On 2016-02-13 15:11, Gareth McCaughan wrote:
On 13/02/2016 18:38, Bill Gosper wrote:
gosper.org/Perigal.gif --rwg
Cute picture, but is proving that it actually proves Pythagoras any easier than just proving Pythagoras[?] Well, no. It's just a sketch of a pure-dissection proof that is a bit harder than the skewy ones sketched by gosper.org/Pythanim.gif, but more visually convincing. --rwg
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.
QED. On Sun, Feb 14, 2016 at 3:32 PM, Bill Gosper <billgosper@gmail.com> wrote:
On 2016-02-13 15:11, Gareth McCaughan wrote:
On 13/02/2016 18:38, Bill Gosper wrote:
gosper.org/Perigal.gif --rwg
Cute picture, but is proving that it actually proves Pythagoras any easier than just proving Pythagoras[?] Well, no. It's just a sketch of a pure-dissection proof that is a bit harder than the skewy ones sketched by gosper.org/Pythanim.gif, but more visually convincing. --rwg
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Brilliant! WFL On 2/15/16, 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.
QED.
On Sun, Feb 14, 2016 at 3:32 PM, Bill Gosper <billgosper@gmail.com> wrote:
On 2016-02-13 15:11, Gareth McCaughan wrote:
On 13/02/2016 18:38, Bill Gosper wrote:
gosper.org/Perigal.gif --rwg
Cute picture, but is proving that it actually proves Pythagoras any easier than just proving Pythagoras[?] Well, no. It's just a sketch of a pure-dissection proof that is a bit harder than the skewy ones sketched by gosper.org/Pythanim.gif, but more visually convincing. --rwg
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Small adjustment: Similar figures in the plane have areas proportional to the *squares* of the lengths of the corresponding sides. 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? —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.
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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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." 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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-- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com
For me the proof of the Pythagorean theorem I've seen that is most immediate is the one where you start with this tessellation based on two arbitrary squares: https://upload.wikimedia.org/wikipedia/commons/5/57/A_Pythagorean_tiling_Vie... <https://upload.wikimedia.org/wikipedia/commons/5/57/A_Pythagorean_tiling_View_4.svg> and then draw the square on the diagonal c of a right triangle whose legs are the sides a, b of the squares. Then just a little mental rearranging of pieces shows that c^2 = a^2 + b^2. —Dan _________ P.S. This still doesn't make the theorem "obvious" as some proofs without words do.
On Feb 15, 2016, at 8:07 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."
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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https://www.youtube.com/watch?v=RksVPKaJU00 --rwg On 2016-02-15 08:44, Dan Asimov wrote:
For me the proof of the Pythagorean theorem I've seen that is most immediate is the one where you start with this tessellation based on two arbitrary squares: https://upload.wikimedia.org/wikipedia/commons/5/57/A_Pythagorean_tiling_Vie... <https://upload.wikimedia.org/wikipedia/commons/5/57/A_Pythagorean_tiling_View_4.svg> and then draw the square on the diagonal c of a right triangle whose legs are the sides a, b of the squares.
Then just a little mental rearranging of pieces shows that c^2 = a^2 + b^2.
—Dan _________ P.S. This still doesn't make the theorem "obvious" as some proofs without words do.
On Feb 15, 2016, at 8:07 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."
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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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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-- Mike Stay - metaweta@gmail.com http://www.cs.auckland.ac.nz/~mike http://reperiendi.wordpress.com
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-- Andy.Latto@pobox.com
If we want a motivation for why one would seek something like the Pythagorean theorem, how about… Take a right triangle with one leg horizontal and fixed in place. Replace the right angle with a hinge, and the hypotenuse with a rubber band. Opening the hinge to a straight angle and drawing squares on each side, the hypotenuse’s square is seen to be bigger than the sum of the other two. Closing the hinge to a zero angle and again drawing squares, the hypotenuse’s square is seen to be smaller than the sum of the other two. So there must be some angle where they are equal, but what is the angle? Investigation discovers it is a right angle, for any length of the legs. Why draw squares and not cubes or just the side lengths themselves? Probably you do try those, but nothing interesting comes of it. Math as engineering, tinkering around, noticing when something useful happens. — Mike
Mike's mental image of a hinge closing to a zero angle and opening to a straight angle was helpful to me. Start side a at the origin extending along the x axis to the right. Start side b where a ends. Theta = 0, c = a+b, c^2 = a^2 + 2ab + b^2. Now rotate b about the "hinge" where a joins b and set theta = pi, c = b - a, c^2 = (b - a)^2 = a^2 - 2ab + b^2. Calculate the midpoint: theta = pi/2, c^2 = a^2 + b^2. Alternatively, consider that for theta = 0, (a + b)^2 > a^2 + b^2. As theta increases as b rotates about the hinge, c = a + b cos theta < a + b cos 0. We want c small enough to get rid of 2ab in (a + b)^2, but no smaller. a^2 + b^2 = (a + b cos theta)^2. Thus c = a + b cos theta = Sqrt( a^2 + b^2). Jeff On Mon, Feb 15, 2016 at 3:00 PM, Mike Beeler <mikebeeler@verizon.net> wrote:
If we want a motivation for why one would seek something like the Pythagorean theorem, how about…
Take a right triangle with one leg horizontal and fixed in place. Replace the right angle with a hinge, and the hypotenuse with a rubber band. Opening the hinge to a straight angle and drawing squares on each side, the hypotenuse’s square is seen to be bigger than the sum of the other two. Closing the hinge to a zero angle and again drawing squares, the hypotenuse’s square is seen to be smaller than the sum of the other two. So there must be some angle where they are equal, but what is the angle? Investigation discovers it is a right angle, for any length of the legs.
Why draw squares and not cubes or just the side lengths themselves? Probably you do try those, but nothing interesting comes of it. Math as engineering, tinkering around, noticing when something useful happens.
— Mike
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Sorry to be the bearer of bad news, but I have seen that clever proof of the Pythagorean theorem a long time ago. -- Gene From: Andy Latto <andy.latto@pobox.com> To: math-fun <math-fun@mailman.xmission.com> Sent: Monday, February 15, 2016 7:44 AM Subject: Re: [math-fun] Perigal's Pythagorean proof 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.
On 2016-02-15 07:44, Andy Latto 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.
I don't think I've seen the exhibit, but know that I've seen this proof before, and I think I know where. I'm very sure I've seen this proof in either Polya "How to Solve it", or "Induction and Analogy in Mathematics". I tried to check, but it appears that I've lent out "How to solve it" years ago (I forget to whom :-(), and I won't be able to get my hands on my copy of "Induction ..." until tonight). That isn't evidence that you didn't come up with it independently, though.
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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See proof #7 here: http://www.cut-the-knot.org/pythagoras/
On Feb 15, 2016, at 12:49 PM, Michael Greenwald <mbgreen@seas.upenn.edu> wrote:
I'm very sure I've seen this proof in either Polya "How to Solve it", or "Induction and Analogy in Mathematics".
On 2016-02-15 13:40, Hans Havermann wrote:
See proof #7 here: http://www.cut-the-knot.org/pythagoras/
Ah, so Andy Latto's proof is actually the proof in Euclid. That predates my Polya reference, for sure. However, the presentation by Andy (and Polya, if I recall correctly) were both much more clearly presented (well, to the extent that I am a proxy for modern audiences). [Incidentally, I actually also really like the proof in the "Remark" on #8. I'm not sure I ever saw that one before.]
On Feb 15, 2016, at 12:49 PM, Michael Greenwald <mbgreen@seas.upenn.edu> wrote:
I'm very sure I've seen this proof in either Polya "How to Solve it", or "Induction and Analogy in Mathematics".
I've heard this proof attributed to Einstein. (But then again, you could argue that it was inevitable that this proof would eventually be attributed to Einstein, whether or not he was the first to come up with it, or indeed, whether or not he ever came up with it at all!) It's my favorite proof too, because the principles of scaling and proportionality apply in lots of other situations. Jim Propp On Sun, Feb 14, 2016 at 11:17 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.
QED.
On Sun, Feb 14, 2016 at 3:32 PM, Bill Gosper <billgosper@gmail.com> wrote:
On 2016-02-13 15:11, Gareth McCaughan wrote:
On 13/02/2016 18:38, Bill Gosper wrote:
gosper.org/Perigal.gif --rwg
Cute picture, but is proving that it actually proves Pythagoras any easier than just proving Pythagoras[?] Well, no. It's just a sketch of a pure-dissection proof that is a bit harder than the skewy ones sketched by gosper.org/Pythanim.gif, but more visually convincing. --rwg
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-- Andy.Latto@pobox.com
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On 15/02/2016 19:43, James Propp wrote:
I've heard this proof attributed to Einstein. (But then again, you could argue that it was inevitable that this proof would eventually be attributed to Einstein, whether or not he was the first to come up with it, or indeed, whether or not he ever came up with it at all!)
Along with its corollary, E = m(a^2+b^2). -- g
This is quite funny, but apparently, not everyone gets the joke. One crank wrote to George Andrews proposing precisely this combination of Pythagorean and Einsteinian wisdom as a genuine contribution to the human race. Actually, it was worse than that: the crank combined "E = m c^2" with "c^2 = a^2 + b^2" to get "E = m c^2 = a^2 + b^2". Jim Propp On Mon, Feb 15, 2016 at 5:03 PM, Gareth McCaughan < gareth.mccaughan@pobox.com> wrote:
On 15/02/2016 19:43, James Propp wrote:
I've heard this proof attributed to Einstein. (But then again, you could
argue that it was inevitable that this proof would eventually be attributed to Einstein, whether or not he was the first to come up with it, or indeed, whether or not he ever came up with it at all!)
Along with its corollary, E = m(a^2+b^2).
-- g
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Many people seem incapable of grasping the notion of a variabl, despite routinely utilising pronouns such as "he", "she", "it" in everyday life. A similar difficulty arises over the distinction between identifier and object referenced. WFL On 2/15/16, James Propp <jamespropp@gmail.com> wrote:
This is quite funny, but apparently, not everyone gets the joke. One crank wrote to George Andrews proposing precisely this combination of Pythagorean and Einsteinian wisdom as a genuine contribution to the human race.
Actually, it was worse than that: the crank combined "E = m c^2" with "c^2 = a^2 + b^2" to get "E = m c^2 = a^2 + b^2".
Jim Propp
On Mon, Feb 15, 2016 at 5:03 PM, Gareth McCaughan < gareth.mccaughan@pobox.com> wrote:
On 15/02/2016 19:43, James Propp wrote:
I've heard this proof attributed to Einstein. (But then again, you could
argue that it was inevitable that this proof would eventually be attributed to Einstein, whether or not he was the first to come up with it, or indeed, whether or not he ever came up with it at all!)
Along with its corollary, E = m(a^2+b^2).
-- g
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So there are many cute proofs of the Pythagorean theorem. I'm convinced it's true, but despite that I've never seen a proof that gives me any intuition for WHY it is true. Why on Earth square the side lengths of a triangle? After all the theorem isn't true if space is slightly negatively or positively curved. I know square root of sum of squares shows up everywhere. Why? On Sun, Feb 14, 2016 at 12:32 PM, Bill Gosper <billgosper@gmail.com> wrote:
On 2016-02-13 15:11, Gareth McCaughan wrote:
On 13/02/2016 18:38, Bill Gosper wrote:
gosper.org/Perigal.gif --rwg
Cute picture, but is proving that it actually proves Pythagoras any easier than just proving Pythagoras[?] Well, no. It's just a sketch of a pure-dissection proof that is a bit harder than the skewy ones sketched by gosper.org/Pythanim.gif, but more visually convincing. --rwg
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Also, the correct generalization to 3D should make the distance from the origin to (3,4,5) be cbrt(27+64+125) = 6. When you make a right-angled tetrahedron by slicing a corner off a cube, the area of the oblique face is sqrt(sum of squares of the areas of the three orthogonal faces). Why not cbrt-of-sum-of-cubes? Rich ------- Quoting Scott Kim <scottekim1@gmail.com>:
So there are many cute proofs of the Pythagorean theorem. I'm convinced it's true, but despite that I've never seen a proof that gives me any intuition for WHY it is true. Why on Earth square the side lengths of a triangle? After all the theorem isn't true if space is slightly negatively or positively curved. I know square root of sum of squares shows up everywhere. Why?
On Sun, Feb 14, 2016 at 12:32 PM, Bill Gosper <billgosper@gmail.com> wrote:
On 2016-02-13 15:11, Gareth McCaughan wrote:
On 13/02/2016 18:38, Bill Gosper wrote:
gosper.org/Perigal.gif --rwg
Cute picture, but is proving that it actually proves Pythagoras any easier than just proving Pythagoras[?] Well, no. It's just a sketch of a pure-dissection proof that is a bit harder than the skewy ones sketched by gosper.org/Pythanim.gif, but more visually convincing. --rwg
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Scott: I think Andy Latto's proof is actually the best answer I've ever come up with to exactly this question. From this point of view, the essence of the Pythagorean theorem is that a right triangle can be cut into two pieces similar to itself, and the ratio of scales is just the ratio of sides of the original triangle. That dissection into similar pieces is an amazing observation about right triangles. --Michael On Mon, Feb 15, 2016 at 1:23 AM, Scott Kim <scottekim1@gmail.com> wrote:
So there are many cute proofs of the Pythagorean theorem. I'm convinced it's true, but despite that I've never seen a proof that gives me any intuition for WHY it is true. Why on Earth square the side lengths of a triangle? After all the theorem isn't true if space is slightly negatively or positively curved. I know square root of sum of squares shows up everywhere. Why?
On Sun, Feb 14, 2016 at 12:32 PM, Bill Gosper <billgosper@gmail.com> wrote:
On 2016-02-13 15:11, Gareth McCaughan wrote:
On 13/02/2016 18:38, Bill Gosper wrote:
gosper.org/Perigal.gif --rwg
Cute picture, but is proving that it actually proves Pythagoras any easier than just proving Pythagoras[?] Well, no. It's just a sketch of a pure-dissection proof that is a bit harder than the skewy ones sketched by gosper.org/Pythanim.gif, but more visually convincing. --rwg
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-- Forewarned is worth an octopus in the bush.
participants (17)
-
Andy Latto -
Bill Gosper -
Dan Asimov -
Eugene Salamin -
Fred Lunnon -
Gareth McCaughan -
Hans Havermann -
James Propp -
Jeff Caldwell -
Michael Greenwald -
Michael Greenwald -
Michael Kleber -
Mike Beeler -
Mike Stay -
rcs＠xmission.com -
rwg -
Scott Kim