P, F, B,M concyclic and EM / EN = BF / BP wanted, incircle related

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P, F, B,M concyclic and EM / EN = BF / BP wanted, incircle related

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Source: 2010 Chile IMO TST p2

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#1 h Aug 23, 2021, 12:04 AM • 1 Y

Y by centslordm

Let $ABC$ be an isosceles triangle at $A$ . The inscribed circle is tangent to $BC$ , $AC$ , $AB$ at $D, E, F$ , respectively. Let $P$ be on the arc $EF$ that does not contain $D$ . Let $Q$ be the point of intersection of $BP$ and the inscribed circle of $ABC$ . Lines $EP$ and $\text{[math]}$ intersect line $BC$ at $M$ and $N$ respectively. Show that the points $P, F, B$ and $M$ lie on a circle and that $\frac{EM}{EN} =\frac{BF}{BP}$

This post has been edited 1 time. Last edited by parmenides51, Aug 23, 2021, 12:04 AM

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#2 h Aug 23, 2021, 1:43 PM • 1 Y

Y by centslordm

Definitely seen this before. Anyway,

$\begin{align*} \measuredangle BNQ&=\measuredangle DQE+\measuredangle NDQ=\measuredangle FED+\measuredangle QED\\&=\measuredangle FEQ=\measuredangle BFQ\implies BQFN \text{ is cyclic.} \end{align*}$ Now, $\begin{align*} \measuredangle FEM&=\measuredangle FEEC+\measuredangle AEP= \measuredangle FPE+\measuredangle EDP\\&=\measuredangle FQP=\measuredangle FQB=\measuredangle FNB=\measuredangle FNM \implies NFEM \text{ is cyclic.} \end{align*}$ Hence, $\begin{align*} \measuredangle FPB&=\measuredangle FPQ=\measuredangle FEQ=\measuredangle FEN\\&=\measuredangle FMN=\measuredangle FMB \implies BFPM \text{ is cyclic.} \end{align*}$ To show that $\frac{EM}{EN} =\frac{BF}{BP}$ , just use law of sines as $\measuredangle BFP=\measuredangle EMN$ and $\measuredangle FPB=\measuredangle BFQ=\measuredangle MNE$ .

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