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NMO Selection Tests for JBMO

Geometry Difficulty 5.6 AIME, harder Prove it Romania

Let $ABC$ be a triangle. An arbitrary circle which passes through the points $B$ and $C$ intersects the sides $AC$ and $AB$ for the second time in $D$ and $E$ , respectively. The line $BD$ intersects the circumcircle of triangle $AEC$ at $P$ and $Q$ , and the line $CE$ intersects the circumcircle of the triangle $ABD$ at $R$ and $S$ , such that $P$ is situated on the segment $BD$ , and $R$ lies on the segment $CE$ . Prove that:

a) the points $P, Q, R$ and $S$ are concyclic;

b) the triangle $APQ$ is isosceles.

Solution

a) Denote by $X$ the intersection of the lines $CE$ and $BD$ . Writing the power of $X$ with respect to all three circles, we obtain:
$XR \cdot XS = XB \cdot XD = XE \cdot XC = XP \cdot XQ.$
Since $\{X\} = PQ \cap RS$ , we infer that the points $P, Q, R$ and $S$ are concyclic.

b) The triangles $ADP$ and $APC$ are similar (A.A.), thus $AP^2 = AD \cdot AC$ . Similarly, we find that $AQ^2 = AE \cdot AB$ .

From the power of the point $A$ with respect to the circumcircle of the quadrilateral $BCDE$ , we infer that $AD \cdot AC = AE \cdot AB = \rho(A)$ , therefore $AP = AQ = \sqrt{\rho(A)}$ .

Similarly, we prove that $AR = AS = \sqrt{\rho(A)}$ , therefore the triangle $APQ$ is isosceles.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.