In the triangle $ABC$, where $<$ $ACB$ $=$ $45$, $O$ and $H$ are the center the circumscribed circle, respectively, the orthocenter. The line that passes through $O$ and is perpendicular to $CO$ intersects $AC$ and $BC$ in $K$, respectively $L$. Show that the perimeter of $KLH$ is equal to the diameter of the circumscribed circle of triangle $ABC$.
Problem
Source: Fourth Saudi Arabia JBMO TST 2019, P4
Tags: geometry
Nymoldin
04.06.2020 18:09
What was the morality of this act?
CrazyInMath
09.04.2023 10:30
Nymoldin wrote: What was the morality of this act? what's wrong? anyways the problem is not in a good shape, let's fix it. Fourth Saudi Arabia JBMO TST 2019, P4 wrote: In the triangle $ABC$, where $\angle ACB=45^\circ$, $O$ and $H$ are the circumcenter, orthocenter, respectively. The line that passes through $O$ and is perpendicular to $CO$ intersects $AC$ and $BC$ in $K,L$, respectively. Show that the perimeter of $KLH$ is equal to the diameter of the circumscribed circle of triangle $ABC$. Now for the solution
We let rays $OK$ and $OL$ intersects $(ABC)$ again at $F,G$ respectively.
Lemma 1: $AHK$ is similar to $CAB$.
Proof: We use complex numbers, denote $s_\theta=e^{i\theta}$.
We let $(ABC)$ be the unit circle, and $a=1, b=i, c=s_\theta$.
We have $f=s_{\theta+\frac{\pi}{2}}$ and $g=s_{\theta-\frac{\pi}{2}}$. Also $h=1+i+s_\theta$.
Now we compute $K$, using the chord intersection formula gives us $k=\frac{s_\theta(s_\theta+1)}{s_\theta-1}$.
So we only need $$\left|\begin{matrix}
s_\theta&1&1\\
1&1+i+s_\theta&1\\
i&\frac{s_\theta(s_\theta+1)}{s_\theta-1}&1\\
\end{matrix}\right|=0$$Expanding and simplifying gives us $s_\theta+s_{2\theta}+\frac{s_\theta-s_{3\theta}}{s_\theta-1}=0$, which is true.
Now we prove that $AHK$ and $AFK$ are congruent. First we have $AK=AK$, and $\angle HAK=45^{\circ}=\angle CGF=\angle CAF=\angle KAF$, and $\angle AKH=\angle ABC=\frac{1}{2}\overarc{AC}=\frac{1}{2}(\overarc{CG}+\overarc{AF})=\angle AKF$, so by ASA congruence we're done.
Now similarly we have $HBL$ is congruent to $FBL$, so $KL+LH+KH=KL+LG+FK=2R$.