Challenge Math Challenge - June 2023

[Throwaway_for_June]

...
I think it's reasonable to claim nonemptiness, though that is also not obvious to me. But I don't think injectivity follows (I mean, in hindsight, it couldn't).

If the $\Delta A$s are all non-empty and disjoint, then taking the least element of each $\Delta A$ will produce an injection. The $\Delta$s have to be disjoint because of the way the chain is ordered, so at most countably many of the elements in one of these chains can have a non-empty $\Delta$ . Infinities are weird.

 

[projective]

4)

Spoiler

From the definition of the commutator $[S_7, S_3] \leq S_7 \cap S_3$ but $S_7 \cap S_3 = 1$ and so $[S_7, S_3] = 1$ but $S_7$ and $S_3$ are abelian and therefore $G$ also is

 

[fresh_42]

@Infrared Again for #2. I think I have a proof for uniformly continuous functions. My proof does not work for only continuous functions because I get a circle argumentation with the quantifiers if continuity depends on the location. Are you sure it isn't necessary? It may not play a role in $P(f)$ but it does in $\mathbb{R}\backslash P(f).$ I even tried tricks with closed intervals but I cannot get rid of the location.

 

[mathwonk]

#2: We may assume P(f) contains more than just 0. Since f is continuous, P(f) is closed, and if P(f) contained an arbitrarily small positive element, it would also be dense. Since f is constant on P(f) with value f(0), this would imply f is constant with value f(0). Thus if f is continuous and non constant, and P(f) ≠ {0}, there is a smallest positive element T of P(f). Since P(f) is closed under addition and subtraction, P(f) = Z.T.

 

[fresh_42]

Here is my proof for #2 under the assumption that $f$ is uniformly continuous.

Spoiler: 2

Lemma: $P(f)$ is a group under addition.

Proof: Clearly, $0\in P(f).$ By the all quantifier $(*)$ in the definition of $P(f)$
\begin{align*}
T\in P(f)&\Longrightarrow f(x)=f(x+T-T)=f((x-T)+T)\stackrel{(*)}{=}f(x-T)\\&\Longrightarrow -T\in P(f)\\
S,T\in P(f)&\Longrightarrow f(x)=f(x+T)\stackrel{(*)}{=}f((x+T)+S)=f(x+(T+S))\\&\Longrightarrow T+S \in P(f)
\end{align*}
Since $0\in P(f)$ in any case, we have $P(f)=\{0\}=0\cdot \mathbb{Z}$ for non-periodic functions $f.$ We may therefore assume that $f$ is periodic and in particular not constant.

If $P(f)$ contains an interval of positive length, then $f$ is constant. If $P(f)$ has a minimal positive element $T_0$ then $T_0\mathbb{Z}\subseteq P(f)$ since $P(f)$ is an additive group. Now assume $0\neq T\in P(f).$ Then we have to show that $T=nT_0$ for some $n\in \mathbb{Z}.$ We assume that $T>T_0$, since there is nothing to show for $T=T_0$ and negative values, can be turned into positive by the above argument. Let $T\in [\,nT_0,(n+1)T_0\,).$ Then
$$
f(x+(T-nT_0))=f((x-nT_0)+T)\stackrel{T\in P(f)}{=}f(x-nT_0)\stackrel{T_0\mathbb{Z}\subseteq P(f)}{=}f(x)
$$
contradicting the minimality of $T_0$ since
$$
T-nT_0 < (n+1)T_0-nT_0=T_0
$$
except for $T=nT_0.$ Equivalently you can write any $T\in P(f)$ as $P(f)\ni T-qT_0=r$ by the Euclidean algorithm. Since $0<r<T_0$ is impossible, we have $r=0$ and $T=qT_0.$

$P(f)$ is closed since $f$ is continuous: Let $(T_n)\subset P(f)$ be a converging sequence. Then
$$
f(x+\lim_{n \to \infty}T_n)=f(\lim_{n \to \infty}(x+T_n))=\lim_{n \to \infty}f(x+T_n)=f(x)
$$
i.e. $\displaystyle{\lim_{n \to \infty}T_n\in P(f).}$ The set $U:=\mathbb{R}\backslash P(f)$ is therefore open and dense as $P(f)$ does not contain an interval of positive length. Assume there is no minimal, positive element in $P(f).$ This means there is a strictly monotone decreasing sequence $(T_n)\subset P(f)$ with $\displaystyle{\lim_{n \to \infty}T_n}=0$ and $T_n>0.$

Let $u\in U$ and $x_u\in \mathbb{R}$ such that $|f(x_u+u)-f(x_u)|=d>0$ and $\varepsilon =d/2.$ We can choose $T_n$ for sufficiently large $n$ and $u$ sufficiently close to $0$ such that $|T_n-u|<\delta(\varepsilon )$ if we want to get $|f(T_n)-f(u)|<\varepsilon .$
Then $|(T_n+x_u)-(u+x_u)|=|T_n-u|<\delta (\varepsilon )$ and if $f$ is uniformly continuous at $T_n$ and $T_n+x_u$ then
$$
d=|f(x_u+u)-f(x_u)|=|f(x_u+u)-f(x_u+T_n)|<\varepsilon =d/2.
$$
This is a contradiction and means that there is no such sequence $(T_n),$ i.e. $P(f)$ has a minimal element.

 

[mathwonk]

#10: If A is path connected then B cannot be connected, hence also not path connected, essentially by the Jordan curve theorem. It is "obvious" but already tricky to prove even that they cannot both be path connected. I do not see whether both can be connected. In fact I only know one connected set that is not path connected, the topologist's sine curve plus (0,0), (or the full closure of the topologist's sine curve).

here is a little more detail on why 3), hence nor 2), cannot be true: if there is a path in A connecting (0,0) to (1,1), then there is a simple such path (every path connected Hausdorff space is "simple path"-connected). Then add on 3/4 of the circle with center (1,0) and radius one (all but the NW quarter), to form a simple closed curve containing (1,0) but not (0,1) in its interior. Then Jordan says the interior and exterior of this curve are disjoint open sets. These open sets will then disconnect any subset B of the square which is disjoint from A and contains both (0,1) and (1,0). QED.

For a slick argument just that A and B cannot both be path connected, one wants to show that any path connecting (0,0) to (1,1) within the square, meets any path connecting (0,1) to (1,0) within the square. But intersection number is a homotopy invariant, so we can replace these 2 paths by the 2 diagonals, which clearly intersect once transversely. QED.

Rmk: One can also do this directly, arguing via various projections within the square, essentially a homotopy argument. In his beautiful book on ODE, Arnol'd uses this fact to solve the following problem: if two cars manage to drive between two points A and B along two different non intersecting roads, while joined by a rope of length < 2L, can two circular wagons of radius L drive along those roads in opposite directions (with their center on the road) without colliding? (He constructs a "phase space" for this problem which is identical to the setup in problem #10. 2).)

 

[mathwonk]

#9: the preimage of [1:0] is the circle consisting of all (z,w) such that w = 0 and |z|^2 = 1, and the preimage of [0:1] is the circle consisting of all (z,w) such that z=0 and |w|^2 = 1. Their linking number is the intersection number of one loop with a disc in S^3 capping off the other loop. Such a disc in S^3 capping off the loop {(z,0): |z|^2 = 1} is the set {(z,t): |z|^2 + t^2 = 1, and 0≤t≤1, t real}. This disc intersects {(0,w): |w|^2 = 1} transversely in the point (0,1). Hence the loops are linked.

 

[Infrared]

@fresh_42 A continous function which is periodic is uniformly continuous.

@mathwonk Everything you said it right, but very concise. Perhaps you would like to elaborate on how the Jordan curve theorem shows that a path from $(0,0)$ to $(1,1)$ disconnects the square? I had a solution in mind without the Jordan curve theorem but perhaps what you have is good too.

@projective This is right, but quite brief, so just to fill in the gaps for anyone else: $S_7$ and $S_3$ are the (Sylow) subgroups of orders $7$ and $9$ respectively. If $x\in S_7$ and $y\in S_3$ because if $x\in S_7$ and $y\in S_3$ then $xyx^{-1}y^{-1}=(xyx^{-1})y^{-1}=x(yx^{-1}y^{-1})$ shows it to be in $S_3\cap S_7=\{1\}$ since both subgroups are normal.

So, every element of $S_3$ commutes with every element of $S_7.$ Together, these subgroups generate the whole group because the map $S_3\times S_7\to G, (x,y)\mapsto xy$ is injective and hence surjective. Since every group of order $9$ and $7$ is already abelian, this means that $G$ is abelian.

 

[fresh_42]

@fresh_42 A continous function which is periodic is uniformly continuous.

I suspected this but couldn't see an immediate argument for locations whose distance isn't in $P(f).$

I was already happy that I recognized that I had to use it. People usually simply write $\forall \varepsilon \exists \delta$ where they actually use $\forall \varepsilon \exists\delta (x_0;\varepsilon ).$ This is a nice example for my mantra to always keep track of dependencies like $\forall \varepsilon \exists N(\varepsilon )$ instead of $N.$

 

[Throwaway_for_June]

Here is my proof for #2 under the assumption that $f$ is uniformly continuous.
...

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.

 

[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.