Advanced book on Mathematics Olympiad

354 Methods of Proof

This is equivalent to

cos

α 1 +β 1 +α 2

2

(

sin

α 1 +α 2

2

sin

β

2

−sin

α 1 +β

2

sin

α 2

2

)

=sin

α 1

2

(

sin

β

2

cos

α 1 +β 1

2

−sin

α 2

2

cos

α 1 +α 2

2

)

,

or

cos

α 1 +α 2 +β

2

(

cos

α 1 +α 2 −β

2

−cos

α 1 −α 2 +β

2

)

=sin

α 1

2

(

sin

(

β 1 +

α 1

2

)

−sin

(

α 2 +

α 1

2

))

.

Using product-to-sum formulas, both sides can be transformed into cos(α 1 +α 2 )
+cosβ 1 −cos(α 1 +β 1 )−cosα 2.

Figure 57

The case of a general polygon follows from the particular case of the quadrilateral.
This is a consequence of the fact that any two dissections can be transformed into one
another by a sequence ofquadrilateral moves(Figure 57). Indeed, any dissection can be
transformed into a dissection in which all diagonals start at a given vertex, by moving
the endpoints of diagonals one by one to that vertex. So one can go from any dissection
to any other dissection using this particular type as an intermediate step. Since the sum
of the inradii is invariant under quadrilateral moves, it is independent of the dissection.

Second solution: This time we use the trigonometric identity

1 +

r

R

=cosX+cosY+cosZ.

We will check therefore that the sum of 1+rRiis invariant, whereriare the inradii of the
triangles of the decomposition. Again we prove the property for a cyclic quadrilateral
and then obtain the general case using the quadrilateral move. Using the fact that the sum
of cosines of supplementary angles is zero and chasing angles in the cyclic quadrilateral
ABCD, we obtain

cos∠DBA+cos∠BDA+cos∠DAB+cos∠BCD+cos∠CBD+cos∠CDB

=cos∠DBA+cos∠BDA+cos∠CBD+cos∠CDB