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Given points $A$ and $B$, between lines $L_1$ and $L_2$ . Illustrate and carefully describe how you would find the shortest path from $A$ to $L_1$ to $L_2$ to $B$.

(The colored lines and points are my drawing) enter image description here

Here's my attempt (I know it's messy, but I still can't spot the mistake):

Let $A'$ be the reflection of $A$ through $L_1$, and similarly for $B'$. I claim that the shortest distance is given by $AX+XY+YB=A'X+XY+B'Y=A'B'$, where $X$ and $Y$ are the points where $\overline{A'B'}$ intercepts $L_1$ and $L_2$, respectively.

I must show that, for any other points $X'$ and $Y'$ on $L_1$ and $L_2$ respectively, we have $AX'+X'Y'+Y'B>A'B'$.

Let $X'$ be a point on $L_1$ not equal to $X$. In triangle $A'X'B'$, we have that $A'X'=AX'>A'B'+X'B'$. Let $Y'$ be a point on $L_2$ not equal to $Y$. In triangle $A'Y'B'$, we have that $Y'B'=Y'B>A'B'+Y'A'$. Therefore, $$AX'+Y'B>2A'B'+X'B'+Y'A' \Rightarrow AX'+Y'B+X'Y'>2A'B'+X'B'+Y'A'+X'Y'>A'B'$$

and this is what was to be shown.

My answer doesn't match the one provided in the book, though. So, what's wrong with my argument? Thanks in advance.

@edit

Here's the answer according to my book:

Reflect $A$ through line $L_1$ and reflect $B$ through line $L_2$. Let $X$ be the point where $\overline{A'B}$ intersects $L_1$ and $Y$ be the point where $\overline{AB'}$ intersects $L_2$. The path would be $A$ to $X$ to $Y$ to $B$.

enter image description here

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  • $\begingroup$ What is the answer in your book? $\endgroup$ – Aretino Apr 5 '18 at 16:43
  • $\begingroup$ I've added it the post. The book doesn't explain why that would be the shortest path though. $\endgroup$ – Sasaki Apr 6 '18 at 0:37
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    $\begingroup$ It seems they 'got the right idea' with using the reflection, then screwed up on the details. $\endgroup$ – Ingix Apr 6 '18 at 7:08
  • $\begingroup$ Your answer is right. $\endgroup$ – Aretino Apr 6 '18 at 8:40
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Your answer is correct, with the exception that in some cases the line segment $A'B'$ may not actually intersect with one of $L_1$ or $L_2$, in which case your argument breaks down. This is only possible if the angle between $L_1$ and $L_2$ of the sector in which $A,B$ lie is obtuse (maybe also possible if it is exatctly 90°, depending on if $A,B$ are allowed to be on $L_1, L_2$).

Similiarly, if that angle is bigger than 60°, it may be that $A'B'$ intersects both $L_1$ and $L_2$, but in the wrong order (so when going from $A'$ to $B'$ it hits $L_2$ first and $L_1$ second, which also breaks down your argument.

In those cases I think the correct answer is to go from A to the intersection point of $L_1$ and $L_2$ and then to B.

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