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Proof of Interesting Geometric Vector Theorem

Date: 09/17/2005 at 09:50:25
From: James
Subject: Proof of a geometric theorem

O is the centre of the circumscribing circle of triangle ABC and H is 
its orthocentre.  Prove that vector OH is equal to the sum of the 
vectors OA, OB and OC.

Combining vectors OB and OC is OL where OL is a diagonal of the 
rhombus OCLB on OB and OC as adjacent sides.  Combining vectors OL and 
OA to give vector OH would complete the proof.  I am unable to do it.

I need to show that the quadrilateral AOLH is a parallelogram.  Now AH
is parallel to OL (both perpendicular to BC) but I cannot show that 
AH = OL which would prove AOLH is a parallelogram.  Nor can I show,
alternatively, that AO is parallel to HL.  Hope you can help.

Date: 09/23/2005 at 11:25:07
From: Doctor George
Subject: Re: Proof of a geometric theorem

Hi James,

I have come up with two proofs of your conjecture.  One uses analytic
geometry, and the other uses vector algebra.

Analytic Geometry Proof
Position the circumscribed circle with O at (0,0).  Now define A as
(a,b), B as (-a,b) and C as (c,d).  This convention is perfectly
general because any triangle can be positioned and oriented to meet
these conditions.

First, add the three vectors.

  OA + OB + OC = (c, d+2b)

Now construct the orthocenter as the intersection of two altitudes of
the triangle.  The altitude from C is on the line x = c.  The altitude
from B is on a line with slope m = (a-c)/(d-b).

To find the orthocenter we need the intersection of

  x = c


  y - b = [(a-c)/(d-b)](x+a)

If we substitute x = c we find that the y coordinate of the 
orthocenter is

  y = [(a-c)/(d-b)](c+a) + b

    = (a^2-c^2)/(d-b) + b

Now substitute a^2 = r^2 - b^2 and c^2 = r^2 - d^2, where 'r' is the
radius of the circumscribed circle.

  y = (r^2- b^2 - r^2 + d^2)/(d-b) + b

    = (d^2 - b^2)/(d-b) + b

    = [(d+b)(d-b)]/(d-b) + b

    = d + 2b

Therefore, the orthocenter is (c, d+2b) as was to be shown, so the
proof is complete.

Vector Algebra Proof
Let's start with the following observation.  For any vectors U and V,
if |U| = |V| then U+V is perpendicular to U-V.  We show this using the
dot product.

  (U+V).(U-V) = |U|^2 - U.V + V.U - |V|^2 = 0

Now for simplicity we will use different notation from the first
proof.  Let the vectors from the center of the circumscribed circle to
the vertices of the triangle be A, B and C.  Note that |A| = |B| = |C|
and vectors A-B, B-C and C-A represent the three sides of the
triangle.  We want to show that the orthocenter is the point A+B+C.

Now construct the orthocenter as the intersection of two altitudes of
the triangle.  Since B+C is perpendicular to B-C the altitude from A
is on the line with parametric equation

  L1 = A + a(B+C)

where 'a' is the parameter.  Also, since A+B is perpendicular to A-B
the altitude from C is on the line with parametric equation

  L2 = C + b(A+B)

where 'b' is the parameter.  When L1 and L2 intersect

  A + a(B+C) = C + b(A+B)

  a(B+C) = C - A + b(A+B)

Now take the dot product of both sides with A-B.  Note that this
causes 'b' to drop out, which allows us to solve for 'a'.

  a(B+C).(A-B) = (C-A).(A-B)

  a(B.A - |B|^2 + C.A - C.B) = C.A - C.B - |A|^2 + A.B

Since |B| = |A| the quantity in the parentheses is equal to the right
hand side.  It can be shown that this quantity is non-zero, so we can
divide by it to arrive at

   a = 1

The orthocenter is the value of L1 when a = 1.  Therefore, the
orthocenter is the point A+B+C as was to be shown, so the proof is

Thanks for writing.  This has been a fun problem.  Write again if you
need more help.

- Doctor George, The Math Forum 

Date: 09/25/2005 at 10:01:59
From: James
Subject: Thank you (Proof of a geometric theorem)

Dr. George - 

Thank you very much indeed for your elegant vector algebra solution
and also your solution by analytical geometry.  I can sleep much more
easily now that I know the conjecture/theorem can be proved true.

I came across the problem on page 32 of the Penguin Dictionary of
Curious and Interesting Geometry by David Wells, first published 1991.
It's well worth a read but it has given me a few headaches doing so!

Thanks again for your help and expertise.

- James

Date: 09/25/2005 at 21:34:25
From: Doctor George
Subject: Re: Thank you (Proof of a geometric theorem)

Hi James,

I'm glad that you were able to follow the proofs.  The vector algebra
proof can actually be shortened.  Instead of solving for 'a' as I did,
the system

  L1 = A + a(B+C)

  L2 = C + b(A+B)

can be solved by inspection to give a = 1 and b = 1.  We now have a
very concise proof for a theorem that is not intuitive.

Thanks for the book reference.  Since not many people read such books
it seems that math is one of your hobbies.  Write again anytime.

- Doctor George, The Math Forum 
Associated Topics:
College Geometry
College Linear Algebra

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