## BdMO National Junior 2020 P6

Mursalin
Posts: 67
Joined: Thu Aug 22, 2013 9:11 pm

### BdMO National Junior 2020 P6

$$ABCD$$ 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
Posts: 17
Joined: Sat Dec 05, 2020 10:45 pm
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### Re: BdMO National Junior 2020 P6

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, $\angle DAP = \angle DCP$

$\angle EAP=\angle HCP$

$\triangle AEP$ 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 $\triangle PFC$

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

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

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

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

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
Posts: 2
Joined: Mon Oct 15, 2018 4:51 pm

### Re: BdMO National Junior 2020 P6

Here point B, P and D are collinear.