AHZOLFAGHARI wrote: We have an equilateral triangle with circumradius $1$. We extend its sides. Determine the point $P$ inside the triangle such that the total lengths of the sides (extended), which lies inside the circle with center $P$ and radius $1$, is maximum. (The total distance of the point P from the sides of an equilateral triangle is fixed ) Proposed by Erfan Salavati Let the triangle be denoted as $ABC$. For some position of $P$, let $a_P,b_P,c_P$ be the distances of $P$ from $BC,CA,AB$ respectively. Since the circumradius is $1$, the side lengths are $\sqrt 3$. Computing the area in two ways, we have $$\frac 12 \sqrt 3\left( a_P+b_P+c_P\right)=\frac{\sqrt{3}}{4}\cdot 3\implies a_P+b_P+c_P=\frac 3 2 .$$We consider two cases: Case 1: $a_P,b_P,c_P\le 1$. In this case the total length of sides lying inside the circle is clearly $2\sqrt{1-a_P^2}+2\sqrt{1-b_P^2}+2\sqrt{1-c_P^2}$. Since the function $x\mapsto 2\sqrt{1-x^2}$ is concave on $[0,1]$, we have by Jensen's inequality that $$2\sqrt{1-a_P^2}+2\sqrt{1-b_P^2}+2\sqrt{1-c_P^2}\le 3\cdot 2\sqrt{1-\left(\frac{a_P+b_P+c_P}{3}\right)^2}=3\sqrt{3}.$$Equality is obtained when $P$ is the circumcenter. Case 2: One of $a_P,b_P,c_P$ is greater than $1$. WLOG suppose $c_P>1$, then the total length of sides lying inside the circle is $2\sqrt{1-a_P^2}+2\sqrt{1-b_P^2}\le 2+2=4$, which is less than the maximum obtained in Case 1. Therefore the required maximum is $3\sqrt{3}$. $\blacksquare$

Let $h_a,h_b,h_c$ be distance of $P$ from segments $a,b,c$. $\frac{h_a.a + h_b.b + h_c.c}{2} = \frac{\frac{3}{2}a}{2} \implies h_a + h_b + h_c = \frac{3}{2}$. Case $1 : h_a,h_b,h_c < 1$. Let circle with center $P$ intersect $a$ at $Z_1,Z_2$ and $b$ at $Y_1,Y_2$ and $c$ at $X_1,X_2$. then we have $X_1X_2 + Y_1Y_2 + Z_1Z_2 = 2(\sin{\frac{X_1PX_2}{2}} + \sin{\frac{Y_1PY_2}{2}} + \sin{\frac{Z_1PZ_2}{2}})$. Let $\frac{X_1PX_2}{2} = \angle k_1$ and $\frac{Y_1PY_2}{2} = \angle k_2$ and $\frac{Z_1PZ_2}{2} = \angle k_3$. Note that $\angle k_1 + \angle k_2 + \angle k_3 \le 180$. Note that $\sin{s+t} \ge \sin{s}$ if $t > 0$ and $s+t \ge 90$ so we can assume $\angle k_1 + \angle k_2 + \angle k_3 = 180$ so now we need to prove in an imaginary triangle with angles $\angle k_1,\angle k_2,\angle k_3$ we have $\sin{k_1} + \sin{k_2} + \sin{k_3} \le \frac{3\sqrt{3}}{2}$ which is true since $\sin{k_1}^2 + \sin{k_2}^2 + \sin{k_3}^2 \le \frac{9}{4}$. case $2 : h_a \ge 1$. In this case circle with center $P$ doesn't cut $a$ so we need to prove $2(\sin{k_1} + \sin{k_2}) \le 3\sqrt{3}$ which is true since $2(\sin {k_1} + \sin {k_2}) \le 4 < 3\sqrt{3}$.