Functions — “s’wat” Math is about !! :-)

Reference: Nordic Mathematical Contest 1987, R. Todev:

Question:

Let f be a function, defined for natural numbers, that is strictly increasing, such that values of the function are also natural numbers and which satisfies the conditions f(2)=a>2 and f(mn)=f(m)f(n) for all natural numbers m and n. Define the smallest possible value of a.

Solution:

Since, f(n)=n^{2} is a function satisfying the conditions of the problem, the smallest possible a is at most 4. Assume that a=3. It is easy to prove by induction that f(n^{k})={f(n)}^{k} for all k \geq 1. So, taking into account that f is strictly increasing, we get

{f(3)}^{4}=f(3^{4})=f(81)>f(64)=f(2^{6})={f(2)}^{6}=3^{6}=27^{2}>25^{2}=5^{4}

as well as {f(3)}^{8}=f(3^{8})=f(6561)<f(8192)=f(2^{13})={f(2)}^{13}=3^{13}<6^{8}.

So, we arrive at 5<f(3)<6. But, this is not possible, since f(3) is an integer. So, a=4.

Cheers,

Nalin Pithwa.

RMO (Regional Mathematics Olympiad) — Plane Geometry Basics

Let us get started on plane geometry, some how or the other, now. Rather than starting with a lecture, let me start in a un-conventional format. A little challenging problem for you. Try to prove the following (Morley’s theorem):

In any arbitrary triangle, draw the trisectors of all the angles. Prove that the triangle formed by the points of intersection of adjacent trisectors (of different angles), taken in pair, is always equilateral.

Important Note: Please do not take look at any solution from the internet or any other source. You can use the basic theorems only presented in the Geometry toolkit, like as in, Problem Primer for the Olympiad, B J Venkatachala and/or in the lecture.

Ref: http://www.amazon.in/Problem-Primer-Olympiad-2Ed-Pranesachar/dp/8172862059/ref=sr_1_1?s=books&ie=UTF8&qid=1491439420&sr=1-1&keywords=problem+primer+for+the+olympiad

Have fun!

Nalin Pithwa.

 

A cute number theory question for Math Olympiads and IITJEE Mathematics

Prove that the equation

x^{3}+y^{3}+z^{3}=2 has an infinity of solutions in Z, the set of integers.

Proof:

Solution by Andrei Stefanescu

For any k \in Z we will consider the numbers x_{k}=1+6^{3k+1}, y_{k}=1-6^{3k+1} and z_{k}=-6^{2k+1}.

Note that the triplet (x_{k},y_{},z_{k}) is a solution of the equation, as

(1+6^{3k+1})^{3}+(1-6^{3k+1})^{3}+(-6^{2k+1})^{3}

= 1 + 3.6^{3k+1}+3.6^{6k+2}+6^{9k+3}+1-3.6^{6k+2}-6^{9k+3}-6^{6k+3}

= 2+ 6^{6k+3}-6^{6k+3}=2.

Since we can define this triplet for all k’s, there will be an infinity of solutions.

QED.

More later,

Nalin Pithwa

A geometry and algebra problem — RMO training

Several Olympiad problems deal with functions defined on certain sets of points. These problems are interesting in that they combine both geometrical and algebraic ideas.

Problem.

Let n>2 be an integer and f:P \rightarrow \Re a function defined on the set of points in the plane, with the property that for any regular n-gon A_{1}A_{2} \ldots A_{n}

f(A_{1})+f(A_{2})+ \ldots + f(A_{n})=0.

Prove that f is the zero function.

Proof:

Core Concept: 

In Euclidean geometry, the only motions permissible are rigid motions — translations, rotations, and reflections.

Solution:

Let A be an arbitrary point. Consider a regular n-gon AA_{1}A_{2}\ldots A_{n-1}. Let k be an integer, 0 \leq k \leq n-1. A rotation with center A of angle \frac{2\pi k }{n} sends the polygon AA_{1}A_{2}\ldots A_{n-1} to A_{k0}A_{k1} \ldots A_{k,n-1}, where A_{k0}=A and A_{ki} is the image of A_{i} for all I=1, 2, \ldots, n-1

From the condition of the statement, we have

\sum_{k=0}^{n-1} \sum_{i=0}^{n-1}f(A_{ki})=0.

Observe that in the sum the number f(A) appears n times, therefore,

nf(A)+ \sum_{k=0}^{n-1} \sum_{i=1}^{n-1}f(A_{kl})=0

On the other hand, we have

\sum_{k=0}^{n-1} \sum_{i=1}^{n-1}f(A_{ki})=\sum_{i=1}^{n-1} \sum_{k=0}^{n-1}f(A_{ki})=0

since the polygons A_{0i}A_{1i} \ldots A_{n-1,i} are all regular n-gons. From the two equalities above we deduce f(A)=0, hence, f is the zero function.

More later,

Nalin Pithwa

 

 

Russian Math Olympiad Problem (2001) — Question and Answer

(Russia 2001).

Let a and b be distinct positive integers such that ab(a+b) is divisible by a^{2}+ab+b^{2}. Prove that

|a-b|> \sqrt[3]{ab}.

Proof.

Let g=gcd(a,b) and write a=xg and b=yg with gcd(x,y)=1. Then,

\frac{ab(a+b)}{a^{2}+ab+b^{2}}=\frac{xy(x+y)g}{x^{2}+xy+y^{2}}

is an integer. N^ote that gcd(x^{2}+xy+y^{2}, x)=gcd(y^{2},x)=1. Similarly, gcd(x^{2}+xy+y^{2},y)=1.

Because gcd(x+y,y)=1, we have

gcd(x^{2}+xy+y^{2},x+y)=gcd(y^{2}, x+y)=1

Now, we apply the following lemma: Let a and b be two coprime numbers. If c is an integer such that a|c, then ab|c.

Hence, we get, x^{2}+xy+y^{2}|g implying that g \geq x^{2}+xy+y^{2}. Therefore,

|a-b|^{3}=|g(x-y)|^{3}=g^{2}|x-y|^{3}.g \geq g^{2}.1.(x^{2}+xy+y^{2})>g^{2}xy=ab.

It follows that |a-b|>\sqrt[3]{ab}. QED.

Note that the key step x^{2}+xy+y^{2} divides g can also be obtained by clever algebraic manipulations such as a^{3}=(a^{2}+ab+b^{2})a-ab(a+b).

More Olympiad problems later,

Nalin Pithwa

Pigeon hole principle — an example for RMO preparation

Problem:

A student has 6 weeks (that is, 42 days) to prepare for her examination and she has decided that during this period she will put in a total of 70 hours towards her preparation for the examination. She decides to study in full hours every day, studying at least one hour each day. We have to prove that no matter how she schedules her studying pattern, she will study for exactly 13 hours during some consecutive days.

Solution:

If a_{i} denotes the number of hours she studies on the ith day, then we are given that each a_{i} is a positive integer and the sum a_{1}+a_{2}+ \ldots + a_{42}=70. To get the required succession of days, we must find some m and j such that m \leq j and a_{m}+ \ldots + a_{j}=13. Trying out all the possibilities, is close to impossible as well as dumb. Let b_{i} denote the partial sum a_{1}+ \ldots + a_{i} which is the number of hours she studies upto the ith day. Our given data then translates into

1 \leq b_{1} < b_{2} < \ldots < b_{41} < b_{42}=70

and we have to find some i<j such that b_{i}+13=b_{j}, that is, b_{j}-b_{i}=a_{i+1}+ \ldots + a_{j}=13 (clearly, then i<j). Hence, besides the 42 numbers in

B= \{ b_{1}, b_{2}, \ldots , b_{42}\} we also look at 42 more numbers in B^{'}= \{ b_{1}+13, b_{2}+13, \ldots, b_{42}+13\} which are also 42 different numbers and the largest among them is b_{42}+13=70+13=83. Hence, the 2 \times 42 are actually among the positive integers from 1 to 83 and hence, by the pigeonhole principle, we see that two numbers in B \bigcup B^{'} must be equal. As we already saw the numbers in B are all distinct and so are the numbers in B^{'}. Hence, we must have for some i and j, b_{i}=b_{j}+13 giving the required succession of days when she studied exactly for 13 hours a day.

More on pigeonhole principle,

Later,

Till then,

Nalin Pithwa