Consider a convex curve in the plane. Let B and C be any two points on it, and let A be the intersection of the tangent to the curve at B and C.

I would like to show, without calculus, that $AB + AC > BC$.

With calculus, it does not seem too bad. I assume we can rotate B and C around A to B' and C' so that, at the point on the curve $B'C'$ below A, the slope of is zero. Call it D.

enter image description here

In the diagram,

$AB' = \int_y^z \sqrt{1+slope(AB')^2} dx$

$B'D = \int_z^w \sqrt{1+slope(B'C')^2} dx$

$AC' = \int_y^z \sqrt{1+slope(AC')^2} dx$

$C'D = \int_z^w \sqrt{1+slope(B'C')^2} dx$

The absolute value of the slope of $B'C'$ is decreasing from the slope of $AB'$ from B' to D, and negative but increasing (up to the absolute value of the slope of $AC'$) from D to C'

This is a general case of this question, which helped me when trying to place an upper limit on Pi.

  • $\begingroup$ I don't see how you can avoid calculus when you talk about tangent segments and curve lengths. $\endgroup$ – TonyK Jan 31 at 23:34

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