# International Mathematical Olympiad '60, First Day

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 at 9:00 pm IST ,8 30 PDT. For more details, see IMO Problems Discussion Group.

Here are the problems from the first day of the 1960 International mathematical Olympiad. They range from easy to harder Number Theory, Algebra and Geometry. Try your hand at them. Don't be discouraged if you can completely solve them. Do post your inspirations and ideas towards the problems. The discussion for these questions will be held soon. Happy Problem Solving!

Q1. (BUL) Find all the three-digit numbers for which one obtains, when dividing the number by 11, the sum of the squares of the digits of the initial number.

Q2. (HUN) For which real numbers $x$ does the following inequality hold: $\dfrac{4x^2}{(1-\sqrt{1+2x})^2} < 2x +9$?

Q3. (ROM) A right-angled triangle $ABC$ is given for which the hypotenuse $BC$ has length $a$ and is divided into $n$ equal segments, where $n$ is odd. Let $\alpha$ be the angle with which the point $A$ sees the segment containing the middle of the hypotenuse. Prove that $\tan{\alpha} = \dfrac{4nh}{ (n^2 -1)a},$ where $h$ is the height of the triangle.

###### This is part of the set International Mathematical Olympiads Note by Sualeh Asif
4 years, 1 month ago

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In third question I am getting an answer (4nh/n^2-1)*b/a^2. Where b is length of other leg. Also if we put n=3 we do not get alpha as 3h/2a. Please post a solution as early as possible. I have checked my solution a lot of times.

- 4 years, 1 month ago

Staff - 4 years, 1 month ago

Q#2 For the L.H.S to be defined, $(1-\sqrt{1+2x})^2\neq 0 \implies x\neq 0$ Also,for the inequality to be defined, $\sqrt{1+2x}\geq 0 \rightarrow 1+2x\geq 0\rightarrow x\geq \frac{-1}{2}$. Moreover, either $\sqrt{1+2x}>1$ or $\sqrt{1+2x}<1$. Case #1 $\sqrt{1+2x}>1\\ 1+2x>1 \\ 2x >0\rightarrow x>0$ Case #2 $\sqrt{1+2x}<1\\ 1+2x<1\\ 2x<0 \rightarrow x <0$ Therefore, either $\frac{-1}{2}\leq x <0$ or $x>0$ Now: $\frac{4x^2}{(1-\sqrt{1+2x})^2}<2x+9\\ \frac{4x^2 \times (1+\sqrt{1+2x})^2}{(1-\sqrt{1+2x})^2\times (1+\sqrt{1+2x})^2}<2x+9\\ \frac{4x^2 \times (1+\sqrt{1+2x})^2}{(1-1-2x)^2} < 2x+9\\ (1+\sqrt{1+2x})^2<2x+9$ Simplifying further,we get: $1+1+2x+2\sqrt{1+2x}<2x+9\\ 2\sqrt{1+2x}<7\\ \sqrt{1+2x}<\frac{7}{2}\\ 1+2x<\frac{49}{4} \\ 2x<\frac{45}{4} \\ x<\frac{45}{8}$ If $x>0$ we get $0 If $\frac{-1}{2}\leq x<0$ ,then $\frac{-1}{2}\leq x<\frac{45}{8},x\neq 0$. Hence,all $x$ where $\frac{-1}{2}\leq x<\frac{45}{8},x\neq 0$ satisfy the inequality. .............................................................................................................................................................................................

- 4 years, 1 month ago

Really good solution other than a few typos in the calculation and rationalisation!

- 4 years, 1 month ago

1. Let the 3 digit number be $\overline{abc}$. The remainder when it is divided by 11 is $a - b + c$. We are told that this is equal to $a^2 + b^2 + c ^2$, thus $a - b + c = a^2 + b^2 + c^2$. Multiplying by 4 and completing the square, we obtain:
$(4a^2 - 4a + 1 ) + ( 4b^2 + 4b + 1 ) + ( 4 c^2 - 4c + 1) = 3$

Hence, we must have $|2a-1 | = |2b+1 | = |2c - 1 | = 1$, or that $a = 1, b = 0 , c = 1$.

Conversely, we can verify that $101$ satisfies the conditions of the question, since $101 \equiv 2 \pmod{11}$ and $1^2 + 0^2 + 1 ^2 = 2$.

Is the above solution correct?

Staff - 4 years, 1 month ago

I think not the remainder , but the quotient since $11 \ | \ \overline{abc}$ (The problem statement is a bit ambiguous)

- 4 years, 1 month ago

I thought so too! And the problem statement seems to imply this too

- 4 years, 1 month ago

In addition to what Nihar said, while it is true that $a-b+c \equiv \overline{abc} \pmod{11}$, that does not necessarily mean that the remainder is $a-b+c$. For example, what is the remainder when $909$ is divided by $11$?

- 4 years, 1 month ago

Great! That's the point I was trying to make. I slid the "seemingly obvious fact" without substantiating why it is true. As it turns out, that is not the remainder and we often have to add / subtract 11 from it.

Didn't realize that I misinterpreted the question lol.

Staff - 4 years, 1 month ago

Third one can be done by applying coordinates

- 4 years, 1 month ago

@Sualeh Asif Can you post what we discussed for Q1? Thanks!

Staff - 4 years, 1 month ago

I was really busy the last whole week! I will In Sha Allah post it in a few days!

- 4 years ago