## 2011 Math Prize for Girls: #1-5

Here are comments and solutions to problems 1-5 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 6-10problems 11-15, and problems 16-20.

Problem #1

There are many approaches to this problem. I’ll show two after the colored text (colored so you can skip to the solutions), but first, let’s talk about contest psychology. You’ve been sitting there waiting for the official start time. You’re in a state of heightened awareness…or maybe you’re filled with adrenalin because you forgot to set your alarm clock and you woke up 5 minutes ago and rushed as fast as you could to the test center and only then realized that you’re still wearing your pajamas. Finally, the proctor says “You may begin!” You flip open the exam and see Problem #1. You’re thinking, “The first problem…should be easy…,” but you read it and no quick solution pops into your mind. “What am I missing?” You think there should be a simple answer but you don’t see it. “It’s only the first problem and I’m already confused!” Panic creeps in. The rest of the exam is one big fog. So much for your dreams of hoisting the trophy.

Hopefully, that hasn’t happened to you. But if it has, the way to prevent it from happening again is to cultivate a good attitude to all these silly contests. You like math and you’re curious to see what kinds of problems the problem committee came up with this time. You’ve studied math and you’ve solved a lot of math problems… you’ve seen dozens of telescoping sums, linear recurrence relations, simultaneous equations, trig identities, roots of unity, greatest integer functions, absolute values, inequalities, binomial coefficients, exponentials, logarithms, polynomials, distances, areas, volumes, counted sets containing all kinds of things… You’re hoping to find something new and a bit exciting! You can’t wait to encounter something that makes you go, “hmmm…never seen that before…” because you can’t wait to play with some math and piece together the solution. And if you can’t get it, so what? The mathematics profession will still be there after the exam is long gone and forgotten, and, believe me, this contest is not an entrance exam to that profession, even if you ace it.

So, I thought, fine, I’ll solve for $m$, substitute into the absolute value expression, and minimize. From $3m + 4n = 100$, I find $m = (100-4n)/3$, so I must minimize $\vert (100-4n)/3 - n \vert = \vert (100 - 7n)/3 \vert$ for values of $n$ that are integers which also make $m = (100-4n)/3$ an integer. Now, $m = (100-4n)/3$ is an integer if and only if $100-4n$ is a multiple of 3, which happens if and only if $4n$ is congruent to 1 modulo 3, which happens if and only if $n$ is congruent to 1 modulo 3. So, we minimize $\vert (100-7n)/3 \vert$ over values of $n = \dots, -5, -2, 1, 4, 7, \dots$. Because the graph of $\vert (100-7n)/3 \vert$ is symmetrically “V”-shaped with apex at $n = 100/7$, I seek the nearest such value to 100/7 = 14 2/7, so the minimum occurs at $n = 13$, and the minimum value is (100 – 7(13))/3 = 3.

Another approach is to parametrize the integral solutions to $3m + 4n = 100$. We recognize that one integral solution is $(m, n) = (0, 25)$. Because 3 and 4 are relatively prime, the other integral solutions will have the form $(4t, 25 -3t)$ for integers $t$. The expression we need to minimize is $\vert m - n \vert = \vert 4t - (25 - 3t) \vert = \vert 7t - 25 \vert$. So we seek an integer $t$ such that $7t$ is as close to 25 as possible, and this happens when $t = 4$ where the value of the expression is $\vert 7(4) - 25 \vert = 3$.

There could be a variety of reasons why one might have had trouble with this problem. Perhaps you aren’t comfortable with absolute values yet, or the integrality condition troubled you. If, during the contest, you really got stuck, don’t fret…just skip the problem and come back to it later. If you stay stuck, you could even try substituting values. With this problem, in doing so, you might well find the answer.

Problem #2

We want $\sqrt{2 + \sqrt{3}} = \frac{a+\sqrt{b}}{\sqrt{c}}$.

My first instinct is to square both sides to reduce the number of radicals in the equation:

$2 + \sqrt{3} = \frac{a^2 + b + 2a\sqrt{b}}{c}$.

Because there is a square root of 3 on the left and a square root of $b$ on the right, I’ll set $b = 3$ and try to find $a$ and $c$ to satisfy the equation. We would need

$\frac{a^2 + 3}{c} = 2$ and $\frac{2a}{c} = 1$.

The second equation means $c=2a$. Substitute this into the first equation and solve for $a$ to find $a = 1$ or $3$. So there are two answers that both express the same number subject to the constraints in the problem statement:

$\frac{1 + \sqrt{3}}{\sqrt{2}}$ and $\frac{3 + \sqrt{3}}{\sqrt{6}}$.

Problem #3

The figure that accompanies the problem statement suggests that the triangle is inscribed in the larger semicircle, i.e., that the triangle is a right triangle. So we check that $20^2 + 21^2 = 29^2$, which is true. So the apex of the triangle does, in fact, lie on all 3 semicircles and the area of the large semicircle equals the sum of the areas of the smaller semicircles. You can find the shaded area by subtracting the area of the larger semicircle from the areas of the smaller semicircles and the triangle which, therefore, leaves behind just the area of the triangle. So the answer is (20)(21)/2 = 210.

There are many other ways to solve this problem, including by computing the areas of the individual segments cut off from the large semicircle by the legs of the right triangle. The only reason those approaches are less desirable is because they are more time-consuming to carry out and more prone to computational error. If you didn’t compute the areas of the segments, you might want to try now to see that you can do it. When you finish a problem, always try to see if you could have solved the problem in a simpler way. For this one, do you think you could solve the whole thing in your head? I bet you can.

If you had trouble, aside from studying circles, triangles (especially, right triangles), and their areas, remember that the Pythagorean theorem implies that if you have 3 similar figures with corresponding lengths that are in proportion as the sides of a right triangle, then the areas of the smaller two will equal the area of the bigger one. (This leads to a nifty one-line proof of the Pythagorean theorem.)

Problem #4

I can’t recall being hit by so many logs before. This borders on contest-taker abuse.

My instincts are to make substitutions to eliminate as many logs as possible. Because $\log \log x$ appears 3 times, I’ll let $a = \log \log x$. The expression becomes

$(10^a)^{\log a} - a^a = a^a - a^a = 0$.

Hm.

If you had trouble, try to show that $(\log^{(m)} x)^{\log^{(n)} x} = (\log^{(n-1)} x)^{\log^{(m+1)} x}$, where $\log^{(n)} x$ denotes $\log \log \log \cdots \log x$, where there are $n$ application of the logarithm. (Distantly relevant is the It Figures! from Volume 5, Number 4 of the Girls’ Angle Bulletin.)

Problem #5

Because the given triangle is so “nice,” the most straightforward approach just might be to compute the desired lengths directly and take their product. We can compute the radius of the incircle by using the formula $A = sr$, where $A$ is the area of the triangle, $s$ is the semiperimeter of the triangle, and $r$ is the radius of the incircle. We find $r = 1$.

The figure illustrates the situation. We apply the Pythagorean theorem 3 times to find the distance from each vertex to the center of the incircle. We get the values indicated in the figure. The product is $\sqrt{2(5)(10)} = 10$.

If you had trouble, study properties of the incircle. Here’s a good problem to try: Prove that in general, the product of the distances of the incenter of a triangle from the 3 vertices is given by $4Rr^2$, where $R$ is the radius of the circumcircle and $r$ is the radius of the incircle.