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 \[|PA|\cdot |PB| = \dfrac {|PT|^2}{|OT|^2}.\]
Problem
Source: Turkey TST 2001 - P2
Tags: geometry, geometric transformation, reflection, geometry proposed
xeroxia
04.04.2013 20:03
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.
thecmd999
15.01.2014 01:37
Let $x=OT$. Furthermore, let $T'$ be the reflection of $T$ over $Q$. Since $Q$ has the same power wrt all three circles, we have $QT^2=QA\cdot QB=QP\cdot QO$; hence $(A, B;T', T)=-1$. It follows that $OT'\cdot OT=1\implies OT'=\frac{1}{x}$. Since $Q$ is the midpoint of $TT'$, we get $OQ=\frac{x+\dfrac{1}{x}}{2}=\frac{x^2+1}{2x}$.
Now remove $T'$ from the diagram; we only needed it to find $OQ$. Similarly to above, we have $(A, B;P, Q)=-1\implies OP\cdot OQ=1\implies OQ=\frac{2x}{x^2+1}$. Since $PA\cdot PB=(1-PO)(1+PO)$, the desired equality is equivalent to $\left(1-\frac{2x}{x^2+1}\right)\left(1+\frac{2x}{x^2+1}\right)=\frac{\left(x-\dfrac{2x}{x^2+1}\right)^2}{x^2}$, which rearranges as \[\left(\frac{(x-1)^2}{x^2+1}\right)\left(\frac{(x+1)^2}{x^2+1}\right)=\left(1-\frac{2}{x^2+1}\right)^2\iff\left(\frac{x^2-1}{x^2+1}\right)^2=\left(\frac{x^2-1}{x^2+1}\right)^2\] which is true. $\blacksquare$
Luis González
15.01.2014 02:32
Any line intersects a pencil of circles at pairs of points in involution, thus $T$ is a double point of the involution $A \mapsto B,$ $P \mapsto O$ $\Longrightarrow$ $(P,O,A,T)=(O,P,B,T)$ $\Longrightarrow$ ${\frac{\overline{PA}}{\overline{PT}} \cdot \frac{ \overline{OT}}{\overline{OA}}=\frac{\overline{OB}}{\overline{OT}} \cdot \frac{ \overline{PT}}{\overline{PB}} \Longrightarrow \frac{\overline{PT}^2}{\overline{OT}^2}}=\frac{ \overline{PA} \cdot \overline{PB}}{ \overline{OA} \cdot \overline{OB}}=\overline{PA} \cdot \overline{PB}.$
xeroxia
24.09.2023 23:04
$CD$ 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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