Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.9 AIME Prove it Romania

Let $ABC$ be a triangle with $AB = AC$ and $\angle BAC = 40^\circ$ . The points $S$ and $T$ lie on the sides $AB$ and $BC$ , respectively, such that $\angle BAT = \angle BCS = 10^\circ$ . The straight lines $AT$ and $CS$ meet at point $P$ .
Show that $BT = 2PT$ .

Solution

Triangle $ABC$ is isosceles, so $\angle ABC = \angle ACB = 70^\circ$ .

Notice that $\angle TAC = 40^\circ - 10^\circ = 30^\circ$ and $\angle ACS = 70^\circ - 10^\circ = 60^\circ$ , hence $\angle APC = 90^\circ$ .

Triangles $ABT$ and $BSC$ are similar, whence $\frac{BS}{BC} = \frac{BT}{AB}$ .

Also, triangles $BST$ and $BCA$ are similar, therefore $TB = TS$ and $\angle TSB = 70^\circ$ .

Since $\angle CSA = \angle SBC + \angle SCB = 70^\circ + 10^\circ = 80^\circ$ , it follows that $\angle PST = 180^\circ - 80^\circ - 70^\circ = 30^\circ$ .

Now, triangle $STP$ has a right angle in $P$ and $\angle PST = 30^\circ$ , so $BT = 2PT$ .

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