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 \(2008\).

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TopNewestb=c square

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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.

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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.

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

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(a)=ad bxqpxmn-1-1X1x(a)(B)2x

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

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Who can prove 4=5

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