Consider a convex quadrilateral $ABCD$. Denote $M, \ N$ the points of tangency of the circle inscribed in $\triangle ABD$ with $AB, \ AD$, respectively and $P, \ Q$ the points of tangency of the circle inscribed in $\triangle CBD$ with the sides $CD, \ CB$, respectively. Assume that the circles inscribed in $\triangle ABD, \ \triangle CBD$ are tangent. Prove that: a) $ABCD$ is circumscriptible. b) $MNPQ$ is cyclic. c) The circles inscribed in $\triangle ABC, \ \triangle ADC$ are tangent.
Problem
Source: Romanian JBTST II 2007, problem 2
Tags: geometry, rectangle, geometry proposed
edriv
22.04.2007 21:45
Invalid URL The two tangents from a poin P to a circle are equal. a) $AD+BC = AN+DN+BP+PC = AM+DT+BT+CQ = AM+DQ+BM+CQ = (AM+MB)+(QC+QD) = AB+CD$ Therefore the sum of opposite sides is constant. This is known as a sufficient condition for a convex quadrilateral to be circumscribed. b) Consider the circles centered at B,D with radii $BM=BT=BP,DN=DT=DQ$. Their distance is the sum of their radii, then they are tangent. Also the two incircles are tangent. And the circles centered at B,D are ortogonal to the incircles. Therefore an inversion with centre T sens this four circles to a rectangle. But M,N,P,Q are the points of the intersection of our circles distinct from T. Then M,N,P,Q are the vertices of a rectangle (after inversion), which is of course cyclic. c) In order to use the same figure, we suppose that ABCD is circumscribed and prove that the incircles of $%Error. "teignale" is a bad command. BDA,%Error. "teigngle" is a bad command. BDC$ are tangent. $\triangle BDA, \triangle BDC$ touch BD recpectively at T,T' We have $DT-BT = DN-BM = DN+NA-BM-MA = AD-AB = AD-AB+(DC+AB)-(AD+BC) = DC-BC = DQ-BP = DT'-BT'$. We have $DT-BT=DT'-BT'$. We add to both sides the same quantity $DT+BT = DT'+BT'$ and get $2DT = 2DT'$, or $DT = DT'$: the circles are tangent.
silouan
23.04.2007 12:16
Your solution to b) is very interesting but you kill a fly with a super-laser bazoukas Here is a simpler one . We have that $DN=DQ$ , $AN=AM$ , $BM=BP$ , $CP=CQ$ . The bisectors of the angles $A,\ B,\ C,\ D$ pass through the same point since the quadrilateral is incriptible .But the bisectors of these angles are also perpendicular bisectors of $NQ,\ QP,\ PM,\ MN$ so we are done .
edriv
23.04.2007 15:54
I whish I had a super-laser bazouka now to kill the fly that is buzzing in my room By the way, with that trick you can kill also IMO 1996 A2.
Panoz93
05.06.2013 17:09
I found a angle-chasing solution to b Let R be the point where the circles are tangent We have $\angle{RDP}=180-2\angle{RQP}$ Also $DN=DR=DN$ therefore $\angle{PNR}=90-\angle{RQP}$ Similar we have $\angle{MQN}=90-\angle{MNR}$ Hence we get $\angle{MQP}+\angle{MNP}=180$