Track / Stage 5 / 287 of 400 #887 of 1964

Problem 887

AIME late

Geometry Difficulty 5.6 Prove it Thai Mathematical Olympiad · Thailand

PA and PB be the tangents to circle $\omega$ from an external point $P$ . Let $M$ and $N$ be the midpoints of $AP$ and $AB$ , respectively. Extend $MN$ to meet $\omega$ at $C$ , where $N$ is between $M$ and $C$ . $PC$ meets $\omega$ at $D$ and extend $ND$ to intersect $PB$ at $Q$ . Show that $MNQP$ is a rhombus.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Observe that $AB \perp NP$ . Thus, $M$ is the circumcenter of $\triangle ANP$ and hence $MN = MP$ .

It can also be seen that $MN \parallel PQ$ .

From the power of the point $M$ , $PM^2 = MA^2 = ME \cdot MC$ .

So, $\frac{PM}{ME} = \frac{MC}{PM}$ and hence $\triangle PME \sim \triangle CMP$ .

Thus, $M\hat{P}E = M\hat{C}P$ .

Since $O$ , $A$ , $P$ , $B$ are cyclic, the power of the point $N$ tells us that
$CN \cdot NE = AN \cdot NB = ON \cdot NP.$
Thus, $C$ , $P$ , $E$ , $O$ are also cyclic and hence $E\hat{P}N = N\hat{C}O$ .

Since $\triangle PAN \sim \triangle POA$ , $\frac{PA}{PN} = \frac{PO}{PA}$ . Thus,
$PN \cdot PO = PA^2 = PD \cdot PC.$
So, $C$ , $D$ , $N$ , $O$ are cyclic and hence $Q\hat{N}P = P\hat{C}O$ .

We can now see that
$Q\hat{N}P = P\hat{C}O = P\hat{C}M + M\hat{C}O = M\hat{P}E + E\hat{P}N = M\hat{P}N.$
Thus, $MP \parallel NQ$ . Therefore, $MNQP$ is a rhombus.

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