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Source: BrMO 2 2016

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#1 h Jan 29, 2016, 8:26 AM • 2 Y

Y by Adventure10, Mango247

Let $ABCD$ be a cyclic quadrilateral. The diagonals $AC$ and $BD$ meet at $P$ , and $DA$ and $CB$ produced meet at $Q$ . The midpoint of $AB$ is $E$ . Prove that if $PQ$ is perpendicular to $AC$ , then $PE$ is perpendicular to $BC$ .

See here: https://bmos.ukmt.org.uk/home/bmo2-2016.pdf

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#2 h Jan 29, 2016, 10:47 AM • 2 Y

Y by AndrewTom, Adventure10

Very nice problem! here is my solution.

Let $AB\cap CD\equiv K$ , $OP\cap QK\equiv L$ , $PE\cap QC\equiv T$ and $AC\cap QK\equiv S$ . Define $\Gamma_1\equiv \odot(ABCD)$ and $\Gamma_2\equiv \odot(OLK)$ .

Lemma 1. $OP$ is perpendicular to $QK$ .
Proof of the lemma. By Brokard's Theorem we have the polar of $Q,K$ both passes through $P$ , and the result immediately follows. Since $OE\perp AB$ we can deduce that $E\in \Gamma_2$ .

Lemma 2. $P$ lies on the radical axis of $\Gamma_1$ and $\Gamma_2$ .
Proof of the lemma. Note that $(A,C;P,S)=-1$ and $OL\perp QK$ , which implies that $\angle ALO=\angle CLO$ , Since $OA=OC$ , so $L,A,O,C$ are concyclic $\implies PA\cdot PC=PL\cdot PO$ .

Lemma 3. $L$ is the Miquel point of complete quadrilateral $ABCD$ .
Proof of the lemma. In fact the Miquel point of complete quadrilateral $ABCD$ lies on $\overline{QK}$ if and only if $A,B,C,D$ are concyclic. It's easy to verify by angle-chasing.

Back to the main problem, by lemma 3 we get $A,L,Q,B \quad \text{are concyclic} \ \ \ (1)$ and $L,K,D,A$ are concyclic as well. Which amounts to $QL\cdot QK=QA\cdot QD=QB\cdot QC\implies P$ lies on the radical axis of $\Gamma_1$ and $\Gamma_2$ . Thus $QP\perp OK \implies OK\parallel AC$ . At last just angle-chasing!

We have $\angle BAC=\angle EKO=\widehat{EO}=\angle ELP \implies L,A,P,E$ are concyclic $\implies \angle TEB=\angle AEP=\angle PLA$ . From $(1)$ we get $\angle KLA=\angle QBE$ , hence $\angle TEB+\angle TBE=\angle PLA+\angle ALK=90^\circ$ , as desired.

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This post has been edited 1 time. Last edited by Complex2Liu, Jan 29, 2016, 11:03 AM

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#3 h Jan 29, 2016, 12:51 PM • 9 Y

Y by rkm0959, Complex2Liu, AndrewTom, Re1gnover, HoseynHeydari, Med_Sqrt, enhanced, Adventure10, Mango247

Let $X,$ $Y$ be the midpoint of $AQ,$ $AP$ , respectively. Since $\measuredangle PXE$ $=$ $\measuredangle (PX,QP)$ $+$ $\measuredangle (QP, BC)$ $=$ $\measuredangle PQC$ $-$ $\measuredangle AQP$ $=$ $\measuredangle PAQ$ $-$ $\measuredangle QCP$ $=$ $\measuredangle CAD$ $-$ $\measuredangle BDA$ $=$ $\measuredangle CPB$ $=$ $\measuredangle PYE$ , so $E,$ $P,$ $X,$ $Y$ are concyclic $\Longrightarrow$ $\measuredangle XEP$ $=$ $90^{\circ}$ , hence we conclude that $PE$ is perpendicular to $BC$ .

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#4 h Jan 29, 2016, 8:02 PM • 3 Y

Y by AndrewTom, songssari, Adventure10

My solution:
Let $E'$ be a point in $AB$ such that $PE'\perp BC$ and let $X=\odot (PAE')\cap PQ$ $\Longrightarrow$ by angle-chasing we get $\triangle XQA\sim \triangle E'PB$ and by law of sines in $\triangle BPC$ and $\triangle QAC$ we get $\frac{BP}{PC}=\frac{QA}{QC}$ . Also that $\triangle XAE'\sim \triangle QCP$ $\Longrightarrow$ $\frac{XA}{AE'}=\frac{QC}{CP}$ but $\frac{XA}{E'B}=\frac{QA}{BP}$ since $\triangle XQA\sim \triangle E'PB$ $\Longrightarrow$ $\frac{XA}{AE'}=\frac{QC}{CP}=\frac{QA}{BP}=\frac{XA}{E'B}$ $\Longrightarrow$ $AE'=E'B$ $\Longrightarrow$ $E'=E$ $\Longrightarrow$ $PE\perp BC$ ...

This post has been edited 1 time. Last edited by FabrizioFelen, Jan 31, 2016, 9:13 PM

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#5 h Jan 30, 2016, 2:25 AM • 4 Y

Y by AndrewTom, songssari, Adventure10, Mango247

Vectors. It suffices to show that $\vec{PE} \cdot \vec{BC}=0$ . Using $\vec{PE}=\frac{1}{2}(\vec{PA}+\vec{PB})$ , we get $\vec{PE} \cdot \vec{BC} = \frac{1}{2}(\vec{PA}+\vec{PB})\cdot \vec{BC}$ $= -\frac{1}{2} \cdot PA \cdot BC \cdot \cos \angle BCA + \frac{1}{2}\cdot PB \cdot BC \cdot \cos \angle QBP$ $= \frac{1}{2} \cdot BC \cdot ( -PA \cos \angle BCA + PB \cos \angle QBP)$
Note that $\angle QAP = 180-\angle PAD = 180-\angle PBC = \angle QBP$ , so we have $\frac{1}{2} \cdot BC \cdot ( -PA \cos \angle BCA + PB \cos \angle QBP)$ $= -\frac{1}{2}BC \cdot (PA \cdot \frac{PC}{QC} - PB \cdot \cos \angle QAP)$ $=-\frac{1}{2}BC \cdot PA \cdot (\frac{PC}{QC}-\frac{BP}{QA})$
It now suffices to show that $\frac{PC}{QC}=\frac{BP}{QA}$ , which is true since $\frac{BP}{QA}=\frac{BP}{PQ} \cdot \frac{PQ}{QA} = \frac{\sin \angle PQC}{\sin \angle PBC} \cdot \sin \angle QAP = \sin \angle PQC = \frac{PC}{QC} \text{ }\blacksquare$

This post has been edited 4 times. Last edited by rkm0959, Jan 30, 2016, 10:03 AM

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#6 h Jan 30, 2016, 9:08 PM • 1 Y

Y by Adventure10

Thanks Complex2Liu, TelvCohl, FabrizioFelen and rkm0959 for your interesting solutions.

Are there other ways of doing this question?

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#7 h Jan 31, 2016, 9:30 PM • 3 Y

Y by AndrewTom, Adventure10, Mango247

There is actually,
Let $PE\cap BC$ = $F$ .
Miquel's point $M$ is the second intersection of circumcircles of $\triangle BPC$ and $\triangle APD$ .
Then $Q$ is radical centar of circumcircles of $BPC$ , $APD$ and $ABCD$ $\Longrightarrow$ $Q$ , $P$ and $M$ are collinear.
Because of $\measuredangle CPM$ = $90^{\circ}$ $\Longrightarrow$ $\measuredangle CBM$ = $90^{\circ}$ .
Let $G$ be midpoint of $AM$ and ray $GP$ $\cap$ $QC$ = $F_1$ .
By Brahmagupta Theorem on cyclic $AQCM$ $\Longrightarrow$ $GF_1\perp QC$ $\Longrightarrow$ $GF_1\parallel BM$ $(1)$ .
Because $EG$ is midsegment of $\triangle ABM$ $\Longrightarrow$ $EG\parallel BM$ $(2)$ . By $(1)$ and $(2)$ $\Longrightarrow$ $P$ , $E$ and $F_1$ are collinear.
This means $F_1$ = $F$ $\Longrightarrow$ $PF \perp BC$ .
$Q.E.D$ .

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#8 h Feb 3, 2016, 10:08 PM • 2 Y

Y by Adventure10, Mango247

Thanks mitevski169: your solution looks amazing. Unfortunately, I don't understand lines 2 and 3, nor do I follow the line on Brahmagupta's theorem.

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#9 h Feb 7, 2016, 2:47 PM • 1 Y

Y by Adventure10

Ok, a clarification without radical axes mentioning.
Let $BCMP$ be cyclic and $Q$ , $P$ , $M$ are collinear.
By power of a point $QP\cdot QM$ = $QB\cdot QC$ . Also we have $QB\cdot QC$ = $QA\cdot QD$ .
So $QA\cdot QD$ = $QP\cdot QM$ $\Longrightarrow$ $APMD$ is cyclic. So $M$ is miquel's point on $ABCD$ .
And for Brahmagupta see https://en.wikipedia.org/wiki/Brahmagupta_theorem.

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#10 h Apr 13, 2016, 2:31 PM • 3 Y

Y by chenguk2001, oolite, Adventure10

Let the reflection of $A$ across $P$ be $F$ , then $PE\parallel FB$ . Since $\angle PAD=\angle PBC$ $\implies \angle QFP=\angle QAP=\angle QBP$ $\implies F,P,Q,B$ cyclic $\implies \angle FBC=90^{\circ}$ , hence $PE\perp BC$ .

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