Difference between revisions of "2001 IMO Problems/Problem 1"
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− | Take <math>D</math> on the circumcircle with <math>AD \parallel | + | Take <math>D</math> on the circumcircle with <math>AD \parallel BC</math>. Notice that <math>\angle CBD = \angle BCA</math>, so <math>\angle ABD \ge 30^\circ</math>. Hence <math>\angle AOD \ge 60^\circ</math>. Let <math>Z</math> be the midpoint of <math>AD</math> and <math>Y</math> the midpoint of <math>BC</math>. Then <math>AZ \ge R/2</math>, where <math>R</math> is the radius of the circumcircle. But <math>AZ = YX</math> (since <math>AZYX</math> is a rectangle). |
Now <math>O</math> cannot coincide with <math>Y</math> (otherwise <math>\angle A</math> would be <math>90^\circ</math> and the triangle would not be acute-angled). So <math>OX > YX \ge R/2</math>. But <math>XC = YC - YX < R - YX \le R/2</math>. So <math>OX > XC</math>. | Now <math>O</math> cannot coincide with <math>Y</math> (otherwise <math>\angle A</math> would be <math>90^\circ</math> and the triangle would not be acute-angled). So <math>OX > YX \ge R/2</math>. But <math>XC = YC - YX < R - YX \le R/2</math>. So <math>OX > XC</math>. |
Revision as of 19:40, 11 July 2012
Problem
Consider an acute triangle . Let
be the foot of the altitude of triangle
issuing from the vertex
, and let
be the circumcenter of triangle
. Assume that
. Prove that
.
Solution
Take on the circumcircle with
. Notice that
, so
. Hence
. Let
be the midpoint of
and
the midpoint of
. Then
, where
is the radius of the circumcircle. But
(since
is a rectangle).
Now cannot coincide with
(otherwise
would be
and the triangle would not be acute-angled). So
. But
. So
.
Hence . Let
be a diameter of the circle, so that
. But
and
, since
is a diameter. Hence
.
See also
2001 IMO (Problems) • Resources | ||
Preceded by First question |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
All IMO Problems and Solutions |