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2010 IMO Shortlist Problems/G1: Difference between revisions

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== Solution ==
== Solution ==
[[File:2010_IMO_Shortlist_G1.png|400px|thumb|right]]
Let <math> \measuredangle</math> denote [[directed angles]] modulo <math>180^{\circ}</math>.
Let <math> \measuredangle</math> denote [[directed angles]] modulo <math>180^{\circ}</math>.
As <math> \measuredangle AFC =  \measuredangle ADC = 90^{\circ}</math>, <math>AFDC</math> is cyclic.
As <math> \measuredangle AFC =  \measuredangle ADC = 90^{\circ}</math>, <math>AFDC</math> is cyclic.
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We deduce that <math>\measuredangle AQP = \measuredangle BCA = \measuredangle QPA</math> , which is enough to apply that <math>\bigtriangleup APQ</math> is isosceles with <math>AP = AQ</math>.
We deduce that <math>\measuredangle AQP = \measuredangle BCA = \measuredangle QPA</math> , which is enough to apply that <math>\bigtriangleup APQ</math> is isosceles with <math>AP = AQ</math>.
   
   
(Note that with directed angles in place, both the two possible configurations are solved.)
(Note that with directed angles in place, both the two possible configurations (shown in graph) are solved.)


{{alternate solutions}}
{{alternate solutions}}
== Resources ==
== Resources ==


Revision as of 15:00, 18 February 2025

Problem

(United Kingdom) Let $ABC$ be an acute triangle with $D$, $E$, $F$ the feet of the altitudes lying on $BC$, $CA$, $AB$ respectively. One of the intersection points of the line $EF$ and the circumcircle is $P$. The lines $BP$ and $DF$ meet at point $Q$. Prove that $AP = AQ$.

Solution

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Let $\measuredangle$ denote directed angles modulo $180^{\circ}$. As $\measuredangle AFC = \measuredangle ADC = 90^{\circ}$, $AFDC$ is cyclic.

As $APBC$ and $AFDC$ are both cyclic,

$\measuredangle QPA = \measuredangle BPA = \measuredangle BCA = \measuredangle DCA = \measuredangle DFA = \measuredangle QFA$.

Therefore, we see $AFPQ$ is cyclic. Then

$\measuredangle AQP = \measuredangle AFP = \measuredangle AFE = \measuredangle AHE = \measuredangle DHE = \measuredangle DCE = \measuredangle BCA$.

We deduce that $\measuredangle AQP = \measuredangle BCA = \measuredangle QPA$ , which is enough to apply that $\bigtriangleup APQ$ is isosceles with $AP = AQ$.

(Note that with directed angles in place, both the two possible configurations (shown in graph) are solved.)

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

Resources

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