Difference between revisions of "1998 USAMO Problems/Problem 2"
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== Solution 2 == | == Solution 2 == | ||
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+ | We will use signed lengths. WLOG let <math>AC = 4</math>, <math>AM = x</math>, and <math>AE = y > 0</math>. Then <math>AB = 2</math> and <math>AD = 1</math>. Therefore, <math>DM = x - 1</math> and <math>MC = 4 - x</math>. By Power of a Point, <math>AE \cdot AF = AB^2 = 4</math>, so <math>AF = \frac{4}{AE} = \frac{4}{y}</math>, and <math>EF = \frac{4}{y} - y = \frac{4 - y^2}{y}</math>. | ||
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+ | By Stewart's Theorem, <cmath>AF(AE \cdot EF + EM^2) = AE \cdot FM^2 + EF \cdot AM^2.</cmath> Since <math>M</math> is on the perpendiculat bisectors of <math>DE</math> and <math>CF</math>, we have <math>|DM| = |EM|</math> and <math>|CM| = |FM|</math>. Therefore, <cmath>AF(AE \cdot EF + DM^2) = AE \cdot MC^2 + EF \cdot AM^2;</cmath> | ||
+ | <cmath>\frac{4}{y}(y(\frac{4-y^2}{y}) + (x-1)^2) = y(4-x)^2 + \frac{4-y^2}{y} \cdot x^2;</cmath> | ||
+ | <cmath>4(4 - y^2 + (x-1)^2) = y^2(4-x)^2 + (4-y^2)x^2;</cmath> | ||
+ | <cmath>16 - 4y^2 + 4x^2 - 8x + 4 = x^2y^2 - 8xy^2 + 16y^2 + 4x^2 - x^2y^2;</cmath> | ||
+ | <cmath>8xy^2 - 8x- 20y^2 + 20 = 0;</cmath> | ||
+ | <cmath>(8x - 20)(y^2 - 1) = 0.</cmath> | ||
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+ | So either <math>8x - 20 = 0</math> implying <math>x = \frac{5}{2}</math>, or <math>y^2 - 1 = 0</math> implying <math>y = 1</math>. If <math>y = 1</math>, then <math>AE = AD = 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 16:05, 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
.
By Stewart's Theorem, Since
is on the perpendiculat bisectors of
and
, we have
and
. Therefore,
So either implying
, or
implying
. 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.