Let $ABC$ be acute triangle. Let $E$ and $F$ be points on $BC$, such that angles $BAE$ and $FAC$ are equal. Lines $AE$ and $AF$ intersect cirumcircle of $ABC$ at points $M$ and $N$. On rays $AB$ and $AC$ we have points $P$ and $R$, such that angle $PEA$ is equal to angle $B$ and angle $AER$ is equal to angle $C$. Let $L$ be intersection of $AE$ and $PR$ and $D$ be intersection of $BC$ and $LN$. Prove that $$\frac{1}{|MN|}+\frac{1}{|EF|}=\frac{1}{|ED|}.$$
Problem
Source: Mediterranean math competition 2018
Tags: geometry, circumcircle, Ugly
L3435
07.06.2018 13:37
Is the competition over in every participating country?
JANMATH111
07.06.2018 17:08
Yes it is over!
sbealing
07.06.2018 19:56
$$\frac{1}{MN}+\frac{1}{EF}=\frac{1}{ED} \Leftrightarrow 1+\frac{MN}{EF}=\frac{MN}{ED}$$Using $EF \parallel MN$ we get: $$\frac{MN}{ED}=\frac{LM}{LE} \quad , \quad \frac{MN}{EF}=\frac{AM}{AE}$$So it suffices to prove: $$\frac{AM}{AE}=\frac{EM}{LE} \Leftrightarrow EM \cdot AE=AM \cdot LE=AM \cdot (AE-AL)$$ Now $P$ is the point on $AB$ such that $\odot BPE$ is tangent to $AE$ giving $AP \cdot AB=AE^2$. Similarly for $R$ we get $AR \cdot AC=AE^2$. This gives $BPRC$ cyclic. Now angle chasing: $$\measuredangle BPL=\measuredangle BPR=\measuredangle BCR=\measuredangle BCA=BML$$So $BPLM$ is cyclic and so by POP $AL \cdot AM=AP \cdot AB=AE^2$. Going back to what we need to prove and using this result: $$\Leftrightarrow EM \cdot AE=AM \cdot AE-AM \cdot AL=AM \cdot AE-AE^2=AE \cdot EM$$So we're done.