Geometry and Trigonometry
573.This is the famous Jacobi identity. Identifying vectors with so( 3 )matrices, we
compute
−→u×(−→v ×−→w)+−→v ×(−→w×−→u)+−→w×(−→u×−→v)=U(VW−WV)−(V W−WV)U+V(WU−UW)−(W U−UW)V
+W(UV−VU)−(U V−VU)W
=UVW−UWV−VWU+WVU+VWU−VUW−WUV+UWV
+WUV−WVU−UVW+VUW.All terms of the latter sum cancel, giving the answer zero.
574.One checks easily that−→u +−→v +−→w=0;hence−→u,−→v,−→wform a triangle. We
compute
−→u·−→c =(−→b ·−→c)(−→a ·−→c)−(−→c ·−→a)(−→b ·−→c)= 0.It follows that−→u and−→c =0 are orthogonal. Similarly, we prove that−→v is orthogonal
to−→a, and−→wis orthogonal to−→
b. Hence the sides of the triangle formed with−→u,−→v,−→w
are perpendicular to the sides of the triangle formed with−→a,−→
b,−→c. This shows that
the two triangles have equal angles hence are similar, and we are done.
(Romanian Mathematical Olympiad, 1976, proposed by M. Chiri ̧ta) ̆
575.Multiply the second equation on the left by−→a to obtain
−→a ×(−→x ×−→b)=−→a ×−→c.Using the formula for the double cross-product, also known as thecab-bacformula, we
transform this into(−→a ·−→
b)−→x −(−→a ·−→x)−→
b =−→a ×−→c.