Oh What Fun

Being me can be somewhat fun, even if it is definitely somewhat weird. For example, I have the day off today because I’ll be heading up to Denver with my sister to pick up our parents from DIA since they’re back in the states for a month. Since I wasn’t working, and since I had some time to kill before heading off to Denver, I was reading a book. And then, right in the middle of a sentence, I suddenly had a thought about geometry. It’s hard for me to put into words what exactly the thought was, but after about thirty seconds I had come to a conclusion.

Namely, if you take an isosceles right triangle and get the hypotenuse, then you construct another isosceles right triangle with the base and arms equal to that hypotenuse, the hypotenuse of the second triangle will be twice the size of the base of your first triangle.

How many other people have that thought in the middle of a stinking novel?

In any case, since I’ve mentioned it here, I can go ahead and prove it to you just for fun. All you need to know is the Pythagorean theorem, which is a² + b² = c². Oh, and I suppose you should also know that in an isosceles right triangle, a = b. So for them, you can say 2a² = c².

So, if a = 1, then c = 2^1/2 (the square root is a number to the 1/2 power)

This follows from both equations, as you have:

1² + 1² = c²
1 + 1 = c²
2 = c²
2^1/2 = c

Or:

2(1²) = c²
2(1) = c²
2 = c²
2^1/2 = c

In any case, now we take 2^1/2 as the base of our second triangle. To keep the variables separate, I’ll call the second triangle 2x² = z², where x = 2^1/2. So:

2[(2^1/2)²] = z²
2(2) = z²
4 = z²
2 = z

So you can see that a = 1, z = 2. So z = 2a.

In any case, this works for all numbers, since we can show that by using variables instead of real numbers, as follows:

Take a triangle ABC where AB = BC, and take a triangle ACD, where AC = CD, then I say that AD = AB + BC.

Let AB be of the length a.
Therefore, BC is also of the length a.
Therefore, AB + BC is of the length 2a.

Also, therefore, AC is of the length (2a²)^1/2.

Which is also 2^1/2a since (xy)^1/2 = x^1/2y^1/2.

Since AC = CD, then CD = 2^1/2a

Therefore, AD² = (2^1/2a) ² + (2^1/2a) ²

AD² = 2a² + 2a²
AD² = 4a²
AD = [4a²]^1/2
AD = (4^1/2)(a²)^1/2
AD = 2a

Therefore, AD = AB + BC.
Q.E.D.

All that from reading a completely non-related novel.