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# Don't know how to solve. Do you?

$$Q.1$$ 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 $$CAB$$.

Note by Aakash Khandelwal
1 year, 3 months ago

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Here 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 $$A = \frac{\pi}{3}$$ · 1 year, 3 months ago

Awesome!! Try to get first also. · 1 year, 3 months ago

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, $$\sin(\frac{A}{2} + B) = \sin(\frac{A}{2} + C)$$ · 1 year, 3 months ago

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. · 1 year, 3 months ago

I think the answer to first is $$a=b=c=d=0$$ or $$a=b=c=d=1/3$$ · 1 year, 3 months ago

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.) · 1 year, 3 months ago