# Difference between revisions of "2004 IMO Shortlist Problems/G3"

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Let <math> \displaystyle O </math> be the circumcenter of an acute-angled triangle <math> \displaystyle ABC </math> with <math> \angle B < \angle C </math>. The line <math> \displaystyle AO </math> meets the side <math> \displaystyle BC </math> at <math> \displaystyle D </math>. The circumcenters of the triangles <math> \displaystyle ABD </math> and <math> \displaystyle ACD </math> are <math> \displaystyle E </math> and <math> \displaystyle F </math>, respectively. Extend the sides <math> \displaystyle BA </math> and <math> \displaystyle CA </math> beyond <math> \displaystyle A </math>, and choose, on the respective extensions points <math> \displaystyle G </math> and <math> \displaystyle H </math> such that <math> \displaystyle AG= AC </math> and <math> \displaystyle AH = AB </math>. Prove that the quadrilateral <math> \displaystyle EFGH </math> is a rectangle if and only if <math> \angle ACB - \angle ABC = 60^{\circ} </math>. | Let <math> \displaystyle O </math> be the circumcenter of an acute-angled triangle <math> \displaystyle ABC </math> with <math> \angle B < \angle C </math>. The line <math> \displaystyle AO </math> meets the side <math> \displaystyle BC </math> at <math> \displaystyle D </math>. The circumcenters of the triangles <math> \displaystyle ABD </math> and <math> \displaystyle ACD </math> are <math> \displaystyle E </math> and <math> \displaystyle F </math>, respectively. Extend the sides <math> \displaystyle BA </math> and <math> \displaystyle CA </math> beyond <math> \displaystyle A </math>, and choose, on the respective extensions points <math> \displaystyle G </math> and <math> \displaystyle H </math> such that <math> \displaystyle AG= AC </math> and <math> \displaystyle AH = AB </math>. Prove that the quadrilateral <math> \displaystyle EFGH </math> is a rectangle if and only if <math> \angle ACB - \angle ABC = 60^{\circ} </math>. | ||

− | (''This was also Problem 2, Day 3 of the 2005 Moldova | + | (''This was also Problem 2 of the 2005 3rd German [[TST]]; Problem 2, Day 3 of the 2005 Moldova TST; and Problem 5 of the 2005 Taiwan 2nd TST final exam.'') |

== Solution == | == Solution == | ||

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* [[2004 IMO Shortlist Problems]] | * [[2004 IMO Shortlist Problems]] | ||

+ | * [[2005 Germany TST Problems]] | ||

* [[2005 Moldova TST Problems]] | * [[2005 Moldova TST Problems]] | ||

* [[2005 Taiwan TST Problems]] | * [[2005 Taiwan TST Problems]] |

## Latest revision as of 10:45, 10 June 2007

## Problem

(*South Korea*)
Let be the circumcenter of an acute-angled triangle with . The line meets the side at . The circumcenters of the triangles and are and , respectively. Extend the sides and beyond , and choose, on the respective extensions points and such that and . Prove that the quadrilateral is a rectangle if and only if .

(*This was also Problem 2 of the 2005 3rd German TST; Problem 2, Day 3 of the 2005 Moldova TST; and Problem 5 of the 2005 Taiwan 2nd TST final exam.*)

## Solution

**Lemma.** In any triangle with circumcenter , the altitude from is the reflection of over the angle bisector of .

*Proof.* This is well-known, but we prove it anyway. Let meet sides at , and let be the foot of the altitude from . Let us denote , , , and let us use the notation for the angles of triangle . By virtue of inscribed arcs in the circumcircle of , we know , , , so , and again by inscribed arcs, . The lemma follows. ∎

We first note that is the reflection of over the exterior angle bisector of . It follows that line is the altitude from in triangle , i.e., . Since both and line on the perpendicular bisector of , it follows that and are always parallel.

We extend and to meet a point . Since , is a convex quadrilateral. In particular, if we use the notation , , , then , , , so . It follows that line makes an angle of with . Now, if is the midpoint of and is the midpoint of , we note that and are perpendicular to . Hence . But if is a rectangle, then , so and . Thus the condition is necessary for to be a rectangle.

We now prove that it is sufficient. From the previous paragraph, we know that if , then is a parallelogram. It is sufficient to show that if is the intersection of line with and is the intersection of line and , then , since is perpendicular to and . Indeed, since the cosine of the angle between lines and is , it is sufficient to show that if is the projection of onto , then . Let be the projection of onto . Since are congruent, . On the other hand, since is the midpoint of , is the midpoint of the projection of onto , namely, , so , as desired. Thus is a rectangle if and only if , as desired.

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