Bariz - Problem

Source: Ukrainian Geometry Olympiad 2017 X p4

Tags: geometry, circumcircle, incircle, midline, Concyclic

parmenides51

12.12.2018 09:15

In the right triangle $ABC$ with hypotenuse $AB$, the incircle touches $BC$ and $AC$ at points ${{A}_{1}}$ and ${{B}_{1}}$ respectively. The straight line containing the midline of $\Delta ABC$ , parallel to $AB$, intersects its circumcircle at points $P$ and $T$. Prove that points $P,T,{{A}_{1}}$ and ${{B}_{1}}$ lie on one circle.

Let $I$ is incenter and $CI$ intersect circumcircle in $E$. Easy to show that $EP=ET,B_1E=A_1E$ $\angle CAE= \frac{\pi}{4}+\angle CAB$ so $ \sin \angle CAE = \frac{\sqrt{2}}{2}( \sin \angle CAB+\cos \angle CAB)= \frac{\sqrt{2}(a+b)}{2c}$ $CE=AB\sin \angle CAE= \frac{\sqrt{2}(a+b)}{2}$ Let $X$ is point of intersection $A_1B_1$ and $CE$ then $B_1X=A_1X=CX=\frac{\sqrt{2}B_1C}{2}=\frac{\sqrt{2}(a+b-c)}{4}$ and $XE=CE-CX = \frac{2\sqrt{2}(a+b)-\sqrt{2}(a+b-c)}{4}=\frac{\sqrt{2}(a+b+c)}{4}$ and so $B_1E^2=XE^2+XB_1^2=\frac{(a+b-c)^2+(a+b+c)^2}{8}=\frac{(a+b)^2+c^2}{4}=\frac{a^2+b^2+ab}{2}$ Let $O$ is circumcenter of $ABC$ and $M$ is midpoint of $PT$. $OM= \frac{ab}{2c}$ Then $PE^2=PM^2+ME^2=PO^2-OM^2+(OE+OM)^2=2R^2+2R *OM=\frac{c^2+ab}{2}=\frac{a^2+b^2+ab}{2}$ So $B_1E=A_1E=PE=TE=\sqrt{\frac{a^2+b^2+ab}{2}}$