數學

本研究是[對於正n 邊形A1A2…An邊上一點P(含頂點)，想像自定點P 朝鄰邊發出一條光線，若依逆(順)時針方向依序與每邊皆碰撞一次，經一圈而可回到P 點，則此路徑稱為「光圈」。過程試著追蹤在正n 邊形內能形成光圈的光線行進路徑及其相關問題。 本研究令，且以逆時針得光圈來討論： 1.根據[光的反射原理]，探討光圈之存在性，發現除定點P 在正2m 邊形或正三角形的頂點外，其餘皆有光圈。 2.將可形成光圈的路徑圖展開成[直線路徑圖]來探討。 3.由[直線路徑圖]，觀察到形成光圈的光線行進路徑，可能存在下列情況： (1)不通過正n 邊形的頂點，且產生路徑循環與不循環問題。 (2)通過正n 邊形的頂點。 4.發現正2m 邊形光圈皆為[完美光圈]。 5.發現正2m+1 邊形光圈之路徑與有理數、無理數之特質有關。即當s 值為有理數時，路徑會循環；當s 值為無理數時，路徑不循環。 The research is about [on Point P (including the angles) on the side of regular polygons A1、A2…An , imagine the light goes from Point P to the closest side, then bumps each side sequentially counterclockwise. After going a circle, it’s back to Point P. The track is called “the circle of light.” I try to trace the light track of the circle of light and other correlative questions.] In this research, we suppose,and we discuss the circle of light according counterclockwise direction:1.According to the light reflective principles, we discuss whether the circle of light exists or not. And then we discover that the circle of light really exists except when Point P is on the angles of regular triangle or regular 2m polygons. 2.Spread out the circle of light’s track to [rectilinear track.] 3.By [the picture of rectilinear track], observing there are two kinds of the circle of light’s track: (1)If the light doesn’t go through the angles of regular polygons, it can be a circulative track or a non-circulative track. (2)When the light goes through the angles, it stops. 4.We discover that all the circles of light in regular 2m polygons are [the perfect circles of light.] 5.We discover the circle of light’s track is correlative with rational numbers and irrantional numbers. When s is a rational number, the track is circulative, if s is a irrantional number, the track is not circulative.

這個題目是源自2003年的TRML思考賽的題目，原題目並不難，它只有用到簡單的排列方法，主要是討論 an 、bn 兩種數字的排列，其中 an 為滿足下列所有條件之N位數A的個數。 I. A中每一個數字為1或2 II. A中至少有相鄰的兩數字是1 而 bn 表示滿足下列所有條件的N位數B的個數 I. B中每一個數字為0或1 II. B中至少有相鄰的兩數字是1 以及探討an 、bn 與費氏數列cn之關係，其中 cn = cn-1 + cn-2 ，n≧3 ,c1=1, c2=2 。 其中 an 如果改成考慮為一數列，其值不變；而 bn 如果改為數列，那麼就不需要考慮0不能為首位數字的情況。如此，讓人聯想到一個用生成函數解的題目「一個N項數列，其中每一項只能是0或1或2，其中0和2永不能相鄰，求這個數列個數的一般式。」，因此，我們嘗試將這個題目改變它的要求繼續做下去，發現其中有某些規則，例如：不只是原來的11相鄰，甚至是排列其它種方式，都可能從其遞迴式看出它排列的意義，甚至這種排列數是可以用遞迴式求出來的。這提供了我們另一種求數字排列的方法，也是我們覺得有趣的地方。 在過程中我們初步得到以下結論： This solution is according to power contest of 2003 TRML. It is composed of two number arrangements, an , bn . First, suppose an is the total number conforming to the following rules. I. Each number is 1 or 2 in A. II. There is a couple of (11) in A at least. Then, suppose bn is the total number conforming to the following conditions. I. Each number is 0 or 1 in B. II. There is a couple of (11) in B at least. Furthermore , we give the thought to the relation among an ， bn ，and cn (Fibonacci Sequence). By the way, if an is changed to a sequence, and the result is the same. But if bn is to arrange number, we have to give thought to the fact that the first number can’t be zero. If it is a sequence, we don’t have to consider it. The problem belongs to combinatorics. After we do this problem, we find not only original question but also other permutation can be understood by its formula. The problem provides us with other means to solve permutation and combination question. Then, we get the conclusion as follows：

本專題的目的是研究以任意實數 a1 、 a2 、 a3 為起始的M&m Sequences 之穩定性質。我們主要關心的問題是：(1) 是否任給定三數a1 、 a2 、 a3 為起始的M&m 數列皆會穩定？(2) 若上述的M&m 數列穩定，則其穩定的長度與a1 、 a2 、 a3的關係為何？(3) 其穩定的值與a1 、 a2 、 a3的關係為何？我們研究的主要步驟及結果如下︰1. 當1 2 3 a 1) 為起始的M&m 數列。3. 我們證明了下列性質：(1) 若M&m 數列中前n 項所成數列的中位數為n m ，則下式成立： (2) 當存在 k > 4 , k ? N ，使得 ?1 ?2 = k k m m 成立時，則此數列穩定，且穩定長度p 滿足：min{ | 4 } ?1 ?2 = > = k k p k k 且m m ，其中p 必為奇數。(3) { n m }為單調遞增且, 5 1 ? ? ? a m n n n4. 如果x ? 41.625，則{?x,1, x}為起始的M&m 數列，其對應的數列有相同的大小次序且此M&m 數列會穩定，穩定值為41.625，且穩定長度為73。5. 我們觀察發現：如果x 1). 3. We prove the following properties: (1) If the median of the former n numbers of the M&m sequence is n m , we obtain (2) There exist k > 4 , k ? N such that ?1 ?2 = k k m m , then the sequence is stable and the stable length min{ | 4 }?1 ?2 = > = k k p k k and m m , where p must be an odd number. (3) { n m } is monotone increasing and , 5 1 ? ? ? a m n n n . 4. Suppose x ? 41.625, then the all M&m Sequences beginning with –x , 1 , x are the same, and the sequences will be stable, the stable value is 41.625 and the stable length is 73. 5. By the computer experiments, we observe that if x is any positive real number less than 41.625, the M&m Sequence starting with –x, 1, x, will be also stable but does not appear to follow any clearly discernible pattern of behavior. However, the stable lengths are much variant and exist some unknown relation with point format of x. Moreover, we have the following properties: (1)If x is a node, then the stable value is x and the stable length equals to the index of median of the node + 2； (2)Near the branch of 41.625, the stable length is almost a constant except at the edge area，the stable length of (-x,1,x) as x around branch 1 is chaos； (3)If x near the node (K= 3, 5, 7, …, 67, 69), then the stable length is l(K)+K?1 where the positive integral l(K) is determined by Prop1 (see Table 6 and 7).

「重複圖形」是本篇報告研究的問題，我們利用「方程式」建立一個尋找重複圖形，並証明其個數的方法。利用此方法得出下面的結論：1.會形成lap 2 的凸多邊形只有2 種，即三角形和四邊形。(1)「lap 2 三角形」只有1 種，即等腰直角三角形。(2)「lap 2 四邊形」只有1 種，即二邊之比為1： 且內角是45°、135°的平行四邊形。2.會形成lap 3 的凸多邊形只有2 種，即三角形和四邊形。(1)「lap 3 三角形」只有1 種，即內角為30°–60°–90°的直角三角形。3.其他的lap k 三角形：(1)任意內角為30°–60°–90°的直角三角形都是lap 3k²，其中k是正整數。(2)邊長比為1：m： 的直角三角形是lap (m²+1)k²三角形，其中m、k是正整數。 To find repeated figures, we construct a method to search them with the help of algebraic equations. Here we arrive at：1. There are only two kinds of lap 2 convex polygons, triangles and quadrilaterals. (1) The only lap 2 triangle is isogonal right-angled. (2) The only lap 2 quadrilateral is the one that contains angles 45°, 90° and two neighboring sides with the ratio 1: . 2. There are also two kinds of lap 3 convex polygons, triangles and quadrilaterals. (1) The only lap 3 triangle is the one with angles 30°, 60° and 90°. 3. Other kinds of lap k triangles are listed as following: (1) A triangle with angles 30o, 60°, 90° is a lap 3k², the k is a natural number. (2) A right-angled triangle whose ratio is 1 : m : is a lap (m2+1)k², the m and the k are natural numbers.

完全圖Kn是指一個圖中有n個點，且任意一個點都跟其它的點有邊相連。兩個圖G和H的卡氏乘積G□H的點集V(G□H)＝{(g,h)| g∈V(G),h∈V(H)}，兩個點(g1,h1)和(g2,h2)有邊相連若且為若g1=g2 且h1~h2，或g1~g2且h1=h2。 三個完全圖Ka、Kb、Kc 的立方乘積是指Ka□Kb□Kc。一個圖G中的一點v所連的其它點稱為這個點v的鄰居，也就是N(v)={x | x~v}。一個點集S中的所有點的鄰居的聯集稱為這個點集的鄰居，也就是N(S)=∪v∈S N(v)。如果一個點集S和它的鄰居N(S)包含了一個圖G的所有的點，也就是S∪N(S)＝V(G)稱這個點集S是這個圖G的一個控制集。我們把圖G的所有控制集中點數最少的稱為最小控制集，並定最小控制集的點數為最小控制數γ(G)，也就是γ(G)＝min { | S |, S是G的控制}。 本文的目的在於研究完全圖立方乘積的最小控制，也就是要給γ(Ka□Kb□Kc)一個上界。特別當 a = b = c = n時，γ(Ka□Kb□Kc) = 。 A complete graph Kn is a graph with n vertices, which any vertex is adjacency to every other vertices. The Cartesian product of two graph G and H which is denoted G□H is define as follow: the vertex set V(G□H)＝{(g,h)| g∈V(G),h∈V(H)}，and two vertices (g1,h1) and (g2,h2) is adjacent if and only if g1=g2 and h1~h2 or g1~g2 and h1=h2. The Cartesian product of three complete graph Ka,Kb,Kc is Ka□Kb□Kc,which is the same with (Ka□Kb)□Kc. In a graph G, the neighbor of a vertex v N(v) is the set of the vertices adjacent to the vertex v, that is N(v)={x | x~v}。 The neighbor of a vertex set S is N(S), which is the union of the neighbors of vertex v over S, that is N(S)=∪v∈SN(v). For a graph G, if a vertex set S unions its neighbor N(S) equal to the vertex set of G, that is S∪N(S)＝V(G), we say that S is a dominating set of G. The domination number of a graph G will be denoted as γ(G), which is the minimum size of all dominating set of G.. We give an upper bound to γ(Ka□Kb□Kc). And when a=b=c, γ(Ka□Kb□Kc) ≦

一、若Mn×n(s)表示在n×n 的正方形棋盤中，排列s 顆棋子在方格內，且每一方格最多只能排1子，其中s 顆棋子的配置需滿足兩個條件：1. 並無任意4 子可以形成矩形框的4 個頂點。（此矩形框的邊需與棋盤的邊平行）2. 在沒有棋子的方格中，無法再加入棋子。二、若Vn×n×n(a1,……,an) 表示在n×n×n 的正方體棋盤中，每層的棋子個數分別為a1,……,an，且s= a1+……+an，其中s 顆棋子的配置需滿足兩個條件：1. 並無任意8 子可以形成長方體的8 個頂點。（此長方體的邊需與立體棋盤的邊平行）2. 在沒有棋子的方格中，無法再加入棋子。本研究即在Mn×n(s)與Vn×n×n(a1,……,an) , s= a1+……+an 中探討s 的最小值、最大值及變化情形，並分析其配置方法。之後推廣至長方形Mn×m(s)及長方體Vn×m×k(a1,……,ak) , s= a1+……+ak。最後根據其研究結果設計一個「避開矩形框棋」，並加以分析出致勝的策略。一．If Mn×n(s) indicates in the n×n square chessboard, we put s chesses to line in the square and each square only can put one chess. Then the station of s chesses must satisfy the following two conditions：1. No any 4 chesses can form the tops of the rectangular frame ( The sides of rectangular frame must be parallel to the sides of chessboard )2. If there’s no chess in the square, we can’t add any chess. 二．If Mn×n×n(a1,……,an) indicates in the n×n×n square chessboard, the chess number in each layer are a1,……,an and s= a1+……+an. The station of s chesses must satisfy the following two conditions： 1. No any 8 chesses can form eight tops of the rectangular cube ( The sides of rectangular cube must be parallel to the sides of cubic chessboard ) 2. If there’s no chess in the square, we can’t add any chess. This research try to explore the minimum, maximum and variation of s which in Mn×n(s) and Mn×n×n(a1,……,an), s= a1+……+an, and analyze its station. Then we will extend the research to rectangle Mn×m(s) and rectangular cube Vn×m×k(a1,……,ak), s= a1+……+ak. Finally, according to the result of research we wish can design one “avert rectangular frame chess“ and analyze the strategies to triumph.

本研究由下述問題開始：將n1 個黑色棋子和n2 個白色棋子排成一列，規定第一個棋子必為黑棋；對於每一種排列方法中，同色棋相鄰處記為1，異色棋相鄰處記為-1，所有1 和-1 的總和記為 t (n1,n2 )。對所有可能的排列方法所算出來的t( n1,n2 ) 值求其平均值，記為a (n1,n2 ) 。我們先由觀察各種n1 和n2 值，得到這平均值的可能公式，隨後並嚴格證明其正確性，證明方法也經過多次精鍊到十分簡潔的方式。以此為基礎，我們並做了各方向的推廣，研究涉及下列各點：(一) 利用組合數探討原來的問題。(二) 在第一個棋子不限定為黑棋的假設下，求平均值a( n1,n2 ) 。(三) 將棋子由兩種增加到多種。(四) 改變棋子排列以及相鄰的方式。經由研究，我們發現，每一次愈將問題推廣時，愈能找出清晰的概念涵蓋並印證先前的想法。Our study starts with the following problem. Suppose n1 black chesses and n2 white chesses are arranged in a line under the condition that the first chess is black. For any arrangement of these chesses, an adjacent pair of chesses having the same (respectively, different) colors is associated with a value of 1 (respectively, -1). Let t(n1,n2 ) denote the sum of these values. The purpose of this problem is to calculate the average value a (n1,n2 ) of these t (n1,n2 )which runs over all possible arrangements of the chesses described above. We begin from observing various values of n1 and n2 and find a possible formula for the solution. We then give a rigorous proof for the formula. After some refinements, simple proofs are also established. Based on this, we also make some generalizations. In summary, the research includes the following: 1. Study the problem by using binomial coefficients. 2. Calculate a(n1,n2 ) when t( n1,n2 ) runs over all possible arrangements in which the first chess can be black or white. 3. Increase the types of chesses from two to many. 4. Variant the arrangement method of the chesses from a line to other configurations. During the study, we find that whenever we extend the problem to a more general case, we make the ideas for the original problem clearer.

給定一平面E，A為平面上一點。取r>0，則我們知道到其距離為定值的點形成一圓，而A為此圓圓心。如果把A改成一平面圖形，則到其距離為定值的點形成的集合會是什麼樣子？類似地，給定平面上兩焦點F1及F2在平面上，則到其距離和為定值的點形成橢圓。同樣的，若把F1及F2改成平面圖形，其圖形會是什麼樣子？藉著GSP的輔助，到目前為止，我們得到了以下的結果： \r 1. 給定一平面E及此平面上的一個凸多邊形, 我們描繪出在此平面上到此凸多邊形之距離為定值的點所形成的圖形。\r 2. 設F1和F2分別為平面E上之點或線段或多邊形（未必是凸多邊形），我們利用包絡線描繪出所有滿足d(P,F1)+d(P,F2)=k（k夠大）的點所形成的圖形。 \r 3. 設C1,C2為平面E上之兩圓，我們討論所有滿足 d(P,C1)+d(P,C2)=k\r （k夠大）的點形成的圖形並討論其性質。 \r 4. 設L1和L2分別為平面E上之兩線段，我們討論所有滿足d(P,L1)+d(P,L2)=k（k夠大）的點形成的圖形並討論其性質。 \r 5. 設A為平面E上之一點，Γ為平面上一凸多邊形，我們討論所有滿足d(P,A)+D(P,Γ)=k（k夠大）的點形成的集合並討論其特性。 \r 6. 藉由和圓作比較，我們研究了變形圓的光學性質；而對變形橢圓也做類似的討論。\r Let E be a plane and A a fixed point on E. Given , it is known that all of the points on E with distance to 0r>rA form a circle and the point A is called the center of this circle. What is the corresponding graph if we replace the point A with a set (for example,a segament or a polygon) contained in FE? Similarly, what is the case when we modify the two focuses and in the definition of an ellcpse to sets and (or example,two segments or two polygons) contained in 1F2F1F2FE ? Taking advantages of GSP and analytic geomety, we research related situations and so far we have obtained the following results:\r 1. Let Γ?E be a segment, a convex polygon or a circle , etc. and r>0 be fixed. We sketch the graph of points on E with distance r to Γ and study properties of such graphs.\r 2. Let F1 and F2 be singletons, line segments , polygons(may not be convex), or circles,etc., on E Taking advantage of envelopes, we sketch the graph of those points P on E satisfying d(P,F1)=k(K>0 is large enough).\r 3. Let C1 and C2 be circles on 1C2CE. We sketch the graph of the points P on E that satisfiy d(P,C1)6d(P,C2)=k (k>0 is large enough) and study properties of this graph.\r 4. Let L1 and L2 be two line segments on E and be a large enough constant. We sketch the graph of points P on E that satisfy d(P,L1)+d(P,L2)=k(k >0is large enough) and research properties of this graph. 0k>\r 5. Let A?E and be a convex polygon on ΓE. We sketch the graph of points on E that satisfy d(P,L1)+d(P,L2)=k(k>0 is large enough) and research properties of this graph.\r 6.We compare the optical properties of metamorphic circles with circles and we deal with metamorphic ellipses similiarly.

剛開始考慮平分物件時，我們從二維的多邊形部分著手，後來發現已經有人做過相關研究，並且得到類似的結論。這個部份顯現出面積平分線與其包絡曲線間的密切關係。我們將其中的方法和結果加以歸納、改善，為了更全面地研究，我們推導出一般性的包絡方程。之後當我們推廣到三維領域時，發現四面體體積平分面與之前的結論有些相似之處，平分的情況卻也更複雜，我們將推導的結果用電腦軟體呈現出來，以便更深入地了解它。最後嘗試了相當抽象的高維積平分，結果仍具有工整的對稱性，讓我們充分領略了數學之美！When considering bisecting a subject, at first we focused our attention on 2-D case, polygons. But afterwards, we found there were already some similar studies conducted by other students, which indicated the close relation between the area-bisecting lines of a polygon and their envelope. We rearranged their methods and results, and then made further improvement. Moreover, in order to study the bisecting problem entirely, we derived the general envelope equation. Then when extending the generalization to the 3-D case, we came to the conclusion that tetrahedrons’ volume-bisecting planes is similar to that in 2-D, but the circumstances are more complex. We tried to show our result with the aid of software, hoping to understand it fully. Finally, we tried to do the case in higher dimension, which is very abstract, and the result was clear-cut symmetrical. During the studying process, we had seen “the beauty of mathematics.”