Modern family: Addition

State level Olympiad (1)

After coming from the Hawaii vacation; days gone by. Finally, day for state level Olympiad exams grew near. Exams were to be held in San Francisco, so I under my parents consultation decided to stay in the hotel nearby in San Francisco for exam period.

We decided that it would be best if dad go with me. So we setoff almost 2 days before the exam date for our destination so that not only can we reach there before time, have our necessary rest but also not be in any foreseeable hurry.

There we checked- in in one of the good hotels named Ritz- Carlton hotel. We obviously had money so there was no point in searching for cheaper alternative, afterall that's what dad would have done if he would have been alone or with mom/my siblings. Not sure what would had been the reason, may I'm making my own Money sizing millions or something else but he himself did all these things without me saying anything. It seems that my parents do consider me something more than their child fitting to my growing status.

Firstly there will be mathematics paper, then physics and then chemistry seperated by days, all within a week. Schedule was neat and short so there was no problem. I had already informed my Principal about my exams so there was no loose end on that part as well.

------

Finally day for state level Math Olympiad exam came. We drove to the exam centre in taxi, after usual checking and verification process; I entered in exam hall.

After waiting for some time, they distributed the papers. I looked at exam paper, it was easy.

Q1 : Determine all pairs of positive integers (m,n) such that (1+x^n+x^{2n}+........+x^{mn}) is divisible by (1+x+x^2+.....+x^{m}).

Solution1 :
Denote the first and larger polynomial to be f(x) and the second one to be g(x). In order for f(x) to be divisible by g(x) they must have the same roots. The roots of g(x) are the (m+1)th roots of unity, except for 1. When plugging into f(x), the root of unity is a root of f(x) if and only if the terms x^n, x^{2n}, x^{3n}, ......
, x^{mn}$ all represent a different (m+1)th root of unity not equal to 1.

Note that if gcd(m+1,n)=1, the numbers n, 2n, 3n, ........ , mn represent a complete set of residues minus 0 modulo m+1. However, if gcd(m+1,n)=a not equal to 1, then {(m+1)(n)}/{a}is congruent to 0 (mod {m+1}) and thus a complete set is not formed. Therefore, f(x) divides g(x) if and only if {gcd(m+1,n)=1}.

-------

This is the solution that I wrote in my exam paper; there are other ways to approach the problem but I did neat, short and direct way.

Q 2 : ABC and A'B'C' are two triangles in the same plane such that the lines AA',BB',CC' are mutually parallel. Let [ABC] denote the area of triangle ABC with an appropriate +ve, -ve sign, etc.; prove that 3([ABC]+ [A'B'C']) = [AB'C'] + [BC'A'] + [CA'B']+ [A'BC]+[B'CA] + [C'AB].

Solution 2 : Let the parallel lines AA', BB', CC' be parallel to the x-axis, and choose arbitrary origin. Then we can define A(x1,a), A'(x2, a), B(y1, b), B'(y2, b), C(z1, c), C'(z2, c), and so, by the area of a triangle formula, it suffices to prove an algebraic statement that is readily shown to be true. (I wrote whole process in the paper.)

Q 3 : If a and b are two of the roots of x^4+x^3-1=0, prove that ab is a root of x^6+x^4+x^3-x^2-1=0.

Solution 3 :
Given the roots a,b,c,d of the equation x^{4}+x^{3}-1=0.

First, Vieta's relations give a+b+c+d = -1 , ab+ac+ad+bc+bd+cd=0, abc+abd+acd+bcd=0,
abcd = -1.

Then cd=- {1}/{ab} and c+d=-1-(a+b).

The other coefficients give ab+(a+b)(c+d)+cd = 0 or ab+(a+b)[-1-(a+b)] - {1}/{ab}=0.

Let a+b=s and ab=p.

Thus, 0=ab+ac+ad+bc+bd+cd=p+s(-1-s)- {1}/{p}. ---------- (1)

Also, 0=abc+abd+acd+bcd=p(-1-s)- s/p.

Solving this equation for s,
s={- p^2}/{p^2+1}.

Substituting into (1): {p^6+p^4+p^3 - p^2 -1}/{p(p^2+1)^2}=0.

Conclusion: p =ab is a root of x^6+x^4+x^3 - x^2 - 1=0.

Q 4 : Prove that if the opposite sides of a skew (non- planar) quadrilateral are congruent, then the line joining the midpoints of the two diagonals is perpendicular to these diagonals, and conversely, if the line joining the midpoints of the two diagonals of a skew quadrilateral is perpendicular to these diagonals, then the opposite sides of the quadrilateral are congruent.

Solution 4 : We first prove that if the opposite sides are congruent, then the line joining the midpoints of the two diagonals is perpendicular to these diagonals.

Let the vertices of the quadrilateral be A, B, C, D, in that order. Thus, the opposite sides congruent condition translates to (after squaring):

AB^2 = CD^2, BC = AD^2
Let the position vectors of A, B, C, D be a, b, c, d with respect to an arbitrary origin O. Define the square of a vector a^2 = a.a . Thus,(a- b)^2 = (c- d)^2 and so a^2 - 2ab + b^2 = c^2 - 2cd + d^2.
Similarly, b^2 - 2bc + c^2 = a^2 - 2ad + d^2.
Subtracting the second equation from the first eventually results in a^2 - ab + cd + bc - ad - c^2 = 0, which factors as (a - b + c - d)(a - c) = 0.

Let M and N be the midpoints of AC and BD, with position vectors m = {a+c}/{2} and n = {b+d}/{2}, respectively. Then notice that the perpendicularity condition for vectors gives MN perpendicular to AC. Similarly (by adding the equations), we find that MN is perpendicular to BD, completing the first part of the proof.

Proving the converse is straightforward also: indeed, MN perpendicular to both diagonals gives

a^2 - ab + cd + bc - ad - c^2 = 0
b^2 - ab + cd - bc + ad - d^2 = 0
Adding and subtracting the equations eventually simplifies to

(a- b)^2 = (c- d)^2
(a- d)^2 = (b- c)^2,
or AB^2 = CD^2 and AD^2 = BC^2.

This reduces to AB = CD and
AD = BC, as desired.