Let $ABC$ be a triangle with $AB > BC$ and $\gamma$ its circumscribed circle. Let $M, N$ be on the sides $[AB]$, $[BC]$, so that $AM = CN$.

Question

Let $ABC$ be a triangle with $AB > BC$ and $\gamma$ its circumscribed circle. Let $M, N$ be on the sides $[AB]$, $[BC]$ respectively, so that $AM = CN$. If $K = MN \cap AC$, $P$ is the center of the circle inscribed in the triangle $AMK$, $Q$ is the center of the circle $K$−exscribed of the triangle $CNK$ (opposite the vertex $K$,tangent to $CN$). If $R$ is the midpoint of the arc $ABC$ of $\gamma$, show that $RP = RQ$.

I've tried constructing some random points (like the intersection of the perpendicular bisector of $QP$ with $\gamma$) to notice any properties, but that did not help. I've also tried redefining the point $R$.

Answer 1

This may be also a picture proof, the reader may want to ignore below everything but the pictures and come with her or his own proof. But for didactic purposes, since we have a NMO problem and want to collect all points, there is also the story around the pics.

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I need a lemma first, some points from the posted question are involved, notations are kept. I will introduce some more points, showing how the picture was realized.

Given is a triangle $\Delta ABC$, we draw its circumcenter $\odot(ABC)=\odot(O)$ with center $O$, the diameter perpendicular on $AC$ is $RS$, with $R$ on the same side of $AC$ as $B$. Consider on $AC$ points $M_1$, $N_1$ on $AC$, reflected w.r.t. the side bisector $RS$ of $AC$, denote by $m$ the common length $m=AM_1=CN_1$. Circles centered in $A$, respectively $C$ with radius $m$ further intersect the segments $AB$ in $M$, and respectively $CB$ in $N$. Let $\Omega$ be the circumcenter of the cirumcircle $\odot(BMN)$ of $\Delta BMN$.

Then:

• $R$ is also on the circle $\odot(\Omega)=\odot(BMN)$,
• $RM=RN$.

Proof: Draw parallels and perpendiculars through the circumcenters $O$ and $\Omega$ to the sides $BA$ and $BC$ of $\Delta ABC$. We obtain as in the picture two orange triangles with same hypotenuse $O\Omega$, and one of the legs is $m/2$ in each, the distance from the mid points of $AB$ and $MB$ for one triangle, respectively mid points of $CB$ and $NB$ for the other one.
These triangles are thus congruent, so $\Omega O$ is an angle bisector for an angle delimited by sides parallel to $BA$ and $BC$. So $\Omega O$ is parallel to the angle bisector $BS$ of $\widehat{ABC}$. Since $\Omega O\|BS\perp BR$ we obtain that $\Omega O$ is also the side bisector of $BR$, thus $\Omega B=\Omega R$, so $R$ is also on the circle $\odot(\Omega)=\odot(BMN)$.

To see that $RM=RN$, compare $\Delta RNC$ and $\Delta RMA$. They have $RC=RA$, $NC=m=NA$, and $\widehat{NCR}=\widehat{BCR}=\widehat{BAR}=\widehat{MAR}$. So the two triangles are congruent. (And a rotation around $R$ with angle $\widehat{NRM}=\widehat{NBM}=\widehat{CBA}=\hat B$ brings one triangle into the other.)

$\square$

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We are now in position to show the claimed result. Here there is a slightly changed way to introduce points, done in an equivalent manner. Since some other points, useful for the proof, are also introduced let us explicitly restate:

Problem: Let $\Delta ABC$ be given, consider $M,N,R,S$ as in the previous Lemma. Let $K=AC\cap MN$. Then $\bbox[yellow]{(1)}$ the quadrilaterals $AMRK$ and $CNRK$ are cyclic. Let $P^*$, $Q^*$ be the mid points of the arcs $\overset\frown{AM}$ of $\odot(AMRK)$, and respectively $\overset\frown{CN}$ of $\odot(CNRK)$, that can be defined alternatively as the intersections of the angle bisector $$\kappa$$ of $\widehat{AKM}=\widehat{CKN}$ with the two mentioned circles. Draw the interior and exterior angle bisectors of $\Delta ABC$ going either through $I$, its incenter, or through $I_B$, its $B$-excenter. Consider the intersections of these bisectors with $\kappa$: $$ \begin{aligned} P'&=\kappa\cap IA \ , & Q'&=\kappa\cap IC\ ,\\ P &=\kappa\cap I_BA\ , & Q &=\kappa\cap I_BC\ . \end{aligned} $$ Then $\bbox[yellow]{(2)}$ $\odot(APMP')$ is a circle centered in $P^*$, and $\odot(CQNQ')$ is a circle centered in $Q^*$.

In particular we have $\bbox[yellow]{(3)}$: $$ \begin{aligned} RA&=RC\ ,\\ RM&=RN\ ,\\ RP^*&=RQ^*\ ,\\ RP&=RQ\ ,\\ RP'&=RQ'\ . \end{aligned} $$

Proof: For $\bbox[yellow]{(1)}$ - using the above Lemma: $$ \widehat{RNK} = \widehat{RNM} = \widehat{RBM} = \widehat{RBA} = \widehat{RCA} \ . $$ So $CNRK$ cyclic. Similarly $AMRK$ cyclic.

$\bbox[yellow]{(2)}$ take first from the picture only the piece involving the points $K,C,N;Q,Q',Q*$ and the angle bisectors in $C,K$. Denote by $x,y$ half of the angle $\hat C$, respectively $\hat K$ in $\Delta CKN$. Then $$ \begin{aligned} \widehat{Q^*CQ'} &= \widehat{Q^*CN} + \widehat{NCQ'} = \widehat{Q^*KN} + \widehat{NCQ'} =x+y=\widehat{Q'CK} + \widehat{Q'KC} \\ &=\widehat{Q^*Q'C}\ . \end{aligned} $$ So $\Delta Q^*Q'C$ isosceles, $Q^*Q'=Q^*C$. The angle bisectors $CQ$, $CQ'$ are perpendicular, so $\Delta QCQ'$ has a right angle in $C$. Its mid point of the hypotenuse is the only point of this side at same distance from $C$ and $Q'$, so it is $Q^*$. We obtain thus $Q^*Q=Q^*C=Q^*Q'$. Since $Q^*$ is the mid point of the arc $\overset\frown{CN}$ of $\odot(CNRK)$, we can add also $Q^*N$ to the chain of equalities. So $Q^*$ is the center of the circle $\odot(CQNQ')$. The same happens also for $\odot(APMP')$.

We can attack $\bbox[yellow]{(3)}$. The obvious $R$-rotation moves $\Delta RNC\to\Delta RMA$, so take in the movement also corresponding objects, circles $\odot(RNC)\to\odot(RMA)$, arcs of them $\overset\frown{NC}\to\overset\frown{MA}$, and their mid points $Q^*\to P^*$. So $$ RQ^*=RP^*\ , $$ so the perpendicular bisector of $P^*Q^*$ goes through $R$, and w.r.t. this bisector we have reflected pairs of points $(P^*,Q^*)$, and $(P,Q)$, and $(P',Q')$, the last two pairs since we know the distances on the line $\kappa=KPP^*P'QQ^*Q$.

$\square$

Answer 2

Hint: May be the figure I constructed is a particular case where following points can be seen:

1-M is mid point of AB, O is the center of circle $\gamma$ so $OM\bot AB$.

2- Draw circle $\alpha$ on diameter PQ, mark it's center as E.

3-Construct isosceles trapezoid ACGM on base AC. Connect O to G and extend it to meet circle $\gamma$ at I.So $OG\bot CG$

4-Draw a tangent from Q, meets circle $\alpha$ at F. You will find $QF||CG$, $EI\bot QF$.

5- Connect F to E, clearly $FE\bot PQ$ and EFQ is isosceles right angled triangle.

6-Draw perpendicular bisector of AC, it intersect circle $\gamma$ at R which is coincident on FE; also R is midpoint of arc ABC.

In this way $RP=RQ$.

In fact the key point is that R is where EF and perpendicular bisector of AC intersect.

I can send you the figure if I have an address.