Friday, August 29, 2014

Mathletes’ Greatest Secrets Finally Revealed Episode 0: Introduction - Why Compete?

Mathletes’ Greatest Secrets Finally Revealed
Episode 0: Introduction - Why Compete? 
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Henry


Everest North Face toward Base Camp Tibet Luca Galuzzi 2006.jpg

To echo a famous mountaineer: it's there -- will you climb?

Preliminaries

If you’re reading this, one of these things has happened:
  • You want to be the very best (in math), that no one ever was!
  • You’ve been admitted into your school math varsity and you want to know what you got yourself into.
  • You’ve accumulated experience in a few contests, but you’re stuck in a rut and you don’t know what the next step is.

From Solvespace to Project Phi

After a satisfactory run as Solvespace and a short hiatus (a student must have priorities) this blog has reached the point where I can rebrand and finally call it what I've always wanted to: Project Phi.

What's New

  • Refurbished About page.
  • An entirely new series, Mathletes' Greatest Secrets Finally Revealed.
  • An entirely new section, The Codex, "premiering" after MGSFR concludes.
  • Posts about math will often involve several sample problems, and generally will be longer.
  • PDF versions will be furnished for MGSFR, Codex articles, and new math posts.
  • A difficulty scale has been implemented and retroactively applied to the older problem posts.

Tuesday, December 31, 2013

Problem 1-9: Triangle Side Expression

Problem Posts

QUESTION
(Hong Kong Team Selection Test 2009) Let $a$, $b$, $c$ be the sides of a triangle. Determine all possible values of $$\frac{a^2+b^2+c^2}{ab+bc+ac}$$


Laconic Solution Sketch
Apply Triangle Inequality, or Ravi Transformation.

Tuesday, December 24, 2013

Problem 1-8 Meticulous Manipulations

Problem Posts

QUESTION
(2012 PMO Area Stage) If $x+y+xy=1$, where $x$, $y$ are nonzero real numbers, find the value of $$xy+\frac{1}{xy}-\frac{y}{x}-\frac{x}{y}$$

Laconic Solution Sketch
Manipulate.

Saturday, December 21, 2013

Side Story 1-2: Museum Curios

There's AM-GM. Pigeonhole. Invariance. Muirhead (to a certain extent). These are tactics and tools to swear by - use them well, and they will grant you powers beyond your wildest dreams. They will bring you places.

Then there are these theorems. Some of them are mere museum curios; I've never encountered them in any question thrown to me.

For someone who collects theorems like stamps, it is the grandest moment when a problem demands that you cockily brandish your new-found weapon of mass deduction. Especially if you've mustered the temperance to prove it yourself without consulting Google, and succeeded. More especially if you forged the masterpiece yourself. (Of course, mindlessly applying theorems is the antithesis of competitive maths.) Hence, I always find it pitiful that a few cool-looking tools end up unused, like some sort of Chekhov's Gun left hanging in the wall as the curtain falls.

Of course, all this is relative. Contests change all the time; tools could rise and fall in utility. Moreover some will say my being a mere tenderfoot in the Math Games makes me fancy rare, endangered beasts out of the standard wildlife. And I will have to grudgingly say that that's plausible.

Lagrange's Identity.
$$ \left(\sum_{k=1}^{n}a^2_{k}\right)  \left(\sum_{k=1}^{n}b^2_{k}\right) - \left( \sum_{k=1}^n a_k b_k \right)^2  = \sum_{i=1}^{n-1} \sum_{j=i+1}^n (a_i b_j - a_j b_i)^2 = \frac{1}{2} \sum_{i=1}^n \sum_{j=1}^n (a_i b_j - a_j b_i)^2  $$
Granted its similarity to the vaunted Cauchy-Schwarz Inequality, I thought this beautiful identity would figure more into algebra problems. Alas, to the best of my memory, I have never had to use it. Still, it is intuitive, and lends much insight into how things multiply out.

Fibonacci is a Square

Ira Gessel's Problem H-187:
A positive integer $n$ is a Fibonacci number if and only if either $5n^2+4$ or $5n^2-4$ is a square.
It is intuitive to surmise that this very beautiful statement has found applications in computer science. I've never had the joy of finding it to be the crux of some problem, though.


Beatty's theorem.
Given two positive irrational reals $\alpha$ and $\beta$ so that $1/\alpha + 1/\beta = 1$, the sets $\left\{ \left\lfloor \alpha\right\rfloor ,\left\lfloor 2\alpha\right\rfloor ,\left\lfloor 3\alpha\right\rfloor \ldots\right\}$ and $\left\{ \left\lfloor \mathbf{\beta}\right\rfloor ,\left\lfloor 2\beta\right\rfloor ,\left\lfloor 3\beta\right\rfloor \ldots\right\} $ form a partition of the real numbers. 
 Who would have thought? It bears a certain likeness with that problem of a walking person with an irrational footstep destined to fall into the single pothole of an otherwise smooth planet.

Routh's theorem.

$$A = \left[ ABC \right]\left( \frac{(xyz - 1)^2}{(xz + x + 1)(yx + y + 1)(zy + z + 1)}\right)$$

This should have been useful. I surmise that only pure chance was behind my never being able to use this gem of a theorem. There's little point in memorizing it, since mass points and area formulae could bring about the same result.

The Hermite Identity.
$$\sum_{k=0}^{n-1} \left\lfloor x+\frac{k}{n} \right\rfloor = \left\lfloor nx \right\rfloor$$

Bordering between "it makes sense" and "really?!", we have the intuitive Hermite identity.


Do comment if you've had similar experiences with some cool but impractical theorems.

Tuesday, December 17, 2013

Problem 1-7: Back-Engineering

Problem Posts

QUESTION
(Adapted from UVA 10784) The number of diagonals of an $n$-gon is not less than $N$. Find a closed-form expression for the minimum possible value of $n$.

Laconic Solution Sketch
A polygon has an integral side number. Use bounds.

Tuesday, December 10, 2013

Problem 1-6 Romance of a Right Triangle

Problem Posts

QUESTION
(Canadian Mathematical Olympiad 2013, Problem 3) Let $G$ be the centroid of a right-angled triangle $ABC$ with $\angle BCA = 90^{\circ}$. Let $P$ be the point on ray $AG$ such that $\angle CPA = \angle CAB$, and let $Q$ be the point on ray $BG$ such that $\angle CQB = \angle ABC$. Prove that the circumcircles of triangles $AQG$ and $BPG$ meet at a point on side $AB$.

Laconic Solution Sketch.
Using similar triangles, we can show that this common point is the foot of the altitude on $AB$ from $C$.