Jessore 2007 HS prob 6

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Labib
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Jessore 2007 HS prob 6

Unread post by Labib » Mon Dec 05, 2011 8:55 pm

Let $AOP$ be a triangle with $AO=2, OP=a, \angle AOP=90^{\circ}$ and $B$ be the midpoint of $OA$.
Find the value of $a$ for which $\angle APB$ be maximum.
Last edited by Labib on Mon Dec 05, 2011 11:17 pm, edited 1 time in total.
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Re: Jessore 2007 HS prob 6

Unread post by nafistiham » Mon Dec 05, 2011 10:50 pm

the value of $\angle APB$ will be the greatest when $a=AO$ because that is the length when $\angle APB= \angle BPO$.so,
\[a=2\]
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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Re: Jessore 2007 HS prob 6

Unread post by Labib » Mon Dec 05, 2011 11:32 pm

nafistiham wrote:the value of $\angle APB$ will be the greatest when $a=AO$ because that is the length when $\angle APB= \angle BPO$.so,
\[a=2\]
কিভাবে?

Trying it your way, $\angle APO =45^{\circ}$.
But $ \angle BPO=tan^{-1} \frac 1 2 >22.5^{\circ}$
and thus $\angle APB> \angle BPO$.

Let me make it easier. I know the answer(not the solution). It's $\sqrt 2$.

BTW, can't $\angle BPO > \angle APB$?
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Re: Jessore 2007 HS prob 6

Unread post by amlansaha » Sun Dec 18, 2011 12:28 am

$\angle APB = tan^{-1} (\frac{2}{a} ) - tan^{-1} (\frac{1}{a} ) =tan^{-1} \frac{\frac{2}{a}-\frac{1}{a}}{1+ \frac{2}{a^2}} = tan^{-1} \frac{a}{a^2+2}$
as, $\angle APB$ সূক্ষ্মকোণ(ইংলিশটা ভুলে গেছি), $\angle APB$ will be maximum if $\frac{a}{a^2+2}$ is maximum.
now, by differntiating $tan^-1 \frac{a}{a^2+2}$ twice, we get that for $a=\sqrt 2$ the second defferantial is negetive (সবার সেকেন্ড ডিফারেনশিয়াল নেগেটিভ হোক :D ).
so, $a=\sqrt 2$ makes $\angle APB$ maximum :) :) :)
Last edited by amlansaha on Sun Dec 18, 2011 1:32 pm, edited 2 times in total.
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Re: Jessore 2007 HS prob 6

Unread post by bristy1588 » Sun Dec 18, 2011 10:32 am

amlansaha wrote:$\angle APB = tan^{-1} (\frac{2}{a} ) - tan^{-1} (\frac{1}{a} ) =tan^{-1} \frac{\frac{2}{a}-\frac{1}{a}}{1+ \frac{2}{a^2}} = tan^{-1} \frac{a}{a^2+1}$
as, $\angle APB$ সূক্ষ্মকোণ(ইংলিশটা ভুলে গেছি), $\angle APB$ will be maximum if $\frac{a}{a^2+1}$ is maximum.
now, by differntiating $tan^-1 \frac{a}{a^2+1}$ twice, we get that for $a=\sqrt 2$ the second defferantial is negetive (সবার সেকেন্ড ডিফারেনশিয়াল নেগেটিভ হোক :D ).
so, $a=\sqrt 2$ makes $\angle APB$ maximum :) :) :)
$\angle APB = tan^{-1} (\frac{2}{a} ) - tan^{-1} (\frac{1}{a} ) =tan^{-1} \frac{\frac{2}{a}-\frac{1}{a}}{1+ \frac{2}{a^2}} = tan^{-1} \frac{a}{a^2+1}$
Ei Shutro tar general formula ki?
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Re: Jessore 2007 HS prob 6

Unread post by amlansaha » Sun Dec 18, 2011 12:49 pm

$tan^{-1} x \pm tan^{-1} y = tan^{-1} \frac{x \pm y}{1 \mp xy}$
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Re: Jessore 2007 HS prob 6

Unread post by bristy1588 » Sun Dec 18, 2011 1:23 pm

Amlan Da,
Thank you,. I think there it should be $a^2+2$ instead of $a^2+1$. I was stuck there, when i later realised this..
Bristy Sikder

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Re: Jessore 2007 HS prob 6

Unread post by amlansaha » Sun Dec 18, 2011 1:33 pm

:oops: :oops: :oops: thanx bristy. i have edited it :)
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Re: Jessore 2007 HS prob 6

Unread post by Labib » Sun Dec 18, 2011 1:37 pm

Trig Rocks!! :p
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Re: Jessore 2007 HS prob 6

Unread post by bristy1588 » Sun Dec 18, 2011 1:53 pm

Amlan Da,
Can you tell me where to get laws of trigonometry like this?
Bristy Sikder

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