==> competition/tests/math/putnam/putnam.1988.s <== % % Layout % \def\layout{\hsize 150truemm % 210mm - 1in * 2 for margins \vsize 246truemm % 297mm - 1in * 2 for margins \advance \vsize by -24 pt \topskip=10pt plus 1 pt} \magnification=1200 \layout \parindent=0pt \parskip=\medskipamount \newdimen\digitindent \digitindent=16truemm \newdimen\solindent \solindent=24truemm \newdimen\problemskip\newdimen\subproblemskip \problemskip=\bigskipamount \subproblemskip=\medskipamount \hoffset=\digitindent \def\problem #1 { \vskip \problemskip\hskip-\digitindent\hbox to\digitindent {\bf #1\hfill}\ignorespaces} \def\solution #1 { \vskip \problemskip\hskip-\solindent\hbox to\solindent {\bf #1\hfill}\ignorespaces} \def\notes #1 { \vskip \problemskip\hskip-\solindent\hbox to\solindent {\bf #1\hfill}\ignorespaces} \def\subproblem #1 {\vskip \subproblemskip\hskip-\digitindent \hbox to\digitindent{\hfill{\bf #1}\hfill}\ignorespaces} \def\endpage {\vfill\supereject\dosupereject} \headline={\headlinemacro} \footline={\hfil} \def\activeheadline{\hglue -\digitindent Chris Long, Rutgers University \hfill January 22, 1989} \let\headlinemacro=\activeheadline \advance \baselineskip by 6pt % % Aliases % \def\streck{\hbox {\vbox {\vfill \hrule width 0.5pt height 6pt \vskip 0.7pt}}} \def\C{\hbox{\hskip 1.5pt \streck \hskip -3.2pt {\rm C}}} \def\Q{\hbox{ \streck \hskip -3pt {\rm Q}}} \def\negativethinspace{\mskip-\thinmuskip} \def\R{\hbox{$\rm I \negativethinspace R $}} \def\Z{\hbox{$\rm Z \negativethinspace \negativethinspace Z $}} \def\N{\hbox{$\rm I \negativethinspace N $}} \def\H{\hbox{$\rm I \negativethinspace H $}} \def\adj{\rm adj} \def\det{\rm det} \def\Matrix#1{\left\lgroup\matrix{#1}\rgroup\right} \def\mod #1 {({\rm mod~}#1)} \def\Mod #1 {\ ({\rm mod~}#1)} \def\ceil#1{\lceil #1 \rceil} \def\floor#1{\lfloor #1 \rfloor} % % Body of article % \problem {A-1:} Let $R$ be the region consisting of the points $(x,y)$ of the cartesian plane satisfying both $|x| - |y| \le 1$ and $|y| \le 1$. Sketch the region $R$ and find its area. \solution {Solution:} The area is $6$; the graph I leave to the reader. \problem {A-2:} A not uncommon calculus mistake is to believe that the product rule for derivatives says that $(fg)' = f'g'$. If $f(x) = e^{x^{2}}$, determine, with proof, whether there exists an open interval $(a,b)$ and a non-zero function $g$ defined on $(a,b)$ such that the wrong product rule is true for $x$ in $(a,b)$. \solution {Solution:} We find all such functions. Note that $(fg)' = f'g' \Rightarrow f'g' = f'g + fg'$ hence if $g(x), f'(x) - f(x) \neq 0$ we get that $g'(x)/g(x) = f'(x)/(f'(x) - f(x))$. For the particular $f$ given, we then get that $g'(x)/g(x) = (2x)e^{x^2}/( (2x-1) (e^{x^2}) ) \Rightarrow g'(x)/g(x) = 2x/(2x-1)$ (since $e^{x^2} > 0$). Integrating, we deduce that $\ln{|g(x)|} = x + (1/2)\ln{|2x-1|} + c$ (an arbitrary constant) $\Rightarrow |g(x)| = e^{c} \sqrt{|2x-1|} e^{x} \Rightarrow g(x) = C \sqrt{|2x-1|} e^{x}, C$ arbitrary $\neq 0$. We finish by noting that any $g(x)$ so defined is differentiable on any open interval that does not contain $1/2$. Q.E.D. \problem {A-3:} Determine, with proof, the set of real numbers $x$ for which $\sum_{n=1}^{\infty} {( {1\over n} \csc ({1\over n}) - 1)}^{x}$ converges. \solution {Solution:} The answer is $x > {1\over 2}$. To see this, note that by Taylor's theorem with remainder $\sin( {1\over{n}} ) = \sum_{i=1}^{k-1} { {(-1)}^{i-1} {n}^{-(2i+1)} } + c { {(-1)}^{k-1} {n}^{-(2k+1)} }$, where $0 \leq c \leq {1\over n}$. Hence for $n\geq 1 ( 1/n )/( 1/n - 1/(3! n^{3}) + 1/(5! n^{5}) - 1 < (1/n) \csc(1/n) - 1 < ( 1/n )/( 1/n - 1/(3! n^{3}) ) - 1 \Rightarrow$ for $n$ large enough, $ (1/2) 1/(3! n^{2}) < (1/n) \csc(1/n) - 1 < 2\cdot 1/(3! n^{2})$. Applying the p-test and the comparison test, we see that $\sum_{n=1}^{\infty} {( {1\over n} \csc({1\over n}) - 1)}^{x}$ converges iff $x > {1\over 2}$. Q.E.D. \problem {A-4:} Justify your answers. \subproblem {(a)} If every point on the plane is painted one of three colors, do there necessarily exist two points of the same color exactly one inch apart? \solution {Solution:} The answer is yes. Assume not and consider two equilateral triangles with side one that have exactly one common face $\Rightarrow$ all points a distance of $\sqrt{3}$ apart are the same color; now considering a triangle with sides $\sqrt{3}, \sqrt{3}, 1$ we reach the desired contradiction. Here is a pretty good list of references for the chromatic number of the plane (i.e., how many colors do you need so that no two points 1 away are the same color) up to around 1982 (though the publication dates are up to 1985). This asks for the chromatic number of the graph where two points in R^2 are connected if they are distance 1 apart. Let this chromatic number be chi(2) and in general let chi(n) be the chromatic number of R^n. By a theorem in [2] this is equivalent to finding what the maximum chromatic number of a finite subgraph of this infinite graph is. [1] H. Hadwiger, ``Ein Ueberdeckungssatz f\"ur den Euklidischen Raum,'' Portugal. Math. #4 (1944), p.140-144 This seems to be the original reference for the problem [2] N.G. de Bruijn and P. Erd\"os, ``A Color Problem for Infinite Graphs and a Problem in the Theory of Relations,'' Nederl. Akad. Wetensch. (Indag Math) #13 (1951), p. 371-373. [3] H. Hadwiger, ``Ungel\"oste Probleme No. 40,'' Elemente der Math. #16 (1961), p. 103-104. Gives the upper bound of 7 with the hexagonal tiling and also a reference to a Portugese journal where it appeared. [4] L. Moser and W. Moser, ``Solution to Problem 10,'' Canad. Math. Bull. #4 (1961), p. 187-189. Shows that any 6 points in the plane only need 3 colors but gives 7 points that require 4 (``the Moser Graph'' see [7]). [5] Paul Erd\"os, Frank Harary, and William T. Tutte, ``On the Dimension of a Graph,'' Mathematika #12 (1965), p. 118-122. States that 34.'' States a question of L. Moser: Let R be large and S a measurable set in the circle of radius R so that no two points of S have distance 1. Denote by m(S) the measure of S. Determine Lim_{R => infty} max m(S)/R^2. Erd\"os conjectures that this limit is less than 1/4. Erd\"os asks the following: ``Let S be a subset of the plane. Join two points of S if their distances is 1. This gives a graph G(S). Assume that the girth (shortest circuit) of G(S) is k. Can its chromatic number be greater than 3? Wormald proved that such a graph exists for k<6. The problem is open for k>5. Wormald suggested that this method may work for k=6, but probably a new idea is needed for k>6. A related (perhaps identical) question is: `Does G(S) have a subgraph that has girth k and chromatic number 4?' '' [7] N. Wormald, ``A 4-chromatic graph with a special plane drawing,'' J. Austr. Math. Soc. Ser. A #28 (1970), p. 1-8. The reference for the above question. [8] R.L. Graham, ``Old and New Euclidean Ramsey Theorems,'' in ``Discrete Geometry and Convexity,'' Edited by Jacob E. Goodman, Erwin Lutwak, Joseph Malkevitch, and Richard Pollack, Annals of the New York Academy of Sciences Vol. 440, New York Academy of Sciences 1985, Pages 20-30. States that the best current bounds are 3 0$ for all $x > 0$. \solution {Solution 1:} Clearly $f(x) = 2x$ is one such solution; we need to show that it is the {\it only} solution. Let $f^{1}(x) = f(x), f^{n}(x) = f(f^{n-1}(x))$ and notice that $f^{n}(x)$ is defined for all $x>0$. An easy induction establishes that for $n>0 f^{n}(x) = a_{n} x + b_{n} f(x)$, where $a_{0} = 0, b_{0} = 1$ and $a_{n+1} = 6 b_{n}, b_{n+1} = a_{n} - b_{n} \Rightarrow b_{n+1} = 6 b_{n-1} - b_{n}$. Solving this latter equation in the standard manner, we deduce that $\lim_{n\to \infty} a_{n}/b_{n} = -2$, and since we have that $f^{n}(x) > 0$ and since $b_{n}$ is alternately negative and positive; we conclude that $2x \leq f(x) \leq 2x$ by letting $n \rightarrow \infty$. Q.E.D. \solution {Solution 2:} (Dan Bernstein, Princeton) As before, $f(x) = 2x$ works. We must show that if $f(x) = 2x + g(x)$ and $f$ satisfies the conditions then $g(x) = 0$ on ${\R}^{+}$. Now $f(f(x)) = 6x - f(x)$ means that $2f(x) + g(f(x)) = 6x - 2x - g(x)$, i.e., $4x + 2g(x) + g(f(x)) = 4x - g(x)$, i.e., $3g(x) + g(f(x)) = 0$. This then implies $g(f(f(x))) = 9g(x)$. Also note that $f(x) > 0$ implies $g(x) > -2x$. Suppose $g(x)$ is not $0$ everywhere. Pick $y$ at which $g(y) \neq 0$. If $g(y) > 0$, observe $g(f(y)) = -3g(y) < 0$, so in any case there is a $y_{0}$ with $g(y_{0}) < 0$. Now define $y_{1} = f(f(y_{0})), y_{2} = f(f(y_{1}))$, etc. We know $g(y_{n+1})$ equals $g(f(f(y_{n}))) = 9g(y_{n})$. But $y(n+1) = f(f(y_{n})) = 6y_{n} - f(y_{n}) < 6y_{n}$ since $f>0$. Hence for each $n$ there exists $y_{n} < 6^{n} y_{0}$ such that $g(y_{n}) = 9^{n} g(y_{0})$. The rest is obvious: $0 > g(y_{0}) = 9^{-n} g(y_{n}) > -2\cdot 9^{-n} y_{n} > -2 (6/9)^{n} y_{0}$, and we observe that as $n$ goes to infinity we have a contradiction. Q.E.D. \problem {A-6:} If a linear transformation $A$ on an $n$-dimensional vector space has $n+1$ eigenvectors such that any $n$ of them are linearly independent, does it follow that $A$ is a scalar multiple of the identity? Prove your answer. \solution {Solution:} The answer is yes. First note that if $x_{1}, \ldots, x_{n+1}$ are the eigenvectors, then we must have that $a_{n+1} x_{n+1} = a_{1} x_{1} + \cdots + a_{n} x_{n}$ for some non-zero scalars $a_{1}, \ldots, a_{n+1}$. Multiplying by $A$ on the left we see that $\lambda_{n+1} a_{n+1} x_{n+1} = \lambda_{1} a_{1} x_{1} + \cdots + \lambda_{n} a_{n} x_{n}$, where $\lambda_{i}$ is the eigenvalue corresponding to the eigenvectors $x_{i}$. But since we also have that $\lambda_{n+1} a_{n+1} x_{n+1} = \lambda_{n+1} a_{1} x_{1} + \cdots + \lambda_{n+1} a_{n} x_{n}$ we conclude that $\lambda_{1} a_{1} x_{1} + \cdots + \lambda_{n} a_{n} x_{n} = \lambda_{n+1} a_{1} x_{1} + \cdots + \lambda_{n+1} a_{n} x_{n} \Rightarrow a_{1} (\lambda_{1} - \lambda_{n+1}) x_{1} + \cdots + a_{n} (\lambda_{n} - \lambda_{n+1}) x_{1} = 0 \Rightarrow \lambda_{1} = \cdots = \lambda_{n+1} = \lambda$ since $x_{1}, \ldots, x_{n}$ are linearly independent. To finish, note that the dimension of the eigenspace of $\lambda$ is equal to $n$, and since this equals the dimension of the nullspace of $A - \lambda I$ we conclude that the rank of $A - \lambda I$ equals $n - n = 0 \Rightarrow A - \lambda I = 0$. Q.E.D. \problem {B-1:} A {\it composite} (positive integer) is a product $ab$ with $a$ and $b$ not necessarily distinct integers in $\{ 2,3,4,\ldots \}$. Show that every composite is expressible as $xy + xz + yz + 1$, with $x, y$, and $z$ positive integers. \solution {Solution:} Let $x = a-1, y = b-1, z = 1$; we then get that $xy + xz + yz + 1 = (a-1)(b-1) + a-1 + b-1 + 1 = ab$. Q.E.D. \problem {B-2:} Prove or disprove: If $x$ and $y$ are real numbers with $y \geq 0$ and $y(y+1) \leq {(x+1)}^2$, then $y(y-1) \leq x^2$. \solution {Solution:} The statement is true. If $x+1 \geq 0$ we have that $\sqrt{y(y+1)} - 1 \leq x \Rightarrow x^{2} \geq y^{2} + y + 1 - 2 \sqrt{y^{2}+y} \geq y^{2} - y$ since $2y + 1 \geq 2 \sqrt{y^{2}+y}$ since ${(2y + 1)}^{2} \geq 4 (y^{2}+y)$ if $y\geq 0$. If $x+1 < 0$, we see that $\sqrt{y(y+1)} \leq -x - 1 \Rightarrow x^{2} \geq y^{2} + y + 1 + 2 \sqrt{y^{2}+y} \geq y^{2} - y$. Q.E.D. \problem {B-3:} For every $n$ in the set ${\Z}^{+} = \{ 1,2,\ldots \}$ of positive integers, let $r(n)$ be the minimum value of $|c-d \sqrt{3}|$ for all nonnegative integers $c$ and $d$ with $c + d = n$. Find, with proof, the smallest positive real number $g$ with $r(n) \leq g$ for all $n$ in ${\Z}^{+}$. \solution {Solution:} The answer is $(1 + \sqrt{3})/2$. We write $|c-d\sqrt{3}|$ as $|(n-d) - d\sqrt{3}|$; I claim that the minimum over all $d, 0 \leq d \leq n$, occurs when $d = e = \floor{n/(1+\sqrt{3})}$ or when $d = f = e+1 = \floor{n/(1+\sqrt{3})} + 1$. To see this, note that $(n-e) - e \sqrt{3} > 0$ and if $e' (n-e) - e \sqrt{3}$, and similarly for $f'>f$. Now let $r = n/(1+\sqrt{3}) - \floor{n/(1+\sqrt{3})}$ and note that $|(n-e) - e \sqrt{3}| = r (1+\sqrt{3})$ and $|(n-f) - f \sqrt{3}| = (1-r) (1+\sqrt{3})$. Clearly one of these will be $\leq (1+\sqrt{3})/2$. To see that $(1+\leq{3})/2$ cannot be lowered, note that since $1+\sqrt{3}$ is irrational, $r$ is uniformly distributed $\mod{1} $. Q.E.D. \notes {Notes:} We do not really need the result that $x$ irrational $\Rightarrow x n - \floor{x n}$ u. d. $\mod{1} $, it would suffice to show that $x$ irrational $\Rightarrow x n - \floor{x n}$ is dense in $(0,1)$. But this is obvious, since if $x$ is irrational there exists arbitrarily large $q$ such that there exists $p$ with $(p,q) = 1$ such that $p/q < x < (p+1)/q$. The nifty thing about the u. d. result is that it answers the question: what number $x$ should we choose such that the density of $\{ n : r(n) < x \}$ equals $t, 0 < t < 1$? The u. d. result implies that the answer is $t (1+\sqrt{3})/2$. The u. d. result also provides the key to the question: what is the average value of $r(n)$? The answer is $(1+\sqrt{3})/4$. \problem {B-4:} Prove that if $\sum_{n=1}^{\infty} a(n)$ is a convergent series of positive real numbers, then so is $\sum_{n=1}^{\infty} {(a(n))}^{n/(n+1)}$. \solution {Solution:} Note that the subseries of terms ${a(n)}^{n\over{n+1}}$ with ${a(n)}^{1\over{n+1}} \leq {1\over 2}$ converges since then ${a(n)}^{n\over{n+1}}$ is dominated by $1/2^{n}$, the subseries of terms ${a(n)}^{n\over{n+1}}$ with ${a(n)}^{1\over{n+1}} > {1\over 2}$ converges since then ${a(n)}^{n\over{n+1}}$ is dominated by $2 a(n)$, hence $\sum_{n=1}^{\infty} {a(n)}^{n\over{n+1}}$ converges. Q.E.D. \problem {B-5:} For positive integers $n$, let $M(n)$ be the $2n + 1$ by $2n + 1$ skew-symmetric matrix for which each entry in the first $n$ subdiagonals below the main diagonal is $1$ and each of the remaining entries below the main diagonal is $-1$. Find, with proof, the rank of $M(n)$. (According to the definition the rank of a matrix is the largest $k$ such that there is a $k \times k$ submatrix with non-zero determinant.) One may note that \break\hfill $M(1) = \left( \matrix{0&-1&1 \cr 1&0&-1 \cr -1&1&0} \right)$ and $M(2) = \left( \matrix{0&-1&-1&1&1 \cr 1&0&-1&-1&1 \cr 1&1&0&-1&-1 \cr -1&1&1&0&-1 \cr -1&-1&1&1&0} \right)$. \solution {Solution 1:} Since $M(n)$ is skew-symmetric, $M(n)$ is singular for all $n$, hence the rank can be at most $2n$. To see that this is indeed the answer, consider the submatrix $M_{i}(n)$ obtained by deleting row $i$ and column $i$ from $M(n)$. From the definition of the determinant we have that $\det(M_{i}(n)) = \sum {(-1)}^{\delta(k)} a_{1 k(1)} \cdots a_{(2n) k(2n)}$, where $k$ is member of $S_{2n}$ (the group of permutations on $\{1,\ldots,2n\}$) and $\delta(k)$ is $0$ if $k$ is an even permutation or $1$ if $k$ is an odd permutation. Now note that ${(-1)}^{\delta(k)} a_{1 k(1)} \cdots a_{(2n) k(2n)}$ equals either $0$ or $\pm 1$, and is non-zero iff $k(i) \neq i$ for all $i$, i.e. iff $k$ has no fixed points. If we can now show that the set of all elements $k$ of $S_{2n}$, with $k(i) \neq i$ for all $i$, has odd order, we win since this would imply that $\det(M_{i}(n))$ is odd $\Rightarrow \det(M_{i}) \neq 0$. To show this, let $f(n)$ equal the set of all elements $k$ of $S_n$ with $k(i) \neq i$ for all $i$. We have that $f(1) = 0, f(2) = 1$ and we see that $f(n) = (n-1) ( f(n-1) + f(n-2) )$ by considering the possible values of $f(1)$ and whether or not $f(f(1)) = 1$; an easy induction now establishes that $f(2n)$ is odd for all $n$. Q.E.D. \notes {Notes:} In fact, it is a well-known result that $f(n) = n! ( 1/2! - 1/3! + \cdots + {(-1)}^{n}/n! )$. \solution {Solution 2:} As before, since $M(n)$ is skew-symmetric $M(n)$ is singular for all $n$ and hence can have rank at most $2n$. To see that this is the rank, let $M_{i}(n)$ be the submatrix obtained by deleting row $i$ and column $i$ from $M(n)$. We finish by noting that ${M_{i}(n)}^{2} \equiv I_{2n} \Mod{2} $, hence $M_{i}(n)$ is nonsingular. Q.E.D. \problem {B-6:} Prove that there exist an infinite number of ordered pairs $(a,b)$ of integers such that for every positive integer $t$ the number $at + b$ is a triangular number if and only if $t$ is a triangular number. (The triangular numbers are the $t(n) = n(n + 1)/2$ with $n$ in $ \{ 0,1,2,\ldots \}$ ). \solution {Solution:} Call a pair of integers $(a,b)$ a {\it triangular pair} if $at + b$ is a triangular number iff $t$ is a triangular number. I claim that $(9,1)$ is a triangular pair. Note that $9 (n(n+1)/2) + 1 = (3n+1)(3n+2)/2$ hence $9t+1$ is triangular if $t$ is. For the other direction, note that if $ 9t+1 = n(n+1)/2 \Rightarrow n = 3k+1$ hence $ 9t+1 = n(n+1)/2 = 9(k(k+1)/2)+1 \Rightarrow t = k(k+1)/2$, therefore $t$ is triangular. Now note that if $(a,b)$ is a triangular pair then so is $(a^{2},(a+1)b)$, hence we can generate an infinite number of triangular pairs starting with $(9,1)$. Q.E.D. \notes {Notes:} The following is a proof of necessary and sufficient conditions for $(a,b)$ to be a triangular pair. I claim that $(a,b)$ is a triangular pair iff for some odd integer $o$ we have that $a = o^{2}, b = (o^{2}-1)/8$. I will first prove the direction $\Leftarrow$. Assume we have $a = o^{2}, b = (o^{2}-1)/8$. If $t = n(n+1)/2$ is any triangular number, then the identity $o^{2} n(n+1)/2 + (o^{2}-1)/8 = (on + (o-1)/2) (on + (o+1)/2)/2$ shows that $at + b$ is also a triangular number. On the other hand if $o^{2} t + (o^{2}-1)/8 = n(n+1)/2$, the above identity implies we win if we can show that $( n - (o-1)/2 )/o$ is an integer, but this is true since $o^{2} t + (o^{2}-1)/8 \equiv n(n+1)/2 \Mod{o^{2}} \Rightarrow 4n^{2} + 4n \equiv -1 \Mod{o^{2}} \Rightarrow {(2n + 1)}^{2} \equiv 0 \Mod{o^{2}} \Rightarrow 2n + 1 \equiv 0 \Mod{o} \Rightarrow n \equiv (o-1)/2 \Mod{o} $. For the direction $\Rightarrow$ assume that $(a,b)$ and $(a,c), c\ge b$, are both triangular pairs; to see that $b = c$ notice that if $at + b$ is triangular for all triangular numbers $t$, then we can choose $t$ so large that if $c>b$ then $at + c$ falls between two consecutive triangular numbers; contradiction hence $b = c$. Now assume that $(a,c)$ and $(b,c)$ are both triangular pairs; I claim that $a = b$. But this is clear since if $(a,c)$ and $(b,c)$ are triangular pairs $\Rightarrow (ab,bc+c)$ and $(ab,ac+c)$ are triangular pairs $\Rightarrow bc+c = ac+c$ by the above reasoning $\Rightarrow bc=ac \Rightarrow$ either $a=b$ or $c=0 \Rightarrow a=b$ since $c=0 \Rightarrow a=b=1$. For a proof of this last assertion, assume $(a,0), a>1$, is a triangular pair; to see that this gives a contradiction note that if $(a,0)$ is a triangular pair $\Rightarrow (a^{2},0)$ is also triangular pair, but this is impossible since then we must have that $a(a^{3}+1)/2$ is triangular (since $a^{2} a (a^{3}+1)/2$ is triangular) but $(a^{2}-1)a^{2}/2 < a(a^{3}+1)/2 < a^{2}(a^{2}+1)/2$ (if $a>1$). We are now done, since if $(a,b)$ is a triangular pair $\Rightarrow a 0 + b = n(n+1)/2$ for some $n\geq 0 \Rightarrow b = ({(2n+1)}^{2} - 1)/8$. Q.E.D. \bye