Competition problem — Geometry — Olympiad Maths Prep

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Geometry Difficulty 5.0 AIME Prove it Ukraine

Point $P$ was chosen inside the triangle $ABC$ so that $BC = AP$ and $\angle APC = 180^\circ - \angle ABC$ . On side $AB$ , there exists point $K$ such that $AK = KB + PC$ . Prove $\angle AKC = 90^\circ$ .

Solution

We extend the ray $AB$ further after $B$ to find the point $T$ such that $BT = PC$ (Fig. 1). Then, $\Delta TBC = \Delta CPA$ due to two equal length sides and the same angle between them. Hence, $TC = CA$ . Analogously,
$AK = KB + PC = KB + BT = KT.$
Therefore, in the isosceles triangle $ATC$ segment $KC$ is a median, which also makes it the altitude. From which follows that $KC \perp AB$ , Q.E.D.

Fig. 1

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