Let $ABCD$ be circumscribed quadrilateral such that the midpoints of $AB$,$BC$,$CD$ and $DA$ are $K$, $L$, $M$, $N$ respectively. Let the reflections of the point $M$ wrt the lines $AD$ and $BC$ be $P$ and $Q$ respectively. Let the circumcenter of the triangle $KPQ$ be $R$. Prove that $RN=RL$
Problem
Source: Turkey JBMO TST 2024 P7
Tags: geometry, geometric transformation, reflection, circumcircle
PROF65
13.05.2024 16:06
As stated below the quadrilateral should be inscribed not circumscribed else the desired result doesn't occur if so: Let $I$ be the center of the parallelogram $KLMN$ , $T=AD\cap BC$, $R'$ be the isogonal of $M$ wrt $TNL$ , let $U,V$ be the respective midpoints of $MP,MQ$ since $\angle TNM= \angle MLT$ then $R'$ is on the perpendicular bisector of $NL$ so $(U VI)$ is the common pedal circle of $M$ and $R'$ hence $R'$ is the circumcenter of $KPQ$ i.e. $R'=R$ and the result follows. Best regards. RH HAS
Assassino9931
13.05.2024 22:20
PROF65 wrote: since $\angle TNM= \angle MLT$ Wait, this holds if and only if $\angle DAC = \angle DBC$, i.e. if $ABCD$ is cyclic, but it is given that $ABCD$ is circumscribed (i.e. has an incircle) in the problem statement? Update: Yes, the problem is wrong for a circumscribed quadrilateral, OP please edit the post.
Assassino9931
13.05.2024 22:47
Here is a simpler solution (too easy for P7, hopefully correct) for $ABCD$ cyclic. We have $\angle KNM = \angle KLM$ from the parallelogram $KLMN$, as well as $\angle PND = \angle MND = \angle DAC = \angle DBC = \angle CLM = \angle CLQ$ from the midsegments and reflections, so $\angle KNP = \angle QLK$. Also $KL = \frac{AC}{2} = MN = NP$ and similarly $NK = LQ$, thus $\triangle KNP \cong \triangle QLK$, implying $KP = KQ$ and $\angle PKN = \angle KQL = x$. Now $KP = KQ$ with the circumcenter $R$ imply $\angle RQK = \angle RKQ = \angle RKP = \varphi$. Therefore $\angle RKN = \angle RQL$, which together with $KN = QL$ and $RK = RQ$ imply $ \triangle RKN \cong \triangle RQL$ and therefore $RN = RL$.
Not needed in the solution. We have $\angle NRK = \angle QRL$ and similarly $\angle KRL = \angle PRN$, which together with $\angle PRQ = 2\angle PKQ = 4\varphi$ implies $\angle NRL = 180^{\circ} - 2\varphi$. Hence $\angle RNL = \angle RLN = \varphi = \frac{1}{2}\angle PKQ$, and with some more angle chasing around $\triangle KNP \cong \triangle QLK$ we see $\angle PNM = \angle QLM = 2\varphi$ and hence $\angle DAC = \angle DBC = \varphi$.)
wizixez
02.01.2025 16:18
One of my favourite geometry problems Set $\omega =(ABCD)\in \mathbb{S}^1$ such that $\omega $ is the set of points $(x,y)$ such that $x^2+y^2=1$ so $|a|_{b,c,d}=1$ $k=\frac{a+b}{2}$,$l=\frac{b+c}{2}$,$n=\frac{c+d}{2}$ finally $m=\frac{d+a}{2}$. So $p=ref(M,AD)$ and $q=ref(M,BC)$. And $R=\frac{\begin{vmatrix} k& k\overline{k} &1 \\ p& p\overline{p} &1 \\ q & q\overline{q} &1 \end{vmatrix}}{\begin{vmatrix} k& k &1 \\ p& p &1 \\ q & q &1 \end{vmatrix}}$ After that just prove $RNL$ is a $NL$-base isosceles triangle... $\lambda $