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I have the following figure

ABC

Where $AB=10$m, $BD=12$m and $DE=12$m. The point C can slide along the segment BD. Now the problem is to minimize the distance from A to D going along the dashed line. The problem can be solved using simple analysis and differentiation. Let $BC=x$ then $\|ACE\| = f(x) = \sqrt{10^2-x^2}+\sqrt{(12-x)^2+12^2}$. in order to find a minima which we know exists one would have to solve $f'(x)=0$ to obtain the solution $x=60/11$.

My question is however can one prove without analysis that x=60/11 of $BD$ in order to minimize the distance $\|ACE\|$?

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    This result shows that light, whose path has equal angles of incidence and reflection, takes the shortest path when being reflected. :-) – 2012-06-20

3 Answers 3

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According to your math, you're minimizing $|AC|+|CE|$, not $|AC|+|CD|$.

Reflect $E$ about the line $BC$ to $E'$. Observe that $|CE| = |CE'|$, so the problem is equivalent to minimizing $|AC| + |CE'|$. Where should $C$ be? (Hint: The answer is not actually the midpoint.)

You can think of $ACE$ as a light ray reflected in the mirror $BD$. From one point of view, the solution to this problem is actually the reason why mirrors reflect light that way in the first place.

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    @robjohn: Heh. I guess this question is simple enough that the Anna Karenina principle applies: All correct answers are alike; every incorrect answer would be incorrect in its own way. – 2012-06-20
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Rahul Narain's and robjohn's answers explained details well how to solve. I just wanted to add a picture to show the steps of the answer. Reflect E across BD to E′. The shortest $|ACE'|$ should be line. It was shown as blue line in the picture below. Others are longer as you can see.

how to find the $|BC|=x$

$\triangle ABC$ is similiar to $\triangle E'DC $

Thus $\frac{10}{12}=\frac{x}{12-x}$ $x=\frac{60}{11}$

enter image description here

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    @robjohn: Thanks for +1 . I believe that Visual answers are always more helpful. – 2012-06-20
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You must mean $|ACE|$.

Reflect $E$ across $BD$ to $E'$, then $|CE|=|CE'|$ and because a straight line is the shortest distance between two points, The shortest distance from $A$ to $C$ to $E'$ is for $C$ to be $\frac{5}{11}$ of the way from $B$ to $D$ which would make $|BC|=\frac{60}{11}$.