BdMO National 2013: Secondary 7

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BdMO
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BdMO National 2013: Secondary 7

Unread post by BdMO » Fri Jan 10, 2014 1:38 am

$ABCD$ is s quadrilateral. $AB||CD$. $P$ is a point on $AB$ and $Q$ is a point on $CD$. A line parallel to $AB$ intersects $AD$, $BC$, $DP$, $CP$, $AQ$, $BQ$ at points $M, N, X, Y, R, S$ respectively. Prove that $MX+NY=RS$.

Siam
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Re: BdMO National 2013: Secondary 7

Unread post by Siam » Mon Jan 13, 2014 3:43 pm

Is the question correct? Will it be $MX+NY=RS$ or $MY+NX=RS$ ?

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*Mahi*
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Re: BdMO National 2013: Secondary 7

Unread post by *Mahi* » Tue Jan 14, 2014 2:50 pm

There was a printing mistake in last year's question, $CP , DP$ would be $DP, CP$. It's been edited accordingly now, thanks for pointing out :)
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Siam
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Re: BdMO National 2013: Secondary 7

Unread post by Siam » Tue Jan 14, 2014 4:55 pm

Since $MX \| AP$, $ \triangle MXD \sim \triangle APD$
Similarly, $ \triangle RSQ \sim \triangle AQB$ and$ \triangle NYX \sim \triangle NPB$
Let the perpendicular distance between $AB$ and $CD$ be $h$ and $MN$ to $CD$ be $i$
Now, we have, $[APD]+[PBC]=.5(h \times AP)+(h \times PB)=.5 \times h\times (AP+BP)=[AQB]$
Let $\frac{h}{i} =k$
So we have, $[APD]=k^{2} [MXD]$, $[AQB]=k^{2} [RSQ]$, $[PBC]=k^{2} [NCY]$

So, $k^{2} [MXD]+k^{2} [NYC]=k^{2} [RSQ]$
Or, $[MXD]+[NYC]=[RSQ]$
Or, $i \times MX+i \times NY=i \times RS$
Or, $MX+NY=RS$

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