MR 6 2013 Problems (Final)
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Transcript of MR 6 2013 Problems (Final)
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5/24/2018 MR 6 2013 Problems (Final)
1/4
Junior problems
J283. Let a, b, cbe positive real numbers. Prove that
2a + 1
b + c +
2b + 1
c + a +
2c + 1
a + b 3 + 9
2(a + b + c) .
Proposed by Zarif Ibragimov , SamSU, Samarkand, Uzbekistan
J284. Find the greatest integer that cannot be written as a sum of distinct prime numbers.
Proposed by Ivan Borsenco, Massachusetts Institute of Technology, USA
J285. Let a, b, cbe the sidelengths of a triangle. Prove that
8< (a + b + c)(2ab + 2bc + 2ca a2 b2 c2)
abc 9.
Proposed by Adithya Ganesh, Plano, USA
J286. Let ABCDbe a square inscribed in a circle. IfPis a point on the arc AB, find the minimumof the expression
PC PDPA PB .
Proposed by Panagiote Ligouras, Noci, Italy
J287. Let n be a positive integer and let a1, a2 . . . , an be real numbers in the interval
0, 1
n. Prove
that log1a1(1 na2) + log1a2(1 na3) + + log1an(1 na1) n2.
Proposed by Titu Andreescu, University of Texas at Dallas, USA
J288. Four points are given in the plane such that no three of them are collinear. These four pointsform six segments. Prove that if five of their midpoints lie on a single circle, then the sixthmidpoint lies on this circle too.
Proposed by Michal Rolinek and Josef Tkadlec, Czech Republic
Mathematical Reflections 6 (2013)
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5/24/2018 MR 6 2013 Problems (Final)
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Senior problems
S283. Let a, b, cbe positive real numbers greater than or equal to 1, such that
5(a2
4a + 5)(b2
4b + 5)(c2
4c + 5)
a + b + c
1.
Prove that(a2 + 1)(b2 + 1)(c2 + 1) (a + b + c 1)3.
Proposed by Titu Andreescu, University of Texas at Dallas, USA
S284. Let Ibe the incenter of triangle ABCand let D,E,F be the feet of the angle bisectors. Theperpendicular dropped fromD ontoBCintersects the semicircle with diameter BC inA, whereAand A lie on the same side of line BC. Then angle bisector ofBACintersects BCat D .Denote by V =BE FD and W =CF ED . Prove that V W is parallel to BC.
Proposed by Ercole Suppa, Teramo, Italy
S285. Let be a, b, cbe positive real numbers such that ab + bc + ca= 1. Prove that
a
b2 + c2 + 2+
b
c2 + a2 + 2+
c
a2 + b2 + 2 1
8
1
a+
1
b+
1
c
.
Proposed by Mircea Lascu and Marius Stanean, Zalau, Romania
S286. Let ABC be a triangle and let P be a point in its interior. Lines AP,BP,CP intersectBC,CA,AB at A1, B1, C1, respectively. Prove that the circumcenters of triangles APB1,APC1, BPC1, BPA1 are concyclic if and only if lines A1B1 and AB are parallel.
Proposed by Mher Mnatsakanyan, Armenia
S287. Let mand n be positive integers such that gcd(m,n) = 1. Prove that
mk=1
nk
m
mk
=
nk=1
mk
n
nk
,
where is the Euler totient function.
Proposed by Marius Cavachi, Romania
S288. Consider triangle ABCwith vertices A, B, Cin the counterclockwise order. On the sides ABandACconstruct outwards similar rectanglesABXY andCAZTin the clockwise order. LetDbe the second point of intersection of the circumcircle of triangle AY Zand , the circumcircleof triangleABC. Denote by U=DY andV =DZ and let Wbe the midpoint ofU V.Prove that AW is perpendicular to Y Z.
Proposed by Marius Stanean, Zalau, Romania
Mathematical Reflections 6 (2013)
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Undergraduate problems
U283. Let a,b,c,d be nonnegative real numbers such that a+ b + c + d = 2. Prove that for anypositive integer n,
an + bn + cn + dn + 23n
3
2 a
3
n
+
2 b
3
n
+
2 c
3
n
+
2 d
3
n
.
Proposed by Titu Andreescu, University of Texas at Dallas, USA
U284. Let an =
n2 + 1
be the sequence of real numbers, where{x} denotes the fractional partofx. Find limn nan.
Proposed by Ivan Borsenco, Massachusetts Institute of Technology, USA
U285. Let (an)n>0 be the sequence defined by an= e
1
n+1
n + 1+ e
1
n+2
n + 2+ +e
1
2n
2n. Prove that sequence(an)n>0 is decreasing and find its limit.
Proposed by Angel Plaza, Universidad de Las Palmas de Gran Canaria, Spain
U286. Let (xn)n1 be a sequence of positive real numbers such that
(xn+1 xn)(xn+1xn 1) 0
for all nand limnxn+1
xn= 1. Prove that (xn)n1 is convergent.
Proposed by Mihai Piticari, Campulung Moldovenesc, Romania
U287. Let u, v : [0 + ) (0,+) be differentiable functions such that u =vv and v =uu. Provethat limx(u(x) v(x)) = 0.
Proposed by Aaron Doman, University of California, Berkeley , USA
U288. Let f : [a, b] R be an antiderivative and integrable function such that for each interval(m,n)[a, b] there is an interval (m, n) contained in (m,n) such that f(x)0 for all x in(m, n). Prove that f(x) 0 for all x in [a, b].
Proposed by Mihai Piticari and Sorin Radulescu, Romania
Mathematical Reflections 6 (2013)
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5/24/2018 MR 6 2013 Problems (Final)
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Olympiad problems
O283. Prove that for all positive real numbersx1, x2, . . . , xn the following inequality holds:
ni=1
x3i
x21+ + x2i1+ x2i+1+ + x2n
x1+
+ xnn 1 .
Proposed by Mircea Becheanu, Bucharest, Romania
O284. Consider a convex hexagon ABCDEF with AB||DE, BC||EF, CD||FA. The distance be-tween the lines AB and DE is equal to the distance between the lines BC and EF and isequal to the distance between the lines CD and FA. Prove thatAD+ BE+ CF does notexceed the perimeter of the hexagon ABCDEF.
Proposed by Nairi Sedrakyan, Yerevan, Armenia
O285. Let an be the sequence of integers defined as a1= 1 and an+1= 2n(2an 1) for n 1. Provethat n! divides an.
Proposed by Ivan Borsenco, Massachusetts Institute of Technology, USA
O286. LetABCbe a triangle with orthocenter H. LetHMbe the median andHSbe the symmedianin triangle BHC. Denote by P the orthogonal projection ofA onto HS. Prove that thecircumcircle of triangle MPSis tangent to the circumcircle of triangle ABC.
Proposed by Marius Stanean, Zalau, Romania
O287. We are given a 6 6 table with 36 unit cells. A 2 2 square with 4 unit cells is called a block.A set of blocks covers the table if each cell of the table is covered by at least one cell of oneblock in the set. Blocks can overlap with each other. Find the largest integer n such thatthere is a cover of the table with n blocks, but if we remove any block from the covering, theremaining set of blocks will no longer cover the table.
Proposed by Nairi Sedrakyan, Yerevan, Armenia
O288. Let ABCD be a square situated in the planeP. Find the minimum and the maximum of thefunction f : P Rdefined by
f(P) = PA + PB
PC+ PD
.
Proposed by Dorin Andrica, Babes-Bolyai University, Romania
Mathematical Reflections 6 (2013)