Sunday 21 December 2008

Solving a Zimbabwe Math Olympiad - Part 1

Today, while reading this entry in bit-player, I saw a link to a Maths Olympiad (an Olympiad is supposedly: period of four years, by which the ancient Greeks reckoned time, being the interval from one celebration of the Olympic games to another. I don't know why they call this an Olympiad...)

There were 20 questions from various areas of math, and they seemed quite easy and that was attractive to me (as my math workouts these days are very less). So, I've started solving them, and am gonna post my modus operandi for each problem as I solve it.

The first problem was,
If log2(log3 a) = 2, what is the value of a?
Quite very easy, if you ask me. First, to remove any logarithm, you've got to raise both sides to the power of the base of the log. So, here, the first step would be, to raise both sides to the power of 2, so we'll end up with 2^(log2(log3a)) = 2^2 which would reduce to log3a = 4 (since any x to the power of a logxy will leave us with y alone.)
Now, formally, I should have used the same method to remove the next log also. Instead, I found myself thinking, 'Which number's log to the base 3 will be 4?'. The answer to this was stored in my memory's cache as '81', so I determined that was the answer. And I did the ordinary computation method too in order to verify my answer, since the answers don't seem to have been given in that page...

The second question was:
In triangle ABD, the angular bisector of angle DAB meets BD at C. If CD=6, BD=10 and AD=9, what is the length of AB?
Ok, this wasn't as easy to me. Geometry has never been a particularly favourite subject of mine, and the theorems were all things I learnt to pass through the exams. I remembered I had read somewhere about angle bisectors and things, but that was all I remembered. :)
No worries, Google to the rescue. I searched for properties of angle bisectors and ended up in this interactive page where they have some Java applet. Fortunately, my NoScript Firefox extension blocked the Java thingies (they slow down my system badly!) and I had what I wanted scrolling down a little itself:

Angle bisectors divide the opposite side in the ratio of the adjacent sides. More accurately, if, in ABC, AD is an angle bisector of angle A, then


AB/AC = DB/DC
So now, I have AD, CD and CB, I just need to apply this formula (note that C and D have reversed roles in the problem and the formula.)

For us, AB/AD = CB/CD which means AB/9 = 4/6 which gives us AB = 6.
I think I cheated by Googling, but what the heck! ;)

to be continued... (as I solve more problems)




1 comment:

Randy A MacDonald said...

Good to see multiple bogs covering Project Euler. This one does as well: http://my.opera.com/arcfide/blog/

Also, the sine law for triangles, and the sine of a supplementary angle seemed like enough to get this one.