Bariz - Problem

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Tags: geometry, incenter, fixed, Fixed point, Equilateral

parmenides51

18.09.2020 16:32

The equilateral triangle $ABC$ is inscribed in the circle $w$. Points $F$ and $E$ on the sides $AB$ and $AC$, respectively, are chosen such that $\angle ABE+ \angle ACF = 60^o$. The circumscribed circle of $\vartriangle AFE$ intersects the circle $w$ at the point $D$ for the second time. The rays $DE$ and $DF$ intersect the line $BC$ at the points $X$ and $Y$, respectively. Prove that the center of the inscribed circle of $\vartriangle DXY$ does not depend on the choice of points $F$ and $E$. (Hilko Danilo)

Given the circle $x^{2}+y^{2}=r^{2}$, midpoint $O(0,0)$. On this circle the points $A(r,0),B(-\frac{r}{2},\frac{r\sqrt{3}}{2}),C(-\frac{r}{2},-\frac{r\sqrt{3}}{2})$. On the line $AB\ :\ y=-\frac{1}{\sqrt{3}}(x-r)$ the point $F(f,-\frac{f-r}{\sqrt{3}})$. On the line $AC\ :\ y=\frac{1}{\sqrt{3}}(x-r)$ the point $E(e,\frac{e-r}{\sqrt{3}})$. With the slopes $m_{AB}=-\frac{1}{\sqrt{3}}$ and $m_{BE}=\frac{2e-5r}{\sqrt{3}(r+2e)}$, we find $\tan \angle ABE=\frac{\sqrt{3}(r-e)}{2r+e}$. With the slopes $m_{AC}=\frac{1}{\sqrt{3}}$ and $m_{CF}=\frac{5r-2f}{\sqrt{3}(r+2f)}$, we find $\tan \angle ACF=\frac{\sqrt{3}(r-f)}{2r+f}$. Calculating the given condition $\angle ABE+ \angle ACF = 60^{\circ}$ or $\tan(\angle ABE+ \angle ACF)=\sqrt{3}$, we have $r=2(e+f)$ or $f=\frac{r}{2}-e$. Midpoint of the circumscribed circle of $\triangle AFE$ is the point $M(\frac{r}{2},\frac{\sqrt{3}(4e-r)}{6})$. This circle cuts the given circle in the point $D(-\frac{r(8e^{2}-4er-r^{2})}{2(4e^{2}-2er+r^{2})},\frac{\sqrt{3}r^{2}(4e-r)}{2(4e^{2}-2er+r^{2})})$. Further calculations show that the center of the inscribed circle of $\triangle DXY$ is the point $O$.