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

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==Problem== | ==Problem== | ||

− | Let <math>D</math>, <math>E</math>, <math>F</math> be the midpoints of the arcs <math>BC</math>, <math>CA</math>, <math>AB</math> on the circumcircle of a triangle <math>ABC</math> not containing the points <math>A</math>, <math>B</math>, <math>C</math>, respectively. Let the line <math>DE</math> | + | Let <math>D</math>, <math>E</math>, <math>F</math> be the midpoints of the arcs <math>BC</math>, <math>CA</math>, <math>AB</math> on the circumcircle of a triangle <math>ABC</math> not containing the points <math>A</math>, <math>B</math>, <math>C</math>, respectively. Let the line <math>DE</math> meet <math>BC</math> and <math>CA</math> at <math>G</math> and <math>H</math>, and let <math>M</math> be the midpoint of the segment <math>GH</math>. Let the line <math>FD</math> meet <math>BC</math> and <math>AB</math> at <math>K</math> and <math>J</math>, and let <math>N</math> be the midpoint of the segment <math>KJ</math>. |

a) Find the angles of triangle <math>DMN</math>; | a) Find the angles of triangle <math>DMN</math>; | ||

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

− | Let <math>FC</math>, <math>EB</math> intersect <math>DE</math>, <math>FD</math> at <math>M'</math>, <math>N'</math> respectively. We will prove first that <math>M' = M, N = N'</math> and that lines <math>AD</math>, <math>BE</math>, <math>FC</math> are altitudes of the <math>\triangle | + | Let <math>FC</math>, <math>EB</math> intersect <math>DE</math>, <math>FD</math> at <math>M'</math>, <math>N'</math> respectively. We will prove first that <math>M' = M, N = N'</math> and that lines <math>AD</math>, <math>BE</math>, <math>FC</math> are altitudes of the <math>\triangle DEF</math>. |

It's easy to see that lines <math>AD</math>, <math>BE</math> and <math>CF</math> form the internal angle bisectors of <math>\triangle ABC</math>. | It's easy to see that lines <math>AD</math>, <math>BE</math> and <math>CF</math> form the internal angle bisectors of <math>\triangle ABC</math>. | ||

− | Consequently, we can determine the <math>\angle | + | Consequently, we can determine the <math>\angle FDE</math> of <math>\triangle DEF</math> as being equal to <math>\angle FDA + \angle ADE = \angle C/2 + \angle B/2 = 90^{\circ} - \angle A/2</math> |

Also we have <math>\angle DFC = \angle A/2</math>, thus <math>\angle FM'D = 90^{\circ}</math>. Similarly <math>\angle EN'D = 90^{\circ}</math>. | Also we have <math>\angle DFC = \angle A/2</math>, thus <math>\angle FM'D = 90^{\circ}</math>. Similarly <math>\angle EN'D = 90^{\circ}</math>. | ||

− | Thus <math>AD</math>, <math>BE</math>, <math>FC</math> are altitudes of the <math>\triangle | + | Thus <math>AD</math>, <math>BE</math>, <math>FC</math> are altitudes of the <math>\triangle DEF</math> with <math>P</math>, <math>N'</math>, <math>M'</math> respectively being the feet of the altitudes. |

Now since <math>M'C</math> is internal bisector of <math>\angle HCG</math> and <math>CM'</math> is perpendicular to <math>GH</math>, we have that | Now since <math>M'C</math> is internal bisector of <math>\angle HCG</math> and <math>CM'</math> is perpendicular to <math>GH</math>, we have that | ||

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Now lines <math>AD</math>, <math>BE</math> and <math>CF</math> intersect at point <math>Q</math>. So <math>Q</math> is the incenter of <math>\triangle ABC</math> and orthocenter of <math>\triangle DEF</math>. | Now lines <math>AD</math>, <math>BE</math> and <math>CF</math> intersect at point <math>Q</math>. So <math>Q</math> is the incenter of <math>\triangle ABC</math> and orthocenter of <math>\triangle DEF</math>. | ||

− | Clearly, <math>QNDM</math> is a cyclic quadrilateral as | + | Clearly, <math>QNDM</math> is a cyclic quadrilateral as <math>N</math>, <math>M</math> are the feet of perpendiculars from <math>E</math> and <math>F</math>. |

So, we have <math>\angle QNM = \angle QDM = \angle ADE = \angle B/2</math>. | So, we have <math>\angle QNM = \angle QDM = \angle ADE = \angle B/2</math>. | ||

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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>. |

## Latest revision as of 22:53, 13 December 2018

## Problem

Let , , be the midpoints of the arcs , , on the circumcircle of a triangle not containing the points , , , respectively. Let the line meet 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 , 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 .