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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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 »
Criss Cross Applesauce. Topic: Geometry. Level: AIME.	March 5th, 2007
Problem: (2007 Mock AIME 6 – #2) Draw in the diagonals of a regular octagon. What is the sum of all distinct angle measures, in degrees, formed by the intersections of the diagonals in the interior of the octagon? Solution: Consider the circumscribed circle of the octagon. Each diagonal is a chord of this circle, and we know that the angle between two chords that intercept arcs of measure $\alpha$ and $\beta$ is $\frac{\alpha+\beta}{2}$ . Now, any pair of diagonals can together intercept , , , , or little arcs (between vertices of the octagon). is not possible, because then on one side there is no little arc which means the intersection is not in the interior. This also means is not possible (they come in supplement pairs – except $90^{\circ}$ of course). Since each little arc is $45^{\circ}$ and we have to account for the division by , the final sum is $45 \cdot \frac{1}{2} \cdot (2+3+4+5+6) = 450$ . QED. ——————– Comment: This wasn’t an extremely hard problem, but it needed some clever thinking. Most people overcounted, which is reasonable given that there are diagonals in an octagon. Pretty hard to draw accurately. Using arcs on a circle proved to be much easier and less prone to careless mistakes. ——————– Practice Problem: Let be a set of points in the plane, no three of which lie on the same line. At most how many ordered triples of points in exist such that $\angle ABC$ is obtuse? Posted in AIME, Geometry \|\| 2 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 »
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 »

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