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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
2 years 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.

Sharky Kesa - 2 years ago

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

Sualeh Asif - 2 years ago

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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 - 2 years ago

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

Sualeh Asif - 2 years ago

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

Sharky Kesa - 2 years ago

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

Sualeh Asif - 2 years 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:

  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.

Sharky Kesa - 2 years ago

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@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!

Sualeh Asif - 2 years ago

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