Australian School of Excellence 2015 Algebra Exam

  • Each question is worth 7 points

  • Time allowed is 4 hours

  • No books, notes or calculators permitted

  • Write full proofs with your answers

1) Find all real numbers aa such that there are four different real numbers xx satisfying the equation

x(x+1)3=(2x+a)(x+a1).x(x+1)^3=(2x+a)(x+a-1).

2) Find all functions f:RRf:\mathbb{R} \rightarrow \mathbb{R} such that for x,yRx, y \in \mathbb{R} we have

f(xf(y)+x)=xy+f(x).f(x f(y) + x) = xy+f(x).

3) Suppose that xx, yy and zz are real numbers satisfying the following three equations.

x+y+z=2x2+y2+z2=6x3+y3+z3=8\begin{aligned} x+y+z &=2\\ x^2 +y^2 + z^2 &=6\\ x^3+y^3+z^3&=8 \end{aligned}

Find all possible values of x2015+y2015+z2015x^{2015} + y^{2015} + z^{2015}.

4) Suppose that a function f:RRf:\mathbb{R} \rightarrow \mathbb{R} has the property that

f(x)2f(y)f(x)^2 \leq f(y)

for every x>yx > y.

Prove that 0f(x)10 \leq f(x) \leq 1 for all xRx \in \mathbb{R}.

5) Prove the following inequality for all a,b,c>0a, b, c > 0.

a2(a3+b3)a2+ab+b2+b2(b3+c3)b2+bc+c2+c2(c3+a3)c2+ca+a22abc\dfrac {a^2(a^3 + b^3)}{a^2 + ab + b^2} +\dfrac {b^2(b^3 + c^3)}{b^2 + bc + c^2} +\dfrac {c^2(c^3 + a^3)}{c^2 + ca + a^2} \geq 2abc

Note by Sharky Kesa
3 years, 9 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](https://brilliant.org)example 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}

Comments

Sort by:

Top Newest

3). let s1,s2,s3s_1,s_2,s_3 be the symmetric sums. let pnp_n denote the nth power sum. then by newtons sum: ap1=s1=2p2=s1p12s2=42s2=6s2=1p3=s1p2s2p1+3s3=14+3s3=8s3=2\begin{array}{c}a p_1=s_1=2\\ p_2=s_1p_1-2s_2=4-2s_2=6\Longrightarrow s_2=-1\\ p_3=s_1p_2-s_2p_1+3s_3=14+3s_3=8\Longrightarrow s_3=-2\end{array} so, x,y,z are roots of x32x2x+2=0x^3-2x^2-x+2=0 x3x2x2+2=x(x21)2(x21)=(x2)(x1)(x+1)=0x^3-x-2x^2+2=x(x^2-1)-2(x^2-1)=(x-2)(x-1)(x+1)=0 since the expression is symmetric, order doesnt matter. we have (x,y,z)=(2,1,1)(x,y,z)=(2,1,-1). put this in to get x2015+y2015+z2015=22015x^{2015}+y^{2015}+z^{2015}=\boxed{2^{2015}} this is the only value possible. are we suppose to expand 220152^{2015}?

Aareyan Manzoor - 3 years, 9 months ago

Log in to reply

Do you think we do? Of course not! Nice solution. 7 out of 7.

Sharky Kesa - 3 years, 9 months ago

Log in to reply

Is the answer to the second question is f(x)=0f(x)=0

Seong Ro - 3 years, 9 months ago

Log in to reply

@Seong Ro Is the answer to the first question

a(124(273),124(27+3))\large{a \in \left(\frac{1}{24} (27-\sqrt{3}),\frac{1}{24}(27+\sqrt{3}) \right)}

Seong Ro - 3 years, 9 months ago

Log in to reply

@Seong Ro No. Also, please give proof with your answers.

Sharky Kesa - 3 years, 9 months ago

Log in to reply

@Sharky Kesa As my answers are wrong , proof would be useless :(

Seong Ro - 3 years, 9 months ago

Log in to reply

@Sharky Kesa Although for the first question, expand the equation and we get a quartic polynomial.

For roots to be real and distinct its derivative must have three distinct roots

4x3+6x2+2x+3(a1)4x^3+6x^2+2x+3(a-1).....(1)(1) And for that to happen , we again take its derivative, we get

12x2+12x+2=012x^2+12x+2=0

For real and distinct roots

f(x1)f(x2)<0f(x_{1}) f(x_{2})<0 where x1,2=3±34x_{1,2}=\frac{-3 \pm \sqrt{3}}{4} in (1)(1)

and then i found interval for aa

Seong Ro - 3 years, 9 months ago

Log in to reply

@Sharky Kesa Now plz tell me if this is correct?

a(1432(1532957),1432(153+2957))\large{a \in \left(\frac{1}{432} (153-29 \sqrt{57}),\frac{1}{432}(153+29 \sqrt{57}) \right)}

Seong Ro - 3 years, 9 months ago

Log in to reply

@Seong Ro No. Also, please give proof.

Sharky Kesa - 3 years, 9 months ago

Log in to reply

Putting y=0y=0 we get f(xf(0)+x)=f(x))f(xf(0)+x)=f(x)) xf(0)+x=xxf(0)+x=x f(0)=0.........(1)\boxed{f(0)=0}.........(1) Now we assume that there exists f(y)=1f(y)=-1 Then putting it in original functional equation we get, f(0)=xy+f(x)f(0)=xy+f(x) xy=f(x)........(2)-xy=f(x)........(2) now put x=yx=y f(y)=y2f(y)=-y^{2} But f(y)=1f(y)=-1 Therefore y=1y=1 or y=1y=-1. Now putting these values in (2)(2) we get f(x)=xf(x)=x or f(x)=xf(x)=-x . since above two functions satisfy the given functional equation and (1)(1) Hence they are the solutions of the given functional equation.

Shivam Jadhav - 3 years, 9 months ago

Log in to reply

You would get 3 out of 7 for this proof. In line 2, how do you know ff is injective?

Sharky Kesa - 3 years, 9 months ago

Log in to reply

Answer is right I guess (function)

Shivam Jadhav - 3 years, 9 months ago

Log in to reply

@Shivam Jadhav Yeah it is. But you have done some assumptions.

Sharky Kesa - 3 years, 9 months ago

Log in to reply

Proving that the function is injective is very easy. For two reals a and b let f(a)=f(b). We take two cases. In the first case put y=a and in second case put y=b. Now note that in both cases the LHS is the same. Therefore comparing the RHS you obtain that a=b. Hence function is injective

Shrihari B - 3 years, 9 months ago

Log in to reply

Using AMGMAM-GM a2+b22aba^{2}+b^{2}\geq2ab 3(a2+b2)2(a2+ab+b2)\frac{3(a^{2}+b^{2})}{2}\geq(a^{2}+ab+b^{2}) This implies a2(a3+b3)a2+ab+b22a2(a3+b3)3(a2+b2)\frac{a^{2}(a^{3}+b^{3})}{a^{2}+ab+b^{2}}\geq\frac{2a^{2}(a^{3}+b^{3})}{3(a^{2}+b^{2})} Using Generalized Power mean inequality We get a2(a3+b3)a2+ab+b22a2(a2+b22)3\frac{a^{2}(a^{3}+b^{3})}{a^{2}+ab+b^{2}}\geq\frac{2a^{2}(\sqrt{\frac{a^{2}+b^{2}}{2}})}{3} Now by RMSAMRMS-AM inequality we get a2(a3+b3)a2+ab+b2a2(a+b)32a2ab3\frac{a^{2}(a^{3}+b^{3})}{a^{2}+ab+b^{2}}\geq\frac{a^{2}(a+b)}{3}\geq\frac{2a^{2}\sqrt{ab}}{3} Therefore cyca,b,ca2(a3+b3)a2+ab+b2cyca,b,c2a2ab3\sum_{cyc}^{a,b,c}\frac{a^{2}(a^{3}+b^{3})}{a^{2}+ab+b^{2}}\geq\sum_{cyc}^{a,b,c}\frac{2a^{2}\sqrt{ab}}{3} Now using simple AMGMAM-GM on the RHS of the above inequality we get cyca,b,ca2(a3+b3)a2+ab+b2cyca,b,c2a2ab32abc\sum_{cyc}^{a,b,c}\frac{a^{2}(a^{3}+b^{3})}{a^{2}+ab+b^{2}}\geq\sum_{cyc}^{a,b,c}\frac{2a^{2}\sqrt{ab}}{3}\geq2abc

cyca,b,ca2(a3+b3)a2+ab+b22abc\sum_{cyc}^{a,b,c}\frac{a^{2}(a^{3}+b^{3})}{a^{2}+ab+b^{2}}\geq2abc Hence proved .

Shivam Jadhav - 3 years, 9 months ago

Log in to reply

Could you provide the solutions to all questions please? So that we can reference and improve. Thanks!

Lai Hong Chiu - 3 years, 9 months ago

Log in to reply

@Sharky Kesa : For question 1 my solution is as follows : The given equation simplifies to x^4+3.x^3+x^2+(3-3a)x +(a-a^2)=0. Let the roots be p,q,r,s. Now, sum of roots two at a time=1, sum of roots three at a time =(3a-3) and all four roots at a time=(a-a^2)

Now we use the identity : (pqr+pqs+prs+qrs)^2=(p^2.q^2.r^2 + .........) + 2pqrs(pq+pr+ps+qr+qs+rs) Thus we obtain the inequality : (pqr+pqs+prs+qrs)^2 >= 3pqrs(pq+pr+ps+qr+qs+rs) Substituting the values we obtain a condition : 4a^2-7a+3 >=0. Solving that we obtain that a should lie outside [3/4,1].

Please tell me whether the answer and the solution is correct or wrong.

Shrihari B - 3 years, 9 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...