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 at Saturday 10/26/2015 at 10pm IST, 930am PDT. For more details, see IMO Problems Discussion Group.

These are the problems from the first ever International Mathematical Olympiad.
They are pretty easy and I ask every one to try and solve them! Remember that the contestants were given **Four and a Half Hours**, and are required to prove these statements completely. So here Goes:

**Q1.** For every integer \(n\), prove that the fraction \(\dfrac{21n+4} {14n+3}\) cannot be reduced any further.**(POL)**

**Q2.** For which real numbers \(x\) do the following equations hold:
\(
(a) \sqrt{x+\sqrt{2x-1}}+ \sqrt{x-\sqrt{2x-1}}=√2, \\\
(b) \sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=1, \\\
(c) \sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}} =2.\)

**(ROM)**

**Q3.** Let \(x\) be an angle and let the real numbers \(a, b, c, \cos x\) satisfy the following equation: \[a\cos^2 {x}+ b \cos {x} + c =0.\]
Write the analogous quadratic equation for \(a, b, c, \cos {2x} \). Compare the given and the obtained equality for \(a = 4,b = 2,c = −1\). **(HUN)**

## Comments

Sort by:

TopNewestQ1. Let g.c.d(\(a,b\)) be represented by (\(a,b\)).

(\(21n+4,14n+3\)) \(=\) (\(7n+1,14n+3\)) \(=\) (\(7n+1, 7n+2\)) \(=\) (\(7n+1,1\)) \(=\) \(1\)

Therefore, it can't it reduced for any integer \(n\). – Surya Prakash · 1 year, 10 months ago

Log in to reply

– Sualeh Asif · 1 year, 10 months ago

This is absolutely correct! From a collary of the Euclidean Algorithm we know that \((a,b)=(b,a-b)\) and this makes the problem trivial!Log in to reply

– Vinayak Verma · 1 year, 8 months ago

Same approach!Log in to reply

At the IMO, this solution is worth 7-. – Calvin Lin Staff · 1 year, 9 months ago

Log in to reply

– Mehul Arora · 1 year, 10 months ago

Simple standard solution.Log in to reply

– Anik Mandal · 1 year, 10 months ago

Same solutionLog in to reply

I'll give my solution to Q3 (which is probably the long way around)

Note that \(\cos 2x = 2 \cos^2 x - 1\). Let \(\cos x = y\) and \(\cos 2x = z\).

We have

\[y = \dfrac {-b \pm \sqrt{b^2 - 4ac}}{2a}\]

\[y^2 = \dfrac{2b^2-4ac \pm 2b\sqrt{b^2 - 4ac}}{4a^2}\]

\[2y^2 = \dfrac{b^2-2ac \pm b\sqrt{b^2 - 4ac}}{a^2}\]

\[2y^2 - 1 = \dfrac{b^2-a^2-2ac \pm b\sqrt{b^2-4ac}}{a^2}\]

\[z = \dfrac{b^2-a^2-2ac \pm b\sqrt{b^2-4ac}}{a^2}\]

Now, we will reverse engineer completing the square (which is used to get the quadratic equation. I just reversed it)

\[\left (z + \dfrac {2ac+a^2-b^2}{a^2} \right )^2 = \dfrac {b^4 - 4ab^2 c}{a^4}\]

\[z^2 + \dfrac {4ac + 2a^2 - 2b^2}{a^2} z + \dfrac {4a^2c^2+a^4 + b^4+4a^3c-4ab^2c-2a^2b^2}{a^4} = \dfrac {b^4 - 4ab^2 c}{a^4}\]

\[z^2 + \dfrac {4ac + 2a^2 - 2b^2}{a^2} z + \dfrac {4a^2c^2+a^4+4a^3c-2a^2b^2}{a^4} = 0\]

\[a^2 z^2 + (2a^2-2b^2+4ac)z + (a+2c)^2-2b^2=0\]

\[a^2 \cos^2 2x + (2a^2-2b^2+4ac) \cos 2x + (a+2c)^2-2b^2=0\]

Therefore, the analogous equation is

\[a^2 \cos^2 2x + (2a^2-2b^2+4ac) \cos 2x + (a+2c)^2-2b^2=0\]

Comparing for \(a=4, b=2, c=-1\), we have

\[4 \cos^2 x + 2 \cos x - 1 = 0\]

\[4 \cos^2 2x + 2 \cos 2x - 1 = 0\]

Clearly, both equations have the same coefficients. The value of \(x\) in this case are certain multiples of \(\dfrac {\pi}{5}\).

I hope that's what I had to do when I had to compare.

Edit: This solution is only valid as long as \(a \neq 0\) because otherwise, we have division by 0. If \(a\) does equal 0, we have

\[by+c=0\]

\[y= \dfrac{-c}{b}\]

\[y^2=\dfrac{c^2}{b^2}\]

\[2y^2=\dfrac{2c^2}{b^2}\]

\[2y^2-1=\dfrac{2c^2-b^2}{b^2}\]

\[z=\dfrac{2c^2-b^2}{b^2}\]

\[b^2 z = 2c^2-b^2\]

\[b^2 z + b^2 - 2c^2 = 0\]

\[b^2 \cos 2x + b^2 - 2c^2 = 0\] – Sharky Kesa · 1 year, 9 months ago

Log in to reply

At the IMO, your solution would be worth a 7-. – Calvin Lin Staff · 1 year, 9 months ago

Log in to reply

This ends our questions for the First (ever) day of the IMO! Great work @Sharky Kesa ,@Surya Prakash @Chew-Seong Cheong and all others! – Sualeh Asif · 1 year, 9 months ago

Log in to reply

Solution for Q1:Now, from repeated use of Lemma-(1), we can write (21n+4, 14n+3) = (7n+1, 14n+3) = (7n+1, 7n+2) = (7n+1,1) = 1. Therefore, the fraction is irreducible for any integer n. – Karthik Venkata · 1 year, 9 months ago

Log in to reply

Answer to Question 3:

\(\color{blue}{a\cos^2{x} + b\cos{x} + c = 0 \quad \quad ...(1) }\) \(\begin{equation} \begin{split} \Rightarrow a\cos^2{x} & = - b\cos{x} - c \\ \frac{a}{2}(\cos{2x}+1) & = - b\cos{x} - c \\ a \cos{2x} & = -a - 2b\cos{x} - 2c \\ a^2 \cos^2{2x} & = a^2 + 4b^2\cos^2{x} + 4c^2 + 2(2ab \cos{x} + 4bc \cos{x} + 2ca) \\ & = \frac{4b^2}{2}(\cos{2x}+1) + 4(a+2c)b\cos{x} + a^2 + 4c^2+4ca \\ & = 2b^2\cos{2x} - 2(a+2c)(-a-2b\cos{x} -2c) - 2a^2 - 8c^2 - 8ca + a^2 + 2b^2 + 4c^2+4ca \\ & = 2b^2\cos{2x} - 2a(a+2c)\cos{2x} - a^2 + 2b^2 - 4c^2 - 4ca \end{split} \end{equation} \)

The required equation is:

\[\color{blue}{a^2 \cos^2{2x}+ (2a^2-2b^2+4ca)\cos{2x} + a^2 - 2b^2 + 4c^2 + 4ca = 0 \quad \quad ...(2)} \]

For \(a=4\), \(b=2\) and \(c=-1\),

\(\begin{cases} (1): & 4\cos^2{x} + 2\cos{x} - 1 = 0 \\ \\ (2): & 16\cos^2{2x} + 8\cos{2x} - 4 = 0 & \Rightarrow 4\cos^2{2x} + 2\cos{2x} - 1 = 0 \end{cases} \)

\(\begin{equation} \begin{split} \text{From }(1): \quad \cos{x} & = \frac{-2\pm\sqrt{20}}{8} = \frac{-1\pm\sqrt{5}}{4} \\ \Rightarrow \cos{2x} & = 2\cos^2{x} - 1 \\ & = 2 \left( \frac{-1\pm\sqrt{5}}{4} \right)^2 - 1 \\ & = 2 \left( \frac{6 \color{red}{\mp} 2 \sqrt{5}}{16} \right) - 1 \\ & = \frac{3 \color{red}{\mp} \sqrt{5} - 4}{4} = \frac{-1 \color{red}{\mp} \sqrt{5}}{4} \end{split} \end{equation} \)

Therefore, both \(\cos{x}\) and \(\cos{2x}\) satisfy the equation \(4\cos^2{x} + 2\cos{x} - 1 = 0 \).

Actually, \(\cos{72^\circ} = \frac{-1 + \sqrt{5}}{4}\) and \(\cos{144^\circ} = \frac{-1 - \sqrt{5}}{4}\). – Chew-Seong Cheong · 1 year, 9 months ago

Log in to reply

– Sualeh Asif · 1 year, 9 months ago

Another typo in the eleventh line. I think the equation in (2) is in \(\cos {2x}\)Log in to reply

– Chew-Seong Cheong · 1 year, 9 months ago

Thanks. I have edited it.Log in to reply

– Sualeh Asif · 1 year, 9 months ago

Another great solution! Do note that the coefficients of \((1),(2)\) are proportional.Log in to reply

Good solution. At the IMO, this would be worth 7-. – Calvin Lin Staff · 1 year, 9 months ago

Log in to reply

– Chew-Seong Cheong · 1 year, 9 months ago

Yes, you are right. Edited it.Log in to reply

Q2. I will just write the solution for the first one. Try the remaining on your own.

Clearly, the root is satisfied iff \(x \geq \dfrac{1}{2}\). Squaring on both sides gives \(2x +2|x-1| = 2\).

So, \(x+ |x-1| = 1\).

If \(x > 1\), then \(2x - 1 = 1\) which gives \(x=1\) which is false since we assumed that \(x>1\).

If \(x \leq 1\), then we get \(1=1\) which is true. So, all reals less than or equal to \(1\) satisfied the above equation. But \(x \geq \dfrac{1}{2}\). So, all reals \(x \in \Big[ \dfrac{1}{2}, 1 \Big] \) satisfy this equation. – Surya Prakash · 1 year, 10 months ago

Log in to reply

– N Solomon · 1 year, 9 months ago

It might seem like a silly question to some, but how do you jump from the original equation to \(2x + 2|x - 1| = 2\)Log in to reply

Furthermore, when you square an equation, you introduce extraneous roots. So, any solution that you obtain still has to be verified that it is a solution to the original problem. So far, we can only conclude that "the solution is a subset of ....".

At the IMO, this solution would be worth 0+. – Calvin Lin Staff · 1 year, 9 months ago

Log in to reply

– Surya Prakash · 1 year, 9 months ago

Got it Sir!! Thanks for suggestion.Log in to reply

– Sualeh Asif · 1 year, 10 months ago

This is a perfect solution. All reals \(\frac{1}{2} \leq x \leq 1\) satisfy the equation for the part one. It can be proved using the same method that no solutions exist for the second part. The main thing is to note that \(x\geq \frac{1}{2}\) and then square to obtain expressions. The third one is just a follow up from this.Log in to reply

– Surya Prakash · 1 year, 10 months ago

Yes. U r right.Log in to reply

Let \(\dfrac{21n + 4}{ 14 n +3 }= k\) – D H · 11 months ago

Log in to reply

Q#2\[(a) \sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=1\] For the L.H.S to be real,\(\sqrt{2x-1}\) must be greater then or equal to zero.So \[\sqrt{2x-1}\geq 0\\2x-1\geq 0\\x\geq \frac {1}{2}\] ............................................................................................................................................................................................. \[\begin{align}\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=1\\ \text{Squaring both sides} 2x+2(\sqrt{x+\sqrt{2x-1}})(\sqrt{x-\sqrt{2x-1}})=1\\2x+2\sqrt{(x-1)^2}=1\\2x+2|x-1|=1\\x+|x-1|=\frac{1}{2}\end{align}\] \[\text{CASE 1:} x>1\\ x+|x-1|=\frac{1}{2}\rightarrow x+x-1=\frac{1}{2}\\2x=\frac{1+2}{2}\\x=\frac{3}{4}\] Here \(\frac{3}{4}<1\),but we assumed that (x>1/),a contradiction.Hence no solutions exist in this case. \[\text{CASE 2:} x<1 \\x+|x-1|=\frac{1}{2}\rightarrow x-x+1=\frac{1}{2}\\1=\frac{1}{2} \text{A contradiction}\] Therefore no solutions can exust in this case. As no solutions can exist in either case,so the equation has no solutions. – Abdur Rehman Zahid · 1 year, 9 months agoLog in to reply

– Sualeh Asif · 1 year, 9 months ago

Great work, Though you made a mistake. In case 1 you assumed \(x>1\) while the answer you got was less than 1. Contradiction. No solutions for the first or second case. Put in a complete solutoon doing the third part too.Log in to reply

– Abdur Rehman Zahid · 1 year, 9 months ago

I just realized it now.sorry!!!! (Ican't edit my solution,can I?)Log in to reply

– Mehul Arora · 1 year, 9 months ago

Sure you can! :D Just click the "EDIT" button at the bottom of your solution :)Log in to reply

– Abdur Rehman Zahid · 1 year, 9 months ago

Thanks!!!!Log in to reply

@Sualeh Asif please open your slack :) – Mehul Arora · 1 year, 9 months ago

Log in to reply

1) gcd{21n + 4, 14n + 3} = 1 so they cant be reduced – Dev Sharma · 1 year, 10 months ago

Log in to reply

– Hrithik Nambiar · 1 year, 10 months ago

Sorry that's not what I meant. Just went out of track!Log in to reply

– Calvin Lin Staff · 1 year, 9 months ago

At the IMO, this solution is a 0+, because you have not proven the statement.Log in to reply

– Hrithik Nambiar · 1 year, 9 months ago

Yes sir! Thanks' a lot.Log in to reply

– Swapnil Das · 1 year, 10 months ago

Ya, this can be proved by Euclidean Algorithm.Log in to reply

– Aditya Chauhan · 1 year, 10 months ago

How did you declare gcd as 1 directly?Log in to reply

@Dev Sharma ! BUT THEN I HAD A SENSE OF DOING IT IN ANOTHER WAY! ANYWAYS MY BAD ! – Hrithik Nambiar · 1 year, 10 months ago

NO ! I USED THE METHOD AS THAT OFLog in to reply

– Nihar Mahajan · 1 year, 10 months ago

Avoid using upper case of letters. It is considered as rude on brilliant.Log in to reply

– Calvin Lin Staff · 1 year, 10 months ago

Please refrain from typing in all caps, as that is considered very rude on the internet. Thanks for your assistance!Log in to reply

– Hrithik Nambiar · 1 year, 10 months ago

yes sir .Log in to reply

– Hrithik Nambiar · 1 year, 10 months ago

Did it the same way .But , Let me ask you guys something - won't the numerator be even always if "n" is not equal to zero and the denominator would always be odd. So , obviously, it can not be further reduced.Log in to reply

I.e. The numerator is 14(yet even) and denominator 7(yet odd). The fraction can be reduced.

And This is to be proved for all primes! – Sualeh Asif · 1 year, 10 months ago

Log in to reply

– Dev Sharma · 1 year, 10 months ago

by using the fact ax + by = 1Log in to reply

– Hrithik Nambiar · 1 year, 10 months ago

exactly! sorryLog in to reply