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How to find the inradius of orthic triangle in terms of side lengths or area or circumdiameter of original triangle?

The incentre of the orthic triangle is the orthocentre of the original triangle. How to proceed further?

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  • $\begingroup$ How to find the inradius of the orthic triangle in terms of what? The side lengths of the original triangle? $\endgroup$ – Jack D'Aurizio Oct 21 '15 at 15:05
  • $\begingroup$ In terms of side lengths or area or circumdiameter of original triangle $\endgroup$ – user272383 Oct 21 '15 at 16:14
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The orthocenter of $ABC$ is the incenter of its pedal triangle, i.e. the orthic triangle. The inradius of the orthic triangle can be computed as the ratio between twice the area of the orthic triangle and the perimeter of the orthic triangle.

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Since $ABH_A H_B$ is a cyclic quadrilateral, we have $H_A H_B = AB\cos C$. Since the circumcircle of $H_A H_B H_C$ is the nine-point circle, the circumradius of $H_A H_B H_C$ is just half the circumradius of $ABC$, and: $$ [H_A H_B H_C]=\frac{H_A H_B\cdot H_A H_C\cdot H_B H_C}{2R} = \frac{abc}{4R}\cdot 2\cos(A)\cos(B)\cos(C) $$ or: $$ 2[H_A H_B H_C]= R^2 \sin(2A)\sin(2B)\sin(2C) $$ that together with: $$ H_A H_B+H_A H_C+H_B H_C = R\left(\sin(2A)+\sin(2B)+\sin(2C)\right) $$ gives: $$ r_H = R\cdot\frac{\sin(2A)\sin(2B)\sin(2C)}{\sin(2A)+\sin(2B)+\sin(2C)}.$$

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  • $\begingroup$ Can we do it by pure geometry? $\endgroup$ – user272383 Oct 22 '15 at 1:12

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