http://www.cut-the-knot.org/pythagoras/
This one is by Michael Hardy from University of Toledo and was published in The Mathematical Intelligencer in 1988. It must be taken with a grain of salt.
Let ABC be a right triangle with hypotenuse BC. Denote
y·dy - x·dx = 0,
which after integration gives y² - x² = const. The value of the constant is
determined from the initial condition for
It is easy to take an issue with this proof. What does it mean for a triangle
to be
x² + a² = y² | |
(x + dx)² + a² = (y + dy)² |
which, after subtraction, gives
y·dy - x·dx = (dx² - dy²)/2.
For small dx and dy, dx² and dy² are even smaller and might be neglected,
leading to the approximate
The trick in Michael's vignette is in skipping the issue of approximation.
But can one really justify the derivation without relying on the Pythagorean
theorem in the first place? Regardless, I find it very much to my enjoyment to
have the ubiquitous equation
An amplified, but apparently independent, version of this proof has been published by Mike Staring (Mathematics Magazine, V. 69, n. 1 (Feb., 1996), 45-46).
Assuming Δx > 0 and detecting similar triangles,
Δf / Δx = CQ/CD > CP/CD = CA/CB = x/f(x).
But also,
Δf / Δx = SD/CD < RD/CD = AD/BD = (x + Δx) / (f(x) + Δf) < x/f(x) + Δx/f(x).
Passing to the limit as Δx tends to 0+, we get
df / dx = x / f(x).
The case of Δx < 0 is treated similarly. Now, solving the differential equation we get
f 2(x) = x² + c.
The constant c is found from the boundary condition