Collinear points

by liberator, Aug 21, 2014, 11:51 AM

Problem: Let $AL$ be an angle bisector of triangle $ABC$ such that $AB>AC$ .Point $K$ on the side $AB$ is such that $AK=AC$ .Let $N$ be the intersection point of circumcircles of triangles $ABC$ and $BKL$ .Prove that points $A,L,N$ are collinear.

$[asy] /* Free script by liberator, 21 August 2014 */ unitsize(3.2cm); /* Initialize objects */ pair T = (-1,1.3); pair A = tangent(T, (0,0), 1, 2); pair B = dir(200); pair C = dir(340); pair L1 = bisectorpoint(B,A,C); pair L2 = rotate(180,L1)*A; pair L = intersectionpoint(A--L2, B--C); pair K = reflect(A,L)*C; pair N = intersectionpoint(L--L2, unitcircle); pair T1 = rotate(180,A)*T; /* Draw objects */ draw(A--B--C--cycle, rgb(0.4,0.6,0.8)+linewidth(1)); draw(K--L, rgb(0.4,0.6,0.8)); draw(A--N, rgb(0.4,0.6,0.8)+dashed+linewidth(1)); draw(T--T1, rgb(0.4,0.6,0.8)); draw(K--N, rgb(0.4,0.6,0.8)); draw(C--N, rgb(0.4,0.6,0.8)); draw(unitcircle, rgb(0,0.6,0)+linewidth(1)); draw(circumcircle(B,K,L), rgb(0.9,0,0)); /* Place dots on and label each point */ dot(A); label("$A$", A, dir(80)); dot(B); label("$B$", B, dir(200)); dot(C); label("$C$", C, dir(340)); dot(K); label("$K$", K, dir(150)); dot(L); label("$L$", L, dir(45)); dot(N); label("$N$", N, dir(-90)); label("$\mathcal{T}_A$", T, dir(160)); [/asy]$
[b]My solution[/b]
[b]Solution due to Tsikaloudakis[/b]

This post has been edited 3 times. Last edited by liberator, Aug 21, 2014, 7:30 PM

geometry

Reim s theorem

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Here is my solution. I will do bary bash.
Since $AC=AK$ so $K=(c-b,b,0)$ and it's well known that $L=(0,b,c)$ .Now I will intersect the circumcircles of $ABC$ and $BKL$ to find coordinates of $N$ .Since $N$ lies on $(ABC)$ so $a^2yz+b^2zx+c^2xy=0$ .It's not difficult to find that $u,v,w$ in circle $BKL$ are $bc,0,\frac{a^2b}{b+c}$ respectively. Since $N$ lies also on $(BKL)$ $\implies$ $bcx+\frac{a^2bz}{b+c}=0$ after opening brackets and clearing similar things we get $\frac{x}{z}=\frac{-a^2}{c(b+c)}$ .Now Putting this in $a^2yz+b^2zx+c^2xy=0$ doing similar things we get $\frac{x}{y}=\frac{-a^2}{b(b+c)}.So$ N=(-a^2,b(b+c),c(b+c)) $. It's well known that the intersection of circumcircle and angle bisector has coordinates like$ N$ has.Done!!!

by tenplusten, Mar 3, 2017, 2:06 PM

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Collinear points

by liberator, Aug 21, 2014, 11:51 AM

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