Return to Answer

Are you able to make a cube? If so, the vertices of a cube are included in the vertices of a regular dodecahedron. Then, you just need to add the remaining points locating them by the compass. Now, I shall not add anything else not to spoil the pleasure to solve this incredibly easy task... (PS: and of course, a cube is easily obtained starting from a tetrahedron, locating the laking vertices by a ruler. So, everything is reduced to the construction of a tetrahedron)

Summary: the details of the construction are described in various people's comments below. One needs a "spheric ruler" (able to draw the great circle passing for two non-antipodal given points) and a compass (C) (able to draw a circle with given center passing for a given point). One first draws three orthogonal great circles (the third one is tricky: see below), that is an octahedron; then one makes a cube, and then the dodecahedron.

Are you able to make a cube? If so, the vertices of a cube are included in the vertices of a regular dodecahedron. Then, you just need to add the remaining points locating them by the compass. Now, I shall not add anything else not to spoil the pleasure to solve this incredibly easy task... (PS: and of course, a cube is easily obtained starting from a tetrahedron, locating the laking vertices by a ruler. So, everything is reduced to the construction of a tetrahedron)

Summary: the details of the construction are described in various people's comments below. One needs a "spherical ruler" (able to draw the great circle passing for two non-antipodal given points) and a compass (C) (able to draw a circle with given center passing for a given point). One first draws three orthogonal great circles (the third one is tricky: see below), that is an octahedron; then one makes a cube, and then the dodecahedron.

Are you able to make a cube? If so, the vertices of a cube are included in the vertices of a regular dodecahedron. Then, you just need to add the remaining points locating them by the compass. Now, I shall not add anything else not to spoil the pleasure to solve this incredibly easy task... (PS: and of course, a cube is easily obtained starting from a tetrahedron, locating the laking vertices by a ruler. So, everything is reduced to the construction of a tetrahedron)

Summary: the compelte construction is described in various people's comments below. One needs a "spheric ruler" (able to draw the great circle passing for two non-antipodal given points) and a compass (C) (able to draw a circle with given center passing for a given point). One first draws three orthogonal great circles (the third one is tricky: see below), that is an octahedron; then one makes a cube, and then the dodecahedron.

Are you able to make a cube? If so, the vertices of a cube are included in the vertices of a regular dodecahedron. Then, you just need to add the remaining points locating them by the compass. Now, I shall not add anything else not to spoil the pleasure to solve this incredibly easy task... (PS: and of course, a cube is easily obtained starting from a tetrahedron, locating the laking vertices by a ruler. So, everything is reduced to the construction of a tetrahedron)

Summary: the details of the construction are described in various people's comments below. One needs a "spheric ruler" (able to draw the great circle passing for two non-antipodal given points) and a compass (C) (able to draw a circle with given center passing for a given point). One first draws three orthogonal great circles (the third one is tricky: see below), that is an octahedron; then one makes a cube, and then the dodecahedron.

Are you able to make a cube? If so, the vertices of a cube are included in the vertices of a regular dodecahedron. Then, you just need to add the remaining points locating them by the compass. Now, I shall not add anything else not to spoil the pleasure to solve this incredibly easy task... (PS: and of course, a cube is easily obtained starting from a tetrahedron, locating the laking vertices by a ruler. So, everything is reduced to the construction of a tetrahedron)

Summary: the compelte construction is described in various people's comments below. One needs a "spheric ruler" (able to draw the great circle passing for two non-antipodal points) and a compass (C) (able to draw a circle passing for a point with given center). One first makes three orthogonal great circles (the third one is tricky: see below), that is an octahedron; then one makes a cube, and then the dodecahedron.

Are you able to make a cube? If so, the vertices of a cube are included in the vertices of a regular dodecahedron. Then, you just need to add the remaining points locating them by the compass. Now, I shall not add anything else not to spoil the pleasure to solve this incredibly easy task... (PS: and of course, a cube is easily obtained starting from a tetrahedron, locating the laking vertices by a ruler. So, everything is reduced to the construction of a tetrahedron)

Summary: the compelte construction is described in various people's comments below. One needs a "spheric ruler" (able to draw the great circle passing for two non-antipodal given points) and a compass (C) (able to draw a circle with given center passing for a given point). One first draws three orthogonal great circles (the third one is tricky: see below), that is an octahedron; then one makes a cube, and then the dodecahedron.