OK=3R wanted, incenter, external bisector, perp. bisector, IL _|_IK

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OK=3R wanted, incenter, external bisector, perp. bisector, IL _|_IK

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Source: 2021 Saudi Arabia IMO TST p5 (2.2)

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#1 h Aug 2, 2021, 10:21 PM • 1 Y

Y by jhu08

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 $OK=3R$ .

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#3 h Aug 3, 2021, 4:29 AM • 1 Y

Y by jhu08

solution

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#4 h Aug 3, 2021, 5:45 AM • 1 Y

Y by jhu08

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. $\mathcal{Q.E.D.}$

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#5 h Aug 3, 2021, 5:08 PM • 1 Y

Y by jhu08

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.

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