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2017 | Buch

Undergraduate Mathematics Competitions (1995–2016)

Taras Shevchenko National University of Kyiv

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SUCHEN

Über dieses Buch

Versatile and comprehensive in content, this book of problems will appeal to students in nearly all areas of mathematics. The text offers original and advanced problems proposed from 1995 to 2016 at the Mathematics Olympiads. Essential for undergraduate students, PhD students, and instructors, the problems in this book vary in difficulty and cover most of the obligatory courses given at the undergraduate level, including calculus, algebra, geometry, discrete mathematics, measure theory, complex analysis, differential equations, and probability theory. Detailed solutions to all of the problems from Part I are supplied in Part II, giving students the ability to check their solutions and observe new and unexpected ideas. Most of the problems in this book are not technical and allow for a short and elegant solution. The problems given are unique and non-standard; solving the problems requires a creative approach as well as a deep understanding of the material. Nearly all of the problems are originally authored by lecturers, PhD students, senior undergraduates, and graduate students of the mechanics and mathematics faculty of Taras Shevchenko National University of Kyiv as well as by many others from Belgium, Canada, Great Britain, Hungary, and the United States.

Inhaltsverzeichnis

Frontmatter

Problems

Frontmatter
1995

(1-year) Prove that for every $$n\in \mathbb {N}$$ there exists a unique $$t(n)>0$$ such that $$(t(n)-1)\ln t(n)=n$$.

Volodymyr Brayman, Alexander Kukush
1996

Let $$a,b, c\in \mathbb {C}.$$ Find $$\limsup \limits _{n\rightarrow \infty }|a^n+b^n+c^n|^{1/n}$$.

Volodymyr Brayman, Alexander Kukush
1997

Let $$1\le k \le n.$$ Consider all possible decompositions of n into a sum of two or more positive integer summands.

Volodymyr Brayman, Alexander Kukush
1998

See William Lowell Putnam Mathematical Competition, 1996, Problem B1.

Volodymyr Brayman, Alexander Kukush
1999

See Problem 4, 1997.

Volodymyr Brayman, Alexander Kukush
2000

Let $$\{a_n,\, n\ge 1\}$$ be an arbitrary sequence of positive numbers.

Volodymyr Brayman, Alexander Kukush
2001

Problems for 1–2-Year Students.

Volodymyr Brayman, Alexander Kukush
2002

Does there exist a function $$F:\mathbb {R}^2 \rightarrow \mathbb {N}$$ such that the equality $$F(x,y)=F(y, z)$$ holds if and only if $$x=y=z?$$.

Volodymyr Brayman, Alexander Kukush
2003

Problems1–8for 1–2-years students and problems5–12for 3–4-years students

Volodymyr Brayman, Alexander Kukush
2004

Prove that for every positive integer n the inequality.

Volodymyr Brayman, Alexander Kukush
2005

Is it true that a sequence.

Volodymyr Brayman, Alexander Kukush
2006

Find all positive integers n such that the polynomial $$(x^4-1)^n+(x^2-x)^n$$ is divisible by $$x^5-1$$.

Volodymyr Brayman, Alexander Kukush
2007

Problems for 1–2-Years Students.

Volodymyr Brayman, Alexander Kukush
2008

Find $$\inf \{a+b+c\, :\, a,b, c>0,\, \sin a\cdot \sin b\cdot \sin c=\frac{3}{\pi }\, abc\}.$$

Volodymyr Brayman, Alexander Kukush
2009

Problems for 1–2-years students.

Volodymyr Brayman, Alexander Kukush
2010

Does there exist a family of functions $$f_{\alpha }:[0,1]\rightarrow \mathbb {R},$$$$\alpha \in \mathbb {R},$$ such that the intersection of graphs of any two distinct functions from the family contains exactly 3 points, while the intersection of graphs of any three distinct functions contains exactly 2 points.

Volodymyr Brayman, Alexander Kukush
2011

Do there exist two different strictly convex functions with domain [0,1] such that their graphs intersect countably many times.

Volodymyr Brayman, Alexander Kukush
2012

Let $$a_{ij}=\tan (i-j)$$, $$i, j=0,1,\ldots , 2012$$. Find $$\det (a_{ij})$$.

Volodymyr Brayman, Alexander Kukush
2013

Find all continuous functions $$f:[1,2]\rightarrow [1,2]$$ such that $$f(1)=2,$$ and $$f(f(x))f(x)=2\; \text {for all}\; x\in [1,2].$$

Volodymyr Brayman, Alexander Kukush
2014

Prove that $$\cos x<e^{-x^2/2}$$ for all $$0<x\le \pi $$.

Volodymyr Brayman, Alexander Kukush
2015

Is it possible that the middle point and one of the endpoints of a segment belong to the hyperbola $$y=\frac{1}{x},$$ while the other endpoint of the segment belongs to the hyperbola $$y=\frac{8}{x}$$.

Volodymyr Brayman, Alexander Kukush
2016

Find minimal possible value of the expression.

Volodymyr Brayman, Alexander Kukush

Solutions

Frontmatter
1995

The function $$f(t)=(t-1)\ln t$$ is continuous and strictly increasing on the interval $$[1,+\infty ),$$ moreover $$f(1)=0$$ and $$f(t) \rightarrow +\infty ,$$ as $$t\rightarrow +\infty $$.

Volodymyr Brayman, Alexander Kukush
1996

Let $$a=a_0 e^{i\alpha }, $$$$b=b_0 e^{i\beta }, $$ and $$c=c_0 e^{i\gamma },$$ where $$a_0, b_0, c_0\ge 0$$ and $$\alpha , \beta , \gamma \in [0,2\pi ).$$

Volodymyr Brayman, Alexander Kukush
1997

Each summand k corresponds to adjacent numbers of the collection which differ by k. Fix the numbers c and $$c+k,$$ and find the number of collections in which the numbers c and $$c+k$$ are adjacent.

Volodymyr Brayman, Alexander Kukush
1998

Denote by $$\mathbb {N}_n$$ the set $$\{1,2,\ldots , n\}$$ and by $$f_n$$ the number of minimal selfish subsets of $$\mathbb {N}_n.$$

Volodymyr Brayman, Alexander Kukush
1999

A solution is similar to the solution of Problem 7, 1997.

Volodymyr Brayman, Alexander Kukush
2000

If the series $$\mathop {\sum }\limits _{n=1}^{\infty }a_n$$ converges to S then $$b_n\le Sn$$.

Volodymyr Brayman, Alexander Kukush
2001

On a circle of the unit length mark the points $$k\pi ,$$$$1\le k\le n_0.$$.

Volodymyr Brayman, Alexander Kukush
2002

Enumerate all the rational numbers $$\{r_i, i\ge 1\}$$.

Volodymyr Brayman, Alexander Kukush
2003

Decompose the fraction $$\dfrac{9n+4}{n(3n+1)(3n+2)}$$ into a sum of partial fractions.

Volodymyr Brayman, Alexander Kukush
2004

If $$n\ne 3k+2$$ then $$2n-1$$ not divides by 3 and covering is impossible. For $$n=3k+2$$ covering is possible if and only if the figure can be divided into $$2\times 2$$ square which contains the erased cell and several $$2\times 3$$ rectangles.

Volodymyr Brayman, Alexander Kukush
2005

Necessity. Let $$x_n\rightarrow a$$, as $$n\rightarrow \infty $$. Then $$\lim \limits _{m\rightarrow \infty }|x_n-x_m|=|x_n-a|\ \text {and}\ \lim \limits _{n\rightarrow \infty }\limsup _{m\rightarrow \infty }|x_n-x_m|=\lim \limits _{n\rightarrow \infty }|x_n-a|=0$$.

Volodymyr Brayman, Alexander Kukush
2006

The condition of the problem is equivalent to each of the following statements: the polynomial $$(x^3+x^2+x+1)^n+x^n$$ is divisible by $$Q(x)=x^4+x^3+x^2+x+1,$$ or $$(-x^4)^n+x^n$$ is divisible by Q(x), or $$(-1)^nx^{3n}+1$$ is divisible by Q(x).

Volodymyr Brayman, Alexander Kukush
2007

It holds $$\prod \limits _{k=1}^n\frac{(k+p)(k+q)}{(k+r)(k+s)}= \frac{r!s!}{p!q!}\cdot \frac{(n+p)!}{(n+r)!}\cdot \frac{(n+q)!}{(n+s)!}$$.

Volodymyr Brayman, Alexander Kukush
2008

It suffices to consider $$0<a,b, c<\frac{\pi }{2}.$$ For $$0<x<\frac{\pi }{2},$$ put $$f(x)=\ln \left( \frac{\sin x}{x}\right) $$ and define by continuity $$f(0)=0.$$ Because for $$0<x<\frac{\pi }{2}$$ it holds $$f'(x)=\cot \, x-\frac{1}{x}<0,\quad f''(x)=-\frac{1}{\sin ^2x}+\frac{1}{x^2}<0,$$ the function f is decreasing and concave.

Volodymyr Brayman, Alexander Kukush
2009

A quadrilateral ABCD is circumscribed if and only if $$AD+BC=AB+CD$$. Let R be the radius of the circle circumscribed about the quadrilateral ABCD, $$\angle ABC=\beta $$ and $$\angle ABD=\varphi \le \beta $$.

Volodymyr Brayman, Alexander Kukush
2010

The graphs of all the functions $$F_{\alpha }$$ pass through the points (0, 0) and (1, 0). For $$\alpha <\beta ,$$ the graphs of the functions $$F_{\alpha }$$ and $$F_{\beta },$$ besides these two points, intersect at a single point $$\left( \frac{\beta }{1-\alpha +\beta },\frac{1}{1-\alpha +\beta }\right) $$.

Volodymyr Brayman, Alexander Kukush
2011

Let $$f_1(x)=x^2$$ and $$f_2(x)=\frac{1}{2}(x^2+g(x)),\ x\in [0,1],$$ where the function g is defined as follows: $$g(0)=0,$$$$g\left( \frac{1}{n}\right) =\frac{1}{n^2},\ n\ge 1,$$ and g is linear on the segment $$\left[ \frac{1}{n+1},\frac{1}{n}\right] $$ for each $$n\ge 1.$$ It is not difficult to verify that g is convex, hence the functions $$f_1$$ and $$f_2$$ are strictly convex and satisfy the condition of the problem.

Volodymyr Brayman, Alexander Kukush
2012

Notice that $$a_{ij}=-a_{ji}$$, $$i, j=0,1,\ldots , 2012$$, i.e., the matrix $$A=(a_{ij})_{i, j=0}^{2012}$$ is skew-symmetric, $$A^\text {T}=-A.$$ Therefore, $$\begin{aligned} \det \,A=\det \, A^\text {T}=\det (-A)=(-1)^{2013}\det \,A=-\det \, A, \end{aligned}$$whence $$\det \, A=0$$.

Volodymyr Brayman, Alexander Kukush
2013

Substitute $$x=1$$ and get $$f(2)=1$$. Because $$f(1)=2$$ and $$f(2)=1,$$ by the intermediate value theorem for each $$y\in [1,2]$$ there exists $$z\in [1,2]$$ such that $$f(z)=y.$$

Volodymyr Brayman, Alexander Kukush
2014

For $$\pi /2\le x\le \pi ,$$ it holds $$\cos x\le 0<e^{-x^2/2},$$ hence it remains to consider $$0<x<\pi /2$$.After taking the logarithm on both sides the inequality turns to $$\ln \,\cos x<-x^2/2,$$ or equivalently $${x^2/2+\ln \,\cos x<0},$$$$0<x<\pi /2$$. The function $$f(x)=x^2/2+\ln \,\cos x$$ is decreasing on $$[0,\pi /2),$$ because $$f'(x)=x-\tan x<0,$$$${0<x<\pi /2.}$$ Therefore, $$f(x)<f(0)=0,$$$$0<x<\pi /2$$.

Volodymyr Brayman, Alexander Kukush
2015

Let the endpoints of the segment have coordinates $$(a,\frac{1}{a})$$ and $$(b,\frac{8}{b}).$$ Then the middle point has coordinates $$(\frac{a+b}{2},\frac{1}{2a}+\frac{4}{b}).$$ If this point lies on the hyperbola $$y=\frac{1}{x}$$ then $$\frac{a+b}{2}(\frac{1}{2a}+\frac{4}{b})=1,$$ whence $$\frac{1}{4}+\frac{b}{4a}+2+\frac{2a}{b}=1,$$$$\frac{b}{4a}+\frac{2a}{b}=-\frac{5}{4},$$$$\left( \frac{b}{a}\right) ^2+5\frac{b}{a}+8=0.$$ But the quadratic equation $$x^2+5x+8=0$$ has no real roots, a contradiction.

Volodymyr Brayman, Alexander Kukush
2016

If n is divisible by 9, then $$\cos \frac{n\pi }{9}=\pm 1$$ and the expression equals $$4-\root 3 \of {5}.$$ If n is divisible by 3 and is not divisible by 9, then $$\cos \frac{n\pi }{9}=\pm \frac{1}{2}$$ and the expression equals $$1+\root 3 \of {4}.$$ Let n be not divisible by 3.

Volodymyr Brayman, Alexander Kukush
Backmatter
Metadaten
Titel
Undergraduate Mathematics Competitions (1995–2016)
verfasst von
Volodymyr Brayman
Alexander Kukush
Copyright-Jahr
2017
Electronic ISBN
978-3-319-58673-1
Print ISBN
978-3-319-58672-4
DOI
https://doi.org/10.1007/978-3-319-58673-1