A triangle has vertices at #A(a,b )#, #C(c,d)#, and #O(0,0)#. What are the endpoints and length of the perpendicular bisector of AC ?

2 Answers
May 22, 2018

Answer:

Hey, this got featured before it was really done. It's still not done.

Explanation:

This is another in my series where I generalize one of these old questions.

"The triangle's perpendicular bisectors" is not a term we hear that often. Each side has a perpendicular bisector, the perpendicular through the midpoint. It will occasionally intersect the opposite vertex (in which case it's the altitude of an isosceles triangle) but usually it will intersect exactly one of the other sides.

I've rewritten the question to put one vertex at the origin and to just ask for the perpendicular bisector of the opposite side.

The sides of the triangle are given parametrically as

#OA: quad quad # #(x,y) = v(a,b) quad quad quad quad 0 le v le 1#

#OC: quad quad # #(x,y) = w(c,d) quad quad quad quad 0 le w le 1#

#AC: quad quad # #(x,y) = (a,b)+u(c-a,d-b) quad quad quad quad 0 le u le 1#

The perpendicular family to #AC# has direction vector given by swapping coordinates and negating one; the bisector is through the midpoint of AC:

#(x,y) = 1/2 (a+c, b+d) + t(d-b,a-c) quad quad quad # for real #t#

That meets #OA # when

#v(a,b)= 1/2 (a+c, b+d) + t(d-b,a-c) #

Two equations in two unknowns.

#va = (a+c)/2+t(d-b)#

#vb = (b+d)/2+t(a-c)#

We're more interested in #v# than #t#:

# 2va(a-c) + 2t(b-d)(a-c) = (a+c)(a-c)#

#2vb(b-d) - 2t(a-c)(b-d) = ( b+d)(b-d)#

# v = 1/2 cdot { (a^2+b^2) - (c^2 + d^2) }/{a^2+b^2 -(ac+bd)} #

The roles aren't quite symmetrical so we repeat the process for the meet of the bisector and #OC#.

#w(c,d) = 1/ 2(a+c, b+d) + t(d-b,a-c)#

#2wc(a-c) = (a+c)(a-c) - 2t(b-d)(a-c) #

#2wd(b-d) = (b+d)(b-d) + 2t(a-c)(b-d)#

# w = 1/2 cdot {(a^2+b^2) - (c^2 +d ^2) }/{(ac+bd)-(c^2+d^2)}#

Yes I know this is getting long Socratic. It's an involved problem.

The numerators are the same in #v# and #w#. We need to show that either #v=w=0# (an isosceles triangle where the perpendicular bisector is an altitude, here to the origin) or exactly one of #0 < v < 1# or #0 < w <1.#

~~~~~~~

# #

We want to solve #v>0#

# { (a^2+b^2) - (c^2 + d^2) }/{a^2+b^2 -(ac+bd)} > 0#

Let's first assume # a^2+b^2 > c^2+d^2,# positive numerator, so we need a positive denominator.

# a^2+b^2 > ac+bd #

# |A|^2 > A cdot C = |A| \ | C| \cos AOC #

# |A| > |C| \ cos AOC #

We're assuming #|A|>|C|# and the cosine is never bigger than one so we've shown #|A|>|C| implies v>0.#

What about when # a^2+b^2 < c^2+d^2,# or #|A|<|C|#? For #v>0# we need a negative denominator.

# a^2+b^2 < ac+bd #

# |A|^2 < A cdot C = |A| \ | C| \cos AOC #

# |A| < |C| \ cos AOC #

# cos AOC > |A|/|C| #

How about #w>0#?

# w = 1/2 cdot {|A|^2-|C|^2}/{ A cdot C - |C|^2 }#

For #|A|>|C|# we need #A cdot C > |C|^2# or

#|A| \ |C| \ cos AOC > |C|^2 #

# cos AOC > |C|/|A| #

Unless #|A|=|C|# exactly one of those fractions #|C| /|A| or |A|/|C|# is in cosine range, between -1 and 1.

Still not there.

~~~~~~~~~

We need to show the denominators have opposite signs.

# ( (a^2+b^2) -(ac+bd) ) ( (ac+bd)-(c^2+d^2) ) #

#= (a^2+b^2 + c^2 + d^2)(ac+bd) + (ac+bd)^2 + (a^2+b^2-c^2-d^2)(ac+bd) #

#= (ac+bd)( a^2+b^2 + c^2 + d^2 + (ac+bd) + a^2+b^2-c^2-d^2) # =(ac+bd)( 2a^2 + 2b^2 + (ac+bd) )

~ ~~~~~~

The triangle inequality says

#|AC| > |OA| + |OC|#

#|AC|^2 > |OA|^2 + |OC|^2 + 2|OA| \ |OC|#

# (a-c)^2+(b-d)^2 > a^2+b^2+c^2+d^2+ 2sqrt{(a^2+b^2)(c^2 + d^2)} #

# (a-c)^2+(b-d)^2 -( a^2+b^2+c^2+d^2) > 2sqrt{(a^2+b^2)(c^2 + d^2)} #

#-2 ac -2bd > 2sqrt{(a^2+b^2)(c^2 + d^2)} #

#ac-bd > 2sqrt{(a^2+b^2)(c^2 + d^2)} #

The dot product and the sum of squares have a close relationship. We expand #|AC|^2:#

# |AC|^2 = (a-c)^2+(b-d)^2 = a^2+b^2 + c^2 + d^2 - 2(ac+bd) #

# (ac+bd) - (c^2+d^2) = (a^2+b^2)- (ac+bd) - |AC|^2 #

# w = 1/2 cdot {|A|^2+|C|^2}/{ |A|^2 - Acdot C - |AC|^2 }#

# v = 1/2 cdot {|A|^2+|C|^2 }/{|A|^2 -A cdot C) #

We still haven't shown we pass through at most one side or the vertex. It remains to show #-1 < v < 1# which ensures that the bisector passes through exactly one side.

Still more to do.

Jun 3, 2018

Answer:

#t=(a-c)^2+(b-d)^2#

#p = 1/t ( b(b+d)-a(a+c))#

#q = 1/t (ad-bc)#

If #ad-bc = 2t,# i.e #q=1/2,# the bisector goes through #O#.

If #q<1/2# the bisector intersects OA at# ((x),(y))=((c-a,b-d),(d-b,c-a)) (( 1/2 - p),( q/{2p} - q )) #

If #q>1/2# the bisector intersects OB at # ((x),(y))=((c-a,b-d),(d-b,c-a)) (( 1/2 - p),( q/{2-2p} - q )) #

Explanation:

My other answer got featured before I finished it. I'm looking for a simpler way that I can actually finish.

Let's start with a simpler problem. Triangle OPI, coordinates #O(0,0), P(p,q), I(1,0)#. What are the endpoints and lengths of the perpendicular bisector of # OI?#

Solution: The foot of the bisector is the midpoint of #OI,# #F(1/2, 0).# There are three possibilities:

If #q=1/2# the bisector goes through #P#.

If #q <1/2# the bisector intersects OP at #Q(1/2 , q/{2p}).# That's the intersection of #x=1/2# with #py=qx.#

If #q>1/2# the bisector intersects IP #(p-1)y=q(x-1)# at #(1/2, q/{2-2p})#

That seems simple enough. We want to map #AOC# to # OPI#. That's #A(a,b) to (0,0),# #C(c,d) to (1,0)# and #O(0,0) to P(p,q)#

The general rotation, scaling, translation is

#x' = rx + sy + p#

#y' =-sx + r y + q #

We mapped O to P. Unknowns #p,q,r,s# The remaining equations are

#0 = ar + bs + p#

#0 = -as + br + q #

#1 = cr + ds + p#

#0 = -sc + rd + q #

Let #t=(a-c)^2+(b-d)^2.# We know that squared distance is one in the translated space, so that's the denominator. I'll skip the tedium.

#p = 1/t ( b(b+d)-a(a+c))#

#q = 1/t (ad-bc)#

# r = 1/t(c-a)#

# s = 1/t(d-b)#

The remaining difficult issue is the inverse mapping so we can get the other endpoints #Q(1/2 , q/{2p})# and #Q'(1/2 , q/{2-2p}).# Sorry to devolve into matrices for a minute.

#((x' - p),( y' - q )) = ((r,s),(-s,r))((x),(y))#

The almost inverse is

#((r,s),(-s,r)) ((r,-s),(s,r))=((r^2+s^2, 0),(0, r^2+s^2))=1/t((1,0),(0,1))#

# ((x),(y)) = t ((r,-s),(s,r)) ((x' - p),( y' - q )) #

# ((x),(y)) = ((c-a,b-d),(d-b,c-a)) ((x' - p),( y' - q )) #

#Q(1/2 , q/{2p})# maps to # ((c-a,b-d),(d-b,c-a)) (( 1/2 - p),( q/{2p} - q )) #

#Q(1/2 , q/{2-2p})# maps to # ((c-a,b-d),(d-b,c-a)) (( 1/2 - p),( q/{2-2p} - q )) #

Transforming our original solution,

#t=(a-c)^2+(b-d)^2#

#p = 1/t ( b(b+d)-a(a+c))#

#q = 1/t (ad-bc)#

If #ad-bc = 2t,# i.e #q=1/2,# the bisector goes through #O#.

If #q<1/2# the bisector intersects OA at# ((x),(y))=((c-a,b-d),(d-b,c-a)) (( 1/2 - p),( q/{2p} - q )) #

If #q>1/2# the bisector intersects OB at # ((x),(y))=((c-a,b-d),(d-b,c-a)) (( 1/2 - p),( q/{2-2p} - q )) #