# A Vietnamese Olympiad Problem: 1968: RMO Training for Algebra

Problem:

Let a and b satisfy $a \geq b >0$ and $a+b=1$.

1. Prove that if m and n are positive integers with $m, then $a^{m}-a^{n} \geq b^{m}-b^{n}>0$.
2. For each positive integer n, consider a quadratic function: $f_{n}(x)=x^{2}-b^{n}x-a^{n}$.

Show that $f(x)$ has two roots that are in between -1 and 1.

Solution:

Let $k=n-m \in (0,n)$. Consider $a^{m}b-ab^{m}=ab(a^{m-1}-b^{m-1})$ with $m-1 \geq 0$. Since $a \geq b >0$, we have $a^{m-1} \geq b^{m-1}$. Hence, $a^{m}b \geq ab^{m}$. Call this relationship I.

On the other hand, notice that $a \geq b$, $a^{2} \geq b^{2}$, $\ldots$, $a^{k-1} \geq b^{k-1}$, which implies

$1+a+a^{2}+\ldots+a^{k-1} \geq 1+b+b^{2}+ \ldots + b^{k-1}$….call this relationship II.

From relationships I and II, it follows that

$a^{m}b(1+a+a^{2}+\ldots+a^{k-1}) \geq ab^{m}(1+b+b^{2}+\ldots+b^{k-1})$

which can be written as

$a^{m}(1-a)(1+a+a^{2}+\ldots+a^{k-1}) \geq b^{m}(1-b)(1+b+b^{2}+\ldots+b^{k-1})$, or equivalently, $a^{m}(1-a^{k}) \geq b^{m}(1-b^{k})$. That is, $a^{m}-a^{n} \geq b^{m}-b^{n}$.

It remains to prove that $b^{m}-b^{n}>0$. Indeed, $b^{m}-b^{n}=b^{m}(1-b^{k})>0$ as $0.

The equality occurs if and only if $a=b=\frac{1}{2}$.

2) Since discriminant $\Delta=b^{2n}+4a^{n}>0$, $f_{n}(x)$ has two distinct real roots $x_{1} \neq x_{2}$. Also, note that if $a, b \in (0,1)$, then the following holds:

$f_{n}(1)=1-b^{n}-a^{n}=a+b-b^{n}-a^{n}=(a-a^{n})+(b-b^{n}) \geq 0$,

$f_{n}(-1)=1+b^{n}-a^{n}=(1-a^{n})+b^{n} \geq 0$,

$\frac{S}{2} = \frac{x_{1}+x_{2}}{2}=\frac{b^{n}}{2} \in (-1,1)$.

We conclude that $x_{1}, x_{2} \in [-1, 1]$.

Cheers,

Nalin Pithwa.

Reference: Selected Problems of the Vietnamese Mathematical Olympiad (1962-2009), Le Hai Chau, Le Hai Khoi.

# Algebra : max and min: RMO/INMO problem solving practice

Question 1:

If $x^{2}+y^{2}=c^{2}$, find the least value of $\frac{1}{x^{2}} + \frac{1}{y^{2}}$.

Solution 1:

Let $z^{'}=\frac{1}{x^{2}} + \frac{1}{y^{2}} = \frac{y^{2}+x^{2}}{x^{2}y^{2}} = \frac{c^{2}}{x^{2}y^{2}}$

$z^{'}$ will be minimum when $\frac{x^{2}y^{2}}{c^{2}}$ will be minimum.

Now, $let z=\frac{x^{2}y^{2}}{c^{2}}=\frac{1}{c^{2}}(x^{2})(y^{2})$….call this equation I.

Hence, z will be maximum when $x^{2}y^{2}$ is maximum but $(x^{2})(y^{2})$ is the product of two factors whose sum is $x^{2}+y^{2}=c^{2}$.

Hence, $x^{2}y^{2}$ will be maximum when both these factors are equal, that is, when

$\frac{x^{2}}{1}=\frac{y^{2}}{1}=\frac{x^{2}+y^{2}}{1}=\frac{c^{2}}{1}$. From equation I, maximum value of $z=\frac{c^{2}}{4}$. Hence, the least value of $\frac{1}{x^{2}} + \frac{1}{y^{2}}=\frac{4}{c^{2}}$.

Some basics related to maximum and minimum:

Basic 1:

Let a and b be two positive quantities, S their sum and P their product; then, from the identity:

$4ab=(a+b)^{2}-(a-b)^{2}$, we have

$4P=S^{2}-(a-b)^{2}$ and $S^{2}=4P+(a-b)^{2}$.

Hence, if S is given, P is greatest when $a=b$; and if P is given, S is least when $a=b$. That is, if the sum of two positive quantities is given, their product is greatest when they are equal; and, if the product of two positive quantities is given, their sum is least when they are equal.

Basic 2:

To find the greatest value of a product the sum of whose factors is constant.

Solution 2:

Let there be n factors $a,b,c,\ldots, k$, and suppose that their sum is constant and equal to s.

Consider the product $abc\ldots k$, and suppose that a and b are any two unequal factors. If we replace the two unequal factors a and b by the two equal factors $\frac{a+b}{2}, \frac{a+b}{2}$, the product is increased, while the sum remains unaltered; hence, so long as the product contains two unequal factors it can be increased without altering the sum of the factors; therefore, the product is greatest when all the factors are equal. In this case, the value of each of the n factors is $\frac{s}{m}$, and the greatest value of the product is $(\frac{s}{n})^{n}$, or $(\frac{a+b+c+\ldots+k}{n})^{n}$.

Corollary to Basic 2:

If $a, b, c, \ldots k$ are unequal, $(\frac{a+b+c+\ldots+k}{n})^{2}>abc\ldots k$;

that is, $\frac{a+b+c+\ldots +k}{n} > (\frac{a+b+c+\ldots + k}{n})^{\frac{1}{n}}$.

By an extension of the meaning of the terms arithmetic mean and geometric mean, this result is usually stated as follows: the arithmetic mean of any number of positive quantities is greater than the geometric mean.

Basic 3:

To find the greatest value of $a^{m}b^{n}c^{p}\ldots$ when $a+b+c+\ldots$ is constant; m,n, p, ….being positive integers.

Solution to Basic 3:

Since m,n,p, …are constants, the expression $a^{m}b^{n}c^{p}\ldots$ will be greatest when $(\frac{a}{m})^{m}(\frac{b}{n})^{n}(\frac{c}{p})^{p}\ldots$ is greatest. But, this last expression is the product of $m+n+p+\ldots$ factors whose sum is $m(\frac{a}{m})+n(\frac{b}{n})+p(\frac{c}{p})+\ldots$, or $a+b+c+\ldots$, and therefor constant. Hence, $a^{m}b^{n}c^{p}\ldots$ will be greatest when the factors $\frac{a}{m}, \frac{b}{n}, \frac{c}{p}, ldots$ are all equal, that is, when

$\frac{a}{m} = \frac{b}{n} = \frac{c}{p} = \ldots = \frac{a+b+c+\ldots}{m+n+p+\ldots}$

Thus, the greatest value is $m^{m}n^{n}p^{p}\ldots (\frac{a+b+c+\ldots}{m+n+p+\ldots})^{m+n+p+\ldots}$.

Some examples using the above techniques:

Example 1:

Show that $(1^{r}+2^{r}+3^{r}+\ldots+n^{r})>n^{n}(n!)^{r}$ where r is any real number.

Solution 1:

Since $\frac{1^{r}+2^{r}+3^{r}+\ldots+n^{r}}{n}>(1^{r}.2^{r}.3^{r}\ldots n^{r})^{\frac{1}{n}}$

Hence, $(\frac{1^{r}+2^{r}+3^{r}+\ldots+n^{r}}{n})^{n}>1^{r}.2^{r}.3^{r} \ldots n^{r}$

that is, $>(n!)^{r}$, which is the desired result.

Example 2:

Find the greatest value of $(a+x)^{3}(a-x)^{4}$ for any real value of x numerically less than a.

Solution 2:

The given expression is greatest when $(\frac{a+x}{3})^{3}(\frac{a-x}{4})^{4}$ is greatest; but, the sum of the factors of this expression is $3(\frac{a+x}{3})+4(\frac{a-x}{4})$, that is, $2a$; hence, $(a+x)^{3}(a-x)^{4}$ is greatest when $\frac{a+x}{3}=\frac{a-x}{4}$, that is, $x=-\frac{a}{7}$. Thus, the greatest value is $\frac{6^{3}8^{4}}{7^{7}}a^{r}$.

Some remarks/observations:

The determination of maximum and minimum values may often be more simply effected by the solution of a quadratic equation than by the foregoing methods. For example:

Question:

Divide an odd integer into two integral parts whose product is a maximum.

Let an odd integer be represented as $2n+1$; the two parts by x and $2n+1-x$; and the product by y; then $(2n+1)x-x^{2}=y$; hence,

$2x=(2n+1)\pm \sqrt{(2n+1)^{2}-4y}$

but the quantity under the radical sign must be positive, and therefore y cannot be greater than $\frac{1}{4}(2n+1)^{2}$, or, $n^{2}+n+\frac{1}{4}$; and since y is integral its greatest value must be $n^{2}+n$; in which case $x=n+1$, or n; thus, the two parts are n and $n+1$.

Sometimes we may use the following method:

Find the minimum value of $\frac{(a+x)(b+x)}{c+x}$.

Solution:

Put $c+x=y$; then the expression $=\frac{(a-c+y)(b-c+y)}{y}=\frac{(a-c)(b-c)}{y}+y+a-c+b-c$

which in turn equals

$(\frac{\sqrt{(a-c)(b-c)}}{\sqrt{y}}-\sqrt{y})^{2}+a-c+b-c+2\sqrt{(a-c)(b-c)}$.

Hence, the expression is a a minimum when the square term is zero; that is when $y=\sqrt{(a-c)(b-c)}$.

Thus, the minimum value is $a-c+b-c+2\sqrt{(a-c)(b-c)}$, and the corresponding value of x is $\sqrt{(a-c)(b-c)}-c$.

Problems for Practice:

1. Find the greatest value of x in order that $7x^{2}+11$ may be greater than $x^{3}+17x$.
2. Find the minimum value of $x^{2}-12x+40$, and the maximum value of $24x-8-9x^{2}$.
3. Show that $(n!)^{2}>n^{n}$ and $2.4.6.\ldots 2n<(n+1)^{n}$.
4. Find the maximum value of $(7-x)^{4}(2+x)^{5}$ when x lies between 7 and -2.
5. Find the minimum value of $\frac{(5+x)(2+x)}{1+x}$.

More later,

Nalin Pithwa.

# Algebra question: RMO/INMO problem-solving practice

Question:

If $\alpha$, $\beta$, $\gamma$ be the roots of the cubic equation $ax^{3}+3bx^{2}+3cx+d=0$. Prove that the equation in y whose roots are $\frac{\beta\gamma-\alpha^{2}}{\beta+\gamma-2\alpha} + \frac{\gamma\alpha-\beta^{2}}{\gamma+\\alpha-2\beta} + \frac{\alpha\beta-\gamma^{2}}{\alpha+\beta-2\gamma}$ is obtained by the transformation $axy+b(x+y)+c=0$. Hence, form the equation with above roots.

Solution:

Given that $\alpha$, $\beta$, $\gamma$ are the roots of the equation:

$ax^{3}+3bx^{2}+3cx+d=0$…call this equation I.

By relationships between roots and co-efficients, (Viete’s relations), we get

$\alpha+\beta+\gamma=-\frac{3b}{a}$ and $\alpha\beta+\beta\gamma+\gamma\alpha=\frac{3c}{a}$, and $\alpha\beta\gamma=-\frac{d}{a}$

Now, $\gamma=\frac{\beta\gamma-\alpha^{2}}{\beta+\gamma-2\alpha}=\frac{\frac{\alpha\beta\gamma}{\alpha}-\alpha^{2}}{(\alpha+\beta+\gamma)-3\alpha}=\frac{-\frac{d}{a\alpha}-\alpha^{2}}{-\frac{3b}{a}-3\alpha}=\frac{d+a\alpha^{3}}{3\alpha(b+a\alpha)}$, that is,

$3xy(b+ax)=d+ax^{3}$, or $ax^{3}-3ayx^{2}-3byx+d=0$…call this equation II.

Subtracting Equation II from Equation I, we get

$3(b+ay)x^{2}+3(c+by)x=0$

$(b+ay)x+c+by=0$ since $x \neq 0$

$axy+b(x+y)+c=0$ which is the required transformation.

Now, $(ay+b)x=-(by+c)$, that is, $x=-\frac{by+c}{ay+b}$

Putting this value of x in Equation I, we get

$-a(\frac{by+c}{ay+b})^{3}+3b(\frac{by+c}{ay+b})^{2}-3c(\frac{by+c}{ay+b})+d=0$, that is,

$a(by+c)^{3}-3b(by+c)^{2}(ay+b)+3c(by+c)(ay+b)^{2}-d(ay+b)^{3}=0$, which is the required equation.

Cheers,

Nalin Pithwa.

# From USSR with love for algebra !

Problem:

Find non-zero distinct integers a, b and c such that the following fourth-degree polynomial with integral coefficients, can be written as the product of two other polynomials with integral coefficients:

$x(x-a)(x-b)(x-c)+1$.

Solution: TBD.

Hint: Of course, the natural thought that comes to mind is the method of undetermined coefficients, and perhaps, some more trial and error to obtain some coefficients, and if we find these, we can find the remaining through their relationship with other coefficients. That brings to mind, Viete’s relations also !! But, there will be two cases here: in one case, the polynomial is a product of two quadratic polynomials, and in the other case, the given fourth degree polynomial is a product of a first degree polynomial and a third degree polynomial.

Problem:

For which integers $a_{1}, a_{2}, \ldots, a_{n}$, where these are all distinct, are the following polynomials with integral coefficients expressible as the product of two polynomials with integral coefficients?

(a) $(x-a_{1})(x-a_{2})(x-a_{3}) \ldots (x-a_{n}) -1$

(b) $(x-a_{1})(x-a_{2})(x-a_{3}) \ldots (x-a_{n}) + 1$

Solution: TBD.

Hint: This is seems to be related to the previous question of fourth degree polynomial ! But, I am not sure if my hint will work ! No spoon-feeding please ! Well, that said, I will think of a clever hint in a couple of days.

Problem:

Prove that if the integers $a_{1}, a_{2}, \ldots, a_{n}$ are all distinct, then the polynomial

$(x-a_{1})^{2}(x-a_{2})^{2}\ldots (x-a_{n})^{2} + 1$

cannot be expressed as a product of two other polynomials with integral coefficients.

Solution: TBD.

Remark: What can be said if the polynomial were

$(x-a_{1})^{2}(x-a_{2})^{2}\ldots (x-a_{n})^{2} -1$

Remark: It is too premature right now for math olympiad students, but polynomial factorization/decomposition has applications in a practical field called control systems.

More later,

Nalin Pithwa

# More Tripos Problems for Practice : RMO and INMO Algebra

If you want to be a Wrangler, you need to tackle the following (so also the problems in previous post :-)):

1. If $x(2a-y) = y(2a-z) = z(2a-u) = u(2a-x) = b^{2}$, show that $x = y = z= u$ unless $b^{2}=2a^{2}$, and that if this condition is satisfied, the equations are not  independent.
2. Out of n straight lines whose lengths are 1, 2, 3, $\ldots , n$ inches respectively, the number of ways in which four may be chosen which will form a quadrilateral in which a circle may be inscribed is $\frac{1}{48} \{ 2n(n-2)(2n-5)-3+3(-1)^{n}\}$.
3. For the expansion of $\frac{1+2x}{1-x^{3}}$, or otherwise, prove that

$1-3n+\frac{(3n-1)(3n-2)}{1.2}-\frac{(3n-2)(3n-3)(3n-4)}{1.2.3}+\frac{(3n-3)(3n-4)(3n-5)(3n-6)}{1.2.3.4} - etc. = (-1)^{n}$, where n is an integer, and the series stops at the first term that vanishes.

4. Find the real roots of the equations:

$x^{2}+v^{2}+w^{2}=a^{2}$, $vw + u(y+z)=bc$,

$y^{2}+w^{2}+u^{2}=b^{2}$, $wu + v(z+x)=ca$,

$z^{2}+u^{2}+v^{2}=c^{2}$, $uv + w(x+y) = ab$.

5. If the equation $\frac{a}{x+a} + \frac{b}{x+b} = \frac{c}{x+c} + \frac{d}{x+d}$ have a pair of equal roots, then either one of the quantities a or b is equal to one of the quantities c or d, or else $\frac{1}{a} + \frac{1}{b} = \frac{1}{c} + \frac{1}{d}$. Prove also that the roots are then $-a, -a, 0$; $-b, -b, 0$; or, $0, 0, \frac{-2ab}{a+b}$.

More challenges are on the way! Are you getting ready for RMO 2016? !

Nalin Pithwa

# Some Tripos Problems Practice for RMO

1. Solve the equation: $\frac{(x-a)(x-b)}{x-a-b} = \frac{(x-c)(x-d)}{x-c-d}$
2. Find the value of a for which the fraction $\frac{x^{3}-ax^{2}+19x-a-4}{x^{3}-(a+1)x^{2}+23x-a-7}$ admits of reduction. Reduce it to lowest terms.
3. An AP, a GP and an HP have a and b for their first two terms; show that their $(n+2)^{th}$ terms will be in GP if $\frac{b^{2m+2}-a^{2m+1}}{ba(b^{2n}-a^{2n})}=\frac{n+1}{n}$

-Nalin Pithwa