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.

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## Comments

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TopNewestIn 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.

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Can you post your solution?

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Q#2For 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 \(0 <x<\frac{45}{8}\) 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. .............................................................................................................................................................................................Log in to reply

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

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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?

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

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

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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 \(11\)?

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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.

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Third one can be done by applying coordinates

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

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

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