The Pythagorean Theorem shows how strange our concept of distance is. Using the rule %%a^2 b^2 = c^2%%, we can trade some "a" to get more "b".
kalid: I’m with you up to “We can only move to neighboring points on the circle (options at the same distance)”. To what circle are you referring? The circle that includes points (13,0), (12,5), (5,12) and (0,13)? Shouldn’t the slopes be negative? Thanks.
I like how you explain concepts intuitively. Pythagorean theorem is quite simple to many, but for young students who still have a vague notion of proofs, we need intuitive explanation.
[…] Points in the direction of greatest increase of a function (intuition on why gradient works) […]
cool,nice…but what do u want to say?
sry kiddng he he…
Wow. These wonderful sparks of intuition just keep getting better and better each time - truly amazing!
As soon as my understanding of the intuition for this concept of pythogorean distance really sinks in, I’ll perhaps be able to add something to the distance concept, through folding-paper thought experiments… there’s really a lot of insights that can already be derived from just “folding paper” in unique ways.
Wonderful, intriguing work, Kalid!! 
Great new look.
[…] education, Without Geometry, Life Is Pointless started a new series on habits of the mind, and at Better Explained there was a nice post on how to enable students to understand (and not just learn) the Pythagorean […]
[…] I’ve gained immense value from upgrading the bow that holds the Pythagorean Theorem. That “arrow” (a2 + b2 = c2) can be launched in so many ways — each year I find a new personal discovery (it’s not about distance; it can apply to any shape; it explains the gradient). […]
How about proving the pythagorean theorem in your hands with a new shape-making ruler. This thing is too cool!
Wow… Extraordinary!
I didn’t know the Pythagorean Theorem could be used up to such extents, and with so many scenarios! Thanks for the elaborations. 
Hi Pat, yep, I’m referring to the circle of radius 13 (which includes the points the points you mentioned). Every point on this circle is the same amount of “effort” (distance) to get to, so it’s a question of which gives us the best payoff for that identical effort.
Some of the slopes (rise/run) are indeed negative, and you can see this as “you give up some North (rise) and gain some East (run)”.
Hi Khalid,
I did not understand
"At (1,2) we have reached the perfect 1 North = 2 East tradeoff."
(1,2) = one unit towards East and 2 units towards North.
But for perfect trade off (equate X and Y or east and North) we would need 2 units east and 1 unit North i.e. (2,1).
Am I really missing a basic point. Would appreciate your help! Thanks!
What a wonderful post, and a truly great blog! I have a very literal mind, and I just love the way you describe abstract concepts in terms of physical things (chopsticks, a porch, pizza…). Thank you SO much for the time and effort you have put into this site! It has helped me so much.
@ Rohan
Kalid writes: Payoff(x,y) = ax + by
Best trajectory = (a, b) [in our case, (East, North) => (1, 2)]
As you go “up” along to the y axis, you go to North, not East (it is a convention to show the North direction as “up”). You can see in the equation of Payoff, b is associated with y axis, and thus North.
Similarly, you can see that a is associated with x, and thus East.
“Psst. Confused about how we got .707? No problem. Taking sides of 1 and 1 means the hypotenuse is 2:”
Surely this is an error? a^2 + b^2 = c^2, yeah? so 1^2 + 1^2 = c^2, then c = sqrt(2).