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> or <math>y^2 - 1 = 0</math>. If <math>y^2 - 1 = 0</math>, then <math>AE = y = 1 = AD</math> and <math>AF = AC = 4</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 = y = 1 = AD</math> and <math>AF = AC = 4</math>, so the perpendicular bisectors of <math>CF</math> and <math>DE</math> are the same line, and they do not intersect at a point. Therefore, <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 16:24, 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
, so the perpendicular bisectors of
and
are the same line, and they do not intersect at a point. Therefore,
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.