an old maths olympiad question

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kenbrooker
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Re: an old maths olympiad question

Post by kenbrooker » Mon Jul 09, 2018 6:29 pm

S_r wrote:
Mon Jul 09, 2018 9:59 am
I take my thanks back
More questions would give us more opportunities to earn your appreciation...
"Good Judgment comes from Experience;
Experience comes from Bad Judgment
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hk
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Re: an old maths olympiad question

Post by hk » Mon Jul 09, 2018 7:45 pm

S_r wrote:
Mon Jul 09, 2018 2:23 pm
So people here can't solve an Olympiad question
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.
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S_r
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Re: an old maths olympiad question

Post by S_r » Tue Jul 10, 2018 2:47 am

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.
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.

another one was:
find all positive integers a,b for which the numbers (21/2+a1/2)/(31/2+b1/2) is a rational number.

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S_r
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Re: an old maths olympiad question

Post by S_r » Tue Jul 10, 2018 4:17 am

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

kenbrooker
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Re: an old maths olympiad question

Post by kenbrooker » Tue Jul 10, 2018 7:12 am

S_r wrote:
Tue Jul 10, 2018 4:17 am
please forgive me for those rude comments. I will never do that again :(
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...
"Good Judgment comes from Experience;
Experience comes from Bad Judgment
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MuthuVeerappanR
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Re: an old maths olympiad question

Post by MuthuVeerappanR » Tue Jul 10, 2018 11:13 am

jaap wrote:
Mon Jul 09, 2018 12:28 pm
...

It is straightforward to see that:
A1A2 = 2sin(x)
A1A3 = 2sin(2x)
A1A4 = 2sin(3x)
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 jaap.

Btw, jaap's cubic should be $8z^3-4z^2-4z+1=0$ I think.

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

Post by jaap » Tue Jul 10, 2018 1:49 pm

MuthuVeerappanR wrote:
Tue Jul 10, 2018 11:13 am
... hence $\sin 4x=\sin 3x$ implies $\sin(\pi-4x)=\sin 3x$
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
NOTE: All this in hindsight knowing that n should be 7. To derive it in an Olympiad is close to impossible for me.
I got to $n=7$ fairly quickly, just by calculating the diagonal lengths for n=6 and n=8 and plugging them into the expression
$\displaystyle\frac{1}{|A_1A_2|}-\frac{1}{|A_1A_3|}-\frac{1}{|A_1A_4|}$
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.

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Re: an old maths olympiad question

Post by MuthuVeerappanR » 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.
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