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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AP Tests.	April 30th, 2006
Good luck to everyone on AP Tests! Posted in Announcements \|\| 1 Comment »
2006 Bellevue BATH Competition.	April 26th, 2006
Everyone in Washington should come to the first ever Bellevue BATH Competition! Details can be found here. Posted in Announcements \|\| 35 Comments »
2006 USAMO.	April 17th, 2006
A three-day break on my blog for the USAMO! Good luck everyone. Posted in Olympiad \|\| 12 Comments »
Perimeter To Angle? Topic: Geometry/Trigonometry. Level: Olympiad.	April 16th, 2006
Problem: (1986 China TST – #5) Given a square whose side length is , and are points on the sides and , respectively. If the perimeter of is find the angle . Solution: Let and $\angle PCB = \alpha, \angle QCD = \beta$ . By the given condition, we have $AP+AQ+QP = x+y+\sqrt{x^2+y^2} = 2$ (1). From this, we find $x+y = 2-\sqrt{x^2+y^2}$ $(x+y)^2 = 4-4\sqrt{x^2+y^2}+(x^2+y^2)$ $xy = 2-2\sqrt{x^2+y^2}$ Substituting $\sqrt{x^2+y^2} = 2-x-y$ from (1), we have $xy = 2-2(2-x-y) \Rightarrow 2-x-y = x+y-xy \Rightarrow \frac{2-x-y}{x+y-xy} = 1$ (2). Note that $\tan{\alpha} = \frac{PB}{BC} = 1-x$ and $\tan{\beta} = \frac{QD}{DC} = 1-y$ . Then $\tan{(\alpha+\beta)} = \frac{\tan{\alpha}+\tan{\beta}}{1-\tan{\alpha} \cdot \tan{\beta}} = \frac{(1-x)+(1-y)}{1-(1-x)(1-y)} = \frac{2-x-y}{x+y-xy}$ . But by (2), we have $\tan{(\alpha+\beta)} = \frac{2-x-y}{x+y-xy} = 1 \Rightarrow \alpha+\beta = 45^{\circ}$ . Hence $\angle PCQ = 90^{\circ}-(\alpha+\beta) = 45^{\circ}$ . QED. ——————– Comment: Trigonometry can come in handy quite often, especially when dealing with angles. Purely geometric solutions to this problem are a lot more complicated in my opinion; when in doubt, use algebra. The following identity comes in handy on quite a few problems. ——————– Practice Problem: Prove that in any triangle we have $\tan{\frac{A}{2}} \cdot \tan{\frac{B}{2}} + \tan{\frac{B}{2}} \cdot \tan{\frac{C}{2}} + \tan{\frac{C}{2}} \cdot \tan{\frac{A}{2}} = 1$ . Posted in Geometry, Olympiad, Trigonometry \|\| 8 Comments »
That’s A Lot Of Numbers. Topic: Logic/NT. Level: Olympiad.	April 15th, 2006
Problem: (ACoPS 3.4.21) Initially, we are given the sequence $1,2, \ldots, 100$ . Every minute, we erase any two numbers and and replace them with the value . Clearly, we will be left with just one number after minutes. Does this number depend on the choices we made? Solution: We claim that the number does not depend on the choices. Let $a_1, a_2, \ldots, a_n$ be the numbers on the board after minutes. Define $\displaystyle P_n = \prod_{i=1}^n (a_i+1)$ . We claim that the sequence $P_{100}, P_{99}, \ldots, P_1$ is constant. Noticing that , we see that replacing with has no effect on the product . Indeed, $P_{k-1} = \frac{P_k(uv+u+v+1)}{(u+1)(v+1)} = P_k$ so we have $P_{100} = P_{99} = \cdots = P_1$ by induction. Since $P_{100}$ is clearly constant, must be as well. Seeing that is equivalent to the remaining number, we conclude that it does not depend on the choices, as desired. QED. ——————– Comment: Problems with invariants require you to discover something (often a sum or product) that doesn’t change with each step. The term should remind you of Simon’s favorite factoring trick, leading to the invariant as described above. ——————– Practice Problem: (ACoPS 3.4.23) Start with the set $\{3,4,12\}$ . You are then allowed to replace any two numbers and with the new pair and . Can you transform the set into $\{4,6,12\}$ ? Posted in Logic, Number Theory, Olympiad \|\| 3 Comments »

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