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#1 h Mar 2, 2025, 5:48 AM

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How difficult is it to get into USA Canada math camp? What should be expected from an accepted applicant in terms of the qualifying quiz, essays and other awards or math context?

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#2 h Mar 2, 2025, 7:53 PM

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From the FAQ:

"How do you select students?

First, we're looking for evidence in the application that a student is mathematically prepared for our curriculum. (If you aren't ready for Mathcamp's classes, then you won't have a good time at the program.) We learn about your mathematical preparation via several avenues in the application, but primarily from your Qualifying Quiz solutions. Once we've established that baseline and we're convinced that a student is mathematically prepared, we're looking for fit between applicants and the program. Your "About You" section is the primary way for us to learn more about you as a person and to understand what moves you to apply to Mathcamp. We're looking for students who will really benefit from Mathcamp, academically and personally."

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deduck

#3 h Mar 2, 2025, 8:03 PM • 1 Y

Y by megarnie

Bread10 wrote:

How difficult is it to get into USA Canada math camp? What should be expected from an accepted applicant in terms of the qualifying quiz, essays and other awards or math context?

its super hard to get in!!!

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abbominable_sn0wman

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abbominable_sn0wman

#4 h Mar 3, 2025, 2:20 AM

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Are we allowed to talk about the problems yet?

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#5 h Mar 3, 2025, 6:23 PM

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abbominable_sn0wman wrote:

Are we allowed to talk about the problems yet?

no

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EGMO

#6 h Mar 4, 2025, 3:55 AM

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when will we be allowed to

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#7 h Mar 4, 2025, 3:58 PM

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April 17th

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#8 h Mar 20, 2025, 12:45 PM • 9 Y

Y by Pengu14, ihatemath123, RaymondZhu, Alex-131, aidan0626, Jack_w, akliu, MathPerson12321, Yiyj1

Update from Mathcamp HQ: I opened up solution chat early this year because we're done early with file review. Discuss away!

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bjump

#9 h Mar 20, 2025, 3:30 PM

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Anyone mind sharing there solution to 6c?

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#11 h Mar 20, 2025, 4:49 PM

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I got the following.
Anna and Benny will not pass within 1 meter of each other infinitely often
Anna and Charlotte will pass within 1 meter of each other infinitely often.
Charlotte and Benny will be of distance 1 from each other in a finite amount of time.

For Anna and Charlotte a got a construction.
In the other 2, I looked at the pattern from $H_n$ to $H_{n+k}.$ I forgot what my $k$ was. Then we can clearly see they will finitely pass each other.

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#12 h Mar 20, 2025, 6:14 PM

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Anna and Charlotte is the only satisfactory pair:

For Charlotte and Anna, since their speeds have a ratio of $3\!:\!1$ , they will be within one meter infinitely often. We have
$L(3,3)=10 \to L(2,3)=11,\quad L(4,4)=32 \to L(4,3)=31.$ When $L(3,3)$ moves toward $L(2,3)$ and $L(4,2)$ moves toward $L(4,3)$ , there will be a moment where their distances are 3 times apart, with a physical separation of $1$ meter (i.e., the same $x$ -value, but their $y$ -values differ by $1$ meter).

Similarly, for all $n=4^k$ , we have
$L(4^k,4^k)=2\cdot4^{2k},$ and
$L(4^k-1,4^k-1)=\sum_{i=0}^{2k-1} 2\cdot4^i.$ Thus,
$L(4^k-1,4^k-1)\cdot 3 = \sum_{i=0}^{2k-1}\Bigl(2\cdot4^i\cdot 3\Bigr) = \sum_{i=0}^{2k-1}\Bigl(2\cdot4^i\,(4-1)\Bigr) = 2\cdot4^{2k}-2.$
Therefore, between
$L(4^k-2,4^k) \quad \text{and} \quad L(4^k-1,4^k),$ and between
$L(4^k-2,4^k-1) \quad \text{and} \quad L(4^k-1,4^k-1),$ there will be a point where Anna and Charlotte have the same $x$ -value and a $y$ -difference of $1$ meter.

Rigorously, we determine the following:

For $n=4^k$ , where $k$ is a positive integer,
$L(n,n)=L(4^k,4^k)=2\cdot4^{2k},$ $L(n-1,n-1)=2\sum_{i=0}^{2k-1} 4^i,$ and
$L(n-1.25,n)=L(n,n)-1.25 = 2\cdot4^{2k}-1.25.$ Also,
$L(n-1.25,n-1)=L(n-1,n-1)+0.25.$ Thus, we obtain:
$3\cdot L(n-1.25, n-1)=3\cdot L(n-1,n-1)+0.75.$ Since
$3\cdot L(n-1,n-1)=3\cdot 2\sum_{i=0}^{2k-1} 4^i,$ we can write
$3\cdot L(n-1,n-1)+0.75 = 2\sum_{i=0}^{2k-1} 4^i\,(4-1)+0.75 = 2\cdot4^{2k}-2+0.75 = 2\cdot4^{2k}-1.25.$ Thus,
$3\cdot L(n-1.25, n-1) = L(n-1.25,n).$
Therefore, for any positive integer $k$ , the points
$(4^k-1.25,\,4^k-1) \quad \text{and} \quad (4^k-1.25,\,4^k)$ will be such that Anna and Charlotte are within 1 meter of each other. Since $k$ can be any positive integer, this event occurs infinitely often.

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#13 h Mar 20, 2025, 6:31 PM

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eugenewang1 wrote:

Anna and Charlotte is the only satisfactory pair:

For Charlotte and Anna, since their speeds have a ratio of $3\!:\!1$ , they will be within one meter infinitely often. We have
$L(3,3)=10 \to L(2,3)=11,\quad L(4,4)=32 \to L(4,3)=31.$ When $L(3,3)$ moves toward $L(2,3)$ and $L(4,2)$ moves toward $L(4,3)$ , there will be a moment where their distances are 3 times apart, with a physical separation of $1$ meter (i.e., the same $x$ -value, but their $y$ -values differ by $1$ meter).

Similarly, for all $n=4^k$ , we have
$L(4^k,4^k)=2\cdot4^{2k},$ and
$L(4^k-1,4^k-1)=\sum_{i=0}^{2k-1} 2\cdot4^i.$ Thus,
$L(4^k-1,4^k-1)\cdot 3 = \sum_{i=0}^{2k-1}\Bigl(2\cdot4^i\cdot 3\Bigr) = \sum_{i=0}^{2k-1}\Bigl(2\cdot4^i\,(4-1)\Bigr) = 2\cdot4^{2k}-2.$
Therefore, between
$L(4^k-2,4^k) \quad \text{and} \quad L(4^k-1,4^k),$ and between
$L(4^k-2,4^k-1) \quad \text{and} \quad L(4^k-1,4^k-1),$ there will be a point where Anna and Charlotte have the same $x$ -value and a $y$ -difference of $1$ meter.

Rigorously, we determine the following:

For $n=4^k$ , where $k$ is a positive integer,
$L(n,n)=L(4^k,4^k)=2\cdot4^{2k},$ $L(n-1,n-1)=2\sum_{i=0}^{2k-1} 4^i,$ and
$L(n-1.25,n)=L(n,n)-1.25 = 2\cdot4^{2k}-1.25.$ Also,
$L(n-1.25,n-1)=L(n-1,n-1)+0.25.$ Thus, we obtain:
$3\cdot L(n-1.25, n-1)=3\cdot L(n-1,n-1)+0.75.$ Since
$3\cdot L(n-1,n-1)=3\cdot 2\sum_{i=0}^{2k-1} 4^i,$ we can write
$3\cdot L(n-1,n-1)+0.75 = 2\sum_{i=0}^{2k-1} 4^i\,(4-1)+0.75 = 2\cdot4^{2k}-2+0.75 = 2\cdot4^{2k}-1.25.$ Thus,
$3\cdot L(n-1.25, n-1) = L(n-1.25,n).$
Therefore, for any positive integer $k$ , the points
$(4^k-1.25,\,4^k-1) \quad \text{and} \quad (4^k-1.25,\,4^k)$ will be such that Anna and Charlotte are within 1 meter of each other. Since $k$ can be any positive integer, this event occurs infinitely often.

What did you do for the others?

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#14 h Mar 20, 2025, 7:10 PM

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which ones

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690 posts

#15 h Mar 20, 2025, 7:26 PM

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@above, for Anna and Benny, and Benny and Charlotte.

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#16 h Mar 20, 2025, 8:03 PM

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both not true, did casework on repeating patterns

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1746 posts

akliu

#17 h Mar 20, 2025, 10:36 PM

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I'd upload my 7-page solution (including diagrams) onto here, but can't find a nice way to do it. Here's what I did for 6c:

The cheese solution:
For Problem 6c, it’s trivial for all three pairs. There always exists a continuous time interval when pairs of these three people are within 1 meter of each other at the start, meaning that all three pairs satisfy this condition infinitely often! (It goes without being said that I don’t need to provide an actual solution after this masterpiece of a solution.)

The actual solution:
I will assume that the problem is asking us to prove whether the number of times that there is some interval of time where two people are exactly $1$ meter apart is finite or infinite. When we consider such intervals, we note that if there were to be an infinite amount of intervals, there must also be an infinite amount of times that the distance between the two people is exactly $1$ ; for each interval, consider the start of each distinct interval. Thus, it is sufficient to prove that there exists a finite or infinite amount of times that two people are exactly $1$ meter apart for each pair.

This is just setup; my answer was Anna + Benny and Benny + Charlotte was finite, while Anna + Charlotte was infinite. The proof overall for all three sections was to prove that the "duplicate" of $H_{n}$ that each of the two important people are in are not orthogonally adjacent. This works for practically everything except the Anna + Charlotte case, which has a bit of some annoying casework. Because of this, I had to invoke Figure 26 (attached below).

My favorite problems were probably Problem 4 and Problem 6! How did you guys do problem 4?

Attachments:

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#18 h Mar 20, 2025, 11:32 PM

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p4)

Let the 12-sided crescent polyhedron be centered at the point $(0,0,0)$ . Because the polyhedron is equilateral and has 12 faces, 30 edges, and 20 vertices, it is also symmetrical and can be represented by the paper model below:

\begin{center}
\includegraphics[width=0.5\linewidth]{Crescent Paper Model.jpg}
\end{center}

Thus, we can inscribe it in a cube, and we have 12 of the following crescents:

\begin{center}
\includegraphics[width=1\linewidth]{Crescents.pdf}
\end{center}

Given equal side lengths, we assume each side length of the crescents to be 1, and by the golden ratio, we have that
$\frac{\phi}{1} = l,$ where $l$ is the length of the cube the polyhedron is inscribed inside.

Thus, we can find the coordinates of each of the 8 outer vertices to be:
$v_{1-8} = (x,y,z) = \left(\pm\frac{\sqrt{5}+1}{4}, \pm\frac{\sqrt{5}+1}{4}, \pm\frac{\sqrt{5}+1}{4}\right).$ Now, we wish to find the remaining symmetrical inner 12 vertices of the polyhedron. Each of these points is marked by one short axis $a$ , one longer axis $b$ , and an axis of length 0.

To find the shorter axis $a$ , we set up similar triangles:

\begin{center}
\includegraphics[width=1\linewidth]{Crescent 2.pdf}
\end{center}

Thus, from similar triangles, we have the equation:
$\frac{x}{1+x} = \frac{a}{1} = a.$
Now, we find $x$ :

First, we do angle chasing to find $x$ as the base of an isosceles triangle (i.e. the golden triangle):

\begin{center}
\includegraphics[width=1\linewidth]{Crescent 3.pdf}
\end{center}

Then, we solve for $x$ :

$\frac{1}{x} = \phi, \quad \text{so} \quad x = \frac{1}{\phi}.$
Finally, we can solve for $a$ :

$a = \frac{\frac{1}{\phi}}{1 + \frac{1}{\phi}} = \frac{1}{\phi+1}.$
From symmetry, the longer axis $b$ has length $s/2 = 1/2$ .

Thus, we can define these 12 inner vertices of the polyhedron to be:

$v_{9-20} = \left(\pm\frac{1}{1+\phi}, \pm\frac{1}{2}, 0\right), \left(0, \pm\frac{1}{1+\phi}, \pm\frac{1}{2} \right), \left(\pm\frac{1}{2}, 0, \pm\frac{1}{1+\phi}\right)$
$= \left(\pm\frac{2}{3+\sqrt{5}}, \pm\frac{1}{2}, 0\right), \left(0, \pm\frac{2}{3+\sqrt{5}}, \pm\frac{1}{2}\right), \left(\pm\frac{1}{2}, 0, \pm\frac{2}{3+\sqrt{5}}\right).$
Each vertex touches 3 pieces, so we have $5 \times 12 / 3 = 20$ vertices, which is consistent with our results.

Since all 12 crescents are symmetrical, we only need to prove the validity of one crescent in the polyhedron in order to prove the existence of the polyhedron.

Let the crescent have vertices:

$\left(\frac{\sqrt{5}+1}{4}, \frac{\sqrt{5}+1}{4}, \frac{\sqrt{5}+1}{4}\right), \quad \left(\frac{2}{3+\sqrt{5}}, \frac{1}{2}, 0\right), \quad \left(\frac{\sqrt{5}+1}{4}, \frac{\sqrt{5}+1}{4}, -\frac{\sqrt{5}+1}{4}\right),$ $\left(0, \frac{2}{3+\sqrt{5}}, -\frac{1}{2}\right), \quad \left(0, \frac{2}{3+\sqrt{5}}, \frac{1}{2}\right)$
First, we need to prove that all 5 side lengths are equal:

Using the distance formula, we see that:

$d_{AB} = \sqrt{\left( \frac{2}{3+\sqrt{5}} - \frac{\sqrt{5}+1}{4} \right)^2 + \left( \frac{1}{2} - \frac{\sqrt{5}+1}{4} \right)^2 + \left( 0 - \frac{\sqrt{5}+1}{4} \right)^2 }$ $= 1$
$d_{BC} = \sqrt{\left( \frac{\sqrt{5}+1}{4} - \frac{2}{3+\sqrt{5}} \right)^2 + \left( \frac{\sqrt{5}+1}{4} - \frac{1}{2} \right)^2 + \left( -\frac{\sqrt{5}+1}{4} - 0 \right)^2 }$ $= 1$
$d_{CD} = \sqrt{\left( 0 - \frac{\sqrt{5}+1}{4} \right)^2 + \left( \frac{2}{3+\sqrt{5}} - \frac{\sqrt{5}+1}{4} \right)^2 + \left( -\frac{1}{2} + \frac{\sqrt{5}+1}{4} \right)^2 }$ $= 1$
$d_{DE} = \sqrt{\left( 0 - 0 \right)^2 + \left( \frac{2}{3+\sqrt{5}} - \frac{2}{3+\sqrt{5}} \right)^2 + \left( \frac{1}{2} + \frac{1}{2} \right)^2 }$ $= 1$
$d_{EA} = \sqrt{\left( \frac{\sqrt{5}+1}{4} - 0 \right)^2 + \left( \frac{\sqrt{5}+1}{4} - \frac{2}{3+\sqrt{5}} \right)^2 + \left( \frac{\sqrt{5}+1}{4} - \frac{1}{2} \right)^2 }$ $= 1$
Thus, we conclude that all sides are equal.

Now, we just must prove that the lengths AD and CE intersect B to check that all the angles match the 108°-36°-252°-36°-108° requirement to form a crescent pentagon, and we have already proved this while angle chasing to find x.

Thus, our proof is complete and it is possible to make a 12-sided polyhedron each of whose faces is a crescent.

\textbf{References:}
Golden Ratio Applications in Pentagons: \url{https://dynamicmathematicslearning.com/parallel-pentagon-golden-ratio.html}

Golden Triangle: \url{https://mathworld.wolfram.com/GoldenTriangle.html}

\item[(b)] A {\em net} for a polyhedron is a connected arrangement of polygons in the plane that can be folded along a subset of the edges to create the polyhedron. For example, here are a few possible nets for a cube:

\begin{center}
\includegraphics[width = 3in]{cube nets.png}
\end{center}

Is it possible to make a net for the polyhedron that you constructed in part (a)? Keep in mind that the twelve crescents forming the net are not allowed to overlap (except at the edges).

\begin{solution} We have already proved that our paper model construction is a valid representation of a 12-sided crescent polyhedron, so we just need to produce a valid paper net configuration so that the twelve crescents forming the net do not overlap (except at the edges).

Experimenting, we find that the net configuration:

\begin{center}
\includegraphics[width=0.75\linewidth]{Net Joints.png}
\end{center}

forms a valid 12-sided crescent polyhedron.

Thus, our proof is complete, and it is possible to make a net for the 12-sided crescent polyhedron.

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#19 h Mar 20, 2025, 11:32 PM

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oops my images didn’t load

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akliu

#20 h Mar 21, 2025, 1:09 AM

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I had a solution using phantom point trig to prove that a crescent face's edge was on the plane perpendicular to another crescent face.

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690 posts

#21 h Mar 21, 2025, 3:44 AM

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This is P4
Let's define $a = \frac{1}{2}, b=\frac{\sqrt{5} - 1}{4}, c =\frac{1+\sqrt{5}}{4}.$ The coordinates are $\begin{align} (a, b, 0), (-a, b, 0), (0, a, b), (c, c, c), (-c, c, c)\\ (a, b, 0), (-a, b, 0), (0, a, -b), (c, c, -c), (-c, c, -c)\\ (a, -b, 0), (-a, -b, 0), (0, -a, b), (c, -c, c), (-c, -c, c)\\ (a, -b, 0), (-a, -b, 0), (0, -a, -b), (c, -c, -c), (-c, -c, -c)\\ (-b, 0, a), (-b, 0, -a), (-a, b, 0), (-c, c, c), (-c, c, -c)\\ (-b, 0, a), (-b, 0, -a), (-a, -b, 0), (-c, -c, c), (-c, -c, -c)\\ (b, 0, a), (b, 0, -a), (a, b, 0), (c, c, c), (c, c, -c)\\ (b, 0, a), (b, 0, -a), (a, -b, 0), (c, -c, c), (c, -c, -c)\\ (0, a, b), (0, -a, b), (b, 0, a), (c, c, c), (c, -c, c)\\ (0, a, b), (0, -a, b), (-b, 0, a), (-c, c, c), (-c, -c, c)\\ (0, a, -b), (0, -a, -b), (b, 0, -a), (c, c, -c), (c, -c, -c)\\ (0, a, -b), (0, -a, b), (-b, 0, -a), (-c, c, -c), (-c, -c, -c) \end{align}$

Attachments:

diagram_p_4.pdf (522kb)

This post has been edited 2 times. Last edited by sharknavy75, Mar 21, 2025, 3:46 AM

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1000 posts

torch

#22 h Mar 21, 2025, 4:27 AM • 2 Y

Y by ihatemath123, dppvlit123

A net existed for 4b) right? Overlapped a lot at the edges

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#23 h Mar 21, 2025, 4:35 PM

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@above yes I made a diagram of it

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akliu

#24 h Mar 21, 2025, 7:36 PM • 2 Y

Y by ihatemath123, williamxiao

Whoops, time to break 1729 posts... Here are some cool drawings of nets that I made in mspaint for problem 4B!

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653 posts

#25 h Mar 21, 2025, 10:30 PM

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oops I thought it was impossible and gave an invalid proof for it

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#26 h Mar 22, 2025, 3:09 AM

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https://www.youtube.com/watch?v=xtHiD4_Za8I

4b animation for folding

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1746 posts

akliu

#27 h Mar 22, 2025, 1:06 PM

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thoomgus wrote:

https://www.youtube.com/watch?v=xtHiD4_Za8I

4b animation for folding

he took my burner youtube account credentials for this and i just found out yesterday lmao

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abbominable_sn0wman

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abbominable_sn0wman

#28 h Mar 24, 2025, 12:41 AM

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thoomgus wrote:

https://www.youtube.com/watch?v=xtHiD4_Za8I

4b animation for folding

very cool

Did anyone get an answer for the bonus? If so what was it? (I just cheesed I didn’t have enough time)

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#29 h Mar 24, 2025, 1:12 AM

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abbominable_sn0wman wrote:

thoomgus wrote:

https://www.youtube.com/watch?v=xtHiD4_Za8I

4b animation for folding

very cool

Did anyone get an answer for the bonus? If so what was it? (I just cheesed I didn’t have enough time)

the closest i got was bounding it to be 13-25 (accidentally mistyped the answer that i boxed but had the write numbers in the line before it

). I think this problem might be bashable with a computer.

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330 posts

#30 h Mar 24, 2025, 6:28 PM

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@akliu
I coded it
I reached char limit for n = 5 lmao
$[asy] size(10cm); pair[] points = { (0, 0), (0, 1), (1, 1), (1, 0), (2, 0), (3, 0), (3, 1), (2, 1), (2, 2), (3, 2), (3, 3), (2, 3), (1, 3), (1, 2), (0, 2), (0, 3), (0, 4), (1, 4), (1, 5), (0, 5), (0, 6), (0, 7), (1, 7), (1, 6), (2, 6), (2, 7), (3, 7), (3, 6), (3, 5), (2, 5), (2, 4), (3, 4), (4, 4), (5, 4), (5, 5), (4, 5), (4, 6), (4, 7), (5, 7), (5, 6), (6, 6), (6, 7), (7, 7), (7, 6), (7, 5), (6, 5), (6, 4), (7, 4), (7, 3), (7, 2), (6, 2), (6, 3), (5, 3), (4, 3), (4, 2), (5, 2), (5, 1), (4, 1), (4, 0), (5, 0), (6, 0), (6, 1), (7, 1), (7, 0), (8, 0), (9, 0), (9, 1), (8, 1), (8, 2), (8, 3), (9, 3), (9, 2), (10, 2), (10, 3), (11, 3), (11, 2), (11, 1), (10, 1), (10, 0), (11, 0), (12, 0), (12, 1), (13, 1), (13, 0), (14, 0), (15, 0), (15, 1), (14, 1), (14, 2), (15, 2), (15, 3), (14, 3), (13, 3), (13, 2), (12, 2), (12, 3), (12, 4), (12, 5), (13, 5), (13, 4), (14, 4), (15, 4), (15, 5), (14, 5), (14, 6), (15, 6), (15, 7), (14, 7), (13, 7), (13, 6), (12, 6), (12, 7), (11, 7), (10, 7), (10, 6), (11, 6), (11, 5), (11, 4), (10, 4), (10, 5), (9, 5), (9, 4), (8, 4), (8, 5), (8, 6), (9, 6), (9, 7), (8, 7), (8, 8), (9, 8), (9, 9), (8, 9), (8, 10), (8, 11), (9, 11), (9, 10), (10, 10), (10, 11), (11, 11), (11, 10), (11, 9), (10, 9), (10, 8), (11, 8), (12, 8), (12, 9), (13, 9), (13, 8), (14, 8), (15, 8), (15, 9), (14, 9), (14, 10), (15, 10), (15, 11), (14, 11), (13, 11), (13, 10), (12, 10), (12, 11), (12, 12), (12, 13), (13, 13), (13, 12), (14, 12), (15, 12), (15, 13), (14, 13), (14, 14), (15, 14), (15, 15), (14, 15), (13, 15), (13, 14), (12, 14), (12, 15), (11, 15), (10, 15), (10, 14), (11, 14), (11, 13), (11, 12), (10, 12), (10, 13), (9, 13), (9, 12), (8, 12), (8, 13), (8, 14), (9, 14), (9, 15), (8, 15), (7, 15), (7, 14), (6, 14), (6, 15), (5, 15), (4, 15), (4, 14), (5, 14), (5, 13), (4, 13), (4, 12), (5, 12), (6, 12), (6, 13), (7, 13), (7, 12), (7, 11), (6, 11), (6, 10), (7, 10), (7, 9), (7, 8), (6, 8), (6, 9), (5, 9), (5, 8), (4, 8), (4, 9), (4, 10), (5, 10), (5, 11), (4, 11), (3, 11), (2, 11), (2, 10), (3, 10), (3, 9), (3, 8), (2, 8), (2, 9), (1, 9), (1, 8), (0, 8), (0, 9), (0, 10), (1, 10), (1, 11), (0, 11), (0, 12), (0, 13), (1, 13), (1, 12), (2, 12), (3, 12), (3, 13), (2, 13), (2, 14), (3, 14), (3, 15), (2, 15), (1, 15), (1, 14), (0, 14), (0, 15), (0, 16), (1, 16), (1, 17), (0, 17), (0, 18), (0, 19), (1, 19), (1, 18), (2, 18), (2, 19), (3, 19), (3, 18), (3, 17), (2, 17), (2, 16), (3, 16), (4, 16), (4, 17), (5, 17), (5, 16), (6, 16), (7, 16), (7, 17), (6, 17), (6, 18), (7, 18), (7, 19), (6, 19), (5, 19), (5, 18), (4, 18), (4, 19), (4, 20), (4, 21), (5, 21), (5, 20), (6, 20), (7, 20), (7, 21), (6, 21), (6, 22), (7, 22), (7, 23), (6, 23), (5, 23), (5, 22), (4, 22), (4, 23), (3, 23), (2, 23), (2, 22), (3, 22), (3, 21), (3, 20), (2, 20), (2, 21), (1, 21), (1, 20), (0, 20), (0, 21), (0, 22), (1, 22), (1, 23), (0, 23), (0, 24), (0, 25), (1, 25), (1, 24), (2, 24), (3, 24), (3, 25), (2, 25), (2, 26), (3, 26), (3, 27), (2, 27), (1, 27), (1, 26), (0, 26), (0, 27), (0, 28), (1, 28), (1, 29), (0, 29), (0, 30), (0, 31), (1, 31), (1, 30), (2, 30), (2, 31), (3, 31), (3, 30), (3, 29), (2, 29), (2, 28), (3, 28), (4, 28), (5, 28), (5, 29), (4, 29), (4, 30), (4, 31), (5, 31), (5, 30), (6, 30), (6, 31), (7, 31), (7, 30), (7, 29), (6, 29), (6, 28), (7, 28), (7, 27), (7, 26), (6, 26), (6, 27), (5, 27), (4, 27), (4, 26), (5, 26), (5, 25), (4, 25), (4, 24), (5, 24), (6, 24), (6, 25), (7, 25), (7, 24), (8, 24), (8, 25), (9, 25), (9, 24), (10, 24), (11, 24), (11, 25), (10, 25), (10, 26), (11, 26), (11, 27), (10, 27), (9, 27), (9, 26), (8, 26), (8, 27), (8, 28), (9, 28), (9, 29), (8, 29), (8, 30), (8, 31), (9, 31), (9, 30), (10, 30), (10, 31), (11, 31), (11, 30), (11, 29), (10, 29), (10, 28), (11, 28), (12, 28), (13, 28), (13, 29), (12, 29), (12, 30), (12, 31), (13, 31), (13, 30), (14, 30), (14, 31), (15, 31), (15, 30), (15, 29), (14, 29), (14, 28), (15, 28), (15, 27), (15, 26), (14, 26), (14, 27), (13, 27), (12, 27), (12, 26), (13, 26), (13, 25), (12, 25), (12, 24), (13, 24), (14, 24), (14, 25), (15, 25), (15, 24), (15, 23), (14, 23), (14, 22), (15, 22), (15, 21), (15, 20), (14, 20), (14, 21), (13, 21), (13, 20), (12, 20), (12, 21), (12, 22), (13, 22), (13, 23), (12, 23), (11, 23), (11, 22), (10, 22), (10, 23), (9, 23), (8, 23), (8, 22), (9, 22), (9, 21), (8, 21), (8, 20), (9, 20), (10, 20), (10, 21), (11, 21), (11, 20), (11, 19), (11, 18), (10, 18), (10, 19), (9, 19), (8, 19), (8, 18), (9, 18), (9, 17), (8, 17), (8, 16), (9, 16), (10, 16), (10, 17), (11, 17), (11, 16), (12, 16), (13, 16), (13, 17), (12, 17), (12, 18), (12, 19), (13, 19), (13, 18), (14, 18), (14, 19), (15, 19), (15, 18), (15, 17), (14, 17), (14, 16), (15, 16), (16, 16), (17, 16), (17, 17), (16, 17), (16, 18), (16, 19), (17, 19), (17, 18), (18, 18), (18, 19), (19, 19), (19, 18), (19, 17), (18, 17), (18, 16), (19, 16), (20, 16), (20, 17), (21, 17), (21, 16), (22, 16), (23, 16), (23, 17), (22, 17), (22, 18), (23, 18), (23, 19), (22, 19), (21, 19), (21, 18), (20, 18), (20, 19), (20, 20), (20, 21), (21, 21), (21, 20), (22, 20), (23, 20), (23, 21), (22, 21), (22, 22), (23, 22), (23, 23), (22, 23), (21, 23), (21, 22), (20, 22), (20, 23), (19, 23), (18, 23), (18, 22), (19, 22), (19, 21), (19, 20), (18, 20), (18, 21), (17, 21), (17, 20), (16, 20), (16, 21), (16, 22), (17, 22), (17, 23), (16, 23), (16, 24), (16, 25), (17, 25), (17, 24), (18, 24), (19, 24), (19, 25), (18, 25), (18, 26), (19, 26), (19, 27), (18, 27), (17, 27), (17, 26), (16, 26), (16, 27), (16, 28), (17, 28), (17, 29), (16, 29), (16, 30), (16, 31), (17, 31), (17, 30), (18, 30), (18, 31), (19, 31), (19, 30), (19, 29), (18, 29), (18, 28), (19, 28), (20, 28), (21, 28), (21, 29), (20, 29), (20, 30), (20, 31), (21, 31), (21, 30), (22, 30), (22, 31), (23, 31), (23, 30), (23, 29), (22, 29), (22, 28), (23, 28), (23, 27), (23, 26), (22, 26), (22, 27), (21, 27), (20, 27), (20, 26), (21, 26), (21, 25), (20, 25), (20, 24), (21, 24), (22, 24), (22, 25), (23, 25), (23, 24), (24, 24), (24, 25), (25, 25), (25, 24), (26, 24), (27, 24), (27, 25), (26, 25), (26, 26), (27, 26), (27, 27), (26, 27), (25, 27), (25, 26), (24, 26), (24, 27), (24, 28), (25, 28), (25, 29), (24, 29), (24, 30), (24, 31), (25, 31), (25, 30), (26, 30), (26, 31), (27, 31), (27, 30), (27, 29), (26, 29), (26, 28), (27, 28), (28, 28), (29, 28), (29, 29), (28, 29), (28, 30), (28, 31), (29, 31), (29, 30), (30, 30), (30, 31), (31, 31), (31, 30), (31, 29), (30, 29), (30, 28), (31, 28), (31, 27), (31, 26), (30, 26), (30, 27), (29, 27), (28, 27), (28, 26), (29, 26), (29, 25), (28, 25), (28, 24), (29, 24), (30, 24), (30, 25), (31, 25), (31, 24), (31, 23), (30, 23), (30, 22), (31, 22), (31, 21), (31, 20), (30, 20), (30, 21), (29, 21), (29, 20), (28, 20), (28, 21), (28, 22), (29, 22), (29, 23), (28, 23), (27, 23), (27, 22), (26, 22), (26, 23), (25, 23), (24, 23), (24, 22), (25, 22), (25, 21), (24, 21), (24, 20), (25, 20), (26, 20), (26, 21), (27, 21), (27, 20), (27, 19), (27, 18), (26, 18), (26, 19), (25, 19), (24, 19), (24, 18), (25, 18), (25, 17), (24, 17), (24, 16), (25, 16), (26, 16), (26, 17), (27, 17), (27, 16), (28, 16), (29, 16), (29, 17), (28, 17), (28, 18), (28, 19), (29, 19), (29, 18), (30, 18), (30, 19), (31, 19), (31, 18), (31, 17), (30, 17), (30, 16), (31, 16), (31, 15), (31, 14), (30, 14), (30, 15), (29, 15), (28, 15), (28, 14), (29, 14), (29, 13), (28, 13), (28, 12), (29, 12), (30, 12), (30, 13), (31, 13), (31, 12), (31, 11), (30, 11), (30, 10), (31, 10), (31, 9), (31, 8), (30, 8), (30, 9), (29, 9), (29, 8), (28, 8), (28, 9), (28, 10), (29, 10), (29, 11), (28, 11), (27, 11), (26, 11), (26, 10), (27, 10), (27, 9), (27, 8), (26, 8), (26, 9), (25, 9), (25, 8), (24, 8), (24, 9), (24, 10), (25, 10), (25, 11), (24, 11), (24, 12), (24, 13), (25, 13), (25, 12), (26, 12), (27, 12), (27, 13), (26, 13), (26, 14), (27, 14), (27, 15), (26, 15), (25, 15), (25, 14), (24, 14), (24, 15), (23, 15), (22, 15), (22, 14), (23, 14), (23, 13), (23, 12), (22, 12), (22, 13), (21, 13), (21, 12), (20, 12), (20, 13), (20, 14), (21, 14), (21, 15), (20, 15), (19, 15), (19, 14), (18, 14), (18, 15), (17, 15), (16, 15), (16, 14), (17, 14), (17, 13), (16, 13), (16, 12), (17, 12), (18, 12), (18, 13), (19, 13), (19, 12), (19, 11), (19, 10), (18, 10), (18, 11), (17, 11), (16, 11), (16, 10), (17, 10), (17, 9), (16, 9), (16, 8), (17, 8), (18, 8), (18, 9), (19, 9), (19, 8), (20, 8), (21, 8), (21, 9), (20, 9), (20, 10), (20, 11), (21, 11), (21, 10), (22, 10), (22, 11), (23, 11), (23, 10), (23, 9), (22, 9), (22, 8), (23, 8), (23, 7), (22, 7), (22, 6), (23, 6), (23, 5), (23, 4), (22, 4), (22, 5), (21, 5), (21, 4), (20, 4), (20, 5), (20, 6), (21, 6), (21, 7), (20, 7), (19, 7), (19, 6), (18, 6), (18, 7), (17, 7), (16, 7), (16, 6), (17, 6), (17, 5), (16, 5), (16, 4), (17, 4), (18, 4), (18, 5), (19, 5), (19, 4), (19, 3), (19, 2), (18, 2), (18, 3), (17, 3), (16, 3), (16, 2), (17, 2), (17, 1), (16, 1), (16, 0), (17, 0), (18, 0), (18, 1), (19, 1), (19, 0), (20, 0), (21, 0), (21, 1), (20, 1), (20, 2), (20, 3), (21, 3), (21, 2), (22, 2), (22, 3), (23, 3), (23, 2), (23, 1), (22, 1), (22, 0), (23, 0), (24, 0), (24, 1), (25, 1), (25, 0), (26, 0), (27, 0), (27, 1), (26, 1), (26, 2), (27, 2), (27, 3), (26, 3), (25, 3), (25, 2), (24, 2), (24, 3), (24, 4), (25, 4), (25, 5), (24, 5), (24, 6), (24, 7), (25, 7), (25, 6), (26, 6), (26, 7), (27, 7), (27, 6), (27, 5), (26, 5), (26, 4), (27, 4), (28, 4), (29, 4), (29, 5), (28, 5), (28, 6), (28, 7), (29, 7), (29, 6), (30, 6), (30, 7), (31, 7), (31, 6), (31, 5), (30, 5), (30, 4), (31, 4), (31, 3), (31, 2), (30, 2), (30, 3), (29, 3), (28, 3), (28, 2), (29, 2), (29, 1), (28, 1), (28, 0), (29, 0), (30, 0), (30, 1), (31, 1), (31, 0) }; for(int i = 0; i < 1023; ++i){ draw((points[i].x, points[i].y) -- (points[i+1].x, points[i+1].y)); } [/asy]$

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This post has been edited 1 time. Last edited by Ilikeminecraft, Mar 24, 2025, 6:29 PM
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1746 posts

akliu

#31 h Mar 24, 2025, 9:51 PM

Y by

Ilikeminecraft wrote:

@akliu
I coded it
I reached char limit for n = 5 lmao
$[asy] size(10cm); pair[] points = { (0, 0), (0, 1), (1, 1), (1, 0), (2, 0), (3, 0), (3, 1), (2, 1), (2, 2), (3, 2), (3, 3), (2, 3), (1, 3), (1, 2), (0, 2), (0, 3), (0, 4), (1, 4), (1, 5), (0, 5), (0, 6), (0, 7), (1, 7), (1, 6), (2, 6), (2, 7), (3, 7), (3, 6), (3, 5), (2, 5), (2, 4), (3, 4), (4, 4), (5, 4), (5, 5), (4, 5), (4, 6), (4, 7), (5, 7), (5, 6), (6, 6), (6, 7), (7, 7), (7, 6), (7, 5), (6, 5), (6, 4), (7, 4), (7, 3), (7, 2), (6, 2), (6, 3), (5, 3), (4, 3), (4, 2), (5, 2), (5, 1), (4, 1), (4, 0), (5, 0), (6, 0), (6, 1), (7, 1), (7, 0), (8, 0), (9, 0), (9, 1), (8, 1), (8, 2), (8, 3), (9, 3), (9, 2), (10, 2), (10, 3), (11, 3), (11, 2), (11, 1), (10, 1), (10, 0), (11, 0), (12, 0), (12, 1), (13, 1), (13, 0), (14, 0), (15, 0), (15, 1), (14, 1), (14, 2), (15, 2), (15, 3), (14, 3), (13, 3), (13, 2), (12, 2), (12, 3), (12, 4), (12, 5), (13, 5), (13, 4), (14, 4), (15, 4), (15, 5), (14, 5), (14, 6), (15, 6), (15, 7), (14, 7), (13, 7), (13, 6), (12, 6), (12, 7), (11, 7), (10, 7), (10, 6), (11, 6), (11, 5), (11, 4), (10, 4), (10, 5), (9, 5), (9, 4), (8, 4), (8, 5), (8, 6), (9, 6), (9, 7), (8, 7), (8, 8), (9, 8), (9, 9), (8, 9), (8, 10), (8, 11), (9, 11), (9, 10), (10, 10), (10, 11), (11, 11), (11, 10), (11, 9), (10, 9), (10, 8), (11, 8), (12, 8), (12, 9), (13, 9), (13, 8), (14, 8), (15, 8), (15, 9), (14, 9), (14, 10), (15, 10), (15, 11), (14, 11), (13, 11), (13, 10), (12, 10), (12, 11), (12, 12), (12, 13), (13, 13), (13, 12), (14, 12), (15, 12), (15, 13), (14, 13), (14, 14), (15, 14), (15, 15), (14, 15), (13, 15), (13, 14), (12, 14), (12, 15), (11, 15), (10, 15), (10, 14), (11, 14), (11, 13), (11, 12), (10, 12), (10, 13), (9, 13), (9, 12), (8, 12), (8, 13), (8, 14), (9, 14), (9, 15), (8, 15), (7, 15), (7, 14), (6, 14), (6, 15), (5, 15), (4, 15), (4, 14), (5, 14), (5, 13), (4, 13), (4, 12), (5, 12), (6, 12), (6, 13), (7, 13), (7, 12), (7, 11), (6, 11), (6, 10), (7, 10), (7, 9), (7, 8), (6, 8), (6, 9), (5, 9), (5, 8), (4, 8), (4, 9), (4, 10), (5, 10), (5, 11), (4, 11), (3, 11), (2, 11), (2, 10), (3, 10), (3, 9), (3, 8), (2, 8), (2, 9), (1, 9), (1, 8), (0, 8), (0, 9), (0, 10), (1, 10), (1, 11), (0, 11), (0, 12), (0, 13), (1, 13), (1, 12), (2, 12), (3, 12), (3, 13), (2, 13), (2, 14), (3, 14), (3, 15), (2, 15), (1, 15), (1, 14), (0, 14), (0, 15), (0, 16), (1, 16), (1, 17), (0, 17), (0, 18), (0, 19), (1, 19), (1, 18), (2, 18), (2, 19), (3, 19), (3, 18), (3, 17), (2, 17), (2, 16), (3, 16), (4, 16), (4, 17), (5, 17), (5, 16), (6, 16), (7, 16), (7, 17), (6, 17), (6, 18), (7, 18), (7, 19), (6, 19), (5, 19), (5, 18), (4, 18), (4, 19), (4, 20), (4, 21), (5, 21), (5, 20), (6, 20), (7, 20), (7, 21), (6, 21), (6, 22), (7, 22), (7, 23), (6, 23), (5, 23), (5, 22), (4, 22), (4, 23), (3, 23), (2, 23), (2, 22), (3, 22), (3, 21), (3, 20), (2, 20), (2, 21), (1, 21), (1, 20), (0, 20), (0, 21), (0, 22), (1, 22), (1, 23), (0, 23), (0, 24), (0, 25), (1, 25), (1, 24), (2, 24), (3, 24), (3, 25), (2, 25), (2, 26), (3, 26), (3, 27), (2, 27), (1, 27), (1, 26), (0, 26), (0, 27), (0, 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2), (22, 3), (23, 3), (23, 2), (23, 1), (22, 1), (22, 0), (23, 0), (24, 0), (24, 1), (25, 1), (25, 0), (26, 0), (27, 0), (27, 1), (26, 1), (26, 2), (27, 2), (27, 3), (26, 3), (25, 3), (25, 2), (24, 2), (24, 3), (24, 4), (25, 4), (25, 5), (24, 5), (24, 6), (24, 7), (25, 7), (25, 6), (26, 6), (26, 7), (27, 7), (27, 6), (27, 5), (26, 5), (26, 4), (27, 4), (28, 4), (29, 4), (29, 5), (28, 5), (28, 6), (28, 7), (29, 7), (29, 6), (30, 6), (30, 7), (31, 7), (31, 6), (31, 5), (30, 5), (30, 4), (31, 4), (31, 3), (31, 2), (30, 2), (30, 3), (29, 3), (28, 3), (28, 2), (29, 2), (29, 1), (28, 1), (28, 0), (29, 0), (30, 0), (30, 1), (31, 1), (31, 0) }; for(int i = 0; i < 1023; ++i){ draw((points[i].x, points[i].y) -- (points[i+1].x, points[i+1].y)); } [/asy]$

That's really cool! My proof used region numbering so that wouldn't have applied for me, but I tried (and failed) to use a python turtle to draw the paths for me lol

This post has been edited 1 time. Last edited by akliu, Mar 24, 2025, 9:52 PM
Reason: grammar

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#32 h Mar 25, 2025, 9:27 PM

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what'd you guys get for P1

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#33 h Mar 25, 2025, 10:44 PM

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I got No and No. The solution was a trivial application of the acklew-thoom theorem, which essentially just nukes the problem.

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akliu

#34 h Mar 26, 2025, 1:09 AM

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mathnerd_101 wrote:

I got No and No. The solution was a trivial application of the acklew-thoom theorem, which essentially just nukes the problem.

I only applied the first half of that theorem... my friend applied only the second half.

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#35 h Mar 26, 2025, 2:15 AM

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Interesting. I used a clever combination of both halves in order to show that we could manipulate palindromes into whatever we wanted to, thus meaning that 42 is the acklew-representation of both 2025 base thoom and 2026 base thoom. Since 42 is not a palindrome, both answers were no.

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#36 h Mar 26, 2025, 3:51 PM

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That's interesting, I just used mods to check every case

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abbominable_sn0wman

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abbominable_sn0wman

#37 h Mar 27, 2025, 2:34 AM • 1 Y

Y by bjump

whats acklew thoom representation

i also got no and no

2025 was pretty trivial, cuz the multiple can't be even and both $5*10^n/2025$ and $6*10^n/2025$ start with a 2.
2026 was mostly mods and narrowing down cases by digit (1st/nth, 2nd/n-1st, 3rd/n-2nd, etc) until i got a contradiction

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torch

#38 h Mar 29, 2025, 1:17 AM

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Did anyone manage to rigorize 5c? I took discrete intervals, hopefully that was intention of the 1x1 grids?

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#39 h Mar 29, 2025, 1:39 AM

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torch wrote:

Did anyone manage to rigorize 5c? I took discrete intervals, hopefully that was intention of the 1x1 grids?

i think the 1 by 1 stuff is to have a minimum "width" which is why 5c works in the first place cause otherwise we can have each being infinitely small and not reach a conclusive finding. it reminded me a lot of why u can only fold a paper like 9 times or something (cause the paper has a quantized minimum width)

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#40 h Mar 29, 2025, 2:13 AM

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well my friend @wikjay solved all of the problems. please search up his youtube channel (it's called wikjay) and subscribe to learn how!

This post has been edited 1 time. Last edited by DarkDementor, Mar 29, 2025, 2:22 AM

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#41 h Mar 29, 2025, 2:47 AM

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torch wrote:

Did anyone manage to rigorize 5c? I took discrete intervals, hopefully that was intention of the 1x1 grids?

yes, here was my solution
did anyone rigorize 5b? I just said something along the lines of it having to be a finite forest

Attachments:

Test.pdf (97kb)

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#42 h Mar 29, 2025, 10:31 PM

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DarkDementor wrote:

well my friend @wikjay solved all of the problems. please search up his youtube channel (it's called wikjay) and subscribe to learn how!

bruh it doesnt even have it

This post has been edited 1 time. Last edited by cowstalker, Mar 29, 2025, 10:36 PM
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#43 h Yesterday at 1:05 AM

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Ilikeminecraft wrote:

torch wrote:

Did anyone manage to rigorize 5c? I took discrete intervals, hopefully that was intention of the 1x1 grids?

yes, here was my solution
did anyone rigorize 5b? I just said something along the lines of it having to be a finite forest

my framweork for it:
the whole problem you can boil down to showing that for a "pipe", it is impossimble to fill the region that the pipe splits the plane into with just more pipes. There are three cases: one is where you add a pipe that is nesting the pipe so you basically get two "U"'s stacked onto each other snugle fit, this does not reduce or simplify the problem so we move on the the next. The second case is where you place another pipe that only barely touches the original pipe. It's easy to see that this actually splits the region into 3 faces which does once again does not simplify the problem. The last case is when you place a tube that does not touch the original tube. Then the space between the two pipes can be filled out with just one pipe. There still remains another region between the new pipe to fill so it doesn't solve anything. Thus, it is impossible to fill it out.

The main logic in my proof is just simplifying the fact that this extends to infinite so finding a case that repeats the original pipe will not help at all. (Sorry for not adding any images im too lazy). Did anyone solve 6d? My proof for it looks really sus so I want to know if I did it right or not.

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akliu

#45 h Yesterday at 1:07 AM

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This reminds me that I didn't actually post a solution for P5:
5a: Infinite pitchfork configuration!
5b: Basically just some logic with "regions to infinity"
5c: Showing that if there exists some infinite set S of irreducible territory, we can only reduce the size of, but not fully eliminate, set S.

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