Midpoint on diagonal of cyclic quadrilateral

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Midpoint on diagonal of cyclic quadrilateral

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Source: Mexico National Olympiad 2017, Problem 5

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#1 h Nov 7, 2017, 10:32 PM • 3 Y

Y by rkm0959, Adventure10, Mango247

On a circle $\Gamma$ , points $A, B, N, C, D, M$ are chosen in a clockwise order in such a way that $N$ and $M$ are the midpoints of clockwise arcs $BC$ and $AD$ respectively. Let $P$ be the intersection of $AC$ and $BD$ , and let $Q$ be a point on line $MB$ such that $PQ$ is perpendicular to $MN$ . Point $R$ is chosen on segment $MC$ such that $QB = RC$ , prove that the midpoint of $QR$ lies on $AC$ .

This post has been edited 1 time. Last edited by juckter, Nov 8, 2017, 3:31 AM

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#2 h Nov 8, 2017, 8:52 AM • 4 Y

Y by rkm0959, AlastorMoody, sunken rock, Adventure10

As $M$ is midpoint of arc $AD$ then for every point $X$ , that lies on the arc $AD$ not containing $M$ it is truem that $XM$ is angle bisector of $AXD$
Let $QC$ intersect $MC$ at $G$ . $\angle APB=180-\angle BAP -\angle ABP$ . $\angle NMC= \frac{\angle BAC}{2},\angle PCM=\frac{\angle ABD}{2}$ , $\angle PGM=90- \angle NMC= 90-\frac{\angle BAC}{2}, \angle CPG=\angle PGM-\angle PCG= 90-\frac{\angle BAC}{2}-\frac{\angle ABD}{2}= \frac{CPD}{2}$ so $GP$ is angle bisector of $CPD$ and $G$ - is incenter of $CPD$
$QP$ is angle bisector of $APB$ and $Q$ lies on $BM$ - angle bisector of $ABP$ , so $Q$ is incenter of $APB$

Let $F$ is point of intersection of $AC$ and $QR$ . $QF= \frac{AQ \sin \angle NAC}{\sin \angle QFA}=\frac{BQ \sin \angle ABM }{\sin \angle QFA}=\frac{RC \sin \angle ACM }{\sin \angle CFR}=FR$

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AlanLG

#3 h Jul 18, 2022, 11:23 PM

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Let $X=QR\cap AC$ , by menelao in $\triangle QMR$

$\frac{MC}{CR}\cdot \frac{RX}{XQ}\cdot \frac{QY}{YM}=1$
So is sufficient to prove that

$\frac{MC}{CR}=\frac{YM}{QY}$
As $\angle BMN=\angle NMC, \angle MBD=\angle ACM$ by exterior angle it´s easy to see that $PQ$ is the bisector of $\angle BPA$ then

$\frac{BQ}{QY}=\frac{PB}{PY}$
Note that $\frac{BP}{PY}=\frac{\sin\angle BYP}{\sin \angle PBY}=\frac{\sin\angle CYM}{\sin\angle YCM}=\frac{MC}{MY}$ , so $\frac{BQ}{QY}=\frac{MC}{MY}$ or $\frac{MC}{CR}=\frac{YM}{QY}$ , as desired.

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#4 h Jul 20, 2022, 9:07 AM

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RagvaloD wrote:

As $M$ is midpoint of arc $AD$ then for every point $X$ , that lies on the arc $AD$ not containing $M$ it is truem that $XM$ is angle bisector of $AXD$
Let $QC$ intersect $MC$ at $G$ . $\angle APB=180-\angle BAP -\angle ABP$ . $\angle NMC= \frac{\angle BAC}{2},\angle PCM=\frac{\angle ABD}{2}$ , $\angle PGM=90- \angle NMC= 90-\frac{\angle BAC}{2}, \angle CPG=\angle PGM-\angle PCG= 90-\frac{\angle BAC}{2}-\frac{\angle ABD}{2}= \frac{CPD}{2}$ so $GP$ is angle bisector of $CPD$ and $G$ - is incenter of $CPD$
$QP$ is angle bisector of $APB$ and $Q$ lies on $BM$ - angle bisector of $ABP$ , so $Q$ is incenter of $APB$

Let $F$ is point of intersection of $AC$ and $QR$ . $QF= \frac{AQ \sin \angle NAC}{\sin \angle QFA}=\frac{BQ \sin \angle ABM }{\sin \angle QFA}=\frac{RC \sin \angle ACM }{\sin \angle CFR}=FR$

Very nice! Just a small typo: $G\in PQ\cap MC$ . To avoid use of trigo calculations, I ended your proof as follows: $\triangle APB\sim\triangle DPC, Q$ and $R$ are their respective incenters, thus $\frac{QP}{GP}=\frac{QB}{GC}=\frac{CR}{GC}$ and, from Menelaos in $\triangle CQG$ with transversal $Q-F-R$ we get $FR=FG$ .

Best regards,
sunken rock

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khina

#5 h Jul 20, 2022, 2:58 PM

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Evidently $Q$ is the incenter of $PAB$ . Hence we may delete some points and reduce the problem to the following:

new problem wrote:

Consider a triangle $MBC$ , and let $S$ and $N$ denote the midpoints of the major and minor arcs $BC$ . Two points $Q$ and $R$ are chosen on segments $MB$ and $MC$ such that $QB = RC$ . If $T$ and $K$ are the respective midpoints of $QR$ and $BC$ , prove that $TC$ and $QN$ meet on $(BMC)$ .

solution

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#6 h Aug 5, 2022, 2:46 PM

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By angle chasing we get $Q$ is the incenter of $PAB$ . Then $\angle BAQ= \angle BAC/2=\angle BAN$ . So $A, Q, N$ collinear.
In a similar way we let $Q'=MC\cap PQ$ . $Q'$ is the incenter of $PDC$ . So $D, Q', N$ collinear.
As $\angle ANM = \angle MND$ and $NM \perp QQ'$ we get $NM$ is angle bisector of $QQ'$ .
By angle chasing we get $\triangle QBP$ is similar to $\triangle Q'CP$ . Now let $R'$ on $AC$ such that $RR' \parallel QQ'$ .
Then $\triangle RCR'$ is also similar, as $QB=RC$ then $RR'=QP$ and since these are parallel they form a parallelogram. So $QT$ and $PR'$ intersect at their midpoints.
In particular $T=PR'\cap QR$ then T lies on AC and is the midpoint of QR. $\square$

This post has been edited 1 time. Last edited by enrique17, Aug 6, 2022, 3:10 PM
Reason: typo

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