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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Pick one, any one! Topic: Algebra. Level: Olympiad	December 31st, 2005
Problem: (1989 USAMO – #5) Let and be real numbers such that $(u + u^2 + u^3 + \cdots + u^8) + 10u^9 = (v + v^2 + v^3 + \cdots + v^{10}) + 10v^{11} = 8$ . Determine, with proof, which of the two numbers, or , is larger Solution: Consider the functions $f(x) = 1+x+\cdots+x^8+10x^9$ and $g(x) = 1+x+\cdots+x^{10}+10x^{11}$ , which are monotonically increasing on the interval $(0,\infty)$ . We are given that . We have $g(x)-f(x) = 10x^{11}+x^{10}-9x^9 = x^9(10x-9)(x+1)$ . $f\left(\frac{9}{10}\right) = 1+\frac{9}{10}+\cdots+\frac{9^8}{10^8}+10 \cdot \frac{9^9}{10^9}$ . By summing the geometric series, $f\left(\frac{9}{10}\right) = \frac{1-\frac{9^9}{10^9}}{1-\frac{9}{10}} + 10 \cdot \frac{9^9}{10^9}= 10-10 \cdot \frac{9^9}{10^9}+10 \cdot \frac{9^9}{10^9} = 10 > 9$ . Since f is an increasing function, $f(u) < f\left(\frac{9}{10}\right) \Rightarrow u < \frac{9}{10}$ . Hence $g(u)-f(u) = u^9(10u-9)(u+1) < 0 \Rightarrow g(u) < f(u) = 9$ . But since is an increasing function as well, we know $g(u) < g(v) = 9 \Rightarrow u < v$ . QED. ——————– Comment: The last problem of the 1989 USAMO, so supposedly it is difficult, but that wasn’t really the case. A simple analysis of and gives us the answer quickly. ——————– Practice Problem: (2002 USAMO – #4) Let $\mathbb{R}$ be the set of real numbers. Determine all functions $f: \mathbb{R} \to \mathbb{R}$ such that for all pairs of real numbers and . Posted in Algebra, Olympiad \|\| 5 Comments »
All Those Digits. Topic: Number Theory. Level: Olympiad.	December 28th, 2005
Problem: (2003 USAMO – #1) Prove that for every positive integer there exists an -digit number divisible by all of whose digits are odd. Solution: We will prove the result by induction. Base Case: – works. Suppose there exists a -digit positive integer with all odd digits that is divisible by . Call this integer . Consider the following numbers: , all of which have digits. Let . Then the above numbers become . Notice that are all different modulo (since they are in an arithmetic sequence with difference $2^{k+1}$ and $(2^{k+1},5) = 1$ ). Therefore one of the numbers must be divisible by . Let that one be . So one of the original numbers is , where is divisible by . Hence the number is divisible by $5^{k+1}$ and has digits, as desired. This completes the induction. QED. ——————– Practice Problem: (1994 USAMO – #1) Let $k_1 < k_2 < k_3 < \cdots$ , be positive integers, no two consecutive, and let $s_m = k_1 + k_2 + \cdots + k_m$ for $m = 1,2,3, \ldots$ . Prove that, for each positive integer , the interval $[s_n, s_{n+1})$ contains at least one perfect square. Posted in Number Theory, Olympiad \|\| 2 Comments »
Rounding Time! Topic: Number Theory. Level: AIME.	December 24th, 2005
Problem: (1995 AIME – #13) Let be the integer closest to $\sqrt[4]{n}$ . Find $\displaystyle \sum_{k=1}^{1995} \frac{1}{f(k)}$ . Solution: So how can we best define “integer closest?” It basically means iff $x-\frac{1}{2} < \sqrt[4]{n} < x+\frac{1}{2}$ . Note we may use strict bounds because we know $\sqrt[4]{n}$ will not have fractional part $\frac{1}{2}$ . So we can find bounds for such that . We must have $\left(x-\frac{1}{2}\right)^4 < n < \left(x+\frac{1}{2}\right)^4$ . Expanding, we get $x^4-2x^3+\frac{3}{2}x^2-\frac{1}{2}x+\frac{1}{16} < n < x^4+2x^3+\frac{3}{2}x^2+\frac{1}{2}x+\frac{1}{16}$ from which we get integral solutions (subtract lower bound from upper). Thus the summation can be divided into separate cases based on the value of . Since there are terms and each has value $\frac{1}{x}$ , they sum to $\frac{1}{x}(4x^3+x) = 4x^2+1$ . Since , we may only take this from to , giving $\displaystyle \sum_{i=1}^6(4i^2+1) = 4 \cdot \frac{(6)(6+1)(2 \cdot 6+1)}{6}+6 = 370$ . But this only accounts for $\displaystyle \sum_{i=1}^6(4i^3+i) = 4\left(\frac{(6)(6+1)}{2}\right)^2+\frac{(6)(6+1)}{2} = 1785$ terms. So we have remaining terms, all of which are $\frac{1}{7}$ , giving the new total of $370+\frac{1}{7}(210) = 400$ . QED. ——————– Comment: At this point, we’re happy. Why? Because it’s an AIME problem and we got an integer answer! And the chances of us doing it wrong and getting an integer answer is not too high, so we are probably right. ——————– Practice Problem: Find the smallest such that $\displaystyle \sum_{k=1}^m \frac{1}{f(k)}$ exceeds . Posted in AIME, Number Theory \|\| 3 Comments »
Back Again? Topic: Inequalities. Level: Olympiad.	December 21st, 2005
Problem: (2003 USAMO – #5) Let be positive real numbers. Prove that $\frac{(2a+b+c)^2}{2a^2+(b+c)^2}+\frac{(2b+c+a)^2}{2b^2+(c+a)^2}+\frac{(2c+a+b)^2}{2c^2+(a+b)^2} \le 8$ . Solution: Well first of all we notice that the inequality is homogeneous. Therefore, we can set to an arbitrary value. Note that using works particularly well. Then the inequality becomes $\displaystyle \sum \frac{(3+a)^2}{2a^2+(3-a)^2} = \sum \frac{a^2+6a+9}{3a^2-6a+9} = \sum \left(\frac{1}{3}+\frac{8a+6}{3(a-1)^2+6}\right)$ . Well, noticing that $\frac{8a+6}{3(a-1)^2+6} \le \frac{8a+6}{6} = 1+\frac{4a}{3}$ we have $\displaystyle \sum \left(\frac{1}{3}+\frac{8a+6}{3(a-1)^2+6}\right) \le \sum \frac{4}{3}(1+a) = \frac{4}{3}(3+a+b+c) = \frac{4}{3}(6) = 8$ as desired. QED. ——————– Comment: This was a pretty cool problem, too. The most important part was splitting the fraction and getting rid of the quadratic term. After both the squared terms are gone, it’s pretty simple to finish off with only linear and constant terms. Posted in Inequalities, Olympiad \|\| No Comments »
See Through It. Topic: Inequalities. Level: Olympiad.	December 20th, 2005
Problem: (2004 USAMO – #5) Let . Prove that $(a^5-a^2+3)(b^5-b^2+3)(c^5-c^2+3) \ge (a+b+c)^3$ . Solution: Note that and always have the same sign. Therefore $(a^2-1)(a^3-1) \ge 0 \Rightarrow a^5-a^2+3 \ge a^3+2$ . Then $(a^5-a^2+3)(b^5-b^2+3)(c^5-c^2+3) \ge (a^3+1+1)(1+b^3+1)(1+1+c^3) \ge (a+b+c)^3$ by Holder. QED. ——————– Comment: This is probably my favorite USAMO problem just because the solution is so short and simple. And the fact that I love inequalities. But honestly, can you get another USAMO problem (especially a #5) to give such a quick solution? ——————– Practice Problem: (2003 USAMO – #5) Let be positive real numbers. Prove that $\frac{(2a+b+c)^2}{2a^2+(b+c)^2}+\frac{(2b+c+a)^2}{2b^2+(c+a)^2}+\frac{(2c+a+b)^2}{2c^2+(a+b)^2} \le 8$ . Posted in Inequalities, Olympiad \|\| 1 Comment »

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