Challenge Math Challenge - June 2023

[mathwonk]

I have edited my post 56 to provide more detail on #10, that 3) hence also 2) fails, as well as thrown in an independent, but terse, proof just that 2) fails.SPOILER: problem #10,
Here is an argument that 1) is actually possible, using the only example I know, the topologist's sine curve. For convenience, change from the given square, to the square with vertices (-1,2), (3,2), (3,-2), (-1,-2). We wish to find a connected subset A of this square containing (-1,-2) and (3,2), and a connected subset B containing (-1,2) and (3,-2), with A and B disjoint.

Take the graph of y = sin(1/x), for 0 < x ≤ 1/π, and add in the line segment from (1/π,0) to (3,2), plus the line segment from (-1,-2) to (0,0), as our set A. Then as is well known, this set is connected, although not path connected. For B take the complement of this subset in the square. We claim that against all (my) intuition, B is also connected.

To prove it, use my favorite definition of connectedness, namely that every continuous function f from B to the set {0,1} is constant. Start by setting f(3,-2) = 0. Then, to be continuous, f must equal zero on every point of the square which is directly below (i.e. has the same x coordinate as, and a smaller y coordinate than) a point of A, since in fact the set of such points is path connected, every point being connected by a vertical segment to the bottom edge of the square. For the same reason, if not constant, f must have value 1 at all points directly above a point of A. But that is a contradiction to continuity at points of the y axis lying between (0,-1) and (0,0), and also those between (0,0) and (0,1), since all such points are in the closure of points where f has value zero, and also of points where f has value 1. So f must be constant, and B is connected.

One can easily shift and scale this square until it is the one given.

what fun! so now I know a second example of a connected, but not path connected, set.

 

[fresh_42]

My attempt for #7:

Spoiler: 7.

Let $GL_n^+(\mathbb{R})$ be the set of invertible $n\times n$ real matrices with positive determinant. Suppose that $A$ and $B$ are elements of $GL_n^+(\mathbb{R})$ and similar in the sense that $A=PBP^{-1}$ for some $P\in GL_n(\mathbb{R})$. Can you necessarily find a matrix $Q\in GL_n^+(\mathbb{R})$ such that $A=QBQ^{-1}?$ In other words, if two matrices in $GL_n^+(\mathbb{R})$ are conjugate as elements of $GL_n(\mathbb{R})$, are they also conjugate as elements of $GL_n^+(\mathbb{R})?$

If $n$ is odd, then ($-I$ is in the center of $GL_n(\mathbb{R})$)
$$
A=PBP^{-1}=(P\cdot (-I))B(P\cdot (-I))^{-1}
$$
and $\det (P\cdot(-I))=(-1)^n\det P=-\det P.$ We can thus always change the sign of the determinant of $P$ if it exists at all.

The statement is not generally true in case $n$ is even.

Set $A=\begin{pmatrix}0&-1\\2&0\end{pmatrix}$ and $B=\begin{pmatrix}0&2\\-1&0\end{pmatrix}.$ Then
\begin{align*}
\begin{pmatrix}0&-1\\2&0\end{pmatrix}&=\begin{pmatrix}0&1\\1&0\end{pmatrix}\cdot\begin{pmatrix}0&2\\-1&0\end{pmatrix}\cdot\begin{pmatrix}0&1\\1&0\end{pmatrix}
\end{align*}
Assume there is a matrix $Q=\begin{pmatrix}p&q\\r&s\end{pmatrix}$ such that $ps-rq>0$ and $A=QBQ^{-1}.$ Then
$$
\begin{pmatrix}p&q\\r&s\end{pmatrix}=\underbrace{\begin{pmatrix}q&p\\s&r\end{pmatrix}}_{=R}\cdot \underbrace{\begin{pmatrix}0&1\\1&0\end{pmatrix}}_{=P}
$$
and
\begin{align*}
A&=PBP^{-1}=QBQ^{-1}=RPB(RP)^{-1}=RPBP^{-1}R^{-1}=RAR^{-1}\\
[A,R]&=AR-RA=0
\end{align*}
but
$$
\left[\begin{pmatrix}0&-1\\2&0\end{pmatrix}\, , \,\begin{pmatrix}q&p\\s&r\end{pmatrix}\right]=\begin{pmatrix}-s-2p&-r+q\\ 2q-2r&2p+s\end{pmatrix}\neq\begin{pmatrix}0&0\\0&0\end{pmatrix}
$$
If it was the zero matrix, then $r=q$ and $s=-2p$ and $ps-rq=-2p^2-q^2<0.$

 

[fresh_42]

Seems like you're using some unnecessary big guns...

Spoiler: 2

Suppose for the sake of contradiction that $f$ is not constant, that $f$ is continuous, and that $P(f)$ has arbitrarily small strictly positive elements.

Since $f$ is not constant, it must take on at least two different values. Let those values be $a=f(x_a)$ and $b=f(x_b)$ so $b-a \neq 0$

Since $f$ is continuous everywhere, it's continuous at $0$. So for $\forall \epsilon > 0 \exists \delta>0 : |x| < \delta \implies \left| f(x) - f(0) \right| < \epsilon$. In particular, there's some $\delta_{ab} > 0$ corresponding to $\epsilon = \left| \frac{ b-a }{2} \right|$.

Since $P(f)$ has arbitrarily small strictly positive elements it contains some element $\iota$ with $0 < \iota < \delta_{ab}$.

Now we have $\delta_{ab} > \iota > \left| x_a - \lfloor \frac{x_a}{\iota} \rfloor \iota \right|$, $\delta_{ab} > \iota > \left| x_b - \lfloor \frac{x_b}{\iota} \rfloor \iota \right$, so $\left| b-a \right| > \left| f(x_a - \lfloor \frac{x_a}{\iota} \rfloor \iota) - f(0) \right| + \left| f(x_b - \lfloor \frac{x_b}{\iota} \rfloor \iota) - f(0) \right| = \left| f(x_a) - f(0) \right| + \left|f(x_b - f(0) \right| \geq \left| f(x_b) - f(x_a) \right| = \left| b-a \right|$, but $\left| b-a \right|$ can't be bigger than itself so we have a contradiction.

You haven't seen my attempts with Urysohn! Anyway, I needed the same delta at two different, arbitrary locations. Otherwise, my quantifiers wouldn't have allowed my conclusion.

 

[mathwonk]

a continuous periodic function is a composition of a distance decreasing function (exp) with a continuous function on a (compact) circle, hence uniformly continuous.

 

[Infrared]

@fresh_42 Nice, looks all right to me for number 7! A geometric example is that rotation by an angle $\theta$ counterclockwise and rotation by $\theta$ clockwise are only conjugate with a negative determinant matrix (because if you make an orientation-preserving transformation, that preserves which basis vector is counterclockwise to the other).

 

[Infrared]

@mathwonk Be careful about invoking homotopy invariance of intersection number- that's true for closed manifolds, but otherwise you'll need to say something about the curves being nice towards the boundary (which fortunately isn't hard to arrange: you can assume that they start off as diagonals and then change inside of a slightly smaller square). This is very similar to the solution I came up with when my friend gave me this problem. I glued the square into a torus along opposite edges and then used homotopy invariance there.

For the case of one connected, and one path-connected, here's a more elementary argument if you've already solved the case of both path-connected: Since $[0,1]$ is compact, the image of a path connecting opposite vertices is a closed subset of the square. Its complement being open means that the path-component of any point in $X=\text{(square) }-\text{ (path connecting opposite vertices)}$ is open. Since you already know that $X$ has at least two path-components, writing it as the union of its (open) path-components shows it to be disconnected.

 

[Infrared]

Thanks to everyone for participating! It looks like all of the problems have been solved, except for the botched #1. Stay tuned next month for some more problems. It looks like I'll have to make them harder given how quickly these were all solved!

 

[fresh_42]

@fresh_42 Nice, looks all right to me for number 7! A geometric example is that rotation by an angle $\theta$ counterclockwise and rotation by $\theta$ clockwise are only conjugate with a negative determinant matrix (because if you make an orientation-preserving transformation, that preserves which basis vector is counterclockwise to the other).

In case some readers are interested in how I have proceeded:

I thought about the FTA. Means: problems (artificial or not) in linear algebra have often to do with the lack of eigenvalues, especially over the real numbers. Polynomials of an odd degree have always a real zero, those of an even degree do not necessarily. The translation of $\boldsymbol i$ into $\begin{pmatrix}0&1\\1&0\end{pmatrix}$ was the next step and finally looking for a matrix that does not commute with it. And matrix multiplication was done by
https://www.symbolab.com/solver/matrix-calculator
listed on
https://www.physicsforums.com/threa...h-physics-earth-and-other-curiosities.970262/

 

[mathwonk]

@Infrared: I extended the curves from maps of intervals to maps of the circle, hence defined on a compact one-manifold without boundary, and with an extra intersection outside the square. Then used definition of intersection number of the maps (mod 2 would do) via inverse image of the diagonal of the target manifold (the plane) under the product map.

Bott gave the short argument in class some 50 odd years ago, just as I did, without details, as was his habit. I like your torus trick.

 

[mathwonk]

Working on problem #10 taught me a little fact about connectedness I had not noticed before. I knew that the union of connected sets which have a common point is also connected. Thus one can describe the connected component of a space X containing a point p as the union of all those connected subsets of X that contain p. I also knew that the closure of a connected set is connected, but somehow I did not realize that the union of a family of connected sets, whose union contains a point common to all their closures, is therefore also connected. I.e. if the sets are all connected and there is a point p that is in the closure of all of them, and actually in at least one of them, then the union is connected. In the construction for problem #10 above in post 61, the set B is the union of two path connected sets whose closures (in B) intersect, hence it is connected. I.e. the point (0,1/2) for instance is a point of B that is above the set A, and also in the closure of points of B that are below the set A. (In fact so is (0,1), although slightly less obviously to me.) Thus the connected component of B containing the point (0,1/2) is the union of all connected subsets of B that have (0,1/2) in their closure, i.e. all of B.

@Infrared: By the way, your nice elementary argument in post 66 seems to prove another case, namely if A is connected and closed, then B cannot be connected, since the complement of A is open, so since the other two vertices are not in the same path component of the complement of A, they are also not in the same connected component. (Of course, as you know, it is not enough just to observe that the complement of A is disconnected, you also want the opposite vertices to be in different connected components, but that also follows from your argument about path components equalling connected components.)

 

[The Fez]

What am I missing on Question 1? It would be easy to construct a line through the centroids of both shapes, which should bisect each one. Centroid of a rectangle is the intersection of the diagonals, and triangle is intersection of the three medians.

 

[mathwonk]

@The Fez: see posts 28, 29.

 

[mathwonk]

Thanks Fresh and Infrared for insight on prob. 7. I finally see it. If the problem were true, the matrix would have to commute with a matrix of negative determinant, but the 2x2 real matrix for multiplication by i, (i.e. 90 degree rotation), only commutes, by definition of complex linearity, with matrices of multiplication by complex numbers, and the matrix for multiplication by a+bi has determinant a^2+b^2.

 

[Throwaway_for_June]

What am I missing on Question 1? It would be easy to construct a line through the centroids of both shapes, which should bisect each one. Centroid of a rectangle is the intersection of the diagonals, and triangle is intersection of the three medians.

Lines through the centroid of the triangle don't always cut the area of the triangle in half. Consider, for example, a triangle with vertices at $(0,0)$, $(2,0)$ and $(0,2)$ in a plane. Then the centroid is at $(\frac{2}{2},\frac{2}{3})$, but the line $y=\frac{2}{3}$ doesn't divide the area of the triangle in half.

There may be other ways to think about it, but, effectively, problem 1 is asking us to find the tangent to a hyperbola that goes through a particular point. I can describe a method for constructing the line, but it's basically "solve this quadratic equation that was developed using differential calculus and analytic geometry" which is pretty unsatisfying as a "compass and straightedge" kind of thing.

 

[mathwonk]

nonetheless if you feel like posting it, i would surely learn something enjoyable.

 

[Throwaway_for_June]

nonetheless if you feel like posting it, i would surely learn something enjoyable.

My capabilities with GeoGebra weren't quite up to making nice illustrations.

Start by constructing the centroid of the rectangle $C_R$ by crossing the diagonals of the rectangle. We know that lines divide the area of the rectangle in two if and only if they go through the centroid. So our eventual solution is going to have to go through that point.

Construct the midpoints of the edges of the triangle and the centroid of the triangle $C_T$. Draw a ray from $C_T$ through $C_R$. If this ray happens to cross the triangle at a vertex or at the midpoint of an edge, then extending it to a line it bisects both the triangle and the rectangle and we're done. Otherwise, the ray crosses exactly one of the edges of the triangle, and one of the vertices on that edge is further from the intersection than the other. Label that vertex $O$. Label the edge that is crossed by the ray $Y$, and the other edge that goes through $O$, $X$.

Now we know that there's a line from $C_R$ through $X$ and $Y$ which divides the area of hte triangle in half.

Construct an altitude from one of the sides of the triangle as the base, and then take the geometric mean of that altitude with half the base. We'll be using this geometric mean as our unit length. (A square with this as the side length will have the same area as the triangle.)

At this point I'm going to assume that $C_R$ is outside the triangle. It doesn't actually make things more difficult, but without oriented distances, it ends up being a case with a sign change.

Extend the edges $X$ and $Y$ to lines. Construct a line parallel to $Y$ through $C_R$ and label the distance along that line from $C_R$ to the extended $X$ edge $D_Y$. Construct the perpendicular distance from $C_R$ to the line $Y$ and label that distance $D_X$.

Using well-known constructions for multiplication, division, addition, and the square root, construct the length $\frac{ 2D_Y } {1 + \sqrt{1+4D_Y D_X}}$. Mark that distance from $O$ along the edge $Y$ and call the point $T$. The line from $C_R$ through $T$ will divide the areas of the rectangle and the triangle in half.

 

[mathwonk]

very nice! but it seems to use only similarity of triangles. i.e. when Dx = Dy = 1, just bh = 1 and b(1+h) = h. From your summary earlier, I was expecting to find a hyperbolic envelope, or some calculus. ?? ( I also thought I had made an error, since I got different answers solving for b and for h, until I realized that [sqrt(5)-1]/2 is the reciprocal of [sqrt(5)+1]/2.)

 

[Throwaway_for_June]

very nice! but it seems to use only similarity of triangles .... From your summary earlier, I was expecting to find a hyperbolic envelope, or some calculus. ??

It was me thinking that everything can work as a nail when I've got a hammer in my hand kind of thing.

I was in a "compass and straightedge" kind of mindset, so I went looking to draw a tangent to some kind of circle that also went through the centroid of the rectangle. Then I remembered (or realized) that the tangents to a particular hyperbola all form triangles with the same area when intersected with the hyperbola's asymptotes. Compass and straightedge don't allow for constructing a hyperbola to draw the tangent against, but I can do that algebraically using analytic geometry and calculus which got me something that worked.

If I had started by thinking in terms of similar triangles then I think I would have ended up with a slightly more elegant construction since we don't actually need the $D_x$ and $D_y$ to be measured along perpendicular axes.

... ( I also thought I had made an error, since I got different answers solving for b and for h, until I realized that [sqrt(5)-1]/2 is the reciprocal of [sqrt(5)+1]/2.)

Yeah. There are lots of subtle identities involving square roots and fractions. Though it's also totally plausible that I made some algebra errors last night.

 

[mathwonk]

actually this sentence puzzled me:

"Now we know that there's a line from C_R
CRthrough XX and Y and Ywhich divides the area of the triangle in half."

and I thought maybe calculus motivated it. I.e. I did not see how the areas cut by the lines through the centroid C_T varied as they rotated. But I didn't draw a picture. (...... Ok I drew a picture and see it now.)

anyway, this is beautiful. thanks for posting!

 

[Throwaway_for_June]

actually this sentence puzzled me:

"Now we know that there's a line from C_R
CRthrough XX and Y and Ywhich divides the area of the triangle in half."

and I thought maybe calculus motivated it. I.e. I did not see how the areas cut by the lines through the centroid C_T varied as they rotated. But I didn't draw a picture.

anyway, this is beautiful. thanks for posting!

Here's a drawing:

N is the centroid of triangle ABC.

If M is the centroid of the rectangle, then triangle ARC has more area than triangle ALC which has half the area of triangle ABC, and triangle IQC has less area than triangle IBC which has half the area of triangle ABC.

So there's some line that goes from M through the segments RQ and AI that produces a triangle that has a vertex at C with the desired area.

The closest this gets to calculus is the idea of continuously sweeping the ray from M.

 

[mathwonk]

thank you! and in reference to the arithmetic identities you remarked, if you do the calculation backwards as I did, you seem to get [sqrt(1+4D_xD_y)-1]/2D_x, which I believe equals your 2D_y/[1 + sqrt(1+4D_xD_y)]. So division is required either way. In particular I believe your arithmetic is correct.