Calculus: Analytic Geometry and Calculus, with Vectors

1.4 Distances, circles, and parabolas 33

(7) shows that ys cannot be an integer. This shows that if two vertices of an

equilateral triangle are lattice points, then the third vertex cannot be a lattice

point.

31 Let n be a positive integer. Let ml, m2, ,m,,, xi, x2,

yl, y2, , yn be numbers for which

(1)

Let

mi+m2+ ... +mn = M> O.

(2) z =

maxi + m2x2 + + m

Y

= 1y1 + m2y2 + + mnYn

M^1 M

For each k = 1, 2, , n, let rk be the distance from (x,y) to (xk,yk) and let

dk be the distance from (z,y) to (xk,yk). A timid person may be comforted by

the special case in which n = 4, ml = m2 = ma = m4 = 1, and the points (xk,yk)

are the vertices (1,1), (-1,1), (-1,-1), (1,-1) of a square. Confining attention

to the special case if this be deemed desirable, prove that

(3) mlri + m2r2 + ... +mnrn = M[(x - z)2 + (Y Y)2]

+ midi + m2d2 + + mndn.

With the aid of this result, let I be a constant and describe the set of points (x,y)

for which

(4) miri + m2r2 + + mnrn = I.

32 Let Pi, P2, P3 have coordinates (xi,yi), (x2y2), (xa,ya), respectively. The

triangle inequality

(1) (xa - x1)2 + (ya - yi)2 (x2 - x1)2 + (y2 - yi)2

+1/(xa-x2)2+(Ya-Y2)

says that the distance fromP1 toPs is less than or equal to the sum of the distance

from Pi to P2 and the distance from P2 to P3. In more advanced mathematics,

analytic proofs of (1) and more or less similar inequalities are very important.

Show that setting

(2) ai = X2 - X1, a2 = Y2 - Y1, bi = xa - x2, b2 = Ya - y2

puts (1) in the more agreeable form

(3) v (ai + bi)2 + (a2 + b2)2 < V ai -f- a2 } b1 b2.

By squaring and simplifying, show that (3) holds if

(4) laibi + aabaj < Vai + a2 bi b2.

By squaring and simplifying again, show that (4) holds if

(5) 0 < (aib2 - a2b1)2.