# Math Challenge - August 2019

• Challenge
• Featured
Making use of induction. Suppose $a,b,c$ is a nonzero solution. Due to $x^2\equiv 0,1\pmod{3}$ it must hold that $a^2,b^2$ are multiples of three. Putting $d:= (a,b)$, if $d\mid c$ holds, then $\left (\frac{a}{d},\frac{b}{d},\frac{c}{d}\right )=:(a',b',c')$ would be a solution with $(a',b')=1$ which is impossible.

Could we show $d\mid c$?

Making use of induction. Suppose $a,b,c$ is a nonzero solution. Due to $x^2\equiv 0,1\pmod{3}$ it must hold that $a^2,b^2$ are multiples of three. Putting $d:= (a,b)$, if $d\mid c$ holds, then $\left (\frac{a}{d},\frac{b}{d},\frac{c}{d}\right )=:(a',b',c')$ would be a solution with $(a',b')=1$ which is impossible.

Could we show $d\mid c$?
In this method, ##a^2=9x^2, b^2=9y^2## .

This implies ##c^2## is a multiple of 3. Then you can continue with the same method I did. Using contradiction to prove that solutions don't exist.

However, I'm not sure how to prove that d divides c (if I'm not wrong, d is the gcd of a,b).

Math_QED
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Sure. First I'll explain why x is even. Since l is less than equal to m and n, we can divide both the sides by ##2^{2l}## . That in any scenario gives either x is even or x and y are even or all x,y,z are even (when l=m=n) by taking mod 4 both sides. A similar reasoning is followed for z.

As to why it is okay to consider only two cases instead of three (by including the case of m being smaller), considering the case for x is the same as considering the case for y. If it hadn't been the same, I would have to consider 3 cases. I don't know how to explain this one any better. You could think that if (a,b,c) satisfy the equation, then (b,a,c) will also satisfy.
I'm satisfied with your explanations. Well done!

In fact, there is a shorter solution.

You got to ##a^2+b^2 +c^2\equiv 0 \bmod 4##, implying that ##2|a,b,c##. Now, if we take a minimal positive solution ##(a,b,c)## this procedure yields the solution ##(a/2,b/2,c/2)##, contradicting minimality.

I think this is in the lines of what @nuuskur was trying.

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• Pi-is-3
Mentor
I was thinking about it in a "give a counter-example" fashion. The question then is rather easy don't you think? I could put forward any function ##f## where ##f(0) \neq f(1)## for example.
... plus ##f'(x)\neq 0## for all ##x##.

If you have three conditions which are needed to draw a single conclusion, then the violation of any of the conditions together with the opposite of the conclusion is a counterexample. The best solution would have been to list three examples, violating one condition after the other. And a fourth example where the conclusion holds without any of the conditions.

• Pi-is-3
I think this is in the lines of what @nuuskur was trying.
Indeed, formally it's known as method of infinite descent - stems from induction (equivalent, too, I think). By the way, I revised #33.

Suppose $g\mapsto g^{-1}$ is an endomorphism and take $g,h\in G$. Then $gh \mapsto (gh)^{-1}$. Due to compatibility $gh \mapsto g^{-1}h^{-1}$, thus
$$(gh)^{-1} = g^{-1}h^{-1} = (hg)^{-1} \iff gh = hg.$$
Suppose $G$ is abelian then for every $g,h\in G$
$$gh \mapsto (gh)^{-1} = h^{-1}g^{-1} = g^{-1} h^{-1}.$$
The map $g\mapsto g^{-1}\sigma (g)$ on $G$ is injective, because if $g^{-1}\sigma (g) = h^{-1}\sigma (h)$, then $hg^{-1} = \sigma (hg^{-1})$. The only fixpoint is $e$, thus $h=g$. The map is also surjective, because $G$ is a finite set.

Now, pick $g\in G$ and write $g= h^{-1}\sigma (h)$, then
$$\sigma (g) = \sigma (h^{-1}\sigma (h)) = (\sigma (h))^{-1} (h^{-1})^{-1} = (h^{-1}\sigma (h))^{-1} = g^{-1}.$$
By 7a $G$ is abelian.

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• Math_QED
Math_QED
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Suppose $g\mapsto g^{-1}$ is an endomorphism and take $g,h\in G$. Then $gh \mapsto (gh)^{-1}$. Due to compatibility $gh \mapsto g^{-1}h^{-1}$, thus
$$(gh)^{-1} = g^{-1}h^{-1} = (hg)^{-1} \iff gh = hg.$$
Suppose $G$ is abelian then for every $g,h\in G$
$$gh \mapsto (gh)^{-1} = h^{-1}g^{-1} = g^{-1} h^{-1}.$$
The map $g\mapsto g^{-1}\sigma (g)$ on $G$ is injective, because if $g^{-1}\sigma (g) = h^{-1}\sigma (h)$, then $hg^{-1} = \sigma (hg^{-1})$. The only fixpoint is $e$, thus $h=g$. The map is also surjective, because $G$ is a finite set.

Now, pick $g\in G$ and write $g= h^{-1}\sigma (h)$, then
$$\sigma (g) = \sigma (h^{-1}\sigma (h)) = (\sigma (h))^{-1} (h^{-1})^{-1} = (h^{-1}\sigma (h))^{-1} = g^{-1}.$$
By 7a $G$ is abelian.
Correct! I will give feedback on the measure theory question soon! Bit busy now!

Making use of induction. Suppose $a,b,c$ is a nonzero solution. Due to $x^2\equiv 0,1\pmod{3}$ it must hold that $a^2,b^2$ are multiples of three. Putting $d:= (a,b)$, if $d\mid c$ holds, then $\left (\frac{a}{d},\frac{b}{d},\frac{c}{d}\right )=:(a',b',c')$ would be a solution with $(a',b')=1$ which is impossible.

Could we show $d\mid c$?
Write
$$a = 3 ^{k_1} \ldots p_n^{k_n} \quad b = 3^{l_1}\ldots p_n^{l_n} \quad c = 3^{r_1} \ldots p_n ^{r_n}$$
where the powers are non-negative. Then $(a,b) = 3 ^{m_1} \ldots p_n ^{m_n}$, where $m_j = \min \{k_j,l_j\}$. The goal is to show $m_j\leq r_j$. The initial equality can be written as
$$3^{2m_1} \ldots p_n^{2m_n} \left ( 3^{2k_1 - 2m_1} \ldots p_n^{2k_n-2m_n} + 3^{2l_1 - 2m_1} \ldots p_n^{2l_n-2m_n}\right ) = 3^{2r_1+1}\ldots p_n^{2r_n}$$
Suppose, for a contradiction, some $m_j>r_j$. The case $j\neq 1$ would run into $2m_j + k = 2r_j$, where $k\geq 0$, which is impossible. So it must be that $j=1$. But then $2m_1 + k = 2r_1 + 1$ for some $k\geq 0$. By assumption $m_1 \geq r_1+1$ so
$$2m_1 + k = 2r_1 + 1 \Rightarrow 2r_1 + 1 - k \geq 2r_1 + 2 \Rightarrow k\leq -1,$$
a contradiction. Thus $(a,b) \mid c$.
I think this is overkill, not sure how to optimise.

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• Pi-is-3
Math_QED
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Let $\Sigma$ be an infinite sigma algebra on a set $X$, choose $\emptyset\subset A_1\subset X$ and pick for every $n\in\mathbb N$
$$A_{n+1} \in \Sigma \setminus \left ( \left\lbrace\emptyset, X\right\rbrace \cup \left\lbrace\bigcup_{k=1}^mA_k\mid m\leq n\right\rbrace\cup\left\lbrace\bigcap _{k=1}^m A_k^c\mid m\leq n\right\rbrace\right ).$$
$\Sigma$ is infinite, so we can do this. Put $S_n := \bigcup _{k=1}^nA_k\in\Sigma,n\in\mathbb N$. Observe that all $S_n\subset X$.

The sequence $S_n^c,n\in\mathbb N,$ is strictly decreasing. Put
$$T_n := S_n^c\setminus S_{n+1}^c\in \Sigma, \quad n\in\mathbb N.$$
The $T_n$ are pairwise disjoint. The map
$$\mathcal P(\mathbb N) \to \Sigma, \quad A\mapsto \begin{cases} \emptyset, &A=\emptyset \\ \bigcup_{x\in A}T_x, &\emptyset\subset A\subset\mathbb N \\ X, &A=\mathbb N \end{cases}$$
is injective. Note that if $A\subset\mathbb N$ then $\bigcup _{x\in A}T_x \subseteq S_1^c \subset X$.
What happens when we take ##X =\mathbb{N}, A_1=\{0,1\},A_2=\{0\}##?

Then ##S_1=S_2 = A_1## and ##T_1=\emptyset## so again injectivity is violated.

What happens when ...
This problem is of the devil mhh, the following is a cheapshot
Take a sequence of pairwise disjoint nonempty subsets $S_n\in\Sigma, n\in\mathbb N$. Then
$$\mathcal P(\mathbb N) \setminus \{\emptyset, \mathbb N\} \to \Sigma, \quad A \mapsto \bigcup _{x\in A} S_x$$
is injective, because of disjointedness. Construction of sequence given in #33. We take pairwise distinct $\emptyset \subset A_n \subset X\in\Sigma, n\in\mathbb N$ and put $S_n := \bigcup _{k=1}^n A_k\in\Sigma$. Then $T_n := S_n^c \setminus S_{n+1}^c\in\Sigma, n\in\mathbb N,$ is a sequence of pairwise disjoint non-empty subsets.

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Math_QED
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This problem is of the devil mhh, the following is a cheapshot
Take a sequence of pairwise disjoint nonempty subsets $S_n\in\Sigma, n\in\mathbb N$. Then
$$\mathcal P(\mathbb N) \setminus \{\emptyset, \mathbb N\} \to \Sigma, \quad A \mapsto \bigcup _{x\in A} S_x$$
is injective, because of disjointedness.
Now we are going somewhere. Or course, the question is why such a sequence exists. How can we construct one?

Math_QED
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This problem is of the devil mhh, the following is a cheapshot
Take a sequence of pairwise disjoint nonempty subsets $S_n\in\Sigma, n\in\mathbb N$. Then
$$\mathcal P(\mathbb N) \setminus \{\emptyset, \mathbb N\} \to \Sigma, \quad A \mapsto \bigcup _{x\in A} S_x$$
is injective, because of disjointedness. Construction of sequence given in #33. We take pairwise distinct $\emptyset \subset A_n \subset X\in\Sigma, n\in\mathbb N$ and put $S_n := \bigcup _{k=1}^n A_k\in\Sigma$. Then $T_n := S_n^c \setminus S_{n+1}^c\in\Sigma, n\in\mathbb N,$ is a sequence of pairwise disjoint non-empty subsets.
Again, it can happen that ##T_i=\emptyset## for some i. See my last counterexample.

Again, it can happen that ##T_i=\emptyset## for some i. See my last counterexample.
You're right, we need a change of gears .. Call a subset $S\in \Sigma$ to have property $(P)$ iff the sub sigma algebra $\Sigma _S := \{S\cap A \mid A\in\Sigma\}$ is infinite. Note that
$$\Sigma = \sigma \left (\Sigma _S \cup \Sigma _{S^c}\right ).\tag{E}$$
On the one hand we have by definition
$$\Sigma _S \cup \Sigma _{S^c} \subseteq \Sigma \Rightarrow\sigma \left (\Sigma _S \cup \Sigma _{S^c}\right ) \subseteq \sigma (\Sigma) = \Sigma.$$
Conversely, take $A\in \Sigma$, then
$$A = A\cap (S\cup S^c) = A\cap S \cup A\cap S^c \in \sigma \left ( \Sigma _S \cup \Sigma _{S^c}\right ).$$
Suffices to show the following: if $S$ has property $(P)$, then there exists a partition $S = T\dot{\cup}T'$ such that $T,T'\in\Sigma _S$ are non-empty and at least one of them has property $(P)$. Since $X$ has property $(P)$ by assumption, we can repeatedly apply this fact and obtain a strictly decreasing sequence of non-empty proper subsets.

Proof of fact. Suppose $S\in\Sigma$ has property $(P)$. Then $\Sigma _S$ is an infinite sub sigma algebra. Pick $T \in \Sigma _S \setminus \{\emptyset, S\}$ Write $S= T\cup (S\cap T^c)$. By (E) at least one of the respective sub sigma algebras must be infinite.

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For the sake of convenience/not cluttering denote

$$\frac{d^n}{dz^n} f(z) =: f_n(z) =: f_n \quad\mbox{and}\quad f^n(z) := (f(z)) ^n =: f^n.$$
Some preliminaries. By the chain and product rules

\begin{align*} (gf)_1 &= g_1(f)f_1 \\ (gf)_2 &= g_2(f)f_1^2 + g_1(f)f_2 \\ (gf)_3 &= g_3(f)f_1^2 + 3g_2(f)f_1f_2 + g_1(f)f_3 \end{align*}

Checking ahead in wiki - the Schwarzian in general would have to be:

$$S_{gf} = (S_g \circ f) \cdot f_1^2 + S_f$$

Now it's pretty straightforward:

\begin{align*} S_{gf} &= \frac{(gf)_3}{(gf)_1} - \frac{3}{2} \left (\frac{(gf)_2}{(gf)_1}\right )^2 \\ &= \frac{g_3(f)f_1^3 + 3g_2(f)f_1f_2 + g_1(f)f_3}{g_1(f)f_1} \\ &-\frac{3}{2} \cdot \frac{g_2^2(f)f_1^4 + 2g_1(f)g_2(f)f_1^2f_2 + g_1^2(f)f_2^2}{g_1^2(f)f_1^2} \\ \end{align*}
Post-reduction:
\begin{align*} \frac{g_3(f)}{g_1(f)}f_1^2 + \frac{f_3}{f_1} - \frac{3}{2} \left (\frac{g_2^2(f)}{g_1^2(f)}f_1^2 + \frac{f_2^2}{f_1^2}\right ) \end{align*}
Re-arrange:
\begin{align*} S_{gf} &= \left (\frac{g_3(f)}{g_1(f)} - \frac{3}{2} \left ( \frac{g_2(f)}{g_1(f)}\right )^2\right ) f_1^2 + \left (\frac{f_3}{f_1} - \frac{3}{2} \left (\frac{f_2}{f_1}\right )^2\right ) \\ &= (S_g \circ f) f_1^2 + S_f \end{align*}
The Schwarzian is said to have negative derivative if $f'(z) \neq 0$ implies $S_f (z) <0$, thus if both Schwarzians are negative, then by definition the sum is a negative Schwarzian.

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Suppose $G$ is of order $2m$, where $m>1$ is odd.

Consider the map
$$\varphi : G \to \mbox{Sym}(G) : \varphi (g)(x) := gx, x\in G.$$
the map $\varphi (g)$ is injective, because $gx = gy$ implies $x=y$ and for a fixed $h\in G$ we have $\varphi (g) (g^{-1}h) = gg^{-1}h = h$. Take $g,h\in G$, then
$$\varphi (gh)(x) = (gh)x = g(hx) = g \varphi (h)(x) = \varphi (g) (\varphi (h)(x)) = (\varphi (g)\varphi (h)) (x)\quad (x\in G)$$
By Sylow's first theorem there exists $a \in G$ of order two. Note that $(g,ga), g\in G,$ are cycles in the permutation $\varphi (a)$, because
$$\varphi (a)(g) = ga \Rightarrow \varphi (a)(ga) = gaa = ge =g.$$
Due to injectivity, the cycles are disjoint, thus there are precisely $m$ such cycles, which implies $\varphi (a)$ is an odd permutation. Consider the signum morphism $\varepsilon : \mbox{Sym}(G) \to \{-1,1\}$ where $\varepsilon (\sigma ) = 1$ iff $\sigma$ is an even permutation. We have $\varepsilon \varphi : G\to \{-1,1\}$ a surjective morphism, because $e\mapsto 1$ ($\varphi (e)$ has no inversions). By first isomorphism theorem
$$G / \mbox{Ker} (\varepsilon\varphi) \cong \{-1,1\}$$
which implies $|\mbox{Ker} (\varepsilon\varphi)| = m$. Also $\mbox{Ker} (\varepsilon\varphi) \triangleleft G$.

summer's almost over .. Last edited:
• Math_QED
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Suppose GG is of order 2m2m, where m>1m>1 is odd.

Consider the map
$$\varphi : G \to \mbox{Sym}(G) : \varphi (g)(x) := gx, x\in G.$$
the map φ(g)φ(g) is injective, because gx=gygx=gy implies x=yx=y and for a fixed h∈Gh∈G we have φ(g)(g−1h)=gg−1h=hφ(g)(g−1h)=gg−1h=h. Take g,h∈Gg,h∈G, then
$$\varphi (gh)(x) = (gh)x = g(hx) = g \varphi (h)(x) = \varphi (g) (\varphi (h)(x)) = (\varphi (g)\varphi (h)) (x)\quad (x\in G)$$
By Sylow's first theorem there exists a∈Ga∈G of order two. Note that (g,ga),g∈G,(g,ga),g∈G, are cycles in the permutation φ(a)φ(a), because
$$\varphi (a)(g) = ga \Rightarrow \varphi (a)(ga) = gaa = ge =g.$$
Due to injectivity, the cycles are disjoint, thus there are precisely mm such cycles, which implies φ(a)φ(a) is an odd permutation. Consider the signum morphism ε:Sym(G)→{−1,1}ε:Sym(G)→{−1,1} where ε(σ)=1ε(σ)=1 iff σσ is an even permutation. We have εφ:G→{−1,1}εφ:G→{−1,1} a surjective morphism, because e↦1e↦1 (φ(e)φ(e) has no inversions). By first isomorphism theorem
$$G / \mbox{Ker} (\varepsilon\varphi) \cong \{-1,1\}$$
which implies |Ker(εφ)|=m|Ker(εφ)|=m. Also Ker(εφ)◃GKer(εφ)◃G.

summer's almost over .. Correct! Well done! Could you maybe give some insight in why you thought about that action?

Summary: 1. - 2. posed and moderated by @QuantumQuest
3. - 8. posed and moderated by @Math_QED
9. - 10. posed and moderated by @fresh_42

keywords: calculus, abstract algebra, measure theory, mechanics, dynamical systems

Questions
9.
Three identical airplanes start at the same time at the vertices of an equilateral triangle with side length ##L##. Let's say the origin of our coordinate system is the center of the triangle. The planes fly at a constant speed ##v## above ground in the direction of the clockwise next airplane. How long will it take for the planes to reach the same point, and which are the flight paths?
Disclaimer: This spoiler may not be a spoiler at all, as I am no sure if the answer is correct :) Too long since I last used calculus to solve anything other than the most basic and even longer since I last worked on solving a physics problem :)

If the solution is incorrect, please explain which assumption or step is incorrect.

I am unable to draw here the diagram I sketched on paper to derive the solution, but the fundamental observation is that if all the 3 planes fly at the same speed, then at any point in time till they meet, the positions of the 3 planes will continue to form an equilateral triangle and this triangle keeps shrinking in size as time progresses.

Suppose the current distance between 2 planes is ##a## and ##\delta## is the small distance traveled by each of them, along the direction of edges of the current triangle, in a very small instant of time. Let ##b## denote the distance between the 2 planes after this instant. Using trigonometry, I found that the following equation determines the value of ##b##:

##b^2 = a^2 - 3 \delta (a - \delta)##

If ##v## is the constant velocity and ##\Delta t## is the small time instant, then ##\delta = v \Delta t##, yielding ##b^2 = a^2 - 3 v \Delta t (a - v\Delta t)##

If ##D^2## is used as the variable denoting the squared length of side of equilateral triangle, then we see that ##\Delta D^2 = -3 v \Delta t (D - v\Delta t)##.

$$\lim_{\Delta t \rightarrow 0} {\frac {\Delta D^2} {\Delta t}} = -3 v D \Rightarrow \frac {dD^2} {dt} = -3 v D \Rightarrow 2 D \frac {dD} {dt} = - 3 v D \Rightarrow \\ \frac {dD} {dt} = - \frac {3} {2} v$$

So the cumulative change in the value of ##D## over a period of time ##T## would be
$$\int_0^T - \frac {3} {2} v \, dt = - \frac {3} {2} v T$$

When the planes meet, the value of D would be 0 whereas its initial value is ##L##, so the cumulative change in ##D## has to be ##-L##. Substituting this in the previous equation gives the value of ##T## to be

$$-L / (- \frac {3} {2} v) = \frac {2L} {3v}$$

• archaic
I think P6 generalises via induction to the case $2^rm$.
Correct! Well done! Could you maybe give some insight in why you thought about that action?
I sort of started at the back. My hopes and dreams were that I could compose the signum morphism with some $\varphi : G\to\mbox{Sym}(G)$ such that I would get $\varepsilon \varphi : G\to \{-1,1\}$ surjective and apply the first isomorphism theorem. Of course, getting a preimage for $1$ is trivial, but why woud there be an odd permutation? Then I thought maybe there is a permutation in $\mbox{Sym}(G)$ which contains an odd number of disjoint cycles of length two making it an odd permutation, this is where the order $2m$ gave a clue. A way to get such cycles is to put $(g,ga)$ where $a$ is of order two and at that point the definition of $\varphi$ became self-evident.

• Math_QED
Math_QED
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I think P6 generalises via induction to the case $2^rm$.

I sort of started at the back. My hopes and dreams were that I could compose the signum morphism with some $\varphi : G\to\mbox{Sym}(G)$ such that I would get $\varepsilon \varphi : G\to \{-1,1\}$ surjective and apply the first isomorphism theorem. Of course, getting a preimage for $1$ is trivial, but why woud there be an odd permutation? Then I thought maybe there is a permutation in $\mbox{Sym}(G)$ which contains an odd number of disjoint cycles of length two making it an odd permutation, this is where the order $2m$ gave a clue. A way to get such cycles is to put $(g,ga)$ where $a$ is of order two and at that point the definition of $\varphi$ became self-evident.
Yes, using actions to find kernels is something that works when all other things seem to fail. You are right btw that this exercise generalises (via induction)

The generalisation is:

If ##G## is a group of order ##2^nm## with ##m## odd and ##G## has a cyclic Sylow 2-subgroup, then ##G## has a normal subgroup of order ##m##.

• nuuskur
Mentor
Disclaimer: This spoiler may not be a spoiler at all, as I am no sure if the answer is correct :) Too long since I last used calculus to solve anything other than the most basic and even longer since I last worked on solving a physics problem :)

If the solution is incorrect, please explain which assumption or step is incorrect.

I am unable to draw here the diagram I sketched on paper to derive the solution, but the fundamental observation is that if all the 3 planes fly at the same speed, then at any point in time till they meet, the positions of the 3 planes will continue to form an equilateral triangle and this triangle keeps shrinking in size as time progresses.

Suppose the current distance between 2 planes is ##a## and ##\delta## is the small distance traveled by each of them, along the direction of edges of the current triangle, in a very small instant of time. Let ##b## denote the distance between the 2 planes after this instant. Using trigonometry, I found that the following equation determines the value of ##b##:

##b^2 = a^2 - 3 \delta (a - \delta)##

If ##v## is the constant velocity and ##\Delta t## is the small time instant, then ##\delta = v \Delta t##, yielding ##b^2 = a^2 - 3 v \Delta t (a - v\Delta t)##

If ##D^2## is used as the variable denoting the squared length of side of equilateral triangle, then we see that ##\Delta D^2 = -3 v \Delta t (D - v\Delta t)##.

$$\lim_{\Delta t \rightarrow 0} {\frac {\Delta D^2} {\Delta t}} = -3 v D \Rightarrow \frac {dD^2} {dt} = -3 v D \Rightarrow 2 D \frac {dD} {dt} = - 3 v D \Rightarrow \\ \frac {dD} {dt} = - \frac {3} {2} v$$

So the cumulative change in the value of ##D## over a period of time ##T## would be
$$\int_0^T - \frac {3} {2} v \, dt = - \frac {3} {2} v T$$

When the planes meet, the value of D would be 0 whereas its initial value is ##L##, so the cumulative change in ##D## has to be ##-L##. Substituting this in the previous equation gives the value of ##T## to be

$$-L / (- \frac {3} {2} v) = \frac {2L} {3v}$$
This seems to be correct and is similar to what @nuuskur did

My solution is slightly different, but of course not basically.

The side length of the triangle at ##t=0## is ##L(0)=L.## For the position ##\vec{r}(t)## of the first airplane we have ##|\vec{r}(0)|=r(0)=\dfrac{2}{3}L\cos \dfrac{\pi}{6}=\dfrac{L}{\sqrt{3}}.## The distance between the airplanes are the same at any point in time, because of the symmetry, i.e. the airplanes will always mark the vertices of an equilateral triangle with its center at the origin. Thus the angle between the velocity ##\vec{v}(t)## and the position ##\vec{r}(t)## is always
$$\sphericalangle (\vec{v}(t),\vec{r}(t))=\psi(t)=\psi(0)=\psi =\pi - \dfrac{\pi}{6}$$
Thus we have
\begin{align*}
\vec{v}(t)&=\dot{\vec{r}}(t)\\
\vec{r}(t))\vec{v}(t)&=\vec{r}(t)\dot{\vec{r}}(t)\\
r\cdot v \cdot \cos \psi &=\dfrac{1}{2}\dfrac{d}{dt}(\vec{r}(t)\vec{r}(t)\\
r\cdot v \cdot \cos \psi &=\dfrac{1}{2}\dfrac{dr^2}{dt}\\
r\cdot v \cdot \cos \psi &=r\dfrac{dr}{dt}\\
\dfrac{dr}{dt}&= -v \dfrac{\sqrt{3}}{2}\\
r(t)&= \dfrac{L}{\sqrt{3}}-v\dfrac{\sqrt{3}}{2}t
\end{align*}
Hence ##r(t_f)=0 ## implies ##t_f=\dfrac{2L}{3v}.##

You're right, we need a change of gears .. Call a subset $S\in \Sigma$ to have property $(P)$ iff the sub sigma algebra $\Sigma _S := \{S\cap A \mid A\in\Sigma\}$ is infinite. Note that
$$\Sigma = \sigma \left (\Sigma _S \cup \Sigma _{S^c}\right ).\tag{E}$$
On the one hand we have by definition
$$\Sigma _S \cup \Sigma _{S^c} \subseteq \Sigma \Rightarrow\sigma \left (\Sigma _S \cup \Sigma _{S^c}\right ) \subseteq \sigma (\Sigma) = \Sigma.$$
Conversely, take $A\in \Sigma$, then
$$A = A\cap (S\cup S^c) = A\cap S \cup A\cap S^c \in \sigma \left ( \Sigma _S \cup \Sigma _{S^c}\right ).$$
Suffices to show the following: if $S$ has property $(P)$, then there exists a partition $S = T\dot{\cup}T'$ such that $T,T'\in\Sigma _S$ are non-empty and at least one of them has property $(P)$. Since $X$ has property $(P)$ by assumption, we can repeatedly apply this fact and obtain a strictly decreasing sequence of non-empty proper subsets.

Proof of fact. Suppose $S\in\Sigma$ has property $(P)$. Then $\Sigma _S$ is an infinite sub sigma algebra. Pick $T \in \Sigma _S \setminus \{\emptyset, S\}$ Write $S= T\cup (S\cap T^c)$. By (E) at least one of the respective sub sigma algebras must be infinite.
Ok, now that we know..
Pick a strictly decreasing sequence $X \supset S_1 \supset S_2 \supset \ldots$ from the sigma algebra. Then put $A_n := S_n\setminus S_{n+1}, n\in\mathbb N$, which is a sequence of pairwise disjoint elements and
$$\mathcal P(\mathbb N) \setminus \{ \emptyset, \mathbb N\} \to \Sigma, \quad A \mapsto \bigcup _{x\in A} A_x,$$
is injective, thus an infinite sigma algebra must be at least of the cardinality of the continuum.

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2019 Award
Ok, now that we know..
Pick a strictly decreasing sequence $X \supset S_1 \supset S_2 \supset \ldots$ from the sigma algebra. Then put $A_n := S_n\setminus S_{n+1}, n\in\mathbb N$, which is a sequence of pairwise disjoint elements and
$$\mathcal P(\mathbb N) \setminus \{ \emptyset, \mathbb N\} \to \Sigma, \quad A \mapsto \bigcup _{x\in A} A_x,$$
is injective, thus an infinite sigma algebra must be at least of the cardinality of the continuum.
Will give feedback tomorrow. Currently busy with abstract algebra and measure theory needs a change of mindset :P.

• nuuskur and fresh_42
The fundamental theorem of abelian groups says that for any finite abelian group

$$A \cong \mathbb Z_{p_1^{k_1}} \oplus \ldots \oplus \mathbb Z_{p_m^{k_m}}$$
Hyperlinking breaks my raw text for some reason. I'll be using Sylow's theorems which I will refer to as theorems 1, 2 and 3 respectively. (https://en.wikipedia.org/wiki/Sylow_theorems)
In the abelian case there are $\mathbb Z_6\cong\mathbb Z_2\oplus\mathbb Z_3$.

Quick remark. Being a Sylow $p-$subgroup is invariant with respect to conjugating, thus if $H$ was a unique Sylow $p-$subgroup, we would have $gHg^{-1} = H, g\in G,$ i.e $H$ would be normal.

Fact. Suppose a non-abelian group $G$ has order $pq$ (primes) with $q\equiv 1 \pmod{p}$, then there are precisely $q$ Sylow $p-$subgroups in $G$.

Proof of fact. Let $n_p,n_q$ be the numbers of Sylow $p,q-$subgroups respectively. By assumption $q>p$. By theorem 3 $n_q \equiv 1 \pmod{q}$ and by theorem 1 $n_q\mid p$, which forces $n_q = 1$, thus we have a (normal) subgroup of order $q$, which makes it cyclic, therefore abelian.

For $n_p$ we thus have two choices: $n_p = 1,q$. If $n_p = 1$, then $G\cong \mathbb Z_p \oplus \mathbb Z_q$ would be abelian, thus it must be that $n_p = q$.

Our non-abelian group $G$ therefore contains three Sylow $2-$subgroups, call them $H_1,H_2,H_3$, which by theorem 2 are all conjugate to each other. We have $\mbox{Sym}(H_1,H_2,H_3) \cong S_3$. We show that $G\cong S_3$. Define
$$\varphi : G\to \mbox{Sym}(H_1,H_2,H_3), \quad\varphi (g)(H_j) := gH_jg^{-1}$$
(also called conjugating)
Firstly, if $H$ is a Sylow $p-$subgroup, then conjugating it gives another Sylow $p-$subgroup, hence it must hold that $gH_ig^{-1} = H_j$ for some $j$.

The map $\varphi (g)$ is injective, because if $gH_ig^{-1} = gH_jg^{-1}$, then
$$h_i \in H_i \Rightarrow gh_ig^{-1} \in gH_ig^{-1} = gH_jg^{-1} \Rightarrow h_i \in H_j.$$
The argument is symmetrical, thus $H_i=H_j$. It is surjective due to finiteness.

Take $g,h\in G$, then
$$\varphi (gh) (H_j) = (gh)H_j(gh)^{-1} = g(hH_jh^{-1})g^{-1} = g (\varphi (h)(H_j)) g^{-1} = \varphi (g) (\varphi (h)(H_j)) = (\varphi (g) \varphi (h))(H_j).$$
To show it's an isomorphism, it suffices to show it's injective (because $|G| = |S_3| = 6$). Suppose $\varphi (g) = \mbox{id}$ i.e $gH_jg^{-1}=H_j$, then by definition $g\in N(H_j)$ the normaliser of $H_j$. If $N(H_j) = G$, then $H_j$ would be normal, which would then make it equal to its conjugates, contradicting the fact that there are three Sylow $2-$subgroups.

Therefore, all $N(H_j) = H_j$ and $g\in H_1\cap H_2\cap H_3 = \{e\}$ i.e $g=e$, which makes $\varphi$ an isomorphism.

Last edited:
DEvens
Gold Member
Regarding question 5, a clarification question.

Would constructing all possible character tables be an acceptable answer? I can get my group theory text and swat this up, but it would take me a couple hours to refresh that stuff since I have not looked at it in some time.

Math_QED