Given is a triangle $ABC$ and a point $X$ inside its circumcircle. If $I_B, I_C$ denote the $B, C$ excenters, then prove that $XB \cdot XC <XI_B \cdot XI_C$.
Problem
Source: All-Russian MO Final stage 2023 9.4
Tags: geometry
frc
24.04.2023 08:35
Bump this
PNT
24.04.2023 14:02
It’s easy to bash. Let $(ABC)$ be the unit circle and write $a=u^2,b=v^2,c=w^2$ Then the excenters are just the reflection of $I$ around the midpoints of the arcs. And of course $|x|<1$
hakN
25.04.2023 00:28
Set $(ABC)$ as the unit circle, let $a^2,b^2,c^2$ be the complex numbers corresponding to $A,B,C$ and $ab+bc-ac , bc+ac-ab$ correspond to $I_B,I_C$ respectively. Then we want to show $|f(x)| := \left|\frac{(x-b^2)(x-c^2)}{(x+ab-bc-ac)(x+ac-ab-bc)}\right| < 1$ for all $|x|<1$. Let $D = \{z\in \mathbb{C} : |z|\le 1\}$. Since $D$ is compact, $|f|$ has a global maximum on $D$, and by maximum modulus principle, this maximum is reached on the boundary of $D$.
So it will suffice to show $|f(x)| \le 1$ for all $|x|=1$, which after expanding is equivalent to $\frac{(a^2-x)^2(bc-x)^2(b-c)^2}{a^2b^2c^2x^2} = \left(\frac{(a^2-x)(bc-x)(b-c)}{abcx}\right)^2 \le 0$, which is obvious since $\frac{(a^2-x)(bc-x)(b-c)}{abcx} \in i\mathbb{R}$.
(Note also that the equality is satisfied only when $X=A$ or $X$ is the midpoint of $I_BI_C$.)
sayheykid
27.10.2023 10:37
Let $W$ be the midpoint of the arc $BAC$. Note that $BCI_BI_C$ is inscribed in circle $\Omega$ with the center $W$. If $X$ lies "under" $BC$ (such that segments $AX,BC$ intersect) then $\angle I_CBX, \angle I_BCX$ both obtuse, and the statement is obvious. Otherwise, let $F$ be the second intersection of $\Omega$ and $CX$. Then $2 \cdot \angle BFX = \angle BWC < \angle BXC = \angle BFX + \angle FBX \implies \angle BFX < \angle FBX \implies BX < FX$, so $XB \cdot XC < XF \cdot XC = |pow(X,\Omega)|$. Now, let $G$ be the second intersection of $I_CX$ and $\Omega$. Then $|pow(X,\Omega)|=I_CX \cdot XG < I_CX \cdot XI_B$ (because $\angle I_BGX = \frac{\pi}{2}$) hence $XB \cdot XC < |pow(X,\Omega)| < I_CX \cdot XI_B$ as desired.