11 ms·

A seventh-grader student found a beautiful proof to Thales' Theorem (2002)

139 points by samdung 2y ago

mikhael28 2y ago
I’m curious what this seventh grader went on to do, twenty years later.
kristopolous 2y ago
I'd really like to read it in the language of the kid. Great mathematicians tend to start young. I don't question it, I'm just curious how the student phrased it.
- tromp 2y ago
  I don't think the student's writing is available anywhere, but here is what Paul Lockhart said about it in his "A Mathematician’s Lament" [1]: > To be fair, I did paraphrase the proof considerably. The original was quite a bit more convoluted, and contained a lot of unnecessary verbiage (as well as spelling and grammatical errors). But I think I got the feeling of it across. And these defects were all to the good; they gave me something to do as a teacher. I was able to point out several stylistic and logical problems, and the student was then able to improve the argument. For instance, I wasn’t completely happy with the bit about both diagonals being diameters— I didn’t think that was entirely obvious— but that only meant there was more to think about and more understanding to be gained from the situation. And in fact the student was able to fill in this gap quite nicely: “Since the triangle got rotated halfway around the circle, the tip must end up exactly opposite from where it started. That’s why the diagonal of the box is a diameter.” So a great project and a beautiful piece of mathematics. I’m not sure who was more proud, the student or myself. This is exactly the kind of experience I want my students to have. [1] https://fermatslibrary.com/s/a-mathematicians-lament https://fermatslibrary.com/s/a-mathematicians-lament
matt3210 2y ago
People seem to forget a lot that child math prodigies are being trained by parents to be math geniuses instead of enjoying being a kid. I had a 12 year old in my freshman CS class and he was the most miserable unhappy kid I've ever seen in my entire life. Edit: Pretty good points in the replies. Probably a lot of latent jealousy on my side in this. I should have thought on this more before commenting.
- saagarjha 2y ago
  You don't have to be miserable to come up with a proof like this.
- fardinahsan 2y ago
  Conversely, you can't come up with a proof like this if you don't love math.
  - croes 2y ago
    Why not? It may be more likely but not impossible.
- Larrikin 2y ago
  What's the take away you want from this comment? Advancement of humanity, feeling bad for missed childhood fulfillments, some mix? Did the child actually meaningfully contribute anything to your class? Were they a real person? Right now there's a lot we've "forgotten" about child geniuses that we should all know from our interactions from them ?
- seszett 2y ago
  Just to provide another data point, I also had a "math genius" friend at school and he was a well-balanced, fun and reasonably happy person.
  - knightoffaith 2y ago
    Some more data - the math geniuses I knew were actually happier, more well-rounded, and fitter than the average person at my former school.
aaron695 2y ago
[dead]
akho 2y ago
Proving that a parallelogram with equal diagonals is a rectangle is an exercise in itself; I’d prove it through Thales’ theorem, myself... Also, Lockhart's Lament is from 2002, so this post probably needs a (2002). It is very unlikely that the proof was new. It was certainly new to the seventh-grader, and a great result at that.
- lloeki 2y ago
  > Proving that a parallelogram with equal diagonals is a rectangle is an exercise in itself; I’d prove it through Thales’ theorem, myself.. IIRC you prove it via angles. a) The sum of angles of any triangle is pi. b) The sum of angles of any quadrilateral is 2*pi. c) Since you rotate the triangle, opposite angles end up adding together. Because of the above, a bit of reasoning around symmetry shows that when diagonals are equal they can only add to pi/2 and that the other ones can only be pi/2. Any other angle leads to a contradiction. I mentioned angle values but (again IIRC) this can all be proven with compass and ruler.
  - vidarh 2y ago
    As the article mentions, in this particular case it's simpler: You have a point on the triangle that is also on the circle. Now you rotate the triangle 180 degrees, with the point following the circle. The point inherently must end up exactly opposite the original point, and so the diagonal formed between the old and new point must be the diameter of the circle.
    - lloeki 2y ago
      This proves that the parallelogram diagonals, both being diameters of the same circle, are equal, akho was pointing out that there's a leap from that to "the parallelogram is a rectangle", which can be proven with Thales. But then, the article is about laying out a proof of Thales so if it's using a lemma that leveraged Thales itself that's circular logic right there and so we're in trouble. One other way to prove Thales is by computing angles, but then you've got yourself a proven Thales so the point of the article is moot.
rocqua 2y ago
I don't think the note "Since the triangle got rotated halfway around the circle, the tip must end up exactly opposite from where it started. That's why the diagonal of the box is a diameter." is needed, nor do you need to show the diagonals of the parallelogram are equally long. Rather, it suffices to simply state, the only parellelogram that is inscribed in a circle is a rectangle. For the less rigorous, that statement is obviously true. For complete rigor, it is sufficient to argue that the center of the circle, combined with the vertices of the parallelogram forms an isoceles triangle. So the centre of the circle must lie on the bisector all edges of a parallelogram. But on a non-rectangle parellelogram the bisectors of opposite edges never intersect.
- lupire 2y ago
  If your informal proof is to draw a picture and say "it's obviously true by lookikg at it", then you don't need a parallelogram at all. Just draw the diametric triangle and look at it.
  - rocqua 2y ago
    Hence the addition of a formal proof based on the bisectors of opposite edges of a parallelogram.
forgotpwd16 2y ago
To be honest, I prefer the unattractive and inelegant proof given alongside that one in Lockhart's essay.
- lupire 2y ago
  Lockhart missed the point a bit. His student's "proof" is an illustration, not a proof. There's no way to know if its circular or simply unfounded, since it is purely an appeal to intuition. That's only a part of mathematics. The correct thing to do, mathematically, is to validate the intuition by formatting it as a proper proof, based on non-circular axioms and theorems. In the paper itself he admits that his student's work was incoherent and needed him to rewrite it. He's right that 2 column geometry proofs are ugly, and could be presented better. This has been known for centuries. https://www.c82.net/euclid/en/book3/#prop31 https://www.c82.net/euclid/en/book3/#prop31 For children and for starting out, the pictures are great. But for mathematics, pictures are extremely limiting. 2D and 3D are notgod models for N-Dimemsions. ("Spiky balls" , for example). Mathematics is far, far more powerful than human eyes. The amazing thing about geometry is that the whole thing works without pictures! A blind person can be a great geometer, because geometry is axiomatizable. Meanwhile, Euclid's Elements, while an incredible achievement in its day, is not well-founded, relying on unstated axioms.
  - JadeNB 2y ago
    > Lockhart missed the point a bit. I think it's fairer to say that his point may not have been what you expected it to be. As a teacher myself, trust me, if a student comes to you and says "I came up with a proof!" and you say "no, you see, what you have is an intuitive explanation that can possibly be turned into a proof," then all that will happen is that that student will not be interested any more in exploring, or at least will not be interested any more in sharing their explorations with you. At that point, you have both lost. Lockhart is well aware of the standard of mathematical proof, and knows that, all else aside, this theorem is not in want of proof. His focus is on the fact that, if we want mathematics to remain a live profession, then we must improve our pedagogy, and help to train students who enjoy and want to pursue mathematics—even if it means occasionally accepting less than maximally rigorous mathematics from a seventh grader.
t_mann 2y ago
> Since the triangle got turned completely around, the sides of the box must be parallel, so it makes a parallelogram. But it can't be a slanted box because both of its diagonals are diameters of the circle, so they're equal, which means it must be an actual rectangle. I'd be careful with such "visual proofs", even more so if accompanied by such handwavy reasoning. Eg, do we know that both diagonals are diameters? Do we know that a parallelogram with equal diagonals is a rectangle? While in this case things do work out nicely, I'd say this is almost more luck than a real proof - it's easy to mistakenly "prove" stuff like Pi=4 with similar reasoning. I believe 3B1B even has a video on the topic.
- pringk02 2y ago
  > do we know that both diagonals are diameters This must be true, because the diagonals are both straight lines that go through the centre and are bound by the edges, so it follows they must be equal to the diameter of the circle by definition. > Do we know that a parallelogram with equal diagonals is a rectangle? As another commenter points out, this is a theorem you can reach for, but proving it by itself is a bit more of a task.
  - HarHarVeryFunny 2y ago
    > This must be true, because the diagonals are both straight lines that go through the centre How do we know the 2nd diagonal goes through the center ? Is it because of the construction by rotation ?
    - surajms 2y ago
      Yes. So, here when we rotate the triangle, we are essentially rotating each of the endpoints. For each endpoint, we rotate it by 180 degrees around the line segment joining the endpoint and the center. This by definition will result in a new position for each endpoint that creates a chord (as the two endpoints lie on the circle) and passes through the center (we rotated around it). A chord that passes through the center is by definition a diameter.
rodneyzeng 2y ago
The normal approach to prove Thale's theorem should be induced from the property of central angle being twice of an inscribed angle that subtends the same arc. Since a diameter has central angle of 180 degrees, its corresponding inscribed angle should be half of 180, that is 90 degrees.
- lupire 2y ago
  No, because if you do it that way, you wouldn't have Thales's Theorem. It would be Thales's Trivial Corollary. Thales's Theroem is a simpler, easier to prove (as in OP), less powerful statement than the inscribed angle theorem.
  - rodneyzeng 2y ago
    Your second sentence denies the first sentence. The proof of the Inscribed angle theorem does not need Thale's Theorem, and it is stronger than Thale's Theorem.
lupire 2y ago
Euclid's proof, with Byrne's beautiful visualization: https://www.c82.net/euclid/en/book3/#prop31 https://www.c82.net/euclid/en/book3/#prop31
ralferoo 2y ago
The corners of the "parallelogram" in the diagram don't touch the circle at the top or bottom. So, those two corners wouldn't be right angles, but instead would be slightly obtuse - rather like this comment!
- tantalor 2y ago
  huh
  - vidarh 2y ago
    One of the illustrations in the article is off by a few pixels. Which of course has no bearing on the text, and so pointing it out above was intentionally obtuse, presumably for the sake of the pun.
abecedarius 2y ago
Perhaps this was essentially Thales's own proof: https://intellectualmathematics.com/blog/first-proofs-thales-and-the-beginnings-of-geometry/ https://intellectualmathematics.com/blog/first-proofs-thales... BTW it's also quite direct using vectors: the legs of the triangle are the sum and difference of radius vectors. Take their dot product, distribute it, it's zero because radii are the same length.
- jacobolus 2y ago
  Thales and Pythagoras are quasi-mythical figures, and we don't actually know anything concrete about any mathematical accomplishments they might have had, which are all apocryphal and date from many centuries after their deaths. Greek deductive mathematics per se dates from at least a century after Thales' time, while many of the basic facts about Euclidean geometry were understood in ancient Egypt and Mesopotamia long before him, and it is most likely that Thales himself never did any of the mathematical or scientific things attributed to him. Viktor Blåsjö's speculation that ancient Greeks began with the same insight as Lockhart's 7th grade student is plausible but is not backed by any evidence whatsoever. (This is an insight that many people have had over the centuries, certainly including anyone deeply investigating cyclic quadrilaterals, but also probably plenty masons or metalworkers working with circles and right angles, etc.) > sum and difference of radius vectors This is a nice one. Another way to use Thales' theorem in characterizing a circle, without involving the center point, is to start with one point P on a circle and a vector d which is a diameter from that point to the antipodal point. Then the vector v from P to any other point Q on the circle satisfies v² = v · d, or equivalently v · (v − d) = 0.
deleted 2y ago
[deleted]
fraukeadeline 2y ago
[dead]
dang 2y ago
Related. Others? A Mathematician's Lament (2002) [pdf] - https://news.ycombinator.com/item?id=35929333 https://news.ycombinator.com/item?id=35929333 - May 2023 (47 comments) A Mathematicians Lament [pdf] - https://news.ycombinator.com/item?id=30829704 https://news.ycombinator.com/item?id=30829704 - March 2022 (3 comments) A Mathematician's Lament [pdf] - https://news.ycombinator.com/item?id=15385104 https://news.ycombinator.com/item?id=15385104 - Oct 2017 (1 comment) A Mathematician’s Lament (2002) [pdf] - https://news.ycombinator.com/item?id=14331752 https://news.ycombinator.com/item?id=14331752 - May 2017 (27 comments) A Mathematician’s Lament (2002) [pdf] - https://news.ycombinator.com/item?id=8845507 https://news.ycombinator.com/item?id=8845507 - Jan 2015 (89 comments) A Mathematician's Lament [pdf] - https://news.ycombinator.com/item?id=6994939 https://news.ycombinator.com/item?id=6994939 - Jan 2014 (1 comment) A Mathematician’s Lament (2002) [pdf] - https://news.ycombinator.com/item?id=6187014 https://news.ycombinator.com/item?id=6187014 - Aug 2013 (119 comments) New book Measurement by author of Lockhart's Lament - https://news.ycombinator.com/item?id=4317199 https://news.ycombinator.com/item?id=4317199 - July 2012 (4 comments) Mathematician's Lament: An essay on math education and on how we view math - https://news.ycombinator.com/item?id=666563 https://news.ycombinator.com/item?id=666563 - June 2009 (18 comments) On Math Teaching: Lockhart's Lament - https://news.ycombinator.com/item?id=256176 https://news.ycombinator.com/item?id=256176 - July 2008 (21 comments) Lockhart's Lament: On Mathematics at School - https://news.ycombinator.com/item?id=130499 https://news.ycombinator.com/item?id=130499 - March 2008 (20 comments)

139 points by samdung 2y ago

I’m curious what this seventh grader went on to do, twenty years later.

I'd really like to read it in the language of the kid. Great mathematicians tend to start young. I don't question it, I'm just curious how the student phrased it.

tromp 2y ago
I don't think the student's writing is available anywhere, but here is what Paul Lockhart said about it in his "A Mathematician’s Lament" [1]: > To be fair, I did paraphrase the proof considerably. The original was quite a bit more convoluted, and contained a lot of unnecessary verbiage (as well as spelling and grammatical errors). But I think I got the feeling of it across. And these defects were all to the good; they gave me something to do as a teacher. I was able to point out several stylistic and logical problems, and the student was then able to improve the argument. For instance, I wasn’t completely happy with the bit about both diagonals being diameters— I didn’t think that was entirely obvious— but that only meant there was more to think about and more understanding to be gained from the situation. And in fact the student was able to fill in this gap quite nicely: “Since the triangle got rotated halfway around the circle, the tip must end up exactly opposite from where it started. That’s why the diagonal of the box is a diameter.” So a great project and a beautiful piece of mathematics. I’m not sure who was more proud, the student or myself. This is exactly the kind of experience I want my students to have. [1] https://fermatslibrary.com/s/a-mathematicians-lament https://fermatslibrary.com/s/a-mathematicians-lament

matt3210 2y ago

People seem to forget a lot that child math prodigies are being trained by parents to be math geniuses instead of enjoying being a kid. I had a 12 year old in my freshman CS class and he was the most miserable unhappy kid I've ever seen in my entire life. Edit: Pretty good points in the replies. Probably a lot of latent jealousy on my side in this. I should have thought on this more before commenting.

saagarjha 2y ago
You don't have to be miserable to come up with a proof like this.
fardinahsan 2y ago
Conversely, you can't come up with a proof like this if you don't love math.
- croes 2y ago
  Why not? It may be more likely but not impossible.
Larrikin 2y ago
What's the take away you want from this comment? Advancement of humanity, feeling bad for missed childhood fulfillments, some mix? Did the child actually meaningfully contribute anything to your class? Were they a real person? Right now there's a lot we've "forgotten" about child geniuses that we should all know from our interactions from them ?
seszett 2y ago
Just to provide another data point, I also had a "math genius" friend at school and he was a well-balanced, fun and reasonably happy person.
- knightoffaith 2y ago
  Some more data - the math geniuses I knew were actually happier, more well-rounded, and fitter than the average person at my former school.

aaron695 2y ago

[dead]

akho 2y ago

Proving that a parallelogram with equal diagonals is a rectangle is an exercise in itself; I’d prove it through Thales’ theorem, myself... Also, Lockhart's Lament is from 2002, so this post probably needs a (2002). It is very unlikely that the proof was new. It was certainly new to the seventh-grader, and a great result at that.

lloeki 2y ago
> Proving that a parallelogram with equal diagonals is a rectangle is an exercise in itself; I’d prove it through Thales’ theorem, myself.. IIRC you prove it via angles. a) The sum of angles of any triangle is pi. b) The sum of angles of any quadrilateral is 2*pi. c) Since you rotate the triangle, opposite angles end up adding together. Because of the above, a bit of reasoning around symmetry shows that when diagonals are equal they can only add to pi/2 and that the other ones can only be pi/2. Any other angle leads to a contradiction. I mentioned angle values but (again IIRC) this can all be proven with compass and ruler.
- vidarh 2y ago
  As the article mentions, in this particular case it's simpler: You have a point on the triangle that is also on the circle. Now you rotate the triangle 180 degrees, with the point following the circle. The point inherently must end up exactly opposite the original point, and so the diagonal formed between the old and new point must be the diameter of the circle.
  - lloeki 2y ago
    This proves that the parallelogram diagonals, both being diameters of the same circle, are equal, akho was pointing out that there's a leap from that to "the parallelogram is a rectangle", which can be proven with Thales. But then, the article is about laying out a proof of Thales so if it's using a lemma that leveraged Thales itself that's circular logic right there and so we're in trouble. One other way to prove Thales is by computing angles, but then you've got yourself a proven Thales so the point of the article is moot.

rocqua 2y ago

I don't think the note "Since the triangle got rotated halfway around the circle, the tip must end up exactly opposite from where it started. That's why the diagonal of the box is a diameter." is needed, nor do you need to show the diagonals of the parallelogram are equally long. Rather, it suffices to simply state, the only parellelogram that is inscribed in a circle is a rectangle. For the less rigorous, that statement is obviously true. For complete rigor, it is sufficient to argue that the center of the circle, combined with the vertices of the parallelogram forms an isoceles triangle. So the centre of the circle must lie on the bisector all edges of a parallelogram. But on a non-rectangle parellelogram the bisectors of opposite edges never intersect.

lupire 2y ago
If your informal proof is to draw a picture and say "it's obviously true by lookikg at it", then you don't need a parallelogram at all. Just draw the diametric triangle and look at it.
- rocqua 2y ago
  Hence the addition of a formal proof based on the bisectors of opposite edges of a parallelogram.

forgotpwd16 2y ago

To be honest, I prefer the unattractive and inelegant proof given alongside that one in Lockhart's essay.

lupire 2y ago
Lockhart missed the point a bit. His student's "proof" is an illustration, not a proof. There's no way to know if its circular or simply unfounded, since it is purely an appeal to intuition. That's only a part of mathematics. The correct thing to do, mathematically, is to validate the intuition by formatting it as a proper proof, based on non-circular axioms and theorems. In the paper itself he admits that his student's work was incoherent and needed him to rewrite it. He's right that 2 column geometry proofs are ugly, and could be presented better. This has been known for centuries. https://www.c82.net/euclid/en/book3/#prop31 https://www.c82.net/euclid/en/book3/#prop31 For children and for starting out, the pictures are great. But for mathematics, pictures are extremely limiting. 2D and 3D are notgod models for N-Dimemsions. ("Spiky balls" , for example). Mathematics is far, far more powerful than human eyes. The amazing thing about geometry is that the whole thing works without pictures! A blind person can be a great geometer, because geometry is axiomatizable. Meanwhile, Euclid's Elements, while an incredible achievement in its day, is not well-founded, relying on unstated axioms.
- JadeNB 2y ago
  > Lockhart missed the point a bit. I think it's fairer to say that his point may not have been what you expected it to be. As a teacher myself, trust me, if a student comes to you and says "I came up with a proof!" and you say "no, you see, what you have is an intuitive explanation that can possibly be turned into a proof," then all that will happen is that that student will not be interested any more in exploring, or at least will not be interested any more in sharing their explorations with you. At that point, you have both lost. Lockhart is well aware of the standard of mathematical proof, and knows that, all else aside, this theorem is not in want of proof. His focus is on the fact that, if we want mathematics to remain a live profession, then we must improve our pedagogy, and help to train students who enjoy and want to pursue mathematics—even if it means occasionally accepting less than maximally rigorous mathematics from a seventh grader.

t_mann 2y ago

> Since the triangle got turned completely around, the sides of the box must be parallel, so it makes a parallelogram. But it can't be a slanted box because both of its diagonals are diameters of the circle, so they're equal, which means it must be an actual rectangle. I'd be careful with such "visual proofs", even more so if accompanied by such handwavy reasoning. Eg, do we know that both diagonals are diameters? Do we know that a parallelogram with equal diagonals is a rectangle? While in this case things do work out nicely, I'd say this is almost more luck than a real proof - it's easy to mistakenly "prove" stuff like Pi=4 with similar reasoning. I believe 3B1B even has a video on the topic.

pringk02 2y ago
> do we know that both diagonals are diameters This must be true, because the diagonals are both straight lines that go through the centre and are bound by the edges, so it follows they must be equal to the diameter of the circle by definition. > Do we know that a parallelogram with equal diagonals is a rectangle? As another commenter points out, this is a theorem you can reach for, but proving it by itself is a bit more of a task.
- HarHarVeryFunny 2y ago
  > This must be true, because the diagonals are both straight lines that go through the centre How do we know the 2nd diagonal goes through the center ? Is it because of the construction by rotation ?
  - surajms 2y ago
    Yes. So, here when we rotate the triangle, we are essentially rotating each of the endpoints. For each endpoint, we rotate it by 180 degrees around the line segment joining the endpoint and the center. This by definition will result in a new position for each endpoint that creates a chord (as the two endpoints lie on the circle) and passes through the center (we rotated around it). A chord that passes through the center is by definition a diameter.

rodneyzeng 2y ago

The normal approach to prove Thale's theorem should be induced from the property of central angle being twice of an inscribed angle that subtends the same arc. Since a diameter has central angle of 180 degrees, its corresponding inscribed angle should be half of 180, that is 90 degrees.

lupire 2y ago
No, because if you do it that way, you wouldn't have Thales's Theorem. It would be Thales's Trivial Corollary. Thales's Theroem is a simpler, easier to prove (as in OP), less powerful statement than the inscribed angle theorem.
- rodneyzeng 2y ago
  Your second sentence denies the first sentence. The proof of the Inscribed angle theorem does not need Thale's Theorem, and it is stronger than Thale's Theorem.

lupire 2y ago

Euclid's proof, with Byrne's beautiful visualization: https://www.c82.net/euclid/en/book3/#prop31 https://www.c82.net/euclid/en/book3/#prop31

ralferoo 2y ago

The corners of the "parallelogram" in the diagram don't touch the circle at the top or bottom. So, those two corners wouldn't be right angles, but instead would be slightly obtuse - rather like this comment!

tantalor 2y ago
huh
- vidarh 2y ago
  One of the illustrations in the article is off by a few pixels. Which of course has no bearing on the text, and so pointing it out above was intentionally obtuse, presumably for the sake of the pun.

abecedarius 2y ago

Perhaps this was essentially Thales's own proof: https://intellectualmathematics.com/blog/first-proofs-thales-and-the-beginnings-of-geometry/ https://intellectualmathematics.com/blog/first-proofs-thales... BTW it's also quite direct using vectors: the legs of the triangle are the sum and difference of radius vectors. Take their dot product, distribute it, it's zero because radii are the same length.

jacobolus 2y ago
Thales and Pythagoras are quasi-mythical figures, and we don't actually know anything concrete about any mathematical accomplishments they might have had, which are all apocryphal and date from many centuries after their deaths. Greek deductive mathematics per se dates from at least a century after Thales' time, while many of the basic facts about Euclidean geometry were understood in ancient Egypt and Mesopotamia long before him, and it is most likely that Thales himself never did any of the mathematical or scientific things attributed to him. Viktor Blåsjö's speculation that ancient Greeks began with the same insight as Lockhart's 7th grade student is plausible but is not backed by any evidence whatsoever. (This is an insight that many people have had over the centuries, certainly including anyone deeply investigating cyclic quadrilaterals, but also probably plenty masons or metalworkers working with circles and right angles, etc.) > sum and difference of radius vectors This is a nice one. Another way to use Thales' theorem in characterizing a circle, without involving the center point, is to start with one point P on a circle and a vector d which is a diameter from that point to the antipodal point. Then the vector v from P to any other point Q on the circle satisfies v² = v · d, or equivalently v · (v − d) = 0.

deleted 2y ago

[deleted]

fraukeadeline 2y ago

[dead]

dang 2y ago

Related. Others? A Mathematician's Lament (2002) [pdf] - https://news.ycombinator.com/item?id=35929333 https://news.ycombinator.com/item?id=35929333 - May 2023 (47 comments) A Mathematicians Lament [pdf] - https://news.ycombinator.com/item?id=30829704 https://news.ycombinator.com/item?id=30829704 - March 2022 (3 comments) A Mathematician's Lament [pdf] - https://news.ycombinator.com/item?id=15385104 https://news.ycombinator.com/item?id=15385104 - Oct 2017 (1 comment) A Mathematician’s Lament (2002) [pdf] - https://news.ycombinator.com/item?id=14331752 https://news.ycombinator.com/item?id=14331752 - May 2017 (27 comments) A Mathematician’s Lament (2002) [pdf] - https://news.ycombinator.com/item?id=8845507 https://news.ycombinator.com/item?id=8845507 - Jan 2015 (89 comments) A Mathematician's Lament [pdf] - https://news.ycombinator.com/item?id=6994939 https://news.ycombinator.com/item?id=6994939 - Jan 2014 (1 comment) A Mathematician’s Lament (2002) [pdf] - https://news.ycombinator.com/item?id=6187014 https://news.ycombinator.com/item?id=6187014 - Aug 2013 (119 comments) New book Measurement by author of Lockhart's Lament - https://news.ycombinator.com/item?id=4317199 https://news.ycombinator.com/item?id=4317199 - July 2012 (4 comments) Mathematician's Lament: An essay on math education and on how we view math - https://news.ycombinator.com/item?id=666563 https://news.ycombinator.com/item?id=666563 - June 2009 (18 comments) On Math Teaching: Lockhart's Lament - https://news.ycombinator.com/item?id=256176 https://news.ycombinator.com/item?id=256176 - July 2008 (21 comments) Lockhart's Lament: On Mathematics at School - https://news.ycombinator.com/item?id=130499 https://news.ycombinator.com/item?id=130499 - March 2008 (20 comments)