Mathematical Food for Thought


			Mathematical Food for Thought

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Serves a Daily Special and an All-You-Can-Eat Course in Problem Solving. Courtesy of me, Jeffrey Wang.

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Get A Tan. Topic: Trigonometry. Level: AMC.	April 9th, 2007
Problem: (2007 MAO State – Gemini) Let be in degrees and $0^{\circ} < x < 45^{\circ}$ . Solve for : $\tan{(4x)} = \frac{\cos{x}-\sin{x}}{\cos{x}+\sin{x}}$ . Solution: Here’s a nice tangent identity that is not very well-known, but rather cool. Start with the regular tangent angle addition identity, $\tan{(x+y)} = \frac{\tan{x}+\tan{y}}{1-\tan{x} \cdot \tan{y}}$ . Letting $x = 45^{\circ}$ and $y = -\theta$ , we obtain $\tan{(45^{\circ}-\theta)} = \frac{1-\tan{\theta}}{1+\tan{\theta}} = \frac{\cos{\theta}-\sin{\theta}}{\cos{\theta}+\sin{\theta}}$ . Well, look at that. It’s the same expression as the RHS of the equation we want to solve. Substituting accordingly, it remains to solve $\tan{(4x)} = \tan{(45^{\circ}-x)} \Rightarrow 4x = 45^{\circ}-x \Rightarrow x = 9^{\circ}$ . QED. ——————– Comment: This identity is a nice one to keep around because it can turn up unexpectedly. Especially when you see that exact form and you’re like “whoa this is such a nice form there must be an identity for it.” So there. ——————– Practice Problem: (2007 MAO State – Gemini) One hundred positive integers, not necessarily distinct, have a sum of . What is the largest possible product these numbers can attain? Posted in AMC, Trigonometry \|\| 3 Comments »
MAO Results.	April 3rd, 2007
Check them out online! http://www.wamath.net/contests/StateMAT/ Posted in Announcements \|\| 2 Comments »
Sumsine. Topic: Trigonometry/Geometry. Level: AMC.	April 1st, 2007
Problem: (2007 MAO State – Gemini) Find the sum of the sines of the angles of a triangle whose perimeter is five times as large as its circumradius. Solution: At first, this problem seems very strange because we do not expect a nice relationship between the perimeter and the circumradius to offer much, but take another look. Sines, circumradius… Law of Sines! In fact, we get $\frac{a}{\sin{A}} = \frac{b}{\sin{B}} = \frac{c}{\sin{C}} = 2R$ so we can rewrite $\sin{A}+\sin{B}+\sin{C}$ as $\sin{A}+\sin{B}+\sin{C} = \frac{a}{2R}+\frac{b}{2R}+\frac{c}{2R} = \frac{a+b+c}{2R} = \frac{5R}{2R} = \frac{5}{2}$ . QED. ——————– Comment: Admittedly, MAO does not exactly have a plethora of cool problems, but this one was not bad. It wasn’t too difficult, but it required some clever use of well-known identities to pull it off. Given that half of the problems were computationally intensive (think AIME I but worse), this was a nice relief in the middle of the contest. ——————– Practice Problem: (2007 MAO State – Gemini) The zeros of form an arithmetic sequence. Let $K = \frac{m}{n}$ , where and are relatively prime positive integers. Find the value of . Posted in Geometry, Trigonometry \|\| 2 Comments »
2007 MAO.	March 31st, 2007
Good job to all Bellevue people at MAO State! We placed 3rd in Sweepstakes! That’s the highest we’ve ever gotten in my four years . Posted in Announcements \|\| 1 Comment »
Condensation Sensation. Topic: Calculus/S&S.	March 27th, 2007
Theorem: (Cauchy Condensation Test) If $\displaystyle \{a_n\}_{n = 0}^{\infty}$ is a monotonically decreasing sequence of positive reals and is a positive integer, then $\displaystyle \sum_{n=0}^{\infty} a_n$ converges if and only if $\displaystyle \sum_{n=0}^{\infty} p^n a_{p^n}$ converges. ——————– Problem: Determine the convergence of $\displaystyle \sum_{n=2}^{\infty} \frac{1}{(\ln{n})^k}$ , where is a positive real. Solution: Well, let’s apply the Cauchy condensation test. Then we know that $\displaystyle \sum_{n=2}^{\infty} \frac{1}{(\ln{n})^k}$ converges if and only if $\displaystyle \sum_{n=2}^{\infty} \frac{p^n}{(\ln{p^n})^k} = \sum_{n=2}^{\infty} \frac{p^n}{(n \ln{p})^k} = \frac{1}{(\ln{p})^k} \sum_{n=2}^{\infty} \frac{p^n}{n^k}$ does. But this clearly diverges due to the fact that the numerator is exponential and the denominator is a power function. QED. ——————– Comment: This is a pretty powerful test for convergence, at least in the situations in which it can be applied. The non-calculus proof for the divergence of the harmonic series is very similar to the Cauchy condensation test; in fact, the condensation test would state that $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n}$ converges iff $\displaystyle \sum_{n=1}^{\infty} \frac{p^n}{p^n}$ converges, which clearly shows that the harmonic series diverges. ——————– Practice Problem: Determine the convergence of $\displaystyle \sum_{n=2}^{\infty} \frac{1}{(\ln{n})^{\ln{n}}}$ . Posted in Calculus, Sequences & Series \|\| 5 Comments »

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