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,x2+y2+z2=b2,xy=z2x + y + z = a,\\ x^2 + y^2 + z^2 = b^2,\\ xy = z^2 where aa and bb are given real numbers. What conditions must hold on aa and bb for the solutions to be positive and distinct?

Q2. (POL) Let a,ba, b, and cc be the lengths of a triangle whose area is SS. Prove that a2+b2+c24S3a^2 + b^2 + c^2 \geq 4S\sqrt{3} . In what case does equality hold?

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

This is part of the set International Mathematical Olympiads

Note by Sualeh Asif
3 years, 11 months ago

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2) Just use Heron's Formula:

Replace SS with s(sa)(sb)(sc)\sqrt{s(s-a)(s-b)(s-c)} where s=a+b+c2s=\dfrac {a+b+c}{2}

We now have

43(a+b+c)(a+bc)(ab+c)(a+b+c)16=3(a4+b4+c42a2b22b2c2c2a2)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

a4+b4+c4+2a2b2+2b2c2+2c2a23a43b43c4+6a2b2+6b2c2+6c2a2a^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

4a4+4b4+4c44a2b2+4b2c2+4c2a24a^4+4b^4+4c^4 \geq 4a^2b^2+4b^2c^2+4c^2a^2

a4+b4+c4a2b2+b2c2+c2a2a^4+b^4+c^4 \geq a^2b^2+b^2c^2+c^2a^2

Since a2b2a4+b42a^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=ca=b=c, which is an equilateral triangle.

Sharky Kesa - 3 years, 11 months ago

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

Sualeh Asif - 3 years, 11 months ago

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

We have x2+y2=(x+y)22xy=(x+y)22z2x^2+y^2=(x+y)^2-2xy=(x+y)^2-2z^2. Putting this in, we now have

(x+y)+z=a(x+y)+z=a

(x+y)2z2=b2(x+y)^2-z^2=b^2

Note that ((x+y)+z)((x+y)z)=(x+y)2z2\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=0a=0 and when a0a \neq 0.

When a=0a=0, we have b=0b=0 and x+y=zx+y=-z, which gives us xy=(x+y)2xy = (x+y)^2. Then, we get x2+xy+y2=0x^2+xy+y^2=0. Multiplying by (xy)(x-y), we have x3y3=0x^3 - y^3 = 0 which means x=yx = y.This means that x2+x2+x2=0x^2 + x^2 + x^2 = 0, which gives x=0x=0. Thus, there are no positive solutions in this case.

When a0a \neq 0, we have

(x+y)+z=a(x+y)+z=a

(x+y)z=b2a(x+y)-z = \dfrac{b^2}{a}

This has solutions (x+y)=a2+b22a(x+y)=\dfrac {a^2+b^2}{2a} and z=a2b22az = \dfrac {a^2 - b^2}{2a}. Thus a2>b2a^2 > b^2. Now, we have (x+y)=a2+b22a(x+y)=\dfrac {a^2+b^2}{2a} and xy=(a2b22a)2xy = \left (\dfrac {a^2 - b^2}{2a} \right )^2. This means that xx and yy are roots of the quadratic k2a2+b22ak+(a2b22a)2k^2 - \dfrac {a^2+b^2}{2a} k + \left (\dfrac {a^2-b^2}{2a} \right )^2. The discriminant of this quadratic is

(a2+b22a)24(a2b22a)2=(3a2b2)(3b2a2)4a2\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 3a2b23a^2 - b^2 and 3b2a23b^2-a^2 are negative or both are positive. If they are both negative, 2(a2+b2)2(a^2+b^2) is negative, which is incorrect. Therefore 3a2>b23a^2 > b^2 and 3b2>a23b^2 > a^2. But we have already replaced the first inequality with the sharper bound a2>b2a^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 (a2+b22a)2>(a2+b22a)2(2a2b22a)2\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 aa is positive and 3b2>a2>b23b^2 > a^2 > b^2.

Sharky Kesa - 3 years, 11 months ago

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

Sualeh Asif - 3 years, 11 months ago

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

Sharky Kesa - 3 years, 11 months ago

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

Sualeh Asif - 3 years, 11 months ago

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3) My solution - hopefully not too short

We have that cos2x+sin2x=1\cos^2 x + \sin^2 x = 1 so we can't have solutions with n2n \neq 2 and 0<cos(x),sin(x)<10<|\cos (x)|, |\sin(x)|<1. Also, we can't have solutions when n=2n=2 since the sign is wrong (needs to be '-', not '+'). So the only solutions have cosx=0cos x = 0 or sinx=0\sin x = 0, and these are:

  1. x=kπx = k\pi for any integral kk and nn is even,

  2. x=2k(π)x=2k(\pi) for any integral kk and nn is odd,

  3. x=2k(π+3π2)x=2k(\pi+\dfrac {3\pi}{2}) for any integral kk and nn is odd.

Sharky Kesa - 3 years, 11 months ago

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@sharky An integral part of the solution is still missing, I.e. Why n2n\not= 2 doesnt have solutions. Though you created the bound, complete your proof by adding that in!

Sualeh Asif - 3 years, 11 months ago

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