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Calculate the area of the spherical triangle defined by the points $(0, 0, 1)$, $(0, 1, 0)$ and $(\dfrac{1}{\sqrt{2}}, 0, \dfrac{1}{\sqrt{2}})$.

I have come up with this:

From the spherical Gauss-Bonnet Formula, where $T$ is a triangle with interior angles $\alpha, \beta, \gamma$. Then the area of the triangle $T$ is $\alpha + \beta + \gamma - \pi$.

How do I work out the interior angles in order to use this formula?

Any help appreciated.

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  • $\begingroup$ Do you know the spherical trig. formulas? $\endgroup$ – user416426 Feb 27 '17 at 12:58
  • $\begingroup$ The only thing that we have been given in our notes is the formula i have written above. @Noah $\endgroup$ – Anna Feb 27 '17 at 13:18
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$A(0, 0, 1)$, $B(0, 1, 0)$ and $C(\dfrac{1}{\sqrt{2}}, 0, \dfrac{1}{\sqrt{2}})$ with $|A|=|B|=|C|=1$ these point lie on unit sphere. These points specify three plane $x=0$, $y=0$ and $x=z$ then the angle between them are $\dfrac{\pi}{2}$, $\dfrac{\pi}{2}$ and $\dfrac{\pi}{4}$, since their normal vectors are $\vec{i}$, $\vec{j}$ and $\vec{i}-\vec{k}$, respectively (by $\cos\theta=\dfrac{u.v}{|u||v|}=u.v$).

At last $\sigma=\dfrac{\pi}{4}+\dfrac{\pi}{2}+\dfrac{\pi}{2}-\pi=\dfrac{\pi}{4}$.

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  • $\begingroup$ @JeanMarie Thanks for your comment. Please check my edit, if it's possible. $\endgroup$ – Nosrati Feb 27 '17 at 18:20
  • $\begingroup$ This time, not only I agree, but I find your solution very well adapted to the situation at hand. See the graphics I just added with a remark that is more or less in the same vein. $\endgroup$ – Jean Marie Feb 27 '17 at 18:43
  • $\begingroup$ @JeanMarie Thanks. $\endgroup$ – Nosrati Feb 27 '17 at 18:45
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Preliminary remark: (see figure below) It is visible that the area of spherical triangle $ABC$ is a half of the area of spherical triangle $ABD$, whose area is the eighth part of the area of the sphere with radius $r=1$ ( triangle $ABD$ is the part of the sphere situated in the positive orthant). Thus, the final result is:

$$\dfrac{1}{16}(4\pi r^2)=\dfrac{\pi}{4}.$$

As the objective is to use Gauss-Bonnet formula, more precisely Girard's theorem (http://www.princeton.edu/~rvdb/WebGL/GirardThmProof.html), instead of using rather opaque spherical trigonometry formulas, it is clearer IMHO to use (double) cross-products. Here is how:

Let $\vec{A}=(0,0,1), \vec{B}=(0,1,0), \vec{C}=(a,0,a)$ with $a=\dfrac{1}{\sqrt{2}}$.

all of them being on the unit sphere.

Cross product $\vec{C_1}=\vec{A} \times \vec{B}=(-1,0,0)$ is a normal vector to plane $OAB$.

In the same way, cross product $\vec{A_1}=\vec{B} \times \vec{C}=(a,0,-a)$ is a normal vector to plane $OBC$.

Thus, cross product $\vec{C_1} \times \vec{A_1}$ is equal to $(0,-a,0)$ with norm $\dfrac{1}{\sqrt{2}}$ is equal to $\|C_1\|\|A_1\|\sin(\pi-\beta)=1 * 1 * \sin(\beta)$.

Thus $\sin(\beta)=\dfrac{1}{\sqrt{2}}$, whence $\beta=\dfrac{\pi}{4}$.

Remark: in fact, $\beta=3\dfrac{\pi}{4}$ is forbidden because points $A,B,C$ are situated in the positive orthant. Had the elimination of ambiguity not been possible between $\dfrac{\pi}{4}$ and $3\dfrac{\pi}{4}$, we would have used the value of $\cos(\beta)$ obtained using the dot product $\vec{C_1}.\vec{A_1}=\|C_1\|\|A_1\|\cos(\beta)$.

The same process can be used to find angles $\alpha=\dfrac{\pi}{2}$ and $\gamma=\dfrac{\pi}{2}$ (see figure), and then use Girard's formula.

enter image description here

(this figure has been built with a Matlab program).

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  • $\begingroup$ Thank you for your help. I just wanted to ask how did you know $\vec{C_1}=\vec{A} \times \vec{B}$ and not $\vec{B}\times\vec{A}$ $\endgroup$ – Anna Feb 28 '17 at 10:51
  • $\begingroup$ Taking $\vec{B} \times \vec{A}$ would give an opposite normal vector (it is true that it provides an outward normal vector which is slightly more natural) but would not change the angle's value. $\endgroup$ – Jean Marie Feb 28 '17 at 10:59
  • $\begingroup$ I have slightly modified my answer, placing in front the remark I made at the end. $\endgroup$ – Jean Marie Feb 28 '17 at 11:00
  • $\begingroup$ Thank you. So I would need to work out $\vec{C_1} \times \vec{B_1}$ and $\vec{A_1} \times \vec{B_1}$. $\endgroup$ – Anna Feb 28 '17 at 11:04
  • $\begingroup$ Yes, exactly. A little tedious, but can be "computerized". $\endgroup$ – Jean Marie Feb 28 '17 at 11:07

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