數學

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.

三角形的邊上取任意多個點，我們可以把這塊大三角形沿著切割線切割成較小塊的三角形，但切割線必須是點（或頂點）和點的連線，而且必須切割三角形，同時可以切任意大小的三角形，如圖（1）與圖（2）。但不可以一開始就取走整個三角形。定義拿到最後一塊三角形的人獲勝，而在多邊型中的玩法與在三角形中相同。 我們分A、B、C三種規則來討論，其中A規則即是上面提到的玩法，B規則大部分的玩法和A規則都相同，唯一不同的地方在於：A規則中，只要有一方取到剩下的圖形為三角形，另一方就可以直接取走剩下的三角形，而B規則規定即使剩下的圖形已經是三角形，也必須取到剩下的圖形邊上都沒有分點為止。C規則是限制玩家一次所能取的三角形數來進行遊戲。 我們完成了A、B、C規則中三角形與多邊形的必勝策略，並找出必勝策略之間的關聯。 ;Given any numbers of points on the sides of a triangle, the players can cut this triangle into pieces. Each cutting line has to be one, linked between two points given from two different sides. And the player can’t have to cut smaller triangles out of the original triangle. The out-cut triangles can be chosen randomly without any restriction in size, just like what’s shown in picture（1）and（2）. Meanwhile the first player can’t cut the original triangle exactly all out in the very beginning process. We define the player as the winner, who gets the last triangle. And the above way we play can be applies to any multi-side shapes. We discussed the question respectively in three rules, A, B, and C. Rule A is what we mention above. Rule B is generally the same as rule A, except for the only difference：The rule A , if there is any triangle left , the next player can get it directly, but while in rule B, the every next player has to cut out smaller triangles until no point is left on sides. Rule C proceeds on conditions that there is a limitation to a certain number of triangles cut out at a time. We has finished the winning tactic respectively in rule A, B, and C in the games with a triangle and multi-side shapes. Furthermore, we find the connection between the winning tactives.

在一個有障礙的平面上，給三個定點，我們探討連接此三點的最佳網絡。我們討論了諸如直線、射線、線段、圓、網格狀、三角形……等類的障礙，當網絡每穿越障礙一次，就必須付出代價，例如「拖延5 分鐘」。所以，設網絡穿越障礙的次數為y ，則網絡除了原本的總長度之外，還額外加入y 倍某固定數值的損耗。我們以費瑪點的各種性質及三角形不等式等方法為工具，就不同的穿越障礙次數綜合比較，而找出最佳網絡。在某些情況下，最佳網絡不是以費瑪點來連接三點，而是在障礙(如：直線)上找出符合某種與餘弦值相關特殊性質的點，以該點來連接三點，而此網絡可用GSP 軟體相當精確地作出。另外，我們也探討在考慮障礙造成損耗的情況下，兩點間的「實際距離」為何。 最後，我們考慮「混合障礙」問題。在此類問題中，除了前面所討論的障礙，還另加了如同「河流」的兩平行直線間區域之障礙，在這種障礙區域中，網絡的長度要乘以數倍來計算。我們發現，此類問題的最佳網絡也可用特定的正弦條件配合GSP 而相當精確地作出來。;Considering various kinds of obstacles in a plane, such as a line, a segment, a ray, a circle, a triangle or chessboard grids, which function like a red light, we research into the problem of finding the optimal network connecting three given points A, B, C in the plane amidst obstacles described above. Each time when the network crosses an obstacle, it will cause losses, such as five minute’s delay or a loss of one hundred dollars. Taking advantage of Fermat points, some basic inequalities concerning triangles and some special qualities about sine or cosine functions, we obtain the optimal networks in different situations. Besides, we consider what the “real distance” between two points is when there are obstacles in a plane. We also put another obstacle, including a line and a weighted region between two parallel lines, into consideration. In the region, like a river or a muddy ground in real life, the length of the network should be multiplied by a fixed time. Furthermore, we can use GSP to make the networks very accurately.

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.

傳統的畢氏定理三元二次不定方程x² + y² = z²有一組漂亮的整數解為(m² - n²、2mn、m² + n² )；中國數學家嚴鎮軍、盛立人所著的從勾股定理談起一書中記載四元二次不定方程x² + y² + z² = w²的整數解為(mn、m² + mn、mn + n²、m²+ mn + n² )，這組解被我們發現有多處遺漏，本文以擴展的畢氏定理做基礎修正了他的整數解公式，並推廣取得N 元二次不定方程的整數解公式。 There is a beautiful integer solution formula for the Pythagorean theorem equation, x² + y² = z² , such as (m² - n² , 2mn ,m² + n² ). The “m＂ and “n＂ of the solution formula are integer number. A book written by two Chinese mathematicians, Yen Chen-chun and Sheng Li-jen who expanded the Pythagorean theorem equation to the four variables squares’ indeterminate equation, x² + y² + z² = w² . They claimed that they found its integer solution formula, such as (mn , m² + mn , mn + n² , m² + mn + n² ) for any integer “m＂ and “n＂. But we found it losses many solutions. This paper corrected their faults due to the expanded Pythagorean theorem built by ourselves. Further more, we derived a general formula of N variables squares’ indeterminate equation. Now, we can get integer solutions of the equation, (for all natural number “n＂) easily by choosing integers m1 , m2 , m3 ,……, mn−1 up to you.

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.

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) ≦

格子點上的三角形表示這個三角形三頂點的座標皆是整數。本研究先探討用平面格子點可以連出哪些三角形的相似形，再推廣到可以連出哪些多邊形的相似形；接著再研究空間中的格子點可以連出哪些三角形及多邊形的相似形，並用研究的結果討論空間中的格子點可以連出哪些正多面體。推廣到四次空間的格子點時，運用一條數論中的恆等式，發現可以做出來的三角形種類(所有相似形為同一種)居然與空間中的格子點一樣，這是個非常神奇的結果。另外，運用四平方和定理可導出，在五維空間中就能夠將所有可能用格子點連出的三角形種類連出來，這也是另一項收穫。When a triangle is formed with grid points, this means the coordinates of the vertices are all integers. This research aims to find out what type of triangles can develop into symmetries with vertices that fall right on 3 grid points on a plane. The same process is further applied to polygons. Based on the results obtained, the researcher moves on to explore what type of triangles can develop into symmetries that can be formed with spatial grid points. By using an equation in number theory to expand the study to a 4-dimension space, it is formed that the kinds of triangles—their symmetries included—which can be formed with 4-dimension grid points can also be formed in a 3-dimension space. In addition, all the possible kinds of triangles which can be formed in a 6-dimension space or up can also be formed in a 5-dimension space.

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.