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Source: Kazakhstan national olympiad 2016 Final Round,grade 11,P4

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#1 h Mar 22, 2016, 10:43 AM • 2 Y

Y by Adventure10, Mango247

In isosceles triangle $ABC$ ( $CA=CB$ ), $CH$ is altitude and $M$ is midpoint of $BH$ .Let $K$ be the foot of the perpendicular from $H$ to $AC$ and $L=BK \cap CM$ .Let the perpendicular drawn from $B$ to $BC$ intersects with $HL$ at $N$ .Prove that $\angle ACB=2 \angle BCN$ .(M. Kunhozhyn)

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#2 h Mar 22, 2016, 5:56 PM • 2 Y

Y by Adventure10, Mango247

It suffices to show that $CM \bot HN$ . Click to reveal hidden text

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This post has been edited 1 time. Last edited by george_54, Mar 22, 2016, 5:57 PM
Reason: figure

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#3 h Mar 22, 2016, 6:29 PM • 1 Y

Y by Adventure10

Using George_54 figure , easy to see that $AB$ is tangent to circumcircle of $\triangle{CKH}$ so let the this circle cut $CM$ again at $L'$
we have $ML'$ x $MC = MH^2 = MB^2$ so $\angle{MHL'} = \angle{HCL'}$ and $\angle{MBL'} = \angle{L'CB}$
thus we get $\angle{HL'B} = 180-w$
but since $\angle{KL'H} = w$ we must have $K,L',B$ collinear so $L' = L$
thus we get $\angle{CLH} = 90$ so $C,N,B,L$ is cyclic , so $\angle{ACB} = 2\angle{KCH} = 2\angle{KLH}= 2\angle{BCN}$

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#4 h Mar 22, 2016, 6:53 PM • 2 Y

Y by Adventure10, Mango247

It is well know that there is a unique angle $\phi$ such that $\frac{sin\phi}{sin(\alpha-\phi)}=k$ for some constant $k$ and $\alpha$ .
Now note that $\triangle CHK$ is similar to $\triangle CBH$ so $\frac{BC}{CH}=\frac{BH}{KH}$ and when we apply sine theorems to $\triangle BCH$ and $\triangle BKH$ (and also the sine theorem for 2 angles built by median in $\triangle BCH$ ) we get that $\angle HKB=\angle HCM=\angle HCL$ so $CKHL$ is cyclic so $\angle CLH=90$ so $CLBN$ is cyclic. Now let $KH$ and $BN$ intersect at $X$ . We have that $CKXB$ is cyclic so $\angle HXN=180-\angle ACB$ . Also note that $HX$ and $BX$ are tangents to the circumcircle of $\triangle CHB$ so $CX$ is symmedian in that triangle so we get $\angle HCX=\angle MCB=\angle XNH$ so $XHCN$ is also cyclic so $\angle HCN=180-\angle HXN=\angle ACB$ so we have $\angle BCN=\angle ACH$ so we are finished.

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mjuk

#5 h Mar 22, 2016, 7:24 PM • 2 Y

Y by Adventure10, Mango247

Obviously $\bigtriangleup HAK\sim \bigtriangleup CBH\rightarrow \frac{HA}{AK}=\frac{BC}{BH}\rightarrow \frac{BA}{AK}=\frac{BC}{BH}$ and $\angle BAK=\angle CBM\rightarrow \bigtriangleup BAK\sim \bigtriangleup CBM$ .
$\angle HKL=\angle AHK-\angle ABK=\angle BCH-\angle BCM=\angle HCL\rightarrow$ $HLCK$ is cyclic $\rightarrow \angle CLN=90^{\circ}\rightarrow CLBN$ is cyclic $\rightarrow \angle ACB=2\angle KCH=2\angle KLH=2\angle BLN=2\angle BCN.$

This post has been edited 2 times. Last edited by mjuk, Mar 22, 2016, 7:26 PM
Reason: typo

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