Math O' Man : The Blog of Mathematics

Theorem about a circle, three chords and a midpoint



Posted by Mathoman, Friday 29 January 2010 at 13:36 - Riddles And Exercises - Tags

Here is a nice exercise in plane geometry. As often in mathematics the statement is quite simple — but the proof is not!

Let \scr{C} be a circle, A,B two distinct points on \scr{C} and M be the midpoint of the chord [AB]. Take two other chords, [PQ] and [SR], that pass through M. Let C (resp. D) be the intersection of [AB] with [PS] (resp. [RQ]).
Prove that M is the midpoint of the chord [CD].

Chords of a circle, midpoint of segments, plane geometry, a theorem in euclidean geometry, butterfly theorem
Suprising! If M is the midpoint of [AB] then also of [CD].

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Christmas Riddle

Posted by Mathoman, Friday 25 December 2009 at 13:57 - Riddles And Exercises - Tags

X-mas is the time to crack nuts. At least that's what normal people do. But mathematicians enjoy cracking very peculiar nuts, like the following one. Show that the equation below is true for every positive integer.

\sum_{k=0}^n\left\(\begin{array}{c}2n+1\\2k+1\end{array}\right\)(2k+1)\:=\:2^{2n-1}(2n+1).

As always in maths, the slogan is short is beautiful, i.e., you should find a solution that is as short and elegant as possible, without using much computations. For this exercise ou may get inspiration from Santa Claus who has to distribute gifts into ChristmasStockings...

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An exercise about groups

Posted by Mathoman, Sunday 20 December 2009 at 12:45 - Riddles And Exercises - Tags

A topological group is a set G with a group structure and a topology such that the binary law

G \times G \rightarrow G ,\;\; (x,y) \rightarrow xy,
and the map of the inverse
G \rightarrow G ,\;\; x \rightarrow x^{-1},
are continuous. In other words both structures, the algebraic and the topological one, are linked in a natural way. So the following question is of interest.

Question

Are there topological groups which are isomorphic as groups and homeomorphic as topological spaces but not isomorphic as topological groups?

Answer

Yes. The proof goes in three stages:

  1. Let G and H be dense subsets of \mathbb{R} and f :\, G \rightarrow H a monotonic bijection. Then f is a homeomorphism.

    Without restriction of generality we can suppose that f is increasing. We prove that it is continuous. Let x_0\in G and \epsilon>0. Since H is dense in \mathbb{R} we have H\cap\,]f(x_0)-\epsilon,f(x_0)[\,\neq\emptyset. Thus there exists y_1\in H\cap\,]f(x_0)-\epsilon,f(x_0)[\,. Similarly there exists y_2\in H\cap\,]f(x_0),f(x_0)+\epsilon[\,. Since f is onto we can write y_k=f(x_k) with x_k\in G, k=1,2. Let \delta=\min(x_0-x_1,x_2-x_0). Then for all x in G

    \begin{align*}x_0-\delta<x<x_0+\delta \;\;\;\Longrightarrow\;\;\;& f(x_0-\delta)<f(x)<f(x_0+\delta)\\ \Longrightarrow\;\;\;&y_1=f(x_1)\leq f(x)\leq f(x_2)=y_2\\ \Longrightarrow\;\;\;&f(x_0)-\epsilon<f(x)<f(x_0)+\epsilon\,, \end{align*}

    which proves that f is continuous in x_0. The proof of the continuity of the inverse function f^{-1} is the same.
     
  2. Let G and H be countable and dense subsets of \mathbb{R}. Then they are homeomorphic.

    First we write

    \begin{align*} G&=\{x_0,x_1,x_2,\ldots\}\;\;\;\;\;(*)\,,&H&=\{y_0,y_1,y_2,\ldots\}\;\;\;\;(**)\,. \end{align*}

    We will now list G and H in a new way, G=\{x'_0,x'_1,x'_2,\ldots\} and H=\{y'_0,y'_1,y'_2,\ldots\}. The aim is to make G \to H, x'_k \mapsto y'_k, a monotonic bijection (then it is automatically a homeomorphism). For this we proceed as follows.
     
    • k=0. We take x'_0=x_0,\;y'_0=y_0
       
    • k=1. We take x'_1=x_1. To choose y'_1 we look at the order of x'_0 and x_1'.
      If x'_1<x'_0 then we take as y'_1 any element in H which is smaller than y'_0.
      If x'_1>x'_0 then we take as y'_1 any element in H which is greater than y'_0.
       
    • k=2. We take as y'_2 the first element of H\setminus\{y'_0,y'_1\} listed in (**). To choose x'_2 we look at the order of y'_0,y'_1,y'_2.
      If y'_2 is smaller than y'_0 and y'_1 we take as x'_2 any element in G which is smaller than x'_0 and x'_1.
      If y'_2 is greater than y'_0 and y'_1 we take as x'_2 any element in G which is greater than x'_0 and x'_1.
      If y'_2 is between y'_0 and y'_1 we take as x'_2 any element in G which is between x'_0 and x'_1.
       
    • k=3. We take as x'_3 the first element of G\setminus\{x'_0,x'_1,x'_2\} listed in (*). To choose y'_3 we look at the order of x'_0,x'_1,x'_2,x'_3. There are 24 possible orders for these four numbers.
      If x'_3<x'_0<x'_1<x'_2 we take as y'_3 any element in H which is smaller than y'_0,y'_1,y'_2.
      If x'_2<x'_3<x'_0<x'_1 we take as y'_3 any element in H which is bewteen y'_2 and y'_0.
      And so forth.
       
  3. The topological groups G=\mathbb{Q}+\mathbb{Q}\sqrt2 and H=\mathbb{Q}+\mathbb{Q}\sqrt3 fulfill our wishes.

    After what has been said before, G and H are homeomorphic as topological spaces. Clearly they are isomorphic groups. But they are not isomorphic topological groups. In fact, suppose that there exists an isomorphism of topological groups f :\, G \to H. One proves easily by induction that f(n)=nf(1) for every integer n, and then that f(r)=rf(1) for every rational r. Therefore by continuity

    f(\sqrt2)=\sqrt2f(1),\;\;\;\;\;\lightning

    a contradiction in H=\mathbb{Q}+\mathbb{Q}\sqrt3.

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An exercise about the temperature distribution on the earth

Posted by Mathoman, Friday 23 October 2009 at 14:55 - Riddles And Exercises - Tags

Heute mal ein bisschen climatology! The following question, funny and useless, is dedicated to my friend A. Wirth, who has left pure mathematics in order to use his mind for applied sciences like oceanography and meteorology ;-)

Exercise: Suppose that the earth is a perfect ball and that the temperature on the earth's surface is a continuous function. Prove that there are infinitely many pairwise disjoint sets {A,B} with A and B points on the earth surface such that the temperature in A and B is the same and such that the distance between A and B is 1000 km.

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Comatrix and adjugate matrix

Posted by Mathoman, Tuesday 14 July 2009 at 13:38 - Riddles And Exercises - Tags

The comatrix com(M) of a n x n-matrix M is the n x n-matrix whose entry in (l,k) is \small{(-1)^{l+k}} times the determinant of the matrix which you get by deleting in M the line l and the row k.
It is the transpose of the comatrix which is of interest to us; it is called adjugate matrix, and it is shown in every lecture on linear algebra that it has the following fundamental property:

^t\text{com}(M)\:M\;=\;M\:^t\text{com}(M)\;=\;\det(M)\:I\:.

If you work with matrix coefficients in a ring R, it follows that the matrix M is invertible in the matrix ring precisely if the scalar det(M) is invertible in the ring R. For example over \small\mathbb{Z} the invertible matrices are those whose determinant is 1 oder -1 ist.

Exercise:  Prove that  com  preserves matrix multiplication, i.e.,

com(I) = I      et      com(MN) = com(M) com(N).

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Dimension of the Commutator Space

Posted by Mathoman, Thursday 9 July 2009 at 14:02 - Riddles And Exercises - Tags

Here is a beautiful exercise from matrix theory.

Let A be a n x n matrix. Show that the dimension of the commutator space of A (i.e., the set of all matrices that commute with A) is at least n.

There is a proof that works on any field. But in the special cases of real or complex numbers there are alternative proofs.

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A sunday child

Posted by Mathoman, Sunday 5 July 2009 at 20:08 - Riddles And Exercises - Tags

Here is a personalised birthday gift for my father, in form of a mathematical riddle.

Today, sunday 5th of july 2009, is my father's birthday. He was born on a sunday of a leap year. What is his age today?

In order to solve that exercise you can use the fact that I am older than twenty-three, that my father was older than twenty-three when he took upon him the responsibility of becoming my father and finally that he isn't hundred years old.

Anyway, happy birthday, Daddy!

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Deserve your dessert!

Posted by Mathoman, Wednesday 1 July 2009 at 11:30 - Riddles And Exercises - Tags

We all know those little riddles that entertain mathematicians at the end of lunch at the cafeteria. Here is one of them:

Connect the following nine points by four straight lines, without lifting your pencil.

°             °             °



°             °             °



°             °             °



Easier said than done. The solution in the video below shows that you should not get contained in your thinking habits...


MathOMan connects 9 points with 4 lines


Smiling Bin is even more witty and takes her revenge by challenging me with the following task, which must be impossible for any specialist on connectivity questions in topology.

Without lifting your pencil draw a circle and its center (but not more).

Here is the video where she shows us her tricky solution:


Bin draws a circle and its midpoint

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Another bet you can win

Posted by Mathoman, Sunday 24 May 2009 at 12:05 - Riddles And Exercises - Tags

The photos to my recent post on the circumference remind me of another bet about a beer glass. Just ask your fellow drinkers the following question:

There are two glasses, one filled with beer and the other one with the same quantity of wine. One takes a spoonful of the beer and pours it into the wine; then one makes the the same in the other direction, i.e., one transfers a spoonful of that new wine-beer mix back into the glass filled with beer. Now in the first glass the beer is spoiled by a small quantity of wine and in the second glass the wine is spoiled by a small quantity of beer. Which glass is more polluted?

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Exercise on the codimension in linear algebra

Posted by Mathoman, Saturday 16 May 2009 at 21:54 - Riddles And Exercises - Tags

I am collecting interesting maths exercises that can be solved by undergraduate students. One can find a lot of them in textbooks, on the internet, in old worksheets of one's own student time... and sometimes one is lucky and invents a new exercise. Here is a linear algebra question that came up in my mind last week end. I like it because the solution I found needs no deep theorem but only some basic understanding of linear algebra:
What is the biggest integer k, such that every affine sub-space of codimension k in the space of n x n matrices contains an invertible matrix?
Recall: The codimension of a sub-space is the difference between the dimension of the ambient space and the one of the sub-space. In other words, it is the number of equations necessary to describe the subspace (every equation taking away one degree of liberty). For example, in our common three-dimenssional space the codimension of a line is 2 and the codimension of a plane is 1.

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Rationnal vs. irrationnal

Posted by Mathoman, Saturday 25 April 2009 at 09:44 - Riddles And Exercises - Tags

Are there rational numbers x, y such that y^x is irrational?
Are there irrational numbers x, y such that y^x is rational?

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The Master And His Dog

Posted by Mathoman, Saturday 21 March 2009 at 12:05 - Riddles And Exercises - Tags

After a walk a man and his dog return home. Their distance from home is 10 km. The man walks with constant speed of 5 km/h. But the dog runs twice as fast and reaches home at the moment when his master has only done half of the way; immediately the dog runs back to his master, then back home, then aganin back to his master, and so forth.

This back and forth finally stops when the man reaches home. What is the total distance the dog has done?

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