# Math Challenge - January 2019

• Challenge
• Featured
Mentor

## Main Question or Discussion Point

Merry Christmas to all who celebrate it today!

Rules:

a)
In order for a solution to count, a full derivation or proof must be given. Answers with no proof will be ignored. Solutions will be posted around 15th of the following month.
b) It is fine to use nontrivial results without proof as long as you cite them and as long as it is "common knowledge to all mathematicians". Whether the latter is satisfied will be decided on a case-by-case basis.
c) If you have seen the problem before and remember the solution, you cannot participate in the solution to that problem.
d) You are allowed to use google, wolframalpha or any other resource. However, you are not allowed to search the question directly. So if the question was to solve an integral, you are allowed to obtain numerical answers from software, you are allowed to search for useful integration techniques, but you cannot type in the integral in wolframalpha to see its solution.

Questions:

1.
Given the surface $$f(t,\varphi) = ((1+t^2)\cos \varphi \; , \;(1+t^2)\sin \varphi\; , \;t)\quad (t\in \mathbb{R}\; , \;0\leq \varphi \leq 2\pi)$$
• Compute the first fundamental form of this surface.
• Compute the second fundamental form and the Gauss curvature of this surface.
• Compute the geodesic curvature ##\kappa_g## and the normal curvature ##\kappa_n## of the circular latitude at ##t=1##.
Only solutions to all three parts will be accepted.

2. (solved by @Young physicist ) Three pirates are stranded on an island and find that there are only a few monkeys besides drinking water and coconuts. After collecting coconuts for a whole day, they want share them the next morning. At night, one of the pirates awakes and hides his third of the coconuts. But since an odd number of nuts is left, he gives one to a monkey. The second pirate awakens shortly afterwards and hides his third of the remaining coconuts. Again an odd number of coconuts remains, so he gives one to a monkey. The third does the same thing a short time later and gives a leftover nut to a monkey. The next morning they divided the few remaining coconuts among each other. Now the question: How many coconuts did the three pirates at least collect the day before and how are they distributed on each?

3. (solved by @PeroK and @Charles Link )and ) A cyclist drives along a railway track. Every ##30## minutes, he is overtaken by a train and every ##20## minutes he is met by a train. At which frequency do the trains travel on this connection?

4. The Heisenberg group ##H=\left\{\begin{bmatrix}1&a&c\\0&1&b\\0&0&1\end{bmatrix}\,:\, a,b,c\in \mathbb{Z}^3 \right\}## operates discontinuously on ##\mathbb{R}^3## by
$$h(p)=h(x,y,z)=\begin{bmatrix}1&a&c\\0&1&b\\0&0&1\end{bmatrix} \cdot \begin{bmatrix}x\\y\\z\end{bmatrix} = \begin{bmatrix}x+a\\y+b\\z+ay+c\end{bmatrix}$$
Show that the Heisenberg manifold ##\mathbb{R}^3/H## is orientable.

5. (solved by @Hiero ) Solve ##x'(t)=\dfrac{2t+2x(t)}{3t+x(t)}\; , \;x(2)=0\,.##

6. (solved by @Hiero ) Show that ##T\, : \,C([1,2])\longrightarrow C([1,2])## defined by
$$T(y)(t) := 1+\int_1^t \dfrac{y(s)}{2s}\,ds$$
has at least one fixed point and determine them.

7. (solved by @Ibix ) Compute ##\exp(tA)## where ##A=\begin{bmatrix}1&0&1\\0&1&0\\-1&0&1\end{bmatrix}## and determine the behavior of ##\det(\exp(tA))## for ##t \to \pm \infty\,.##

8. (solved by @Periwinkle ) Let ##G## be a group generated by ##\sigma,\varepsilon,\delta## with ##\sigma^7=\varepsilon^{11}=\delta^{13}=1 ##.
• Show that there is no transitive operation of ##G## on a set with ##8## elements.
• Is there are group ##G## with the above properties, that operates transitively on a set with ##12## elements?

9. (solved by @Math_QED ) Let ##R,S## be rings and ##\varphi\, : \,R \longrightarrow S## a ring epimorphism. Further let ##J \subseteq S## be an ideal.
• Define an ideal ##I \subseteq R## such that ##R/I \cong S/J\,.##
• Is the preimage of the center of ##S## equal to the center of ##R\,?##

10. A lie algebra ##\mathfrak{g}## is called reductive, if ##\mathfrak{g}=\mathfrak{Z(g)} \oplus [\mathfrak{g},\mathfrak{g}]## is the direct sum of its center and its derived algebra. (This is an important class of Lie algebras, as they are exactly those whose representations split into a direct sum of irreducible representations. Semisimple and in particular the simple, classical matrix Lie algebras are reductive.)

Show that the Lie algebra ##\mathfrak{gl}(V)## of all endomorphisms of a finite dimensional complex vector space is reductive.

Last edited:
Math_QED and Greg Bernhardt

Problem 5,
The equation is not separable. I’m not sure if there’s some methodical approach to solve this, but I just happened to guess a substitution which makes it separable, namely, for t ≠ 0, let ##u(t)≡x(t)/t##

Then (leaving out the arguments) we have:
$$\frac{dx}{dt} = \frac{d}{dt}(ut) = u + \frac{du}{dt}t = \frac{2+2u}{3+u}$$
The last equality is the right side of the differential equation in question. Subtracting u over gives:
$$\frac{du}{dt}t = \frac{2+2u}{3+u}-u = \frac{2-u-u^2}{3+u}=-\frac{(u+2)(u-1)}{3+u}$$
Now we separate the equation:
$$\int \frac{-1}{t}dt = \int \frac{3+u}{(u+2)(u-1)}du$$
For the u integral we use the method of ‘partial fraction decomposition,’ so we want to find A and B such that ##\frac{A}{u+2}+\frac{B}{u-1} =\frac{3+u}{(u+2)(u-1)}## from which we find A = -1/3, B = 4/3, and so (multiplying the 3 to the other side) we get:
$$-3\ln t = \int \Big(\frac{4}{u-1}-\frac{1}{u+2} \Big)du = 4\ln (u-1)-\ln(u+2) +C_0$$
Which can be re-written as:
$$\ln(t^{-3}) = \ln \Big(C\frac{(u-1)^4}{u+2}\Big)$$
Which of course implies:
$$t^{-3} = C\frac{(u-1)^4}{u+2}$$
The initial condition x(2) = 0 implies u(2) = 0 which gives ##2^{-3} = C\frac{(-1)^4}{2}## which implies C = 1/4, so we have:
$$t^{-3} = \frac{(u-1)^4}{4(u+2)}$$
Putting x(t) back in the equation, we get:
$$4(x+2t)= (x-t)^4$$
Which is my final (implicit) answer.
(I think it only works for t > 0 since we used log t, maybe some further restrictions too, I haven’t checked my answer.)

Problem 6,
If I’m understanding correctly, a fixed “point” (or fixed function) would be one such that T(y) = y, which means that:
$$y(t) = 1 + \int_1^t \frac{y(s)}{2s}ds$$
The most natural thing to do is to differentiate both sides with respect to t, yielding:
$$\frac{dy(t)}{dt} = \frac{y(t)}{2t}$$
This is separable and easy to solve:
$$\int \frac{dy}{y} = \int \frac{dt}{2t}$$
$$\ln y = \ln (t)/2 +C_0$$
$$y = C\sqrt t$$
(Funny how differentiating then integrating an equation is fruitful!)
To find permissible values of C, we go back to the original equation:
$$C\sqrt t = 1 + \int_1^t \frac{C\sqrt s}{2s}ds = 1 + \int_1^t \frac{C}{2\sqrt s}ds = 1 + C(\sqrt t - 1)$$
Hence we require C = 1 and the only stationary point is ##y(t) = \sqrt t##
That took tooooo long to type out on mobile

(Edited to fix problem 5)

Last edited:
Mentor
Problem 5,
The equation is not separable. I’m not sure if there’s some methodical approach to solve this, but I just happened to guess a substitution which makes it separable, namely, for t ≠ 0, let ##u(t)≡x(t)/t##

Then (leaving out the arguments) we have:
$$\frac{dx}{dt} = \frac{d}{dt}(ut) = u + \frac{du}{dt}t = \frac{2+2u}{3+u}$$
The last equality is the right side of the differential equation in question. Subtracting u over gives:
$$\frac{du}{dt}t = \frac{2+2u}{3+u}-u = \frac{2-u-u^2}{3+u}=-\frac{(u+2)(u-1)}{3+u}$$
Now we separate the equation:
$$\int \frac{-1}{t}dt = \int \frac{3+u}{(u+2)(u-1)}du$$
For the u integral we use the method of ‘partial fraction decomposition,’ so we want to find A and B such that ##\frac{A}{u+2}+\frac{B}{u-1} =\frac{3+u}{(u+2)(u-1)}## from which we find A = -1/3, B = 4/3, and so (multiplying the 3 to the other side) we get:
$$-3\ln t = \int \Big(\frac{4}{u-1}-\frac{1}{u+2} \Big)du = 4\ln (u-1)-\ln(u+2) +C_0$$
Which can be re-written as:
$$\ln(t^{-3}) = \ln \Big(C\frac{(u-1)^4}{u+2}\Big)$$
Which of course implies:
$$t^{-3} = C\frac{(u-1)^4}{u+2}$$
The initial condition x(2) = 0 implies u(2) = 0 which gives ##2^{-3} = C\frac{1^4}{4}## which implies C = 1/2, so we have:
$$t^{-3} = \frac{(u-1)^4}{2(u+2)}$$
Putting x(t) back in the equation, we get:
$$2(x+2t)= (x-t)^4$$
Which is my final (implicit) answer.
(I think it only works for t > 0 since we used log t, maybe some further restrictions too, I haven’t checked my answer.)
You made a mistake in the resubstitution step which also affects the constant. And where has your constant gone to?
Problem 6,
If I’m understanding correctly, a fixed “point” (or fixed function) would be one such that T(y) = y, which means that:
$$y(t) = 1 + \int_1^t \frac{y(s)}{2s}ds$$
The most natural thing to do is to differentiate both sides with respect to t, yielding:
$$\frac{dy(t)}{dt} = \frac{y(t)}{2t}$$
This is separable and easy to solve:
$$\int \frac{dy}{y} = \int \frac{dt}{2t}$$
$$\ln y = \ln (t)/2 +C_0$$
$$y = C\sqrt t$$
(Funny how differentiating then integrating an equation is fruitful!)
Indeed! That's the trick here.
To find permissible values of C, we go back to the original equation:
$$C\sqrt t = 1 + \int_1^t \frac{C\sqrt s}{2s}ds = 1 + \int_1^t C\frac{1}{2\sqrt s}ds = 1 + C(\sqrt t - 1)$$
Hence we require C = 1 and the only stationary point is ##y(t) = \sqrt t##
You deduced a necessary condition for a fixed point, but why are there any fixed points at all?

You made a mistake in the resubstitution step which also affects the constant.
I was hoping you weren’t online! Check again, I’ve fixed it before your reply.

You deduced a necessary condition for a fixed point, but why are there any fixed points at all?
Hmm, because the solution works? Not sure what you’re asking

Edit:
If you’re asking for some formal argument (maybe involving the intermediate value theorem?) then it’s beyond me!

I’ll leave it for someone else to argue. I’m not good at being rigorous

Mentor
Hmm, because the solution works? Not sure what you’re asking

Edit:
If you’re asking for some formal argument (maybe involving the intermediate value theorem?) then it’s beyond me!

I’ll leave it for someone else to argue. I’m not good at being rigorous
Either you perform the test that it is actually a fixed point, or, the elegant way, use Banach's fixed point theorem.

The main reason why I'm complaining is, that we are used to solve equations by ##f(x)=g(x) \Longrightarrow \ldots \Longrightarrow x=c## and always ignore, that this does not imply, that ##x## actually fulfills the equation. This is o.k. if we only use equivalence relations, but differentiation and integration are not equivalent. So necessity is only half of a proof, existence has to be shown, too. I'm not saying that it's difficult, I'm just fighting the (general) sloppiness.

Hiero
Mentor
That took tooooo long to type out on mobile
Not sure it works on a mobile, but I installed a little script program (AutoHotkey) which allows me to have e.g. \left. \dfrac{d}{d}\right|_{} on Alt+U. That saves a lot of time, especially with the non letter codes like brackets.

Either you perform the test that it is actually a fixed point, or, the elegant way, use Banach's fixed point theorem.

The main reason why I'm complaining is, that we are used to solve equations by ##f(x)=g(x) \Longrightarrow \ldots \Longrightarrow x=c## and always ignore, that this does not imply, that ##x## actually fulfills the equation. This is o.k. if we only use equivalence relations, but differentiation and integration are not equivalent. So necessity is only half of a proof, existence has to be shown, too. I'm not saying that it's difficult, I'm just fighting the (general) sloppiness.
I’ve never studied analysis or general metric spaces, so reading the wiki on Banach’s fixed point theorem makes my eyes glaze over.

But doesn’t my last step (where C was determined) count as a direct test of existence? (We would just do the same steps with C = 1)

Also, about problem 5; I presume you solved it by the same substitution? Is there any reasoning behind it or is it just a trial and error type of substitution? I just got lucky that it turned out separable; I didn’t fully expect it to work.
Not sure it works on a mobile, but I installed a little script program (AutoHotkey) which allows me to have e.g. \left. \dfrac{d}{d}\right|_{} on Alt+U. That saves a lot of time, especially with the non letter codes like brackets.
That would be nice! I just use a lot of copying and pasting

Problem 2:
The least amount of coconuts they can split in the end is 3
3+1 = 4
4 x (3/2) = 6. //Last pirate taking the 1/3
6+1 = 7
7 x (3/2) = 10.5. // Not a whole number, 3 is impossible

6,9,12,15,18 is impossible for the same reason.

21+1 = 22
22 x (3/2) = 33
33+1 = 34
34 x (3/2) = 51
51 +1 = 52
52 x (3/2) = 78

ANS:78

Mentor
I’ve never studied analysis or general metric spaces, so reading the wiki on Banach’s fixed point theorem makes my eyes glaze over.
Here is it:
\begin{align*}
|T(y)(t)-T(z)(t)|&\leq \int_1^t\dfrac{2s}{|y(s)-z(s)|}\,ds\leq \int_1^t \dfrac{||y-z||_\infty}{2}\\[8pt] &\leq \dfrac{t-1}{2}||y-z||_\infty \leq \dfrac{1}{2}||y-z||_\infty
\end{align*}
But doesn’t my last step (where C was determined) count as a direct test of existence? (We would just do the same steps with C = 1)
I'm not sure. The initial condition could make the steps reversible in which case we only had equivalent transformations, but I don't think that this is obvious. A simple integration
$$T(\sqrt{t})=1+\int_1^t \dfrac{\sqrt{s}}{2s}\,ds = 1+\dfrac{1}{2}\int_1^t s^{-\frac{1}{2}}\,ds =1+ \dfrac{1}{2}\cdot 2 \cdot \left[s^{\frac{1}{2}}\right]_1^t = 1+\sqrt{t}-1=\sqrt{t}$$
would have probably been the easiest way to check existence.

Honestly? I was fighting math jokes here! You know, the dozens of jokes with the pattern:
An engineer, a physicist and a mathematician were asked to solve ...
...
The engineer replied after five minutes: ##100##.
The physicist needed a quarter of an hour and replied: ##99.9985## with a ##5\sigma ## uncertainty.
The mathematician calculated for an hour and found: There is a solution!
Also, about problem 5; I presume you solved it by the same substitution? Is there any reasoning behind it or is it just a trial and error type of substitution? I just got lucky that it turned out separable; I didn’t fully expect it to work.
It's always difficult to answer "How did you get that?". A golden rule is: "Take away whatever is disturbing!" In this case it is the single variable "t", so cancelling the quotient by "t" is worth a try.

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Hiero
Mentor
Problem 2:
The least amount of coconuts they can split in the end is 3
3+1 = 4
4 x (3/2) = 6. //Last pirate taking the 1/3
6+1 = 7
7 x (3/2) = 10.5. // Not a whole number, 3 is impossible

6,9,12,15,18 is impossible for the same reason.

21+1 = 22
22 x (3/2) = 33
33+1 = 34
34 x (3/2) = 51
51 +1 = 52
52 x (3/2) = 78

ANS:78
##78## is correct. But how are they distributed?

A simple integration
$$T(\sqrt{t})=1+\int_0^t \dfrac{\sqrt{s}}{2s}\,ds = \dfrac{1}{2}\int_0^t s^{-\frac{1}{2}}\,ds = \dfrac{1}{2}\cdot 2 \cdot \left[s^{\frac{1}{2}}\right]_0^t = \sqrt{t}$$
would have probably been the easiest way to check existence.
Agreed. (Just a couple typos: should be integrating from 1 not 0, and your +1 term disappeared; but it works out the same.)
Honestly? I was fighting math jokes here! You know, the dozens of jokes with the pattern:
An engineer, a physicist and a mathematician were asked to solve ...
...
The engineer replied after five minutes: ##100##.
The physicist needed a quarter of an hour and replied: ##99.9985## with a ##5\sigma ## uncertainty.
The mathematician calculated for an hour and found: There is a solution!
An engineer must’ve written that one!

scottdave
Mentor
Agreed. (Just a couple typos: should be integrating from 1 not 0, and your +1 term disappeared; but it works out the same.)
Thanks. I corrected it. This happens if you integrate within LaTeX code instead of writing it down first

##78## is correct. But how are they distributed?
The first pirate takes 1/3 of 78 which is 26.
78-26 = 52
52-1 = 51 // To the monkey
The second pirate takes 1/3 of 51 which is 17.
51-17 = 34
34-1 = 33 // To the monkey
The third pirate takes 1/3 of 33 which is 11.
33-11 = 22
22-1 = 21.
Then, they spilt the 21 coconut and each of them gets 7 coconuts.
In the end:

The first porate got 26+7 = 33 coconuts.
The second porate got 17+7 = 24 coconuts.
The third porate got 11+7 = 18 coconuts.
Monky get 3.

fresh_42
PeroK
Homework Helper
Gold Member
Re question 3:

Assume the trains travel at speed ##v## and the cyclist at speed ##u##. Let the time between trains be ##\Delta t## (in both directions), relative to the track and ##\Delta t', \Delta t''## for the cyclist. Then we have, for trains moving in the same direction as the cyclist:

##\Delta t' = \Delta t + \frac{u\Delta t}{v-u} = \Delta t(\frac{v}{v-u})##

And, for trains moving in the opposite direction:

##\Delta t'' = \Delta t - \frac{u\Delta t}{v+u} = \Delta t(\frac{v}{v+u})##

The frequency of the trains are ##f = \frac{1}{\Delta t}## etc., and:

##f' = f(1 - \frac{v}{u}), \ \ f'' = f(1 + \frac{v}{u})##

And:

##f' + f'' = 2f##

Hence, in this case, ##f = 2.5## trains per hour, or a train every ##24## minutes.

Ibix
Ibix
Learning what the exponential of a matrix is has been on my to do list for a while. To that end - question 7
You can write ##tA## in diagonal form as ##tA=UDU^{-1}##. Then ##\exp(tA)=U\exp(D)U^{-1}##, and the exponential of a diagonal matrix ##D=\mathrm{diag}(d_1,d_2,\ldots,d_n)## is ##\mathrm{diag}(e^{d_1},e^{d_2},\ldots,e^{d_n})##. Grinding through the algebra gives me
$$\pmatrix{\frac 12\left(e^{t+it}+e^{t-it}\right)& 0& \frac i2\left(e^{t-it}-e^{t+it}\right)\cr 0& e^{t}& 0\cr \frac i2\left(e^{t+it}-e^{t-it}\right)& 0& \frac 12\left(e^{t+it}+e^{t-it}\right) \cr}$$You can easily calculate the determinant of this (particularly since we got the diagonal form en route). That turns out to be $$\det(\exp(tA))=e^{3t}$$So it goes to infinity in the limit ##t\rightarrow\infty## and goes to zero in limit ##t\rightarrow-\infty##. Since ##tA## represents a 45° rotation and a non-uniform scaling proportional to ##t##, I think this seems reasonable.

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Mentor
Learning what the exponential of a matrix is has been on my to do list for a while. To that end - question 7
You can write ##tA## in diagonal form as ##tA=UDU^{-1}##. Then ##\exp(tA)=U\exp(D)U^{-1}##, and the exponential of a diagonal matrix ##D=\mathrm{diag}(d_1,d_2,\ldots,d_n)## is ##\mathrm{diag}(e^{d_1},e^{d_2},\ldots,e^{d_n})##. Grinding through the algebra gives me
$$\pmatrix{\frac 12\left(e^{t+it}+e^{t-it}\right)& 0& \frac i2\left(e^{t-it}-e^{t+it}\right)\cr 0& e^{t}& 0\cr \frac i2\left(e^{t+it}-e^{t-it}\right)& 0& \frac 12\left(e^{t+it}+e^{t-it}\right) \cr}$$You can easily calculate the determinant of this (particularly since we got the diagonal form en route). That turns out to be $$\det(\exp(tA))=e^{3t}$$So it goes to infinity in the limit ##t\rightarrow\infty## and goes to zero in limit ##t\rightarrow-\infty##. Since ##tA## represents a 45° rotation and a non-uniform scaling proportional to ##t##, I think this seems reasonable.
I wished you had written ##e^t\cdot\begin{bmatrix}\cos t & 0 & \sin t \\ 0 & 1 & 0 \\ -\sin t &0 & \cos t\end{bmatrix}## because this way I didn't have to look up this horrible sign distribution ... but yes, that's right. Grinding through the algebra is fastest, if we write ##\exp(tA)=e^t \cdot \exp(t(E_{13}-E_{31}))## with matrices ##E_{ij}## that have a ##1## on position ##(i,j) ## and ##0's## otherwise, since they are easily to take to powers.

Ibix
I wished you had written ##e^t\cdot\begin{bmatrix}\cos t & 0 & \sin t \\ 0 & 1 & 0 \\ -\sin t &0 & \cos t\end{bmatrix}## because this way I didn't have to look up this horrible sign distribution ... but yes, that's right.
I did notice that on a re-read, but didn't have a chance to edit at the time. However, note that our matrices look to be transposed with respect to each other. Which of us has flipped a sign?
Grinding through the algebra is fastest, if we write ##\exp(tA)=e^t \cdot \exp(t(E_{13}-E_{31}))## with matrices ##E_{ij}## that have a ##1## on position ##(i,j) ## and ##0's## otherwise, since they are easily to take to powers.
Shouldn't that be
##\exp(tA)=e^t \cdot \exp(t(I+E_{13}-E_{31}))##, or am I missing something?

Mentor
I did notice that on a re-read, but didn't have a chance to edit at the time. However, note that our matrices look to be transposed with respect to each other. Which of us has flipped a sign?
None of us. It all depends on how you write the sine function: ##\dfrac{i}{2}## or ##\dfrac{1}{2i}##. I checked it several times, that's why I made the comment. Now please don't tell me I was wrong ...
Shouldn't that be
##\exp(tA)=e^t \cdot \exp(t(I+E_{13}-E_{31}))##, or am I missing something?
No, the ##I## is in the leading factor ##e^t\,.## ## tA=t(1+tB)## with ##B=E_{13}-E_{31}##. Since ##[I,B]=0## we have ##\exp(tA)=\exp(t\cdot I)\cdot \exp(tB)=e^t \cdot \exp(t(E_{13}-E_{31}))\,.##

Ibix
Ibix
None of us. It all depends on how you write the sine function: ##\dfrac{i}{2}## or ##\dfrac{1}{2i}##. I checked it several times, that's why I made the comment. Now please don't tell me I was wrong ...
Nope - you're right, I got the ##i## in the wrong place in the definition of sine. Should have kept my mouth shut
No, the ##I## is in the leading factor ##e^t\,.## ## tA=t(1+tB)## with ##B=E_{13}-E_{31}##. Since ##[I,B]=0## we have ##\exp(tA)=\exp(t\cdot I)\cdot \exp(tB)=e^t \cdot \exp(t(E_{13}-E_{31}))\,.##
Neat - thanks.

Homework Helper
Gold Member
A slightly simpler way for problem 3: There is a (unknown) distance D between the trains. Let ## v_t ## be the speed of the trains and ## v_c ## the speed of the bicycle. ## \\ ## We have ## 30= \frac{D}{v_t-v_c} ## and ## 20=\frac{D}{v_t+v_c} ##. The problem is to solve for ## T=\frac{D}{v_t} ##. The frequency will be the inverse of that. ## \\ ## Rewriting: ## \frac{v_t}{D}-\frac{v_c}{D}=\frac{1}{30} ##, and ## \frac{v_t}{D}+\frac{v_c}{D}=\frac{1}{20} ##. ## \\ ## Let ## \frac{v_t}{D}=x ## and ## \frac{v_c}{D}=y ##. ## \\ ## This gives ## x-y=\frac{1}{30} ## and ## x+y=\frac{1}{20} ##. ## \\ ## Adding these two equations: ## 2x=\frac{1}{12} ##. ## \\ ## The result is ## x=\frac{1}{T}=\frac{1}{24} ## (in frequency per minute), or ## \frac{60 \, min/hr}{24 \, min}= 2.5 ## trains each hour.

Ibix and PeroK
PeroK
Homework Helper
Gold Member
There is a (unknown) distance D between the trains. Let ## v_t ## be the speed of the trains and ## v_c ## the speed of the bicycle. ## \\ ## We have ## 30= \frac{D}{v_t-v_c} ## and ## 20=\frac{D}{v_t+v_c} ## ...

... ## \frac{v_t}{D}-\frac{v_c}{D}=\frac{1}{30} ##, and ## \frac{v_t}{D}+\frac{v_c}{D}=\frac{1}{20} ##
... ## f= \frac{1}{T} = \frac{v_t}{D} = \frac{1}{24}##

Even simpler!

Mentor

Imagine she rides one hour in one direction and one hour in the other. Then she meets three trains in the first hour and is overtaken by two in the second hour. So the frequency is thus ##5## trains per direction in ##120## minutes, i.e. every ##24## minutes a train.

Math_QED, Ibix, StoneTemplePython and 2 others
Bilal Rajab Abbasi
Gold Member
Can we upload a picture of solution here?

Mentor
Can we upload a picture of solution here?
Please don't, type it out. Here's an info page on how to do this: https://www.physicsforums.com/help/latexhelp/
Pictures are almost always hard to read and often don't even have the correct orientation. Both forces the readers to download and edit it. I don't like being forced to do that.