sum of inradius + radius of mixtiinear excircle is fixed, start with isosceles

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sum of inradius + radius of mixtiinear excircle is fixed, start with isosceles

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mixtilinear excircle

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Source: 2006 All-Ukrainian Correspondence MO of magazine ''In the World of Mathematics'', grades 5-11 p10

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#1 h Apr 28, 2021, 7:56 AM • 1 Y

Y by HWenslawski

Let $ABC$ be an isosceles triangle ( $AB=AC$ ). An arbitrary point $M$ is chosen on the extension of the $BC$ beyond point $B$ . Prove that the sum of the radius of the circle inscribed in the triangle $AMB$ and the radius of the circle tangent to the side $AC$ and the extensions of the sides $AM, CM$ of the triangle $AMC$ does not depend on the choice of point $M$ .

This post has been edited 1 time. Last edited by parmenides51, Apr 29, 2021, 12:34 PM

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#2 h Nov 12, 2021, 4:33 PM • 1 Y

Y by parmenides51

Given $\triangle ABC\ :\ A(0,a),B(-b,0),C(b,0)$ ; choose $M(-\lambda,0)$ .

The bisector of $\angle AMB\ :\ ax-y(\lambda+u)+a\lambda=0$ (with $u=\sqrt{a^{2}+\lambda^{2}}$ ) cuts
the bisector of $\angle ABM\ :\ y=-\frac{b+w}{a}(x+b)$ (with $w=\sqrt{a^{2}+b^{2}}$ ) in the point $O_{1}$
and $y_{O_{1}}=r_{1}=\frac{a(\lambda-b)(b+w)}{a^{2}+(b+w)(b+v)}$ .

The bisector of $\angle AMB\ :\ ax-y(\lambda+u)+a\lambda=0$ cuts the external bisector
of $\angle ACB\ :\ ax+y(b-w)-ab=0$ in the point $O_{2}$ and $y_{O_{2}}=r_{2}=\frac{a(b+\lambda)}{b+\lambda+v-w}$ .

Then $r_{1}+r_{2}=a$ .

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