Nice geometry problem

Let ABCABC be a triangle. DD and EE lie on ABAB such that AD=AC,BE=BCAD = AC, BE = BC and the points D,A,B,ED, A, B, E are collinear in that order. The bisectors of angle AA and BB intersect BC,ACBC, AC at PP and QQ respectively, and the circumcircle of ABCABC at MM and NN respectively. Let O1O_1 be the circumcenter of BMEBME and O2O_2 be the circumcenter of ANDAND. AO1AO_1 and BO2BO_2 intersect at XX. Prove that CXCX is perpendicular to PQPQ.

Source : Serbia 20082008.

Note by Zi Song Yeoh
5 years, 3 months ago

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b=c square

Sergio VIlla - 5 years, 3 months ago

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Solution (SPOILER) :

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

Note that AEAQ=(a+c)bca+c=bcAE \cdot AQ = (a + c) \cdot \frac{bc}{a + c} = bc, by Angle Bisector Theorem. Also, APAM=AP(AP+PM)=AP2+APPM=AP2+BPCPAP \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, AP2+BPCP=AB2CP+AC2BPBC=c2(abb+c)+b2(acb+c)a=bc(b+c)b+c=bcAP^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 APAM=bcAP \cdot AM = bc.

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

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

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

So, it follows immediately that EE and QQ map to each other. Similarly, PP and MM map to each other. Thus, the circumcircle of BMEBME is mapped to the circumcircle of CPQCPQ, with center OO. Therefore, AOAO is isogonal to AO1AO_1 in BAC\angle BAC, since the reflection of O1O_1 across the angle bisector of BAC\angle BAC is collinear with AA and OO. Similarly, BOBO is isogonal to BO2BO_2 in ABC\angle ABC. Thus, OO and XX are isogonal conjugates with respect to ABC\triangle ABC. Finally, since in triangle CQPCQP, CXCX is isogonal with COCO, C,H,XC, H, X are collinear and thus CXPQCX \perp PQ. (HH is the orthocenter of CPQCPQ)

The last line follows from the fact that OO and HH are isogonal conjugates.

Zi Song Yeoh - 5 years, 3 months ago

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

Jubayer Nirjhor - 5 years, 3 months ago

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

Shehanaaz Sk - 5 years, 3 months ago

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

Sergio VIlla - 5 years, 3 months ago

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

Shreya Hardas - 5 years, 3 months ago

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Who can prove 4=5

Atif Qureshi - 5 years, 3 months ago

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