Chapter 8.1, Problem 45WE

a.

To determine

To prove: $\overline{CM}$ is the arithmetic mean between AH and BH, and that CH is the geometric mean between AH and BH and arithmetic mean is greater than geometric mean using the diagram.

Explanation of Solution

Given:

Following diagram is given

  

Concept used:

Sum of angles of a triangle are ${180}^{\circ }$

For arithmetic mean, following construction are made in the figure,

  

As CM is the median at right angle of $\Delta ABC$ , if a perpendicular is drawn from the median at hypotenuse to any other side it will also be the median of triangle formed.

That is $\overline{PM}$ is the altitude as well as median of $\Delta BMC$ at point P.

Now consider $\Delta BMP$ and $\Delta CMP$ ,

Here MP is the common side, $\angle CPM=\angle BPM={90}^{\circ }$ and $\overline{CP}=\overline{BP}$ .

Thus by side-angle-side theorem of congruency, $\Delta CMP\cong \Delta BMP$

Hence, $\overline{CM}=\overline{BM}$ ….(i) as they are corresponding sides of triangle considered.

Also, $\angle MCP=\angle MBP=\alpha$ .

Now, consider $\Delta ACM$ , here $\angle ACM=90-\alpha$

In $\Delta ABC$ sum of all angles in a triangle is ${180}^{\circ }$ , therefore

  $\begin{array}{c}\angle A+\angle B+\angle C={180}^{\circ }\\ \angle A+{90}^{\circ }+\alpha ={180}^{\circ }\\ \angle A={90}^{\circ }-\alpha \end{array}$

As a result, it can be said that $\angle A=\angle ACM.$

Now, in $\Delta ACM$ , as $\angle A=\angle ACM$ , therefore the side opposite to these angle will also be equal.

Hence, $\overline{CM}=\overline{AM}$ …(ii)

Now adding equation (i) and (ii),

  $\begin{array}{c}2CM=AM+BM\\ 2CM=AB\\ 2CM=AH+BH\\ CM=\frac{AH+BH}{2}\end{array}$

Thus, arithmetic mean of AH and BH can be defined as CM

For geometric mean,

Here, consider $\Delta AHC$ and $\Delta CHB$ ,

  $\angle AHC=\angle CHB={90}^{\circ }$

Side AH is common.

Also if $\angle A=x$ , then $\angle ACH={90}^{\circ }-x$ and $\angle HCB=x$

Thus $\angle HCB=\angle ACH$

Hence by angle-side-angle theorem of similarity, $\Delta AHC\approx \Delta CHB$

Therefore the ratio of their sides will be

  $\begin{array}{c}\frac{CH}{AH}=\frac{BH}{CH}\\ {(}^C=(AH)(BH)\\ (CH)=\sqrt{(AH)(BH)}\end{array}$

Thus, it is proved that CH is the geometric mean of AH and BH.

Now, to prove that arithmetic mean is greater than the geometric mean using diagram one can simply refer to the diagram as

  

Here as proved CM is the arithmetic mean and CH is geometric mean.

To prove this by diagram, the given triangle is a right angle triangle with altitude and median of different length thus it is not an isosceles triangle.

The definition of altitude is that it is the smallest distance from an angle to the side opposite to it or it is the height of the triangle from the angle through which the altitude passes.

Whereas median is line joining a vertex of a triangle and the mid-point of the side opposite to it.

As clearly indicated in the figure that altitude and median are different for the given triangle, median is larger than altitude.

That is CM is larger than CH, which proves that arithmetic mean is larger than geometric mean.

Conclusion:

Arithmetic mean is larger than geometric mean can be proved by geometry using basic properties of a triangle and can also be proved by observing the figure given and definition of altitude and median.

b.

To determine

To show: Algebraically that the arithmetic mean between two different numbers $r$ and $s$ is greater than the geometric mean.

Explanation of Solution

Given:

Two positive numbers $r$ and $s.$

Concept used:

Arithmetic mean: $\frac{r+s}{2}$

Geometric mean: $\sqrt{rs}$

For two numbers $r$ and $s$ ,

  ${(r-s)}^2$ is always positive.

Consider,

  ${(r-s)}^2\ge 0$

  $\Rightarrow r^2-2rs+s^2\ge 0$

Since $r$ and $s$ are positive,

Adding $4rs$ on both sides,

  $r^2-2rs+s^2+4rs\ge 0+4rs$

  $r^2+2rs+s^2\ge 4rs$

  ${(r+s)}^2\ge 4rs$

Taking square root on both sides,

  $r+s\ge 2\sqrt{rs}$

  $\frac{r+s}{2}\ge \sqrt{rs}$

Conclusion:

Therefore, arithmetic mean is greater than the geometric mean.