Regular polygons, exercise.

With a drawing program yesterday's pictures are easy to fake, of course. But these drawing programs don't give you the numbers.

Exercise.

Place a decagon edge-to-edge on a square with sides of length 1 ( see figure ). What is the distance between the two marked points?

( Answer: $ \frac{1}{4} \left(3 \sqrt{10-2 \sqrt{5}}+\sqrt{50-10 \sqrt{5}}+4\right) $ )

Solution to the sequence.

8, 72, 46, 521, ... ?

612.

What's the next number in this sequence ,,, ?

8, 72, 46, 521, ... ?

Answer follows.

Polynomial exercise - Solution.

In a recent post I proposed the following exercise.

Let $$x^3 + bx^2 + cx + d$$ be a polynomial with coefficients in $\mathbf{Q}$. We ask which condition(s) $b,c,d$ must satisfy in order that one ( any ) root be the average of the other two roots?

Solution.

Let the roots of $x^3 + bx^2 + cx + d = 0$ be $\alpha_1, \alpha_2, \alpha_3$. Clearly, the roots must satisfy the following equations:
\begin{align*}
\alpha_1=& \frac{1}{2}\alpha_2 + \frac{1}{2}\alpha_3 \\
\alpha_2=& \frac{1}{2}\alpha_1 + \frac{1}{2}\alpha_3 \\
\alpha_3=& \frac{1}{2}\alpha_1 + \frac{1}{2}\alpha_2 \\
\end{align*}
This is a rank $2$ system of linear equations with solution $( \alpha_1, \alpha_2, \alpha_3) = \lambda (1,1,1) $, and implies that the equation has three equal roots and can thus be written as follows:$$(x-\alpha)^3 = x^3-3\alpha x^2 + 3\alpha^2 x - \alpha^3$$.
So if the root is $\alpha$ then $b=-3\alpha, c=3\alpha^2$ and $d=-\alpha^3$.

Polynomials

I am studying some more about polynomials, the topic of symmetric polynomials for example, is an interesting one.

Let $$x^3 + bx^2 + cx + d$$ be a polynomial with coefficients in $\mathbf{Q}$. We ask...

...which condition(s) $b,c,d$ must satisfy in order that one ( any ) root be the average of the other two roots?

To be continued.

M381 'Challenge Exercise' - Revisited

Sorry, Paddy and Jobidaker for the late reply. ( Auto-accepting comments has its drawbacks too. )

My solution for the 1/2, 1/3, 1/5 and 1/7 case is the following. Follow the pattern for the solution to the general case. Click to enlarge pic.

Please let me know if you think a smaller number qualifies.

Numbers divisible by seven

Sometimes, doing an exercise runs out of control, slurps time and leads to posts like this:

\begin{array}{l}
111111101101010101001001 \\
111111101101001001010101 \\
111111010101101101001001 \\
111111010101001001101101 \\
111111001001101101010101 \\
111111001001010101101101 \\
101101111111010101001001 \\
101101111111001001010101 \\
101101010101111111001001 \\
101101010101001001111111 \\
101101001001111111010101 \\
101101001001010101111111 \\
10101111111101101001001 \\
10101111111001001101101 \\
10101101101111111001001 \\
10101101101001001111111 \\
10101001001111111101101 \\
10101001001101101111111 \\
1001111111101101010101 \\
1001111111010101101101 \\
1001101101111111010101 \\
1001101101010101111111 \\
1001010101111111101101 \\
1001010101101101111111
\end{array}

you might assume they are binary numbers. They might be, but they are just ordinary decimal numbers consisting of only ones and zeros. They are all divisible by seven. For your convenience I'll show them again, divided by seven this time.

\begin{array}{l}
15873014443001443000143 \\
15873014443000143001443 \\
15873001443014443000143 \\
15873001443000143014443 \\
15873000143014443001443 \\
15873000143001443014443 \\
14443015873001443000143 \\
14443015873000143001443 \\
14443001443015873000143 \\
14443001443000143015873 \\
14443000143015873001443 \\
14443000143001443015873 \\
1443015873014443000143 \\
1443015873000143014443 \\
1443014443015873000143 \\
1443014443000143015873 \\
1443000143015873014443 \\
1443000143014443015873 \\
143015873014443001443 \\
143015873001443014443 \\
143014443015873001443 \\
143014443001443015873 \\
143001443015873014443 \\
143001443014443015873
\end{array}

Even mathematics tries to steal your time, sometimes.

An easy problem

A wise man rode into a desert village one evening as the sun was setting. Dismounting from his camel, he asked one of the villagers for a drink of water.‘Of course,’ said the villager and gave him a cup of water. The traveller drank the whole cupful. ‘Thank you,’ he said. ‘Can I help you at all before I travel on?’‘Yes,’ said the young man. ‘We have a dispute in our family. I am the youngest of three brothers. Our father died recently, God rest his soul, and all he possessed was a small herd of camels. Seventeen, to be exact. He decreed in his will that one half of the herd was to go to my oldest brother, one third to the middle brother and one ninth to me. But how can we divide a herd of 17? We do not want to chop up any camels, they are worth far more alive.’ ‘Take me to your house,’ said the sage. When he entered the house he saw the other two brothers and the man’s widow sitting around the fire arguing. The youngest brother interrupted them and introduced the traveller.

‘Wait,’ said the wise man, ‘I think I can help you. Here, I give you my camel as a gift. Now you have 18 camels. One half goes to the eldest, that’s nine camels. One third goes to the middle son, that’s six camels. And one ninth goes to my friend here, the youngest son. That’s two.’ ‘That’s only 17 altogether,’ said the youngest son. ‘Yes. By a happy coincidence, the camel left over is the one I gave to you. If you could possibly give it back to me, I will continue on my journey.’ And he did. 

What went wrong ?

Exercise ( algebra )

Given that $$x^n-y^n = (x-y) (\sum_{k=1}^n x^{n-k}y^{k-1} )$$ with for example: $$x^4-y^4 = (x-y)(x^3 +x^2y +xy^2 + y^3)$$.
a) How would you factorize $x^5 + y^5$?
b) Generalize.
c) Prove the identity above for $x^n-y^n$ using mathematical induction.

Exercise ( logic ).

Knowledge of mathematics is not required to solve the following exercise, but it will sure help ;-)

Dr. Who asked you for a ride in the Tardis. Naturally, you couldn't decline, it might be a matter of national, if not global, importance. The Tardis landed on the Planet of Truth which is is inhabited by people who always tell the truth. A minority however decided to lie, always. It is Doctor Who's mission to seek and destroy all liars. First you must get to the Capital of the Planet. Two roads fork out. Should you go left, or right? An inhabitant approaches, greets you and gives you the privilege of asking him one Yes/No question.

Ask him where the capital is, left or right on the fork, but beware he might be a liar!

Credit follows with the answer.

Take the challenge, test your ability to think logically.

Exercise

Exercise:

Find $x, y$ such that $$\frac{1}{x} + \frac{1}{y} = \frac{1}{pq}$$ where $x,y \in \mathbf{Z}$ and $p,q$ are prime.

Hint: there are nine different solutions. I'll publish the method and solution on request ( comment ).

Exercise in geometry

This exercise is about observation and perception. The exercise is simple. ( Come on, geometry of triangles: piece of cake for MST121-ers and above. ) Observe your observation, perceive your perception. Draw your own conclusions.

Enjoy!

[Exercise] - 2

This is a famous problem. I found it on the Internet by searching for "mathematics, monkey, coconut, problem". - My version is in a Dutch book called Algebra by M. Riemersma.

Five men and a monkey were shipwrecked on a desert island, and they spent the first day gathering coconuts for food. Piled them all up together and then went to sleep for the night. But when they were all asleep one man woke up, and he thought there might be a row about dividing the coconuts in the morning, so he decided to take his share. So he divided the coconuts into five piles. He had one coconut left over, and he gave that to the monkey, and he hid his pile and put the rest all back together.

By and by the next man woke up and did the same thing. And he had one left over, and he gave it to the monkey. And all five of the men did the same thing, one after the other; each one taking a fifth of the coconuts in the pile when he woke up, and each one having one left over for the monkey. And in the morning they divided what coconuts were left, and they came out in five equal shares. Of course each one must have known there were coconuts missing ; but each one was guilty as the others, so they did not say anything.

How many coconuts were there in the beginning?”

To be continued ( i.e. answer and comment )

[Exercise] - 1

Let $n \in \mathbb{N}$, show that $$f(n) = \frac{(2+\sqrt{3})^{1+2n}+(2-\sqrt{3})^{1+2n}+2}{6}$$ is a square.

For $n=1$ to $5$ we have ($n, \ \sqrt{f(n)}, \ f(n)$):
$\begin{array}{lll}
1. & 3. & 9. \\
2. & 11. & 121. \\
3. & 41. & 1681. \\
4. & 153. & 23409. \\
5. & 571. & 326041.
\end{array}$

An exercise in arithmetic

This tomb holds Diophantus. Ah, what a marvel! And the tomb tells scientifically the measure of his life. God vouchsafed that he should be a boy for the sixth part of his life; when a twelfth was added, his cheeks acquired a beard; He kindled for him the light of marriage after a seventh, and in the fifth year after his marriage He granted him a son. Alas! late-begotten and miserable child, when he had reached the measure of half his father's life, the chill grave took him. After consoling his grief by this science of numbers for four years, he reached the end of his life.

No tricks. A (simple) exercise in arithmetic encoded in a story about Diophantus. What is your answer?

Group Theory - Exercise ( 24/9-'10 )

Which group is represented by the following representation :
$(a,b|a^5=1,b^2=1,(a \circ b)^3=1)$.

Although I don't expect a question like this on the M208 or MS221 exams on Group Theory due to the ugly '5-min-to-think' constraint, candidates for M208 ( and possibly MS221 ) are well prepared to solve it.

Group Theory - Exercise - Continued

I have been working on the problem that I published last week....

S3
Element - Order - Permutation Sign - Transpositions
() - 1 - +1 - ()
(1,2,3) - 3 - +1 - (1,2)(1,3)
(1,3,2) - 3 - +1 - (1,3)(1,2)
(1,2) - 2 - -1 - (1,2)
(1.3) - 2 - -1 - (1,3)
(2,3) - 2 - -1 - (1,2)

Now take the following subgroup of A5:
() - 1 - +1 - ()
(3,4,5) - 3 - +1 - (3,4)(4,5)
(3,5,4) - 3 - +1 - (3,5)(3,4)
(1,2)(4,5) - 2 -  +1 - (1,2)(4,5)
(1,2)(3,4) - 2 -  +1 - (1,2)(3,4)
(1,2)(3,5) - 2 -  +1 - (1,2)(3,5)
This is a group with S3 structure but conisting entirely of even permutations.

If any group of n elements is a subgroup of A(n+2) it must have an isomorphic copy consisting of all positive permutations.

Group Theory - Exercise

Just had the pleasure of watching an episode of Poirot. What we share with Poirot is the need to exercise our gray cells...

If $G$ is a finite group of order $n$, prove that $G$ is isomorphic to a subgroup

of the alternating group $A_{n+2}$'

From 'Groups and Symmetry (Springer UTM) by M. A. Armstrong'

Enjoy!

Exercise - comments

A reader ( Paddy ) surprised me with an excellent answer to the exercise I posted. The series I posted was in fact the running total of the squares of the Fibonacci numbers. Let me put that in a table.

$ \begin{matrix}
n & F(n) & (F(n))^2  & \Sigma\\
1 & 1 & 1 & 1\\
2 & 1 & 1 & 2\\
3 & 2 & 4 & 6\\
4 & 3 & 9 & 15\\
5 & 5 & 25 & 40\\
6 & 8 & 64 & 104\\
7 & 13 & 169 & 273
\end{matrix} $

Paddy's answer...

$ \begin{matrix}
n & F(n) & F(n) * F(n+1)\\
1 & 1 & 1 \\
2 & 1 & 2 \\
3 & 2 & 6 \\
4 & 3 & 15 \\
5 & 5 & 40 \\
6 & 8 & 104 \\
7 & 13 & 273
\end{matrix} $

So $ \sum_{k=1}^{n} F_n^{2} = F_n * F_{n+1} $
An interesting conjecture to prove formally.

Exercise

Readers,
I would like to propose a challenge.

What follows ...
a) 1 - 2 - 6 - 15 - 40 - 104 - 273 - ?
b) If the sequence in a) is generated by a function f: N->R then what is f[n] ? Or... what is the closed form for the sequence in a).
c) Modify the polynomial involved in creating f such that at least one power of ten becomes a member of the image of f.

Remarks:
a) Should be simple for those with a pass on MS221;
b) Not solvable with the standard MS221 / M208 tools but I'll expect 221+ folk will accept the challenge;
c) No hints as it is the bonus question;

Solvers are nominated for the gallery of excellence.

Update:
( 1/5-'10)
What sort of a person are you? Do you give it a try? Or don't you think it isn't worth the time? Then may I ask, why read my blog in the first place? Do you only like the " TMA-sort-of-exercices " which are easy to solve and get marked? Then... mathematics is not for you. Think about it. It concerns: -you-.
Anyway, I designed parts b) and c) of the exercise for myself. I haven't find the answer yet. It's more difficult than I thought. But I am learning... I'll solve it.l