Olympiad Maths Prep

Track / Stage 7 / 273 of 300 #1673 of 2000

Problem 1673

National olympiad second round; IMO P1/P4
Geometry Difficulty 7.7 Prove it

Let ABCDABCD be a parallelogram with DAB<90\angle DAB < 90 Let EE be the point on the line BCBC such that AE=ABAE = AB and let FF be the point on the line CDCD such that AF=ADAF = AD. The circumcircle of the triangle CEFCEF intersects the line AEAE again in PP and the line AFAF again in QQ. Let XX be the reflection of PP over the line DEDE and YY the reflection of QQ over the line BFBF. Prove that A,X,YA, X, Y lie on the same line.

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

1. Identify the given elements and their properties:
- ABCDABCD is a parallelogram with DAB<90\angle DAB < 90^\circ.
- EE is a point on the line BCBC such that AE=ABAE = AB.
- FF is a point on the line CDCD such that AF=ADAF = AD.
- The circumcircle of CEF\triangle CEF intersects the line AEAE again at PP and the line AFAF again at QQ.
- XX is the reflection of PP over the line DEDE.
- YY is the reflection of QQ over the line BFBF.

2. **Prove that ADPFADPF and ABQEABQE are isosceles trapezoids:**
- Since AE=ABAE = AB and AF=ADAF = AD, triangles ABEABE and ADFADF are isosceles.
- In ADF\triangle ADF, since AF=ADAF = AD, AFD=ADF\angle AFD = \angle ADF.
- In ABE\triangle ABE, since AE=ABAE = AB, AEB=ABE\angle AEB = \angle ABE.
- Since PP lies on the circumcircle of CEF\triangle CEF and intersects AEAE again, AEP=AFP\angle AEP = \angle AFP.
- Similarly, since QQ lies on the circumcircle of CEF\triangle CEF and intersects AFAF again, AFQ=AEQ\angle AFQ = \angle AEQ.

3. **Show that the circumcircles of ADPFADPF and ABQEABQE meet at AA and another point ZZ:**
- Let the circumcircles of ADPFADPF and ABQEABQE intersect at AA and another point ZZ.
- Since ADPFADPF is an isosceles trapezoid, ADP=AFP\angle ADP = \angle AFP.
- Since ABQEABQE is an isosceles trapezoid, ABQ=AEQ\angle ABQ = \angle AEQ.
- Therefore, AZE=AQB=AFD\angle AZE = \angle AQB = \angle AFD.

4. **Prove that E,Z,DE, Z, D and F,Z,BF, Z, B are collinear:**
- Since AZE=AFD\angle AZE = \angle AFD, E,Z,DE, Z, D must be collinear.
- Similarly, since AZF=AEB\angle AZF = \angle AEB, F,Z,BF, Z, B must be collinear.

5. **Determine the positions of XX and YY:**
- Let the perpendicular from PP to DEDE intersect AZAZ at XX'.
- Since DZX=AFD=PZD\angle DZX = \angle AFD = \angle PZD and AF=AD=DPAF = AD = DP, X=XX = X'.
- Similarly, let the perpendicular from QQ to BFBF intersect AZAZ at YY'.
- Since BZY=AEB=QZB\angle BZY = \angle AEB = \angle QZB and AE=AB=BQAE = AB = BQ, Y=YY = Y'.

6. **Conclude that A,X,YA, X, Y lie on the same line:**
- Since XX and YY are reflections of PP and QQ over DEDE and BFBF respectively, and both lie on AZAZ, A,X,YA, X, Y are collinear.

\blacksquare

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.