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Difference between revisions of "2010 IMO Shortlist Problems/G1"

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== Problem ==
 
== Problem ==
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(United Kingdom) Let <math>ABC</math> be an acute triangle with <math>D</math>, <math>E</math>, <math>F</math> the feet of the altitudes lying on <math>BC</math>, <math>CA</math>, <math>AB</math> respectively. One of the intersection points of the line <math>EF</math> and the circumcircle is <math>P</math>. The lines <math>BP</math> and <math>DF</math> meet at point <math>Q</math>. Prove that <math>AP = AQ</math>.
 
(United Kingdom) Let <math>ABC</math> be an acute triangle with <math>D</math>, <math>E</math>, <math>F</math> the feet of the altitudes lying on <math>BC</math>, <math>CA</math>, <math>AB</math> respectively. One of the intersection points of the line <math>EF</math> and the circumcircle is <math>P</math>. The lines <math>BP</math> and <math>DF</math> meet at point <math>Q</math>. Prove that <math>AP = AQ</math>.
  
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== Solution ==
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== Solution 1 ==
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[[File:2010_IMO_Shortlist_G1.png|400px|thumb|right]]  
 
[[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>.
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(Note that with directed angles in place, both the two possible configurations (shown in graph) are solved.)
 
(Note that with directed angles in place, both the two possible configurations (shown in graph) are solved.)
  
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{{alternate solutions}}
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== See Also ==
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== Resources ==
 
  
 
* [[2010 IMO Shortlist Problems]]
 
* [[2010 IMO Shortlist Problems]]
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* [https://artofproblemsolving.com/community/c6h418633p2361970| Discussion on AoPS]
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* [//artofproblemsolving.com/community/c6h418633p2361970 Discussion on AoPS]
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[[Category:Olympiad Geometry Problems]]
 
[[Category:Olympiad Geometry Problems]]

Latest revision as of 16:36, 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 1

2010 IMO Shortlist G1.png

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.)

See Also

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