162 Mr. Davies Gilbert on the Regular or Platonic Solids. 



into one plane. The same would take place with four squares 

 and with six equilateral triangles. 



The solid angles or points of all solids bounded by regular 

 equal and similar sides, must then consist either of three, four, 

 or five equilateral triangles. Of three squares or tetragons, 

 or of three pentagons. And consequendy the number of these 

 finite bodies cannot exceed five. 



That five such actually exist, is thus proved. 



When three equilateral triangles meet in a point, the three 

 equal spherical angles must, to complete space, amount each 

 to one-third of four right angles, or to 120°. The three an- 

 gles therefore of each spherical triangle will equal four right 

 angles, or be two right angles above the two right angles of a 

 plane triangle. The solid angle therefore at the centre of 

 each spherical triangle will subtend one quarter part of the 

 surface of the sphere, and the solid must be a tetraedron. 



If four equilateral triangles meet at a point, the angle of 

 each spherical triangle must, for the same reason, be in this 

 case a right one. Consequently, the three angles of each sphe- 

 rical triangle will equal three right angles, or be one right angle 

 in excess of the three angles of a plane triangle. The solid 

 angle therefore, at the centre of each spherical triangle, will 

 subtend one-eighth part of the surface of the sphere, and the 

 solid must be an octaedron. 



If five equilateral triangles meet in a point, the angle of each 

 spherical triangle must be one-fifth of four right angles, or 

 72°; that is, four-fifths of one right angle : the three angles of 

 each spherical triangle will amount, therefore, to two right 

 angles and two-fifths of aright angle, exceeding the two right 

 angles of a plane triangle by two-fifths of a right angle, or 36°, 

 being the twentieth part of eight right angles ; the solid angle 

 therefore at the centre of each spherical triangle is one-twen- 

 tieth part of the spherical surface, and the solid must be the 

 icosaedron. 



If three squares or tetragons meet in a point, each spherical 

 angle must be 1 J right angle, and the four spherical angles of 

 the tetragon will amount to 5^ right angles, exceeding (2n— 4) 

 right angles, or the four right angles of a plane tetragon by 

 l£ right angle, or by one-sixth of eight right angles : the solid 

 must therefore be the hexaedron or cube. 



If three pentagons meet in a point, each spherical angle 

 must be 1^ right angle, as in the last instance; and the five 

 spherical angles of the pentagon will be 6§ right angles, ex- 

 ceeding the (2n — 4) right angles of the plane pentagon by f ds 

 of a right angle, or by T ^th of eight right angles. The solid 

 must therefore be the dodecaedron. 



The 



