Problem 10:
$H$ is the orthocenter of acute triangle $ABC$. The triangle is inscribed in a circle with center $K$ with radius $R = 1$. Let $D$ is the intersection of the lines passing through $HK$ and $BC$. Also, $DK\cdot (DK - DH) = 1$. Find the area of the region $ABHC$.
BdMO National Higher Secondary 2009/10
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Tahmid Hasan
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- Joined:Thu Dec 09, 2010 5:34 pm
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Re: BdMO National Higher Secondary 2009/10
$DK(DK-DH)>0 \Rightarrow DK>DH$.
So $DK-DH=HK$, hence $DK.HK=1=KB^2=KC^2$.
$KD.KH=KB^2 \Rightarrow KB$ is tangent to $\odot BDH \Rightarrow \angle HBK=\angle HDB$.....$(1)$
$KD.KH=KC^2 \Rightarrow KC$ is tangent to $\odot CDH \Rightarrow \angle HCK=\angle HDC$.....$(2)$
From $(1),(2)$ we get $\angle HBK=\angle HCK \Rightarrow HBCK$ is cyclic $\Rightarrow \angle BHC=\angle BKC \Rightarrow 180^{\circ}-\angle A=2\angle A \Rightarrow \angle A=60^{\circ}$.
So $AH=2R \cos A=1$
Now area of $AHBC=(AHB)+(AHC)=\frac12(AH.BH.\sin AHB+AH.CH.\sin AHC)$
$=\frac12(1.2R\cos B\sin C+1.2R\cos C\sin B)=(\sin B\cos C+\cos B\sin C)=\sin(B+C)=\sin A=\frac{\sqrt3}{2}$.
[Solved a BdMO problem after a very long time
]
So $DK-DH=HK$, hence $DK.HK=1=KB^2=KC^2$.
$KD.KH=KB^2 \Rightarrow KB$ is tangent to $\odot BDH \Rightarrow \angle HBK=\angle HDB$.....$(1)$
$KD.KH=KC^2 \Rightarrow KC$ is tangent to $\odot CDH \Rightarrow \angle HCK=\angle HDC$.....$(2)$
From $(1),(2)$ we get $\angle HBK=\angle HCK \Rightarrow HBCK$ is cyclic $\Rightarrow \angle BHC=\angle BKC \Rightarrow 180^{\circ}-\angle A=2\angle A \Rightarrow \angle A=60^{\circ}$.
So $AH=2R \cos A=1$
Now area of $AHBC=(AHB)+(AHC)=\frac12(AH.BH.\sin AHB+AH.CH.\sin AHC)$
$=\frac12(1.2R\cos B\sin C+1.2R\cos C\sin B)=(\sin B\cos C+\cos B\sin C)=\sin(B+C)=\sin A=\frac{\sqrt3}{2}$.
[Solved a BdMO problem after a very long time
]বড় ভালবাসি তোমায়,মা