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.


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

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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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