Difference between revisions of "2001 IMO Problems/Problem 1"
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Hence <math>\angle COX < \angle OCX</math>. Let <math>CE</math> be a diameter of the circle, so that <math>\angle OCX = \angle ECB</math>. But <math>\angle ECB = \angle EAB</math> and <math>\angle EAB + \angle BAC = \angle EAC = 90^\circ</math>, since <math>EC</math> is a diameter. Hence <math>\angle COX + \angle BAC < 90^\circ</math>. | Hence <math>\angle COX < \angle OCX</math>. Let <math>CE</math> be a diameter of the circle, so that <math>\angle OCX = \angle ECB</math>. But <math>\angle ECB = \angle EAB</math> and <math>\angle EAB + \angle BAC = \angle EAC = 90^\circ</math>, since <math>EC</math> is a diameter. Hence <math>\angle COX + \angle BAC < 90^\circ</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | Notice that because <math>\angle{PCO} = 90^\circ - \angle{A}</math>, it suffices to prove that <math>\angle{POC} < \angle{PCO}</math>, or equivalently <math>PC < PO.</math> | ||
+ | |||
+ | Suppose on the contrary that <math>PC > PO</math>. By the triangle inequality, <math>2 PC > PC + PC = PC + PO = CO = R</math>, where <math>R</math> is the circumradius of <math>ABC</math>. But the Law of Sines and basic trigonometry gives us that <math>PC = 2R \sin B \cos C</math>, so we have <math>2 \sin B \cos C \ge 1</math>. But we also have <math>2 \sin B \cos C < 2 \sin B \cos B = \sin 2B \le 1</math> because <math>\angle{C} > \angle{B}</math>, and so we have a contradiction. Hence <math>PC < PO</math> and so <math>\angle{PCO} + \angle{A} < 90^\circ</math>, as desired. | ||
== See also == | == See also == |
Revision as of 22:24, 16 May 2015
Contents
[hide]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
.
Solution 2
Notice that because , it suffices to prove that
, or equivalently
Suppose on the contrary that . By the triangle inequality,
, where
is the circumradius of
. But the Law of Sines and basic trigonometry gives us that
, so we have
. But we also have
because
, and so we have a contradiction. Hence
and so
, as desired.
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 |