On January 10, 2024 at 5:39PM EST Hanna wrote:

I thought I had it… but… I need to try again.

On December 25, 2023 at 8:00PM EST remy585 wrote:

And here I was treating the letters like variables and trying to figure out: if x1x1 = oo + hh and hh + ff = x2x2 + x2x2 etc then perhaps x1 is the word “oh” and the others are… was way more fun when I facepalmed into the hints

On December 13, 2023 at 11:43AM EST vanillaicee wrote:

I found this quite hard for an "easy" theorem.

On September 16, 2022 at 9:50PM EST Not_X wrote:

how do you solve this? I've gone as far as finding all the numbers...where do the letters come into play?

On January 30, 2022 at 5:49PM EST TheJhonny wrote:

This was a fun one. I like math. I need to work on the more advanced stuff.

On January 13, 2022 at 10:39AM EST kiko13 wrote:

I made the mistake of thinking each x resulted in a word instead of a single letter. Once I realized that approach was wrong, it was pretty straight forward for me. Interesting theorem. :-)

On December 26, 2021 at 12:40PM EST EMC2 responded to hiabc:

You got me curious when you said this, so I plotted it out on an online geometry solver, but I think it does in fact work.

See here (spoiler warning):

https://www.geogebra.org/classic/fd42nmze

Take a look and the segment length and angle calculations at left. It all seems to match up.

On December 26, 2021 at 1:47AM EST hiabc wrote:

Not sure where to put this, so I put it here:

The diagram is technically impossible with the numbers that you end up with. In particular, it isn't drawn this way, but the diagonals of the square are necessarily parallel to the two edges labeled h and f. This means that the bottom right triangle (the one involving c and x_3) must have equal side lengths, since that triangle should "look like" one of the four triangles formed by connecting the center of the square to the vertices of the square. Not sure how best to explain this all, but I guess the point is simply that forcing the square to be a square makes the whole diagram not really work.

On December 26, 2021 at 1:47AM EST hiabc wrote:

Not sure where to put this, so I put it here:

The diagram is technically impossible with the numbers that you end up with. In particular, it isn't drawn this way, but the diagonals of the square are necessarily parallel to the two edges labeled h and f. This means that the bottom right triangle (the one involving c and x_3) must have equal side lengths, since that triangle should "look like" one of the four triangles formed by connecting the center of the square to the vertices of the square. Not sure how best to explain this all, but I guess the point is simply that forcing the square to be a square makes the whole diagram not really work.

On April 14, 2021 at 4:05PM EST chilludio wrote:

This isn't as much of a spoiler as it is a hint but still:

Having a unit grid was very clever, M

On March 31, 2021 at 11:19AM EST EMC2 responded to TragicKnavery:

You mean when checking the solution on the website? Uppercase seems to work for me... are you sure you have the right answer?

On March 31, 2021 at 11:11AM EST TragicKnavery wrote:

Is there a reason why uppercase does not work? It would seem that would be the preferred format for this answer.

On March 31, 2021 at 11:11AM EST TragicKnavery wrote:

Is there a reason why uppercase does not work? It would seem that would be the preferred format for this answer.

On March 31, 2021 at 1:57AM EST Loopykd responded to themselves:

Ok ok this is why I got this in the first place! Thank you. I have to wake up my old brain!!!🤪

On March 31, 2021 at 1:04AM EST Loopykd wrote:

I haven't been so frustrated by something since elementary school. Not fun! 🥺

On March 31, 2021 at 1:04AM EST Loopykd wrote:

I haven't been so frustrated by something since elementary school. Not fun! 🥺

On January 17, 2021 at 12:17PM EST ukla responded to EMC2:

Ha, ha! Yes, I'm starting to learn a bit about that! Thanks!

On January 16, 2021 at 10:07AM EST EMC2 responded to ukla:

@ukla You are way overthinking this 😅. Keep it simple.

On January 16, 2021 at 9:29AM EST ukla wrote:

Okay, I'm going to give in to my need to talk about this. Based on the shape of the math problem, I searched online to see if there was a pythagorean cipher, there was. And when what I found(just an image, really)about the pythagorean cipher online told me that all numbers MUST be reduced to a single digit. I believed it.

Using 6 for O (O=15, 1+5=6), results in all sorts of interesting possibilities and a need for further research. When you reduce, each of the x's has 3 letters associated with it. AJS, ENW, DMV, respectively. I'll skip a bunch about reasoning out which letters it ought to be, letter-combining into phrases, searching for names of theorem masters, etc., and skip to "Masters of Many Theorems" as a support for what the answer might be, as well as the words "novel" and "steering" near the end of M's discussion. Some searching yields a novel* where the driver named Euclid* Madden* steers* his streetcar into Teddy Roosevelt. If you constrain your answers to the book(a review of)and to beginning with the x-letters, yields Sherlock, Euclid, Madden - but the red lines M gives are all the same length - it would be convenient if there were a Jacobi in the book... even more awesome if Sherlock and Watson could be replaced by Hay, Bly and ? A much more "National Treasure" line of solution-finding, which I admit, though incomplete here, appeals to me.

..

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On January 16, 2021 at 10:07AM EST EMC2 responded to ukla:

@ukla You are way overthinking this 😅. Keep it simple.

On January 16, 2021 at 9:29AM EST ukla wrote:

Okay, I'm going to give in to my need to talk about this. Based on the shape of the math problem, I searched online to see if there was a pythagorean cipher, there was. And when what I found(just an image, really)about the pythagorean cipher online told me that all numbers MUST be reduced to a single digit. I believed it.

Using 6 for O (O=15, 1+5=6), results in all sorts of interesting possibilities and a need for further research. When you reduce, each of the x's has 3 letters associated with it. AJS, ENW, DMV, respectively. I'll skip a bunch about reasoning out which letters it ought to be, letter-combining into phrases, searching for names of theorem masters, etc., and skip to "Masters of Many Theorems" as a support for what the answer might be, as well as the words "novel" and "steering" near the end of M's discussion. Some searching yields a novel* where the driver named Euclid* Madden* steers* his streetcar into Teddy Roosevelt. If you constrain your answers to the book(a review of)and to beginning with the x-letters, yields Sherlock, Euclid, Madden - but the red lines M gives are all the same length - it would be convenient if there were a Jacobi in the book... even more awesome if Sherlock and Watson could be replaced by Hay, Bly and ? A much more "National Treasure" line of solution-finding, which I admit, though incomplete here, appeals to me.

..

On January 16, 2021 at 9:29AM EST ukla wrote:

Okay, I'm going to give in to my need to talk about this. Based on the shape of the math problem, I searched online to see if there was a pythagorean cipher, there was. And when what I found(just an image, really)about the pythagorean cipher online told me that all numbers MUST be reduced to a single digit. I believed it.

Using 6 for O (O=15, 1+5=6), results in all sorts of interesting possibilities and a need for further research. When you reduce, each of the x's has 3 letters associated with it. AJS, ENW, DMV, respectively. I'll skip a bunch about reasoning out which letters it ought to be, letter-combining into phrases, searching for names of theorem masters, etc., and skip to "Masters of Many Theorems" as a support for what the answer might be, as well as the words "novel" and "steering" near the end of M's discussion. Some searching yields a novel* where the driver named Euclid* Madden* steers* his streetcar into Teddy Roosevelt. If you constrain your answers to the book(a review of)and to beginning with the x-letters, yields Sherlock, Euclid, Madden - but the red lines M gives are all the same length - it would be convenient if there were a Jacobi in the book... even more awesome if Sherlock and Watson could be replaced by Hay, Bly and ? A much more "National Treasure" line of solution-finding, which I admit, though incomplete here, appeals to me.

..

On January 14, 2021 at 10:58AM EST ukla wrote:

For M:

1) Clearly, I need to find a better resource for learning ciphers than the internet. It told me to reduce the number for O to a single digit. Using 6 for O was great for a variety of interesting wrong answers. Hours of thinking there must be another hidden layer or two and intricacy to the final answer.

2) Did you not know there was a NOVEL about a Euclid Madden who STEERED his trolley into Teddy Roosevelt? Or was that meant to mislead?

Thank you!

On January 14, 2021 at 10:44AM EST ukla wrote:

Oh! Maybe I'll read the answer. A posting from Dec. 20 says the online solution board doesn't acknowledge the correct answer.

On January 14, 2021 at 10:42AM EST ukla wrote:

Do I need to read the book by B.S. to answer this?

On January 6, 2021 at 10:50AM EST GeekMoJoe wrote:

I used Hint 2 to help guide me. This was a fun one, and helped me relearn things I had mostly brain dumped.