Return to Answer

A solution is in effect an arc-length parametrization of a space curve. Let $\vec v =(x,y,z) = (\sin \alpha, \sin \beta, \sin \gamma)$. The first equation is then the somewhat complicated algebraic surface, call it $S_1$: $$ S_1: 2(y^2z^2+z^2x^2+x^2y^2) - (x^4+y^4+z^4) = 4(xyz)^2. $$ More precisely, it's a quarter of $S_1$, because $S_1$ also contains the loci where one of $\alpha,\beta,\gamma$ is the sum of the other two. You may recognize the left-hand side from Hero(n)'s formula: it factors as $(x+y+z)(-x+y+z)(x-y+z)(x+y-z)$. The second equation then intersects this $S_1$ with the sphere $\Sigma_c: \|\vec v\| = c$. The final equation says that $\vec v$ depends on $t$ and its derivative has norm $1$. This makes $\vec v(t)$ the arc-length parametrization of the curve.

The following might give you a handle on what happens for small $c$. Scaling by $c$ yields the intersection of the unit sphere $\Sigma_1$ with the varying surface $$ S_c: (x+y+z)(-x+y+z)(x-y+z)(x+y-z) = 4c^2(xyz)^2. $$ Now $S_0 \cap \Sigma_1$ is the union of the four great circles $\{ x \pm y \pm z = 0 \} \cap \Sigma_1$, each of which has an easy arc-length parametrization. The one that corresponds o $\alpha + \beta + \gamma = 0$ is $x+y+z=0$. To get at your problem with small $c$, you might start from these parametrizations of $S_0 \cap \Sigma_1$ and consider the curves $S_c \cap \Sigma_1$ as deformations of that great circles, then at the end speed up the resultion arc-length parametrizations by a factor $1/c$ to undo the scaling.

A solution is in effect an arc-length parametrization of a space curve. Let $\vec v =(x,y,z) = (\sin \alpha, \sin \beta, \sin \gamma)$. The first equation is then the somewhat complicated algebraic surface, call it $S_1$: $$ S_1: 2(y^2z^2+z^2x^2+x^2y^2) - (x^4+y^4+z^4) = 4(xyz)^2. $$ More precisely, it's a quarter of $S_1$, because $S_1$ also contains the loci where one of $\alpha,\beta,\gamma$ is the sum of the other two. You may recognize the left-hand side from Hero(n)'s formula: it factors as $(x+y+z)(-x+y+z)(x-y+z)(x+y-z)$. The second equation then intersects this $S_1$ with the sphere $\Sigma_c: \|\vec v\| = c$. The final equation says that $\vec v$ depends on $t$ and its derivative has norm $1$. This makes $\vec v(t)$ the arc-length parametrization of the curve.

The following might give you a handle on what happens for small $c$. Scaling by $c$ yields the intersection of the unit sphere $\Sigma_1$ with the varying surface $$ S_c: (x+y+z)(-x+y+z)(x-y+z)(x+y-z) = 4c^2(xyz)^2. $$ Now $S_0 \cap \Sigma_1$ is the union of the four great circles $\{ x \pm y \pm z = 0 \} \cap \Sigma_1$, each of which has an easy arc-length parametrization. The one that corresponds to $\alpha + \beta + \gamma = 0$ is $x+y+z=0$. To get at your problem with small $c$, you might start from these parametrizations of $S_0 \cap \Sigma_1$ and consider the curves $S_c \cap \Sigma_1$ as deformations of that great circle, then at the end speed up the resulting arc-length parametrizations by a factor $1/c$ to undo the scaling.

A solution is in effect an arc-length parametrization of a space curve. Let $\vec v =(x,y,z) = (\sin \alpha, \sin \beta, \sin \gamma)$. The first equation is then the somewhat complicated algebraic surface, call it $S_1$: $$ S_1: 2(y^2z^2+z^2x^2+x^2y^2) - (x^4+y^4+z^4) = 4(xyz)^2. $$ More precisely, it's a quarter of $S_1$, because $S+1$ also contains the loci where one of $\alpha,\beta,\gamma$ is the sum of the other two. You may recognize the left-hand side from Hero(n)'s formula: it factors as $(x+y+z)(-x+y+z)(x-y+z)(x+y-z)$. The second equation then intersects this $S$ with the sphere $\Sigma_c: \|\vec v\| = c$. The final equation says that $\vec v$ depends on $t$ and its derivative has norm $1$. This makes $\vec v(t)$ the arc-length parametrization of the curve.

The following might give you a handle on what happens for small $c$. Scaling by $c$ yields the intersection of the unit sphere $\Sigma_1$ with the varying surface $$ S_c: (x+y+z)(-x+y+z)(x-y+z)(x+y-z) = 4c^2(xyz)^2. $$ Now $S_0 \cap \Sigma_1$ is the union of the four great circles $\{ x \pm y \pm z = 0 \} \cap \Sigma_1$, each of which has an easy arc-length parametrization. The one that corresponds o $\alpha + \beta + \gamma = 0$ is $x+y+z=0$. To get at your problem with small $c$, you might start from these parametrizations of $S_0 \cap \Sigma_1$ and consider the curves $S_c \cap \Sigma_1$ as deformations of that great circles, then at the end speed up the resultion arc-length parametrizations by a factor $1/c$ to undo the scaling.

A solution is in effect an arc-length parametrization of a space curve. Let $\vec v =(x,y,z) = (\sin \alpha, \sin \beta, \sin \gamma)$. The first equation is then the somewhat complicated algebraic surface, call it $S_1$: $$ S_1: 2(y^2z^2+z^2x^2+x^2y^2) - (x^4+y^4+z^4) = 4(xyz)^2. $$ More precisely, it's a quarter of $S_1$, because $S_1$ also contains the loci where one of $\alpha,\beta,\gamma$ is the sum of the other two. You may recognize the left-hand side from Hero(n)'s formula: it factors as $(x+y+z)(-x+y+z)(x-y+z)(x+y-z)$. The second equation then intersects this $S_1$ with the sphere $\Sigma_c: \|\vec v\| = c$. The final equation says that $\vec v$ depends on $t$ and its derivative has norm $1$. This makes $\vec v(t)$ the arc-length parametrization of the curve.

The following might give you a handle on what happens for small $c$. Scaling by $c$ yields the intersection of the unit sphere $\Sigma_1$ with the varying surface $$ S_c: (x+y+z)(-x+y+z)(x-y+z)(x+y-z) = 4c^2(xyz)^2. $$ Now $S_0 \cap \Sigma_1$ is the union of the four great circles $\{ x \pm y \pm z = 0 \} \cap \Sigma_1$, each of which has an easy arc-length parametrization. The one that corresponds o $\alpha + \beta + \gamma = 0$ is $x+y+z=0$. To get at your problem with small $c$, you might start from these parametrizations of $S_0 \cap \Sigma_1$ and consider the curves $S_c \cap \Sigma_1$ as deformations of that great circles, then at the end speed up the resultion arc-length parametrizations by a factor $1/c$ to undo the scaling.

A solution is in effect an arc-length parametrization of a space curve. Let $\vec v =(x,y,z) = (\sin \alpha, \sin \beta, \sin \gamma)$. The first equation is then the somewhat complicated algebraic surface, call it $S$: $$ S_1: 2(y^2z^2+z^2x^2+x^2y^2) - (x^4+y^4+z^4) = 4(xyz)^2. $$ More precisely, it's a quarter of $S$, because $S$ also contains the loci where one of $\alpha,\beta,\gamma$ is the sum of the other two. You may recognize the left-hand side from Hero(n)'s formula: it factors as $(x+y+z)(-x+y+z)(x-y+z)(x+y-z)$. The second equation then intersects this $S$ with the sphere $\Sigma_c: \|\vec v\| = c$. The final equation says that $\vec v$ depends on $t$ and its derivative has norm $1$. This makes $\vec v(t)$ the arc-length parametrization of the curve.

The following might give you a handle on what happens for small $c$. Scaling by $c$ yields the intersection of the unit sphere $\Sigma_1$ with the varying surface $$ S_c: (x+y+z)(-x+y+z)(x-y+z)(x+y-z) = 4c^2(xyz)^2. $$ Now $S_0 \cap \Sigma_1$ is the union of the four great circles $\{ x \pm y \pm z = 0 \} \cap \Sigma_1$, each of which has an easy arc-length parametrization. The one that corresponds o $\alpha + \beta + \gamma = 0$ is $x+y+z=0$. To get at your problem with small $c$, you might start from these parametrizations of $S_0 \cap \Sigma_1$ and consider the curves $S_c \cap \Sigma_1$ as deformations of that great circles, then at the end speed up the resultion arc-length parametrizations by a factor $1/c$ to undo the scaling.

A solution is in effect an arc-length parametrization of a space curve. Let $\vec v =(x,y,z) = (\sin \alpha, \sin \beta, \sin \gamma)$. The first equation is then the somewhat complicated algebraic surface, call it $S_1$: $$ S_1: 2(y^2z^2+z^2x^2+x^2y^2) - (x^4+y^4+z^4) = 4(xyz)^2. $$ More precisely, it's a quarter of $S_1$, because $S+1$ also contains the loci where one of $\alpha,\beta,\gamma$ is the sum of the other two. You may recognize the left-hand side from Hero(n)'s formula: it factors as $(x+y+z)(-x+y+z)(x-y+z)(x+y-z)$. The second equation then intersects this $S$ with the sphere $\Sigma_c: \|\vec v\| = c$. The final equation says that $\vec v$ depends on $t$ and its derivative has norm $1$. This makes $\vec v(t)$ the arc-length parametrization of the curve.

The following might give you a handle on what happens for small $c$. Scaling by $c$ yields the intersection of the unit sphere $\Sigma_1$ with the varying surface $$ S_c: (x+y+z)(-x+y+z)(x-y+z)(x+y-z) = 4c^2(xyz)^2. $$ Now $S_0 \cap \Sigma_1$ is the union of the four great circles $\{ x \pm y \pm z = 0 \} \cap \Sigma_1$, each of which has an easy arc-length parametrization. The one that corresponds o $\alpha + \beta + \gamma = 0$ is $x+y+z=0$. To get at your problem with small $c$, you might start from these parametrizations of $S_0 \cap \Sigma_1$ and consider the curves $S_c \cap \Sigma_1$ as deformations of that great circles, then at the end speed up the resultion arc-length parametrizations by a factor $1/c$ to undo the scaling.