Advanced book on Mathematics Olympiad

232 4 Geometry and Trigonometry

tanx=

sin 20◦

2 cos 40◦−cos 20◦

.

The tangent function is one-to-one on the interval( 0 , 90 ◦), which implies that the solution
to the original equation is unique. 

Example.

(a) Prove that if cosπa=^13 thenais an irrational number.
(b) Prove that a regular tetrahedron cannot be dissected into finitely many regular tetra-
hedra.

Solution.(a) Assume thatais rational,a=mn. Then cosnaπ=±1. We will prove by
induction that for allk>0, coskaπ=m 3 kk, withmkan integer that is not divisible by 3.
This will then contradict the initial assumption.
The property is true fork=0 and 1. The product-to-sum formula for cosines gives
rise to the recurrence

cos(k+ 1 )aπ=2 cosaπcoskaπ−cos(k− 1 )aπ, k≥ 1.

Using the induction hypothesis, we obtain cos(k+ 1 )aπ=m 3 kk++ 11 , withmk+ 1 = 2 mk−
3 mk− 1. Sincemkis not divisible by 3, neither ismk+ 1 , and the claim is proved.
Part (b) is just a consequence of (a). To see this, let us compute the cosine of the
dihedral angle of two faces of a regular tetrahedronABCD.IfAHis an altitude of the
tetrahedron andAEis an altitude of the faceABC, then∠AEHis the dihedral angle of the
facesABCandBCD(see Figure 35). In the right triangleHAE, cosAEH=EHAD=^13.

E

C

B

H

D

A

Figure 35

Now assume that there exists a dissection of a regular tetrahedron into regular tetra-
hedra. Several of these tetrahedra meet along a segment included in one of the faces
of the initial tetrahedron. Their dihedral angles must add up toπ, which implies that
the dihedral angle of a regular tetrahedron is of the formπn, for some integern. This
was shown above to be false. Hence no dissection of a regular tetrahedron into regular
tetrahedra exists.