Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus
During AMC testing, the AoPS Wiki is in read-only mode and no edits can be made.

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

(→‎Solution)
(Formatting)
 
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
 +
 
(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>.
  
−
== Solution ==
+
== Solution 1 ==
 +
 
 
[[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>.
Line 19: Line 21:
 
(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.)
  
−
{{alternate solutions}}
+
== See Also ==
−
 
 
−
== Resources ==
 
  
 
* [[2010 IMO Shortlist Problems]]
 
* [[2010 IMO Shortlist Problems]]
−
* [https://artofproblemsolving.com/community/c6h418633p2361970| Discussion on AoPS]
+
* [//artofproblemsolving.com/community/c6h418633p2361970 Discussion on AoPS]
−
 
 
−
 
 
  
 
[[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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2010_IMO_Shortlist_Problems/G1&oldid=243120"