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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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I am a little confused. You want the shortest path from A to D along the dashed line? Do you mean from A to E? –  N.U. Jun 20 '12 at 20:30
    
This result shows that light, whose path has equal angles of incidence and reflection, takes the shortest path when being reflected. :-) –  robjohn Jun 20 '12 at 20:41
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3 Answers

up vote 3 down vote accepted

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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I tried to check whether this question had a duplicate before posting this answer, but I couldn't find one. –  Rahul Jun 20 '12 at 20:31
    
I updated my question, silly me! –  N3buchadnezzar Jun 20 '12 at 20:32
    
@RahulNarain: I seem to remember answering a question like this before, but I haven't found it yet. –  robjohn Jun 20 '12 at 20:35
    
@robjohn, it certainly feels like a question that must have been asked before already. –  Rahul Jun 20 '12 at 20:37
    
Heh, we think very much alike. Our answers are almost identical and I just added a comment about the reflection of light to the question. :-) –  robjohn Jun 20 '12 at 20:42
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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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I was going to add one, but got distracted. Thanks! (+1). –  robjohn Jun 20 '12 at 21:13
    
@robjohn: Thanks for +1 . I believe that Visual answers are always more helpful. –  Mathlover Jun 20 '12 at 21:19
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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 $|BC|=\frac{60}{11}$.

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