USA Mathematical Talent Search

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Beverly Phelps
5 years ago
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1 8 3 We claim that there are eactly 4 distinguishable colorings. We briefly describe them: one cube has each set of 4 parallel edges colored one color. There are three such sets, and it is not difficult to verify that this coloring is indeed valid. Call this cube type I. Now, the second type of cube has one set of 4 parallel edges colored one color. The two other faces are colored in an alternating fashion with the other two colors so that we end up with eactly one set of 4 parallel edges. Call this cube type II. There are three cubes of this type, so there are 4 total. To simplify the casework, we fi the orientation of the cube and consider three sets of 4 edges covering the whole cube. Let A be the set of edges on the top face, B the set of edges perpendicular to this face, and C the set of edges on the bottom face. Label the edges of A as a, a, a 3, a 4 in that order and similarly the respective edges of C directly below c, c, c 3, c 4. Let b, b 3, b 34, b 4 be the edges of B, where b meets both a, a, c, c, etc. Now that we have enough notation to tackle the problem, let us without loss of generality) color a red and a blue. Then, b must be yellow. Now we consider two general cases. If a 3 is red, then b 3 is yellow. Now, for even more casework. If a 4 is yellow, then b 34 and b 4 are blue. This implies that c, c 3 are red, c is blue, and that c 4 is yellow. This is eactly a type II cube with the set of 4 parallel edges being red. If a 4 is blue, then all the edges in B are yellow. Then, the edges of C must be colored red, blue, red, blue. There are two ways to do this, one is type I and the other type II. Similarly, if a 3 is yellow, then b 3 is red. Also, a 4 must be blue. This means that b 34 is red and b 4 is yellow. Now, we must have c, c 4 both blue. Finally, this means that our only option is coloring c red and c 3 yellow. This is eactly a type II cube with the set of 4 parallel edges blue. Seeing that we have ehausted every possible case, we can safely say that we re done! Page of Problem

2 8 3 The crucial observation in this problem is that the given equation is actually an inequality! For any real number 0, we claim that +. Indeed, without loss of generality, we have > 0 ) and + = 0 with equality if and only if =. If < 0, similarly, we have equality if and only if =. We rewrite our equation as tan 7 + cot 7 = cos 4 + sin 6. By our previous observation, tan 7 + cot 7 since tan = / cot. Also, we have sin 6 + cos 4 so this means that we must have equality! In other words, we are left with two cases: all four terms are either equal to or. Case : sin 6 = cos 4 = tan 7 = cot 7 = Since cos 4 =, we must have 4 = kπ for some integer k. Thus, { π, π, 3π }. Testing these values, we see that sin 6 = 0 for all these, so there are no solutions in this case. Case : sin 6 = cos 4 = tan 7 = cot 7 = Similarly, considering cos 4 = drastically reduces our casework. We have 4 = k + )π for some integer k, so { π 4, 3π 4, 5π 4, 7π 4 }. Testing these values for, we see that sin 6 = only when { π 4, 5π 4 }. Now, we see that tan 7 = cot 7 = for both values of. Thus, our two solutions are = π 4 or 5π 4. Page of Problem

3 8 3 3 Position the three circles in the coordinate plane so that A =, 0), B =, 0) and let O be the origin. We can easily calculate CO = CA OA = 3 so let C = 0, 3). Now, notice that the locus of all X such that AX +XB = AC +CB is an ellipse E. More specifically, E is centered at O and has foci at A and B. If we let A = 4, 0) and B = 4, 0) then AA +A B = AB +B B = 8 so A and B are on the ellipse as well. The ellipse passes through C too so the equation of E is given by 6 + y = = + 4y 3 = 6. To locate X, we want to find the point at which this intersects the circle centered at C, with equation + y 3) = 4. Substituting, we obtain 4 y 3) + 4y 3 = 6 y y = 4. Solving this and taking the positive root yields y = This is equal to the altitude of triangle AXB, so the area of this triangle is simply 4 y = 5 3. Page of Problem 3

4 8 3 4 First of all, it only makes sense to consider 0 k < n. If k = 0, then it is clear that m = 0. Otherwise, we claim that the answer to the general case is m = n + k so it will follow that Alice needs at least + 8 = 9 points for part a). The proof is in two parts. First we show that we can construct a tournament in which Alice needs at least n + k to beat k other players, in other words m n + k. Then, we will show that m n + k indeed implies that Alice beats at least k other players. Let s s s n be the scores of the n players P, P,..., P n. Fi a k > 0 and consider the following tournament: let X = {P, P,..., P k } and Y = {P k+, P k+,..., P n }. Say everyone in Y beats everyone in X and everyone in X ties with everyone else in X. Also, assume that everyone in Y ties with everyone else in Y. Then, we have s = s = = s k = k, s k+ = s k+ = = s n = k + n k ) = n + k. In this case, Alice needs at least n + k points for Bob to be sure that she beat at least k other players. Now, we show that if Alice scores at least n + k points, then Bob will know that she beat at least k other players. We proceed by contradiction. Assume that Alice scores at least n + k points and that she beats at most k other players. If m is Alice s score, we have m s k so s k + s k+ + s k+ + + s n n k + )m n k + )n + k ). However, can this happen? Let s see how many points the set of players {P k, P k+,..., P n } can earn in total. Within the set itself, there is a point net gain regardless, and we might as well assume that they beat everyone else, i.e. all the players P, P,..., P k. Thus, they can earn at most ) n k + + k )n k + ) = n k + )n k) + k )n k + ) = n k + )n + k ) < n k + )n + k ), which is a contradiction! Thus, if Alice scores at least n + k points, then it is guaranteed that she beat at least k other players. Page of Problem 4

5 8 3 5 We show that in fact f) = 5 for all > 0. Then, we will have f) = 5. First, a useful observation: f) is nonzero for all > 0, or else f)f f) + ) cannot equal. Also, f) + > 0 is given. Suppose > 0. We use our given equation and replace with f) + to obtain f)f f) + ) = f f) + ) f f f) + ) ) + f) + =. Now, because f f) + ) is nonzero, we have f) = f f ) f) + ) +. Since f) is f)+ strictly increasing, we must also have = f f) + ) + f)+ so it follows that = f) + f) + f) f) + ) = f) + f) f) + 4 = 5 4 f) ) = 5 4 Thus, f) = ± 5 for all > 0. We show that f) + 5 contrary and suppose that there eists c > 0 such that fc) = + 5 c > c, so f) = + 5 a contradiction! Thus, f) = 5. We note that f f) + ) = f), for every > 0. Assume the. Then, f) > fc) for all when > c, but this function is strictly decreasing when > c. We have for all > 0, as claimed. It is not difficult to verify that this indeed satisfies the conditions of the problem, so we are done. Page of Problem 5

Graphing Review Part 1: Circles, Ellipses and Lines

Graphing Review Part 1: Circles, Ellipses and Lines Graphing Review Part : Circles, Ellipses and Lines Definition The graph of an equation is the set of ordered pairs, (, y), that satisfy the equation We can represent the graph of a function by sketching