Difference between revisions of "1997 AIME Problems/Problem 4"
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== Solution 2 == | == Solution 2 == | ||
We may also use Descartes' theorem, <math>k_4=k_1+k_2+k_3\pm 2\sqrt{k_1k_2+k_2k_3+k_3k_1}</math> where each of <math>k_i</math> is the curvature of a circle with radius <math>r_i</math>, and the curvature is defined as <math>k_i=\frac{1}{r_i}</math>. The larger solution for <math>k_4</math> will give the curvature of the circle externally tangent to the other circles, while the smaller solution will give the curvature for the circle internally tangent to each of the other circles. Using Descartes' theorem, we get <math>k_4=\frac15+\frac15+\frac18+2\sqrt{\frac{1}{40}+\frac{1}{40}+\frac{1}{25}}=\frac{21}{40}+2\sqrt{\frac{45}{500}}=\frac{45}{40}</math>. Thus, <math>r_4=\frac{1}{k_4}=\frac{40}{45}=\frac89</math>, and the answer is <math>\boxed{017}</math> | We may also use Descartes' theorem, <math>k_4=k_1+k_2+k_3\pm 2\sqrt{k_1k_2+k_2k_3+k_3k_1}</math> where each of <math>k_i</math> is the curvature of a circle with radius <math>r_i</math>, and the curvature is defined as <math>k_i=\frac{1}{r_i}</math>. The larger solution for <math>k_4</math> will give the curvature of the circle externally tangent to the other circles, while the smaller solution will give the curvature for the circle internally tangent to each of the other circles. Using Descartes' theorem, we get <math>k_4=\frac15+\frac15+\frac18+2\sqrt{\frac{1}{40}+\frac{1}{40}+\frac{1}{25}}=\frac{21}{40}+2\sqrt{\frac{45}{500}}=\frac{45}{40}</math>. Thus, <math>r_4=\frac{1}{k_4}=\frac{40}{45}=\frac89</math>, and the answer is <math>\boxed{017}</math> | ||
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| + | ==Video Solution 1== | ||
| + | https://www.youtube.com/watch?v=KUs44jwBLD8&t=118s | ||
| + | -amritvignesh0719062.0 | ||
== See also == | == See also == | ||
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[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
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Latest revision as of 21:33, 2 February 2021
Problem
Circles of radii 
and 
are mutually externally tangent, where 
and 
are relatively prime positive integers. Find 
Solution
If (in the diagram above) we draw the line going through the centers of the circles with radii 
and 
, that line is the perpendicular bisector of the segment connecting the centers of the two circles with radii 
. Then we form two right triangles, of lengths 
and 
, wher 
is the distance between the center of the circle in question and the segment connecting the centers of the two circles of radii . By the Pythagorean Theorem, we now have two equations with two unknowns:
So 
.
NOTE: It can be seen that there is no apparent need to use the variable x as a 5,12,13 right triangle has been formed.
Solution 2
We may also use Descartes' theorem, 
where each of 
is the curvature of a circle with radius 
, and the curvature is defined as 
. The larger solution for 
will give the curvature of the circle externally tangent to the other circles, while the smaller solution will give the curvature for the circle internally tangent to each of the other circles. Using Descartes' theorem, we get 
. Thus, 
, and the answer is 
Video Solution 1
https://www.youtube.com/watch?v=KUs44jwBLD8&t=118s -amritvignesh0719062.0
See also
| 1997 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 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. 
