Solutions to review problems

This page is going to contain some of the solutions I submit to mathematical periodicals with problem sections such as Mathematical Reflections and The American Mathematical Monthly. Problems are usually rewritten in order to avoid excessive quoting and often to homogenize the notations used in a problem and its solution.

I don't put up solutions on this site prior to the discussion of the problem in the respective magazine, so usually they won't appear here directly after the deadline is over.

Mathematical Reflections

• Solution to: Darij Grinberg, Problem U228, Mathematical Reflections 2/2012. (Link to the solution. For the problem see 2/2012.)
My solution as a PDF file, including two unpublished alternative solutions and a generalization.
Let L/K be a separable algebraic extension of fields and let V, W and U be L-vector spaces.
Let h : V × W -> U be a K-bilinear map satisfying
h(xa, xb) = x2 h(a, b) for every x ∈ L, a ∈ V and b ∈ W.
Prove that h is L-bilinear.
Notes: 1. The solution file contains three solutions, one of which generalizes the problem from separable algebraic field extensions to commutative separable algebras over commutative rings.
2. The problem can be discussed on MathLinks.

• Solution to: Darij Grinberg, Problem O222, Mathematical Reflections 1/2012. (Link to the solution. For the problem see 1/2012.)
My solution (also the second published solution) as a PDF file.
Let n be a nonnegative integer. Consider two n-tuples (a1, a2, ..., an) and (b1, b2, ..., bn) of nonnegative reals, and a permutation σ of the set {1,2,...,n}. For every k in the set {1,2,...,n}, let ck be the maximum of the set
{a1bk, a2bk, ..., akbk} ∪ {akb1, akb2, ..., akbk}.
Prove that a1bσ(1) + a2bσ(2) + ... + anbσ(n) ≤ c1 + c2 + ... + cn.
Note: The solution file contains some background on this problem.

• Solution to: Gabriel Dospinescu, Problem O219, Mathematical Reflections 1/2012. (Link to the solution. For the problem see 1/2012.)
My solution (also the second published solution) as a PDF file.
Let a, b, c, d be positive reals such that (1-c)/a + (1-d)/b + (1-a)/c + (1-b)/d ≥ 0. Prove that a(1-b) + b(1-c) + c(1-d) + d(1-a) ≥ 0.

• Solution to: Titu Andreescu, Problem U111, Mathematical Reflections 1/2009. (Link to the solution. For the problem see 1/2009.)
My solution (also the second published solution) as a PDF file.
Let n be a positive integer. For every k in {0, 1, ..., n-1}, let ak = 2 cos(π / 2n-k). Prove that product_{k=0}^{n-1} (1-ak) = (-1)n-1 / (1 + a0).

• Solution to: Cezar Lupu and Valentin Vornicu, Problem U112, Mathematical Reflections 1/2009. (Link to the solution. For the problem see 1/2009.)
My solution (also the first published solution) as a PDF file.
Let x, y, z be real numbers greater or equal to 1. Prove that x^{x³+2xyz} y^{y³+2xyz} z^{z³+2xyz} ≥ (xxyyzz)yz+zx+xy.

• Solution to: Titu Andreescu, Problem O111, Mathematical Reflections 1/2009. (Link to the solution. For the problem see 1/2009.)
My solution (also the second published solution) as a PDF file.
Prove that, for each integer n ≥ 0, the number (binom(n,0) + 2 binom(n,2) + 22 binom(n,4) + ...)² (binom(n,1) + 2 binom(n,3) + 22 binom(n,5) + ...)² is triangular.
Here, binom(n,m) means the (n,m)-th binomial coefficient (that is, n(n-1)...(n-m+1) / m! if m ≥ 0, and 0 otherwise).

• Solution to: Cezar Lupu and Pham Huu Duc, Problem O112, Mathematical Reflections 1/2009. (Link to the solution. For the problem see 1/2009.)
My solution (not published) as a PDF file.
Let a, b, c be positive real numbers. Prove that
(a³+abc) / (b+c)² + (b³+abc) / (c+a)² + (c³+abc) / (a+b)² ≥ 3/2 · (a³+b³+c³)/(a²+b²+c²).

• Solution to: Gabriel Dospinescu, Problem O114, Mathematical Reflections 1/2009. (Link to the solution. For the problem see 1/2009.)
My solution (also the first published solution) as a PDF file.
Prove that for all real numbers x, y, z, the following inequality holds:
(y²+yz+z²) (z²+zx+x²) (x²+xy+y²) ≥ 3(x²y+y²z+z²x) (xy²+yz²+zx²).

The American Mathematical Monthly

• Solution to: Anon, Problem A5715, AMM February 1970, in a generalized version.
My solution as a PDF file.
(Generalized statement:) Let R be a commutative ring with unity. Let a1, a2, ..., an as well as b1, b2, ..., bn be elements of R such that a1 b1 + a2 b2 + ... + an bn = 1, and such that any two indices i and j with i < j satisfy bi bj = 0.
Let T be a free R-module with basis (e1, e2, ..., en). Let S be the R-submodule of T generated by b1 e1, b2 e2, ..., bn en. Show that T/S is a free R-module of rank n-1.

• Solution to: M. Farrokhi D.G., Problem 11395, AMM November 2008.
My solution as a PDF file.
Let G be the group of all continuous bijections of [0,1] to itself. If H is a finite subgroup of G, then prove that |H| = 1 or |H| = 2.

• Solution to: Stanley Huang, Problem 11398, AMM December 2008.
My solution as a PDF file.
If the incenter I of a triangle ABC is equidistant from the orthocenter and the circumcenter, then prove that the middle-sized angle of triangle ABC is 60 degrees, and that the circumradius of triangle IBC equals to that of triangle ABC (if A is the vertex of triangle ABC with the middle-sized angle).

• Solution to: Marius Cavachi, Problem 11401, AMM December 2008.
My solution as a PDF file.
Let A be a nonsingular square matrix whose entries are integers. Suppose that for every positive integer k, there is a matrix X with integer entries such that A = Xk.
Prove that A is the identity matrix.
• Solution to: Marian Tetiva, Problem 11391, AMM November 2008.
My solution as a PDF file. Longer (more detailed) version.
Let p be some prime, s some positive integer. Also, let k and n be integers such that n ≥ k ≥ ps - ps-1. Let x1, x2, ..., xn be integers. Assume that (p,s,k) ≠ (2,2,2) and (s,k) ≠ (1,p-1). Prove that both of the sums
sum of ( (-1)j · (binomial coefficient (n-k+j) over j) · (number of all (k-j)-element subsets T of {1,2,...,n} such that the sum of xt over all t in T is divisible by p) ) over j = 0, 1, ..., k
and
sum of ( (-1)j · (binomial coefficient (n-k+j) over j) · (number of all (k-j)-element subsets T of {1,2,...,n} such that the sum of xt over all t in T is not divisible by p) ) over j = 0, 1, ..., k
are divisible by ps.

• Solution to: Omran Kouba, Problem 11392, AMM November 2008.
My solution as a PDF file.
A point M in the plane of a regular n-gon P is orthogonally projected onto each side of this n-gon. Assume that these projections all lie inside the respective sides of the n-gon. Then, the n segments joining M to the vertices of P, as well as the n segments joining M to these projections, subdivide the n-gon P into 2n triangles. We label these triangles by 1, 2, ..., 2n in a counterclockwise way. Prove that the sum of the areas of the triangles with odd labels is half the area of P.

• Solution to: Cosmin Pohoata, Problem 11393, AMM November 2008.
My solution as a PDF file.
In triangle ABC, let M and Q be points on the segment AB, let N and R be points on the segment CA, and let P and S be points on the segment BC. Denote m = BM / MA, n = CN / NA, p = CP / PB, q = AQ / QB, r = AR / RC, s = BS / SC. Prove that the lines MN, PQ and RS are concurrent if and only if mpr + nqs + mq + nr + ps = 1.

• Solution to: M. Farrokhi D.G., Problem 11395, AMM November 2008.
My solution as a PDF file.
Let G be the group of all continuous bijections of [0,1] to itself. If H is a finite subgroup of G, then prove that |H| = 1 or |H| = 2.

• Solution to: Stanley Huang, Problem 11398, AMM December 2008.
My solution as a PDF file.
If the incenter I of a triangle ABC is equidistant from the orthocenter and the circumcenter, then prove that the middle-sized angle of triangle ABC is 60 degrees, and that the circumradius of triangle IBC equals to that of triangle ABC (if A is the vertex of triangle ABC with the middle-sized angle).

• Solution to: Marius Cavachi, Problem 11401, AMM December 2008.
My solution as a PDF file.
Let A be a nonsingular square matrix whose entries are integers. Suppose that for every positive integer k, there is a matrix X with integer entries such that A = Xk.
Prove that A is the identity matrix.
• Solution to: Catalin Barboianu, Problem 11402, AMM December 2008.
My solution as a PDF file.
Given a continuous function f from [0, 1] to the set of nonnegative reals such that f(0) = f(1) = 0 and every 0 < x < 1 satisfies f(x) > 0. Prove that there exists a square which has two of its vertices in the interval (0,1) on the x-axis and its other two vertices on the graph of f.

• Solution to: Yaming Yu, Problem 11403, AMM December 2008.
My solution as a PDF file. Longer (more detailed) version.
For every nonnegative integer n, find the degree of the polynomial
sum of ( (binomial coefficient n over i) (-x)n-i (x+0) (x+1) ... (x+i-1) ) over i = 0, 1, ..., n
in the variable x over the field of rational numbers.

• Solution to: A. A. Dzhumadil'daeva, Problem 11406, AMM January 2009.
My solution as a PDF file. Longer (more detailed) version.
For any positive integer n, prove that
sum of ( (binomial coefficient n over i) · (2i-1)!! · (2(n-i)-1)!! ) over (i = 0, 1, ..., n)  =  2n n!,
where k!! denotes the product of all elements of the set {1, 2, ..., k} which have the same parity as k (in particular, 0!! = (-1)!! = 1).

• Solution to: Erwin Just, Problem 11407, AMM January 2009.
My solution as a PDF file. Longer (more detailed) version.
If p is a prime greater than 3, show that every ring R (not necessarily commutative, and possibly even without unity) which satisfies x1 + x3 + x5 + ... + x2p-1 = 0 for every x in R must be the trivial ring (i. e. the ring with one element only).

• Solution to: Paolo Perfetti, Problem 11409, AMM January 2009.
My solution as a PDF file.
For positive reals α and β satisfying β > α, prove that the limit of
sum of ( n (log n) (-1)n (product of (α + k log k) / (β + (k+1) log (k+1)) over k = 2, 3, ..., n) over n = 2, 3, ... N)
for N → ∞ exists.
• Solution to: Finbarr Holland, Problem 11415, AMM February 2009.
My solution as a PDF file. Warning: it's very long and boring (because written down with maximal detail - I really had to do this to get sure it was right).
Show that the determinant of the sum of n positive-definite Hermitian complex 2x2 matrices is greater or equal to the sum of their largest eigenvalues times the sum of their smallest eigenvalues. (Actually, the problem states this in a slightly different way, but that's the main point.)
If my solution is correct, it renders the "positive-definite" condition useless. In order to make sure it is correct, I have written out every single step of the proof, but a residual uncertainty remains. You should best scroll down to Lemma 2 (it's on page 14) immediately after reading Lemma 1, because the proof of Lemma 1 is well-known.

• Solution to: Cezar Lupu and Tudorel Lupu, Problem 11417, AMM February 2009.
My solution as a PDF file.
Given a real-valued continuously differentiable function f on the interval [0,1] such that the integral of f(x)dx from 1/3 to 2/3 equals 0. Prove that
(integral of (f '(x))2dx from 0 to 1) ≥ 27 (integral of f(x)dx from 0 to 1)2.
• Solution to: Christopher Hillar, Problem 11422, AMM March 2009.
My solution as a PDF file.
Let H be a real symmetric square matrix whose eigenvalues are pairwise distinct. Let A be a real matrix of the same size. Let H0 = H, let H1 = AH0 - H0A and let H2 = AH1 - H1A. Assume that the matrices H1 and H2 are symmetric. Prove that AAt = AtA (in other words, prove that the matrix A is normal).

• Solution to: Emeric Deutsch, Problem 11424, AMM March 2009.
My solution as a PDF file. Longer (more detailed) version.
Given some integer n > 1, compute the number of all elements (a1,a2,...,an) of {0,1}n for which the number of all i in {1,2,...,n-1} satisfying ai = ai+1 = 0 equals the number of all i in {1,2,...,n-1} satisfying ai = ai+1 = 1.

• Solution to: M. L. Glasser, Problem 11426, AMM April 2009.
My solution as a PDF file.
Simplify Γ(1/14) Γ(9/14) Γ(11/14) / (Γ(3/14) Γ(5/14) Γ(13/14)), where Γ denotes the gamma function.

• Solution to: Marius Cavachi, Problem 11433, AMM May 2009.
My solution as a PDF file.
Let n be a positive integer. Let A1, A2, ..., An, B1, B2, ..., Bn, C1, C2, ..., Cn be 3n points on the unit sphere S2. Prove that there exists a point P on S2 satisfying
(sum of |P-Ak|2 over k = 1, 2, ..., n) = (sum of |P-Bk|2 over k = 1, 2, ..., n) = (sum of |P-Ck|2 over k = 1, 2, ..., n).

• Solution to: Slavko Simic, Problem 11434, AMM May 2009.
My solution as a PDF file. Longer (more detailed) version.
Let n be an integer such that n ≥ 2. Let x1, x2, ..., xn be distinct real numbers, and let p1, p2, ..., pn be positive numbers such that p1 + p2 + ... + pn = 1. Prove that the fraction
((p1x13 + p2x23 + ... + pnxn3) - (p1x1 + p2x2 + ... + pnxn)3) / (3 ((p1x12 + p2x22 + ... + pnxn2) - (p1x1 + p2x2 + ... + pnxn)2))
is not smaller than the smallest of the numbers x1, x2, ..., xn, but not greater than the greatest of these numbers either.

• Solution to: Panagiote Ligouras, Problem 11435, AMM May 2009.
My solution as a PDF file.
For every triangle with sidelengths a, b, c, inradius r and circumradius R, prove that
a2bc / ((a+b)(a+c)) + b2ca / ((b+c)(b+a)) + c2ab / ((c+a)(a+b)) ≤ 9/2 r R.

• Solution to: Greg Oman, Problem 11451, AMM 2009.
My solution as a PDF file.
Let k and n be positive integers satisfying k > 1. Let R be a ring (not necessarily having an identity, and not necessarily commutative). Assume that:
(i) There exists an element of R which is not nilpotent.
(ii) If x1, x2, ..., xk are any k nonzero elements of R, then x1n + x2n + ... + xkn = 0.
Prove that R is a division ring, that is, the nonzero elements of R form a group under multiplication.

• Solution to: Donald Knuth, Problem 11452, AMM 2009.
My solution as a PDF file.
Let n be a positive integer.
We identify every permutation π ∈ Sn with the n-letter string π(1) π(2) ... π(n) over the alphabet {1,2,...,n}. Let us say that two permutations a1a2...ak-1akak+1...an and akak-1...a2a1ak+1...an (written as n-letter strings) are pre-equivalent when k = n or when ak+1 exceeds all of a1, a2, ..., ak. Also we say that two permutations in Sn are equivalent whenever they can be obtained from each other by a sequence of such flips (i. e., we define the relation ”equivalent” to be the reflexive, symmetric and transitive closure of the relation ”pre-equivalent”). For example, 321 ≡ 123 ≡ 213 ≡ 312 and 132 ≡ 231 in S3.
Show that the number of equivalence classes of permutations in Sn is equal to the n-th Euler secant-and-tangent number for every n. (The n-th Euler secant-and-tangent number is defined as the number of ”up-down” permutations of length n, by which we mean permutations like 25341 (written as strings) that alternately rise and fall beginning with a rise.)

• Solution to: Richard Stanley, Problem 11453, AMM 2009.
My solution as a PDF file. Longer (more detailed) version.
(Generalized version of the problem:) Let Δ be a finite collection of finite sets. Assume that every subset of every set which lies in Δ must itself lie in Δ. Let k be a nonnegative integer such that every G in Δ such that |G| ≤ k satisfies
2k+1-|G| | (sum of (-1)|F| over all F in Δ such that G is a subset of F).
Then, prove that 2k+1 | |Δ|.

• Solution to: Cezar Lupu and Vicentiu Radulescu, Problem 11458, AMM October 2009.
My solution as a PDF file.
Let a1, a2, ..., an be nonnegative reals, and r be a positive integer. Prove that
(sum of (irjraiaj) / (i+j-1) over all 1 ≤ i,j ≤ n)2 ≤ (sum of mr-1am over all m = 1,2,...,n) · (sum of (irjrkraiajak) / (i+j+k-2) over all 1 ≤ i,j,k ≤ n).

One day I might add a solution to #11397 (I am still pondering whether to write up a proof of Karamata...).

Project PEN (Problems in Elementary Number Theory)

• Solution to: Problem E16, a. k. a.: Mathematics Magazine, Problem 1392 by George Andrews.
(This links to my solution on the PEN server. A local version can be found here: Solution to Project PEN Problem E 16.)

If n is a positive integer, and p is a prime lying in the interval ]n, 4n/3], then prove that p divides sum_{j=0}^{n} binom(n,j) 4, where binom(n,j) means n! / (j! (n-j)!).

My solution (which generalizes the problem three times) is an edited and extended version of my posting in MathLinks topic #150539 (which only generalizes it one time).

• Other PEN solutions posted by me on MathLinks. You can find them all on the MathLinks forum in a better readable format; I am just including them here for the sake of completeness.

Octogon

This was a Vietnamese project at creating a journal for mathematical problems. Sadly, it has not kicked off.
• A modulus and square root inequality.
PDF file. Also, the sourcecode.

For any four reals a, b, c, d, prove that

|a + b - c - d| + |a + c - b - d| + |a + d - b - c| ≥ |sqrt(a2 + b2) - sqrt(c2 + d2)| + |sqrt(a2 + c2) - sqrt(b2 + d2)| + |sqrt(a2 + d2) - sqrt(b2 + d2)|.

Problems from the Book (diverse mathematical problems)

• Solution to: Problem 19.9.
PDF file.

Let n be a positive integer. Let w1, w2, ..., wn be n reals. Prove the inequality

sum_{i=1}^{n} sum_{j=1}^{n} ijwiwj / (i+j-1) ≥ (sum_{i=1}^{n} wi)².

Click here for a longer and more complete version of the solution. (Warning: the completeness comes at the cost of readability.)

Solutions to several problems in German

• Solution to 4th QEDMO, Problem 13.
PDF file.
This is a more detailed and (in my opinion) more readable version of
the solution which I posted on MathLinks. Click on this link to see other solutions as well.

Let n and k be integers such that 0
k n. Prove that

sum_{u=0}^{k} ((n+u-1) binom u) (n binom (k-2u)) = ((n+k-1) binom k).

Here, (a binom b) means the binomial coefficient a! / (b! (a-b)!) if 0
b a, and 0 otherwise.

Other problems and solutions

All sources: Mathematical Reflections, American Mathematical Monthly, QEDMO, PFTB and PEN.

Solutions to review problems

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Darij Grinberg