616 Geometry and Trigonometry
=
(
−
1
x 1 x 2 x 3 x 4+
1
x^21 x 22 x^23 x 42)
∣
∣∣
∣∣
∣∣
∣
x^31 x 12 x 11
x^32 x 22 x 21
x^33 x 32 x 31
x^34 x 42 x 41∣
∣∣
∣∣
∣∣
∣
.
One of the factors is a determinant of Vandermonde type, hence it cannot be 0. Thus the
other factor is equal to 0. From this we infer thatx 1 x 2 x 3 x 4 =1, which is what had to be
proved.
(A. Myller,Geometrie Analitica ̆(Analytical Geometry), 3rd ed., Editura Didactica ̧ ̆si
Pedagogica, Bucharest, 1972) ̆
598.Consider complex coordinates with the originOat the center of the circle. The
coordinates of the vertices, which we denote correspondingly byα, β, γ , δ, η, φ, have
absolute value|r|. Moreover, because the chordsAB,CD, andEFare equal to the
radius,∠AOB=∠COD=∠EOF=π 3. It follows thatβ=αeiπ/^3 ,δ=γeiπ/^3 , and
φ=ηeiπ/^3. The midpointsP,Q,RofBC,DE,FA, respectively, have the coordinates
p=1
2
(αeiπ/^3 +γ), q=1
2
(γ eiπ/^3 +η), r=1
2
(ηeiπ/^3 +α).We compute
r−q
p−q=
αeiπ/^3 +γ( 1 −eiπ/^3 )−η
α−γeiπ/^3 +η(eiπ/^3 − 1 )=
αeiπ/e−γe^2 iπ/^3 +ηe^3 iπ/^3
α−γeiπ/^3 +e^2 iπ/^3 η=eiπ/^3.It follows thatRQis obtained by rotatingPQaroundQby 60◦. Hence the triangle
PQRis equilateral, as desired.
(28th W.L. Putnam Mathematical Competition, 1967)
599.We work in complex coordinates such that the circumcenter is at the origin. Let
the vertices bea, b, con the unit circle. Since the complex coordinate of the centroid is
a+b+c
3 , we have to show that the complex coordinate of the orthocenter isa+b+c.By
symmetry, it suffices to check that the line passing throughaanda+b+cis perpendicular
to the line passing throughbandc. This is equivalent to the fact that the argument of
b−c
b+cis±
π
2. This is true because the vectorb+cis constructed as one of the diagonals of
the rhombus determined by the vectors (of the same length)bandc, whileb−cis the
other diagonal of the rhombus. And the diagonals of a rhombus are perpendicular. This
completes the solution.
(L. Euler)
600.With the convention that the lowercase letter denotes the complex coordinate of the
point denoted by the same letter in uppercase, we translate the geometric conditions from
the statement into the algebraic equations