# International Mathematical Olympiad '61, Day 1

The goal of this set of notes is to improve our problem solving and proof writing skills. You are encouraged to submit a solution to any of these problems, and join in the discussion in #imo-discussion on Saturday 7 at 9:00 pm IST ,8 30 PDT. For more details, see IMO Problems Discussion Group.

Hello Problem Solvers, I have lately been absolutely busy with the academics. Really sorry for the weekly delay! Here are the problems from the 3rd IMO. I recommend everyone to try them and post any progress they make!

Q1. (HUN) Solve the following system of equations: $x + y + z = a,\\ x^2 + y^2 + z^2 = b^2,\\ xy = z^2$ where $a$ and $b$ are given real numbers. What conditions must hold on $a$ and $b$ for the solutions to be positive and distinct?

Q2. (POL) Let $a, b$, and $c$ be the lengths of a triangle whose area is $S$. Prove that $a^2 + b^2 + c^2 \geq 4S\sqrt{3}$ . In what case does equality hold?

Q3. (BUL) Solve the equation $\cos^{n} x- \sin^{n} x = 1$, where $n$ is a given positive integer.

###### This is part of the set International Mathematical Olympiads

Note by Sualeh Asif
5 years, 8 months ago

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2) Just use Heron's Formula:

Replace $S$ with $\sqrt{s(s-a)(s-b)(s-c)}$ where $s=\dfrac {a+b+c}{2}$

We now have

$4\sqrt{3}\sqrt{\dfrac{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}{16}}=\sqrt{3}\sqrt{-(a^4+b^4+c^4-2a^2b^2-2b^2c^2-c^2a^2)}$

Squaring both sides of the original inequality, we have

$a^4+b^4+c^4 + 2a^2b^2+2b^2c^2+2c^2a^2 \geq -3a^4-3b^4-3c^4+6a^2b^2+6b^2c^2+6c^2a^2$

$4a^4+4b^4+4c^4 \geq 4a^2b^2+4b^2c^2+4c^2a^2$

$a^4+b^4+c^4 \geq a^2b^2+b^2c^2+c^2a^2$

Since $a^2 b^2 \leq \dfrac {a^4+b^4}{2}$ by AM-GM, we have the above inequality is true with equality at $a=b=c$, which is an equilateral triangle.

- 5 years, 8 months ago

One of the great many proofs the problem has! Nice short and effective

- 5 years, 8 months ago

My solution to 1 (Probably the long way around):

We have $x^2+y^2=(x+y)^2-2xy=(x+y)^2-2z^2$. Putting this in, we now have

$(x+y)+z=a$

$(x+y)^2-z^2=b^2$

Note that $\left ((x+y)+z \right ) \left ((x+y)-z \right ) = (x+y)^2 - z^2$. We can now split this into 2 cases: when $a=0$ and when $a \neq 0$.

When $a=0$, we have $b=0$ and $x+y=-z$, which gives us $xy = (x+y)^2$. Then, we get $x^2+xy+y^2=0$. Multiplying by $(x-y)$, we have $x^3 - y^3 = 0$ which means $x = y$.This means that $x^2 + x^2 + x^2 = 0$, which gives $x=0$. Thus, there are no positive solutions in this case.

When $a \neq 0$, we have

$(x+y)+z=a$

$(x+y)-z = \dfrac{b^2}{a}$

This has solutions $(x+y)=\dfrac {a^2+b^2}{2a}$ and $z = \dfrac {a^2 - b^2}{2a}$. Thus $a^2 > b^2$. Now, we have $(x+y)=\dfrac {a^2+b^2}{2a}$ and $xy = \left (\dfrac {a^2 - b^2}{2a} \right )^2$. This means that $x$ and $y$ are roots of the quadratic $k^2 - \dfrac {a^2+b^2}{2a} k + \left (\dfrac {a^2-b^2}{2a} \right )^2$. The discriminant of this quadratic is

$\left (\dfrac {a^2 + b^2}{2a} \right )^2 - 4 \left (\dfrac {a^2 - b^2}{2a} \right )^2 = \dfrac {(3a^2-b^2)(3b^2-a^2)}{4a^2}$

This final value must be positive, which means that either both $3a^2 - b^2$ and $3b^2-a^2$ are negative or both are positive. If they are both negative, $2(a^2+b^2)$ is negative, which is incorrect. Therefore $3a^2 > b^2$ and $3b^2 > a^2$. But we have already replaced the first inequality with the sharper bound $a^2 > b^2$. It is clear that both roots of the quadratic must be positive if the discriminant is positive (we can see this either from $\left(\frac{a^2 + b^2}{2a}\right)^2 > \left(\frac{a^2 + b^2}{2a}\right)^2 - \left(2\frac{a^2 -b^2}{2a}\right)^2$). We have now found the solutions to the system, and determined that it has positive solutions if and only if $a$ is positive and $3b^2 > a^2 > b^2$.

- 5 years, 8 months ago

The solution is correct (if not short!) Great work @Sharky Kesa

- 5 years, 8 months ago

I posted solution to Q2. Need to sleep now. Let me think about 3 later.

- 5 years, 8 months ago

@Sharky Kesa @Abdur Rehman Zahid @Xuming Liang This time its Algebra!

- 5 years, 8 months ago

3) My solution - hopefully not too short

We have that $\cos^2 x + \sin^2 x = 1$ so we can't have solutions with $n \neq 2$ and $0<|\cos (x)|, |\sin(x)|<1$. Also, we can't have solutions when $n=2$ since the sign is wrong (needs to be '-', not '+'). So the only solutions have $cos x = 0$ or $\sin x = 0$, and these are:

1. $x = k\pi$ for any integral $k$ and $n$ is even,

2. $x=2k(\pi)$ for any integral $k$ and $n$ is odd,

3. $x=2k(\pi+\dfrac {3\pi}{2})$ for any integral $k$ and $n$ is odd.

- 5 years, 8 months ago

@sharky An integral part of the solution is still missing, I.e. Why $n\not= 2$ doesnt have solutions. Though you created the bound, complete your proof by adding that in!

- 5 years, 8 months ago