intersection of circumcirlce with line lies on angle bisector, <A=60^o

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intersection of circumcirlce with line lies on angle bisector, <A=60^o

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Source: 2012 Swiss IMO TST p4

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#1 h May 21, 2020, 7:12 PM • 3 Y

Y by Mango247, Mango247, Mango247

Let $ABC$ be a triangle with $\angle BAC = 60^o$ . Let $E$ be a point on the line $AB$ such that $B$ is between $A$ and $E$ and such that $BE = BC$ . Similarly, let $F$ be a point on $AC$ such that $C$ is between $A$ and $F$ and such that $CF = BC$ . Let $K$ be the intersection point of the circumscribed circle of $ACE$ with the line $EF$ . Show that $K$ lies on the bisector of the angle $\angle BAC$ .

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#2 h May 21, 2020, 11:19 PM

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It's actually an easy but a little bit long computation.
Let $BC = a, CA = b, AB = c$ , then by cosine law, we get $a^2 = b^2+c^2-b \cdot c$ and $EF^2 = b^2+2ab+a^2+c^2+2ac+a^2-bc-ac-ab-a^2 = 2a^2+ab+ac$ Since $E,K,C,A$ are cyclic, we see that $FK^2 \cdot FE^2 = FC^2 \cdot FA^2$ , so $a^2 \cdot (a+b)^2 = FK^2 \cdot (2a^2+ab+ac)$ From this, we get $FK^2 = \frac{a^2 \cdot (a+b)^2}{(2a^2+ab+ac)}$ Also, $FE-FK = EK = \frac{a \cdot (a+c)}{\sqrt{2a^2+ab+ac}}$ so, $\frac{EK}{FK} = \frac{a+c}{a+b} = \frac{AE}{AF}$ so $K$ lies on the bisector of $\angle BAC$
On second thought, it's not long at all.

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#3 h May 21, 2020, 11:58 PM • 1 Y

Y by Mango247

Let $K'= (ACE)\cap (ABF)$ , notice there's an obvious spiral similarity centered at $K'$ so $\triangle K'CF\sim \triangle K'EB$ but $CF=BE$ then $\triangle K'CF \cong \triangle K'EB \implies K'E=K'C$ and $K'B=K'F$ . This means that $BK', CK'$ are the perpendicular bisectors of $CE, BF$ resp. But notice that $BF$ is parallel to the $B-$ angle bisector in $\triangle ABC$ so $BK'$ is the $B-$ external bisector of $\triangle ABC$ similarly $CK'$ is the $C-$ external bisector of $\triangle ABC$ , then $K'$ is the $A-$ excenter of $\triangle ABC$ . Now $\angle CK'F=BK'E=60^{\circ}$ and if $I$ is the incenter of $\triangle ABC$ $\implies \angle BK'C=180^{\circ}-\angle BIC=60^{\circ}$ then $K'$ lies on $EF$ and $K \equiv K'$ and the conclusion follows

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