Dhaka Higher Secondary 2011/10

Problem for Higher Secondary Group from Divisional Mathematical Olympiad will be solved here.
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BdMO
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Dhaka Higher Secondary 2011/10

Unread post by BdMO » Fri Jan 28, 2011 10:35 pm

A point $P$ is chosen inside a right angled triangle $ABC$, perpendicular lines $PS$, $PQ$ and $PR$ are drawn from $P$ on $AB, BC$ and $AC$. $PR = 1, PQ = 2$ and $PS = 3$ and $\angle RPS = 150^{\circ} $. The length of $ AB$ can be written in the form $x \sqrt y +z$ where $x, y, z$ are integers. Find $ x + y + z$.

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TIUrmi
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Re: Dhaka Higher Secondary 2011/10

Unread post by TIUrmi » Sun Jan 30, 2011 12:23 pm

$\triangle ABC$ = $\triangle APC$ + $\triangle BPC$ + $\triangle BPA$ = $\frac {1}{2}$ $AC.1$ + $\frac{1}{2}.BC.2$ + $\frac{1}{2}AB.3$= $\frac{1}{2}AB.BC$
So, $AB$ =$\frac {AC}{BC}$ + 2$\frac {BC}{BC}$ + 3$\frac{AB}{AC}$

By simple angle chasing we will get $\angle ACB$ = 60
= sec 60 +2 + 3 tan 60
= 2 + 2 + 3 $\sqrt {3}$
= 3 $\sqrt {3}$ + 4

$x$ +$y$ + $z$ = $10$
"Go down deep enough into anything and you will find mathematics." ~Dean Schlicter

ibrahim
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Location: Dhaka

Re: Dhaka Higher Secondary 2011/10

Unread post by ibrahim » Sat Sep 03, 2011 9:23 pm

I've got 8.
ABC=APC+BPC+BPA

1/2 AB*AC=1/2 AC*1+1/2*2*BC+1/2*3*AB

AB=1+ 2*BC/AC+3*AB/AC

=1+2cos60+3tan60

=1+1+3*(3)^(1/2)

so x+y+z=8


akashe shurjota othe nirobe

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Avik Roy
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Re: Dhaka Higher Secondary 2011/10

Unread post by Avik Roy » Sat Sep 03, 2011 10:34 pm

Ibrahim, it's expected that you use LATEX in this forum

And draw a diagram and you should have the explanation for why your answer is incorrect
"Je le vois, mais je ne le crois pas!" - Georg Ferdinand Ludwig Philipp Cantor

ibrahim
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Location: Dhaka

Re: Dhaka Higher Secondary 2011/10

Unread post by ibrahim » Sat Sep 03, 2011 10:48 pm

I am using on mobile, so i can't using latex. Plz explain....

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