# Difference between revisions of "2003 JBMO Problems/Problem 3"

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<math>Part 1</math>: Angles of <math>\triangle DMN</math>: | <math>Part 1</math>: Angles of <math>\triangle DMN</math>: | ||

− | + | Since <math>\angle QNM = \angle B/2, \angle DNM = 90^{\circ} - \angle B/2</math>. | |

Similarly <math>\angle NMD = 90^{\circ} - \angle C/2</math>. | Similarly <math>\angle NMD = 90^{\circ} - \angle C/2</math>. | ||

Finally <math>\angle NDM = 90^{\circ} - \angle A/2</math>. | Finally <math>\angle NDM = 90^{\circ} - \angle A/2</math>. |

## Revision as of 23:30, 12 December 2018

## Problem

Let , , be the midpoints of the arcs , , on the circumcircle of a triangle not containing the points , , , respectively. Let the line meets and at and , and let be the midpoint of the segment . Let the line meet and at and , and let be the midpoint of the segment .

a) Find the angles of triangle ;

b) Prove that if is the point of intersection of the lines and , then the circumcenter of triangle lies on the circumcircle of triangle .

## Solution

Let , intersect , at , respectively. We will prove first that and that lines , , are altitudes of the .

It's easy to see that lines , and form the internal angle bisectors of .

Consequently, we can determine the of as being equal to

Also we have , thus . Similarly .

Thus , , are altitudes of the with , , respectively being the feet of the altitudes.

Now since is internal bisector of and is perpendicular to , we have that is the perpendicular bisector of . Hence .

Similarly it can be shown that is the perpendicular bisector of , and hence .

Now lines , and intersect at point . So is the incenter of and orthocenter of .

Clearly, is a cyclic quadrilateral as the , are the feet of perpendiculars from and .

So, we have .

Similarly, since is also a cyclic-quadrilateral, reasoning as above, .

Thus we have that and so is an internal bisector of . Reasoning in a similar fashion it can be proven that and are internal bisectors of other 2 angles of .

Thus also happens to be the incenter of in addition to being that of .

: Angles of :

Since . Similarly . Finally .

:

Let circumcircle of cut line at point . Since is a cyclic quadrilateral, we have .

Similarly, . Thus = .

Now,

and . Thus = .

Thus we have = = . So is the circumcenter of .