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B ′ C ′ MCA ′2R b2R aA2R cBHowever, |MA ′ | is the diameter of the circumcircle for △BAC, that is, |MA ′ | = 2R a .Therefore, using this in (1) givesSimilarly,2R a ≥ |C ′ A ′ ||B ′ C ′ | |MC| + |A′ B ′ ||MB|. (2)|B ′ C ′ |2R b ≥ |A′ B ′ ||C ′ A ′ | |MA| + |B′ C ′ ||MC|, (3)|C ′ A ′ |2R c ≥ |B′ C ′ ||A ′ B ′ | |MB| + |C ′ A ′ ||MA|. (4)|A ′ B ′ |Multiplying (2) by λ 2 1 , (3) by λ2 2 , and (4) by λ2 3, then adding, we obtain( λ2λ 2 1 R a + 2λ 2 2 R b + 2λ 2 23 R c ≥ 2|A ′ B ′ |+ λ2 3 |C )′ A ′ ||MA||C ′ A ′ | |A ′ B ′ |( ) ( )λ2+ 1|A ′ B ′ |+ λ2 3 |B′ C ′ | λ2|MB| + 1|C ′ A ′ |+ λ2 2 |B′ C ′ ||MC|. (5)|B ′ C ′ | |A ′ B ′ ||B ′ C ′ | |C ′ A ′ |For any real x and y, x 2 + y 2 ≥ 2xy. Applying this to the coefficient of |MA| on theright of (5), we getλ 2 2 |A′ B ′ | 2 + λ 2 3 |C ′ A ′ | 2|A ′ B ′ ||C ′ A ′ |≥ 2λ 2 λ 3 .Similar inequalities follow for the other two terms in (5), and so (5) implies2λ 2 1 R a + 2λ 2 2 R b + 2λ 2 3 R c ≥ 2λ 2 λ 3 |MA| + 2λ 1 λ 3 |MB| + 2λ 1 λ 2 |MC|.This is the required inequality, namelyλ 2 1 R a + λ 2 2 R b + λ 2 3 R c ≥ λ 1 λ 2 λ 3( |MA|λ 1+ |MB|λ 2+ |MC|λ 3).Equality holds when △ABC is equilateral, M is the circumcenter, and λ 1 = λ 2 = λ 3 .Also solved by M. Bataille (France), P. P. Dályay (Hungary), O. Faynshteyn (Germany), O. Geupel (Germany),O. Kouba (Syria), J. H. Smith, T. Smotzer, R. Stong, M. Tetiva (Romania), Z. Vörös (Hungary), J. B. Zacharias& K. T. Greeson, and the proposer.June–July 2012] PROBLEMS AND SOLUTIONS 525
Zero–Nonzero Matrices11534 [2010, 835]. Proposed by Christopher Hillar, Mathematical Sciences ResearchInstitute, Berkeley, CA. Let k and n be positive integers with k < n. Characterize then × n real matrices M with the property that for all v ∈ R n with at most k nonzeroentries, Mv also has at most k nonzero entries.Solution by Richard Stong, Center for Communications Research, San Diego, CA. Weshow that either M has at most k nonzero rows or M has at most 1 nonzero entry inevery column. Such a matrix has the desired property: in the first case Mv has at mostk nonzero entries for any v, and in the second case Mv has at most as many nonzeroentries as v.Now suppose that some matrix M with the desired property is not of the formstated. Thus M has at least k + 1 rows with nonzero entries and at least one columnwith at least two nonzero entries. Build a list of columns of M as follows: say a listof columns represents a row if and only if at least one of the columns in the set hasa nonzero entry in that row. Start with a column w 1 with at least two nonzero entries.If w 1 , . . . , w r have been chosen, and together they represent fewer than k + 1 rows,then choose w r+1 to be any column that represents a new row and append it to the list.Stop when w 1 , . . . , w r represent at least k + 1 rows. Now we started with a columnrepresenting two rows, and each time we added a new column we got at least one newrow. Hence r ≤ k. Thus any linear combination ∑ rj=1 a jw j is of the form Mv, wherev has at most k nonzero entries. Fix k + 1 rows represented by the w j . For the ith suchrow, let V i be the set of all r-tuples (a 1 , . . . , a r ) such that ∑ rj=1 a jw j has a nonzeroentry in that ith row. Since w 1 , . . . , w r do represent this row, V i is the nullspace of anontrivial linear equation on r-tuples and therefore is a codimension-1 subspace of R r .However, the required property of M says that for any r-tuple (a 1 , . . . , a r ), the linearcombination ∑ rj=1 a jv j has at most k nonzero entries; thus (a 1 , . . . , a r ) must lie inone of these k + 1 subspaces. But of course R r cannot be covered by finitely manycodimension-1 subspaces. This contradiction shows that such an M cannot exist.Editorial comment. Several solvers noted that the same result holds for any field ofcharacteristic 0. John Smith (Needham, MA) noted that the result holds for rectangularmatrices.Also solved by P. Budney, N. Caro (Brazil), P. P. Dályay (Hungary), E. A. Herman, Y. J. Ionin, J. H. Lindsey II,O. P. Lossers (Netherlands), R. E. Prather, J. Simons (U. K.), J. H. Smith, M. Tetiva (Romania), E. I. Verriest,Barclays Capital Problems Solving Group (U. K.), NSA Problems Group, and the proposer.How Closely Does This Sum Approximate the Integral?11535 [2010, 835]. Proposed by Marian Tetiva, Bîrlad, Romania. Let f be a continuouslydifferentiable function on [0, 1]. Let A = f (1) and let B = ∫ 1x −1/2 f (x) dx.0Evaluate( ∫ 1n∑( ) ( ) )k2lim n (k − 1)2 (k − 1)2f (x) dx − − fn→∞0n2 n 2 n 2in terms of A and B.k=1Solution by Eugene A. Herman, Grinnell College, Grinnell, IA. Answer: A − B/2.Proposition. If h is continuously differentiable on [0, 1], then( ∫ 1lim n h(x) dx − 1 n∑( ) ) k − 1 h(1) − h(0)h = .n→∞0n n2k=1526 c○ THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 119
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