SOME CHEMISTRY INVOLVED
We discuss a certain category of molecules called fullerenes. These are molecules composed of only carbon atoms in the shape of a hollow sphere, ellipsoid (a 3-d oblong shape) or tube. Their construction has been proven to be significant in research areas involving patents for material science, electronics, and nanotechnology. However, fullerenes are difficult to observe naturally. Thus, what we do instead is study their structure using theoretical tools (hint hint: graph theory) before performing experiments. To do so, we consider each bond between atoms as edges and the atoms themselves as vertices.
What we want to do at this point is determine a Hamiltonian path or cycle inside the graph we've just constructed. A Hamiltonian path is a set of edges that, taken in order, visits each vertex exactly once. A Hamiltonian cycle also visits each vertex exactly once with the added condition that the last edge should link back to the first vertex visited. The existence of Hamilton paths and cycles in fullerene graphs can be related to energy levels and chirality, which has something to do with the non-symmetry in molecules.
Leapfrog-fullerenes
Consider a graph \(F\). Let's call each region enclosed by a set of vertices with no vertex inside it a face. We now add a vertex inside each face and connect it to the vertices associated to the face we place it in. This process is called omnicapping, and results in a new graph which we call \(\text{Omni}(F)\).
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| Figure 1: The graphs \(F\) and \(\text{Omni}(F)\). |
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| Figure 2: The graphs \(\text{Omni}(F)\) and \(\text{Le}(F)\). |
The goal is to create a Hamiltonian path in \(\text{Le}(F)\). Specifically, we want to choose \(\frac{\left(3n-4\right)}{4}\) vertices that will correspond to faces in \(\text{Le}(F)\). Let \(S\) be the set consisting of our chosen vertices, with \(S^{c}\) referring to the complement of \(S\) or the vertices we didn't choose. A series of construction procedures can be done to arrive at \(S\). For example, consider the planar projection of the molecule \(C_{20}\), which is a fancy way of calling the flattened representation of \(C_{20}\) on paper (the plane).
What should always be kept in mind during construction are the following properties:
- No vertices in \(S^{c}\) are adjacent, except for the starting edge;
- the distance from a vertex in \(S^{c}\) to another vertex in the same set should be at most 3;
- For every face of the graph there should be at least one vertex in \(S^{c}\); and
- For every face of the graph there should be at least two vertices in \(S\). These vertices need not be adjacent.
The figure below is the result of the correspondence between the vertices of \(C_{20}\) which are members of \(S\) and the faces of \(C_{60}\) which are shaded, the boundary of which is a cycle on 58 vertices. We connect the shaded region to 2 adjacent vertices to complete the Hamiltonian path in \(C_{60}\).
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| Figure 3: \(C_{20}\) and its leapfrog-fullerene \(C_{60}\) containing a Hamiltonian path. |
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| Figure 4: \(C_{26}\) and its leapfrog-fullerene \(C_{78}\) containing a Hamiltonian cycle. |
ABOUT THE AUTHOR:
Victor Andrew Antonio is a Lecturer at the Ateneo de Manila University. He obtained his B.S. in Mathematics at the Ateneo de Manila University in 2012 and is currently taking his M.S. in Mathematics at the same university.
REFERENCES:
[1] Grünbaum, B. and T. S. Motzkin. The number of hexagons and the simplicity of geodesics on certain polyhedra. Canad: J: Math: 1963, 15. 744-751.[2] Johnson, C. Molecular Graph Theory. Master's Thesis. 2010. 1-23.
[3] Marušič, D. Hamilton Cycles and Paths in Fullerenes. J: Chem: Inf: Model: 2007, 47. 732-736.
[4] Rouvray, D.H. and D. Bonchev. Chemical Topology Introduction and Fundamentals. Gordon and Breach Science. Amsterdam. 1999.
OLYMPIAD CORNER
from the Czech and Slovak Republic National Contest, 2000
Problem: Show that \[ \sqrt[3]\frac{a}{b}+\sqrt[3]\frac{b}{a} \leq 2\sqrt[3](a+b)\left(\frac{1}{a}+\frac{1}{b}\right) \] for all positive real numbers \(a\) and \(b\), and determine when equality occurs.
Solution:
Multiplying both sides of the desired inequality by \(\sqrt[3]{ab}\) gives the equivalent inequality \[ \sqrt[3]{a^2}+\sqrt[3]{b^2} \leq \sqrt[3]{2(a+b)^2}. \] Setting \(\sqrt[3]{a}=x\) and \(\sqrt[3]{b}=y\), we see that it suffices to prove that \[ x^2+y^2 \leq \sqrt[3]{2(x^3+y^3)^2} \qquad \ast \] for \(x,y>0\). By the AM-GM inequality, \[ 3x^4y^2 \leq x^6+x^3y^3+x^3y^3 \text{ and } 3x^2y^4 \leq y^6+x^3y^3+x^3y^3, \] with equality if and only if \(x^6=x^3y^3=y^6\), or equivalently if and only if \(x=y\). Adding these two inequalities and adding \(x^6+y^6\) to both sides yields \[ x^6+y^6+3x^2y^2(x^2+y^2) \leq 2(x^6+y^6+2x^3y^3). \] Taking the cube root of both sides yields \(\ast\), as desired. Equality occurs when \(x=y\), or equivalently when \(a=b\).
Multiplying both sides of the desired inequality by \(\sqrt[3]{ab}\) gives the equivalent inequality \[ \sqrt[3]{a^2}+\sqrt[3]{b^2} \leq \sqrt[3]{2(a+b)^2}. \] Setting \(\sqrt[3]{a}=x\) and \(\sqrt[3]{b}=y\), we see that it suffices to prove that \[ x^2+y^2 \leq \sqrt[3]{2(x^3+y^3)^2} \qquad \ast \] for \(x,y>0\). By the AM-GM inequality, \[ 3x^4y^2 \leq x^6+x^3y^3+x^3y^3 \text{ and } 3x^2y^4 \leq y^6+x^3y^3+x^3y^3, \] with equality if and only if \(x^6=x^3y^3=y^6\), or equivalently if and only if \(x=y\). Adding these two inequalities and adding \(x^6+y^6\) to both sides yields \[ x^6+y^6+3x^2y^2(x^2+y^2) \leq 2(x^6+y^6+2x^3y^3). \] Taking the cube root of both sides yields \(\ast\), as desired. Equality occurs when \(x=y\), or equivalently when \(a=b\).
PROBLEMS
- Find all pairs \((x,y)\) of two positive integers such that \(N = 11x + 99y\) is a perfect square number less than or equal to \(1199\) and \(x+y\) is also a perfect square.
- Show that for all positive integers \(x, y, z\) \[ \left(\frac{x+y}{x+y+z}\right)^{\frac{1}{2}} + \left(\frac{x+z}{x+y+z}\right)^{\frac{1}{2}} +\left(\frac{y+z}{x+y+z}\right)^{\frac{1}{2}} \leq 6^{\frac{1}{2}}. \]
- For a fixed positive integer \(n\), find the minimum value of the sum \[ x_1 + \frac{{x_2}^2}{2} + \frac{{x_2}^3}{3} + \ldots + \frac{{x_2}^n}{n}, \]given that \(x_1, x_2, \ldots, x_n\) are positive numbers satisfying the property that the sum of their reciprocals is \(n\).
We welcome readers to submit solutions to the problems posed below for publication consideration. Solutions may be submitted to the PEM facilitators on the deadline date or online via mactolentino@math.admu.edu.ph. Solutions must be preceded by the solver's name, school affiliation and year level. The deadline for submission is 4:00 PM February 2, 2013.
SOLUTIONS
(for January 19, 2013)
- Let \(n \geq 2\) be an integer and \(a_1, a_2, \ldots, a_n\) be real numbers. Prove that for any nonempty subset \(S \subset \{1, 2, \ldots, n\}\), the following inequality holds: \[ \left(\sum_{i \in S}a_i\right)^2 \leq \sum_{1 \leq i \leq j \leq n}(a_i + \ldots + a_j)^2. \] (Romanian Mathematical Olympiad, 2004)(Solved by Farrell Eldrian Wu [MGC New Life Christian Academy])
SOLUTION:Let \(S = \{i_1, i_1+1, \ldots, j_1, i_2, i_2+1, \ldots, j_2, \ldots, i_p, \ldots, j_p \}\) be the ordering of \(S\), where \(j_k < i_{k+1}\) for \(k = 1, 2, \ldots, p-1\). Take \(S_p = a_1 + a_2 + \ldots + a_p, S_0 = 0.\) Then \[ \sum_{i \in S}{a_i} = S_{j_p} - S_{i_p-1}+S_{j_{p-1}}-S_{i_{p-1}-1}+\ldots+S_{j_1}-S_{i_1-1} \]and \[ \sum_{1 \leq i \leq j \leq n}(a_i+\ldots+a_j)^2 = \sum_{0 \leq i \leq j \leq n}(S_i - S_j)^2. \]By this construction, it is enough to prove an inequality of the following form: \[(x_1 - x_2 + \ldots + (-1)^{p+1}x_p)^2 \leq \sum_{1 \leq i \leq j \leq p}(x_j - x_i)^2 + \sum_{i=1}^p{x_i^2}, \]in effect ignoring the same non-negative terms on the right-hand side. This now reduces to \[4\sum_{\substack{1 \leq i \leq j \leq n \\ j-i \text{ even}}}x_ix_j \leq (p-1)\sum_{i=1}^p{x_i^2}, \]obtained by adding together inequalities of the form \[ 4x_ix_j \leq 2(x_i^2 + x_j^2), i < j, j - i = \text{even}. \]For odd \(i\), \(x_i\) appears in an inequality of this form \(\left[\frac{p-1}{2}\right]\) times and for even \(i\) it appears \(\left[\frac{p}{2}-1\right]\) times, where \([x]\) refers to the integer part of \(x\). - Show that among all boxes with a given surface area, the cube has the largest volume. (Taken from The Cauchy-Schwartz Master Class by J. Michael Steele)(Solved by Clyde Wesley Ang [Chiang Kai Shek College], Jecel Manabat [Valenzuela City Sci HS], Hans Jarett Ong [Chiang Kai Shek College], Jan Kedrick Ong [Chiang Kai Shek College], Terence Tsai [Chiang Kai Shek College] and Farrell Eldrian Wu [MGC New Life Christian Academy])SOLUTION:Let \(a\), \(b\) and \(c\) be the dimensions of the box, with surface area \(A = 2ab + 2ac + 2bc\). Note that if the box is a cube, then the edge length can be expressed in terms of \(A\) as \(\sqrt{\frac{A}{6}}\). Thus, what we wish to prove is that \[ abc \leq \left(\frac{A}{6}\right)^{\frac{3}{2}}. \]By the AM-GM inequality, we have \[ 2(a^2b^2c^2)^{\frac{1}{3}} = [(2ab)(2ac)(2bc)]^{\frac{1}{3}} \leq \frac{2ab+2ac+2bc}{3} = \frac{A}{3}, \]where equality holds if and only if \(ab = ac = bc\). But this equality holds if and only if \(a = b = c\) and we are done.
- Three circles with the same radius pass through a common point. Let \(S\) be the set of points which are interior to at least two of the circles. How should the three circles be placed so that the area of \(S\) is minimized? (French Mathematical Olympiad, 1995)(Solved by Clyde Wesley Ang [Chiang Kai Shek College], Hans Jarett Ong [Chiang Kai Shek College], Jan Kedrick Ong [Chiang Kai Shek College] and Farrell Eldrian Wu [MGC New Life Christian Academy])
SOLUTION:Suppose the circles have unit radius. Let \(P\) be the common point of the circles and let \(A\), \(B\), \(C\) be the second intersection points of each pair of circles. We wish to minimize the common area between any two pairs of circles, which will happen when \(P\) is an interior point of \(\Delta ABC\).
Should this be the case, the area of \(S\) is equal to \(\pi - (\sin \alpha + \sin \beta + \sin \gamma)\), where \(\alpha\), \(\beta\) and \(\gamma\) are the central angles of the common arcs of the circles. Note that \(\alpha + \beta + \gamma - 180^\circ\). Since \(\sin x\) is a concave function the area is minimized when \(\alpha = \beta = \gamma = 60^\circ\), which implies that the centers of the circles form an equilateral triangle.
- A line that contains the centroid \(G\) of the triangle \(ABC\) intersects the side \(AB\) at \(P\) and the side \(CA\) at \(Q\). Prove that \[ \frac{PB}{PA}\cdot\frac{QC}{QA} \leq \frac{1}{4}. \](Spanish Mathematical Olympiad, 1998)(Solved by Farrell Eldrian Wu [MGC New Life Christian Academy]; partial credit for Clyde Wesley Ang [Chiang Kai Shek College], Hans Jarett Ong [Chiang Kai Shek College] and Jan Kedrick Ong [Chiang Kai Shek College])SOLUTION:Note that by the inequality \[ \frac{PB}{PA}\cdot\frac{QC}{QA} \leq \frac{1}{4}\left(\frac{PB}{PA}+\frac{QC}{QA}\right)^2 \]all we have to do is show that \(\frac{PB}{PA}+\frac{QC}{QA} = 1\).
We draw \(BB'\) and \(CC'\) such that they are both parallel to the median \(AA'\) and \(B'\) and \(C'\) are on the line containing \(PQ\). Then \(\Delta APG\) is similar to \(\Delta BPB'\) and \(\Delta AQG\) is similar to \(\Delta CQC'\), implying that \(\frac{PB}{PA} = \frac{BB'}{AG}\) and \(\frac{QC}{QA} = \frac{CC'}{AG}\). Furthermore, \(AG = 2GA' = BB' + CC'\), and the conclusion follows.
ERRATA
Clyde Wesley Ang was not properly credited for answering Problem #14 and having a partial solution to Problem #15.
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