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### Vectors of Parallelograms and Octagons

```
Date: 07/28/98 at 05:49:41
From: Ivan Li
Subject: Vectors

I cannot get started with the following problem. Hope you can help.

ABCDEFGH is a regular octagon and AB = p and BC = q. Express AH in
terms of p and q and show that:

AE + BH + CG + DF = 2(2 + sqrt2)(q - sqrt2*p)

Thank you.
```

```
Date: 07/28/98 at 17:05:04
From: Doctor Anthony
Subject: Re: Vectors

You must draw a fairly accurate figure so that you can draw in the
various vectors and see how they are related. We start by writing in a
few vectors; from these, others can be calculated:

AC = p+q
CD = sqrt(2)q - p

AE = AD + q - sqrt(2)p = q(2+sqrt(2)) - sqrt(2)p

BH = -p(1+sqrt(2)) + q
AH = -sqrt(2)p + q

CG = -q - p + AH + CD  (since HG = CD)
= -q - p - sqrt(2)p + q + sqrt(2)q - p
= -p(2 + sqrt(2)) + sqrt(2)q

DF = BH = -p(1+sqrt(2)) + q

So now we require:

AE + BH + CG + DF
= q(2+sqrt(2)) - sqrt(2)p - 2p(1+sqrt(2)) + 2q - p(2+sqrt(2)) +
sqrt(2)q

= 2q+sqrt(2)q - sqrt(2)p - 2p - 2sqrt(2)p + 2q - 2p - sqrt(2)p +
sqrt(2)q

= 4q - 4p + 2sqrt(2)q - 4sqrt(2)p

= q(4+2sqrt(2)) - p(4 + 4sqrt(2))

= 2q(2+sqrt(2)) - 2sqrt(2)p(sqrt(2) + 2)

= 2(2+sqrt(2))(q - sqrt(2)p)

- Doctor Anthony, The Math Forum
Check out our web site! http://mathforum.org/dr.math/
```

```
Date: 08/24/98 at 00:00:13
From: Ivan Li
Subject: vectors

I hope you can help me solve the following question.

ABCDEFGH is a regular octagon. The vector AB = p, the vector BC = q.
Express the vector AH in terms of p and q.

Thank you.
```

```
Date: 08/24/98 at 09:59:50
From: Doctor Anthony
Subject: Re: Vectors

If you start at A and follow a vector q from A you will be on a line
CH. Let the point on CH be J. Then JH is parallel to -p and since
triangle AJH is right-angled at A, the distance JH is p.sqrt(2). And
so we have:

AH = AJ + JH
= q - p sqrt(2)

- Doctor Anthony, The Math Forum
Check out our web site! http://mathforum.org/dr.math/
```

```
Date: 08/24/98 at 23:05:13
From: Ivan Li
Subject: Re: Vectors

I hope you can solve the following question for me:

The points P and R have position vectors 2a + b and a - 3b
respectively, relative to an origin O. Given that OPQR is a
parallelogram, express the vectors OQ and RP in terms of a and b.
By evaluating two scalar products, show that if OPQR is a square,
then  |a|^2 = 2|b|^2

I got vector OQ = 3a - 2b and vector RP = a + 4b. Unfortunately, I
cannot continue from here.

```

```
Date: 08/25/98 at 08:19:10
From: Doctor Anthony
Subject: Re: Vectors

I agree with you that:

OQ  = 3a - 2b
RP  =  a + 4b

We require to show that if OPQR is a square then |a|^2 = 2|b|^2.

If OPQR is a square then |2a+b| = |a-3b|, so squaring both sides:

4a^2 + 4ab + b^2 = a^2 - 6ab + 9b^2

3a^2 + 10ab - 8b^2 = 0           (Equation 1)

Also:

|3a-2b| = |a+4b|

9a^2 -12ab + 4b^2 = a^2 + 8ab + 16b^2

8a^2 -20ab - 12b^2 = 0

4a^2 - 10ab - 6b^2 = 0           (Equation 2)

7a^2 - 14b^2 = 0

a^2 = 2b^2

|a|^2 = 2|b|^2

You could also have done this problem using the fact that the scalar
product of OQ and RP is 0, since the diagonals of a square are at
right angles.

- Doctor Anthony, The Math Forum
Check out our web site! http://mathforum.org/dr.math/
```
Associated Topics:
High School Euclidean/Plane Geometry
High School Geometry
High School Linear Algebra
High School Triangles and Other Polygons

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