Let A, B, and C be three points on the curve xy = 1 (which is a hyperbola). Prove that the orthocenter of triangle ABC also lies on the curve xy = 1.

Guest Apr 17, 2017

#1**+1 **

Let the points be A =(1/a,a), B = (1/b, b) and C = (1/c, c)

The slope between B and C = [c - b] / [1/c - 1/b] = [c - b] / [b - c] / bc = -bc

So..... the equation of a perpendicular line to BC passing through A will be

y = (1/bc) (x - 1/a) + a (1)

Likewise......the equation through B perpendicular to AC is

y = (1/ac)(x - 1/b) + b (2)

And the equation through C perpendicular to AB is

y = (1/ab)(x - 1/c) + c (3)

Setting (1) = (2) we have

(1/bc) (x - 1/a) + a = (1/ac)(x - 1/b) + b simplify

x/bc - 1/abc + a = x/ac - 1/abc + b

x/bc + a = x/ac + b

x ( 1/bc - 1/ac) = b - a

x [ ( a - b] / (abc) ] = b - a

x [ (a - b) / (abc) ] = - (a - b)

x = -abc

Sub this into (3)

y = (1/ab) (-abc - 1/c) + c

y = -abc/ ab - 1/abc + c

y = - c - 1/abc + c

y = -1/abc

So.....the orthocenter is at ( x, 1/x) = ( -abc, - 1/abc)

To prove this point is on the function xy = 1, we have

( x ) ( y ) =

(-abc) (-1/abc) =

(-1) (-1) = 1

As an aside, the triangle formed is obtuse, so the orthocenter will fall outside the triangle

Here is an example.....points A,B, C are on xy = 1 .....D is the orthocenter

CPhill
Apr 18, 2017