2010 Math Prize for Girls, Problems 11-15

The official solution to problem #11 doesn’t actually prove that the “snug” circle is largest possible; it just claims that it is “clear.” If it isn’t clear to you, you could proceed by showing that any circle contained inside the rhombus can be slid until it is tangent to two adjacent sides while remaining entirely within the rhombus. If such a circle were bigger than the “snug” one, it would then have to contain points outside the rhombus, which is a contradiction.

Problem #11 Since all circles with a given area are congruent and every circle can be placed inside any other circle with a larger area, this problem (or any problem asking for the areas of all circles contained inside a figure) reduces to finding the largest circle contained within. To find the largest circle, some people find it helpful to visualize growing a circle inside the region until it can’t be enlarged any further.

If you’re having trouble computing the radius of the “snug” circle, remember that lines that are tangent to a circle at a point P are perpendicular to the radius to point P as well. With that, you can use right triangle geometry to get an equation involving the unknown radius.

Problem #12 A general strategy for counting the elements of a set is to partition the set into pieces that can readily be counted. The provided solution shows one way to do this by partitioning according to the value of b. Another way is to treat this as a “balls in urns” problem and use the principle of inclusion and exclusion: Mark with an “X” the numbers representing a, b, and c and an “O” the other numbers between 1 and 17 (inclusive). Every configuration of X’s and O’s can be achieved by filling 8 O’s into the pattern


(We place 3 O’s between the X’s to guarantee the difference conditions.) However, the constraint on the problem tells us that between some two of the X’s there must be exactly 3 O’s. So, once we’ve chosen which pair of X’s will have exactly 3 O’s between them, the 8 remaining O’s (the balls) have only 3 places to go (the urns), and there are

{8+2} \choose {2}

ways to do that. Since we have two choices for which pair of X’s are separated by 3 O’s, we double this, and by the principle of inclusion and exclusion, we need to subtract the number of ways in which both the first and last pair of X’s are separated by 3 O’s (they’ve been double counted). So the answer is 45 + 45 – 9 = 81.

If you’re having trouble, the relevant subject is combinatorics. A good place to start is Ivan Niven’s book, Mathematics of Choice: How to Count Without Counting.

Problem #13 This problem exploits two key facts.  One is that the limit, as n tends to infinity, of x^n is equal to 0 if -1 < x < 1 and equal to 1 if x = 1.  The other is that the exponent of the prime p in the prime factorization of k! is given by

\lfloor \frac{k}{p} \rfloor + \lfloor \frac{k}{p^2} \rfloor + \lfloor \frac{k}{p^3} \rfloor + \cdots

Since \cos(x) = 1 only when x is an integral multiple of 2\pi, the problem reduces to determining which values of k are such that 4020 divides evenly into k!.  (Note also that in this problem, the cosine will never evaluate to -1.)

If you’re having trouble with this one, a good question to ponder is, what is the prime factorization of n!? Also, it’s fun to learn about limits.

Problem #14 If you’re having trouble, one topic that you might want to think about more is the relationship between tangency and angles, both in general, and with respect to circles. The geometric conditions given in the problem imply an algebraic condition on the radii, and the problem is to find this algebraic condition in order to solve for the radii.

Problem #15 When I see problems like this, I immediately suspect that there is some trigonometric identity that will result in massive simplification, with terms cancelling out or adding up to simple numbers all over the place. In this case, the relevant trigonometric fact is that the tangents of complementary angles are reciprocals. This can readily be seen straight from the definition of tangent as the ratio of “opposite” over “adjacent” in a right triangle. Notice, by the way, that it hardly matters what the exponent on the tangent function is. They could have made it 17 instead of 3; the answer doesn’t change.

If you’re having trouble, just remember that this type of problem is representative of a class of “cleverly constructed” contest problems that hide some slick identity which, if you see it, everything simplifies nicely, but if you don’t, you can be wandering around for hours unable to make heads or tails of it. In a contest setting, if things aren’t simplifying nicely soon, you should stop and try to think of some other identity to apply or try to find some other symmetry to exploit.


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