Sunday, November 27, 2005

Circular Reasoning. Topic: Geometry. Level: AMC/AIME.

Problem #1: (2006 Mock AIME 1 - #1) $2006$ points are evenly spaced on a circle. Given one point, find the number of other points that are less than one radius distance away from that point.

Solution: We know that one radius distance is equivalent to $ \frac{1}{6} $ of the circumference (by constructing an equilateral triangle with the center and two points on the circle).

We know each point is $\frac{1}{2006} $ of the circumference away from each other, so the $k$th point is $ \frac{k}{2006} $ arc distance away. We want $ \frac{k}{2006} < \frac{1}{6} \Rightarrow k \le 334 $. But since it can be on either side, we double that, to get $668.$. QED.


Comment: Note that this following AMC-12 problem is considerably more difficult than the preceding AIME problem, though it happens quite often that a AMC-12 #25 is more difficult than an AIME #1.


Problem #2: (2003 AMC 12B - #25) Three points are chosen randomly and independently on a circle. What is the probability that all 3 pairwise distances between the points are less than the radius of the circle?

Solution : Now on the surface it looks pretty easy, but remember, this is a #25 - chances are it won't come too quickly for an average problem-solver (it's easy to jump into a bunch of flawed arguments). First convert everything to arc lengths - one radius becomes $\frac{1}{6}$ circumference again. Consider the following argument:

2003 AMC-12B #25

The circumference is set to $1$. Let point $A$ be the "center" of the three points, the "center" being the point that both the other points are closest to (note that the existence of the "center" is guaranteed - think about it).

Call the shortest distance between two points to be $x$ (first assume $AB = x$) - we then have the other length $1-x$. Now think about the possibilities... we want to maintain

1. The minimum distance of $x$. So the arc $AC$ has to be greater than $x$ - $AC > x$.

2. The "center" status of $A$. So the arc $BC$ has to be greater than $AC$ so $ AC < \frac{1-x}{2} $ for a given $x$. Also, $ AB $ must be less than $ BC $ (which can be as small as $\frac{1-x}{2}$), so we have $ x < \frac{1-x}{2} \Rightarrow x < \frac{1}{3}$.

(1) We can graph this such that $AB$ is on the $x$-axis and $AC $ is on the $y$-axis following these rules: $ 0 < AB < \frac{1}{3} $ and $ AB < AC < \frac{1-AB}{2} $.

(2) Now recall that we assumed $AB = x$. But $ AC $ could be $x$ as well, so we apply the same argument, flipping $AB$ and $AC$. But the graph would be the same, only flipped over the line $y=x$, creating this:

2003 AMC-12B #25 Graph

where (1) creates the red region and (2) creates the blue region (both theoretically extending into the yellow region).

Those generate the total "number" of unique possibilites, denoted by the area beneath the graph. Now we have to find the ones that satisfy our given condition: the maximum arc length is less than $ \frac{1}{6} $.

However, we notice by introducing the center status of $A$ we automatically show that our maximum arc length is $AB+AC$ because it is always greater than the independent lengths ($AB+AC > AB$ and $AB+AC > AC$ - of course this is only when $AB$ and $AC$ are both less than $\frac{1}{6}$, which is all we care about). So we add to our graph the yellow region, $AB+AC < \frac{1}{6}$.

Thus the probability is (including the extensions of the red and blue regions into the yellow) $ \frac{[yellow]}{[red]+[blue]} = \frac{\frac{1}{72}}{\frac{1}{6}} = \frac{1}{12} $. QED.


Comment: On the actual contest, the argument would probably be loosely made in the interest of saving time and you could probably finish something like this in 5-10 minutes at most if you knew where you were headed. Of course, given that this is the last problem of an AMC-12, I wouldn't be surprised if most people didn't have those 5-10 minutes to spare.


Practice Problem #1: Find the probability that two points placed on a circle randomly and independently are within one radius distance of each other.

Practice Problem #2: In an acute angled triangle $ABC$, $\angle A = 30^{\circ}$. $H$ is the orthocenter and $M$ is the midpoint of $BC$. On the line $HM$, take point $T$ such that $HM=MT$. Show that $AT= 2BC$. (hint: Use complex numbers - see Post 13: November 24th, 2005).

Practice Problem #3: Three congruent circles of radius 1 intersect at a common point. A larger circle is circumscribed about them, tangent to each of them at one point. Consider the midpoint of one of the diameters of the smaller circles; call it $P$. What is the length of the arc along the bigger circle containing all the points that are within 2 units of $P$?

1 comment:

  1. [...] 6 : For a revamped solution, see here. The same idea, hopefully better explained. This entry was posted in math. Bookmark the [...]