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Source: Turkey TST 2001 - P2

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#1 h Apr 4, 2013, 4:58 PM • 2 Y

Y by Adventure10, Mango247

A circle touches to diameter $AB$ of a unit circle with center $O$ at $T$ where $OT>1$ . These circles intersect at two different points $C$ and $D$ . The circle through $O$ , $D$ , and $C$ meet the line $AB$ at $P$ different from $O$ . Show that
$|PA|\cdot |PB| = \dfrac {|PT|^2}{|OT|^2}.$

This post has been edited 1 time. Last edited by xeroxia, Jan 16, 2014, 8:13 AM

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#2 h Apr 4, 2013, 5:03 PM • 2 Y

Y by Adventure10, Mango247

Let perpendicular from $P$ to $OA$ meet the unit circle at $S$ . Let $ST$ meet the unit circle at $U$ .
In fact, it is $\dfrac {PA\cdot PB}{AO^2} = \dfrac {PT^2}{OT^2} \Rightarrow \dfrac {SP}{AO} = \dfrac {PT}{OT} \Rightarrow SP \parallel OU.$
See https://artofproblemsolving.com/community/c6h528123p3003099 for the rest.

This post has been edited 2 times. Last edited by xeroxia, Sep 24, 2023, 8:09 PM

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#3 h Jan 14, 2014, 11:37 PM • 1 Y

Y by Adventure10

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#4 h Jan 15, 2014, 12:32 AM • 3 Y

Y by thecmd999, Adventure10, Mango247

Any line intersects a pencil of circles at pairs of points in involution, thus $T$ is a double point of the involution $A \mapsto B,$ $P \mapsto O$ $\Longrightarrow$ $(P,O,A,T)=(O,P,B,T)$ $\Longrightarrow$

${\frac{\overline{PA}}{\overline{PT}} \cdot \frac{ \overline{OT}}{\overline{OA}}=\frac{\overline{OB}}{\overline{OT}} \cdot \frac{ \overline{PT}}{\overline{PB}} \Longrightarrow \frac{\overline{PT}^2}{\overline{OT}^2}}=\frac{ \overline{PA} \cdot \overline{PB}}{ \overline{OA} \cdot \overline{OB}}=\overline{PA} \cdot \overline{PB}.$

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#5 h Sep 24, 2023, 8:04 PM

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$CD$ intersects $AB$ at point $S$ . We have $ST^2 = SD \cdot SC$ .

Let $K$ be the point where the tangent to the circle drawn from point $S$ touches the circle (for ease of drawing, assume that points $K$ and $C$ are on opposite sides of the line $AB$ ). Thus, $SK^2 = SD \cdot SC$ .

From the power of point $S$ with respect to the circle $(O, C, D, P)$ , we have $SP \cdot SO = SD \cdot SC = SK^2$ , and we also know that $\angle OKS = 90^\circ$ , which means $KP \perp OS$ .

Let the line $TK$ intersect the circle with diameter $AB$ again at point $L$ . Since $SK = ST$ , we have $\angle STK = \angle SKT$ .

Now, $\angle OLK = \angle OKL = 180^\circ - (90^\circ + \angle SKT) = 90^\circ - \angle SKT$ . Therefore, $LO \perp OT$ .

As a result, $PK \parallel OL$ . Using similarity, we can conclude that $\dfrac {PK}{OL} = \dfrac {PT}{OT}$ .

From the power of point $K$ with respect to the circle with diameter $AB$ , we have $PK^2 = PA \cdot PB$ . So, $\dfrac {PA \cdot PB}{OL^2} = \dfrac {PT^2}{OT^2}$ .

Since $OL = 1$ , the proof is complete.

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