INMO 2008 problem 1

Let $ABC$ be a triangle, $I$ its in-centre; $ A_1, B_1, C_1 $ be the reﬂections of $I$ in BC, CA, AB respectively. Suppose the circumcircle of triangle $ A_1 B_1 C_1 $ passes through A. Prove that $ B_1, C_1, I, I_1 $ are concyclic, where $ I_1 $ is the incentre of triangle $ A_1 B_1 C_1 $.

Hint 1: $I$ is indeed the circumcenter of $ A A_1 B_1 C_1 $ with circum radius = $2r$. $ B_1 C_1 $ is the radical axis of the two circles concerned hence the other center has to lie of $IA$ (since $IA$ is perpendicular to radical axis and I is one of the centers hence $IA$ is the line joining the centers).

Hint 2: We prove that A is the center of the circle $ B_1, C_1, I, I_1 $ . Using cosine we show that $ \Delta AIH $ is a $30-60-90$ triangle as $IH = r$ and $IA = 2r$. This implies $ \angle B_1 I C_1 $ is $ 120^o $ . Hence it is sufficient to show $ \angle B_1 I_1 C_1 $ is $ 120^o $ . A simple angle chasing in triangle $ A_1 I_1 C_1 $ and $ B_1 I_1 C_1 $ completes the proof (we observe that $ \angle A_1 C_1 B_1 = \frac {A+B} {2} $ and similarly with the other angles).

INMO 2008 Question Paper | Math Olympiad Problems

This post contains the problems from Indian National Mathematics Olympiad, INMO 2008 Question Paper. Do try to find their solutions.

Let $ABC$ be a triangle, $I$ its in-centre; $ A_1, B_1, C_1 $ be the reﬂections of $I$ in BC, CA, AB respectively. Suppose the circum-circle of triangle $ A_1 B_1 C_1 $ passes through A. Prove that $ B_1, C_1, I, I_1 $ are concyclic, where $ I_1 $ is the in-centre of triangle $ A_1 B_1 C_1 $.
Find all triples $ (p, x, y) $ such that $ p^x = y^4 + 4 $, where $p $ is a prime and $ x, y $ are natural numbers.
Let $A$ be a set of real numbers such that $A$ has at least four elements. Suppose $A$ has the property that $a^2+b c$ is a rational number for all distinct numbers $a, b, c$ in $A$. Prove that there exists a positive integer $M$ such that $a \sqrt{M}$ is a rational number for every $a$ in $A$.
All the points with integer coordinates in the xy-plane are coloured using three colours, red, blue and green, each colour being used at least once. It is known that the point (0, 0) is coloured red and the point (0, 1) is coloured blue. Prove that there exist three points with integer coordinates of distinct colours which form the vertices of a right-angled triangle.
Let $A B C$ be a triangle; $\Gamma_A, \Gamma_B, \Gamma_C$ be three equal, disjoint circles inside $A B C$ such that $\Gamma_A$ touches $A B$ and $A C ; \Gamma_B$ touches $A B$ and $B C$; and $\Gamma_C$ touches $B C$ and $C A$. Let $\Gamma$ be a circle touching circles $\Gamma_A, \Gamma_B, \Gamma_C$ externally. Prove that the line joining the circum-centre $O$ and the in-centre $I$ of triangle $A B C$ passes through the centre of $\Gamma$.
Let $ P(x) $ be a given polynomial with integer coefficients. Prove that there exist two polynomials $ Q(x) $ and $ R(x) $, again with integer coefficients, such that
1. $ P(x)Q(x) $ is a polynomial in $ x^2 $; and
2. $ P(x)R(x) $is a polynomial in $ x^3 $.

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INMO 2020 Problems, Solutions and Hints

INMO 2018 Problem 6 Part 1 – Video

Inequality of square root function

This post contains a problem from TIFR 2013 Math paper D based on Inequality of square root function.

The inequality $ \sqrt {n+1} - \sqrt n < \frac {1}{\sqrt n } $ is false for all in n such that $ 101 \le n \le 2000 $

False

Discussion:

$ \sqrt {n+1} - \sqrt n < \frac {1}{\sqrt n } $

By cross multiplying we have $ \sqrt {(n+1)n} - n < 1 $. That is $ \sqrt {n(n+1)} < (n+1) $ or $ n(n+1) < (n+1) ^2 $ or $n < n+1$

This is true for all $n$.

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Automorphism of the Additive Group of Rationals

Any automorphism of the group Q under addition is of the form x → qx for some q ∈ Q.

True

Discussion: Suppose f is an automorphism of the group Q. Let f(1) = m (of course 'm' will be different for different automorphisms). Now $f(x+y) = f(x) + f(y)$ implies $f(x) = mx$ where m is a constant and x belongs to set of integers (Cauchy's functional equation).

Now suppose x is rational. Then x = p/q where p and q are integers. Hence $f(p) = mp$. But $p = qx$ hence $f(p) = f(qx) = f(x+x+ ... + x) = qf(x)$

Therefore, $mp = qf(x)$ implies $ m \times {\frac{p}{q} }= f(x) \implies f(x) = mx $ where $m = f(1)$

Test of Mathematics Solution Subjective 35 - Divisibility by 16

Test of Mathematics Solution Subjective 35 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.

Also see: Cheenta I.S.I. & C.M.I. Entrance Course

Problem

(a) Prove that, for any odd integer n, $ n^4 $ when divided by $16$ always leaves remainder $1$.

(b) Hence or otherwise show that we cannot find integers $n_1 , n_2 , ... , n_8 $ such that $n_1^4 + n_2^4 + ... + n_8^4 = 1993 $.

Solution

For part (a) we consider $n = 2k +1$ and expand it's fourth power binomially to get $ (2k+1)^4 = {{4} \choose {0}} (2k)^4 +{{4} \choose {1}} (2k)^3 + {{4} \choose {2}} (2k)^2 + {{4} \choose {3}} (2k)^1 + {{4} \choose {4}} (2k)^0 = 16k^4 + 32k^3 + 24k^2 + 8k + 1 $

Now $24k^2 + 8k = 8k(3k+1) $ ; if $k$ is even then $8k$ is divisible by $16$ and if $k$ is odd $3k+1$ is even and product of $8k$ and $3k+1$ is divisible by $16$. Since $16k^4 + 32 k^3 $ is already divisible by $16$ we conclude $ (2k+1)^4 $ when divided by $16$ gives $1$ as remainder.

For part (b) we note that $1993$ when divided by $16$, produces $9$ as the remainder. Each of the eight of fourth powers when divided by $16$ produces either $0$ (when $n_i $ is even) or $1$ (when $n_i $ is odd using part (a)) as remainder. Thus they can add up to at most $8$ (modulo $16$) hence can never be equal to $9$ (which $1993$ is modulo $16$).

Singapore Math Olympiad (Senior) 2013

Problem 1 . A shop sells two kind of products A and B. One day a salesman sold both A and B at the same price, $2100$ to a customer. Suppose A makes a profit of 20% and B makes a loss of 20%. Then the deal
(A) make a profit of $70$; (B) make a loss of $70$;(C) make a profit of $175$; (D) make a loss of $70$.

Problem 2 . Let f and g be functions such that for all real numbers x and y $g(f(x+y))$ = $f(x)$ + $(x+y)g(y)$...

Find the value of $g(0)$ + $g(1)$ + ... + $g(2013)$

Problem 3 . Each chocolate costs 1 dollar, each licorice stick costs 50 cents and each lolly costs 40 cents. How many different combinations of these three items cost a total of 10 dollars.

Problem 4 . Let $A= {1, 2, 3, 4, 5, 6}$. Find the number of distinct functions f: A → A such that $f(f(f(n))) = n$ for all $n ∈ A$.

Problem 5 . Find the number of triangles whose sides are formed by the sides of the diagonals of a regular heptagon (7 sided polygon). (Note: the vertices of the triangle need not be vertices of the heptagon).

Problem 6 . Six seats are arranged in a circular table. Each seat is to be painted in red, blue or green such that any two adjacent seats have different colors. How many ways are there to paint the seats?

Problem 7 . ΔABC is an equilateral triangle with side length 30. Fold the triangle so that A touches a point X on BC. If BX = 6, find the value of k, where √k is the length of the crease obtained from folding.

Problem 8 . As shown in the figure below, circles C₁ and C₂ of radius 360 are tangent to each other, and both tangent to the straight line l. If the circle C₃ is tangent to C₁, C₂ and l, and circle C₄ is tangent to C₁, C₃ and l, find the radius of C₄.

Problem 9 . Set {x} = x - ⌊x⌋, where ⌊x⌋ denotes the largest integer less than or equal to x. Find the number of real solutions to the equation {x} + {x²} = 1 , |x| ≤ 10.

Problem 10 . Find the value of $ ( \frac {3 + \sqrt 17 }{2} )^6 $.

Problem 11 . A regular dodecagon (12 sided polygon) is inscribed in a circle of radius 10. Find it's area.

Problem 12 . to be continued...

Synthesis 2013 (Reunion of Cheenta) ... revisited

What motivates research in Non-Linear Partial Differential Equation? Swarnendu Sil, presently a Ph.D. student in Ecole polytechnique de federale de lausannee (one of the leading universities of the world located in Switzerland), delivered a talk (through video conference) on this topic this Sunday in the reunion of Cheenta.
The seminar began with an analysis of linear algebra (which deals with finite dimensional vector space). Sil explained why the mathematical machinery of linear algebra fails in the case of infinite dimensional vector space. Functional Analysis was developed as a discipline to address this issue and by 1970 linear partial differential equations could be understood in totality using it.
Sil mentioned that partial differential equations come up in various real life scenario. A large portion of problems are motivated by physics. The problem of non-linear partial differential equation (where partial derivatives are in multiplied form) is much more daunting than it's linear counterpart. Most mathematicians today agree that no general theory is possible to 'crack' non-linear partial differential equations.
The seminar concluded with questions from the audience.

Western Blot is one of the most important techniques for protein detection. Raikamal Paul, a master of science in microbiology from Vellore Institute of Technology, participated in extensive lab work at Indian Institute of Chemical Biology this year. She delivered a seminar on this technique on Sunday which took us to the world of practical research work in genetics.

The seminar was supplemented by presentation and board work.

The Character of Physical Law are a series of seven lectures by physicist Richard Feynman concerning the nature of the laws of physics. The talks were delivered by Feynman in 1964 at Cornell University, as part of the Messenger Lectures series. Their text was published by the BBC in 1965 in a book by the same name.

The lectures covered the following topics:

The law of gravitation, an example of physical law
The relation of mathematics and physics
The great conservation principles
Symmetry in physical law
The distinction of past and future
Probability and uncertainty - the quantum mechanical view of nature
Seeking new laws

At Cheenta Reunion, we watched the first lecture (as recorded by BBC). It discussed "The law of gravitation, an example of physical law".

Anubhav Chatterjee presented a Grand Quiz Show, which included Puzzle Round, Audio/Video Round, Straight Questions and Rapid Fire Round. Subhajit and Akash won the tournament with 187 points. The puzzles were given to the teams at the very beginning of the day (at 10 AM). The rest of the rounds happened in the evening.

There were some short 'talks'. Somrik spoke on Inner Workings of L'Hopital's rule. Subhajit discussed his experience in I.S.I. B.Math interview.

Last but not the least, there was good food. Beginning with cold drinks and potato chips, we ended up with Biriyani, Chicken Tikka Masala, Chicken Tanduri Masala, Lachha Paratha, Rumali Roti, Paneer, Mutton Kasa, Fried Rice. A local restaurant, Chhota Elaichi, served us lunch.

The Reunion concluded at 7 PM. We hope to see you at the reunion next summer (and coming winter).

ISI 2013 B.Math and B.Stat Subjective Solutions

1. For how many values of N (positive integer) N(N-101) is a square of a positive integer?

Solution:
(We will not consider the cases where N = 0 or N = 101)

$N(N-101) = m^2$

=> $N^2 - 101N - m^2 = 0$

Roots of this quadratic in N is
=> $\frac{101 \pm\ sqrt { 101^2 + 4m^2}}{2}$

The discriminant must be square of an odd number in order to have integer values for N.

Thus $101^2 + 4m^2 = (2k + 1)^2$
=> $101^2 = (2k +1)^2 - 4m^2$
=> $101^2 = (2k +2m + 1)(2k - 2m + 1)$

Note that 101 is a prime number

Hence we have two possibilities

Case 1:

$( 2k + 2m + 1 = 101^2; 2k - 2m + 1 = 1)$
Subtracting this pair of equations we get $(4m = 101^2 - 1)$ or $(4m = 100 \times 102$) or m = 50 × 51

This gives N = 2601 (ignoring extraneous solutions)

Case 2:

$(2k + 2m + 1 = 101 ; 2k - 2m + 1 = 101)$ which gives m = 0 or N = 101. This solution we ignore as it makes $N(N- 101) = 0$ (a non positive square).

Hence the only solution is $N = 2601$ and there are no other values of N which makes $N(N-101)$ a perfect square.

ISI Entrance Paper 2013 - B.Stat, B.Math Subjective

Here, you will find all the questions of ISI Entrance Paper 2013 from Indian Statistical Institute's B.Stat Entrance. You will also get the solutions soon of all the previous year problems.

B.Stat. (Hons.) & B.Math. (Hons.) Admission Test: 2013

Multiple-Choice Test

Problem 1:

Let $i=\sqrt{-1}$ and $S=\{i+i^{2}+\cdots+i^{n}: n \geq 1\}$. The number of distinct real numbers in the set $S$ is
(A) $1$
(B) $2$
(C) $3$
(D) infinite.

Problem 2:

From a square of unit length, pieces from the corners are removed to form a regular octagon. Then, the value of the area removed is
(A) $1 / 2$
(B) $1 / \sqrt{2}$
(C) $\sqrt{2}-1$
(D) $(\sqrt{2}-1)^{2}$.

Problem 3:

We define the dual of a line $y=m x+c$ to be the point $(m,-c)$. Consider a set of $n$ non-vertical lines, $n>3$, passing through the point $(1,1)$ . Then the duals of these lines will always
(A) be the same
(B) lie on a circle
(C) lie on a line
(D) form the vertices of a polygon with positive area.

Problem 4:

Suppose $\alpha, \beta$ and $\gamma$ are three real numbers satisfying

$\cos \alpha+\cos \beta+\cos \gamma= 0 =\sin \alpha+\sin \beta+\sin \gamma$.

Then the value of $\cos (\alpha-\beta)$ is
(A) $-\frac{1}{2}$
$(\mathrm{B})-\frac{1}{4}$
(C) $\frac{1}{4}$
(D) $\frac{1}{2}$

Problem 5:

The value of $\lim _{x \rightarrow \infty}(3^{x}+7^{x})^{\frac{1}{x}}$ is
(A) $7$
(B) $10$
(C) $e^{7}$
(D) $\infty$.

Problem 6:

The distance between the two foci of the rectangular hyperbola defined by $x y=2$ is

(A) $2$
(B) $2 \sqrt{2}$
(C) $4$
(D) $4 \sqrt{2}$.

Problem 7:

Suppose $f$ is a differentiable and increasing function on $[0,1]$ such that

$f(0)<$ $0<f(1) .$ Let $F(t)=\int_{0}^{t} f(x) d x .$ Then

(A) $F$ is an increasing function on $[0,1]$
(B) $F$ is a decreasing function on $[0,1]$
(C) $F$ has a unique maximum in the open interval $(0,1)$
(D) $F$ has a unique minimum in the open interval $(0,1)$.

Problem 8:

In an isosceles triangle $\Delta A B C,$ the angle $\angle A B C=120^{\circ} .$ Then the ratio of two sides $A C: A B$ is
(A) $2: 1$
(B) $3: 1$
(C) $\sqrt{2}: 1$
(D) $\sqrt{3}: 1$

Problem 9:

Let $x, y, z$ be positive real numbers. If the equation
$x^{2}+y^{2}+z^{2}=(x y+y z+z x) \sin \theta$ has a solution for $\theta,$ then $x, y$ and $z$ must satisfy

(A) $x=y=z$
(B) $x^{2}+y^{2}+z^{2} \leq 1$
(C) $x y+y z+z x=1$
(D) $0<x, y, z \leq 1$

Problem 10:

Suppose $\sin \theta=\frac{4}{5}$ and $\sec \alpha=\frac{7}{4}$ where $0 \leq \theta \leq \frac{\pi}{2}$ and $-\frac{\pi}{2} \leq \alpha \leq 0 .$ Then
$\sin (\theta+\alpha)$ is

(A) $\frac{3 \sqrt{33}}{35}$
(B) $-\frac{3 \sqrt{33}}{35}$
(C) $\frac{16+3 \sqrt{33}}{35}$
(D) $\frac{16-3 \sqrt{33}}{35}$

Problem 11:

Let $i=\sqrt{-1}$ and $z_{1}, z_{2}, \ldots$ be a sequence of complex numbers defined by $z_{1}=i$ and $z_{n+1}=z_{n}^{2}+i$ for $n \geq 1$. Then $\left|z_{2013}-z_{1}\right|$ is

(A) $0$
(B) $1$
(C) $2$
(D) $\sqrt{5}$.

Problem 12:

The last digit of the number $2^{100}+5^{100}+8^{100}$ is

(A) $1$
(B) $3$
(C) $5$
(D) $7$ .

Problem 13:

The maximum value of $|x-1|$ subject to the condition $|x^{2}-4| \leq 5$ is

(A) $2$
(B) $3$
(C) $4$
(D) $5$ .

Problem 14:

Which of the following is correct?

(A) $e x \leq e^{x}$ for all $x$.

(B) $e x<e^{x}$ for $x<1$ and $e x \geq e^{x}$ for $x \geq 1$

(D) $e x<e^{x}$ for $x>1$ and $e x \geq e^{x}$ for $x \leq 1$.

Problem 15:

The area of a regular polygon of $12$ sides that can be inscribed in the circle $x^{2}+y^{2}-6 x+5=0$ is

(A) $6$ units
(B) $9$ units
(C) $12$ units
(D) $15$ units.

Problem 16:

Let $f(x)=\sqrt{\log _{2} x-1}+\frac{1}{2} \log _{\frac{1}{2}} x^{3}+2$. The set of all real values of $x$ for which the function $f(x)$ is defined and $f(x)<0$ is

(A) $x>2$
(B) $x>3$
(C) $x>e$
(D) $x>4$

Problem 17:

Let $a$ be the largest integer strictly smaller than $\frac{7}{8} b$ where $b$ is also an integer. Consider the following inequalities:
(1) $\frac{7}{8} b-a \leq 1$
(2) $\frac{7}{8} b-a \geq \frac{1}{8}$

and find which of the following is correct.
(A) Only (1) is correct.
(B) Only (2) is correct.
(C) Both (1) and (2) are correct.
(D) None of them is correct.

Problem 18:

The value of $\lim _{x \rightarrow-\infty} \sum_{k=1}^{1000} \frac{x^{k}}{k !}$ is

(A) $-\infty$
(B) $\infty$
(C) $0$
(D) $e^{-1}$.

Problem 19:

For integers $m$ and $n$, let $f_{m . n}$ denote the function from the set of integers to itself, defined by $f_{m, n}(x)=m x+n$
Let $\mathcal{F}$ be the set of all such functions,
$\mathcal{F}=\{f_{m, n}: m, n \text { integers }\}$

Call an element $f \in \mathcal{F}$ invertible if there exists an element $g \in \mathcal{F}$ such that $g(f(x))=f(g(x))=x$ for all integers $x$. Then which of the following is true?

(A) Every element of $\mathcal{F}$ is invertible.
(B) $\mathcal{F}$ has infinitely many invertible and infinitely many non-invertible elements.
(C) $\mathcal{F}$ has finitely many invertible elements.
(D) No element of $\mathcal{F}$ is invertible.

Problem 20:

Consider six players $P_{1}, P_{2}, P_{3}, P_{4}, P_{5}$ and $P_{6}$. A team consists of two players. (Thus, there are 15 distinct teams.) Two teams play a match exactly once if there is no common player. For example, team $\{P_{1}, P_{2}\}$ can not play with $\{P_{2}, P_{3}\}$ but will play with $\{P_{4}, P_{5}\}$. Then the total number of possible matches is

(A) $36$
(B) $40$
(C) $45$
(D) $54$

Problem 21:

The minimum value of $f(\theta)=9 \cos ^{2} \theta+16 \sec ^{2} \theta$ is

(A) $25$
(B) $24$
(C) $20$
(D) $16$ .

Problem 22:

The number of $0^{\prime}$ s at the end of the integer $100 !-101 !+\cdots-109 !+110 !$ is
(A) $24$
(B) $25$
(C) $26$
(D) $27$

Problem 23:

We denote the largest integer less than or equal to $z$ by $[z]$. Consider the identity $(1+x)(10+x)\left(10^{2}+x\right) \cdots\left(10^{10}+x\right)=10^{a}+10^{b} x+a_{2} x^{2}+\cdots+a_{11} x^{11}$
Then
(A) $[a]>[b]$
(B) $[a]=[b]$ and $a>b$
(C) $[a]<[b]$
(D) $[a]=[b]$ and $a<b$.

Problem 24:

The number of four tuples $(a, b, c, d)$ of positive integers satisfying all three equations

$a^{3} =b^{2} \\ c^{3} =d^{2} \\ c-a =64$ is

(A) $0$
(B) $1$
(C) $2$
(D) $4$ .

Problem 25:

The number of real roots of $e^{x}=x^{2}$ is

(A) $0$
(B) $1$
(C) $2$
(D) $3$.

Problem 26:

Suppose $\alpha_{1}, \alpha_{2}, \alpha_{3}$ and $\alpha_{4}$ are the roots of the equation $x^{4}+x^{2}+1=0$. Then the value of $\alpha_{1}^{4}+\alpha_{2}^{4}+\alpha_{3}^{4}+\alpha_{4}^{4}$ is

(A) $-2$
(B) $0$
(C) $2$
(D) $4$.

Problem 27:

Among the four time instances given in the options below, when is the angle between the minute hand and the hour hand the smallest?

(A) $5: 25 \mathrm{p.m}$.
(B) $5: 26 \mathrm{p.m}$
(C) $5: 29 \mathrm{p.m}$
(D) $5: 30 \mathrm{p.m}$.

Problem 28:

Suppose all roots of the polynomial $P(x)=a_{10} x^{10}+a_{9} x^{9}+\cdots+a_{1} x+a_{0}$ are real and smaller than $1 .$ Then, for any such polynomial, the function
$f(x)=a_{10} \frac{e^{10 x}}{10}+a_{9} \frac{e^{9 x}}{9}+\cdots+a_{1} e^{x}+a_{0} x, x>0$

(A) is increasing
(B) is either increasing or decreasing
(C) is decreasing
(D) is neither increasing nor decreasing.

Problem 29:

Consider a quadrilateral $A B C D$ in the XY-plane with all of its angles less than $180^{\circ} .$ Let $P$ be an arbitrary point in the plane and consider the six triangles each of which is formed by the point $P$ and two of the points $A, B, C, D .$ Then the total area of these six triangles is minimum when the point $P$ is

(A) outside the quadrilateral
(B) one of the vertices of the quadrilateral
(C) intersection of the diagonals of the quadrilateral
(D) none of the points given in (A), (B) or (C).

Problem 30:

The graph of the equation $x^{3}+3 x^{2} y+3 x y^{2}+y^{3}-x^{2}+y^{2}=0$ comprises
(A) one point
(B) union of a line and a parabola
(C) one line
(D) union of a line and a hyperbola.

B.Stat. (Hons.) & B.Math. (Hons.) Admission Test: 2013

Short-Answer Type Test

Problem 1:

Let $a, b, c$ be real numbers greater than $1$. Let $S$ denote the sum

$S =\log_{a}bc + \log_{b}ca + \log_{c}ab$. Find the smallest possible value of $S$.

Problem 2:

For $x>0$ define $f(x)=\frac{1}{x+2 \cos (x)}$. Determine the set ${y \in \mathbb{R}: y=f(x), x \geq 0}$

Problem 3:

Let $f: \mathbb{R} \longrightarrow \mathbb{R}$ be a function satisfying $|f(x+y)-f(x-y)-y| \leq y^{2}$ for all $x, y \in \mathbb{R}$. Show that $f(x)=\frac{x}{2}+c$, where $c$ is a constant.

Problem 4:

In a badminton singles tournament, each player played against all the others exactly once and each game had a winner. After all the games, each player listed the names of all the players she defeated as well as the names of all the players defeated by the players defeated by her. For instance, if $A$ defeats $B$ and $B$ defeats $C$. then in the list of $A$ both $B$ and $C$ are included. Prove that at least one player listed the names of all other players.

Problem 5:

Let $A D$ be a diameter of a circle of radius $r$. Let $B, C$ be points on the semicircle (with $C$ distinct from $A$ ) so that $A B=B C=\frac{r}{2}$. Determine the ratio of the length of the chord $C D$ to the radius.

Problem 6:

Let $p(x), q(x)$ be distinct polynomials with real coefficients such that the sum of the coefficients of each of the polynomials equals s. If $(p(x))^{3}-(q(x))^{3}=$ $p\left(x^{3}\right)-q\left(x^{3}\right),$ then prove the following:

(1) $p(x)-q(x)=(x-1)^{a} r(x)$ for some integer $a \geq 1$ and a polynomial $r(x)$ with $r(1) \neq 0$.

(2) $s^{2}=3^{a-1}$ where $a$ is as given in $(a)$.

Problem 7:

Let $N$ be a positive integer such that $N(N-101)$ is the square of a positive integer. Then determine all possible values of $N$. (Note that $101$ is a prime number).

Problem 8:

Let $A B C D$ be a square with the side $A B$ lying on the line $y=x+8$. Suppose $C, D$ lie on the parabola $x^{2}=y$. Find the possible values of the length of the side of the square.

Some useful Links:

Download Pdf: ISI Entrance 2013 Paper

AMC 10 (2013) Solutions

12. In $ (\triangle ABC, AB=AC=28)$ and BC=20. Points D,E, and F are on sides $ (\overline{AB}, \overline{BC})$, and $ (\overline{AC})$, respectively, such that $ (\overline{DE})$ and $ (\overline{EF})$ are parallel to $ (\overline{AC})$ and $ (\overline{AB})$, respectively. What is the perimeter of parallelogram ADEF?$ (\textbf{(A) }48\qquad\textbf{(B) }52\qquad\textbf{(C) }56\qquad\textbf{(D) }60\qquad\textbf{(E) }72\qquad )$

Solution: Perimeter = 2(AD + AF). But AD = EF (since ABCD is a parallelogram).
Hence perimeter = 2(AF + EF).
Now ABC is isosceles (AB = AC = 28). Thus angle B = angle C. But EF is parallel to AB. Thus angle FEC = angle B which in turn is equal to angle C.
Hence triangle CEF is isosceles. Thus EF = CF.
Perimeter = 2(AF + EF) = 2(AF + EF) =2AC = $(2 \times 28)$ = 56.

Ans. (C) 56