Kvant Math Problem 746
Let the square have side length $2$ and center $O$.
Verified: yes
Verdicts: PASS + PASS
Solve time: 30m44s
Source on kvant.digital
Problem
Fig. 1
A paper square is folded in half along some line $l$ passing through its center, forming a (non-convex) nonagon.
- How should the line $l$ be drawn so that the resulting nonagon has the greatest area?
- How should the line $l$ be drawn so that the largest possible circle can be inscribed in it (Fig. 1)?
K. Vulfson
Exploration
Let the square have side length $2$ and center $O$. Let the folding line $l$ pass through $O$ and make an angle $\theta$ with a side of the square. By symmetry it is enough to assume
$0\le \theta\le \frac{\pi}{4}.$
Let $A$ be the half of the square that remains fixed and $B$ the image of the other half after folding. Both have area $2$, and the folded nonagon is
$N=A\cup B.$
Hence
$$=4-|A\cap B|.$$
Let $S$ be the original square and let $S'$ be its reflection in $l$. Since $A=S\cap H$ and $B=S'\cap H$, where $H$ is one of the half-planes bounded by $l$,
$$$$
The set $S\cap S'$ is symmetric with respect to $l$, because reflection in $l$ interchanges $S$ and $S'$. Thus $l$ divides $S\cap S'$ into two congruent halves, and $A\cap B$ is exactly one of them. Consequently
$$$$
The reflection of a square in a line through its center is equivalent to rotating the square through angle
$\alpha=2\theta.$
The area problem becomes the problem of minimizing $|S\cap S'|$.
1. Maximizing the area of the nonagon
Place the square as
$$$$
For $0\le\alpha\le\pi/4$, the intersection $S\cap S'$ is an octagon obtained from $S$ by cutting off four congruent corner triangles.
Consider the corner near $(1,1)$. One side of the rotated square is
$$$$
Its intersections with $x=1$ and $y=1$ are
$$\qquad \left(\frac{1-\cos\alpha}{\sin\alpha},1\right).$$
Using
$$=\tan\frac{\alpha}{2},$$
the leg length of the removed right triangle is
$$$$
Hence each corner triangle has area
$$$$
Subtracting the four triangles from the square gives
$$=4-2\left(1-\tan\frac{\alpha}{2}\right)^2.$$
Since $\tan(\alpha/2)$ is increasing on $[0,\pi/4]$, the quantity above is decreasing. Its minimum occurs at
$$\qquad \theta=\frac{\pi}{8}.$$
Because
$$$$
we obtain
$$=4-2(2-\sqrt2)^2 =8(\sqrt2-1).$$
Therefore
$$=\frac12|S\cap S'| =4(\sqrt2-1),$$
and
$$=4-4(\sqrt2-1) =8-4\sqrt2.$$
The fold line giving the largest-area nonagon is the bisector of the angle between a side and a diagonal of the square.
2. Maximizing the radius of an inscribed circle
Choose coordinates so that the folding line $l$ is the $x$-axis. Before folding, the square is rotated through angle $\theta$ with respect to these axes. Its sides are
$$\qquad -x\sin\theta+y\cos\theta=\pm1.$$
After folding, the nonagon is the part of $S\cup S'$ lying in the half-plane
$$$$
where $S'$ is the reflection of $S$ in the $x$-axis.
The figure is symmetric about the $y$-axis. If an inscribed circle has center $(a,b)$, reflecting it in the $y$-axis produces another inscribed circle of the same radius. The midpoint of the two centers lies on the $y$-axis, and the convexity of the distance function to every supporting line shows that a circle of the same radius may be placed there. Hence a largest inscribed circle may be assumed to have center
$$$$
Since the top side of the nonagon is the fold line $y=0$, tangency to this side gives the vertical coordinate $-r$.
The two edges of the nonagon nearest the center are
$$$$
$$$$
The distance from $C$ to either line equals
$$$$
For a circle of radius $r$ to fit,
$$$$
hence
$$$$
This gives an upper bound. The remaining task is to prove that it is attained.
Set
$$$$
Then
$$$$
so the circle centered at $(0,-r_\theta)$ is tangent to the two lines above and also tangent to $y=0$.
It remains to check every other boundary component of the nonagon.
The lower pair of sides of the original square are
$$\qquad -x\sin\theta+y\cos\theta=1.$$
Their distances from $C$ are
$$>r_\theta.$$
The two remaining sides of the original square are
$$$$
Their distances from $C$ are
$$$$
and
$$$$
Since
$$=\frac{\cos\theta-\sin\theta}{1+\cos\theta} \ge0 \qquad \left(0\le\theta\le\frac{\pi}{4}\right),$$
both distances are at least $r_\theta$.
The reflected square contributes the sides
$$$$
Their distances from $C$ are again
$$\qquad 1+r_\theta\sin\theta,$$
which are also at least $r_\theta$.
Thus every boundary line of the two squares lies at distance at least $r_\theta$ from $C$. The disk of radius $r_\theta$ centered at $C$ is contained simultaneously in both strips determined by every pair of parallel sides. Consequently it lies inside $S\cup S'$, and because it is tangent to $y=0$ and lies entirely below that line, it is contained in the folded nonagon.
Hence
$$$$
Since $\cos\theta$ decreases on $[0,\pi/4]$, $r(\theta)$ increases. The maximum is attained at
$$$$
that is, when the fold line is a diagonal of the square.
The maximal radius is
$$=\frac1{1+\cos(\pi/4)} =\frac1{1+\frac{\sqrt2}{2}} =2-\sqrt2.$$
Solution
For the area problem, let $\theta$ be the angle between the fold line and a side of the square. If $\alpha=2\theta$, then
$$=4-\frac12|S\cap S'|,$$
where $S'$ is the square rotated through angle $\alpha$.
For $0\le\alpha\le\pi/4$,
$$=4-2\left(1-\tan\frac{\alpha}{2}\right)^2.$$
This is minimized when $\alpha=\pi/4$, equivalently
$$$$
The maximal area of the folded nonagon is
$$$$
For the inscribed-circle problem, a largest circle may be centered on the symmetry axis of the folded figure. If its center is $(0,-r)$, comparison with the two central edges gives
$$$$
For
$$$$
the circle is tangent to those two edges and to the fold line. Direct distance calculations show that every other boundary line of the nonagon is at distance at least $r$ from the center, so the circle is indeed contained in the nonagon. Hence
$$$$
This increases on $[0,\pi/4]$, and the maximum occurs at
$$$$
The largest inscribed circle has radius
$$$$
Thus the extremal folds are:
$$$$
for maximum area, giving
$$$$
and
$$$$
for the largest inscribed circle, giving
$\boxed{r=2-\sqrt2}.$