Κυριακή, 11 Ιουνίου 2017

Number Empire - Math Tools

Number Empire - Math Tools

The Mathematics of Juggling | Quanta Magazine

The Mathematics of Juggling | Quanta Magazine

Trigonometric substitution in inequalities | My Two Cents

Trigonometric substitution in inequalities | My Two Cents




Trigonometric substitution in inequalities




 
 
 
 
 
 
5 Votes

Trigonometry
and inequalities are my 2 favourite topics in MO. It gets better when
both of them are involved together. Before we begin, let us attempt some
simple warm up questions that may even appear in our Sec 4 mathematics
syllabus:
For triangle ABC, prove that
1) \tan{A}+\tan{B}+\tan{C}=\tan{A}\tan{B}\tan{C}
2) \tan{\dfrac{A}{2}}\tan{\dfrac{B}{2}}+\tan{\dfrac{B}{2}}\tan{\dfrac{C}{2}}+\tan{\dfrac{C}{2}}\tan{\dfrac{A}{2}}=1
3) \sin^2 {\dfrac{A}{2}}+\sin^2 {\dfrac{B}{2}}+\sin^2 {\dfrac{C}{2}}+2\sin{\dfrac{A}{2}}\sin{\dfrac{B}{2}}\sin{\dfrac{C}{2}}=1
One can get the above results upon
application of compound angle formulas. However, don’t expect to see
these sort of the problems in SMO since these identities are widely
publicised in Olympiad materials already. Instead, these identities will
be applied in other fields of MO, most notably in solving inequalities.
So for example, if the constraint x+y+z=xyz is given, one can substitute x=\tan{A}, y=\tan{B}, z=\tan{C} such that A+B+C=\pi.
After which, one can use trigonometric identities to simplify the
equation or apply Jensen’s inequality on the trigonometric functions.
Let us see an application of the strategy in SMO(O)2011 2nd Round P3:
Suppose x,y,z>0 and \dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}<\dfrac{1}{xyz}, prove that
\dfrac{2x}{\sqrt{1+x^2}}+\dfrac{2y}{\sqrt{1+y^2}}+\dfrac{2z}{\sqrt{1+z^2}}<3
Here’s my solution to the problem during
the competition itself. If you have tried using trigonometric
substitution before, you will realise that this question is SCREAMING
for you to substitute the variables using tangent functions (possible to
use cotangent function too). So let’s do that. From the condition, we
can rearrange the constraint into xy+yz+zx<1. Now let’s introduce our tangent functions, substitute and obtaining \tan{\dfrac{A}{2}}\tan{\dfrac{B}{2}}+\tan{\dfrac{B}{2}}\tan{\dfrac{C}{2}}+\tan{\dfrac{C}{2}}\tan{\dfrac{A}{2}}<1, which suggests that A+B+C<\pi.
The inequality that we want to prove can will be simplified into an elegant expression after substitution. We have
\dfrac{2x}{\sqrt{1+x^2}} = 2\sin{\dfrac{A}{2}}
So the inequality transforms into 2\sin{\dfrac{A}{2}}+2\sin{\dfrac{B}{2}}+2\sin{\dfrac{C}{2}}<3, which is easily proven using Jensen’s inequality since the sine function is concave for \theta<\pi.

In addition to this example,
trigonometric substitution may be useful if a bounded constraint is
given. For example, if it is stated in the question that x_i \in [-1,1], consider substituting x using a sine or cosine function.
There are also situations where you do
not even need any constraints at all to use trigonometric substitution!
Let us look at the following problem:
Prove that (ab+bc+ca-1)^2 \le (a^2+1)(b^2+1)(c^2+1) for real numbers a, b, c.
It doesn’t look feasible to apply mean
inequalities here because the variables take negative values as well. On
the other hand, the expression a^2+1 looks familiar. Sometimes, it is helpful to use trigonometric substitution in inequalities that contain the term a^2+1 as it can be used to simplify trigonometric expressions. Using tangent function, we obtain
(ab+bc+ca-1)^2 \le \sec^2 x\sec^2 y \sec^2 z
\Rightarrow (ab+bc+ca-1)^2\cos^2 x\cos^2 y\cos^2 z \le 1
By using compound angle formulas, we are able to simplify the left hand side of the equation into \cos^2(x+y+z) (try it!), which is obviously smaller than 1.

I shall conclude this post with one of my favourite MO problems:
Prove that \dfrac{x_1}{1+x_1^2}+\dfrac{x_2}{1+x_1^2+x_2^2}+\cdots+\dfrac{x^n}{1+x_1^2+x_2^2+\cdots+x_n^2}<\sqrt{n} where x_i are real numbers.
This problem can be solved beautifully
using using trigonometric substitution. Will give reward to HC students
who can find the correct substitution and solve the inequality 🙂 (Well it can be solved using other methods too but the trigo method is really nice)
Cheers,
ksj

Crux Mathematicorum

Crux Mathematicorum

CRUXv27n1.pdf

CRUXv27n1.pdf

Mathematical Excalibur

Mathematical Excalibur

In 1994, the International Mathematical Olympiad (IMO) was held
in Hong Kong. In January of that year, we began to post a math
newsletter called Mathematical Excalibur on our department webpage.
You can find all the issues up to now in the box below.
In each issue, you can find a math olympiad paper, an article
or two on math olympiad topics and a problem corner. Over the years,
we received many contributions from readers all over the world. Many
readers submitted brilliant solutions to the problem corner. We thank
each of them whole-heartedly.



Other than Math Excalibur, we also started an undergraduate math
competition open to all undergraduates of our university. For more
details, click http://www.math.ust.hk/ug/competition/ .
The winners were high-achievers who often have taken our advanced
stream courses. Each year a number of them participate in exchange
program (see the webpage http://www.math.ust.hk/ug/exchange.php ) and
go on to top graduate programs oversea (see the lower part of the
webpage http://www.math.ust.hk/ug/prospect.php ). Many of them
received big scholarships and graduated with the highest academic
honour. For our university, our medium of instruction is English
and we are very international with students from all over the world.
You can click http://www.math.ust.hk/welcome.php to know more about our
math department and click www.ust.hk to know more about our university.

Our School of Science is offering scholarship for IMO/APMO
medalists. Please see the link http://science.ust.hk/scholarship_for_olympiad_medalists.html