Let $ABC$ be a non isosceles triangle with incenter $I$ . The circumcircle of the triangle $ABC$ has radius $R$. Let $AL$ be the external angle bisector of $\angle BAC $with $L \in BC$. Let $K$ be the point on perpendicular bisector of $BC$ such that $IL \perp IK$.Prove that $OK=3R$.
Problem
Source: 2021 Saudi Arabia IMO TST p5 (2.2)
Tags: geometry, right angle, circumradius, incenter
Kamran011
03.08.2021 08:45
We use complex numbers with the assumption that $(ABC)$ is the unit circle. Let the points $A, B ,C$ have complex coordinates $a^2, b^2, c^2$ respectively. Let $N$ be the midpoint of arc $BAC$. Now \begin{align*} l=\frac{a^{2}bc(b^2+c^2)-b^{2}c^{2}(bc+a^2)}{a^{2}bc-b^{2}c^{2}}\hspace{7 mm} \bar{l}=\frac{a^2-b^2+bc-c^2}{a^{2}bc-b^{2}c^{2}} \end{align*}Redefine $K$ as the point on the perpendicular bisector of $BC$ with $|OK|=3R$ \begin{align*}k=-3bc\hspace{5 mm} \bar{k}=-\frac{3}{bc} \end{align*}It remains to prove $IL \perp IK$. But $$(l-I)(\bar{I}-\bar{k})=-\frac{(a+b)(a+c)(-2a+b+c)(-ab-ac+2bc)}{abc(bc-a^2)}=(I-k)(\bar{I}-\bar{l})$$This finishes the proof. $\mathcal{Q.E.D.}$
PROA200
03.08.2021 20:08
Let $M$ and $N$ be the midpoints of $\overarc{BC}$ and $\overarc{BAC}$, respectively. Let $I_A$ be the $A$-excenter. Claim. $I$ is the orthocenter of $\triangle NLI_A$. Proof. We already know $AI_A\perp AI$. Also, negative inversion at $A$ swapping $B$ and $C$ swaps $(N,L)$, $(I, I_A)$. Thus, $AN\cdot AL=AI\cdot AI_A$ so $I$ is the orthocenter, of $\triangle NLI_A$, as desired. This gives $IL\parallel NI_A$, but since $MI=MI_A$ by Incenter-Excenter lemma, we have $MK=MN$. This gives $OK=MK+OM=MN+OM=3R$, as desired.