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I can solve it analytically, but I need some school-like elementary-geometry solution for this.

Let w = width and h = height of the rectangle ABCD. For any point E inside rectangle, except its boundary, let $f(E)=|AE| \cdot |EC| + |BE| \cdot |ED|$, question is: how to find point E inside this rectangle, except the points on the sides of this rectangle such that f(E) will be minimum. And find this minimum.

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It's a minimization of a continuous real valued function, so it really makes a lot of sense to solve it analytically. If there is a geometric solution, it would have to rely on some minimization principle. (None immediately spring to mind for me :( ) –  rschwieb Aug 29 '12 at 19:57
    
Oh, I guess I can think of one: "use straight lines" But who knows if that applies here... –  rschwieb Aug 29 '12 at 19:57
    
Are you looking for local minimum or a global minimum? –  no identity Aug 29 '12 at 20:02
    
Minimum $f(E)$ over all rectangle points. I think we should talk about global minimum –  user39023 Aug 29 '12 at 20:07

1 Answer 1

up vote 4 down vote accepted

Answer is min $f(E)=w \cdot h.$

Solution:

enter image description here

$E$ any point inside $ABCD$

  1. $2(S_{AEB}+S_{DEC})=|AB|\cdot |EH| +|DC|\cdot |EI|=|AB|\cdot|AD|=S_{ABCD}.$

  2. $S_{ABCD}=2(S_{AEF}+S_{FEB}).$

  3. For any triangle we have $S_\Delta=\frac{1}{2}a\cdot b \cdot \text{sin} \alpha \leq \frac{1}{2}a b \Rightarrow 2 S_\Delta \leq a b$. Because of $|\text{sin} \alpha| \leq 1 $.

$$ S_{ABCD}=2(S_{AEF}+S_{FEB}) \leq (|AE| \cdot |AF| + | BE| \cdot |BF|) $$

$$ S_{ABCD} \leq (|AE| \cdot |EC| + | BE| \cdot |ED|). $$

And finally

$$ \text{min} (|AE| \cdot |EC| + | BE| \cdot |ED|) = S_{ABCD} = w \cdot h. $$

An example of such point E: two circles with radius $|\frac{AB}{2}|$.

enter image description here

For this point E : $|AE| \cdot |EC| + | BE| \cdot |ED|=w \cdot h$

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Nice. Can you explain your example? why f(E)=wh in this case. –  user39023 Aug 29 '12 at 20:27
1  
Two right triangles $\Rightarrow 2(S_{DEC}+S_{AEB})=|AE|\cdot|EB|+|DE|\cdot|EC|$. As $AE=DE$ and $EB=EC$ we can write that $2(S_{DEC}+S_{AEB})=|AE|\cdot|EC|+|BE|\cdot|ED|=S_{ABCD}=w\cdot h$ –  Mike Aug 29 '12 at 20:35
    
Thanks to all, guys! –  user39023 Aug 29 '12 at 20:54

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