Difference between revisions of "1998 USAMO Problems/Problem 2"
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<math>\frac{AM}{MC}=\frac{AC-MC}{MC}=\frac{AC-\frac{CD}{2}}{\frac{CD}{2}}=\frac{AC-\frac{AC-AD}{2}}{\frac{AC-AD}{2}}=\frac{AC-\frac{3AC}{8}}{\frac{3AC}{8}}=\frac{\frac{5AC}{8}}{\frac{3AC}{8}}=\frac{5}{3}</math> | <math>\frac{AM}{MC}=\frac{AC-MC}{MC}=\frac{AC-\frac{CD}{2}}{\frac{CD}{2}}=\frac{AC-\frac{AC-AD}{2}}{\frac{AC-AD}{2}}=\frac{AC-\frac{3AC}{8}}{\frac{3AC}{8}}=\frac{\frac{5AC}{8}}{\frac{3AC}{8}}=\frac{5}{3}</math> | ||
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+ | Borrowed from https://mks.mff.cuni.cz/kalva/usa/usoln/usol982.html | ||
== See Also == | == See Also == |
Revision as of 09:46, 21 October 2014
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
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.