Some random problems in algebra (part b) for RMO and INMO training

1) Solve in real numbers the system of equations:

$y^{2}+u^{2}+v^{2}+w^{2}=4x-1$

$x^{2}+u^{2}+v^{2}+w^{2}=4y-1$

$x^{2}+y^{2}+v^{2}+w^{2}=4u-1$

$x^{2}+y^{2}+u^{2}+w^{2}=4v-1$

$x^{2}+y^{2}+u^{2}+v^{2}=4w-1$

Hints: do you see some quadratics ? Can we reduce the number of variables? …Try such thinking on your own…

2) Let $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ be real numbers such that $a_{1}+a_{2}+a_{3}+a_{4}+a_{5}=0$ and $\max_{1 \leq i <j \leq 5} |a_{i}-a_{j}| \leq 1$ . Prove that $a_{1}^{2}+a_{2}^{2}+a_{3}^{2}+a_{4}^{2}+a_{5}^{2} \leq 10$ .

3) Let a, b, c be positive real numbers. Prove that

$\frac{1}{2a} + \frac{1}{2b} + \frac{1}{2d} \geq \frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a}$

More later

Nalin Pithwa.

Some random assorted (part A) problems in algebra for RMO and INMO training

Posted by Nalin Pithwa

You might want to take a serious shot at each of these. In the first stage of attack, apportion 15 minutes of time for each problem. Do whatever you can, but write down your steps in minute detail. In the last 5 minutes, check why the method or approach does not work. You can even ask — or observe, for example, that if surds are there in an equation, the equation becomes inherently tough. So, as a child we are tempted to think — how to get rid of the surds ?…and so on, thinking in math requires patience and introversion…

So, here are the exercises for your math gym today:

1) Prove that if x, y, z are non-zero real numbers with $x+y+z=0$ , then

$\frac{x^{2}+y^{2}}{x+y} + \frac{y^{2}+z^{2}}{y+z} + \frac{z^{2}+x^{2}}{x+z} = \frac{x^{3}}{yz} + \frac{y^{3}}{zx} + \frac{z^{3}}{xy}$

2) Let a b, c, d be complex numbers with $a+b+c+d=0$ . Prove that

$a^{3}+b^{3}+c^{3}+d^{3}=3(abc+bcd+adb+acd)$

3) Let a, b, c, d be integers. Prove that $a+b+c+d$ divides

$2(a^{4}+b^{4}+c^{4}+d^{4})-(a^{2}+b^{2}+c^{2}+d^{2})^{2}+8abcd$

4) Solve in complex numbers the equation:

$(x+1)(x+2)(x+3)^{2}(x+4)(x+5)=360$

5) Solve in real numbers the equation:

$\sqrt{x} + \sqrt{y} + 2\sqrt{z-2} + \sqrt{u} + \sqrt{v} = x+y+z+u+v$

6) Find the real solutions to the equation:

$(x+y)^{2}=(x+1)(y-1)$

7) Solve the equation:

$\sqrt{x + \sqrt{4x + \sqrt{16x + \sqrt{\ldots + \sqrt{4^{n}x+3}}}}} - \sqrt{x}=1$

8) Prove that if x, y, z are real numbers such that $x^{3}+y^{3}+z^{3} \neq 0$ , then the ratio $\frac{2xyz - (x+y+z)}{x^{3}+y^{3}+z^{3}}$ equals $2/3$ if and only if $x+y+z=0$ .

9) Solve in real numbers the equation:

$\sqrt{x_{1}-1} = 2\sqrt{x_{2}-4}+ \ldots + n\sqrt{x_{n}-n^{2}}=\frac{1}{2}(x_{1}+x_{2}+ \ldots + x_{n})$

10) Find the real solutions to the system of equations:

$\frac{1}{x} + \frac{1}{y} = 9$

$(\frac{1}{\sqrt[3]{x}} + \frac{1}{\sqrt[3]{y}})(1+\frac{1}{\sqrt[3]{x}})(1+\frac{1}{\sqrt[3]{y}})=18$

More later,
Nalin Pithwa

PS: if you want hints, do let me know…but you need to let me know your approach/idea first…else it is spoon-feeding…

Another Romanian Mathematical Olympiad problem

Posted by Nalin Pithwa

Ref: Romanian Mathematical Olympiad — Final Round, 1994

Ref: Titu Andreescu

Problem:

Let M, N, P, Q, R, S be the midpoints of the sides AB, BC, CD, DE, EF, FA of a hexagon. Prove that

$RN^{2}=MQ^{2}+PS^{2}$ if and only if MQ is perpendicular to PS.

Proof:

Let a, b, c, d, e, f be the coordinates of the vertices of the hexagon. The points M, N, P, Q, R, and S have the coordinates

$m=\frac{a+b}{2}$ , $n=\frac{b+c}{2}$ , $=\frac{c+d}{2}$ ,

$q=\frac{d+e}{2}$ , $r=\frac{e+f}{2}$ , $s=\frac{f+a}{2}$ , respectively.

Using the properties of the real product of complex numbers, (please fill in the gaps here), we have

$RN^{2}=MQ^{2}+PS^{2}$

if and only if

$(e+f-b-c).(e+f-b-c) = (d+e-a-b).(d+e-a-b)+(f+a-c-d).(f+a-c-d)$

That is,

$(d+e-a-b).(f+a-c-d)=0$

hence, MQ is perpendicular to PS, as claimed. QED.

More later,

Nalin Pithwa

Practice problems involving moduli and conjugates

Posted by Nalin Pithwa

Problem 1.

Let $z_{1}$ , $z_{2}$ , $\ldots$ , $z_{2n}$ be complex numbers such that $|z_{1}|$ = $|z_{2}| = \ldots = |z_{2n}|$ and $\arg {z_{1}}\leq \arg {z_{2}} \leq \ldots \leq \arg {z_{2n}} \leq \pi$ . Prove that

$|z_{1}+z_{2n}| \leq |z_{2}+z_{2n-1}| \leq \ldots \leq |z_{n}+z_{n+1}|$

Problem 2:

(Vietnamese Mathematical Olympiad, 1996)

Find all positive real numbers x and y satisfying the system of equations:

$\sqrt{3x}(1+\frac{1}{x+y})=2$

$\sqrt{7y}(1-\frac{1}{x+y})=4\sqrt{2}$

Problem 3:

Let $z_{1}$ , $z_{2}$ , $z_{3}$ be complex numbers. Prove that $z_{1}+z_{2}+z_{3}=0$ if and only if $|z_{1}|=|z_{2}+z_{3}|$ , $|z_{2}|=|z_{3}+z_{1}|$ and $z_{3}=|z_{1}+z_{2}|$ .

You are most welcome to send your comments, discuss, etc.

Nalin Pithwa.

A Romanian mathematical olympiad problem

Posted by Nalin Pithwa

Let $z_{1}$ , $z_{2}$ , $\ldots$ , $z_{n}$ be complex numbers such that $|z_{1}|=|z_{2}|=|z_{3}|= \ldots = |z_{n}|>0$ . Prove that

$\Re (\Sigma_{j=1}^{n}\Sigma_{k=1}^{n}\frac{z_{j}}{z_{k}})=0$

if and only if $\Sigma_{k=1}^{n}z_{k}=0$

(Romanian Mathematical Olympiad — Second Round, 1987)

Proof:

Let $S=\Sigma_{j=1}^{n}\Sigma_{k=1}^{n}\frac{z_{j}}{z_{k}}$ .

Then, $S=(\Sigma_{k=1}^{n}z_{k}).(\Sigma_{k=1}^{n}\frac{1}{z_{k}})$ and since $z_{k}.\overline{z_{k}}=r^{2}$ for all k, we have

$S=(\Sigma_{k=1}^{n}z_{k}).(\Sigma_{k=1}^{n}\frac{\overline{z_{k}}}{r^{2}})$

which equals $\frac{1}{r^{2}}(\Sigma_{k=1}^{n}z_{k}).(\overline{\Sigma_{k=1}^{n}z_{k}})=\frac{1}{r^{2}}|\Sigma_{k=1}^{n}z_{k}|^{2}$

Hence, S is a real number, so $\Re{S}=S=0$ , if and only if $\Sigma_{k=1}^{n}z_{k}=0$ .

More later,

Nalin Pithwa

Algebraic equations and polynomials

Posted by Nalin Pithwa

Problem:

Consider the quadratic equation

$a^{2}z^{2}+abz+c^{2}=0$

where a, b, $c \in C^{*}$ and denote by $z_{1}$ , $z_{2}$ its roots. Prove that if $\frac{b}{c}$ is a real number then $|z_{1}|=|z_{2}|$ or $\frac{z_{1}}{z_{2}} \in \Re$ .

Proof:

Let $t = \frac{b}{c} \in \Re$ . Then, $b=tc$ and

$\delta = (ab)^{2}-4a^{2}.c^{2}=a^{2}c^{2}(t^{2}-4)$ .

If $|t| \geq 2$ , the roots of the equation are

$z_{1,2}=-\frac{-tac \pm ac \sqrt{t^{2}-4}}{2a^{2}}=\frac{c}{2a}(-t \pm \sqrt{t^{2}-4})$ and it is obvious that $\frac{z_{1}}{z_{2}}$ is a real number.

If $|t|<2$ ,, the roots of the equation are

$z_{1,2}=\frac{c}{2a}(-t \pm i \sqrt{4-t^{2}})$

hence, $|z_{1}|=|z_{2}|=\frac{|c|}{|a|}$ , as claimed.

QED.

More later,

Nalin Pithwa

Modulli and complex conjugates

Posted by Nalin Pithwa

Problem:

Let $z_{1}$ , $z_{2}$ , $z_{3}$ be complex numbers such that

$|z_{1}|=|z_{2}|=|z_{3}|=r>0$

and $z_{1}+z_{2}+z_{3} \neq 0$ . Prove that

$|\frac{z_{1}z_{2}+z_{2}z_{3}+z_{3}z_{1}}{z_{1}+z_{2}+z_{3}}|=r$

Proof:

Observe that $z_{1}.\overline{z_{1}}=z_{2}.\overline{z_{2}}=z_{3}.\overline{z_{3}}=r^{2}$

Then,

$|\frac{z_{1}z_{2}+z_{2}z_{3}+z_{3}z_{1}}{z_{1}+z_{2}+z_{3}}|^{2}=\frac{z_{1}z_{2}+z_{2}z_{3}+z_{3}z_{1}}{z_{1}+z_{2}+z_{3}}.\frac{\overline{z_{1}z_{2}}+\overline{z_{2}z_{3}}+\overline{z_{3}z_{1}}}{\overline{z_{1}}+\overline{z_{2}}+\overline{z_{3}}}$

This equals

$\frac{z_{1}z_{2}+z_{2}z_{3}+z_{3}z_{1}}{z_{1}+z_{2}+z_{3}}.\frac{\frac{r^{2}.r^{2}}{z_{1}.z_{2}}+\frac{r^{2}.r^{2}}{z_{2}.z_{3}}+\frac{r^{2}.r^{2}}{z_{3}.z_{1}}}{\frac{r^{2}}{z_{1}}+\frac{r^{2}}{z_{2}}+\frac{r^{2}}{z_{3}}}=r^{2}$

as desired.

QED.

More later,

Nalin Pithwa

A Balkan Mathematical Olympiad problem

Posted by Nalin Pithwa

Reference: Complex Numbers from A to …Z by Titu Andreescu and Dorin Andrica

Balkan Mathematical Olympiad, 1985.

Problem:

Let O be the circumcenter of the triangle ABC, let D be the mid-point of the segment AB, and let E be the centroid of the triangle ACD. Prove that lines CD and OE are perpendicular if and only if $AB=AC$ .

Solution:

Let O be the origin of the complex plane and let a, b, c, d, e be the coordinates of points A, B, C, D, E respectively. Then,

$d=\frac{a+b}{2}$ and $e=\frac{a+c+d}{3}=\frac{3a+b+2c}{6}$

Using the real product of complex numbers, if R is the circumradius of triangle ABC, then

$a.a=b.b=c.c=R^{2}$

Lines CD and DE are perpendicular if and only if $(d-c).e=0$ . That is,

$(a+b-2c).(3a+b+2c)=0$ .

The last relation is equivalent to

$3a.a+a.b+2a.c+3a.b+b.b+2b.c-6a.c-2b.c-4c.c=0$ , that is, $a.b=a.c$ — call this equation I.

On the other hand, $AB=AC$ is equivalent to $|b-a|^{2}=|c-a|^{2}$ . That is,

$(b-a).(b-a)=(c-a).(c-a)$

or, $b.b-2a.b+a.a=c.c-2a.c+a.a$ , hence, $a.b=a.c$ —- equation II

The relations (1) and (2) show that CD is perpendicular to OE, if and only if $AB=AC$ .

Ref: Complex Numbers from A to Z by Titu Andreescu and Dorin Andrica

Thanks Prof Andreescu!

from,

Nalin Pithwa

A Bulgarian Mathematical Olympiad Problem

Posted by Nalin Pithwa

Problem:

Two unit squares $K_{1}$ ,, $K_{2}$ with centers M, N are situated in the plane so that $MN=4$ . Two sides of $K_{1}$ are parallel to the line MN, and one of the diagonals of $K_{2}$ lies on MN. Find the locus of the midpoint of XY as X, Y vary over the interior of $K_{1}$ , $K_{2}$ respectively.

(1997 Bulgarian mathematical olympiad)

Solution:

Introduce complex numbers with $M=-2$ , $N=2$ . Then, the locus is the set of points of the form $-(w+xi)+(y+zi)$ , where $|w|$ , $|x|<\frac{1}{2}$ , and $|x+y|$ , $|x-y| < \frac{\sqrt{2}}{2}$ . The result is an octagon with vertices $\frac{1+\sqrt{2}}{2}+\frac{i}{2}$ , $\frac{1}{2}+\frac{(1+\sqrt{2})i}{2}$ , and so on.

Ref: Complex Numbers from A to …Z by Titu Andreescu and Dorin Andrica.

Thanks Prof. Andreescu !

Nalin Pithwa

A William Lowell Putnam Problem and its Solution

Posted by Nalin Pithwa

Problem:

Curves A, B, C and D are defined in the plane as follows:

$A = \{ (x, y): x^{2}-y^{2}=\frac{x}{x^{2}+y^{2}}\}$ –

$B = \{ (x, y): 2xy + \frac{y}{x^{2}+y^{2}}=3\}$

$C = \{ (x, y): x^{3}-3xy^{2}+3y=1\}$

$D = \{ (x, y): 3x^{2}y -3x - y^{3}=0\}$ .

Prove that $A \bigcap B = C \bigcap D$

(1987 William Lowell Putnam Mathematics Competition)

Solution:

Let $z= x+iy$ . The equations defining A and B are the real and imaginary parts of the equation $z^{2}=z^{-1}+3i$ , and similarly the equations defining C and D are the real and imaginary parts of $z^{3}-3iz=1$ . Hence, for all real x and y, we have $(x,y) \in A \bigcap B$ if and only if $z^{2}=z^{-1}+3i$ . This is equivalent to $z^{3}-3iz=1$ , that is, $(x, y) \in C \bigcap D$ .

Thus, $A \bigcap B = C \bigcap D$ .

Isn’t that an elegant solution? What do you think?

Nalin Pithwa

PS: Solution published in “Complex Numbers from A to …Z” by Titu Andreescu and Dorin Andrica