# Problem 5 from the IMO 2004

Here is Problem 5 from the IMO 2004:

In a convex quadrilateral ABCD the diagonal BD does not bisect the angles ABC and CDA. The point P lies inside ABCD and satisfies

The solution below is by Vo Duc Dien who dedicated it to his former teacher Le Chi De.

**The condition AP = CP is necessary.** Indeed, assume point B is already on the circumcircle of ΔACD.

Extend DP to intercept the circle at Q and BP to intercept the circle at E. Consider two quadrilaterals ABPD and CQPE. Since

∠ABE = ∠DBC = ∠DQC

∠DPB = ∠EPQ

∠ADQ = ∠BDC = ∠BEC.

Therefore ∠ DAB = ∠ECQ, for the sum of the angles of a quadrilateral is 360°.

Two triangles DAB and ECQ are identical since

It follows that triangles DPB and EPQ are also identical (two angles on each side of DB and EQ are equal which gives us

Therefore triangles ABP and CQP are identical since

**The condition AP = PC is sufficient.** Indeed, assume

Consider the circumcircle of ΔADC. From C draw a line that intercepts the circle at E and that

Choose arbitrary point P and draw a line to connect E and P and extend it to cut the circle at B to satisfy the first condition

Now let O be the circumcenter of ΔABC. We note that OP is the axis of symmetry of AD and CE. Extend DP to intercept the circle at Q.

Let's show that the assumption

So P has to be on Ox for B to be on the circle so that

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny65971246