FRESHMAN PRIZE EXAM 2017

Transcription

1 FRESHMAN PRIZE EXAM 27 Full reasoning is expected. Please write your netid on your paper so we can let you know of your result. You have 9 minutes. Problem. In Extracurricula all of the high school students are very into clubs. Each club contains at least three students, and there are at least two clubs. After a survey of clubs, an interesting property was noticed: for every two students there is exactly one club that both students are in, and every two clubs have exactly one student in common. What is the minimal number of students at the high school? Solution: There are at least two clubs, with three students each, and they share one. So we have clubs {a, a 2, a } and {a, a 4, a 5 }. We need two more clubs to contain {a, a 4 } and {a, a 5 }. Each of these needs to have one more student, and it can t be any of the original ones, so we have clubs {a, a 4, a } and {a, a 5, a 7 }. So we must have at least 7 students. If we add in the clubs {a 2, a 4, a 7 }, {a 2, a 5, a } and {a, a, a 7 } this gives the desired property, so 7 students is the minimum. Problem 2. Suppose that f is continuous and that Given that fπ = π, find f. [fx + f x] sin x dx = 2. Solution: Use integration by parts: f x sin x dx = = f x sin x π f x cos x dx { = fx cos x π = fπ + f f x cos x dx } fx sin x dx fx sin x dx Date: March 5, 27.

2 2 FRESHMAN PRIZE EXAM 27 So and fπ + f = [fx + f x] sin x dx = 2 f = 2 fπ = 2 π. Problem. Consider all configurations of four lines l, l 2, l, l 4 in R such that the intersections l l 2 and l l 4 both consist of one point each, and all other pairs of lines are disjoint. What are all nonnegative integers k such that for some such configuration of four lines in R there are precisely k lines l that intersect each line l i in exactly one point? Explain your answer. Solution: The possible values of k are and 2. Certainly one such line l always exists: it is the unique line passing through l l 2 and l l 4. Generically there will be a second line: let P be the plane spanned by l and l 2, and let P be the plane spanned by l and l 4. Then, generically, the planes P and P 2 intersect in a line that intersects all l i. This doesn t happen if P P 2 = or the line P P 2 is parallel to one of the l i in the corresponding plane. The only way that a line can intersect both l and l 2 is by passing through l l 2 or by lying in the plane P similarly for l and l 4. So another nongeneric possibility is that either l l 2 P or l l 4 P, in which case there are infinitely many solutions, since in the first case, say any line in P passing through l l 2 and not parallel to l or l 4 will do. Problem 4. Show, for x, that the error in replacing by x n+ is at most. = sin x 2 sin... 2 n sin 2 2 n. Solution: By Taylor s theorem, for x [, π 2 ], x, x such that sin x = x cos x x. Note cos x is the third derivative of sin x. This also implies sin x x in this range.

3 FRESHMAN PRIZE EXAM 27 Applying this to x, we see x k, x, π such that 2 And so which is of the form sin x = x cos x k 2 k. [ = x cos x ] k 2 2k n = x [ n+ cos x k 2 2k x 2 ] x n+ n [ ɛ k ] with all ɛ k [, ] as long as x [, ]. We now use [ ɛ k ] which is readily established by any of mathematical induction, telescoping sums, σ n MacLaurin s identity σ n for elementary symmetric polynomials in n n variables, or Jensen s inequality applied to ln x. This inequality is the generalization of the statement that a 7% + 8% = 5% discount results in a lower price than successive 7% and 8% ones. So x n+ = x n+ [ x n+ xn+ n n n ɛ k [ cos x k 2 2k x 2 ] ] cos x k 2 2k x 2 2 2k xn+ 2 = xn+. Remark : A direct inductive proof on n does not work, because in trying to inductively establish x n+ xn+

4 4 FRESHMAN PRIZE EXAM 27 the coefficient of xn+ will increase beyond. However if you try this, you may well notice that a more refined still not optimal statement like x n+ x n+ 2 2n x n+ is readily established, and this implies what is needed. Here are the details: Please keep in mind x. For n =, this follows from which is true since < cos x <. x sin x = x cos x x x x 2 Inductively assuming the case n, to establish the case n +, we have x n+2 = x n+ n 2 n+ sin 2 n+ x cos x n+ x n+ 2 2n x n+ = x n+2 x as desired. x n+2 x x = x n+2 x 2 2n+ x n+ 2 2n x n+ 2 2n n+ 2n 2 n n 2 2 n n+ 2 2 n x n+ Remark 2: A number of people tried to establish the inequality based on properties of alternating series; in particular the observation in the proof of the alternating series test that the successive partial sums alternate being above and below the series sum. This ordinarily requires that the terms decrease in absolute value. While that is true for the Maclaurin series for sin x on [, ], and so for each factor in the product, it is not so clearly true for the product we have here. In particular, the

5 FRESHMAN PRIZE EXAM 27 5 example.9t+.9 2 t 2.9t+.9 2 t 2 =.9+.9t t t t 4 shows the product of alternating series with terms decreasing in absolute value does not necessarily produce a series with this same property of terms decreasing in absolute value. Problem 5. We say that an n -tuple a = a,..., a n 2 Z/n n is dominated by,,..., n 2 if there is a reordering a π,..., a πn 2 of a such that a πi i for all i. Here we think of the elements of Z/n as represented by,,..., n and thus makes sense. Show that for any a,..., a n 2 Z/n n there is a k Z/n such that a + k,..., a n 2 + k is dominated by,,..., n 2. Solution: Suppose a,..., a n 2 is in ascending order. Let i be the smallest index such that a i i a j j for all j, and let k = n a i. First bring a +k,..., a n 2 +k into ascending order to obtain:, a i+ a i,..., a n 2 a i, a a i + n,..., a i a i + n. The entry in this n -tuple at position j, starting the count from, is a i+j a i if i + j n 2. This entry cannot exceed j since a i+j a i > j implies a i+j j > a i and thus a i+j i + j > a i i, in contradiction to the choice of i. If i + j > n 2 then the entry at position j = n i + t for some t < i is a t a i + n. Now a t a i + n > n i + t is equivalent to a t t a i i but t < i, in contradiction to the choice of i.