Difference between revisions of "2009 IMO Problems/Problem 2"
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''Author: Sergei Berlov, Russia'' | ''Author: Sergei Berlov, Russia'' | ||
+ | |||
+ | == Solution == | ||
+ | ===Diagram=== | ||
+ | <asy> | ||
+ | dot("O", (50, 38), N); | ||
+ | dot("A", (40, 100), N); | ||
+ | dot("B", (0, 0), S); | ||
+ | dot("C", (100, 0), S); | ||
+ | dot("Q", (24, 60), W); | ||
+ | dot("P", (52, 80), E); | ||
+ | dot("L", (62, 30), SE); | ||
+ | dot("M", (38, 70), N); | ||
+ | dot("K", (27, 42), W); | ||
+ | draw((100, 0)--(24, 60), dotted); | ||
+ | draw((0, 0)--(52, 80), dashed); | ||
+ | draw((0, 0)--(100, 0)--(40, 100)--cycle); | ||
+ | draw((24, 60)--(52, 80)); | ||
+ | draw((27, 42)--(38, 70)--(62, 30)--cycle); | ||
+ | draw(circle((49, 49), 23)); | ||
+ | label("$\Gamma$", (72, 49), E); | ||
+ | draw(circle((50, 38), 63)); | ||
+ | label("$\omega$", (-13, 38), NW); | ||
+ | </asy> | ||
+ | Diagram by qwertysri987 | ||
+ | ---------------------------- | ||
+ | By parallel lines and the tangency condition, <cmath>\angle APM\cong \angle LMP \cong \angle LKM.</cmath> Similarly, <cmath>\angle AQP\cong \angle KLM,</cmath> so AA similarity implies <cmath>\triangle APQ\sim \triangle MKL.</cmath> Let <math>\omega</math> denote the circumcircle of <math>\triangle ABC,</math> and <math>R</math> its circumradius. As both <math>P</math> and <math>Q</math> are inside <math>\omega,</math> | ||
+ | |||
+ | <cmath>\begin{align*} | ||
+ | R^2-QO^2&=\text{Pow}_{\omega}(Q)\\ | ||
+ | &=QB\cdot AQ \\ | ||
+ | &=2AQ\cdot MK\\ | ||
+ | &=2AP\cdot ML\\ | ||
+ | &=AP\cdot PC\\ | ||
+ | &=\text{Pow}_{\omega}(P)\\ | ||
+ | &=R^2-PO^2. | ||
+ | \end{align*}</cmath> It follows that <math>OP=OQ.</math> <math>\blacksquare</math> | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{IMO box|year=2009|num-b=1|num-a=3}} |
Latest revision as of 01:16, 19 November 2023
Contents
Problem
Let be a triangle with circumcentre
. The points
and
are interior points of the sides
and
respectively. Let
and
be the midpoints of the segments
and
, respectively, and let
be the circle passing through
and
. Suppose that the line
is tangent to the circle
. Prove that
.
Author: Sergei Berlov, Russia
Solution
Diagram
Diagram by qwertysri987
By parallel lines and the tangency condition, Similarly,
so AA similarity implies
Let
denote the circumcircle of
and
its circumradius. As both
and
are inside
It follows that
See Also
2009 IMO (Problems) • Resources | ||
Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
All IMO Problems and Solutions |