\(Q.1\) 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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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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