Let $K$ and $L$ be the centers of the excircles of a non-isosceles triangle $ABC$ opposite $B$ and $C$ respectively. Let $B_1$ and $C_1$ be the midpoints of the sides $AC$ and $AB$ respectively Let $M$ and $N$ be symmetric to $B$ and $C$ about $B_1$ and $C_1$ respectively. Prove that the lines $KM$ and $LN$ meet on $BC$.
Problem
Source: 2016 Belarus Team Selection Test 7.2
Tags: symmetry, geometry
Jjesus
13.06.2020 03:00
India Postal 2016 https://artofproblemsolving.com/community/c6h1369968_three_lines_meet__or_actually_they_dont
MP8148
13.06.2020 09:47
We use barycentrics wrt $ABC$. Check that $K = (a:-b:c)$ and $M = (1:-1:1)$, so if $P = \overline{KM} \cap \overline{BC}$ has coordinate $P = (0,t,1-t)$: $$0 = \begin{vmatrix} 0 & t & 1-t \\ a & -b & c \\ 1 & -1 & 1 \end{vmatrix} = -a+b+t(c-b) \implies t=\dfrac{a-b}{c-b},$$whence $P = (0:a-b:c-a)$ which is symmetric in $b$ and $c$ as desired.
MathLuis
08.12.2021 06:43
Let $I_A$ be the $A$-excenter and let $LN \cap AC=D$ and $KM \cap AB=E$ Claim: $D,I_A,E$ are colinear. Proof: We use barycentric cordinates, its quite easy to get $M=(1,-1,1)$ and $N=(1,1,-1)$ and remember that $K=(a,-b,c)$ and $L=(a,b,-c)$ and $I_A=(-a,b,c)$ and here $A=(1,0,0)$, $B=(0,1,0)$ and $C=(0,0,1)$. Let $D=(x,0,y)$ and $E=(x_0,y_0,0)$ and now we have: $$\text{det} \; \begin{vmatrix} x_0 & y_0 & 0 \\ a & -b & c \\ 1 & -1 & 1 \end{vmatrix}=0 \implies x_0(b-c)=y_0(c-a) \implies E=(c-a,b-c,0)$$$$\text{det} \; \begin{vmatrix} x & 0 & y \\ a & b & -c \\ 1 & 1 & -1 \end{vmatrix}=0 \implies x(b-c)=y(a-b) \implies D=(a-b,0,b-c)$$Now since its easy to see that this determinant (next one i'm showing below) its 0 we are done. $$\text{det} \; \begin{vmatrix} a-b & 0 & b-c \\ -a & b & c \\ c-a & b-c & 0 \end{vmatrix}=0 \implies D,I_A,E \; \text{colinear}$$Main proof: By Pappus theorem on $L,A,K,D,I_A,E$ we have the desired result thus we are done