You and your research or you and your studies for competitive math exams

You and your research ( You and your studies) : By Richard Hamming, AT and T, Bell Labs mathematician;

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.

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

Comments, observations are welcome 🙂

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

RMO and Pre RMO Geometry Tutorial Worksheet 1: Based on Geometric Refresher

1) Show that quadrilateral ABCD can be inscribed in a circle iff \angle B and \angle D are supplementary.

2) Prove that a parallelogram having perpendicular diagonals is a rhombus.

3) Prove that a parallelogram with equal diagonals is a rectangle.

4) Show that the diagonals of an isosceles trapezoid are equal.

5) A straight line cuts two concentric circles in points A, B, C and D in that order. AE and BF are parallel chords, one in each circle. If CG is perpendicular to BF and DH is perpendicular to AE, prove that GF = HE.

6) Construct triangle ABC, given angle A, side AC and the radius r of the inscribed circle. Justify your construction.

7) Let a triangle ABC be right angled at C. The internal bisectors of angle A and angle B meet BC and CA at P and Q respectively. M and N are the feet of the perpendiculars from P and Q to AB. Find angle MCN.

8) Three circles C_{1}, C_{2},  C_{3} with radii r_{1}, r_{2}, r_{3}, with r_{1}<r_{2}<r_{3}. They are placed such that C_{2} lies to the right of C_{1} and touches it externally; C_{3} lies to the right of C_{2} and touches it externally. Further, there exist two straight lines each of which is a direct common tangent simultaneously to all the three circles. Find r_{2} in terms of r_{1} and r_{3}.

Cheers,

Nalin Pithwa

Basic Geometry: Refresher for pre-RMO, RMO and IITJEE foundation maths

1) Assume that in a \bigtriangleup ABC and \bigtriangleup RST, we know that AB=RS, AC=RT, BC=ST. Prove that \bigtriangleup ABC \cong \bigtriangleup RST without using the SSS congruence criterion.

2) Let \bigtriangleup ABC be isosceles with base BC. Then, \angle B = \angle C. Also, the median from vertex A, the bisector of \angle A, and the altitude from vertex A are all the same line. Prove this.

3) If two triangles have equal hypotenuses and an arm of one of the triangles equals an arm of the other, then the triangles are congruent. Prove.

4) An exterior angle of a triangle equals the sum of the two remote interior angles. Also, the sum of all three interior angles of a triangle is 180 degrees.

5) Find a formula for the interior angles of an n-gon.

6) Prove that the opposite sides of a parallelogram are equal.

7) In a quadrilateral ABCD, suppose that AB=CD and AD=BC. Then, prove that ABCD is a parallelogram.

8) In a quadrilateral ABCD, suppose that AB=CD and AB is parallel to CD. Then, prove that ABCD is a parallelogram.

9) Prove that a quadrilateral is a parallelogram iff its diagonals bisect each other.

10) Given a line segment BC, the locus of all points equidistant from B and C is the perpendicular bisector of the segment. Prove.

11) Corollary to problem 10 above: The diagonals of a rhombus are perpendicular. Prove.

12) Let AX be the bisector of \angle A in \triangle ABC. Then, prove \frac{BX}{XC} = \frac{AB}{AC}. In other words, X divides BC into pieces proportional to the lengths of the nearer sides of the triangle. Prove.

13) Suppose that in \triangle ABC, the median from vertex A and the bisector of \angle A are the same line. Show that AB=AC.

14) Prove that there is exactly one circle through any three given non collinear points.

15) An inscribed angle in a circle is equal in degrees to one half its subtended arc. Equivalently, the arc subtended by an inscribed angle is measured by twice the angle. Prove.

16) Corollary to above problem 15: Opposite angles of an inscribed quadrilateral are supplementary. Prove this.

17) Another corollary to above problem 15: The angle between two secants drawn to a circle from an exterior point is equal in degrees to half the difference of the two subtended arcs. Prove this.

18) A third corollary to above problem 15: The angle between two chords that intersect in the interior of a circle is equal in degrees to half the sum of the two subtended arcs. Prove this.

19) Theorem (Pythagoras): If a right triangle has arms of lengths a and b and its hypotenuse has length c, then a^{2}+b^{2}=c^{2}. Prove this.

20) Corollary to above theorem: Given a triangle ABC, the angle at vertex C is a right angle iff side AB is a diameter of the circumcircle. Prove this.

21) Theorem: The angle between a chord and the tangent at one of its endpoints is equal in degrees to half the subtended arc. Prove.

22) Corollary to problem 21: The angle between a secant and a tangent meeting at a point outside a circle is equal in degrees to half the difference of the subtended arcs.

23) Fix an integer, n \geq 3. Given a circle, how should n points on this circle be chosen so as to maximize the area of the corresponding n-gon?

24) Theorem: Given \bigtriangleup ABC and \bigtriangleup XYZ, suppose that \angle A = \angle X and \angle B= \angle Y. Then, prove that \angle C = \angle Z and so \bigtriangleup ABC \sim \bigtriangleup XYZ. Prove this theorem.

25) Theorem: If \bigtriangleup ABC \sim \bigtriangleup XYZ, then the lengths of the corresponding sides of these two triangles are proportional. Prove.

26) The following lemma is important to prove the above theorem: Let U and V be points on sides AB and AC of \bigtriangleup ABC. Then, UV is parallel to BC if and only if \frac{AU}{AB} = \frac{AV}{AC}. You will have to prove this lemma as a part of the above proof.

27) Special case of above lemma: Let U and V be the midpoints of sides AB and AC, respectively in \bigtriangleup ABC. Then, UV is parallel to BC and UV = \frac{1}{2}BC.

28) Suppose that the sides of \bigtriangleup ABC are proportional to the corresponding sides of \bigtriangleup XYZ. Then, \bigtriangleup ABC \sim \bigtriangleup XYZ.

29) Given \bigtriangleup ABC and \bigtriangleup XYZ, assume that \angle X = \angle A and that \frac{XY}{AB} = \frac{XZ}{AC}. Then, \bigtriangleup ABC \sim \bigtriangleup XYZ.

30) Consider a non-trivial plane geometry question now: Let P be a point outside of parallelogram ABCD and \angle PAB = \angle PCB. Prove that \angle APD = \angle CPB.

31) Given a circle and a point P not on the circle, choose an arbitrary line through P, meeting the circle at points X and Y. Then, the quantity PX.PY depends only on the point P and is independent of the choice of the line through P.

32) You can given an alternative proof of Pythagoras’s theorem based on the following lemma: Suppose \bigtriangleup ABC is a right triangle with hypotenuse AB and let CP be the altitude drawn to the hypotenuse. Then, \bigtriangleup ACP \sim \bigtriangleup ABC \sim \bigtriangleup CBP. Prove both the lemma and based on it produce an alternative proof of Pythagorean theorem.

33) Prove the following: The three perpendicular bisectors of the sides of a triangle are concurrent at the circumcenter of the triangle.

34) Prove the law of sines.

35) Let R and K denote the circumradius and area of \bigtriangleup ABC, respectively and let a, b and c denote the side lengths, as usual. Then, 4KR = abc.

36) Theorem: The three medians of an arbitrary triangle are concurrent at a point that lies two thirds of the way along each median from the vertex of the triangle toward the midpoint of the opposite side.

37) Time to ponder: Prove: Suppose that in \bigtriangleup ABC, medians BY and CZ have equal lengths. Prove that AB=AC.

38) If the circumcenter and the centroid of a triangle coincide, then the triangle must be equilateral. Prove this fact.

39) Assume that \bigtriangleup ABC is not equilateral and let G and O be its centroid and circumcentre respectively. Let H be the point on the Euler line GO that lies on the opposite side of G from O and such that HG = 2GO. Then, prove that all the three altitudes of \bigtriangleup ABC pass through H.

40) Prove the following basic fact about pedal triangles: The pedal triangles of each of the four triangles determined by an orthic quadruple are all the same.

41) Prove the following theorem: Given any triangle, all of the following points lie on a common circle: the three feet of the altitudes, the three midpoints of the sides, and the three Euler points. Furthermore, each of the line segments joining an Euler point to the midpoint of the opposite side is a diameter of this circle.

42) Prove the following theorem and its corollary: Let R be the circumradius of triangle ABC. Then, the distance from each Euler point of \bigtriangleup ABC to the midpoint of the opposite side is R, and the radius of the nine-point circle of \bigtriangleup ABC is R/2. The corollary says: Suppose \bigtriangleup ABC is not a right angled triangle and let H be its orthocentre. Then, \bigtriangleup ABC, \bigtriangleup HBC, \bigtriangleup AHC, and \bigtriangleup ABH have equal circumradii.

43) Prove the law of cosines.

44) Prove Heron’s formula.

45) Express the circumradius R of \bigtriangleup ABC in terms of the lengths of the sides.

46) Prove that the three angle bisectors of a triangle are concurrent at a point I, equidistant from the sides of the triangle. If we denote the by r the distance from I to each of the sides, then the circle of radius r centered at I is the unique circle inscribed in the given triangle. Note that in order to prove this, the following elementary lemma is required to be proved: The bisector of angle ABC is the locus of points P in the interior of the angle that are equidistant from the sides of the triangle.

47) Given a triangle with area K, semiperimeter s, and inradius r, prove that rs=K. Use this to express r in terms of the lengths of the sides of the triangle.

Please be aware that the above set of questions is almost like almost like a necessary set of pre-requisites for RMO geometry. You have to master the basics first.

Regards,

Nalin Pithwa.