MathDB

3PI+IT=AC wanted, incenter in 90-60-30 (2021 Kyiv City MO Round2 8.2)

Source:

February 14, 2021

geometryincenterright triangle

Problem Statement

In a triangle

ABC

\angle B=90^o

and

\angle A=60^o

I

is the point of intersection of its angle bisectors. A line passing through the point

I

parallel to the line

AC

, intersects the sides

AB

and

BC

at the points

P

and

T

respectively. Prove that

3PI+IT=AC

.
(Anton Trygub)