# Surprising Uses of the Pythagorean Theorem

The Pythagorean theorem is a celebrity: if an equation can make it into the Simpsons, I'd say its well-known.

This is a companion discussion topic for the original entry at http://betterexplained.com/articles/surprising-uses-of-the-pythagorean-theorem/

[…] Turns out Pythagorean theorem isn’t only useful for triangle calculation only. This article explains it in details.read more | digg story […]

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…

Einstein found this method of proof for Pythagorean Theorem when he was 12 or so; though he probably wasn’t the first (or last) to have done so.

@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.

[…] Surprising Uses of the Pythagorean Theorem | BetterExplained […]

Spyros,

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.

[…] Surprising Uses of the Pythagorean Theorem | BetterExplained.  Whoa.  This just completely blew my mind. […]

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.

Wow nice article. That puts the 'ol Pythagorean theorem into a whole new perspective.

What program do you use to draw pictures in your articles?

Thanks, glad you liked the article. Yep, I’ve discovered that there are many gems to be found when we revisit concepts we “know” :).

I used PowerPoint 2007 to make the diagrams.

“The Pythagorean theorem is a celebrity: if an equation can make it into the Simpsons, I’d say its well-known.”

Stopped reading right there.

Ah, that’s too bad – you may have found it interesting! [Both the article and the Simpsons episode ]

Your site is triggering a virus warning. Some java class it’s loading has a worm.

Hi, thanks for the info. Avast appears to be have a false positive (many AV companies mis-detect javascript as Feebs, see more here: http://isc.sans.org/diary.html?storyid=2319&rss).

I split my files into smaller chunks and it should be ok now. Appreciate the tip.

Usos sorprendentes del teorema de pitágoras…

Haciendo esto

resulta que…

[…] 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?