Express OQ/OP in terms of t

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Express OQ/OP in terms of t

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#1 h Jun 8, 2005, 4:07 AM • 2 Y

Y by Adventure10, Mango247

Let $M$ be the intersection of two diagonal, $AC$ and $BD$ , of a rhombus $ABCD$ , where angle $A<90^\circ$ . Construct $O$ on segment $MC$ so that $OB<OC$ and let $t=\frac{MA}{MO}$ , provided that $O \neq M$ . Construct a circle that has $O$ as centre and goes through $B$ and $D$ . Let the intersections between the circle and $AB$ be $B$ and $X$ . Let the intersections between the circle and $BC$ be $B$ and $Y$ . Let the intersections of $AC$ with $DX$ and $DY$ be $P$ and $Q$ , respectively. Express $\frac{OQ}{OP}$ in terms of $t$ .

This post has been edited 2 times. Last edited by WakeUp, Mar 25, 2012, 10:53 AM
Reason: LaTeXed. (& changed the title)

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#2 h Jun 8, 2005, 9:23 AM • 2 Y

Y by Adventure10, Mango247

Denote $\alpha = \angle DAB = \angle BCD,\ \beta = \angle ABC = \angle CDA,\ \alpha + \beta = 180^o$ . The quadrilateral $XBYD$ is cyclic, hence the sums of the opposite angles are equal to $180^o$ :

$\angle BYD = 180^o - \angle BXD$

$\angle XDY = 180^o - \angle XBY = 180^o - \beta = \alpha$

Thus the angle $\angle XDY \equiv \angle PDQ = \alpha$ is constant. Using the sine theorem for the triangles $\triangle BDX, \triangle BDY$ ,

$\frac{DX}{BC} = \frac{\sin{\widehat{XBD}}}{\sin{\widehat{BXD}}} = \frac{\sin{\frac \beta 2}}{\sin{\widehat{BXD}}}$

$\frac{DY}{BC} = \frac{\sin{\widehat{YBD}}}{\sin{\widehat{BYD}}} = \frac{\sin{\frac \beta 2}}{\sin{(180^o - \widehat{BXD})}} = \frac{\sin{\frac \beta 2}}{\sin{\widehat{BXD}}}$

Consequently, the segments $DX = DY$ are equal, which implies that the triangle $\triangle XYD$ is isosceles, the perpendicular bisector of the segment $XY$ passes through the vertex $D$ and it is identical with the bisector of the angle $\angle XDY \equiv \angle PDQ$ . Since $(O)$ is the circumcircle of this isosceles triangle, the circumcenter $O$ lies on this perpendicular bisector and the angle bisector. Denote $\phi = \angle MDO$ . Then

$MO = MD \tan \phi = MA \tan{\frac \alpha 2} \tan \phi$

$t = \frac{MA}{MO} = \frac{1}{\tan{\frac \alpha 2} \tan \phi} = \cot{\frac \alpha 2} \cot \phi$

$\frac{MD}{PD} = \cos{\left(\frac \alpha 2 - \phi\right)},\ \ \ \frac{MD}{QD} = \cos{\left(\frac \alpha 2 + \phi\right)}$

The angle bisector $DO$ divides the opposite side of the triangle $\triangle DPQ$ in the ratio of the adjacent sides $OP, OQ$ . As a result,

$\frac{OQ}{OP} = \frac{QD}{PD} = \frac{\cos{\left(\frac \alpha 2 - \phi\right)}}{\cos{\left(\frac \alpha 2 + \phi\right)}} = \frac{\cos{\frac \alpha 2} \cos \phi + \sin{\frac \alpha 2} \sin \phi}{\cos{\frac \alpha 2} \cos \phi - \sin{\frac \alpha 2} \sin \phi} =$

$= \frac{\cot{\frac \alpha 2} \cot \phi + 1}{\cot{\frac \alpha 2} \cot \phi - 1} = \frac{t + 1}{t - 1}$

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#3 h Jun 9, 2005, 6:15 PM • 2 Y

Y by Adventure10, Mango247

We have $\angle{AXD}=\angle{BYD}=\angle{BOA}$ hence $BOPX$ is cyclic.
Consider the inversion I of pole $O$ and power $OB^2$ . Then the circle circumscribed to $BOPX$ will transform to line $AB$ . Hence $I(P)=A$ , consequently $OP\cdot{OA}=OB^2$ .
Similarly we obtain $OQ\cdot{OC}=OB^2$ .
Hence $\frac{OQ}{OP}=\frac{OA}{OC}=\frac{MA+MO}{MA-MO}=\frac{\frac{MA}{MO}+1}{\frac{MA}{MO}-1}=\frac{t+1}{t-1}$

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