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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Something To Think About. Topic: Probability. Level: AMC/AIME.	April 18th, 2007
Problem: A coin is repeatedly flipped. What is the expected number of flips to get two heads in a row? heads in a row? ——————– I will be out of town for the next four days, so have fun with the problem! Posted in AIME, AMC, Probability & Combinatorics \|\| 4 Comments »
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 »
Pythagorean x3. Topic: Geometry/NT. Level: AMC.	February 22nd, 2007
Problem: (2007 AMC12B – #23) How many non-congruent right triangles with positive integer leg lengths have areas that are numerically equal to times their perimeters? Solution: Well, basically, you should know the Pythagorean triple generating formula, i.e. , , . Substitute accordingly and we have to solve the diophantine equation $\frac{1}{2} \cdot k(u^2-v^2) \cdot 2kuv = 3 \cdot (k(u^2-v^2)+2kuv+k(u^2+v^2))$ which conveniently simplifies to . Obviously then so look at these cases: : We can take to get the triples . : We can take to get the triples both of which are already counted. : We can take to get the triples . : We can take to get the triple , which is already counted. So we have triangles. QED. ——————– Comment: Not too hard if you knew the generating formula for Pythagorean triples. It was a little annoying having to check for repeated triples, but at least there weren’t that many. ——————– Practice Problem: (2007 AMC 12B – #24) How many pairs of positive integers are there such that $\text{gcd}(a, b) = 1$ and $\frac{a}{b}+\frac{14b}{9a}$ is an integer? Posted in AMC, Geometry, Number Theory \|\| 2 Comments »
Don’t You Wish You Remembered Those Trig Identities. Topic: Trigonometry. Level: AMC/AIME.	February 9th, 2007
Problem: (1962 IMO – #4) Solve the equation $\cos^2{(x)}+\cos^2{(2x)}+\cos^2{(3x)} = 1$ . Solution: Subtract the $\cos^2{(3x)}$ from both sides, and replace the LHS by $\sin^2{(3x)}$ , so we now have $\cos^2{(x)}+\cos^2{(2x)} = \sin^2{(3x)}$ . Recalling the sine angle addition identity, we write $\sin{(3x)} = \sin{(2x+x)} = \sin{(2x)}\cos{(x)}+\sin{(x)}\cos{(2x)}$ . So our equation is $\cos^2{(x)}+\cos^2{(2x)} = [\sin{(2x)}\cos{(x)}+\sin{(x)}\cos{(2x)}]^2$ . Expanding, collecting terms, and simplifying using $1-\sin^2{(2x)} = \cos^2{(2x)}$ and $1-\sin^2{(x)} = \cos^2{(x)}$ , we get $2\cos^2{(x)} \cos^2{(2x)} = 2\sin{(x)}\cos{(x)}\sin{(2x)}\cos{(2x)}$ . Divide by , rearrange, and factor: $\cos{(x)}\cos{(2x)}[\cos{(x)}\cos{(2x)}-\sin{(x)}\sin{(2x)}] = 0$ . Substitute using the double-angle identities $\cos{(2x)} = 1-2\sin^2{(x)}$ and $\sin{(2x)} = 2\sin{(x)}\cos{(x)}$ to finally find $\cos^2{(x)} \cos{(2x)} [1-4\sin^2{(x)}] = 0$ . Solving yields $x = \frac{\pi}{2}+k\pi \mbox{ ; } \frac{\pi}{4}+\frac{k \pi}{2} \mbox{ ; } \frac{\pi}{6}+k\pi \mbox{ ; } \frac{5 \pi}{6}+k \pi$ for all integers . QED. ——————– Comment: Pretty standard and slightly ugly manipulation of trig identities, but overall not too bad. Considering you get like an hour or more to do this, I don’t feel too bad about it. Other solutions taking into account symmetry or using DeMoivre’s and stuff are also out there, but this seemed much more straightforward and easier to develop. ——————– Practice Problem: (2007 AMC 12A – #24) For each integer , let be the number of solutions of the equation $\sin{(x)} = \sin{(nx)}$ on the interval $[0, \pi]$ . What is $\displaystyle \sum_{n=2}^{2007} F(n)$ ? Posted in AIME, AMC, Trigonometry \|\| 3 Comments »
Spacy. Topic: S&S/Sets. Level: AMC.	February 7th, 2007
Problem: (2007 AMC 12A – #25) Call a set of integers spacy if it contains no more than one out of any three consecutive integers. How many subsets of $\{1, 2, 3, \ldots, 12\}$ , including the empty set, are spacy? Solution: We will solve this problem with a recursion on , which we will define as the number of spacy subsets of $\{1, 2, 3, \ldots, n\}$ . We can easily calculate , , , and . Now we will think about how to evaluate in terms of previous terms. Obviously, we can keep all the subsets in $S_{n-1}$ . These will account for any space subsets that don’t contain the element . Now we just need to count the ones that do contain the element . If a spacy subset contains , it cannot have or . So take all the spacy subsets of $\{1, 2, \ldots, n-3\}$ and add the element to them to get all the spacy subsets that contain . Since these are both of the cases, we obtain $S_n = S_{n-1}+S_{n-3}$ . Some calculations yield $S_{12} = 129$ . QED. ——————– Comment: In all honesty, a pretty standard recursion problem, which is probably better off on an easy olympiad-style contest because it’s too easy to guess on the AMC. After just calculating a few terms, it wasn’t hard to see the recursion and assume its validity. A more interesting problem would be to find the closed form for that recursion (eww?). Maybe generating functions will save the day. ——————– Practice Problem: Can you find a closed form for ? Posted in AMC, Sequences & Series, Sets \|\| 1 Comment »

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