# A series question : pre RMO, RMO, IITJEE

Question 1:

Let $x_{1}, x_{2}, \ldots, x_{2014}$ be real numbers different from 1, such that

$x_{1}+x_{2}+\ldots+x_{2014}=1$ and

$\frac{x_{1}}{1-x_{1}} + \frac{x_{2}}{1-x_{2}} + \ldots + \frac{x_{2014}}{1-x_{2014}} = 1$ also.

Then, what is the value of

$\frac{x_{1}^{2}}{1-x_{1}} + \frac{x_{2}^{2}}{1-x_{2}} + \frac{x_{3}^{2}}{1-x_{3}} + \ldots + \frac{x_{2014}^{2}}{1-x_{2014}}$ ?

Solution 1:

Note that

$\sum \frac{x_{i}^{2}}{1-x_{i}} = \sum \frac{x_{i} + (x_{i} - x_{i}^{2})}{1-x_{i}} = \sum (\frac{x_{i}}{1-x_{i}} - x_{i}) = \sum \frac{x_{i}}{1-x_{i}} -\sum x_{i} = 1 - 1 = 0$ which is required answer.

Note that the maximum index 2014 plays no significant role here.

Question 2:

Let f be a one-to-one function from the set of natural numbers to itself such that

$f(mn) = f(m)f(n)$ for all natural numbers m and n.

What is the least possible value of $f(999)$ ?

From elementary number theory, we know that given f is a multiplicative function and hence, the required function is such that if p and q are prime, then

$f(pq)=f(p)f(q)$

That is we need to decompose 999 into its unique prime factorization.

So, we have $999 = 3 \times 333 = 3^{2} \times 111 = 3^{3} \times 97$ where both 3 and 97 are prime.

We have $f(999) = f(3)^{3} \times f(97)$ and we want this to be least positive integer. Clearly, then f(3) cannot be greater than 97. Also, moreover, we need both f(3) and f(97) to be as least natural number as possible. So, $f(3)=2$ and $f(97)=3$ so that required answer is 24.

Question 3:

HW :

What is the number of ordered pairs (A,B) where A and B are subsets of $\{ 1,2,3,4,5\}$ such that neither $A \subset B$ nor $B \subset A$ ?

Cheers,

Nalin Pithwa

# Compilation of elementary results related to permutations and combinations: Pre RMO, RMO, IITJEE math

1. Disjunctive or Sum Rule:

If an event can occur in m ways and another event can occur in n ways and if these two events cannot occur simultaneously, then one of the two events can occur in $m+n$ ways. More generally, if $E_{i}$ ($i=1,2,\ldots,k$) are k events such that no two of them can occur at the same time, and if $E_{i}$ can occur in $n_{i}$ ways, then one of the k events can occur in $n_{1}+n_{2}+\ldots+n_{k}$ ways.

2. Sequential or Product Rule:

If an event can occur in m ways and a second event can occur in n ways, and if the number of ways the second event occurs does not depend upon how the first event occurs, then the two events can occur simultaneously in mn ways. More generally, \$if $E_{1}$ can occur in $n_{1}$, $E_{2}$ can occur in $n_{2}$ ways (no matter how $E_{1}$ occurs), $E_{3}$ can occur in $n_{3}$ ways (no matter how $E_{1}$ and $E_{2}$ occur), $\ldots$, $E_{k}$ can occur in $n_{k}$ ways (no matter how the previous $k-1$ events occur), then the k events can occur simultaneously in $n_{1}n_{2}n_{3}\ldots n_{k}$ ways.

)3. Definitions and some basic relations:

Suppose X is a collection of n distinct objects and r is a nonnegative integer less than or equal to n. An r-permutation of X is a selection of r out of the n objects but the selections are ordered.

An n-permutation of X is called a simply a permutation of X.

The number of r-permutations of a collection of n distinct objects is denoted by $P(n,r)$; this number is evaluated as follows: A member of X can be chosen to occupy the first of the r positions in n ways. After that, an object from the remaining collections of $(n-1)$ objects can be chosen to occupy the second position in $(n-1)$ ways. Notice that the number of ways of placing the second object does not depend upon how the first object was placed or chosen. Thus, by the product rule, the first two positions can be filled in $n(n-1)$ ways,….and all r positions can be filled in

$P(n,r) = n(n-1)\ldots (n-r+1) = \frac{n!}{(n-r)}!$ ways.

In particular, $P(n,n) = n!$

Note: An unordered selection of r out of the n elements of X is called an r-combination of X. In other words, any subset of X with r elements is an r-combination of X. The number of r-combinations or r-subsets of a set of n distinct objects is denoted by $n \choose r$ (read as ” n ‘choose’ r). For each r-subset of X there is a unique complementary $(n-r)$-subset, whence the important relation ${n \choose r}$ = $n \choose {n-r}$.

To evaluate $n \choose r$, note that an r-permutation of an n-set X is necessarily a permutation of some r-subset of X. Moreover, distinct r-subsets generate r-permutations each. Hence, by the sum rule:

$P(n,r)=P(r,r)+P(r,r)+\ldots + P(r,r)$

The number of terms on the right is the number of r-subsets of X. That is, $n \choose r$. Thus, $P(n,r)=P(r,r) \times {n \choose r}$=$r! \times {n \choose r}$.

The following is our summary:

1. $P(n,r) = \frac{n!}{(n-r)!}$
2. ${n \choose r}=\frac{P(n,r)}{r!}=\frac{n!}{r! (n-r)!}$=$n \choose {n-r}$

4. The Pigeonhole Principle: Basic Version:

If n pigeonholes (or mailboxes) shelter n+1 or more pigeons (or letters), at least 1 pigeonhole (or mailbox) shelters at least 2 pigeons (or letters).

5. The number of ways in which $m+n$ things can be divided into two groups containing m and n equal things respectively is given by : $\frac{(m+n)!}{m!n!}$

Note: If $m=n$, the groups are equal (and hence, indistinguishable), and in this case the number of different ways of subdivision is $\frac{(2m)!}{2!m!m!}$

6. The number of ways in which m+n+p things can be divided into three groups containing m, n, p things severally is given by: $\frac{(m+n+p)!}{m!n!p!}$

Note: If we put $m=n=p$, we obtain $\frac{(3m)!}{m!m!m !}$ but this formula regards as different all the possible orders in which the three groups can occur in any one mode of subdivision. And, since there are 3! such orders corresponding to each mode of subdivision, the number of different ways in which subdivision into three equal groups can be made in $\frac{(3m)!}{m!m!m!3!}$ ways.

7. The number of ways in which n things can be arranged amongst themselves, taking them all at a time, when p of the things are exactly alike of one kind, q of them are exactly alike of a another kind, r of them are exactly alike of a third kind, and the rest are all different is as follows: $\frac{n!}{p!q!r!}$

8. The number of permutations of n things r at a time, when such things may be repeated once, twice, thrice…up to r times in any arrangement is given by: $n^{r}$. Cute quiz: In how many ways, can 5 prizes be given away to 4 boys, when each boy is eligible for all the prizes? (Compare your answers with your friends’ answers :-))

9. The total number of ways in which it is possible to make a selection by taking some or all of n things is given by : $2^{n}-1$

10. The total number of ways in which it is possible to make a selection by taking some or all out of $p+q+r+\ldots$ things, whereof p are alike of one kind, q alike of a second kind, r alike of a third kind, and so on is given by : $(p+1)(q+1)(r+1)\ldots-1$.

Regards,

Nalin Pithwa.

# A fifth primer: plane geometry tutorial for preRMO and RMO: core stuff

1. Show that three straight lines which join the middle points of the sides of a triangle, divide it into four triangles which are identically equal.
2. Any straight line drawn from the vertex of a triangle to the base is bisected by the straight line which joins the middle points of the other sides of the triangle.
3. ABCD is a parallelogram, and X, Y are the middle points of the opposite sides AD, BC: prove that BX and DY trisect the diagonal AC.
4. If the middle points of adjacent sides of any quadrilateral are joined, the figure thus formed is a parallelogram. Prove this.
5. Show that the straight lines which join the middle points of opposite sides of a quadrilateral bisect one another.
6. From two points A and B, and from O the mid-point between them, perpendiculars AP, and BQ, OX are drawn to a straight line CD. If AP, BQ measure respectively 4.2 cm, and 5.8 cm, deduce the length of OX. Prove that OX is one half the sum of AP and BQ. or $\frac{1}{2}(AP-BQ)$ or $\frac{1}{2}(BQ-AP)$ according as A and B are on the same side or on opposite sides of CD.
7. When three parallels cut off equal intercepts from two transversals, prove that of three parallel lengths between the two transversals the middle one is the Arithmetic Mean of the other two.
8. The parallel sides of a trapezium are a cm and b cm respectively. Prove that the line joining the middle points of the oblique sides is parallel to the parallel sides, and that its length is $\frac{1}{2}(a+b)$ cm.
9. OX and OY are two straight lines, and along OX five points 1,2,3,4,5 are marked at equal distances. Through these points parallels are drawn in any direction to meet OY. Measure the lengths of these parallels : take their average and compare it with the lengths of the third parallel. Prove geometrically that the third parallel is the mean of all five.
10. From the angular points of a parallelogram perpendiculars are drawn to any straight line which is outside the parallelogram : prove that the sum of the perpendiculars drawn from one pair of opposite angles is equal to the sum of those drawn from the other pair.  (Draw the diagonals,and from their point of intersection suppose a perpendicular drawn to the given straight line.)
11. The sum of the perpendiculars drawn from any point in the base of an isosceles triangle to the equal to the equal sides is equal to the perpendicular drawn from either extremity of the base to the opposite side. It follows that the sum of the distances of any point in the base of an isosceles triangle from the equal sides is constant, that is, the same whatever point in the base is taken).
12. The sum of the perpendiculars drawn from any point within the an equilateral triangle to the three sides is equal to the perpendicular drawn from any one of the angular points to the opposite side, and is therefore, constant. Prove this.
13. Equal and parallel lines have equal projections on any other straight line. Prove this.

More later,

Cheers,

Nalin Pithwa.

# A primer for preRMO and RMO plane geometry with basic exercises

Plane geometry is axiomatic deductive logic. I present a quick mention/review of “proofs” which can be “derived” in sequence….building up the elementary theorems …so for example, if there is a question like: prove that the three medians of a triangle are concurrent, please do not use black magic complicated machinery like Ceva’s theorem,etc; or even if say, the question asks you to prove Ceva’s theorem only, you have to prove it using elementary theorems like the ones presented below:

For the present purposes, I am skipping axioms and basic definitions and hypothetical constructions. I am using straight away the reference (v v v old text) : A School Geometry, Metric Edition by Hall and Stevens. (available almost everywhere in India):

Theorem 1:

The adjacent angles which one straight line makes with another straight line on one side of it are together equal to two right angles.

Corollary 1 of Theorem 1:

if two straight lines cut another, the four angles so formed are together equal to four right angles.

Corollary 2 of Theorem 1:

When any number of straight lines meet at a point, the sum of the consecutive angles so formed is equal to four right angles.

Corollary 3 of Theorem 1:

(a) Supplements of the same angle are equal. (ii) Complements of the same angle are equal.

Theorem 2 (converse of theorem 1):

If, at a point in a straight line, two other straight lines, on opposite sides of it, make the adjacent angles together equal to two right angles, then these two straight lines are in one and the same straight line.

Remark: this theorem can be used to prove stuff like three points are in a straight line.

Theorem 3:

If two straight lines cut one another, the vertically opposite angles are equal.

Theorem 4: SAS Test of Congruence of Two Triangles:

If two triangles have two sides of the one equal to two sides of the other, each to each, and the angles included by those sides equal, then the triangles are equal in all respects.

Theorem 5:

The angles at the base of an isosceles triangle are equal.

Corollary 1 of Theorem 5:

If the equal sides AB, AC of an isosceles triangle are produced, the exterior angles EBC, FCB are equal, for they are the supplements of the equal angles at the base.

Corollary 2 of Theorem 5:

If a triangle is equilateral, it is also equiangular.

Theorem 6:

If two angles of a triangle are equal to one another, then the sides which are opposite to the equal angles are equal to one another.

Corollary of Theorem 6:

Hence, if a triangle is equiangular, it is also equilateral.

Theorem 7 (SSS Test of Congruence of Two Triangles):

If two triangles have the three sides of the one equal to the three sides of the others, each to each, they are equal in all respects.

Theorem 8:

If one side of a triangle is produced then the exterior angle is greater than either of the interior opposite angles.

Corollary 1 to Theorem 8:

Any two angles of a triangle are together less than two right angles.

Corollary 2 to Theorem 8:

Every triangle must have at least two acute angles.

Corollary 3 to Theorem 8:

Only one perpendicular can be drawn to a straight line from a given point outside it.

Theorem 9 :

If one side of a triangle is greater than another, then the angle opposite of the greater side is greater than the angle opposite to the less.

Theorem 10:

If one angle of a triangle is greater than another, then the side opposite to the greater angle is greater than the side opposite to the less.

Theorem 11: Triangle Inequality:

Any two sides of a triangle are together greater than the third side.

Theorem 12: Another inequality sort of theorem:

Of all straight lines drawn from a given point to a given straight line the perpendicular is the least.

Corollary 1 to Theorem 12:

Hence, conversely, since there can be only one perpendicular and one shortest line from O to AB: if OC is the shortest straight line from O to AB, then OC is perpendicular to AB.

Corollary 2 to Theorem 12:

Two obliques OP, OQ which cut AB at equal distance from C, the foot of the perpendicular are equal.

Corollary 3 to Theorem 12:

Of two obliques OQ, OR, if OR cuts AB at the greater distance from C. the foot of the perpendicular, then OR is greater than OQ.

Theorem 13 :

If a straight line cuts two other straight lines so as to make: (i) the alternate angles equal or (ii) an exterior angle equal to the interior opposite angle on the same side of the cutting line or (iii) the interior angles on the same side equal to two right angles, then in each case, the two straight lines are parallel.

Theorem 14:

If a straight line cuts two parallel lines, it makes : (i) the alternate angles equal to one another; (ii) the exterior angle equal to the interior opposite angle on the same side of the cutting line (iii) the two interior angles on the same side together equal to two right angles.

Theorem 15:

Straight lines which are parallel to the same straight line are parallel to one another.

Theorem 16:

Sum of three interior angles of a triangle is 180 degrees.

Also, if a side of a triangle is produced the exterior angle is equal to the sum of the two interior opposite angles.

Corollary 1:

All the interior angles of one rectilinear figure, together with four right angles are equal to twice as many right angles as the figure has sides.

Corollary 2:

If the sides of a rectilinear figure, which has no reflex angle, are produced in order, then all the exterior angles so formed are together equal to four right angles.

Theorem 17: AAS test of congruence of two triangles:

If two triangles have two angles of one equal to two angles of the other, each to each, and any side of the first equal to the corresponding side of the other, the triangles are equal in all respects.

Theorem 18:

Two right angled triangles which have their hypotenuses equal, and one side of one equal to one side of the other are equal in all respects.

Theorem 19:

If two triangles have two sides of the one equal to two sides of the other, each to each, but the angle included by the two sides of one greater than the angle included by the two corresponding sides of the other, then the base of that which has the greater angle is greater than the base of the other.

Conversely,

if two triangles have two sides of the one equal to two sides of the other, each to each, but the base of one greater than the base of the other, then the angle contained by the sides of that which has the greater base is greater than the angle contained by the corresponding sides of the other.

Theorem 20:

The straight lines which join the extremities of two equal and parallel straight lines towards the same parts are themselves equal and parallel.

Theorem 21:

The opposite sides and angles of a parallelogram are equal to one another, and each diagonal bisects the parallelogram.

Corollary 1 to Theorem 21:

If one angle of a parallelogram to a right angle, all its angles are equal.

Corollary 2 to Theorem 21:

All the sides of a square are equal and all its angles are right angles.

Corollary 3 to Theorem 21:

The diagonals of a parallelogram bisect each other.

Theorem 22:

If there are three or more parallel straight lines, and the intercepts made by them on any transversal are equal, then the corresponding intercepts on any other transversal are also equal.

Tutorial exercises based on the above:

Problem 1: In the triangle ABC, the angles ABC, ACB are given equal. If the side BC is produced both ways, show that the exterior angles so formed are equal.

Problem 2: In the triangle ABC, the angles ABC, ACB are given equal. If AB and AC are produced beyond the base, show that the exterior angles so formed are equal.

Problem 3: Prove that the bisectors of the adjacent angles which one straight line makes with another contain a right angle. That is to say, the internal and external bisectors of an angle are at right angles to one another.

Problem 4: If from O a point in AB two straight lines OC, OD are drawn on opposite sides of AB so as to make the angle COB equal to the angle AOD, show that OC and OD are in the same straight line.

Problem 5: Two straight lines AB, CD cross at O. If OX is the bisector of the angle BOD, prove that XO produced bisects the angle AOC.

Problem 6: Two straight lines AB, CD cross at O. If the angle BOD is bisected by OX, and AOC by OY, prove that OX, OY are in the same straight line.

Problem 7: Show that the bisector of the vertical angle of an isosceles triangle (i) bisects the base (ii) is perpendicular to the base.

Problem 8: Let O be the middle point of a straight line AB, and let OC be perpendicular to it. Then, if P is any point in OC, prove that PA=PB.

Problem 9: Assuming that the four sides of a square are equal, and that its angles are all right angles, prove that in the square ABCD, the diagonals AC, BD are equal.

Problem 10: Let ABC be an isosceles triangle: from the equal sides AB, AC two equal parts AX, AY are cut off, and BY and CX are joined. Prove that BY=CX.

Problem 11: ABCD is a four-sided figure whose sides are all equal, and the diagonal BD is drawn : show that (i) the angle ABD = the angle ADB (ii) the angle CBD = the angle CDB (iii) the angle ABC = the angle ADC.

Problem 12: ABC, DBC are two isosceles triangles drawn on the same base BC, but on opposite sides of it: prove that the angle ABD = the angle ACD.

Problem 13: ABC, DBC are two isosceles triangles drawn on the same base BC, but on the same side of it: prove that the angle ABD = the angle ACD.

Problem 14: AB, AC are the equal sides of an isosceles triangle ABC, and L, M, N are the middle points of AB, BC and CA respectively; prove that (i) LM = NM (ii) BN = CL (iii) the angle ALM = the angle ANM.

Problem 15: Show that the straight line which joins the vertex of an isosceles triangle to the middle points of the base (i) bisects the vertical angle (ii) is perpendicular to the base.

Problem 16: If ABCD is a rhombus, that is, an equilateral four sided figure, show by drawing the diagonal AC that (i) the angle ABC = the angle ADC (ii) AC bisects each of the angles BAD and BCD.

Problem 17: If in a quadrilateral ABCD the opposite sides are equal, namely, AB = CD and AD=CB, prove that the angle ADC = the angle ABC.

Problem 18: If ABC and DBC are two isosceles triangles drawn on the same base BC, prove that the angle ABD = the angle ACD, taking (i) the case where the triangles are on the same side of BC (ii) the case where they are on the opposite sides of BC.

Problem 19: If ABC, DBC are two isosceles triangles drawn on opposite sides of the same base BC, and if AD be joined, prove that each of the angles BAC, BDC will be divided into two equal parts.

Problem 20: Show that the straight lines which join the extremities of the base of an isosceles triangle to the middle points of the opposite sides are equal to one another.

Problem 21: Two given points in the base of an isosceles triangle are equidistant from the extremities of the base: show that they are also equidistant from the vertex.

Problem 22: Show that the triangle formed by joining the middle points of the sides of an equilateral triangle is also equilateral.

Problem 23: ABC is an isosceles triangle having AB equal to AC, and the angles at B and C are bisected by BC and CO: prove that (i) BO = CO (ii) AO bisects the angle BAC.

Problem 24: Show that the diagonals of a rhombus bisect one another at right angles.

Problem 25: The equal sides BA, CA of an isosceles triangle BAC are produced beyond the vertex A to the points E and F, so that AE is equal to AF and FB, EC are joined: prove that FB is equal to EC.

Problem 26: ABC is a triangle and D any point within it. If BD and CD are joined, the angle BDC is greater than the angle BAC. Prove this (i) by producing BD to meet AC (ii) by joining AD, and producing it towards the base.

Problem 27: If any side of a triangle is produced both ways, the exterior angles so formed are together greater than two right angles.

Problem 28: To a given straight line, there cannot be drawn from a point outside it more than two straight lines of the same given length.

Problem 29: If the equal sides of an isosceles triangle are produced, the exterior angles must be obtuse.

Note: The problems 30 to 43 are based on triangle inequalities:

Problem 30: The hypotenuse is the greatest side of a right angled triangle.

Problem 31: The greatest side of any triangle makes acute angles with each of the other sides.

Problem 32: If from the ends of a side of a triangle, two straight lines are drawn to a point within the triangle, then those straight lines are together less than the other two sides of the triangle.

Problem 33: BC, the base of an isosceles triangle ABC is produced to any point D; prove that AD is greater than either of the equal sides.

Problem 34: If in a quadrilateral the greatest and least sides are opposite to one another, then each of the angles adjacent to the least side is greater than its opposite angle.

Problem 35: In a triangle, in which OB, OC bisect the angles ABC, ACB respectively: prove that if AB is greater than AC, then OB is greater than OC.

Problem 36: The difference of any two sides of a triangle is less than the third side.

Problem 37: The sum of the distances of any point from the three angular points of a triangle is greater than half its perimeter.

Problem 38: The perimeter of a quadrilateral is greater than the sum of its diagonals.

Problem 39: ABC is a triangle, and the vertical angle BAC is bisected by a line which meets BC in X, show that BA is greater than BX, and CA greater than CX. Obtain a proof of the following theorem : Any two sides of a triangle are together greater than the third side.

Problem 40: The sum of the distance of any point within a triangle from its angular points is less than the perimeter of the triangle.

Problem 41: The sum of the diagonals of a quadrilateral is less than the sum of the four straight lines drawn from the angular points to any given point. Prove this, and point out the exceptional case.

Problem 42: In a triangle any two sides are together greater than twice the median which bisects the remaining side.

Problem 43: In any triangle, the sum of the medians is less than the perimeter.

Problem 44: Straight lines which are perpendicular to the same straight line are parallel to one another.

Problem 45: If a straight line meets two or more parallel straight lines, and is perpendicular to one of them, it is also perpendicular to all the others.

Problem 46: Angles of which the arms are parallel each to each are either equal or supplementary.

Problem 47: Two straight lines AB, CD bisect one another at O. Show that the straight line joining AC and BD are parallel.

Problem 48: Any straight line drawn parallel to the base of an isosceles trianlge makes equal angles with the sides.

More later. Get cracking. This perhaps the simplest introduction, step by step, to axiomatic deductive logic…discovered by Euclid about 2500 years before ! Hail Euclid !

Cheers,

Nalin Pithwa

# Ratio and proportion: practice problems: set II: pRMO, preRMO or IITJEE foundation maths

Problem 1:

If $\frac{y+z}{pb+qc} = \frac{z+x}{pc+qa} = \frac{x+y}{pa+qb}$, then show that $\frac{2(x+y+z)}{a+b+c} = \frac{(b+c)x+(c+a)y+(a+b)z}{bc+ca+ab}$

Problem 2:

If $\frac{x}{a} = \frac{y}{b} = \frac{z}{b}$, show that $\frac{x^{3}+a^{3}}{x^{2}+a^{2}} +\frac{y^{3}+b^{3}}{y^{2}+b^{2}} + \frac{z^{3}+c^{3}}{z^{2}+c^{2}} = \frac{(x+y+z)^{3}+(a+b+c)^{3}}{(x+y+z)^{2}+(a+b+c)^{2}}$

Problem 3:

If $\frac{2y+2z-x}{a} = \frac{2z+2x-y}{b} = \frac{2x+2y-z}{c}$, show that $\frac{x}{2b+2c-a} = \frac{y}{2c+2a-b} = \frac{z}{2a+2b-c}$

Problem 4:

If $(a^{2}+b^{2}+c^{2})(x^{2}+y^{2}+z^{2}) = (ax+by+cz)^{2}$, prove that $x:a = y:b = z:c$

Problem 5:

If $l(my+nz-lx) = m(nz+lx-my) = n(lx+my-nz)$, prove that $\frac{y+z-x}{l} = \frac{z+x-y}{m} = \frac{x+y-z}{n}$

Problem 6:

Show that the eliminant of

$ax+cy+bz=0$

$cx+by+az=0$

$bx+ay+cz=0$

is $a^{3}+b^{3}+c^{3}-3abc=0$

Problem 7:

Eliminate x, y, z from the equations:

$ax+hy+gz=0$

$hx+by+fz=0$

$gx+fy+cz=0$.

This has significance in co-ordinate geometry. (related to conics).

Problem 8:

If $x=cy+bz$, $y=az+cx$, $z=bx+cy$, show that $\frac{x^{2}}{1-a^{2}} = \frac{y^{2}}{1-b^{2}} = \frac{z^{2}}{1-c^{2}}$.

Problem 9:

Given that $a(y+z)=x$, $b(z+x)=y$, $c(x+y)=z$, prove that $bc+ab+ca+2abc=1$

Problem 10:

Solve the following system of equations:

$3x-4y+7z=0$

$2x-y-2z=0$

$3x^{3}-y^{3}+z^{3}=18$

Problem 11:

Solve the following system of equations:

$x+y=z$

$3x-2y+17z=0$

$x^{3}+3y^{3}+2z^{3}=167$

Problem 12:

Solve the following system of equations:

$7yz+3zx=4xy$

$21yz-3zx=4xy$

$x+2y+3z=19$

Problem 13:

Solve the following system of equations:

$3x^{2}-2y^{2}+5z^{2}=0$

$7x^{2}-3y^{2}-15z^{2}=0$

$5x-4y+7z=0$

Problem 14:

If $\frac{l}{\sqrt{a}-\sqrt{b}} + \frac{m}{\sqrt{b}-\sqrt{c}} + \frac{n}{\sqrt{c}-\sqrt{a}} =0$,

and $\frac{l}{\sqrt{a}+\sqrt{b}} + \frac{m}{\sqrt{b}+\sqrt{c}} + \frac{n}{\sqrt{c}+\sqrt{c}} = 0$,

prove that $\frac{l}{(a-b)(c-\sqrt{ab})} = \frac{m}{(b-c)(a-\sqrt{ab})} = \frac{n}{(c-a)(b-\sqrt{ac})}$

Problem 15:

Solve the following system of equations:

$ax+by+cz=0$

$bcx+cay+abz=0$

$xyz+abc(a^{3}x+b^{3}y+c^{3}z)=0$

Cheers,

Nalin Pithwa

# Method of undetermined coefficients for PreRMO, PRMO and IITJEE Foundation maths

1. Find out when the expression $x^{3}+px^{2}+qx+r$ is exactly divisible by $x^{2}+ax+b$

Solution 1:

Let $x^{3}+px^{2}+qx+r=(x^{2}+ax+b)(Ax+B)$ where A and B are to be determined in terms of p, q, r, a and b. We can assume so because we know from the fundamental theorem of algebra that the if the LHS has to be of degree three in x, the remaining factor in RHS has to be linear in x.

So, expanding out the RHS of above, we get:

$x^{3}+px^{2}+qx+r=Ax^{3}+aAx^{2}+bAx+Bx^{2}+Bax+bB$

$x^{3}+px^{3}+qx+r=Ax^{3}+(aA+B)x^{2}+x(bA+aB)+bB$

We are saying that the above is true for all values of x: hence, coefficients of like powers of x on LHS and RHS are same; we equate them and get a system of equations:

$A=1$

$p=aA+B$

$bA+aB=q$

$bB=r$

Hence, we get $p=a+\frac{r}{b}$ and $bp-ba=r$ or that $b(p-a)=r$

Also, $b+aB=q$ so that $q=b+\frac{ar}{b}$ which means $q-b=\frac{a}{b}r$

but $\frac{r}{b}=B=p-a$ and hence, $q-b=\frac{a}{b}(p-a)$

So, the required conditions are $b(p-a)=r$ and $q-b=\frac{a}{b}(p-a)$.

2) Find the condition that $x^{2}+px+q$ may be a perfect square.

Solution 2:

Let $x^{2}+px+q=(Ax+B)^{2}$ where A and B are to be determined in terms of p and q; finally, we obtain the relationship required between p and q for the above requirement.

$x^{2}+px+q=A^{2}x^{2}+B^{2}+2ABx$ which is true for all real values of x;

Hence, $A^{2}=1$ so $A=1$ or $A=-1$

Also, $B^{2}=q$ and hence, $B=\sqrt{q}$ or $B=-\sqrt{q}$

Also, $2AB=p$ so that $2\sqrt{q}=p$ so $q=\frac{p^{2}}{4}$, which is the required condition.

3) To prove that $x^{4}+px^{3}+qx^{2}+rx+s$ is a perfect square if $(q-\frac{p^{2}}{4})^{2}=4s$ and $r^{2}=p^{2}s$.

Proof 3:

Let $x^{4}+px^{3}+qx^{2}+rx+s=(Ax^{2}+Bx+C)^{2}$

$x^{4}+px^{3}+qx^{2}+rx+s=A^{2}x^{4}+B^{2}x^{2}+C^{2}+2ABx^{3}+2BCx+2ACx^{2}$

$A^{2}=1$

$2AB=p$

$q=B^{2}+2AC$

$2BC=r$

$C^{2}=s$

$A=1$ or $A=-1$

$2AB=p \longrightarrow 2B=p \longrightarrow B=\frac{p}{2}$

$q=B^{2}+2AC=\frac{p^{2}}{4}+2\times \sqrt{s} \longrightarrow (q-\frac{p^{2}}{4})^{2}=4s$

$2 \times \frac{p}{2} \times \sqrt{s}=r \longrightarrow r^{2}=p^{2}s$

More later,

Nalin Pithwa.

PS: Note in the method of undetermined coefficients, we create an identity expression which is true for all real values of x.

# Check your mathematical induction concepts

Discuss the following “proof” of the (false) theorem:

If n is any positive integer and S is a set containing exactly n real numbers, then all the numbers in S are equal:

PROOF BY INDUCTION:

Step 1:

If $n=1$, the result is evident.

Step 2: By the induction hypothesis the result is true when $n=k$; we must prove that it is correct when $n=k+1$. Let S be any set containing exactly $k+1$ real numbers and denote these real numbers by $a_{1}, a_{2}, a_{3}, \ldots, a_{k}, a_{k+1}$. If we omit $a_{k+1}$ from this list, we obtain exactly k numbers $a_{1}, a_{2}, \ldots, a_{k}$; by induction hypothesis these numbers are all equal:

$a_{1}=a_{2}= \ldots = a_{k}$.

If we omit $a_{1}$ from the list of numbers in S, we again obtain exactly k numbers $a_{2}, \ldots, a_{k}, a_{k+1}$; by the induction hypothesis these numbers are all equal:

$a_{2}=a_{3}=\ldots = a_{k}=a_{k+1}$.

It follows easily that all $k+1$ numbers in S are equal.

*************************************************************************************

Regards,

Nalin Pithwa

# Miscellaneous Algebra: pRMO, IITJEE foundation maths 2019

For the following tutorial problems, it helps to know/remember/understand/apply the following identities (in addition to all other standard/famous identities you learn in high school maths):

$a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$

By the way, I hope you also know how to derive the above.Let me mention two methods to derive the above :

Method I: Using polynomial division in three variable, divide the dividend $a^{3}+b^{3}+c^{3}-3abc$ by the divisor $a+b+c$.

Method II: Assume that $P(X)$ is a polynomial with roots a, b and c. So, we know by the fundamental theorem of algebra that $P(X)=(X-a)(X-b)(X-c)$. Now, we also know that a, b and c satisfy P(X). Now, proceed further and complete the proof.

Let us now work on the tutorial problems below:

1) If $2s=a+b+c$, prove that $\frac{1}{s-a} + \frac{1}{s-b} + \frac{1}{s-c} = \frac{abc}{s(s-a)(s-b)(s-c)}$

2) If $x^{2}+a^{2}=2(xy+yz+zu-y^{2}-z^{2})$, prove that $x=y=z=u$.

Prove the following identities:

3) $b(x^{3}+a^{3})+ax(x^{2}-a^{2})+a^{3}(x+a)=(a+b)(x+a)(x^{2}-ax+a^{2})$

4) $(ax+by)^{2}+(ay-bx)^{2}+c^{2}x^{2}+c^{2}y^{2}=(x^{2}+y^{2})(a^{2}+b^{2}+c^{2})$

5) $(x+y)^{3}+ 3(x+y)^{2}z+3(x+y)z^{2}+z^{3}=(x+z)^{3}+3(x+z)^{2}y+3(x+z)y^{2}+y^{3}$

6) $(a+b+c)(ab+bc+ca)-abc=(a+b)(b+c)(c+a)$

7) $(a+b+c)^{2}-a(b+c-a)-b(a+c-b)-c(a+b-c)=2(a^{2}+b^{2}+c^{2})$

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

9) $x^{2}(y-z)+y^{2}(z-x)+z^{2}(x-y)+(y-z)(z-x)(z-y)=0$

10) $a^{3}(b-c)+b^{3}(c-a)+c^{3}(a-b)=-(b-c)(c-a)(a-b)(a+b+c)$

11) Prove that $(b-c)^{3}+(c-a)^{3}+(a-b)^{3}=3(b-c)(c-a)(a-b)$

12) If3 $2s=a+b+c$, prove that $(s-a)^{2}+(s-b)^{2}+(s-c)^{2}+s^{2}=a^{2}+b^{2}+c^{2}$

13) If $2s=a+b+c$, prove that $(s-a)^{3}+(s-b)^{3}+(s-c)^{3}+3abc=s^{3}$

14) If $2s=a+b+c$, prove that $16s(s-a)(s-b)(s-c)=2b^{2}c^{2}+2c^{2}a^{2}+2a^{2}b^{2}-a^{4}-b^{4}-c^{4}$

15) If   $2s=a+b+c$, then prove that  $2(s-a)(s-b)(s-c)+a(s-b)(s-c)+b(s-c)(s-a)+c(s-a)(s-b)=abc$

16) If $a+b+c=0$, then prove that $(2a-b)^{3}+(2b-c)^{3}+(2c-a)^{3}=3(2a-b)(2b-c)(2c-a)$

17) If $a+b+c=0$, then prove that $\frac{a^{2}}{2a^{2}+bc} + \frac{b^{2}}{2b^{2}+ca} + \frac{c^{2}}{2c^{2}+ab} =1$

18) Prove that $(x+y+z)^{3}+(x+y-z)^{3}+(x-y+z)^{3}+(x-y-z)^{3}=4x(x^{2}+3y^{2}+3z^{2})$

19) If $a+b+c=0$ prove that $(s+3a)^{3}-(s-3b)^{3}-(s-3c)^{3}-3(s-3a)(s-3b)(s-3c)=0$

20) If $X=b+c-2a$, $Y=c+a-2b$, $Z=a+b-2c$, find the value of $X^{2}+Y^{2}+Z^{2}-3XYZ$

21) Prove that $(a-b)^{2}+(b-c)^{2}+(c-a)^{2}=2(c-b)(c-a)+2(b-a)(b-c)+2(a-b)(a-c)$

22) Prove that $a^{2}(b^{3}-c^{3})+b^{2}(c^{3}-a^{3})+c^{2}(a^{3}-b^{3})=(a-b)(b-c)(c-a)(ab+bc+ca)=a^{2}(b-c)^{3}+b^{2}(c-a)^{3}+c^{2}(a-b)^{3} = -[a^{2}b^{2}(a-b)+b^{2}c^{2}(b-c)+c^{2}a^{2}(c-a)]$

23) if $(a+b)^{2}+(b+c)^{2}+(c+a)^{2}=4(ab+bc+cd)$, prove that $a=b=c=d$.

24) If $x=a+d$, $y=b+d$, $z=c+d$, prove that $x^{2}+y^{2}+z^{2}-yz-zx-xy=a^{2}+b^{2}+c^{2}-bc-ca-ab$

25) If $a+b+c=3$, prove that $\frac{1}{b^{2}+c^{2}-a^{2}}+ \frac{1}{c^{2}+a^{2}-b^{2}} + \frac{1}{a^{2}+b^{2}-c^{2}}=0$

26) If $a+b+c=0$, simplify: $\frac{b+c}{bc}(b^{2}+c^{2}-a^{2}) + \frac{c+a}{ca} (c^{2}+a^{2}-b^{2})+ \frac{a+b}{ab}(a^{2}+b^{2}-c^{2})$

27) Prove that the equation $(x-a)^{2}+(y-b)^{2}+(a^{2}+b^{2}-1)(x^{2}+y^{2}-1)=0$ is equivalent to the equation $(ax+by-1)^{2}+(bx-ay)^{2}=0$, hence show that the only possible values of x and y are: $\frac{a}{a^{2}+b^{2}}$, $\frac{b}{a^{2}+b^{2}}$

28) If $2(x^{2}+a^{2}-ax)(y^{2}+b^{2}-by)=x^{2}y^{2}+a^{2}b^{2}$, prove that $(x-a)^{2}(y-b)^{2}+(bx-ay)^{2}=0$ and therefore that $a=x$ and $y=b$ are the only possible solutions.

Good luck for the PreRMo August 2019 !!

Regards,

Nalin Pithwa

# Prof. Tim Gowers’ on functions, domains, etc.

https://gowers.wordpress.com/2011/10/13/domains-codomains-ranges-images-preimages-inverse-images/

Thanks a lot Prof. Gowers! Math should be sans ambiguities as far as possible…!

I hope my students and readers can appreciate the details in this blog article of Prof. Gowers.

Regards,
Nalin Pithwa