Suppose $a$, $b$, $c$ are the lengths of the three sides of a triangle. When studying rational Bézier curves inscribed in the triangle, a mysterious quantity $w$ emerges in the algebra: $$ w = \frac{c}{\sqrt{2(a^2 + b^2)}} $$ I'd like to get a better understanding of $w$. Specifically:

  1. Does it represent anything, geometrically? A ratio of lengths or areas, maybe, or some trig function of some angle??
  2. Can we say anything about its range of values? For example, is it true that $w \le 1$?

A couple of special cases I already figured out:

  1. If $a=b$ (the triangle is isosceles), then $w$ is the sine of the base angle (the angle between the sides of lengths $a$ and $c$).
  2. If the sides of length $a$ and $b$ form a right angle, then $w^2 = \tfrac12$.

It looks as if $w^2$ is more tractable. So $$2w^2=\frac{c^2}{a^2+b^2}$$ and $$1-2w^2=\frac{a^2+b^2-c^2}{a^2+b^2}=\frac{2ab\cos C}{a^2+b^2}.$$ For fixed $a$ and $b$ as $C$ varies, this varies between $\pm2ab/(a^2+b^2)$. So $2w^2$ varies between $|a-b|^2/(a^2+b^2)$ and $|a+b|^2/(a^2+b^2)$. Thus $$2w^2\le\frac{a^2+2ab+b^2}{a^2+b^2}\le2$$ by AM/GM so indeed $w\le1$.

I don't see any nice geometry so far.

  • $\begingroup$ Sorry to be dense, but I don't see how you got the last inequality. How does $AM \ge GM$ allow us to conclude that the fraction is $\le 2$. $\endgroup$ – bubba Apr 30 '17 at 8:24
  • $\begingroup$ $2ab\le a^2+b^2$ is AM/GM $\endgroup$ – Lord Shark the Unknown Apr 30 '17 at 8:55
  • $\begingroup$ OK. I thought AM/GM said $\sqrt{ab} \le \tfrac12 (a+b)$, so $2ab \le \tfrac12(a+b)^2$. But $2ab \le a^2 + b^2$ is obvious, too, regardless of what we call it. $\endgroup$ – bubba Apr 30 '17 at 9:44

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