On Mon, Feb 15, 2016 at 11:07 AM, Mike Stay <metaweta@gmail.com> wrote:
I've seen lots that prove the Pythagorean theorem using similar triangles, but they use algebra to go from the similar triangles to squares rather than just say "clearly these right triangles are all the same fraction of a square, so we can factor out that constant."
I don't think the 'clearly' sweeps anything complicated under the rug. If the hypotenuse is h, and the angle is theta, then the sides are h sin theta and h cos theta, so the area is h^2 sin theta cos theta/2, so proportional to h^2 as long as theta is the same. Now all that is being swept under the rug is that cos and sin are well defined, that is, that the ratio of the side lengths of a right triangle depend only on the angle, not on the side lengths, which is just the fundamental similar triangle property. Anyway, the algebraic proof I learned in high school, and most of the other algebraic proofs I've seen, have always felt unmotivated, and this one is doing essentially the same algebraic calculation as the others. I find the proof as a direct consequence of the fact that the areas of the two small triangles add up to the big one to be more intuitive than the other algebraic proofs I've seen, which, like the one at https://www.khanacademy.org/math/geometry/right-triangles-topic/pythagorean_... seem to me to just take similar triangle formulas and calculate with them until something interesting happens. They seem more like "well, since the Pythagorean theorem is true, this sort of proof must work out", rather than providing any intuition of why this should be true. Andy
https://www.khanacademy.org/math/geometry/right-triangles-topic/pythagorean_...
On Mon, Feb 15, 2016 at 7:44 AM, Andy Latto <andy.latto@pobox.com> wrote:
On Mon, Feb 15, 2016 at 10:22 AM, Dan Asimov <dasimov@earthlink.net> wrote:
Small adjustment: Similar figures in the plane have areas proportional to the *squares* of the lengths of the corresponding sides.
Yes, sorry, that was a typo.
But: Is it safe to say that the theorem that a right triangle is split into two similar ones by the altitude to the hypotenuse does not rely on the Pythagorean theorem for its proof?
The fact that a small triangle and a large triangle share two angles is immediate; one angle is shared, and the other is a right angle. The fact that the third angle is the same comes from the fact that the sum of the angles of a triangle is two right angles, which is where we use zero curvature, and is certainly an earlier theorem in Euclid than the Pythagorean. As to the fact that lengths of corresponding sides of similar triangles are proportional, this gets used in pretty much every proof of the Pythagorean theorem I've seen.
Has anyone seen this proof elsewhere? I'm in the odd position of not knowing whether the proof was original with me or not, since I saw it in "Proof without words" form, and don't know whether the proof was intended.
The version I saw was in a science museum (Boston Museum of Science? Exploratorium?). There were right triangles with various figures (semicircles, regular hexagons) on their sides, mounted vertically on a wheel, with colored liquid in them, so you could rotate them and verify that the sum of the small semicircles' area was the area of the large semicircle, and similarly for the hexagons. I don't remember whether they had the case of a triangle similar to the original triangle built on each side or not. If they did, they probably had the idea that you can just "fold the triangles over" to see that the areas of the two small triangles sum to the area of the large triangle, and I give them credit for the proof. If they didn't have this case, then I think they inspired the proof, but I came up with it.
So has anyone else seen this proof? Has anyone seen this museum exhibit?
Andy Latto andy.latto@pobox.com
—Dan
On Feb 14, 2016, at 8:17 PM, Andy Latto <andy.latto@pobox.com> wrote:
Did I ever post my favorite proof of the Pythagorean Theorem to math-fun? It deserves to be better known.
Start by observing that there's nothing special about squares; similar figures have areas proportional to the length of a corresponding side. So the Pythagorean theorem, about the areas of 3 squares with sides the lengths of the sides of a right triangle, is equivalent to theorem where instead of squares, we put equilateral triangles, or regular pentagons, or semicircles, on each side of a right triangle; if the sum of the areas of the two smaller figures is equal to the area of the figure on the hypotenuse for any figure, it's true for all sets of three similar figures.
So the proof, once you accept the lemma above, is simple enough that I don't even have to bother to draw the diagram, I'll just describe it. From the right angle of the triangle, draw an altitude, perpendicular to the hypotenuse. This divides the triangle into two smaller ones, and all 3 are similar.
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