634 Geometry and Trigonometry
Recalling thats=x+y, we have two curves:x+y=1 and(x+y)^2 +x+y+ 1 − 3 xy=0.
The last equality is equivalent to
1
2[
(x−y)^2 +(x+ 1 )^2 +(y+ 1 )^2]
= 0 ,
i.e.,x=y=−1. Thus the curve in the problem consists of the linex+y=1 and the
point(− 1 ,− 1 ), which we will callA. PointsBandCare on the linex+y=1 such
that they are symmetric to one another with respect to the pointD(^12 ,^12 )and such that
BC
√ 3
2 =AD. It is clear that there is only one set{B, C}with this property, so we have
justified the uniqueness of the triangleABC(up to the permutation of vertices). Because
AD=
√(
1
2
+ 1
) 2
+
(
1
2
+ 1
) 2
=
3
2
√
2 ,
it follows thatBC=
√
6; hence Area(ABC)=^6√
3
4 =3
√
3
2.
(49th W.L. Putnam Mathematical Competition, 2006, proposed by T. Andreescu)624.View the parametric equations of the curve as a linear system in the unknownstn
andtp:
a 1 tn+b 1 tp=x−c 1 ,
a 2 tn+b 2 tp=y−c 2 ,
a 3 tn+b 3 tp=z−c 3.This system admits solutions; hence the extended matrix is singular. We thus have
∣∣
∣∣
∣∣a 1 b 1 x−c 1
a 2 b 2 y−c 2
a 3 b 3 y−c 3∣∣
∣∣
∣∣=^0.
This is the equation of a plane that contains the given curve.
(C. Ionescu-Bujor, O. Sacter,Exerci ̧tii ̧si probleme de geometrie analitica ̧si difer-
en ̧tiala(Exercises and problems in analytic and differential geometry), Editura Didactic ̆a
̧si Pedagogica, Bucharest, 1963) ̆
625.Let the equation of the curve bey(x). LetT(x)be the tension in the chain at the
point(x, y(x)). The tension acts in the direction of the derivativey′(x). LetH(x)and
V(x)be, respectively, the horizontal and vertical components of the tension. Because the
chain is in equilibrium, the horizontal component of the tension is constant at all points
of the chain (just cut the chain mentally at two different points). ThusH(x)=H. The
vertical component of the tension is thenV(x)=Hy′(x).