Wow, thats really cool.
So if I’m getting this right, when a car is traveling at 100 mph, the energy that is required to take it to 110 mph (or the difference in its speed vector) is the same as that of two other cars travelling at 100 and 45,8 mph.
So, if they travel for 1 hour, they would have covered 100 miles (car A) and 110 miles (car C) respectively, but car A would have enough fuel to go for another 45 miles.
Makes you think about fuel economy/efficiency…

@Spyros: I think that’s right from a pure energy point of view: the energy needed to go from 100-110mph equals the energy to go from 0-45.8 mph.

I’m not sure how well the pure energy analogy works (drag, efficiency of engine, etc.), but there is an ideal speed for fuel efficiency cars, and it’s not around 110mph :).

@Kaizyn: Thanks for the info. Yep, I like this particular proof compared to others because it focuses on the larger concepts, not just re-arrangement of area.

Thats the energy required to accelerate the vehicle to that speed, not the energy required to maintain the speed, nearly all of which is required to overcome wind resistance at high speeds.

The coolest thing I’ve ever discovered about Pythagorean’s Theorem is an alternate way to calculate it. If you write a program that uses the distance form c = sqrt(a^2 + b^2) you will suffer from the lose of half of your available precision because the square root operation is last. A more accurate calculation is c = a * sqrt(1 + b^2 / a^2). If a is less than b, you should swap them and of course handle the special case of a = 0.

Hi Mccoyn, thanks for the great info! The note about precision is especially useful, I hadn’t thought about the impact of order-of-operations on the calculation. I’m sure that comes in useful in graphics programming, etc.

[…] We’ve underestimated the Pythagorean theorem all along. It’s not about triangles; it can apply to any shape. It’s not about a, b and c; it applies to any formula with a squared term. […]

[…] The Pythagorean Theorem is not just about triangles. It is about the relationship between similar shapes, the distance between any set of numbers, and much more. […]

Your proof of the Pythagorean theorem is very appealing. However, to be complete, you’d need to prove your supposition about similar triangle areas: concretely, the special case that the area of a right triangle can be computed as constant * hypotenuse^2. I tried a few different approaches, but they all end up having to apply Pythagoras in the end, which makes the whole exercise circular and thus ultimately pointless. If you try to establish the result by concrete computation it seems you are bound to run into this same difficulty–how do you get around it?