USA Canada math camp

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Bread10

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Bread10

#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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lovematch13

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lovematch13

#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

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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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lovematch13

653 posts

lovematch13

#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

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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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lovematch13

653 posts

lovematch13

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

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April 17th

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MarisaD

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MarisaD

#8 h Mar 20, 2025, 12:45 PM • 7 Y

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

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

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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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sharknavy75

687 posts

sharknavy75

#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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eugenewang1

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eugenewang1

#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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sharknavy75

687 posts

sharknavy75

#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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eugenewang1

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eugenewang1

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

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which ones

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sharknavy75

687 posts

sharknavy75

#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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eugenewang1

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eugenewang1

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

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both not true, did casework on repeating patterns

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akliu

1731 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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eugenewang1

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eugenewang1

#18 h Thursday at 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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eugenewang1

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eugenewang1

#19 h Thursday at 11:32 PM

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oops my images didn’t load

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akliu

1731 posts

akliu

#20 h Yesterday at 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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sharknavy75

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sharknavy75

#21 h Yesterday at 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 $\begin{align} (a, b, 0), (-a, b, 0), (0, a, b), (c, c, c), (-c, c, c)\\ (a, b, 0), (-a, b, 0), (0, a, -b), (c, c, -c), (-c, c, -c)\\ (a, -b, 0), (-a, -b, 0), (0, -a, b), (c, -c, c), (-c, -c, c)\\ (a, -b, 0), (-a, -b, 0), (0, -a, -b), (c, -c, -c), (-c, -c, -c)\\ (-b, 0, a), (-b, 0, -a), (-a, b, 0), (-c, c, c), (-c, c, -c)\\ (-b, 0, a), (-b, 0, -a), (-a, -b, 0), (-c, -c, c), (-c, -c, -c)\\ (b, 0, a), (b, 0, -a), (a, b, 0), (c, c, c), (c, c, -c)\\ (b, 0, a), (b, 0, -a), (a, -b, 0), (c, -c, c), (c, -c, -c)\\ (0, a, b), (0, -a, b), (b, 0, a), (c, c, c), (c, -c, c)\\ (0, a, b), (0, -a, b), (-b, 0, a), (-c, c, c), (-c, -c, c)\\ (0, a, -b), (0, -a, -b), (b, 0, -a), (c, c, -c), (c, -c, -c)\\ (0, a, -b), (0, -a, b), (-b, 0, -a), (-c, c, -c), (-c, -c, -c) \end{align}$