So it's been a long time since I posted anything
on this blog, mostly because I've had my Grade 10
board exams.
Last Sunday, the 8th of October, I gave the Regional
Mathematical Olympiad. It was the second level on a
4-step process to the IMO, a challenging and prestigious
international examination. My results come out on the
6th of November, and in the meantime, I decided to
share a little bit about the problems.
The exam consists of six questions, to be attempted
in three hours. The questions are not exercises, as the
approach to solving them is not immediately obvious.
The problems are mostly proof-style, with justification
from theorems and mathematical logic.
Coming out of the exam, I felt I had properly
attacked only three of the six questions (not the greatest
result, I know). I share my solutions below.
Q1. Let AOB be a given angle less than 180° and let P be an interior point of the angular region determined by ∠AOB. Show, with proof, how to construct, using only ruler and compasses, a line segment CD passing through P such that C lies on the ray OA and D lies on the ray OB, and CP:PD = 1:2.
A1. Let the base of the perpendicular from P onto ray OB
be called Q, and similarly the base of the perpendicular
from P onto ray OA be called R.
Construct a line equidistant between P and Q, and parallel
to ray OB. By the Triangle Midpoint Theorem, this line
is the locus of midpoints of all line segments drawn
from P to OB. Call this line Locus1.
Construct a second line (called Locus2) such that P is
equidistant between the line and R. By the Triangle
Midpoint Theorem, P is the midpoint of any line
segment drawn between Locus2 and OA that passes
through P.
Call the point where Locus1 and Locus2 intersect, S.
Draw the line PS and let it cut OA in C and OB in D.
Then, due to the constructions and properties of the
loci, CP = PS = PD => CP:PD = 1:2. Thus, C and D
fit the question's requirement.
__________________________________________________________________________
Q2. Show that the equation
a3 + (a+1)3 + (a+2)3 + (a+3)3 + (a+4)3 + (a+5)3 + (a+6)3 = b4 + (b+1)4
has no solutions in integers a, b.
A2. If we expand the left hand side, we get 7a3 + 63a2 + 273a + 441.
Since all coefficients are divisible by 7, we get
7(a3 + 9a2 + 39a + 63) = b4 + (b+1)4
If we show that the right hand side can never be
divisible by 7, we are done.
We do that by taking the residue classes modulo
7 raised to the fourth power.
has no solutions in integers a, b.
A2. If we expand the left hand side, we get 7a3 + 63a2 + 273a + 441.
Since all coefficients are divisible by 7, we get
7(a3 + 9a2 + 39a + 63) = b4 + (b+1)4
If we show that the right hand side can never be
divisible by 7, we are done.
We do that by taking the residue classes modulo
7 raised to the fourth power.
Modulo 7
|
|
b
|
b4
|
0
|
0
|
1
|
1
|
2
|
2
|
3
|
4
|
4
|
4
|
5
|
2
|
6
|
1
|
We see that no consecutive residue classes add up to 0 (mod 7).
Therefore, we conclude that b4 + (b+1)4 cannot be divisible
by 7.
__________________________________________________________________________
Q3. Let P(x) = x2 + ½ x + b and Q(x) = x2 + cx + d be two polynomials with real coefficients such that P(x)Q(x) = Q(P(x)) for all real x. Find all the real roots of P(Q(x)) = 0.
A3. This question comes down to just a lot of algebraic
manipulation.
Using the equality given, we get
x4 + (c + ½)*x3 + (b + c⁄2 + d)*x2 + (bc + d⁄2)*x + bd = x4 + x3 + (2b + c + ¼)*x2 + (b + c⁄2)*x + (b2 + bc + d)
Equating coefficients, we get c = ½, b = -½, d = 0.
This leaves us with a quartic equation to be solved.
x4 + x3 + (¾)x2 + (½)x - ½ = 0
Using the Rational Root Theorem, we get x = ½ as a root.
(x - ½)(x3 + (3⁄2)x2 + (3⁄2)x + 1) = 0
Using the Rational Root Theorem again, we get x = -1 as a root.
(x - ½)(x - 1)(x2 + (½)x + 1) = 0
The quadratic has no real roots, so the only solutions are x = ½ and -1.
I hope you enjoyed this post. The questions in Olympiads
are of a much higher standard than any high school maths,
even though they require no extra knowledge - only ingenuity.
If anything, I hope this gave someone a little insight into
such questions.
If you found a better solution or an error in any of the
working, please leave a comment pointing it out.
A3. This question comes down to just a lot of algebraic
manipulation.
Using the equality given, we get
x4 + (c + ½)*x3 + (b + c⁄2 + d)*x2 + (bc + d⁄2)*x + bd = x4 + x3 + (2b + c + ¼)*x2 + (b + c⁄2)*x + (b2 + bc + d)
Equating coefficients, we get c = ½, b = -½, d = 0.
This leaves us with a quartic equation to be solved.
x4 + x3 + (¾)x2 + (½)x - ½ = 0
Using the Rational Root Theorem, we get x = ½ as a root.
(x - ½)(x3 + (3⁄2)x2 + (3⁄2)x + 1) = 0
Using the Rational Root Theorem again, we get x = -1 as a root.
(x - ½)(x - 1)(x2 + (½)x + 1) = 0
The quadratic has no real roots, so the only solutions are x = ½ and -1.
__________________________________________________________________________
I hope you enjoyed this post. The questions in Olympiads
are of a much higher standard than any high school maths,
even though they require no extra knowledge - only ingenuity.
If anything, I hope this gave someone a little insight into
such questions.
If you found a better solution or an error in any of the
working, please leave a comment pointing it out.
This shared experience about the Olympiad Exam is really useful. Thanks for your efforts for the questions and solutions. Please keep on posting such insightful posts.
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