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.

## Comments

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TopNewest2) 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. – Sharky Kesa · 10 months, 3 weeks ago

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– Sualeh Asif · 10 months, 3 weeks ago

One of the great many proofs the problem has! Nice short and effectiveLog in to reply

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 \(3b^2 > a^2 > b^2\). – Sharky Kesa · 10 months, 3 weeks ago

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@Sharky Kesa – Sualeh Asif · 10 months, 3 weeks ago

The solution is correct (if not short!) Great workLog in to reply

– Sharky Kesa · 10 months, 3 weeks ago

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

@Sharky Kesa @Abdur Rehman Zahid @Xuming Liang This time its Algebra! – Sualeh Asif · 10 months, 3 weeks ago

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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:

\(x = k\pi\) for any integral \(k\) and \(n\) is even,

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

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

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– Sualeh Asif · 10 months, 3 weeks 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!Log in to reply