Purely Geometrical Average of Two Numbers

Draw 2 parallel lines, the lengths of which correspond to 2 numbers A and B whose average you wish to find. Connect the endpoints of these lines so as to form a trapezoid/trapezium.

Bisect each of the legs of this trapezoid/trapezium and connect their midpoints. (This line is called the median)

The length of this line is the average of the two numbers A and B.

Though relatively simple, I thought it was so cool how a mathematical concept like average could be represented completely geometrically, without any knowledge of math!

Anybody want to take a stab at proving that the median is indeed the average of the 2 lines?

#Geometry

Note by David Stiff
3 years, 5 months ago

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Comments

Visually if you just cut by the red line the trspezium take upper region rotate 180 degrees and glue the two part putting base B with base A. You will obtain a parallelogram one side measure A+B and the other 2m therefore m=(A+B)/2. Old people figured out this way

Mara Jares - 3 years, 5 months ago

Suppose ABDCABDC is a trapezium, so that the vectors a=AB a = \overrightarrow{AB} and b=CD b = \overrightarrow{CD} are parallel. So, let a=λv a = \lambda v and b=μv b = \mu v , for some vector vv. Also suppose that c=AC c = \overrightarrow{AC} and d=DB d = \overrightarrow{DB} . Let M,NM,N be the respective midpoints of ACAC and BDBD; this means that AM=MC=12c \overrightarrow{AM} = \overrightarrow{MC} = \frac{1}{2} c and NB=DN=12d \overrightarrow{NB} = \overrightarrow{DN} = \frac{1}{2} d .

So, we have that

MN=12(c+d)+μv=12(c+d)+λv. \overrightarrow{MN} = \frac{1}{2} (c + d) + \mu v = - \frac{1}{2} (c + d) + \lambda v .

Hence,

2MN=(λ+μ)v 2 \overrightarrow{MN} = (\lambda + \mu) v

and as a result

MN=λ+μ2v. \overrightarrow{MN} = \frac{\lambda + \mu}{2} v.

We understand the quantity λ+μ2 \frac{\lambda + \mu}{2} to be the arithmetical mean of λ \lambda and μ \mu , so we have to understand that the line segment connecting the midpoints of the non-parallel sides of a trapezium is half the "sum" of the line segments of the parallel sides of a trapezium; note the lack of use of "lengths" in not only the proof, but also the result obtained.

P.S. The origins of our understanding of mathematics are purely geometric (aside from the idea of counting objects); the Greeks had the right idea in thinking geometrically, and it has carried on until the turn of the last century, when mathematics started to go deeply south.

A Former Brilliant Member - 3 years, 4 months ago
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