Problema O321. Each diagonal of the convex hexagon ABCDEF divides its
area in half. Prove that
AB 2 + CD2 + EF 2 = BC 2 + DE 2 + F A2
Proposed by Nairi Sedrakyan, Yerevan, Armenia
Solution by Ercole Suppa, Teramo, Italy
A
F
B
Q
E
P
D
C
We begin by proving the following preliminary results:
Claim 1. We have AC/DF = CE/F B = EA/BD.
Proof of Claim 1. Let P = AD ∩ CF . Observe that the areas of triangles
4AP F and 4DP C are equal because
[AP F ] = [ADEF ] − [DEF P ] = [CDEF ] − [DEF P ] = [DP C]
Therefore AP · P F = CP · P D, or AP/P D = CP/P F .
Thus 4AP C ∼ 4DP F (SAS) whence ∠CAP = ∠F DP and AC k DF .
In a similar way, we can prove that BF k CE and AE k BD, so that
4ACE ∼ 4DF B and the Claim 1 follows.
Claim 2. The three diagonals AD, BE, CF are concurrent.
Proof of Claim 2. Let P = AD ∩ CF , Q = AD ∩ BE. As proved in Claim
1 we have 4AP C ∼ 4DP F , so
AP
AC
=
PD
DF
(1)
In a similar way, we can prove that
AQ
EA
=
QD
BD
(2)
From (1) and (2), taking into account of Claim 1, we have AP/P D =
AQ/QD so that P = Q. Hence, the three diagonals AD, BE, CF concur
1
in P , as claimed.
Coming back to the proposed problem, note first that the pairs of triangles
4F AP and 4CDP , 4ABP and 4DEP , 4BCP and 4EF P are equivalent.
A
F
a
f
B
b


P



e
E
c
d
C
D
Therefore denoting by a, b, c, d, e, f the lengths of P A, P B, P C, P D, P E, P F
respectively, we have
a · f = c · d,
a · b = d · e,
b·c=e·f
(3)
Finally, putting α = ∠F P A, β = ∠AP B, γ = ∠BP C, the cosinus law yields
AB 2 + CD2 + EF 2 = a2 + b2 − 2ab cos β + c2 + d2 − 2cd cos α + e2 + f 2 − 2ef cos γ
BC 2 + DE 2 + F A2 = b2 + c2 − 2bc cos γ + d2 + e2 − 2de cos β + f 2 + a2 − 2f a cos α
The above equalities and (3) gives
AB 2 + CD2 + EF 2 = BC 2 + DE 2 + F A2
which is precisely what we want to prove.
2