AoPS - Problem

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Tags: geometry, trapezoid, incenter, circumcircle, Triangle, IMO Shortlist, IMO Longlist

Goutham

29.09.2010 21:42

$(FRA 4)$ A right-angled triangle $OAB$ has its right angle at the point $B.$ An arbitrary circle with center on the line $OB$ is tangent to the line $OA.$ Let $AT$ be the tangent to the circle different from $OA$ ( $T$ is the point of tangency). Prove that the median from $B$ of the triangle $OAB$ intersects $AT$ at a point $M$ such that $MB = MT.$

Luis González

30.09.2010 07:56

For convenience rename $O \equiv C$ and denote the subject circle as $(U).$ Let $P$ be the tangency point of $(U)$ with $AC$ and $D$ the midpoint of $AC.$ Since $\angle APU,$ $\angle ATU$ and $\angle ABU$ are right, then $A,B,T,U,P$ are concyclic on a circle $\omega.$ Let $N$ be the second intersection of $BD$ with $\omega.$ Since $\triangle BDA$ is isosceles with apex $D,$ it follows that $BPNA$ is an isosceles trapezoid with legs $AN=BU$ $\Longrightarrow$ $BN=AP,$ but $AP=AT.$ Thus, $BTNA$ is an isosceles trapezoid with legs $BA=TN$ $\Longrightarrow$ $\triangle MBT$ is isosceles with apex $M.$

kalantzis

01.10.2010 18:16

I have a simplier solution only with angle chase Again, using luisgeometria's notation,we get that $A,B,T,U,P$ are concyclic. Using this (the next are angles): $MBT=MBU+UBT=DBC+UAT=DCB+UAP=(90-BAC)+CAU=90-BAU$ and $MTB=ATB=AUB=90-BAU$ QED

dgreenb801

02.10.2010 22:57

I have another solution. Like before, $ABTUP$ is cyclic, and the center of the circumscribed circle is the midpoint $R$ of $AU$ . Since $\triangle BDA$ is isoceles, $R$ lies on the bisector of $\angle BDA$ . Also, $R$ lies on the bisector of $\angle PAT$ . So $R$ is the incenter of $\triangle MDA$ , so $RM$ bisects $\angle TMB$ , and since $R$ is the center of the circle, $MT=MB$ .