angles wanted, 50^o,BM=BH, AC=2HK (2020 YuMSh Olympiad Corresp. 8.6)

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angles wanted, 50^o,BM=BH, AC=2HK (2020 YuMSh Olympiad Corresp. 8.6)

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#1 h Mar 4, 2021, 7:23 AM

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In triangle $ABC$ , angle $A$ is $50^o$ , $BH$ is altitude, point $M$ lies on $BC$ such that $BM = BH$ . The perpendicular bisector of the segment $MC$ meets $AC$ at the point $K$ . It turns out that $AC = 2 HK$ . Find the angles of triangle $ABC$ .

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#2 h Mar 7, 2021, 1:53 PM

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Given $A(-a,0),B(0,at),C(c,0)$ with $t=\tan 50^{\circ}$ .

Point $M(\frac{act}{w},at-\frac{a^{2}t^{2}}{w})$ with $w=\sqrt{c^{2}+a^{2}t^{2}}$ .

Point $K(\frac{atw+c^{2}-a^{2}t^{2}}{2c},0)$ .
The condition $AC = 2 HK$ gives $a=\frac{c \cdot \cos 80^{\circ}}{1+\cos 80^{\circ}}$ .

$\tan \angle BCA=\frac{at}{c}=\frac{\tan 50^{\circ}\cos 80^{\circ}}{1+\cos 80^{\circ}}=\frac{\cot 40^{\circ}\cos 80^{\circ}}{1+\cos 80^{\circ}}=\cot 80^{\circ}=\tan 10^{\circ}$ .

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#3 h Mar 7, 2021, 4:09 PM

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Let $D$ reflection of $A$ aross $H$ ; as $HK=AH+CK$ , we infer $DK=KM=KC$ and $MD\bot BC, BMDH$ is cyclic; with $BM=BH$ it is actually a kite, so $\angle CBH=2\angle ABH=80^\circ$ , therefore $\angle ABC=120^\circ$ .

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