# Nice geometry problem

Let $ABC$ be a triangle. $D$ and $E$ lie on $AB$ such that $AD = AC, BE = BC$ and the points $D, A, B, E$ are collinear in that order. The bisectors of angle $A$ and $B$ intersect $BC, AC$ at $P$ and $Q$ respectively, and the circumcircle of $ABC$ at $M$ and $N$ respectively. Let $O_1$ be the circumcenter of $BME$ and $O_2$ be the circumcenter of $AND$. $AO_1$ and $BO_2$ intersect at $X$. Prove that $CX$ is perpendicular to $PQ$.

Source : Serbia $2008$. Note by Zi Song Yeoh
5 years, 3 months ago

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b=c square

- 5 years, 3 months ago

Solution (SPOILER) :

We let $AB = c, BC = a, CA = b$ for simplicity.

Note that $AE \cdot AQ = (a + c) \cdot \frac{bc}{a + c} = bc$, by Angle Bisector Theorem. Also, $AP \cdot AM = AP \cdot (AP + PM) = AP^2 + AP \cdot PM = AP^2 + BP \cdot CP$, by Power of Point. Since by Stewart's Theorem, $AP^2 + BP \cdot CP = \frac{AB^2 \cdot CP + AC^2 \cdot BP}{BC} = \frac{c^2(\frac{ab}{b + c}) + b^2(\frac{ac}{b + c})}{a} = \frac{bc(b + c)}{b + c} = bc$, we have $AP \cdot AM = bc$.

Consider the following transformation (commonly known as $\sqrt{bc}$-inversion.) :

1) Reflect every point $X$ across the angle bisector of $\angle BAC$.

2) Invert about point $A$ with radius $\sqrt{bc}$.

So, it follows immediately that $E$ and $Q$ map to each other. Similarly, $P$ and $M$ map to each other. Thus, the circumcircle of $BME$ is mapped to the circumcircle of $CPQ$, with center $O$. Therefore, $AO$ is isogonal to $AO_1$ in $\angle BAC$, since the reflection of $O_1$ across the angle bisector of $\angle BAC$ is collinear with $A$ and $O$. Similarly, $BO$ is isogonal to $BO_2$ in $\angle ABC$. Thus, $O$ and $X$ are isogonal conjugates with respect to $\triangle ABC$. Finally, since in triangle $CQP$, $CX$ is isogonal with $CO$, $C, H, X$ are collinear and thus $CX \perp PQ$. ($H$ is the orthocenter of $CPQ$)

The last line follows from the fact that $O$ and $H$ are isogonal conjugates.

- 5 years, 3 months ago

An easy way to establish $AP\cdot AM=bc$: we have $\angle BAM=\angle CAP$ (since $AP$ angle bisector) and $\angle ACB=\angle AMB$ since $ABMC$ cyclic. It follows that $\triangle ABM\sim \triangle APC$ so side ratios are equal.

- 5 years, 3 months ago

Can you provide a diagramatical representation . I coudnt.draw a figure for the given information

- 5 years, 3 months ago

- 5 years, 3 months ago

I m not able to figure out the diagram. Can someone help please?

- 5 years, 3 months ago

Who can prove 4=5

- 5 years, 3 months ago