Growing up in Spata (no, not Sparta – but feel free to ignore this remark) there was not much to do in the evenings. After school was done and volleyball practice was over (with my two brothers we made up half of the school team) my dad would come pick us up for a fun three hours of track and field practice. Just another lazy evening. Who am I kidding… It was exhausting! But, throwing a javelin with exuberant fury was also therapeutic (it’s a Greek thing). Yet, here lied the problem: The adrenaline high from a good five hours of sports every day would not dissipate simply because of physical exhaustion. I don’t know about my brothers, but my brain was on fire and the two pounds of pasta my mom would put on my plate (almost) every night, could not induce a strong enough food coma. Even working on the next day’s homework did not do the trick of putting me to sleep (though it did help significantly). By then, it was past midnight and I was wide awake.

Up until I was about 13, the answer to my insomnia was TV. I watched so much television back then that Roger Ebert calls me up these days to discuss the subtleties of modern superhero movies (like Hitchcock and Stan Lee, I also have cameos in all the movies I have produced – true fact). Yet, when everyone around me started falling in love with their classmates, I started falling in love with Math. My hormones were playing a cruel joke on me, because, even though I found myself attracted to every living female between the ages of 13-30 (I was definitely a teenager), my mind was mainly preoccupied with numbers and triangles. I wasn’t even a bona-fide nerd, but if someone mentioned a puzzle, or a problem that was too hard for adults to solve, my ears perked up like a rabbit in heat (actually, I am not sure their ears perk up, but who cares – it sounds right).

So what was the source of my mental disease? Around the time I hit puberty, a math teacher from my high school (Athens College in Psychico, Athens, was arguably the premier high school in Greece, educating the children of the Greek aristocracy, as well as “scholarship kids” like myself) decided to form a team of 7th – 12th graders who seemed to be good at math (chosen by our GPA, I guess?) It was terrifying being chosen to participate in the group activities of super-smart students from the top high school in the country. I was a jock, not a mathlete! I had competed in several national and international sports events by that time, but I had never faced the nerve racking silence of 90 minutes of feeling inadequately stupid against 6 elementary math problems; especially when I could hear all the other geniuses in the room writing furiously on their pads. What were they calculating? How did they even know where to begin? Well, they didn’t. It turned out that they were as lost as I was during our first problem solving exam. In fact, I solved the most problems that day (it was a number greater than 1), so I became the *de facto* leader of the nerds (they were the coolest nerds in the world, if I may say so myself.) And I was hooked. The idea that an average guy like me could solve impossible-looking problems just through sheer stubbornness (oh well, you work with what you are given) was exhilarating.

But enough about me and my crazy childhood. This post is supposed to be an introduction to the insanely beautiful world of problem solving. It is not a world ruled by Kings and Queens. It is a world where commoners like you and me can become masters of their domain and even build an empire. But as every aspiring ruler-of-the-world knows, you have to start small. So here are two problems to set you on your path to greatness:

**Problem 1: ***Of the two numbers, $latex 17^{14}$ and $latex 31^{11}$, which one is larger?*

**Problem 2 (harder): ***If $latex a+A=b+B=c+C=k$ and all numbers are positive, show that $latex aB+bC+cA le k^2$.*

Let the games begin!

Konstantin KraynAugust 30, 2012 at 9:21 am1. >, consider log on base 2 of both parts (56 plus something > 55 minus something).

2. Let a <= b <= c. Then left part is a(k-b) + b(k-c) +c(k-a) <= a(k-a) + c(k-c) + c(k-a) = -a^2 + (k-c)a +2ck – c^2; max at a=(k-c)/2 is (-3c^2 +6kc+ k^2)/4; max at c=k is k^2. QED.

spirosAugust 30, 2012 at 9:49 amThank you Konstantin! Now that you have found a solution to the problems, the real challenge begins: Can you solve the first problem without a calculator and the second in less than one line?

Konstantin KraynAugust 30, 2012 at 10:01 amMy solution of #1 is without a calculator — no need for a calculator to see that log(2)17 > 4 (as 2^4 = 16) and log(2)31 < 5 (as 2^5 = 32).

My solution of #2 is indeed much longer than "less than one line", so here the real fun begins…

Konstantin KraynAugust 30, 2012 at 10:40 am#1 can be re-written without logs to make it look shorter:

17^14 > 16^14 = (2^4)^14 = 2^56 > 2^55 = (2^5)^11 = 32^11 > 31^11

#2: I think I got it:

Left part is simply (k^3 – abc – ABC)/k <= k^2

spirosAugust 30, 2012 at 1:24 pmYou are the genius of the week Konstantin! Short and elegant solutions to both problems. For those wondering, the product (a+A)(b+B)(c+C) is the thing to look at for problem #2. I posted this problem on my Facebook page and after a few iterations, Greg Kuperberg posted the shortest answer (though I like the solution involving a single square very much!) Can you find it?

DebbieAugust 30, 2012 at 12:18 pmSpiros: for Q2, do you call 1 diagram more or less than 1 line ðŸ™‚ ?

Konstantin KraynAugust 30, 2012 at 1:09 pmI had the same question in mind and hypothesised that this diagram is like three lines — one line for each dimension. ðŸ™‚

(If we are talking about the same diagram.)

spirosAugust 30, 2012 at 1:12 pmI would say that a diagram is less than one line if its fractal dimension is less than one ðŸ™‚

DebbieAugust 30, 2012 at 6:54 pm6 lines in my diagram.

spirosAugust 31, 2012 at 6:15 amSend me your diagram and I will post it next week in the solutions section of this week’s problems.

Dexter Kim (@AstralDexter)August 30, 2012 at 9:33 pm#1 17^3>2^11=2048>(31/17)^11

I’ve memorised powers of 2 up to the tenth so this was easy.

#2 well, this uses inequality of arithmetic/geometric means

2.1 (inequality) (a+A)/2 = k/2 >= sqrt(aA)

2.2 (square 2.1) (k^2)/4 >= aA

2.3 (write for b&c, then add up) 3(k^2)/4 >= aA + bB + cC

Dexter Kim (@AstralDexter)August 30, 2012 at 9:39 pmhmm. seems like I read the problem wrongly.

spirosAugust 31, 2012 at 12:56 amDon’t worry Dexter. I read the problem the same way the first time and gave the same answer. Since there was noone around at the time, I convinced myself it never happened.

ConstantinosAugust 31, 2012 at 10:55 amGreat piece Spyro! It sprang dear memories from some 15 years ago!

Welcome to the math olympics | Quantum FrontiersSeptember 6, 2012 at 5:26 pm[…] Greek Math Olympiad and I convinced a bunch of my friends from school (remember the cool nerds from Geniuses wanted?) to wake up early on a Saturday and go test our mettle against the rest of the Greek population […]