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 »
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 »
Common Law. Topic: Geometry/Trigonometry. Level: AIME.	February 16th, 2007
Problem: (2003 AIME1 – #12) In convex quadrilateral , $\angle A = \angle C$ , , and $AD \neq BC$ . The perimeter of is . Find $\lfloor 1000 \cos{\angle A} \rfloor$ . Solution: Well, let’s say that and . We know $x+y+180+180 = 640 \Rightarrow x+y = 280$ . Consider the two triangles and . Using the Law of Cosines on $\angle A$ and $\angle C$ , which we know are equal (let them both equal $\theta$ ), we have $BD^2 = AB^2+AD^2-2 \cdot AB \cdot AD \cos{\theta}$ $BD^2 = CB^2+CD^2-2 \cdot CB \cdot CD \cos{\theta}$ . Substituting and , , it becomes $BD^2 = 180^2+y^2-360y \cos{\theta}$ $BD^2 = x^2+180^2-360x \cos{\theta}$ . Equating the two, $180^2+y^2-360y \cos{\theta} = x^2+180^2-360x \cos{\theta} \Rightarrow x^2-y^2 = 360(x-y)\cos{\theta}$ . Recall that , so $x^2-y^2 = (x+y)(x-y) = 280(x-y) = 360(x-y)\cos{\theta}$ . Finally, since $x \neq y \Rightarrow x-y \neq 0$ , we divide through to get $\cos{\theta} = \frac{280}{360} = \frac{7}{9} \Rightarrow \lfloor 1000 \cos{\theta} \rfloor = 777$ . QED. ——————– Comment: This is a demonstration of the power of something as simple as the Law of Cosines. Never underestimate what a little experimentation can do for you on the AIME; play around with equations. If at first you do not see an approach, look at the question itself for hints. Since you have to find the cosine, it should immediately trigger the Law of Cosines because it relates sides and angles. Then apply the Law of Cosines to the important angles (the ones you know something about), in this case $\angle A$ and $\angle C$ , from which the result falls quite nicely. ——————– Practice Problem: (2003 AIMEI – #6) The sum of the areas of all triangles whose vertices are also vertices of a $1\times 1 \times 1$ cube is $m+\sqrt{n}+\sqrt{p}$ , where , , and are integers. Find . Posted in AIME, Geometry, Trigonometry \|\| No Comments »
Return Of The Triangle. Topic: Geometry/Inequalities/Trigonometry. Level: AIME.	February 11th, 2007
Problem: (1961 IMO – #2) Let be the sides of a triangle, and its area. Prove: $a^2+b^2+c^2 \ge 4T\sqrt{3}$ . In what case does equality hold? Solution: We begin with the trivial inequality, $(a-b)^2 \ge 0$ , which has equality at . Rearrange to get $a^2+b^2 \ge 2ab$ . Let $\theta$ be the angle between the sides with lengths . Since $2 \ge \cos{\theta}+\sqrt{3}\sin{\theta}$ (can be proved by combining RHS) with equality at $\theta = \frac{\pi}{3}$ , we know $a^2+b^2 \ge ab(\cos{\theta}+\sqrt{3}\sin{\theta})$ $2(a^2+b^2) \ge 2ab(\cos{\theta}+\sqrt{3}\sin{\theta})$ $a^2+b^2+(a^2+b^2-2ab\cos{\theta}) \ge 2\sqrt{3} \cdot ab\sin{\theta}$ . Recalling the Law of Cosines, we know $c^2 = a^2+b^2-2ab\cos{\theta}$ . Also, $T = \frac{1}{2}ab\sin{\theta}$ , so substituting we obtain $a^2+b^2+c^2 \ge 4T\sqrt{3}$ as desired. Equality holds when and $\theta = \frac{\pi}{3}$ , which means the triangle must be equilateral. QED. ——————– Comment: There are lots of ways to prove this, but this is one of the more elementary ones, requiring only basic knowledge of inequalities and trigonometry. Which is always good because I don’t know any geometry. We see that this inequality is in general pretty weak, with equality only when the triangle is equilateral – there is a stronger version that states $a^2+b^2+c^2 \ge 4T\sqrt{3}+(a-b)^2+(b-c)^2+(c-a)^2$ . See if you can prove that… ——————– Practice Problem: Let be the sides of a triangle, and its area. Prove: $a^2+b^2+c^2 \ge 4T\sqrt{3}+(a-b)^2+(b-c)^2+(c-a)^2$ . Posted in AIME, Geometry, Inequalities, Trigonometry \|\| 4 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 »

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