Geometric mean and a basic theorem

geomeanandatheorem

MEANS.

The average of two lengths (or numbers) is called their arithmetic mean: \frac{1}{2}(a+b) is the arithmetic mean of a and b. The geometric mean is the square root of their product: \sqrt{ab}. The arithmetic mean of 8 and 2 is 5; their geometric mean is 4.

It is easy to find the geometric mean of two positive numbers algebraically, by solving for x the equation a/x=x/b, for  this says x^{2}=ab. For this reason, the geometric mean is also called the mean proportional between a and b.

Is there any way to find these quantities with ruler and compass? The arithmetic mean is easy enough: simply lay off the  two lengths end to end on the same straight line and bisect the total segment. A method for the geometric mean is suggested by the equation: \frac{AB}{x}=\frac{x}{BC}.

Using AB+BC as diameter, draw a circle and the chord perpendicular to  that diameter at B. (Fig 1). This chord is bisected by the diameter, and we recognize a special case of Theorem 2,

https://madhavamathcompetition.wordpress.com/2015/09/01/high-school-geometry-chapter-i-basic-theorem/

namely, x.x=AB.BC, which says that x is the mean proportional between AB and BC.

In practice, one uses only half the figure. (Fig 2).

Inasmuch as MO, the perpendicular from O, the mid-point of AC, is the longest perpendicular to the semicircle from its diameter, and is also a radius equal to \frac{AC}{2}=\frac{AB+BC}{2}, we have:

THEOREM 3.

The geometric mean of two unequal positive numbers is always less than their arithmetic mean.

Why do we need the word “unequal”?

We now restate the part of Theorem 2,

https://madhavamathcompetition.wordpress.com/2015/09/01/high-school-geometry-chapter-i-basic-theorem/

THEOREM 4.

If, from an external point, a tangent and a secant are drawn, the tangent is the mean proportional between the whole secant and its external segment.

Of course, the two segments PA and PB become the same length when the chord is moved into the limiting position of tangency, bringing A and B together. If you are bothered by this, note that there is a proof possible, which is quite independent of any limiting process.

if we relabel the diagram as in Fig 3 here, and then call OT by the single letter t, we have

t^{2}=OC.OD

if and only if t is a tangent. These labels are chosen to fit the lettering of the next blog on this topic.

I trust you will agree that we have done nothing difficult so far. Yet you may soon be surprised at the structure we are about to build on this small but sturdy foundation. We are now ready to look at some of the geometry they didn’t teach you.

More later,

Nalin Pithwa

 

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