$Q.1$ Find all quadruples $(a,b,c,d)$ such that : $(b+c+d)^{2010}$=$3a$ ;

$(a+c+d)^{2010}$=$3b$ ;

$(b+a+d)^{2010}$=$3c$ ;

$(b+c+a)^{2010}$=$3d$ ;

$Q.2$ In an acute triangle $ABC$ , the segment $CD$ is an altitude and $H$ is orthocentre. Given that circumcentre of triangle lies on angle bisector of angle $DHB$, determine possible values of angle $CAB$.

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## Comments

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TopNewestHere is the diagram: Triangle ABC

$O$ is the circumcentre of $\triangle ABC$.

Now, do you know these properties? They are simple to prove, so if you don't, I can show you how.

1. $\angle DHB = A$

2. $\angle DHA = B$

3. $\angle OAB = \frac{\pi}{2} - C$

4. $\angle HAB = \frac{\pi}{2} - B$

5. $HA = 2R \cos A$ where $R = OA$ is the circumradius.

From these we get,

$\angle HAO = \angle HAB - \angle OAB = C - B$

$\angle OHA = \angle DHA + \angle DHO = \frac{A}{2} + B$

So I'm aiming for a sine rule, but I need $\angle HOA$ for that.

We have $\angle HOA = \pi - (\angle HAO + \angle OHA$)

Now in $\triangle AHO$,

$\dfrac{2R \cos A}{\sin \angle HOA} = \dfrac{R}{\sin \angle OHA}$

Using the values we've got,

$2 \cos A = \dfrac {\sin (\frac{A}{2} + C)}{\sin (\frac{A}{2} + B)}$

$\dfrac {\sin (\frac{A}{2} + C)}{\sin (\frac{A}{2} + B)} = \dfrac {2 \sin (\frac{A}{2} + C) \cos \frac{A}{2}}{2 \sin (\frac{A}{2} + B)\cos \frac{A}{2}} = \dfrac{\sin (A + C) + \sin C}{\sin (A + B) + \sin B}$

But, $A + C = \pi - B$ and $A + B = \pi - C$

So, $\sin (A + C) = \sin B$ and $\sin (A + B) = \sin C$

$\dfrac{\sin (A + C) + \sin C}{\sin (A + B) + \sin B} = \dfrac{\sin B + \sin C}{\sin C + \sin B} = 1$

So, $2 \cos A = 1$ and $A = \frac{\pi}{3}$

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Awesome!! Try to get first also.

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A faster way (sigh, how did I miss this?) would be to note that, $(\frac{A}{2} + B) + (\frac{A}{2} + C) = \pi$

and hence, $\sin(\frac{A}{2} + B) = \sin(\frac{A}{2} + C)$

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Note that $\angle DHB = \angle A$. Use the sine rule and then trig bash. I'm searching for a nicer solution.

EDIT: Turns out I was using the wrong triangle. Use triangle $HOA$ where $O$ is the circumcentre. The angles are all nice. You get $A = \pi/3$. Will post full solution tomorrow.

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I think the answer to first is $a=b=c=d=0$ or $a=b=c=d=1/3$

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Yeah, you're right. Sorry, I solved the question, but didn't post the solution.

This is just an outline:

1. Since the conditions are symmetric, assume $a \geq b \geq c \geq d \geq e$ without loss of generality.

2. Then use the conditions given to show equality for all variables. (But I think it should be mentioned that $a, b, c, d, e$ are positive.)

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No , it wasn't mentioned from where I picked it up

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