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kjhgyuio   1
N Apr 21, 2025 by vanstraelen
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kjhgyuio
Apr 21, 2025
vanstraelen
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kjhgyuio
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#1 h  Apr 21, 2025, 8:27 AM
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vanstraelen
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#2 h  Apr 21, 2025, 2:10 PM
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Given the trapezium $ABCD\ :\ A(0,0),B(b,0),C(b,4),D(d,4)$, choose $F(b,\lambda)$.
Given $CD+DF=4 \Rightarrow b-d+\sqrt{(b-d)^{2}+(4-\lambda)^{2}}=4$ or $b-d=\frac{8\lambda-\lambda^{2}}{8}$.

The slope of the line $DF\ :\ m_{DF}=\frac{\lambda-4}{b-d}$,
then the equation of the line $FE\ :\ y-\lambda=\frac{b-d}{4-\lambda}(x-b)$.
This line intersects the x-axis in the point $E(b+\frac{\lambda(\lambda-4)}{b-d},0)$.

The perimeter $p=\lambda+b-(b+\frac{\lambda(\lambda-4)}{b-d})+\sqrt{\lambda^{2}+\frac{\lambda^{2}(\lambda-4)^{2}}{(b-d)^{2}}}$,
$p=\lambda-\frac{\lambda(\lambda-4)}{b-d}+\frac{\lambda}{b-d}\sqrt{(b-d)^{2}+(\lambda-4)^{2}}$,
$p=\lambda-\frac{\lambda(\lambda-4)}{b-d}+\frac{\lambda}{b-d}[4-(b-d)]$,
$p=\frac{-\lambda^{2}+8\lambda}{b-d}=8$.
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