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Given the helicoid $$ \boldsymbol{r} = (u\sin v, u\cos v,v)$$ in three-dimensional Euclidean space, consider the triangle $T$ defined by $$ 0 \leq u \leq \sinh v, \qquad 0 \leq v \leq v_0.$$

The angles of the triangle are $\theta_1=\pi/4=\theta_2,\ \theta_3=\pi/2.$

Question: is the triangle geodesic? For a geodesic triangle, we have the equality (http://en.wikipedia.org/wiki/Gaussian_curvature#Total_curvature) $$ \sum_{i=1}^3 \theta_i - \pi = \int_T K \ \mathrm{d}A,$$ and given that the Gaussian curvature is $K=-1/(u^2+1)^2$ using the above data we get $$0 = \log\cosh v_0.$$ So the answer would seem to be NO: is that OK ?

jj_p
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    I don't know whether the sides of your triangle are geodesics on the helicoid; see the link given here: http://math.stackexchange.com/questions/262393/geodesic-of-helicoid . At any rate you can compute their geodesic curvature, and then Gauss' formula should hold true. – Christian Blatter Jun 25 '13 at 15:37
  • Thanks; so it is not a geodesic triangle, likely; can you expand on your last statement? – jj_p Jun 25 '13 at 15:54
  • Where you found the formula to compute $K$ in terms of the parameters there you should also find formulas for $\kappa_g$ along a curve $t\mapsto{\bf r}\bigl(u(t),v(t)\bigr)$. – Christian Blatter Jun 27 '13 at 09:11

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