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

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

− | Its easy to see that <math>B'</math>, <math>C'</math>, <math>D</math> are collinear (since <math>\angle B'DC = \angle C'DC | + | Its easy to see that <math>B'</math>, <math>C'</math>, <math>D</math> are collinear (since <math>\angle B'DC = \angle C'DC = 90^\circ</math>). Applying the sine rule in triangle <math>ABC</math>, we get <cmath>\frac{\sin BAD }{ \sin CAD} = \frac{BD }{ DC}.</cmath> Since <math>BAB'D</math> and <math>CC'AD</math> are cyclic quadrilaterals, <math>\angle BAD \angle BB'D</math> and <math>\angle CAD = \angle CC'D.</math> So, <cmath>\frac{\sin(BB'D)}{\sin(CC'D)} = \frac{BD}{DC}</cmath> and <cmath>\frac{BD}{\sin BB'D} = \frac{DC }{ \sin CC'D}.</cmath> Thus, <math>BB' = CC'</math> (the circumcircles <math>\mathcal{C}_1,\mathcal{C}_2</math> are congruent). |

− | From right triangles < | + | 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 < | + | <cmath>AC'^{2} = CC'^{2} - AC^{2} = BB'^{2} - AB^{2} = AB'^{2}.</cmath> 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>. So area of <math>[MBC] = [ABC]</math> and hence is independent of position of <math>D</math> on <math>BC</math>. |

## Latest revision as of 15:04, 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 ). Applying the sine rule in triangle , we get Since and are cyclic quadrilaterals, and So, and 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 .