1) What is the ratio of the area of a right triangle to the area of its incircle in the form $\frac{a \pi r}{br+ch}$ where $r$ is the inradius of the triangle and $h$ is the hypotenuse of the triangle.

2a) Let $\bigtriangleup ABC$ be an acute triangle with orthocenter $H$. Evaluate $\angle BHC + \angle BAC$.

2b) Let $H^`$ be the reflection of $H$ over $BC$. Prove that $ABH^`C$ is cyclic.


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  • $\begingroup$ Welcome to MSE. It helped if you showed your attempts to solve those, and where you got stuck in particular. This is not a blank-cheque homework solving site. $\endgroup$ – dxiv Nov 29 '16 at 7:37
  • $\begingroup$ OK. Here's my work for #1. So I basically took r= [ABC]/s and made it into a=pi[ABC]^2/s^2. Then I assigned arbitrary 3-4-5 to triangle ABC. I know that a is 1, but I don't know how to proceed from there. $\endgroup$ – jonyoung2002 Nov 29 '16 at 7:51
  • $\begingroup$ For #2a I just arbitrarily assigned ABC to be equilateral. Then, I just added it and got 180, not sure if this is right though. $\endgroup$ – jonyoung2002 Nov 29 '16 at 7:54
  • $\begingroup$ For #2b I again assigned ABC as an equilateral triangle, but then I realized that I have to prove it for all triangles, so now I don't know where/how to start. $\endgroup$ – jonyoung2002 Nov 29 '16 at 7:55
  • $\begingroup$ is $h$ realy the hypotenuse? i think $h$ is the hight and $c$ the hypotenuse? $\endgroup$ – Dr. Sonnhard Graubner Nov 29 '16 at 7:56

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For 2a) Firstly, A, M, H, N are con-cyclic. This makes the color shaded angles are equal in pairs.

Applying vertically opposite angles, the value of the required is then obvious.

For 2b) It is just the direct consequence of (2a).


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