UkraineMO

Geometry Difficulty 5.1 AIME, harder Prove it Ukraine

Bisector of the angle $\angle ABC$ of the triangle $ABC$ intersects the circumcircle of triangle $ABC$ at the point $K$ . Point $N$ belongs to the segment $AB$ and $NK \perp AB$ . Let $P$ be a midpoint of the segment $NB$ . Consider the line through $P$ that is parallel to $BC$ and intersects line $BK$ at the point $T$ . Prove that the line $NT$ passes through the midpoint of the segment $AC$ .

(Nagel Igor)

Solution

Let $M = NT \cap AC$ (fig. 23). We note that since $BK$ is a bisector of $ABC$ then $\angle KBC = \angle KBA = \alpha$ , and since $PT$ is parallel to $BC$ then $\angle KBC = \angle PTB$ . Thus, $PT = PB = PN$ . It follows that $\triangle BNT$ is right-angled with hypotenuse $BN$ . Since triangle $\triangle BNK$ is also right-angled then $\angle KNT = \alpha$ . Moreover, we get $\angle KAC = \angle KBC = \alpha$ . Thus, quadrilateral $KANM$ is cyclic. Since $\angle ANK = 90^\circ$ , then $AK$ is a diameter of the circumcircle of the $KANM$ . Hence, $\angle AMK = 90^\circ$ . Since $\triangle AKC$ is isosceles we get that altitude $KM$ also is a median. Hence, $AM = MC$ .

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.