RMO 2017 Goa and Maharashtra

Let's solve the Regional Mathematics Olympiad Problem, RMO 2017 from Goa and Maharashtra. Try the problems and check your solutions here.

(\ 1).((\ 16) marks)Consider a chessboard of size (\ 8) units(\ \times8) units (i.e., each small square on the board has a side length of (\ 1) unit).Let (\ S) be the set of all the (\ 81) vertices of all the squares on the board. What is the number of line segments whose vertices are in (\ S), and whose length is a positive integer. (The segments need not be parallel to the sides of the board.)

(\ 2).((\ 16) marks)For any positive integer (\ n), let (\ d(n)) denotes the number of positive divisors of (\ n); and let (\ \phi(n)) denotes the number of elements from the set (\ {1,2,...,n}) that are co-prime to (\ n).(For example (\ d(12)=6 ) and (\ \phi(12)=4).)

Find the smallest positive integer (\ n) such that (\ d(\phi(n))=2017).

(\ 3).((\ 16) marks)Let (\ P(x)) and (\ Q(x)) be polynomials of degree (\ 6) and degree (\ 3) respectively,such that:
(\ P(x)>Q(x)^2+Q(x)+x^2-6), for all (\ x\in\mathbb{R}).

If all the roots of (\ P(x)) are real numbers, then prove that there exist two roots of (\ P(x)), say (\ \alpha,\beta), such that (\ |\alpha-\beta|<1).

(\ 4).((\ 16) marks)Let (\ l_1,l_2,l_3,\dots,l_{40}) be forty parallel lines.As shown in the diagram, let m be another line that intersects the line (\ l_1) to (\ l_{40}) in the points (\ A_1,A_2.A_3,\dots,A_{40}) respectively.Similarly let n be another line that intersects the lines (\ l_1) to (\ l_{40}) in the points (\ B_1,B_2,B_3,\dots,B_{40}) respectively.

Given that (\ A_1B_1=1), (\ A_{40}B_{40}=14), and the areas of the (\ 39) trapeziums (\ A_1B_1B_2A_2),(\ A_2B_2B_3A_3,\dots),(\ A_{39}B_{39}B_{40}A_{40}) are all equal; then count the number of segments (\ A_iB_i) whose length is a positive integer; where (\ i\in{1,2,\dots,40}).

(\ 5).((\ 18) marks)If (\ a,b,c,d\in\mathbb{R}) such that (\ a>b>c>d>0) and (\ a+d=b+c);

then prove that :

$$\frac{(a+b)-(c+d)}{\sqrt{2}}>\sqrt{a^2+b^2}-\sqrt{c^2+d^2}$$

(\ 6).((\ 18) marks)Let (\ \triangle{ABC}) be acute-angled; and let (\ \Gamma) be its circumcircle.Let (\ D) be a point on minor arc (\ BC) of (\ \Gamma).Let (\ E) and (\ F) be points on line (\ AD) and (\ AC) respectively, such that (\ BE\perp AD) and (\ DF\perp AC).Prove that (\ EF\parallel BC) if and only if (\ D) is the midpoint of (\ BC).

Differentiability at origin | I.S.I. B.Stat, B.Math Subjective 2017

Try this problem from ISI B.Stat, B.Math Subjective Entrance Exam, 2017 Problem no. 3 based on Differentiability at origin.

Problem: Differentiability at origin

Suppose $ f : \mathbb{R} \to \mathbb{R} $ is a function given by $$
f(x) = \left\{\def\arraystretch{1.2}%
\begin{array}{@{}c@{\quad}l@{}}
1 & \text{if x=1}\\ e^{(x^{10} -1)} + (x-1)^2 \sin \left (\frac {1}{x-1} \right ) & \text{if} x \neq 1\ \end{array}\right.
$$

Find f'(1)
Evaluate $ \displaystyle{\lim_{u \to \infty } \left [ 100 u - u \sum_{k=1}^{100} f \left (1 + \frac {k}{u} \right ) \right ] }$

Discussion:

a)First of all we need to check whether $f'(1)$ exists or not.

We will proceed with the first principle.

Let us check the Right hand derivative(RHD) and Left hand derivative(LHD) of $f$ at $x=1$.

RHD at $x=1$ is

$\lim_{h\to0}\frac{f(1+h)-f(1)}{h}\\=\lim_{h\to0}\frac{e^{((1+h)^{10}-1)}+h^2\sin( \frac{1}{h})-1}{h}\\=\lim_{h\to0}\frac{e^{((1+h)^{10}-1)}-1}{h}+\lim_{h\to0}h\sin( \frac{1}{h})\\=\lim_{h\to0}\frac{e^{((1+h)^{10}-1)}-1}{(1+h)^{10}-1}\frac{(1+h)^{10}-1}{h}+0=10$

LHD at $x=1$ is

$\lim_{h\to0}\frac{f(1-h)-f(1)}{-h}\\=\lim_{h\to0}\frac{e^{((1-h)^{10}-1)}+h^2\sin( \frac{1}{-h})-1}{-h}\\=\lim_{h\to0}\frac{e^{((1-h)^{10}-1)}-1}{-h}+\lim_{h\to0}(-h)\sin( \frac{1}{-h})\\=\lim_{h\to0}\frac{e^{((1-h)^{10}-1)}-1}{(1-h)^{10}-1}\frac{(1-h)^{10}-1}{-h}+0=10$

Thus,LHD=RHD.

Hence $f'(1)$ exists and it is equal to $10$.

(b)

$ \displaystyle{\lim_{u \to \infty } \left [ 100 u - u \sum_{k=1}^{100} f \left (1 + \frac {k}{u} \right ) \right ] }$

As u becomes infinitely large k/u becomes arbitrarily small for finite value of k (clearly k is finite as we are interested in k=1 to 100).

Hence $ f \left ( 1 + \frac {k}{u} \right )$ is nothing but f of (1 plus an infinitesimal positive quantity). This tells us $ f \left ( 1 + \frac {k}{u} \right )$ is almost waiting to become the derivative of f at x=1. And we already know that such a derivative exists from part (a).

With this motivation, divide and multiply by $ \frac{k}{u} $.

$ \displaystyle{\lim_{u \to \infty } \left [ 100 u - u \sum_{k=1}^{100} f \left (1 + \frac {k}{u} \right ) \right ] \\ =\lim_ {u \to \infty} \left [ 100 u -\sum_{k=1}^{100} k \frac{f\left (1+\frac{k}{u} \right ) } {\frac{k}{u}} \right ]\\=\lim_ {u \to \infty} \left [ \sum_{k=1}^{100} k \frac{1-f\left (1+\frac{k}{u} \right ) } {\frac{k}{u}} \right ]\\ =\sum_{k=1}^{100} k \lim_ {u \to \infty}\frac{f(1)-f\left (1+\frac{k}{u} \right ) } {\frac{k}{u}} \\ =\sum_{k=1}^{100} k \times(- f'(1)) \\ = -10\times \left(\frac{100\times101}{2} \right )\\ =-50500 }$

Back to the question paper

Some Useful Links:

RMO 16-OCT-2016-1 Solution

wb-rmo-2016-1-google-docs

RMO 2016 Karnataka, Assam, Andhra Pradesh (except Telangana), West Bengal Region

Problems

Problem 1

Let $ a, b, c $ be positive real numbers such that $$ \frac{a}{1+a} + \frac{b}{1+b} + \frac{c}{1+c} = 1 $$

Prove that $ abc \leq \frac{1}{8} $

RMO 2016 North Bihar Region

RMO 2016 Maharashtra Region

RMO 2016 Delhi Region

WB PRE-RMO 2016 PAPER AND ANSWERS

prmo2016

CLICK ON THE ABOVE LINK to get the WB PRE-RMO 2016 PAPER AND ANSWERS.

Solving equations | Tomato objective 20

This is a beautiful problem based on Solving Equations from Test of Mathematics Subjective Problem no. 20.

Problem : Solving equations

If $\ a,b,c,d$ satisfy the equations

$$a+7b+3c+5d=0,$$

$$8a+7b+6c+2d=-16,$$

$$2a+6b+4c+8d=16,$$

$$5a+3b+7c+d=-16,$$

then $\ (a+d)(b+c)$ equals

$\ (A)16 \quad (B)-16\quad (C)0 \quad$ (D)none of the foregoing numbers

Solution:

$$a+7b+3c+5d=0\dots(1),$$

$$8a+7b+6c+2d=-16\dots(2),$$

$$2a+6b+4c+8d=16\dots(3),$$

$$5a+3b+7c+d=-16\dots(4),$$

$\ (1)-(3)$, and $\ (2)-(4)$, we get

$$-a+b-c-3d=-16\dots(5),$$

$$3a+b-c+d=0\dots(6),$$

$\ (6)-(5)$, we get

$$a+d=4\dots(7),$$

$\ (2)+(3)$,we get

$$a+b+c+d=0\dots(8),$$

$\ (8)-(7)$,we get

$$b+c=-4\dots(9),$$

$\ (7)\times(9)$,we get

Therefore,$$(a+d)(b+c)=-16$$

Thus,$\ (B)$ is the correct option.

Some Useful Links:

A Cauchy Schwarz Problem

Cauchy Schwarz Problem: Let $P(x)$ be a polynomial with non-negative coefficients.Prove that if $P(x)\cdot P(\frac{1}{x})\ge1$ for $x=1$ ,then the same inequality holds for each $\mathbb{R^+} x$ .

Discussion: Cauchy Schwarz's Inequality: Suppose for real numbers (\ a_{i},b_{i}), where (\ i\in{1,2,\dots,n}) we can say that $${\sum_{i=1}^{n}a_{i}^2}{\sum_{i=1}^{n}b_{i}^2}=\sum_{i=1}^{n}{a_{i}b_{i}}^2$$.

Titu's Lemma: Let (\ a_{i},b_{i}\in{\mathbb{R}}) and let (\ a_{i},b_{i}>0) for (\ i\in{1,2,\dots,n})

$$\sum_{i=1}^{n}\frac{a_{i}^2}{b_{i}}\ge\frac{{\sum_{i=1}^{n}a_{i}}^2}{\sum_{i=1}^{n}b_{i}}$$

Proof of Cauchy Schwarz's Inequality: We can write (\sum_{i=1}{n}a_{i}^2=\sum_{i=1}^{n}\frac{a_{i}^2 b_{i}^2}{b_{i}^2}\ge\frac{{\sum_{i=1}^{n}a_{i}b_{i}}^2}{\sum_{i=1}^{n}b_{i}^2}) (using Titu's lemma)

(=>{\sum_{i=1}^{n}a_{i}^2}{\sum_{i=1}^{n}b_{i}^2}\ge {\sum_{i=1}^{n}{a_{i}b_{i}}^2)

Solution: Let $P(x)=a_{n}x^n+a_{n-1}x^{n-1}+\dots+a_{1}x+a_{0}$ ,then $P(\frac{1}{x})=a_{n}\frac{1}{x^n}+a_{n-1}\frac{1}{x^{n-1}}+\dots+a_{1}\frac{1}{x}+a_{0}$

now $P(x)\cdot P(\frac{1}{x})=\{a_{n}x^n+a_{n-1}x^{n-1}+\dots+a_{1}x+a_{0}\}\{a_{n}\frac{1}{x^n}+a_{n-1}\frac{1}{x^{n-1}}+\dots+a_{1}\frac{1}{x}+a_{0}\}=\{\{\sqrt{a_{n}x^n}\}^2+\{\sqrt{a_{n-1}x^{n-1}}\}^2+\dots+\{\sqrt{a_{1}x}\}^2+\{\sqrt{a_{0}}\}^2\} \{\{\sqrt{a_{n}\frac{1}{x^n}}\}^2+\{\sqrt{a_{n-1}\frac{1}{x^{n-1}}}\}^2+\dots+\{\sqrt{a_{1}\frac{1}{x}}\}^2+\{\sqrt{a_{0}}\}^2\} \ge\{a_{n}+a_{n-1}+\dots+a_{1}+a_{0}\}^2$
(Cauchy Schwarz's Inequality)

now it is given that $P(x)\cdot P(\frac{1}{x})\ge1$ for $x=1$ ,so $P(1)^2\ge 1=>\{a_{n}+a_{n-1}+\dots+a_{1}+a_{0}\}^2\ge1$

therefore, $P(x)\cdot P(\frac{1}{x})\ge1$ $\forall x\in{\mathbb{R^+}}$