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Q.1Q.1 Find all quadruples (a,b,c,d)(a,b,c,d) such that : (b+c+d)2010(b+c+d)^{2010}=3a3a ;
(a+c+d)2010(a+c+d)^{2010}=3b3b ;
(b+a+d)2010(b+a+d)^{2010}=3c3c ;
(b+c+a)2010(b+c+a)^{2010}=3d3d ;

Q.2Q.2 In an acute triangle ABCABC , the segment CDCD is an altitude and HH is orthocentre. Given that circumcentre of triangle lies on angle bisector of angle DHB DHB, determine possible values of angle CABCAB.

Note by Aakash Khandelwal
3 years, 7 months ago

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Here is the diagram: Triangle ABC Triangle ABC
O O is the circumcentre of ABC \triangle ABC .
Now, do you know these properties? They are simple to prove, so if you don't, I can show you how.
1. DHB=A \angle DHB = A
2. DHA=B \angle DHA = B
3. OAB=π2C \angle OAB = \frac{\pi}{2} - C
4. HAB=π2B \angle HAB = \frac{\pi}{2} - B
5. HA=2RcosA HA = 2R \cos A where R=OA R = OA is the circumradius.

From these we get,
HAO=HABOAB=CB \angle HAO = \angle HAB - \angle OAB = C - B
OHA=DHA+DHO=A2+B \angle OHA = \angle DHA + \angle DHO = \frac{A}{2} + B

So I'm aiming for a sine rule, but I need HOA \angle HOA for that.
We have HOA=π(HAO+OHA \angle HOA = \pi - (\angle HAO + \angle OHA )
 

Now in AHO\triangle AHO,
2RcosAsinHOA=RsinOHA \dfrac{2R \cos A}{\sin \angle HOA} = \dfrac{R}{\sin \angle OHA}
 

Using the values we've got,
2cosA=sin(A2+C)sin(A2+B) 2 \cos A = \dfrac {\sin (\frac{A}{2} + C)}{\sin (\frac{A}{2} + B)}
sin(A2+C)sin(A2+B)=2sin(A2+C)cosA22sin(A2+B)cosA2=sin(A+C)+sinCsin(A+B)+sinB \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=πB A + C = \pi - B and A+B=πC A + B = \pi - C
So, sin(A+C)=sinB \sin (A + C) = \sin B and sin(A+B)=sinC \sin (A + B) = \sin C

sin(A+C)+sinCsin(A+B)+sinB=sinB+sinCsinC+sinB=1 \dfrac{\sin (A + C) + \sin C}{\sin (A + B) + \sin B} = \dfrac{\sin B + \sin C}{\sin C + \sin B} = 1
 

So, 2cosA=1 2 \cos A = 1 and A=π3 A = \frac{\pi}{3}

Ameya Daigavane - 3 years, 7 months ago

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Awesome!! Try to get first also.

Aakash Khandelwal - 3 years, 7 months ago

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A faster way (sigh, how did I miss this?) would be to note that, (A2+B)+(A2+C)=π (\frac{A}{2} + B) + (\frac{A}{2} + C) = \pi
and hence, sin(A2+B)=sin(A2+C) \sin(\frac{A}{2} + B) = \sin(\frac{A}{2} + C)

Ameya Daigavane - 3 years, 7 months ago

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Note that DHB=A\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 HOA where O O is the circumcentre. The angles are all nice. You get A=π/3 A = \pi/3 . Will post full solution tomorrow.

Ameya Daigavane - 3 years, 7 months ago

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

Aakash Khandelwal - 3 years, 7 months ago

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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 abcde 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 a, b, c, d, e are positive.)

Ameya Daigavane - 3 years, 7 months ago

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

Aakash Khandelwal - 3 years, 6 months ago

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