Advanced book on Mathematics Olympiad

210 4 Geometry and Trigonometry

Example.LetABCandBCDbe two equilateral triangles sharing one side. A line
passing throughDintersectsACatMandABatN. Prove that the angle between the
linesBMandCNisπ 3.

Solution.In the complex plane, letBhave the coordinate 0, andCthe coordinate 1. Then
AandDhave the coordinateseiπ/^3 ande−iπ/^3 , respectively, andNhas the coordinate
teiπ/^3 for some real numbert.
The parametric equations ofNDandACare, respectively,

z=αteiπ/^3 +( 1 −α)e−iπ/^3 and z=βeiπ/^3 +( 1 −β), α,β∈R.

To find their intersection we need to determine the real numbersαandβsuch that

αteiπ/^3 +( 1 −α)e−iπ/^3 =βeiπ/^3 +( 1 −β).

Explicitly, this equation is

αt

1 +i

√

3

2

+( 1 −α)

1 −i

√

3

2

=β

1 +i

√

3

2

+( 1 −β).

Setting the real and imaginary parts equal, we obtain the system

αt+( 1 −α)=β+ 2 ( 1 −β),

αt−( 1 −α)=β.

By adding the two equations, we obtainα =^1 t. So the complex coordinate ofMis
eiπ/^3 +( 1 −^1 t)e−iπ/^3.
The angle between the linesBMandCNis the argument of the complex number

eiπ/^3 +

(

1 −^1 t

)

e−iπ/^3

teiπ/^3 − 1

=

(

eiπ/^3 +e−iπ/^3

)

−^1 te−iπ/^3

teiπ/^3 − 1

=

1 −^1 te−iπ/^3

teiπ/^3 − 1

=

1

t

e−iπ/^3.

The angle is thereforeπ 3 , as claimed.
During the Mathematical Olympiad Summer Program of 2006, J. Bland discovered
the following simpler solution:
Place the figure in the complex plane so that the coordinates ofA, B, C, Dare,
respectively,i

√

3,−1, 1, and−i

√

3. LetMChave length 2t, wheretis a real parameter
   (positive ifCis betweenAandMand negative otherwise). The trianglesMCDand
   NBDhave parallel sides, so they are similar. It follows thatBN=^2 t (positive ifBis
   betweenAandNand negative otherwise). The coordinates ofMandNare

m=−

(

1 +

1

t

)

−

1

t

i

√

3 and n=(t+ 1 )−ti

√

3.