Skip to main content
Became Hot Network Question
added 11 characters in body
Source Link
domotorp
  • 18.7k
  • 3
  • 57
  • 125

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 $S$.

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 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 $S$.

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 $S$.

Source Link
domotorp
  • 18.7k
  • 3
  • 57
  • 125

Arcular triangle inequality

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 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 $S$.