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# 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
2 years, 5 months ago

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b=c square · 2 years, 5 months ago

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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. · 2 years, 5 months ago

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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. · 2 years, 5 months ago

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Can you provide a diagramatical representation . I coudnt.draw a figure for the given information · 2 years, 5 months ago

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(a)=ad bxqpxmn-1-1X1x(a)(B)2x · 2 years, 5 months ago

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I m not able to figure out the diagram. Can someone help please? · 2 years, 5 months ago

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Who can prove 4=5 · 2 years, 5 months ago

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