## BdMO National 2013: Junior 8

BdMO
Posts: 134
Joined: Tue Jan 18, 2011 1:31 pm

### BdMO National 2013: Junior 8

$ABCD$ is a rectangle where $AB= \sqrt{2}$ and $BC= \sqrt6$ . $P$ and $Q$ are two points on $AC$ such that $AP=CQ$. From $P$ and $Q$, two perpendicular $PR$ and $QS$ are drawn on $BC$ and $AD$ respectively. If $PRQS$ is a rhombus and $PR=CQ$, then find the length of $PQ$.

Tahmid
Posts: 110
Joined: Wed Mar 20, 2013 10:50 pm

### Re: BdMO National 2013: Junior 8

from pythagoras's theorem,
$AC^{2}=AB^{2}+BC^{2}$
$AC^{2}=\sqrt{2}^{2}+\sqrt{6}^{2}$
$AC^{2}=2+6=8$
$AC=\sqrt{8}=2\sqrt{2}$

now for $\Delta ABC\Leftrightarrow sin\angle ACB=\frac{AB}{AC}=\frac{\sqrt{2}}{2\sqrt{2}}=\frac{1}{2}$
$\therefore \angle ACB=30$

now for $\Delta PRC\Leftrightarrow sin 30=\frac{PR}{PC}=\frac{1}{2}\Leftrightarrow PC=2PR\Leftrightarrow PQ=PR$

$\therefore PQ=PR=CQ=AP$
$\therefore PQ=\frac{AC}{3}=\frac{2\sqrt{2}}{3}.......(ans:)$ ......

Labib
Posts: 411
Joined: Thu Dec 09, 2010 10:58 pm

### Re: BdMO National 2013: Junior 8

Hi Tahmid. Nice job solving the problem. Just an advice.
Many haven't solved the problem yet. So if you can, please "hide" the solution.
That way, other's fun will not be spoiled and they can always check your solution if they cannot solve it.
Please Install $L^AT_EX$ fonts in your PC for better looking equations,
Learn how to write equations, and don't forget to read Forum Guide and Rules.

"When you have eliminated the impossible, whatever remains, however improbable, must be the truth." - Sherlock Holmes

Tahmid
Posts: 110
Joined: Wed Mar 20, 2013 10:50 pm

### Re: BdMO National 2013: Junior 8

ok labib vaia ...... i'll try that .....