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 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 $u=ax^{4}+4bx^{3}+6cx^{2}+4ax+c=0$Equation 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 $s^{2}-3cs^{2}+(4bd-ae)s+(3ace-2ad^{2}-2eb^{2})=0$Equation D

The second term can be removed by the substitution $s=2t+c$Equation E

and the equation D becomes $4t^{3}-It+J=0$Equation 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