Line Segment AB....Challenge Question

[mathdad]

The point P is located on line segment AB in such a way that AB/AP = AP/AB. In words: P divides line segment AB into two segments such that the ratio of AB to the longer segment is equal to the ratio of the longer segment to the shorter segment.

(A) Show that this ratio is equal to (1/2)(1 + sqrt{5}).

Hint: Let AP = x and PB = y.

(B) The ratio is part (A) is denoted by k and

k = (1 + sqrt{5})/2

Verify that the number k satisfies the following algebraic properties.

1. k^2 = k + 1

2. k^3 = 2k + 1

3. k^(-1) = k - 1

4. k^(-2) = - k + 2

What on Earth is this all about?

 

[MarkFL]

I think the problem should state:

"The point P is located on line segment AB in such a way that AB/AP = AP/PB"

If we use the given hint, then we may state:

\(\displaystyle \frac{x+y}{x}=\frac{x}{y}\)

Or:

\(\displaystyle x^2-xy-y^2=0\)

What do you get when you apply the quadratic formula to solve for $x$ and discard any negative roots?

 

[mathdad]

I think the problem should state:

"The point P is located on line segment AB in such a way that AB/AP = AP/PB"

If we use the given hint, then we may state:

\(\displaystyle \frac{x+y}{x}=\frac{x}{y}\)

Or:

\(\displaystyle x^2-xy-y^2=0\)

What do you get when you apply the quadratic formula to solve for $x$ and discard any negative roots?

I'll get back to you regarding the rest of this question.

 

[mathdad]

I think the problem should state:

"The point P is located on line segment AB in such a way that AB/AP = AP/PB"

If we use the given hint, then we may state:

\(\displaystyle \frac{x+y}{x}=\frac{x}{y}\)

Or:

\(\displaystyle x^2-xy-y^2=0\)

What do you get when you apply the quadratic formula to solve for $x$ and discard any negative roots?

To use the quadratic formula in your equation, we need a, b, and c.

Do we let a = 1?

Do we let b xy?

Do we let c = y^2

 

[MarkFL]

To use the quadratic formula in your equation, we need a, b, and c.

Do we let a = 1?

Do we let b xy?

Do we let c = y^2

To solve for $x$ here, we let:

\(\displaystyle a=1\)

\(\displaystyle b=-y\)

\(\displaystyle c=-y^2\)

 

[mathdad]

To solve for $x$ here, we let:

\(\displaystyle a=1\)

\(\displaystyle b=-y\)

\(\displaystyle c=-y^2\)

Ok. I will work on this and provide solution for x.

 

[mathdad]

x = +- sqrt{(-y)^2 - 4(1)(-y^2)}/2(1)

x = +- sqrt{y^2 + 4y^2}/2

x = +-sqrt{y^2(1 + 4)}/2

x = +- y*sqrt{5}/2

x = - y*sqrt{5}/2

and

x = y*sqrt{5}/2

 

[MarkFL]

That's not quite right:

\(\displaystyle x=\frac{-(-y)\pm\sqrt{(-y)^2-4(1)(-y^2)}}{2(1)}=y\frac{1\pm \sqrt{5}}{2}\)

Discarding the negative root, we obtain:

\(\displaystyle x=y\frac{1+\sqrt{5}}{2}\)

Now:

\(\displaystyle k=\frac{x}{y}=\)?

 

[mathdad]

That's not quite right:

\(\displaystyle x=\frac{-(-y)\pm\sqrt{(-y)^2-4(1)(-y^2)}}{2(1)}=y\frac{1\pm \sqrt{5}}{2}\)

Discarding the negative root, we obtain:

\(\displaystyle x=y\frac{1+\sqrt{5}}{2}\)

Now:

\(\displaystyle k=\frac{x}{y}=\)?

I see that I forgot to include the -b part of the quadratic formula. Solving for k will let me completely solve the problem. I see that k = x/y.

The x value is what you found using the quadratic formula. I am not clear on the y value.
As you said, x = y[(1 + sqrt{5})/2]. Do I solve this equation for y?

 

[MarkFL]

We don't need to solve for $y$. We need only to have shown that:

\(\displaystyle \frac{x}{y}=\frac{1+\sqrt{5}}{2}\)

to satisfy part A) of the problem. For part B), consider:

\(\displaystyle \frac{x+y}{x}=\frac{x}{y}\)

If we write this in terms of $k$, we find:

\(\displaystyle 1+\frac{1}{k}=k\)

This will help you with the four questions of part B). :D

 

[mathdad]

We don't need to solve for $y$. We need only to have shown that:

\(\displaystyle \frac{x}{y}=\frac{1+\sqrt{5}}{2}\)

to satisfy part A) of the problem. For part B), consider:

\(\displaystyle \frac{x+y}{x}=\frac{x}{y}\)

If we write this in terms of $k$, we find:

\(\displaystyle 1+\frac{1}{k}=k\)

This will help you with the four questions of part B). :D

Ok. I just have to replace every k with k = 1 + 1/k to verify all four parts of part (B). So, lots of algebra is basically what needs to be done.

 

[MarkFL]

Ok. I just have to replace every k with k = 1 + 1/k to verify all four parts of part (B). So, lots of algebra is basically what needs to be done.

Now necessarily...for example if we multiply through by $k$, we get:

\(\displaystyle k+1=k^2\)

And BOOM! we have question 1 of part B) done.

 

[mathdad]

Now necessarily...for example if we multiply through by $k$, we get:

\(\displaystyle k+1=k^2\)

And BOOM! we have question 1 of part B) done.

Thanks. Your help in my study of precalculus is greatly appreciated.