Difference between revisions of "2001 IMO Problems/Problem 1"
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== Solution == | == Solution == | ||
− | {{ | + | 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 = YP</math> (since <math>AZYP</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>OP > YP \ge R/2</math>. But <math>PC = YC - YP < R - YP \le R/2</math>. So <math>OP > PC</math>. | ||
+ | |||
+ | Hence <math>\angle COP < \angle OCP</math>. Let <math>CE</math> be a diameter of the circle, so that <math>\angle OCP = \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 COP + \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>4 \sin B \cos C > 1</math>. But we also have <math>4 \sin B \cos C \le 4 \sin B \cos (B + 30^\circ) = 2 (\sin (2B + 30^\circ) - \sin 30^\circ) \le 2 (1 - \frac{1}{2}) = 1</math> because <math>\angle{C} \ge \angle{B} + 30^\circ</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 == |
Latest revision as of 17:24, 21 July 2018
Contents
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 |