PROBLEMS AND SOLUTIONS

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Proof. Let H denote an antiderivative of h. By Taylor’s theorem,∫ 1h(y) dy − 1 n∑( ) k − 1n∑(∫ k/nh =h(y) dy − 1 ( )) k − 1n nn h n0=n∑k=1(Hk=1( kn)− H( k − 1nk=1(k−1)/n)− 1 ( )) k − 1n H ′ n= 12n 2n∑H ′′ (y k ),for some list (y 1 , . . . , y n ) with y k ∈ ((k − 1)/n, k/n) for 1 ≤ k ≤ n. Therefore,( ∫ 1n∑( ) )k − 1 1lim n h(y) dy − h= 1n→∞0n n 2 lim 1n∑H ′′ (y k )n→∞ n= 1 2∫ 10k=1h ′ (y) dy =h(1) − h(0).2For this problem, let h(y) = 2y f (y 2 ). Now (h(1) − h(0))/2 = (2 f (1))/2 = A. Usingthe substitution y = √ x, we have∫ 1∫ 1∫ 1∫ 1f (x) dx = h(y)dy, B = x −1/2 f (x) dx = 2 f (y 2 ) dy.00Therefore, by the proposition,n∑( k2f (x) dx −( ∫ 1lim nn→∞0= limn→∞n[= lim nn→∞=0k=1( ∫ 1h(y) dy −00)(k − 1)2− fn2 n 2n∑k=1( ∫ 1h(y) dy − 1 n0h(1) − h(0)2∫ 1−0( (k − 1)2n 2 ) )k=1k=1( 2(k − 1)+ 1 ) ( ) )(k − 1)2fn 2 n 2 n 2n∑( ) ) k − 1h − 1 n nk=1f (y 2 ) dy = A − B 2 .n∑k=1f( (k − 1)2n 2 ) ]Also solved by P. Bracken, N. Caro (Brazil), H. Chen, D. Constales (Belgium), P. P. Dályay (Hungary), Y. Dumont(France), P. J. Fitzsimmons, D. Fleischman, J.-P. Grivaux (France), F. Holland (Ireland), S. Kaczkowski,P. Khalili, O. Kouba (Syria), W. C. Lang, J. H. Lindsey II, R. Nandan, M. Omarjee (France), K. Schilling, J.Schlosberg, J. Simons (U. K.), N. C. Singer, Z. Song & L. Yin (China), A. Stenger, R. Stong, T. Tam, J. A. VanCasteren (Belgium), E. I. Verriest, P. Xi (China), J. B. Zacharias & K. T. Greeson, Barclays Capital ProblemsSolving Group (U. K.), GCHQ Problem Solving Group (U. K.), NSA Problems Group, and the proposer.Use Hayashi’s Inequality11536 [2010, 835]. Proposed by Mihaly Bencze, Brasov, Romania. Let K , L, and Mdenote the respective midpoints of sides AB, BC, and CA in triangle ABC, and let P bea point in the plane of ABC other than K , L, or M. Show that|AB||PK| + |BC||PL| + |CA| |AB| · |BC| · |CA|≥|PM| 4|PK| · |PL| · |PM| ,where |UV| denotes the length of segment UV.June–July 2012] PROBLEMS AND SOLUTIONS 527
Solution by D. Marinescu, Colegiul Naţional “Iancu de Hunedoara”, Hunedoara, Romania,and M. Monea, Colegiul Naţional“Decebral”, Deva, Romania. In 1913, T.Hayashi proved Hayashi’s Inequality: For any triangle ABC with opposite sides oflengths a, b, c respectively, and for and an arbitrary point M in its plane,a|MB| · |MC| + b|MC| · |MA| + c|MA| · |MB| ≥ abc.See D. M. Mitronović, J. E. Peĉarić, V. Volenec, Recent Advances in Geometric Inequalities,(Kluwer, 1989), p. 297. We now apply Hayashi’s Inequality with triangleKLM and point P to get|KL| · |PK| · |PL| + |KM| · |PK| · |PM| + |ML| · |PM| · |PL| ≥ |KL| · |ML| · |MK|.Since |KL| = |AC|/2, |KM| = |BC|/2, and|ML| = |AC|/2,|CA| · |PK| · |PL| + |BC| · |PK| · |PM| + |AB| · |PM| · |PL| ≥ 1 |CA| · |BC| · |AB|,4which is equivalent to the inequality to be proved.Also solved by G. Apostolopoulos (Greece), M. Bataille (France), D. Beckwith, M. Caragiu, P. P. Dályay (Hungary),O. Geupel (Germany), O. Kouba (Syria), N. Minculete (Romania), B. Mulansky (Germany), C. R. Pranesachar(India), J. Schlosberg, T. Smotzer, M. Vowe (Switzerland), GCHQ Problem Solving Group (U. K.),Northwestern University Math Problem Solving Group, and the proposer.A Circumradius Inequality11541 [2010, 929]. Proposed by Nicuşor Minculete, “Dimitrie Cantemir” University,Brasov, Romania. Let M be a point in the interior of triangle ABC. Let R a , R b , and R cbe the circumradii of triangles MBC, MCA, and MAB, respectively. Let |MA|, |MB|,and |MC| be the distances from M to A, B, and C. Show that|MA|R b + R c+|MB|R a + R c+|MC|R a + R b≤ 3 2 .Solution by Oleh Faynshteyn, Leipzig, Germany. Let ϕ 1 = ∠CAM, ϕ 2 = ∠MAB, ϕ 3 =∠ABM, ϕ 4 = ∠MBC, ϕ 5 = ∠BCM, and ϕ 6 = ∠BCM. Observe that ∑ 6i=1 ϕ i = π.From triangles AMC and ABM, it follows thathence|MA|R b + R c=|MA| = 2R b sin ϕ 6 = 2R c sin ϕ 3 ,2csc ϕ 3 + csc ϕ 6≤ 1 2 (sin ϕ 3 + sin ϕ 6 ) ,where the inequality is a consequence of the arithmetic-harmonic mean inequality.Similarly we get|MB|R c + R a≤ 1 2 (sin ϕ 2 + sin ϕ 5 ) ,Adding these three inequalities, we obtain|MA|R b + R c+|MB|R a + R c+|MC|R a + R b≤ 1 2 (sin ϕ 1 + sin ϕ 4 ) .|MC|R a + R b≤ 1 26∑sin ϕ i .528 c○ THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 119i=1
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