# Miscellaneous questions: part II: tutorial practice for preRMO and RMO

Problem 1:

Let $a_{1}, a_{2}, \ldots, a_{10}$ be ten real numbers such that each is greater than 1 and less than 55. Prove that there are three among the given numbers which form the lengths of the sides of a triangle.

Problem 2:

In a collection of 1234 persons, any two persons are mutual friends or enemies. Each person has at most 3 enemies. Prove that it is possible to divide this collection into two parts such that each person has at most 1 enemy in his subcollection.

Problem 3:

A barrel contains 2n balls numbered 1 to 2n. Choose three balls at random, one after the other, and with the balls replaced after each draw. What is the probability that the three element sequence obtained has the properties that the smallest element is odd and that only the smallest element, if any, is repeated?

That’s all, folks !!

You will need to churn a lot…!! In other words, learn to brood now…learn to think for a long time on a single hard problem …

Regards,
Nalin Pithwa

# Miscellaneous questions: Part I: tutorial practice for preRMO and RMO

Problem 1:

The sixty four squares of a chess board are filled with positive integers one on each in such a way that each integer is the average of the of the integers on the neighbouring squares. (Two squares are neighbours if they share a common edge or vertex. Thus, a square can have 8,5 or 3 neighbours depending on its position). Show that all sixty four entries are in fact equal.

Problem 2:

Let T be the set of all triples (a,b,c) of integers such that $1 \leq a < b < c \leq 6$. For each triple (a,b,c) in T, take the product abc. Add all these products corresponding to all triples in I. Prove that the sum is divisible by 7.

Problem 3:

In a class of 25 students, there are 17 cyclists, 13 swimmers, and 8 weight lifters and no one in all the three. In a certain mathematics examination, 6 students got grades D or E. If the cyclists, swimmers and weight lifters all got grade B or C, determine the number of students who got grade A. Also, find the number of cyclists, who are swimmers.

Problem 4:

Five men A, B, C, D, E are wearing caps of black or white colour without each knowing the colour of his cap. It is known that a man wearing a black cap always speaks the truth while a man wearing a white cap always lies. If they make the following statements, find the colour of the cap worn by each of them:

A: I see three black and one white cap.
B: I see four white caps.
C: I see one black and three white caps.
D: I see four black caps.

Problem 5:

Let f be a bijective (one-one and onto) function from the set $A=\{ 1,2,3,\ldots,n\}$ to itself. Show that there is a positive integer $M>1$ such that $f^{M}(i)=f(i)$ for each $i \in A$. Note that $f^{M}$ denotes the composite function $f \circ f \circ f \ldots \circ f$ repeated M times.

Problem 6:

Show that there exists a convex hexagon in the plane such that:
a) all its interior angles are equal
b) its sides are 1,2,3,4,5,6 in some order.

Problem 7:

There are ten objects with total weights 20, each of the weights being a positive integer. Given that none of the weights exceed 10, prove that the ten objects can be divided into two groups that balance each other when placed on the pans of a balance.

Problem 8:

In each of the eight corners of a cube, write +1 or -1 arbitrarily. Then, on each of the six faces of the cube write the product of the numbers written at the four corners of that face. Add all the fourteen numbers so writtein down. Is it possible to arrange the numbers +1 and -1 at the corners initially so that this final sum is zero?

Problem 9:

Given the seven element set $A = \{ a,b,c,d,e,f,g\}$ find a collection T of 3-element subsets of A such that each pair of elements from A occurs exactly in one of the subsets of T.

Try these !!

Regards,
Nalin Pithwa

# A fifth degree equation in two variables: a clever solution

Question:

Verify the identity: $(2xy+(x^{2}-2y^{2}))^{5}+(2xy-(x^{2}-2y^{2}))^{5}=(2xy+(x^{2}+2y^{2})i)^{5}+(2xy-(x^{2}+2y^{2})i)^{5}$

let us observe first that each of the fifth degree expression is just a quadratic in two variables x and y. Let us say the above identity to be verified is:

$P_{1}+P_{2}=P_{3}+P_{4}$

Method I:

Use binomial expansion. It is a very longish tedious method.

Method II:

Factorize each of the quadratic expressions $P_{1}, P_{2}, P_{3}, P_{4}$ using quadratic formula method (what is known in India as Sridhar Acharya’s method):

Now fill in the above details.

You will conclude very happily that :

The above identity is transformed to :

$P_{1}=(x+y+\sqrt{3}y)^{5}(x+y-\sqrt{3}y)^{5}$

$P_{2}=(-1)^{5}(x-y-\sqrt{3}y)^{5}(x-y+\sqrt{3}y)^{5}$

$P_{3}=(i^{2}(x-y-\sqrt{3}y)(x-y+\sqrt{3}y))^{5}$

$P_{4}=((-i^{2})(x+y+\sqrt{3}y)(x-y-\sqrt{3}y))^{5}$

You will find that $P_{1}=P_{4}$ and $P_{2}=P_{4}$

Hence, it is verified that the given identity $P_{1}+P_{2}=P_{3}+P_{4}$. QED.

Regards,
Nalin Pithwa.

# Set Theory, Relations, Functions: preliminaries: part 10: more tutorial problems for practice

Problem 1:

Prove that a function f is 1-1 iff $f^{-1}(f(A))=A$ for all $A \subset X$. Given that $f: X \longrightarrow Y$.

Problem 2:

Prove that a function if is onto iff $f(f^{-1}(C))=C$ for all $C \subset Y$. Given that $f: X \longrightarrow Y$.

Problem 3:

(a) How many functions are there from a non-empty set S into $\phi$\?

(b) How many functions are there from $\phi$ into an arbitrary set $S$?

(c) Show that the notation $\{ X_{i} \}_{i \in I}$ implicitly involves the notion of a function.

Problem 4:

Let $f: X \longrightarrow Y$ be a function, let $A \subset X$, $B \subset X$, $C \subset Y$ and $D \subset Y$. Prove that

i) $f(A \bigcap B) \subset f(A) \bigcap f(B)$

ii) $f^{-1}(f(A)) \supset A$

iii) $f(f^{-1}(C)) \subset C$

Problem 5:

Let I be a non-empty set and for each $i \in I$, let $X_{i}$ be a set. Prove that

(a) for any set B, we have $B \bigcap \bigcup_{i \in I}X_{i}=\bigcup_{i \in I}(B \bigcap X_{i})$

(b) if each $X_{i}$ is a subset of a given set S, then $(\bigcup_{i \in I}X_{i})^{'}=\bigcap_{i \in I}(X_{i})^{'}$ where the prime indicates complement.

Problem 6:

Let A, B, C be subsets of a set S. Prove the following statements:

(i) $A- (B-C)=(A-B)\bigcup(A \bigcap B \bigcap C)$

(ii) $(A-B) \times C=(A \times C)-(B \times C)$

🙂 🙂 🙂

Nalin Pithwa

# Set Theory, Relations, Functions: Preliminaries: Part IX: (tutorial problems)

Reference: Introductory Real Analysis, Kolmogorov and Fomin, Dover Publications.

Problem 1:

Prove that if $A \bigcup B=A$ and $A \bigcap B=A$, then $A=B$.

Problem 2:

Show that in general $(A-B)\bigcup B \neq A$.

Problem 3:

Let $A = \{ 2,4, \ldots, 2n, \ldots\}$ and $B= \{ 3,6,\ldots, 3n, \ldots\}$. Find $A \bigcap B$ and $A - B$.

Problem 4:

Prove that (a) $(A-B)\bigcap (C)=(A \bigcap C)-(B \bigcap C)$

Prove that (b) $A \Delta B = (A \bigcup B)-(A \bigcap B)$

Problem 5:

Prove that $\bigcup_{a}A_{\alpha}-\bigcup_{a}B_{\alpha}=\bigcup_{\alpha}(A_{\alpha}-B_{\alpha})$

Problem 6:

Let $A_{n}$ be the set of all positive integers divisible by $n$. Find the sets (i) $\bigcup_{n=2}^{\infty}A_{n}$ (ii) $\bigcap_{n=2}^{\infty}A_{n}$.

Problem 7:

Find (i) $\bigcup_{n=1}^{\infty}[n+\frac{1}{n}, n - \frac{1}{n}]$ (ii) $\bigcap_{n=1}^{\infty}(a-\frac{1}{n},b+\frac{1}{n})$

Problem 8:

Let $A_{\alpha}$ be the set of points lying on the curve $y=\frac{1}{x^{\alpha}}$ where $(0. What is $\bigcap_{\alpha \geq 1}A_{\alpha}$?

Problem 9:

Let $y=f(x) = $ for all real x, where $$ is the fractional part of x. Prove that every closed interval of length 1 has the same image under f. What is the image? Is f one-to-one? What is the pre-image of the interval $\frac{1}{4} \leq y \leq \frac{3}{4}$? Partition the real line into classes of points with the same image.

Problem 10:

Given a set M, let $\mathscr{R}$ be the set of all ordered pairs on the form $(a,a)$ with $a \in M$, and let $aRb$ if and only if $(a,b) \in \mathscr{R}$. Interpret the relation R.

Problem 11:

Give an example of a binary relation which is:

• Reflexive and symmetric, but not transitive.
• Reflexive, but neither symmetric nor transitive.
• Symmetric, but neither reflexive nor transitive.
• Transitive, but neither reflexive nor symmetric.

We will continue later, 🙂 🙂 🙂

PS: The above problem set, in my opinion, will be very useful to candidates appearing for the Chennai Mathematical Institute Entrance Exam also.

Nalin Pithwa

# Set Theory, Relations, Functions: Preliminaries: part VIIIA

(We continue from part VII of the same blog article series with same reference text).

Theorem 4:

A set M can be partitioned into classes by a relation R (acting as a criterion for assigning two elements to the same class) if and only R is an equivalence relation on M.

Proof of Theorem 4:

Every partition of M determines a binary relation on M, where $aRb$ means that “a belongs to the same class as b.” It is then obvious that R must be reflexive, symmetric and transitive, that is, R is an equivalence relation on M.

Conversely, let R be an equivalence relation on M, and let $K_{a}$ be the set of all elements $x \in M$ such that $xRa$ (clearly, $a \in K_{a}$, since R is reflexive). Then, two classes $K_{a}$ and $K_{b}$ are either identical or disjoint. In fact, suppose that an element c belongs to both $K_{a}$ and $K_{b}$, so that $cRa$ and $cRb$. But by symmetry of R, being an equivalence relation, we can infer that $aRc$ also and, further by transitivity, we say that $aRb$. If now, $x \in K_{a}$ then we have $xRa$ and hence, $xRb$ (since we already have $aRb$ and using transitivity).

Similarly, we can prove that $x \in K_{b}$ implies that $x \in K_{a}$.

Therefore, $K_{a}=K_{b}$ if $K_{a}$ and $K_{b}$ have an element in common. Therefore, the distinct sets $K_{a}$ form a partition of M into classes.

QED.

Remark:

Because of theorem 4, one often talks about the decomposition of a set M into equivalence classes.

There is an intimate connection between mappings and partitions into classes, as illustrated by the following examples:

Example 1:

Let f be a mapping of a set A into a set B and partition A into sets, each consisting of all elements with the same image $b=f(a) \in B$. This gives a partition of A into classes. For example, suppose f projects the xy-plane onto the x-axis by mapping the point $(x,y)$ into the point $(x,0)$. Then, the preimages of the points of the x-axis are vertical lines, and the representation of the plane as the union of these lines is the decomposition into classes corresponding to f.

Example 2:

Given any partition of a set A into classes, let B be the set of these classes and associate each element $a \in A$ with the class (that is, element of B) to which it belongs. This gives a mapping of A into B. For example, suppose we partition three-dimensional space into classes by assigning to the same class all points which are equidistant from the origin of coordinates. Then, every class is a sphere of a certain radius. The set of all these classes can be identified with the set of points on the half-line $[0, \infty)$ each point corresponding to a possible value of the radius. In this sense, the decomposition of 3-dimensional space into concentric spheres corresponds to the mapping of space into the half-line $[0,\infty)$.

Example 3:

Suppose that we assign all real numbers with the same fractional part to the same class. Then, the mapping corresponding to this partition has the effect of “winding” the real line onto a circle of unit circumference. (Note: The largest integer $\leq x$ is called the integral part of x, denoted by [x], and the quantity $x -[x]$ is called the fractional part of x).

In the next blog article, let us consider a tutorial problem set based on last two blogs of this series.

🙂 🙂 🙂

Nalin Pithwa

# A quadratic and trigonometry combo question: RMO and IITJEE maths coaching

Question:

Given that $\tan {A}$ and $\tan {B}$ are the roots of the quadratic equation $x^{2}+px+q=0$, find the value of

$\sin^{2}{(A+B)}+ p \sin{(A+B)}\cos{(A+B)} + q\cos^{2}{(A+B)}$

Solution:

Let $\alpha=\tan{A}$ and $\beta=\tan{B}$ be the two roots of the given quadratic equation: $x^{2}+px+q=0$

By Viete’s relations between roots and coefficients:

$\alpha+\beta=\tan{A}+\tan{B}=-p$ and $\alpha \beta = \tan{A}\tan{B}=q$ but we also know that $\tan{(A+B)}=\frac{\tan{A}+\tan{B}}{1-\tan{A}\tan{B}}=\frac{-p}{1-q}=\frac{p}{q-1}$

Now, let us call $E=\sin^{2}{(A+B)}+p\sin{(A+B)\cos{(A+B)}}+\cos^{2}{(A+B)}$ which in turn is same as

$\cos^{2}{(A+B)}(\tan^{2}{(A+B)}+p\tan{(A+B)}+q)$

We have already determined $\tan{(A+B)}$ in terms of p and q above.

Now, again note that $\sin^{2}{\theta}+\cos^{2}{\theta}=1$ which in turn gives us that $\tan^{2}{\theta}+1=\sec^{2}{\theta}$ so we get:

$\sec^{2}{(A+B)}=1+\tan^{2}{(A+B)}=1+\frac{p^{2}}{(q-1)^{2}}=\frac{p^{2}+(q-1)^{2}}{(q-1)^{2}}$ so that

$\cos^{2}{(A+B)}=\frac{1}{\sec^{2}{(A+B)}}=\frac{(q-1)^{2}}{p^{2}+(q-1)^{2}}$

Hence, the given expression E becomes:

$(\frac{(q-1)^{2}}{p^{2}+(q-1)^{2}})(\frac{p^{2}}{(q-1)^{2}}+\frac{p^{2}}{q-1}+q)$, which is the desired solution.

🙂 🙂 🙂

Nalin Pithwa.