# Difference between revisions of "1999 JBMO Problems/Problem 4"

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Applying Sine rule in triangle <math>ABC</math>, we get: | Applying Sine rule in triangle <math>ABC</math>, we get: | ||

− | <math> | + | <math>\sin(BAD) / \sin(CAD) = BD / DC</math> |

− | Since <math>BAB'D</math> and <math>CC'AD</math> are cyclic quadrilaterals, | + | Since <math>BAB'D</math> and <math>CC'AD</math> are cyclic quadrilaterals, <math>\angle BAD \angle BB'D</math> |

− | and <math>angle | + | and <math>\angle CAD = \angle CC'D.</math> |

− | So, <math> | + | So, <math>\frac{\sin(BB'D)}{\sin(CC'D)} = \frac{BD}{DC}.</math> |

− | So <math>BD | + | So <math>\frac{BD}{\sin BB'D} = \frac{DC }{ \sin CC'D}.</math> |

− | Thus, <math>BB' = CC'</math> (the | + | Thus, <math>BB' = CC'</math> (the circumcircles <math>\mathcal{C}_1,\mathcal{C}_2</math> are congruent). |

− | From right | + | From right triangles <math>BB'A</math> and <math>CC'A</math>, we have |

− | < | + | <cmath>AC'^{2} = CC'^{2} - AC^{2} = BB'^{2} - AB^{2} = AB'^{2}</cmath> |

− | So <math>AC' = AB'</math> | + | So <math>AC' = AB'.</math> |

Since <math>M</math> is the midpoint of <math>B'C'</math>, <math>AM</math> is perpendicular to <math>B'C'</math> and hence <math>AM</math> is parallel to <math>BC</math>. | Since <math>M</math> is the midpoint of <math>B'C'</math>, <math>AM</math> is perpendicular to <math>B'C'</math> and hence <math>AM</math> is parallel to <math>BC</math>. | ||

− | So area of | + | So area of <math>[MBC] = [ABC]</math> and hence is independent of position of <math>D</math> on <math>BC</math>. |

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## Revision as of 15:00, 17 December 2018

## Problem 4

Let be a triangle with . Also, let be a point such that , and let be the circumcircles of the triangles and respectively. Let and be diameters in the two circles, and let be the midpoint of . Prove that the area of the triangle is constant (i.e. it does not depend on the choice of the point ).

## Solution

Its easy to see that , , are collinear (since angle = = ).

Applying Sine rule in triangle , we get:

Since and are cyclic quadrilaterals, and

So,

So Thus, (the circumcircles are congruent).

From right triangles and , we have
So

Since is the midpoint of , is perpendicular to and hence is parallel to .

So area of and hence is independent of position of on .