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Orthocenter question

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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
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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
edited by CPhill  Apr 18, 2017
edited by CPhill  Apr 18, 2017
edited by CPhill  Apr 18, 2017

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