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SAUDI ARABIAN IMO Booklet, 2023

Geometry Difficulty 8.5 Shortlist Prove it Saudi Arabia

Let $ABC$ be a right-angled triangle with $\angle BAC = 90^\circ$ , and let $E$ be the foot of the perpendicular from $A$ to $BC$ . Let $Z \neq A$ be a point on the line $AB$ with $AB = BZ$ . Let $(c_1)$ be the circumcircle of the triangle $BEZ$ and $(c_2)$ be an arbitrary circle passing through the points $A$ and $E$ . Suppose $(c_1)$ meets the line $CZ$ again at the point $F$ , and meets $(c_2)$ again at the point $N$ . If $P$ is the other point of intersection of $(c_2)$ with $AF$ , prove that the points $N, B, P$ are collinear.

Solution

Since triangles $AEB$ and $CAB$ are similar, we have $\frac{AB}{EB} = \frac{CB}{AB}$ . Note that $AB = BZ$ , thus
$\frac{BZ}{EB} = \frac{CB}{AB},$
from which it follows that triangles $ZBE$ and $CBZ$ are also similar. Since $FEBZ$ is cyclic, then $\angle BEZ = \angle BFZ$ . So by the similarity of triangles $ZBE$ and $CBZ$ , we get
$\angle BFZ = \angle BEZ = \angle BZC = \angle BZF$
and thus $BFZ$ is isosceles. Since $BF = BZ = AB$ , triangle $AFZ$ is right with $\angle AFZ = 90^\circ$ . It follows that points $A, E, F, C$ are concyclic. Since $A, P, E, N$ are also concyclic, then
$\angle ENP = \angle EAP = \angle EAF = \angle BCZ = \angle BZE,$
by using the similarity of triangles $ZBE$ and $CBZ$ . Since $N, B, E, Z$ are concyclic, then $\angle ENP = \angle BZE = \angle ENB$ , which implies that $N, B, P$ are collinear. $\square$

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