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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.

#Algebra

Note by Aakash Khandelwal
5 years, 2 months ago

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Comments

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=π2−C \angle OAB = \frac{\pi}{2} - C
4. ∠HAB=π2−B \angle HAB = \frac{\pi}{2} - B
5. HA=2Rcos⁡A HA = 2R \cos A where R=OA R = OA is the circumradius.

From these we get,
∠HAO=∠HAB−∠OAB=C−B \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,
2Rcos⁡Asin⁡∠HOA=Rsin⁡∠OHA \dfrac{2R \cos A}{\sin \angle HOA} = \dfrac{R}{\sin \angle OHA}
 

Using the values we've got,
2cos⁡A=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)cos⁡A22sin⁡(A2+B)cos⁡A2=sin⁡(A+C)+sin⁡Csin⁡(A+B)+sin⁡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=π−B A + C = \pi - B and A+B=π−C A + B = \pi - C
So, sin⁡(A+C)=sin⁡B \sin (A + C) = \sin B and sin⁡(A+B)=sin⁡C \sin (A + B) = \sin C

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

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

Ameya Daigavane - 5 years, 2 months ago

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Awesome!! Try to get first also.

Aakash Khandelwal - 5 years, 2 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 - 5 years, 2 months ago

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 - 5 years, 2 months ago

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 - 5 years, 2 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 a≥b≥c≥d≥e 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 - 5 years, 2 months ago

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

Aakash Khandelwal - 5 years, 2 months ago
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