Don't know how to solve. Do you?

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
4 years, 6 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link]( link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}


Sort by:

Top Newest

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 - 4 years, 6 months ago

Log in to reply

Awesome!! Try to get first also.

Aakash Khandelwal - 4 years, 6 months ago

Log in to reply

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 - 4 years, 6 months ago

Log in to reply

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 - 4 years, 6 months ago

Log in to reply

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 - 4 years, 5 months ago

Log in to reply

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 - 4 years, 5 months ago

Log in to reply

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

Aakash Khandelwal - 4 years, 5 months ago

Log in to reply


Problem Loading...

Note Loading...

Set Loading...