AK=KC wanted, AM=BN, <ACB = 60^o, midpoints (2019 Euler Olympiad Regional 2.3)

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AK=KC wanted, AM=BN, <ACB = 60^o, midpoints (2019 Euler Olympiad Regional 2.3)

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#1 h Dec 28, 2020, 8:33 AM

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Point $N$ is the midpoint of side $BC$ of triangle $ABC,$ in which $\angle ACB = 60 ^\circ$ . Point $M$ on side $AC$ is such that $AM = BN.$ Point $K$ is the midpoint of segment $BM.$ Prove that $AK = KC.$

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#2 h Feb 23, 2021, 4:03 AM

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Let $A = (a, a\sqrt 3), B = (b,0),$ and $C = (0,0)$ .
$BN = NC$ , so $N = (\frac{b}{2}, 0)$
The circle centered at $A$ with radius $\frac{b}{2}$ has equation $(x-a)^2 + (y-a\sqrt 3)^2 = \frac{b^2}{4}$
$M$ is the point of intersection of the circle and the line $y = \sqrt 3$ that is closer to the origin. Substituting and expanding, we have
$(x-a)^2 + (\sqrt{3} x - \sqrt{3} a)^2 = \frac{b^2}{4}$
$\Rrightarrow x^2 - 2ax + a^2 + 3x^2 + 3a^2 - 6ax = \frac{b^2}{4}$
$\Rrightarrow 4x^2 - 8ax + 4a^2 = \frac{b^2}{4}$
$\Rrightarrow 16x^2 - 32ax + 16a^2 - b^2 = 0$
$\Rrightarrow x = \frac{32a \pm \sqrt{1024a^2 - 64(16a^2 - b^2)}}{32}$
$\Rrightarrow x = \frac{4a - b}{4}$
From $y = \sqrt 3 x$ , we get that
$M = \frac{4a - b}{4}, \frac{\sqrt 3}{4} (4a - b)$
$\Rrightarrow M = (a - \frac{b}{4}, a\sqrt 3 - \frac{\sqrt 3}{4} b)$
We get that $K = (\frac{a - \frac{b}{4}}{2}, \frac{a\sqrt 3 - \frac{\sqrt 3}{4} b}{2}) \Rrightarrow (\frac{4a + 6b}{8}, \frac{4a\sqrt 3 + b \sqrt 3}{8})$
Using the distance formula, we get $AK = KC = \frac{1}{8} \sqrt{64a^2 + 72ab + 39b^2}$
$\blacksquare$

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