Pre RMO Algebra problems for practice

Question 1:

Find the square root of 49x^{4}+\frac{1051x^{2}}{25} - \frac{14x^{3}}{5} - \frac{6x}{5} + 9

Question 2:

The surface area of a circular cone is given by A= {\pi}r^{2}+{\pi}rs, where s cm is the slant height, r cm is the radius of the base and \pi is \frac{22}{7}. Find the radius of the base if a cone of surface area 93.5 square cm has a slant height of 5 cm.

Question 3:

Solve:

\frac{a+x}{a^{2}+ax+x^{2}} +\frac{a-x}{a^{2}-ax+ x^{2}} = \frac{3a}{x(a^{4}+a^{2}x^{2}+x^{4})}

Question 4:

Subtract \frac{x+3}{x^{2}+x-12} + \frac{x+4}{x^{2}-x-12} and divide the difference by 1 + \frac{2(x^{2}-12)}{x^{2}+7x+12}

Question 5:

Solve the following for the unknown x:

\frac{x}{2(x+3)} - \frac{53}{24} = \frac{x^{2}}{x^{2}-9} - \frac{8x-1}{4(x-3)}

Question 6:

Find the square root of the following:

a^{6}+ \frac{1}{a^{6}} -6(a^{4}+\frac{1}{a^{4}}) +12(a^{2}+\frac{1}{a^{2}})-20; also, cube the result.

More later,
Nalin Pithwa.

Solution to a RMO algebra practice question

Refer to question posted on blog of Mar 13 2018, reproduced here for your convenience. Compare your solution with the one given here:

Question:

Show that the following expression: [4-3x+\sqrt{16+9x^{2}-24x-x^{3}}]^{1/3}+[4-3x-\sqrt{16+9x^{2}-24x-x^{3}}]^{1/3} remains constant in the interval 0 \leq x \leq 1. Find this constant value.

Solution/proof:

Let y equal the given expression for x in the prescribed interval. Then, taking cube of both sides, we write

y^{3}=8-6x+3xy

y^{3}-8-3x(y-2)=0

(y-2)(y^{2}+2y+4-3x)=0

The only real value of y=2, a constant. The roots of the quadratic equation for 0<x<1 are complex. It is easy to check that y=2 for both x=0 and 1.

Alternately, derive the square roots of the expression within the radical; you can use the method of undetermined coefficients for this.

Cheers,

Nalin Pithwa.

 

Monic Polynomials — a problem for RMO and INMO

Problem:

Prove that if the polynomial X^{n}+X^{n-1}+ \ldots + X +1 can be written as the product of two monic polynomials with real non-negative coefficients, then those coefficients are all 0 or 1.

Proof by Andrei Stefanescu:

Definition: 

We call any polynomial P(X)=a_{n}X^{n}+\ldots+a_{0} symmetrical if and only if a_{i}=a_{n-i} for all i. We will use the following lemma:

LEMMA: If P is a polynomial which can be written as P=Q.R, where Q and R are symmetrical polynomials, then P is symmetrical too.

Let Q(X)=b_{s}X^{s}+ \ldots + b_{0} and R(X)=c_{t}X^{t}+\ldots+c_{0}. We have that a_{i}=\sum_{j=0}^{i}b_{j}c_{i-j} and a_{n-i}=\sum_{j=0}^{i}b_{s-j}c_{t-(i-j)}. Since Q and R are symmetrical, it follows that b_{j}=b_{s-j} and c_{i-j}=c_{t-(i-j)}. Hence, a_{i}=a_{n-i} for any i, and thus P is symmetrical.

Let f(X)=X^{n}+X^{n-1}+ \ldots+1 and assume that f=g.h, where g and h are nonconstant polynomials with non-negative coefficients. Let g(X)=b_{s}X^{s}+\ldots + b_{0} and h(X)=c_{t}X^{t}+\ldots + c_{0}. Since f=g.h, all the complex roots of  g and h have absolute value 1. Also, if g(\in)=0, then g(\overline{\in}) = g(\frac{1}{\in})=0. It follows that both g and h are a product of symmetric terms of the form (X-\in)(X-\frac{1}{\in}) = X^{2}-2\Re (\in)+1 or X+1, hence by lemma, g and h are symmetrical.

If all the coefficients of g and h are in \{0,1\} we are done. Otherwise, let k be the least number such that \{ b_{k},c_{k} \} is not a subset of \{ 0,1\}. Since b_{s}=c_{t}=1, it follows that b_{0}=c_{0}=1, thus k \geq 1. As g is symmetrical, b_{k}=b_{s-k}. Computing the coefficient of X^{s} gives us 1=b_{s}c_{0}+\ldots+b_{s-k}c_{k}+\ldots+b_{0}c_{s}. Since b_{s}=c_{0}=1 and all the terms of the sum are non-negative, we obtain b_{s-k}c_{k}=b_{k}c_{k}=0, thus one of b_{k} and c_{k} (say, c_{k})is 0. Computing the coefficient of X^{k}, we have 1=b_{k}c_{0}+\ldots + b_{0}c_{k}. As \{ b_{i},c_{i}\} \subset \{ 0,1\} for i < k and c_{k}=0, all the terms but b_{k}c_{0} must be in \{ 0,1\}. It follows that b_{k}c_{0}=b_{k} must be in \{ 0,1\}, which leads to a contradiction with the definition of k.

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

 

 

Ferrari’s solution of a biquadratic

Ferrari’s solution of the Biquadratic. 

Writing the equation u=ax^{4}+4bx^{3}+6cx^{2}+4ax+c=0Equation A

we assume that au=(ax^{2}+2bx+s)^{2}-(2mx+n)^{2}Equation B

Expanding and equating coefficients, we have

2m^{2}=as+2b^{2}-3ac and mn=bs-ad and n^{2}=s^{2}-ae ….Equation C

Eliminating m, n,

(s^{2}-ae)(as+2b^{2}-3ac)=2(bs-ad)^{2} which reduces to

s^{2}-3cs^{2}+(4bd-ae)s+(3ace-2ad^{2}-2eb^{2})=0Equation D

The second term can be removed by the substitution s=2t+cEquation E

and the equation D becomes 4t^{3}-It+J=0Equation F, which is the “reducing cubic”. 

Equations C become

m^{2}=at+b^{2}-ac=at-H

mn=2bt+bc-ad

n^{2}=(2t+c)^{2}-ae

Call the above three equations as G.

Thus, if t_{1} is a root of F and

m_{1}=\sqrt{at_{1}+b^{2}-ac} and n_{1}=\frac{(2bt_{1}+bc-ad)}{m_{1}}, the equation

u=0 can be put in the form

(ax^{2}+2bx+c+2t_{1})^{2}-(2m_{1}x+n_{1})^{2}=0,

and its roots are the roots of the quadratics

ax^{2}+2b^{x}+c+2t_{1}=\pm (2m_{1}x+n_{1}).

It should be noted that the three roots of F correspond to the three ways of expressing u as the product of two quadratic factors.

Homework: Solve the equation u=x^{4}+3x^{3}+x^{2}-2=0.

Hope you enjoyed it…

More later,

Nalin Pithwa