Solution

Also complex numbers kills it

prove ASBK is cyclice

GoJensenOrGoHome wrote: Also complex numbers kills it Pleas show your solution

Dear Mathlinkers, by a converse of the Reim's theorem, ABKS is cyclic. Sincerely Jean-Louis

sasanineq wrote: GoJensenOrGoHome wrote: Also complex numbers kills it Pleas show your solution Let $\omega$ be the unit cicrcle and the points $A,B,C$ be respectively $1,b,c$. It's well known that $ k=\frac{2ab}{a+b}=\frac{2b}{b+1} $. (You can easy prove the fact that the point of intersection of tangents at $A$ and $B$ is $ \frac{2ab}{a+b} $ by simply intersecting tangents at $A$ and $B$ ) Now the line parallel to $BC$ and that passes through $K$ (i'll name it $r$) is $ w=t(b-c)+k $ where $ t\in R $ , so we have $ \frac{w-k}{b-c}\in{R} $ but for any real number $c$ we have $\overline{c}=c $ so in $r$ we have $ \frac{w-k}{b-c}=\frac{\overline{w}-\overline{k}}{\overline{b}-\overline{c}}=\frac{\overline{w}-\overline{k}}{\frac{1}{b}-\frac{1}{c}}=\frac{\overline{w}-\overline{k}}{\frac{-(b-c)}{bc}} $ focussing on $ \frac{w-k}{b-c}=\frac{\overline{w}-\overline{k}}{\frac{-(b-c)}{bc}} $ we have it's equivalent to $ w-k=-bc\overline{w}+bc\overline{k} $ now isolating $\overline{w}$ we have finally $r:$ $ \overline{w}=\frac{bc\overline{k}+k-w}{bc} $. But we know that the line passing through $A$ and $C$ (i'll rename $l$) $l:\overline{w}=\frac{a+c-w}{ac} $ but here we have set $a=1$ so $l:\overline{w}=\frac{c+1-w}{c} $. Now to find $s$ we have to intersect $r$ and $l$, so we have $ \frac{c+1-w}{c}=\frac{bc\overline{k}+k-w}{bc} $, first we need $ \overline{k}=\frac{2\overline{b}}{\overline{b}+1}=\frac{2/b}{(b+1)/b} =\frac{2}{b+1}$, now we can find $w$ from $ \frac{c+1-w}{c}=\frac{bc\overline{k}+k-w}{bc} $ , a little computation: $ c+1-w=\frac{bc\frac{2}{b+1}+k-w}{b} $ so $bc+b-wb=\frac{2bc}{b+1}+\frac{2b}{b+1}-w $ so $ w(b-1)=b(c+1)-\frac{2bc}{b+1}-\frac{2b}{b+1} $ so $ w(b-1)=(c+1)(b-\frac{2b}{b+1})=(c+1)(\frac{b^2+b-2b}{b+1})=(c+1)(\frac{b^2-b}{b+1})=(c+1)(\frac{b(b-1)}{b+1}) $ so we have $w=s=\frac{b(c+1)}{b+1} $. Now we want to prove $ BS=CS$ , this means that $S$ lies on the perpendicular bisector of $BC$ so we have that the projection of $S$ on $BC$ is the midpoint of $BC$. In complex numbers the projection of a point $z$ on the line passing throug $a$ and $b$ is $ proj_{ab}(z)=\frac{1}{2}(a+b+z-ab\overline{z}) $ so we apply this on $s$ (first we need to find $\overline{s}=\frac{1/b(1/c+1)}{1/b+1}=\frac{c+1}{c(b+1)}$) to get: $ proj_{bc}(s)=\frac{1}{2}(b+c+\frac{b(c+1)}{b+1}-bc\frac{c+1}{c(b+1)})=\frac{1}{2}(b+c+\frac{b(c+1)}{b+1}-\frac{b(c+1)}{b+1})= $ $=\frac{b+c}{2} $, so projection of $s$ on line $bc$ is the midpoint of $bc$ and we are done.