Determine all possible values of angle CAB

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Determine all possible values of angle CAB

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Source: Baltic Way 2010

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WakeUp

#1 h Nov 19, 2010, 7:24 PM • 2 Y

Y by Negar, Adventure10

In an acute triangle $ABC$ , the segment $CD$ is an altitude and $H$ is the orthocentre. Given that the circumcentre of the triangle lies on the line containing the bisector of the angle $DHB$ , determine all possible values of $\angle CAB$ .

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jgnr

#2 h Nov 21, 2010, 1:45 AM • 2 Y

Y by Adventure10, Mango247

Let line $CD$ intersect $(ABC)$ again at $E$ . Since $\angle OHE=\angle OHB$ , then $B$ and $E$ are symmetric wrt line $OH$ , thus we get $HE=HB$ . So $\angle HEB=\angle HBE$ . Since $\angle HEB=\angle A$ and $\angle HBE=180^{\circ}-2\angle A$ , we get $\angle A=60^{\circ}$ . This value is obviously possible, for example when $ABC$ is equilateral.

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#3 h Nov 27, 2010, 9:18 AM • 2 Y

Y by Adventure10, Mango247

Let $P$ be on $AC$ , $Q$ on $AB$ such that $PQ$ is the angle bisector of $\angle{DHB}$ . Let $BH$ meet $AC$ at $E$ . Let $BE$ and $CD$ meet the circumcircle of $ABC$ at $M$ and $N$ . Since $O$ lies on the bisector of $\angle{DHB}$ , then $O$ is equidistant from lines $DH$ and $BH$ , therefore $O$ is equidistant from chords $CN$ and $BM$ which implies they have the same length. Now, since points $B,C,M,N$ lie on the same circle and $BM=CN$ we must have $BC$ parallel to $MN$ or $CM$ parallel to $BN$ .

(1) If $MN$ is parallel to $BC$ , $\angle{NCB}= \angle{MBC}$ which implies that $ABC$ is isosceles. Since $\angle{ACB}= 90-\frac{A}{2}$ and by an easy angle chase $\angle{APQ}=90-\frac{A}{2}$ then $PQ$ is paralel to $BC$ . Then, since $O$ lies in $AH$ , and $AH$ is perpendicular to $BC$ , $AO$ is perpendicular to $PQ$ and therefore $O$ would lie on the external bisector of angle $\angle{DHB}$ which can't happen.

(2) If $CM$ is parallel to $BN$ , $\angle{MBN}= \angle{BMC}=\angle{BAC}$
Since $\angle{BAC}= A$ and by an easy angle chase we obtain $\angle{MBN}= \angle{MBA}+\angle{ABN}=90-A+90-A= 180-2A$ we get:

$A= 180-2A$ $=>$ $A=60$

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#4 h Aug 30, 2011, 1:24 PM • 3 Y

Y by Valencia_Ann, Adventure10, Mango247

Let $X,Y$ be the projections of $O$ on CH and BH, $OX=OY$ , we know $\frac{S_{COD}}{S_{BOE}}=\frac{h_c}{h_b}\frac{sin{(C+2A)}}{sin{(B+2C)}}=\frac{h_c}{h_b}$ , where $E$ is the projection of $B$ on AC, so $\sin{(C+2A)}=\sin{(B+2C)}$ , and both angles can't be greater than 90, so we have $2A=B+C$ , so $A=60$ . I used $2S_{XYZ}=XY*XZ*sin{\widehat{YXZ}}=YZ*h_X$

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