area chasing, square, rhombus, symmetric (2018 Romanian NMO VII P2)

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area chasing, square, rhombus, symmetric (2018 Romanian NMO VII P2)

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#1 h Jun 3, 2020, 8:09 PM • 3 Y

Y by Mango247, Mango247, Mango247

In the square $ABCD$ the point $E$ is located on the side $[AB]$ , and $F$ is the foot of the perpendicular from $B$ on the line $DE$ . The point $L$ belongs to the line $DE$ , such that $F$ is between $E$ and $L$ , and $FL = BF$ . $N$ and $P$ are symmetric of the points $A , F$ with respect to the lines $DE, BL$ , respectively. Prove that:

a) The quadrilateral $BFLP$ is square and the quadrilateral $ALND$ is rhombus.
b) The area of the rhombus $ALND$ is equal to the difference between the areas of the squares $ABCD$ and $BFLP$ .

This post has been edited 1 time. Last edited by parmenides51, Aug 15, 2024, 7:41 AM

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#2 h Mar 22, 2025, 8:41 PM

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Given the square $ABCD\ :\ A(0,0),B(b,0),C(b,b),D(0,b)$ .
Choose $E(\lambda,0)$ .

The line $DE\ :\ y-b=-\frac{b}{\lambda}x$ intersects the line $BF\ :\ y=\frac{\lambda}{b}(x-b)$ in the point $F(\frac{b\lambda(b+\lambda)}{b^{2}+\lambda^{2}},\frac{b\lambda(\lambda-b)}{b^{2}+\lambda^{2}})$ .
$BF=FL$ and $L \in DE \Rightarrow L(\frac{2b^{2}\lambda}{b^{2}+\lambda^{2}},-\frac{b(b^{2}-\lambda^{2})}{b^{2}+\lambda^{2}})$ .
The line $DE$ intersects the line $AS \bot DE\ :\ y=\frac{\lambda}{b}x$ in the point $S(\frac{b^{2}\lambda}{b^{2}+\lambda^{2}},\frac{b\lambda^{2}}{b^{2}+\lambda^{2}})$ .
Then $N(\frac{2b^{2}\lambda}{b^{2}+\lambda^{2}},\frac{2b\lambda^{2}}{b^{2}+\lambda^{2}})$ .

$AL=LN=ND=DA=b$ and $AL // DN$ .
Area $ALND\ =\ b \cdot \frac{2b^{2}\lambda}{b^{2}+\lambda^{2}}$ , area $ABCD\ =\ b^{2}$ , area $BFLP\ =\ BF^{2}=(\frac{b(b-\lambda)}{\sqrt{b^{2}+\lambda^{2}}})^{2}$ .

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