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Suppose there are two points, $p_1$ and $p_2$, inside a circle of radius $R$. You must travel from $p_1$ to $p_2$ but you must first "touch" a point on the circle before arriving at $p_2$. Assuming you always use the shortest path possible, can your path ever be longer than $2R$?

After trying a few examples it appears the answer is NO. But I'm finding it tough to prove this. Any help appreciated.

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  • $\begingroup$ I think that in your description of the problem you are using the word "circle" in two different senses. Let us say that "circle" means only the set of points at distance R from the center (meaning, only the circumference), and "disk" means set of points at distance at most R (meaning, the circumference + the inside). Then, what I think you mean is the $p_1$ and $p_2$ belong to the disk, but you must touch a point of the circle along the way. $\endgroup$ – Gabriel Nivasch Oct 15 '16 at 18:37
  • $\begingroup$ @Gabriel I have edited my OP so it says "inside" instead of "in". Hope this clarifies the situation. $\endgroup$ – Jens Oct 15 '16 at 18:39
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Extend the segment $\overline {p_1p_2}$ to the chord $\overline {AB}$ ordered as $\{A,p_1,p_2,B\}$ (though we might have $p_1=A$ or $p_2=B$ or both). Suppose $d(p_1,A)≤d(p_2,B)$. Then the path $p_1\to A\to p_2$ has length no greater than the length of $\overline {AB}$ which, in turn, has length no greater than a diameter.

Note: In general, this need not be the minimal satisfactory path.

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    $\begingroup$ Thank you for this very nice proof. $\endgroup$ – Jens Oct 15 '16 at 18:47

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