concyclic wanted, incircle, circumcircle, midline related in a right triangle

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Source: Ukrainian Geometry Olympiad 2017 X p4

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#1 h Dec 12, 2018, 6:15 AM • 1 Y

Y by Adventure10

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.

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#2 h Dec 12, 2018, 9:31 AM • 2 Y

Y by Adventure10, Mango247

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 $B_1E=A_1E=PE=TE=\sqrt{\frac{a^2+b^2+ab}{2}}$

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jayme

#3 h Dec 12, 2018, 10:26 AM • 2 Y

Y by Adventure10, Mango247

Dear Mathlinkers,

http://jl.ayme.pagesperso-orange.fr/Docs/Quelques%20theoremes%20oublies.pdf (in French)

http://jl.ayme.pagesperso-orange.fr/Docs/Algunos%20teoremas%20olvidados.pdf (in Spain traduction by professor Francisco Bellot-Rosado)

Sincerely
Jean-Louis

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zuss77

#4 h Dec 13, 2018, 6:12 PM • 2 Y

Y by Adventure10, Mango247

see here:
https://artofproblemsolving.com/community/c6t48f6h1673979

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