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Is it true that if inside a circular segment $S$, with vertices $a$ and $b$, we take two circular arcs, one from $a$ to $c$ and the other from $c$ to $b$, then the sum of the lengths of these two arcs will be always at most the length of the arc from $a$ to $b$ on the boundary of the original circular segment?

Note that I do not require the two smaller arcs to have the same center, but I want both to be strictly contained in $S$.

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If the arcs bend the other way, then "yes"; otherwise there is an obvious counterexample.

Note that the statement holds if $c\in [a,b]$. Moreover, if the arcs $ac$ and $bc$ are tangent to the arc $ab$ at $a$ and $b$ respectively, then equality holds. Furthermore, note that we can assume that $ac$ and $bc$ are tangent to $ab$.

On the following diagram, if arcs look tangent, then they are.

enter image description here

Observe that $|ab|=|ac|+|cx|+|xb|\geqslant |ac|+|cb|$, where $|{*}{*}|$ denotes the length of the arc.

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  • $\begingroup$ I assume you mean that first you assume that the arcs are tangent, then claim that equality holds if $c\in [a,b]$. Does your solution also work if one or both of the arcs $ac$ and $cb$ bend the other way? $\endgroup$
    – domotorp
    Commented Oct 18, 2023 at 8:14
  • $\begingroup$ @domotorp I will fix it, thank you. If arcs can bend the other way, then the statement does not hold. Take $c$ on the arc $ab$ closer to $a$; choose arcs $ac$ that are tangent to the line segment $ab$ and let $bc$ be subarc of $ab$. $\endgroup$
    – Anton Petrunin
    Commented Oct 18, 2023 at 14:38
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    $\begingroup$ Indeed, thank you! Let me confess something. I knew a proof for the case when the arcs bend as in your solution, but it was a tedious computation, see page 17 here: doi.org/10.1007/BF02187821 Your solution is much nicer. $\endgroup$
    – domotorp
    Commented Oct 18, 2023 at 15:12

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