Library / Croatian Mathematical Competitions /14 of 114

Croatian Mathematical Competitions

Geometry Difficulty 4.7 AIME Prove it Croatia

Let $ABC$ be an acute triangle such that $|AC| > |BC|$ . Let $H$ be the orthocentre of that triangle, $N$ the foot of the altitude from $B$ , and $P$ the midpoint of the side $AB$ . The circumcircles of the triangles $ABC$ and $CHN$ intersect in $C$ and $D$ . Prove that the points $B, D, N$ and $P$ lie on the same circle.

Solution

Denote $\angle BAC = \alpha$ .

Since $ABN$ is a right-angled triangle and $P$ is the midpoint of its hypotenuse, we have $\angle BPN = 2\alpha$ . On the other hand,
$\begin{align*} \angle NDB &= \angle CDB - \angle CDN \\ &= (180^\circ - \angle BAC) - \angle CHN \quad (\text{cyclic quadrilaterals } ABDC \text{ and } NHDC) \\ &= (180^\circ - \alpha) - \alpha = 180^\circ - 2\alpha. \end{align*}$
Hence $\angle BPN + \angle NDB = 180^\circ$ , so the quadrilateral $BDNP$ is cyclic.

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

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