thaithuan_GC wrote: Let $ A, B$ and $ C$ be non-collinear points. Prove that there is a unique point $ X$ in the plane of $ ABC$ such that \[ XA^2 + XB^2 + AB^2 = XB^2 + XC^2 + BC^2 = XC^2 + XA^2 + CA^2. \] It's easily killed with complex numbers. Consider the complex plane with center at the circumcenter of $ \triangle ABC$, and let $ a,b,c,x$ be the corresponding complex numbers to $ A,B,C,X$ respectively. (Then $ |a| = |b| = |c|$) Besides let $ X'$ be a $ 180 ^\circ$ rotation of $ X$, so $ x' = - x$. Let $ \cdot$ be the real product, then we have: $ XA^2 + XB^2 + AB^2 = XB^2 + XC^2 + BC^2 \iff$ $ XA^2 + AB^2 - XC^2 - BC^2 = 0 \iff$ $ (x - a) \cdot (x - a) + (a - b) \cdot (a - b) - (x - c) \cdot (x - c) - (b - c) \cdot (b - c) = 0\iff$ (Using that $ |a| = |b| = |c|$) $ (a - c) \cdot (b - ( - x)) = (a - c) \cdot (b - x') = 0 \iff$ $ AC \perp BX'$ In the same way we get, $ AB \perp CX'$, $ BC \perp AX'$, so there is equality iff $ X'$ is the orthocenter. So the equation is fulfilled if and only if $ X$ is the orthocenter rotated $ 180 ^\circ$ around the circumcenter

The first condition $ XA^2 + XB^2 + AB^2 = XB^2 + XC^2 + BC^2$ is equivalent to $ XA^2 - XC^2 = BC^2 - AB^2.$ Which clearly means that $ X$ belows to the reflection of the B-altitude with respect to the perpendicular bisector of $AC.$ By similar reasoning, we conclude that $ X$ is the reflection of the orthocenter $H$ with respect to the circuncenter $O.$ In other words, $ X$ is the De Longchamps point $X_{20}$ of $\triangle ABC.$

Draw the following circles: Center $ C$, radius segment AB Center $ B$, radius segment AC Center $ A$, radius segment BC. From the equations, it is clear that the unique point is the radical center of these circles.

Dear kyyuanmathcount, this is the way that de Longchamps quoted by Luis has taken... Sincerely Jean-Louis

I think maybe a problem which can be so easily done with coordinates should not be on ISL. WLOG $B(-1,0),C(1,0),A(a,b),X(x,y)$. Since $A,B,C$ not collinear we have $b \neq 0$. We have $BC^2=4$ $AB^2=(a+1)^2+b^2$ $AC^2=(a-1)^2+b^2$ $XB^2=(x+1)^2+y^2$ $XC^2=(x-1)^2+y^2$ $XA^2=(x-a)^2+(y-b)^2$ The given condition is there exists a real number $K$ with: I. $K=(x-a)^2+(y-b)^2+(x+1)^2+y^2+(a+1)^2+b^2$ II. $K=(x+1)^2+y^2+(x-1)^2+y^2+4$ III. $K=(x-1)^2+y^2+(x-a)^2+(y-b)^2+(a-1)^2+b^2$ Define $N=K-2(x^2+y^2)$. From (II), we have $N=6$. Then (I) and (III) are of the form: I. $\alpha_1x+\beta_1y+\gamma_1=6$ III. $\alpha_3x+\beta_3y+\gamma_3=6$ Note that $\beta_3=-2b \neq 0$. Multiply (I) by $\beta_3$ and (III) by $\beta_1$, and subtract: $x(\alpha_1\beta_3-\alpha_3\beta_1)+(\gamma_1\beta_3-\gamma_3\beta_1)=6(\beta_3-\beta_1)$ Therefore there exists a unique solution to this system iff $\alpha_1\beta_3 \neq \alpha_3\beta_1$. But $\alpha_1=2-2a$,$\beta_1=-2b$,$\alpha_3=-2a-2$, $\beta_3=-2b$, so this is obviously the case.

cant we just assume X is origin, and all that's left is to prove that there exist points A,B,C that satisfy the conditions(also that form a triangle) because can't you just translate point x to the origin for any valid formation, also transforming ABC