### Re: an old maths olympiad question

Posted:

**Mon Jul 09, 2018 6:29 pm**A website dedicated to the puzzling world of mathematics and programming

https://projecteuler.chat/

Page **2** of **2**

Posted: **Mon Jul 09, 2018 6:29 pm**

Posted: **Mon Jul 09, 2018 7:45 pm**

I think it is enough now.

In your first post I read that your teacher has given you and your mates a set of problems to solve.

I gather that it can't be his intention that you get the solution from a forum on the web.

Then if some people take the time to look into your problems you start to be nasty.

So I think it best not to pay attention anymore to these problems of yours.

Posted: **Tue Jul 10, 2018 2:47 am**

i didnt have any intention to be nasty. it was the complete helplessness. i couldnt solve them four years ago, i cant solve them now and if ask help from others what would i expectt. i have always considerd people here to be maths geniuses. especially those who have solved 400+ questions of projecteuler. I checked the answer on internet but I don't know the solution. if you don't wanna pay attention its okay but if you have the solution you can post it.Then if some people take the time to look into your problems you start to be nasty.

So I think it best not to pay attention anymore to these problems of yours.

another one was:

find all positive integers a,b for which the numbers (2

Posted: **Tue Jul 10, 2018 4:17 am**

please forgive me for those rude comments. I will never do that again

Posted: **Tue Jul 10, 2018 7:12 am**

As Catch-22 author Joseph Heller said -- Just because you think everyone's out to get you, doesn't mean they aren't -- and... Just because nobody says they forgive you,

doesn't mean they don't...

Posted: **Tue Jul 10, 2018 11:13 am**

From here, we have,

$\displaystyle\frac{1}{\sin x}=\frac{1}{\sin 2x}+\frac{1}{\sin 3x}$

$\displaystyle\sin 2x=\frac{\sin x \sin 3x}{\sin 3x-\sin x}=\frac{\sin x \sin 3x}{2\sin x-4 \sin^3 x}=\frac{\sin x\sin 3x}{2 \sin x \cos 2x}$

Therefore, $2\cos 2x \sin 2x=\sin 3x$ and hence $\sin 4x=\sin 3x$ implies $\sin(\pi-4x)=\sin 3x$

which gives $x=\pi/7$ as noted by

Btw,

NOTE: All this in hindsight knowing that n should be 7. To derive it in an Olympiad is close to impossible for me.

Posted: **Tue Jul 10, 2018 1:49 pm**

Nicely done. This is the trick that I missed. I thought we would have to go all the way up to $\sin(7x)$, which would be horrible.MuthuVeerappanR wrote: ↑Tue Jul 10, 2018 11:13 am... hence $\sin 4x=\sin 3x$ implies $\sin(\pi-4x)=\sin 3x$

I got to $n=7$ fairly quickly, just by calculating the diagonal lengths for n=6 and n=8 and plugging them into the expressionMuthuVeerappanR wrote: ↑Tue Jul 10, 2018 11:13 amNOTE: All this in hindsight knowing that n should be 7. To derive it in an Olympiad is close to impossible for me.

$\displaystyle\frac{1}{|A_1A_2|}-\frac{1}{|A_1A_3|}-\frac{1}{|A_1A_4|}$

and seeing that the sign changed. This makes $n=7$ a candidate. With the further insight that the diagonals increase in length relative to the side as $n$ grows larger, we know that $n=7$ is the only candidate. When I reformulated the equation to use sines, it became easy to confirm.

Posted: **Tue Jul 10, 2018 1:59 pm**

@**jaap**, I think you did all the heavy lifting leaving just the algebra to me. The key to solving this problem is to pick the 'central angle' as the variable of interest.

Though the 'cubic' approach ended nowhere, it reminded me of the trick of evaluating $\sin(\pi/5)$ exactly, which I kinda used to arrive at the final answer.

Though the 'cubic' approach ended nowhere, it reminded me of the trick of evaluating $\sin(\pi/5)$ exactly, which I kinda used to arrive at the final answer.

Posted: **Sun May 12, 2019 7:00 pm**

it is so easy to write anything on internet. i promise i will never write any stupid post or a nasty comment anywhere ever again. Please forgive me. i am sorry.