2 - 1} - \sqrt{r^2-4}\right)^{\!2} + 7^2,</cmath>which simplifies to <math>r^2 = \frac{160}{13}</math>. The requested sum is <math>160+13 = 173</math>. The value of <math>r</math> is approximately <math>3.5082.</math> | Let <math>A</math>, <math>C</math>, and <math>E</math> be, respectively, the center of the hole with radius <math>1,</math> the center of the sphere resting in that hole, and a point on the edge of that hole. Let <math>B</math>, <math>D</math>, and <math>F</math> be the corresponding points for the hole with radius <math>2.</math> Let <math>G</math> be the point on <math>\overline{AC}</math> such that <math>\overline{AC} \perp \overline{GD}</math>. Let the radius of the spheres be <math>r = CE = DF</math>. Because <math>r^2 = AE^2 + AC^2 = 1 + AC^2</math> and <math>r^2 = BF^2 + BD^2 = 4 + BD^2</math>, it follows that<cmath>CG = AC - AG = AC - BD = \sqrt{r^2 - 1} - \sqrt{r^2-4}.</cmath>Because <math>DG = 7</math>, <math>CD = 2r</math>, and <math>CD^2 = CG^2 +GD^2</math>, it follows that<cmath>4r^2 = \left(\sqrt{r^2 - 1} - \sqrt{r^2-4}\right)^{\!2} + 7^2,</cmath>which simplifies to <math>r^2 = \frac{160}{13}</math>. The requested sum is <math>160+13 = 173</math>. The value of <math>r</math> is approximately <math>3.5082.</math> | ||
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+ | ==Video solution (With Animation)== | ||
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+ | https://youtu.be/cOf9uTJ9J40 | ||
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==Video Solution== | ==Video Solution== | ||
Revision as of 13:15, 13 December 2021
Contents
Problem
A flat board has a circular hole with radius and a circular hole with radius
such that the distance between the centers of the two holes is
Two spheres with equal radii sit in the two holes such that the spheres are tangent to each other. The square of the radius of the spheres is
where
and
are relatively prime positive integers. Find
Solution
Set the common radius to . First, take the cross section of the sphere sitting in the hole of radius 1. If we draw the perpendicular bisector of the chord (the hole) through the circle, this line goes through the center. Connect the center also to where the chord hits the circle, for a right triangle with hypotenuse
and base
. Therefore, the height of this circle outside of the hole is
.
The other circle follows similarly for a height (outside the hole) of . Now, if we take the cross section of the entire board, essentially making it 2-D, we can connect the centers of the two spheres, then form another right triangle with base
, as given by the problem. The height of this triangle is the difference between the heights of the parts of the two spheres outside the holes, which is
. Now we can set up an equation in terms of
with the Pythagorean theorem:
Simplifying a few times,
Therefore, our answer is
.
-molocyxu
Solution 2 (Official MAA)
Consider a cross section of the board and spheres with a plane that passes through the centers of the holes and centers of the spheres as shown.
Let ,
, and
be, respectively, the center of the hole with radius
the center of the sphere resting in that hole, and a point on the edge of that hole. Let
,
, and
be the corresponding points for the hole with radius
Let
be the point on
such that
. Let the radius of the spheres be
. Because
and
, it follows that
Because
,
, and
, it follows that
which simplifies to
. The requested sum is
. The value of
is approximately
Video solution (With Animation)
Video Solution
https://www.youtube.com/watch?v=qCTq8KhZfYQ
See Also
2020 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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.