BdMO National Junior 2020 P6

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Mursalin
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BdMO National Junior 2020 P6

Unread post by Mursalin » Thu Feb 04, 2021 12:25 am

\(ABCD\) বর্গের ভেতরে একটি বিন্দু \(P\) এমনভাবে নেওয়া হলো যেন \(AP+CP = 27, BP-DP = 17\) এবং \(\angle DAP = \angle DCP\) হয়। \(ABCD\) বর্গের ক্ষেত্রফল কত হবে?

Point \(P\) lies inside square \(ABCD\) such that \(AP+CP = 27, BP-DP = 17\) and \(\angle DAP = \angle DCP\). Compute the area of the square \(ABCD\).
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Enthurelxyz
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Re: BdMO National Junior 2020 P6

Unread post by Enthurelxyz » Tue Feb 09, 2021 8:21 pm

Draw two lines $l_1$ and $l_2$ on P such that they are perpendicular to $AD,BC$ and $AB,CD$ respectively. $l_1$ intersects $AD,BC$ at $E,F$ respectively and $l_2$ intersects $AB,CD$ at $G,H$ respectively. As, $\angle DAP = \angle DCP$

:arrow: $ \angle EAP=\angle HCP$

:arrow: $ \triangle AEP$ is similar to $\triangle CHP$. So, $\frac{AP}{CP}=\frac{EP}{HP}=\frac{EP}{ED}=tan(\angle EDP)$ . Again we can prove that $\triangle APG$ is similar to $\triangle PFC$

:arrow: $ \frac{AP}{CP}=\frac{PG}{PF}=\frac{PG}{PF}=\frac{PG}{GB}=tan(\angle PBG)$

:arrow: $ tan(\angle EDP)=tan(\angle PBG)$ as $\angle EDP, \angle PBG<90$

:arrow: $ \angle EDP=\angle PBG$. So, $\angle EDP=\angle PBG=45$ that means $A,P,D$ are collinear

:arrow: $ AP=CP=\frac{27}{2}$

We need to know the value of $(AG+GB)^2$. It is easy to compute that $AG^2+BG^2=\frac{27^2}{4}$ and $\sqrt{2}*AG-\sqrt{2} *BG=17$ so $AG-BG=\frac{17}{\sqrt{2}}$ so $(AG-BG)^2=\frac{17^2}{2}$ so $AG^2+BG^2-(AG-BG)^2=\frac{27^2}{4}-\frac{17^2}{2}$ so $2*AG*BG=\frac{27^2}{4}-\frac{17^2}{2}$.

So, $AG^2+BG^2+2*AG*BG=\frac{27^2}{4}+\frac{27^2}{4}-\frac{17^2}{2}=220=(AG+BG)^2$
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Mubin Hasan
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Re: BdMO National Junior 2020 P6

Unread post by Mubin Hasan » Fri Feb 19, 2021 11:31 am

Here point B, P and D are collinear.

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