2011 Math Prize for Girls: #6-10

Here are comments and solutions to problems 6-10 on the 2011 Math Prize for Girls contest that took place at MIT on September 17, 2011. Earlier I blogged comments and solutions for problems 11-15 and problems 16-20.

Note: For problems on the 2011 Math Prize for Girls contest, it isn’t necessary to provide proofs that one’s answer is correct. That makes problems like #6 and #8 simpler. After the exam it can be worthwhile to produce any missing proofs and I’ll indicate how some of these proofs can go here.

Problem #6

Figure for Math Prize for Girls 2011 Problem #6

a is the measure of the indicated angle.

To minimize the area of the square, we’ll want the circles touching, because if the circles aren’t touching, one can imagine shrinking the square and squeezing the circles until they do touch.  So we need to find the area of the smallest square that will contain two touching unit circles.

The sides of the square go in two perpendicular directions. We need to measure the distance across the two circles in each of these directions. The larger distance corresponds to the smallest side length for a containing square oriented in that direction.

The distances across the two circles are given by 2 - 2 \cos a and 2 + 2 \sin a (see figure). By symmetry, we only have to consider values of a between 135 and 180 degrees. Over this range, 2 - 2 \cos a \ge 2 + 2 \sin a. The minimal value of 2 - 2\cos a over this range occurs when a is 135 degrees and the value at 135 degrees is 2 + \sqrt{2}. A square with this side length has area (2 + \sqrt{2})^2 = 6 + 4\sqrt{2}.

Another way to find the solution is to fix the square and try to fill it with the largest pair of congruent circles you can, and then scale the result so that the circles have unit radii. (Or, think this way to find the optimizing layout and then compute the area of the square on the assumption that the contained circles are unit circles.) One way this approach might go is to argue that any two congruent circles contained in the square can be moved so that they fit snugly into opposite corners of the square and that, therefore, the largest such circles would be ones that not only fit snugly in opposite corners but also touch in the center.

If you had trouble, here are some tips: make a good illustration of the situation, exploit the symmetry in the problem, and make sure you take advantage of and observe the various characteristics of the geometric shapes involved.

Problem #7

A straightforward approach is to solve the given equation for z and substitute into the desired expression.

two back-to-back equilateral triangles with unit side length

A possible image one might imagine that could make the solutions to the quadratic apparent without having to do any computation. If this image doesn't help, figure out the solutions using whatever method you prefer and then plot them on a complex plane. Compare to this figure.

The given condition, z + z^{-1} = \sqrt{3} is a quadratic condition and can be solved with the quadratic formula. But operations involving complex numbers have beautiful geometric analogues in the complex plane and it is often worth visualizing the situation on the complex plane before chugging away with algebra. You might be able to see the solution or see something that greatly simplifies the algebra and save yourself time. Here, you need two complex numbers which are reciprocals of each other and whose sum is \sqrt{3}. Can you see the solutions? If you can’t, not to worry… just solve the equation however you can. Then, after solving, plot your solutions on a complex plane and see if you can see that they are the solutions. By doing this, the next time you encounter an equation involving complex numbers that lends itself to such “visual solution” you will be more likely to see it.

In any case, the solutions are two roots of unity: z = e^{i\pi/6}, e^{-i\pi/6}.

Because the desired expression is invariant when you interchange z and z^{-1}, it doesn’t matter which root you substitute in for z. We use the fact that z^{12} = 1 to compute:

z^{2010} + z^{-2010} = z^{6} + z^{-6} = -1 + (-1) = -2.

If you aren’t familiar with complex numbers, you can solve this problem purely algebraically by computing z^n + z^{-n} recursively. For instance, notice that (z + z^{-1})(z^n + z^{-n}) = z^{n+1} + z^{-(n+1)} + z^{n-1} + z^{-(n-1)}. Also, notice that (z^n + z^{-n})^2 = z^{2n} + z^{-2n} + 2. By computing z^n+z^{-n} for various values of n, you will eventually see that the value only depends on the remainder that n leaves upon division by 12. (Or you could use identities like these to directly compute the desired expression.)

If you had trouble, study complex numbers and, especially, roots of unity. On contests, it often happens that when you see a very large exponent, there is periodicity.

Problem #8

Figure for 2011 Math Prize for Girls Problem #8We can imagine a dog attached to a pole (at A) via a leash of length 8 units. There’s an impenetrable fence stretching from B to C. Where can the dog go? The region of reachability can be deduced by applying the principle that, in a (Euclidean) plane, the shortest distance between two points is a straight line.

For example, how do we know that P cannot be reached? Any path from A to P must cross either the ray from A through B (and beyond B) or the ray from A through C (and beyond C). By symmetry, we can assume that the path crosses the ray from A through B at the point X. Because the shortest distance between two points is a straight line, the path is at least as long as the piece-wise straight path from A to X to P. Now XP cannot exceed 8 – AX. Notice that AX \ge 6, so 8 - AX \le 2. Thus, the circle of radius 8 – AX centered at X is fully contained inside the circle of radius 2 centered at B. So P is not reachable.

Now that we’ve determined the region, we can compute its area. The region consists of 3 circular sectors and an equilateral triangle. The gray circular sector has area (5/6)64\pi = 160\pi/3 square units. The two pink circular sectors each have area (1/3)4\pi for a total of 8\pi/3 square units. And the equilateral triangle has area (\sqrt{3}/4)36 = 9\sqrt{3} square units.

Thus, the total area of the reachable points is 168\pi/3 + 9\sqrt{3} = 56\pi + 9\sqrt{3} square units.

Problem #9

This problem has many points and lines defined by specifying ratios of lengths of segments along a line. A useful technique to solve such problems exploits the fact that affine transformations preserve such ratios. Because any triangle can be mapped onto any other triangle via an affine transformation, we can compute the desired ratio for any specific triangle to get the answer. (The problem statement itself implies this fact because the statement does not give any specific measurements for the triangle.)

So we’re free to specify whatever non-collinear coordinates we wish for the vertices A, B, and C. Because there are 3 halving processes and because the desired ratio pertains to segments along AB, I’ll pick the vertices so that A and B lie along a coordinate axis and all coordinates are multiples of 2^3 = 8. So set A = (0, 0), B = (0, 8), and C = (8, 0).

Then D = (4, 4), E = (2, 2), and F = (1, 5). The x-coordinate of G is 0 and its y-coordinate is the y-intercept of the line through C = (8, 0) and F = (1, 5). The equation of this line is y = -5/7(x - 8) = -5x/7 + 40/7. So G = (0, 40/7).

Thus, \frac{AG}{AB} = \frac{40/7}{8} = \frac{5}{7}.

The principle used is quite useful. For example, you can use it to show that the medians of a triangle intersect in a common point. Because midpoints define medians, one need only verify that the medians intersect in a single point for one triangle to get the result for all triangles, and by symmetry, the 3 medians of an equilateral triangle are concurrent. (For a different proof of the concurrence of a triangle’s medians, check out Volume 3, Number 2 of the Girls’ Angle Bulletin.)

Problem #10

This problem tests your algebraic skill and facility with the arctangent function. The more you know about the arctangent, the less algebra you have to do. In particular, if you know the arctangent sum formula and the relationship between arctangents of complementary angles, then we can get a solution like this:

By combining the second and third arctangents into one using the arctangent sum formula and using the fact that the arctangents of complementary angles are reciprocals of each other, we can transform the given identity to

\tan^{-1}(\frac{1}{x} - \frac{x}{8}) = \tan^{-1} \frac{1 - abx^2}{(a+b)x}.

So we seek a and b so that

\frac{8-x^2}{8x} = \frac{1-abx^2}{(a+b)x},

for all x. This can be achieved if ab = 1/8 and a+b = 1.

The problem wants the value of a^2 + b^2 which is a symmetric function in a and b, so can be expressed in terms of ab and a+b. That is, we need not solve for a and b. Instead, we note that

a^2 + b^2 = (a+b)^2 - 2ab = 1^2 - 2(1/8) = 3/4.

If you had trouble, here are the relevant topics, all worth studying: algebraic manipulation (especially for rational functions), the arctangent function, and symmetric polynomials.

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3 Responses to 2011 Math Prize for Girls: #6-10

  1. Pingback: 2011 Math Prize for Girls: #11-15 | Girls' Angle

  2. Pingback: 2011 Math Prize for Girls: #16-20 | Girls' Angle

  3. Pingback: 2011 Math Prize for Girls: #1-5 | Girls' Angle

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