Chapter 2.3, Problem 85E

Solution

a.

To show that $\frac{8}{x}=\frac{y-8}{y}$ by observing the similar triangles given in the diagram and using the properties of proportions.

Given the diagram:

  

Proof:

To begin with, name the triangles as follows:

  

The triangles $\text{HKL}$ and $\text{HJI}$ are similar.

Thus, the corresponding side lengths of the triangles are proportional.

That is:

  $\frac{D-u}{\left(D-u\right)+u}=\frac{8}{x}$ .

That is:

  $\frac{D-u}{D}=\frac{8}{x}$ .

Also, the triangles $\text{GHI}$ and $\text{KLI}$ are similar.

Then, it follows that:

  $\frac{u}{\left(D-u\right)+u}=\frac{8}{y}$ .

That is:

  $\frac{u}{D}=\frac{8}{y}$ .

Now, $\frac{u}{D}=\frac{8}{y}\Rightarrow \frac{u-D}{D}=\frac{8-y}{y}$ .

Then:

  $\begin{array}{l}\frac{u-D}{D}=\frac{8-y}{y}\\ \frac{D-u}{D}=\frac{y-8}{y}\end{array}$ .

Combine the equations $\frac{D-u}{D}=\frac{8}{x}$ and $\frac{D-u}{D}=\frac{y-8}{y}$ :

  $\frac{8}{x}=\frac{y-8}{y}$ .

Hence, it is found that $\frac{8}{x}=\frac{y-8}{y}$ .

b.

To show that $y=\frac{8x}{x-8}$ .

Proof:

It is found in the first part:

  $\frac{8}{x}=\frac{y-8}{y}$

Solve the equation $\frac{8}{x}=\frac{y-8}{y}$ for $y$ :

  $\begin{array}{l}\frac{8}{x}=\frac{y-8}{y}\\ 8y=x\left(y-8\right)\\ 8y=xy-8x\\ 8y-xy=-8x\\ y\left(8-x\right)=-8x\\ y=\frac{-8x}{\left(8-x\right)}\\ y=\frac{8x}{x-8}\end{array}$ .

Thus, it is found that:

  $y=\frac{8x}{x-8}$ .

c.

To show that $y^2-x^2=500$ and to show that $x^4-16x^3+500x^2-8000x+32000=0$ .

Proof:

Consider the right angled triangle $\text{GHI}$ and apply the Pythagoras theorem to it:

  $D^2={30}^2-y^2$ .

Now, consider the right angled triangle $\text{HJI}$ and apply the Pythagoras theorem to it:

  $D^2={20}^2-x^2$ .

Combine the equations $D^2={30}^2-y^2$ and $D^2={20}^2-x^2$ :

  ${30}^2-y^2={20}^2-x^2$ .

Simplify the equation ${30}^2-y^2={20}^2-x^2$ :

  $\begin{array}{l}{30}^2-y^2={20}^2-x^2\\ {30}^2-{20}^2=y^2-x^2\\ 500=y^2-x^2\end{array}$ .

Hence, it is found that $y^2-x^2=500$ .

Now, substitute $y=\frac{8x}{x-8}$ in the equation $y^2-x^2=500$ :

  $\begin{array}{l}{\left(\frac{8x}{x-8}\right)}^2-x^2=500\\ \left(\frac{64x^2}{x^2-16x+64}\right)-x^2=500\\ \left(\frac{64x^2}{x^2-16x+64}\right)=500+x^2\\ 64x^2=\left(x^2-16x+64\right)\left(500+x^2\right)\\ 64x^2=500x^2-8000x+32000+x^4-16x^3+64x^2\\ 0=x^4-16x^3+500x^2-8000x+32000\end{array}$ .

Hence, it is found that $x^4-16x^3+500x^2-8000x+32000=0$ .

d.

To find the value of $D$ .

It is found that:

  $D\approx 16.21$ .

It is found in the previous parts:

  $x^4-16x^3+500x^2-8000x+32000=0$ .

  $D^2={20}^2-x^2$ .

Claculation:

Since $D$ is a distance, it is always positive.

Then, it follows that:

  $D=\sqrt{{20}^2-x^2}$ .

Now, solve the equation $x^4-16x^3+500x^2-8000x+32000=0$ for positive real roots with the help of a calculator:

  $x\approx 5.94$ and $x\approx 11.71$ .

Now, the length of $x$ is greater than the length of the altitude of the right triangle $\text{HKL}$ .

That is, $x\ge 8$ .

Then, it follows that, $x\approx 11.71$ .

Then, $D\approx \sqrt{{20}^2-{11.71}^2}$ .

That is, $D\approx 16.21$ .

Conclusion:

It is found that:

  $D\approx 16.21$ .