What is the largest possible radius of a circle contained in a $4$-dimensional hypercube of side length $1?$

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In summary, a hypercube is a geometric shape in 4 or more dimensions, which makes it different from a regular cube that only has 3 dimensions. The largest possible radius of a circle in a hypercube can be found by calculating the longest diagonal of the hypercube using the Pythagorean theorem. There is a formula for finding the maximum radius, which is √(n/2), where n is the number of dimensions. The maximum radius cannot be greater than the side length of the hypercube, and it increases as the number of dimensions increases.
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Here is this week's POTW:

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What is the largest possible radius of a circle contained in a $4$-dimensional hypercube of side length $1?$

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$\newcommand{\RR}{\mathbb{R}}\newcommand{\Trace}{\operatorname{tr}}$ No one answered this week's POTW. The solution, attributed to Kiran Kedlaya and associates, follows.

[sp]The largest possible radius is $\frac{\sqrt{2}}{2}$. It will be convenient to solve the problem for a hypercube of side length 2 instead, in which case we are trying to show that the largest radius is $\sqrt{2}$.

Choose coordinates so that the interior of the hypercube is the set $H = [-1,1]^4$ in $\RR^4$. Let $C$ be a circle centered at the point $P$. Then $C$ is contained both in $H$ and its reflection across $P$; these intersect in a rectangular paralellepiped each of whose pairs of opposite faces are at most 2 unit apart. Consequently, if we translate $C$ so that its center moves to the point $O = (0,0,0,0)$ at the center of $H$,
then it remains entirely inside $H$.

This means that the answer we seek equals the largest possible radius of a circle $C$ contained in $H$ \emph{and centered at $O$}. Let $v_1 = (v_{11}, \dots, v_{14})$ and $v_2 = (v_{21},\dots,v_{24})$ be two points on $C$ lying on radii perpendicular to each other. Then the points of the circle can be expressed as $v_1 \cos(\theta) + v_2 \sin(\theta)$ for $0 \leq \theta < 2\pi$. Then $C$ lies in $H$ if and only if for each $i$, we have
\[
|v_{1i} \cos(\theta) + v_{2i} \sin(\theta)|
\leq 1 \qquad (0 \leq \theta < 2\pi).
\]
In geometric terms, the vector $(v_{1i}, v_{2i})$ in $\RR^2$ has dot product at most 1 with every unit vector. Since this holds for the unit vector in the same direction as $(v_{1i}, v_{2i})$, we must have
\[
v_{1i}^2 + v_{2i}^2 \leq 1 \qquad (i=1,\dots,4).
\]
Conversely, if this holds, then the Cauchy-Schwarz inequality and the above analysis imply that $C$ lies in $H$.

If $r$ is the radius of $C$, then
\begin{align*}
2 r^2 &= \sum_{i=1}^4 v_{1i}^2 + \sum_{i=1}^4 v_{2i}^2 \\
&= \sum_{i=1}^4 (v_{1i}^2 + v_{2i}^2) \\
&\leq 4,
\end{align*}
so $r \leq \sqrt{2}$. Since this is achieved by the circle through $(1,1,0,0)$ and $(0,0,1,1)$, it is the desired maximum.

Remark:
One may similarly ask for the radius of the largest $k$-dimensional ball inside an $n$-dimensional unit hypercube; the given problem is the case $(n,k) = (4,2)$. Daniel Kane gives the following argument to show that the maximum radius in this case is $\frac{1}{2} \sqrt{\frac{n}{k}}$. (Thanks for Noam Elkies for passing this along.)

We again scale up by a factor of 2, so that we are trying to show that the maximum radius $r$ of a $k$-dimensional ball contained in the hypercube $[-1,1]^n$ is $\sqrt{\frac{n}{k}}$. Again, there is no loss of generality in centering the ball at the origin. Let $T: \RR^k \to \RR^n$ be a similitude carrying the unit ball to this embedded $k$-ball. Then there exists a vector $v_i \in \RR^k$ such that for $e_1,\dots,e_n$ the standard basis of $\RR^n$, $x \cdot v_i = T(x) \cdot e_i$ for all $x \in \RR^k$. The condition of the problem is equivalent to requiring $|v_i| \leq 1$ for all $i$, while the radius $r$ of the embedded ball is determined by the fact that for all $x \in \RR^k$,
\[
r^2 (x \cdot x) = T(x) \cdot T(x) = \sum_{i=1}^n x \cdot v_i.
\]
Let $M$ be the matrix with columns $v_1,\dots,v_k$; then $MM^T = r^2 I_k$, for $I_k$ the $k \times k$ identity matrix. We then have
\begin{align*}
kr^2 &= \Trace(r^2 I_k) = \Trace(MM^T)\\
&= \Trace(M^TM) = \sum_{i=1}^n |v_i|^2 \\
&\leq n,
\end{align*}
yielding the upper bound $r \leq \sqrt{\frac{n}{k}}$.

To show that this bound is optimal, it is enough to show that one can find an orthogonal projection of $\RR^n$ onto $\RR^k$ so that the projections of the $e_i$ all have the same norm (one can then rescale to get the desired configuration of $v_1,\dots,v_n$). We construct such a configuration by a ``smoothing'' argument. Start with any projection. Let $w_1,\dots,w_n$ be the projections of $e_1,\dots,e_n$. If the desired condition is not achieved, we can choose $i,j$ such that
\[
|w_i|^2 < \frac{1}{n} (|w_1|^2 + \cdots + |w_n|^2) < |w_j|^2.
\]
By precomposing with a suitable rotation that fixes $e_h$ for $h \neq i,j$, we can vary $|w_i|, |w_j|$ without varying $|w_i|^2 + |w_j|^2$ or $|w_h|$ for $h \neq i,j$. We can thus choose such a rotation to force one of $|w_i|^2, |w_j|^2$ to become equal to $\frac{1}{n} (|w_1|^2 + \cdots + |w_n|^2)$. Repeating at most $n-1$ times gives the desired configuration.
[/sp]
 

FAQ: What is the largest possible radius of a circle contained in a $4$-dimensional hypercube of side length $1?$

What is a hypercube and how is it different from a regular cube?

A hypercube is a geometric shape in higher dimensions, specifically in 4 or more dimensions. It is different from a regular cube in that it has more than 3 dimensions, making it difficult to visualize in our 3-dimensional world.

How do you determine the largest possible radius of a circle in a hypercube?

The largest possible radius of a circle in a hypercube can be determined by finding the longest diagonal of the hypercube, which will also be the diameter of the circle. This can be calculated using the Pythagorean theorem.

Is there a formula for finding the maximum radius of a circle in a hypercube?

Yes, the formula for finding the maximum radius of a circle in a hypercube is:
r = √(n/2)
Where n is the number of dimensions of the hypercube.

Can the largest possible radius of a circle in a hypercube be greater than the side length of the hypercube?

No, the largest possible radius of a circle in a hypercube cannot be greater than the side length of the hypercube. This is because the diagonal length of the hypercube, which is also the diameter of the circle, cannot be greater than the side length.

How does the maximum radius of a circle in a hypercube change as the number of dimensions increases?

As the number of dimensions increases, the maximum radius of a circle in a hypercube also increases. This is because higher dimensions allow for more space and therefore, a larger maximum diameter for the circle. The formula for the maximum radius also shows this relationship, as it includes the number of dimensions in its calculation.

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