Difference between revisions of "1998 USAMO Problems/Problem 2"
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<cmath>(8x - 20)(y^2 - 1) = 0.</cmath> | <cmath>(8x - 20)(y^2 - 1) = 0.</cmath> | ||
− | So either <math>8x - 20 = 0</math> | + | So either <math>8x - 20 = 0</math> or <math>y^2 - 1 = 0</math>. If <math>y^2 - 1 = 0</math>, then <math>AE = AD = y = 1</math> and <math>AF = AC = 4</math>. Therefore, the perpendicular bisectors of <math>CF</math> and <math>DE</math> are actually the same line, so they do not intersect at one point. So <math>AM = x = \frac{5}{2}</math> and <math>MC = \frac{3}{2}</math>, so <math>\frac{AM}{MC} = \boxed{\frac{5}{3}}</math>. |
== See Also == | == See Also == |
Revision as of 16:21, 20 June 2019
Contents
[hide]Problem
Let and be concentric circles, with in the interior of . From a point on one draws the tangent to (). Let be the second point of intersection of and , and let be the midpoint of . A line passing through intersects at and in such a way that the perpendicular bisectors of and intersect at a point on . Find, with proof, the ratio .
Solution
First, . Because , and all lie on a circle, . Therefore, , so . Thus, quadrilateral is cyclic, and must be the center of the circumcircle of , which implies that . Putting it all together,
Borrowed from https://mks.mff.cuni.cz/kalva/usa/usoln/usol982.html
Solution 2
We will use signed lengths. WLOG let , , and . Then and . Therefore, and . By Power of a Point, , so , and . Since is on the perpendicular bisectors of and , we have and .
By Stewart's Theorem on and cevian ,
So either or . If , then and . Therefore, the perpendicular bisectors of and are actually the same line, so they do not intersect at one point. So and , so .
See Also
1998 USAMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.