數學

This study was to explore the nature of two basic constitutes of the regular pentagon,With these two constitutes, the regular pentagon could be multiplied into any times in size. We used four multiplication methodsto show how the regular pentagon enlarge and to verify that the enlarged regular pentagons derived from computer did exist. By integrating these four multiplication rules, we were able to arrange regular pentagon of any length of side, and evidenced the equation was ( If m,n is the number of A,B of a regular pentagon respectively ) When we tried to verify if any regular pentagon could be constituted by other smaller regular pentagons, we found that it was un-dividable only if the length of pentagon side were (the number of A, B were the 2n and 2n-1 item of Lucas Sequence), otherwise, any regular pentagon is able to be constituted by other smaller regular pentagons. The divided forms could be multiple. We also found that any pentagon could be divided by two successive un-dividable pentagons, which is called “standard division rule”. We expected to derive all kinds of division by analysis of two successive un-dividable pentagons in standard division rule. 這個研究起源於一個拼圖玩具：利用兩種黃金三角形排出指定大小的正五邊形。我們的研究動機是：一、 假如無限量供應A 和B，能夠拼出哪些邊長的正五邊形？二、 哪些拼好的正五邊形不能拆成一些較小的正五邊形？我們將研究的主要結果分述如下：

對4 × 4 方格表中計數問題的二個解題方法(1..解方程式的方法， 2.分割圖形的方法)作分析和研究後，首先我推廣分割圖形的方法來証明 : “好的n × n 方格表” 存在若且惟若n 為偶數。同時証明這種“好的n × n 方格表”內所有n2 個數的總和f(n) 為n(n+2)/4。當討論一般的n×m 方格表時，發現分割圖形的方法盲點，無法繼續推廣來証明。再經過深入分析與推廣解方程式的方法，藉由n×m 變數方格表，我們終於找到構造所有“好的n × m方格表”的方法。同時計算“好的n × m 方格表” (n≦m)內所有mn 個數的總和f(n,m), n≦7和証明好的nxm 方格表會有2(n+1)行一個循環的現象。We first studied two solution methods (1.solving equations,2.dissecting diagrams.) for calculations on 4x4 checkboard. Using the method of dissecting diagrams, we proved that``good nxn checkboard'' exists if and only if n is even. Furthermore, the sum f(n) of those n2 numbers in a ``good'' nxn checkboard is equal to n(n+2)/4.In studying the more general nx m checkboards, we found that the method of dissecting diagrams does not work, However, by extending the method of solving equations, and by considering nx m variable checkboards, we obtained a way of obtaining all ``good nxm checkboards.'' By way of computing the sum f(n,m) (n≦7) of those mn numbers in a ``good nxm checkboards,'' periodicity in every 2(n+1) rows is observed.

有二系統A和B，A中一開始有2k個物體，，B中有0個物體。在一個單位時間內，兩系統可以互相轉移最多一個物體。當B中物體的個數為 i-1，i∈{1,2,...,k+1}，我們稱其為狀態 i，從狀態1﹝初態﹞開始計時，到達狀態 k+1﹝相同態﹞便即刻停止實驗，經過之時間為一隨機變數T，稱之為實驗時間。問當兩個系統的物體數剛好相等時，經過的實驗時間之分佈為何？本文將以上述問題為核心，分別探討不同條件下系統的實驗時間所反映出來的現象，如機率、期望值、變異數等等。 Define two systems, A includes 2k objects, and B has none. They can transfer at most one object from one system to another in a time unit. When the number of objects in B is i-1, i∈{1,2,...,k+1} , we say the system is at state i. As soon as system transfer form state 1 ( initial state ) to state k+1 ( the same state ), the experiment stop. Random variable T, called the experiment time, is the time before stop. What would be the distribution of the experiment time if all systems have the same amount of objects within? This article will focus on the described question and discuss what property the experiment time of the system under various conditions has, such as probability, mean, and variance.

本研究是藉由數學手法探討；如何由一疊36 張四種花色的撲克牌中，尋找出保證可猜中最多張花色的方法。研究過程是以在適當的猜牌時機，以鴿籠原理、邏輯推理、二進位、分析與歸納……等數學原理與方法，搭配巧妙的策略運用而達到目的。猜牌方法：先約定好猜牌規則，助手將36 張牌背圖樣相同但非對稱的撲克牌，以旋轉牌背的方向傳達訊息。在本研究中得出利用數學原理與方法可「經由巧妙的猜牌方法保證可以猜中26 張花色」，並提供後續研究者利用本研究之結果繼續深入探討與研究。 The study is mathematically based with reasonable explanations behind it. We are to correctly guess as many cards as possible from a deck of 36 cards, with random numbers and four different suits. We will apply mathematical methods, such as pigeonhole principle, logic inference, binary system, and analytical reduction, upon right timing. Using careful arrangement of the principles and reasoning, we can reach our ultimate goal. To state guessing: Conference between the guesser and the assistant about the guessing rules, the assistant will have 36 cards with the same exact pattern on the back but not symmetrical. The pattern of the cards will be different when rotated 180o. The only communication between the two is by rotating cards. In the process we will obtain mathematical theory and methods assuring 26 cards correctly guessed, and the study is for further and deeper discussion.

在此研究中，我們用類似反演變換的方法，以一個定圓創立並證明了一種新的幾何變換，稱為 「調和變換」 · 我們得到點、直線、圓與圓錐曲線經過變換的關係 ·。1 .直線可以映射成原直線或一圓錐曲線 · 2.圓可以映射成一種特殊曲線。 3 .圓錐曲線可以映射成兩條圓錐曲線或一條圓錐曲線和一直線。此外我們還發現調和變換和反演變換的特殊關係 · 最後，由於調和變換可以簡化圓錐曲線的關係，我們將調和變換應用在行星輾些的證明上，並得到了良好的結果。In this research, we use a method similar to the inversion to establish a new geometric transformation, called harmonic transformation, by a fixed circle O, we prove some of its properties. We have gotten the relationship among points. lines, circs, conies and their images: 1 .The image of a line is a conic or a line itself. 2.Thc image of a circle is a special category of curve. 3.The image of a conic with its focus at the center of O is two conies or a line and a conic. Further mote, we also find the special connection between harmonic transformation and inversion. Finally, since the harmonic transformation can simplify the conic, we apply the harmonic transformation to identify the orbit of a planet, and obtain a nice conclusion.

將一個集合{1,2,...,n}分成數個非空的集合（組，區塊），稱為此集合的一個分割。如果可以找到1 ≦ a 已知無孤立點無交錯分割以Riordan 數{rn}n≥0 =1,0,1,1,3,6,15,36,... 來計數。在這篇文章中我們研究無孤立點無交錯分割的一些性質。 首先我們考慮無孤立點的無交錯分割按區塊的細分。我們得出：集合{1,2,...,n}恰含k個區塊的無孤立點的無交錯分割的個數為： 其次，我們證明bn,k和多邊形的剖分有令人訝異的關連。令dn,k是用不相交對角線將凸n 邊形分成k 塊的方法。我們用代數方法證出 bn,k = dn+2−k ,k，也給了一個新的組合證明。 最後，透過對應的方法，我們找出了七個嶄新的組合結構，這些結構都是以Riordan 數來計數。 Partition the set {1,2,...,n} into several nonempty sets (blocks) and call it a partition. If there exists 1 ≦ a It is known that the nonsingleton noncrossing partitions are counted by Riordan numbers {rn}n≥0 =1,0,1,1,3,6,15,36,... In this paper we study the properties of them. First we consider the enumeration of nonsingleton noncrossing partitions in respect to the blocks. We prove that the number of nonsingleton noncrossing partitions of {1,2,...,n} with k blocks is Then we give a connection between nonsingleton noncrossing partitions and polygon dissections. Let dn,k be the ways to dissect an n –gon with noncrossing diagonals. We prove that bn,k = dn+2−k ,k We also give a combinatorial proof. Furthermore, by way of the technic of bijection, we find 7 new combinatorial structures counted by Riordan numbers.

In this study, we mainly explore the geometric construction in 3D. By conducting some problems about constructing circles, we define the PLC construction in 2D as constructing a circle, either passing through a given point (P), tangent to a given line (L) or tangent to a given circle (C). Besides, we aim to discuss the properties of the PLC construction and the relations between each other. We discover if we find a plane satisfying certain conditions in space, the properties in the PLC construction can apply to such a plane. Furthermore, we extend the properties in PLC to the PES construction in 3D, defined as constructing a sphere, either passing through a given point (P), tangent to a given plane (E) or tangent to a given sphere (S). Also we discuss the relations among them.這個研究主要在探討3D 的尺規實作。藉由歸納某些有關作圓的題目，我們定義2D 中的PLC作圖─作圓，過已知點（P）、切已知線（L）、切已知圓（C）。並探討PLC 作圖的性質及彼此的關聯性。而我們發現：在空間中只要找到滿足特定條件的平面，則2D 幾何作圖性質在該平面仍能沿用。此外，運用PLC 作圖性質，我們進一步推廣到空間中的PES 作圖─作球，過已知點（P）、切已知面（E）、切已知球（S），並探討各個類型間的關聯性。

本文由‘‘分數7/17是否能表示成兩個相異的埃及分數之和’’這個問題出發，藉由簡單數論的性質以及反證法，得到一個真分數可表示成兩個相異埃及分數之和的定理檢驗法(定理1)。有了這個基礎，我們進ㄧ步推廣定理1 的結果，做出了嶄新的結果(定理2、定理3) 。此定理分別可以用來檢驗真分數表示成三個、四個相異埃及分數之和的存在性; 至於將真分數表示為5 項、6 項….k 項相異埃及分數之和的部分尚在嘗試。利用定理1、2，我們寫了兩個Matlab 軟體工具的電腦程式，使得我們可以檢驗任意真分數是否可以表示成兩項及三項的和，並可把所有的解列出來; 最後我們研究的是一個有關埃及分數的猜想(Erdos-Strauss Conjecture)問題，當分子為4，且分母為4k、4k+2、4k+3 時，猜想皆成立。對於分母為4k+1 而言，當k 為3r+1、3r+2 猜想亦成立，k=3r 且r 為奇數時也是成立的，因此目前需解決的問題只剩分母為24t+1 的情況了。值得一提的是，我們用Matlab 的程式檢驗出當分母為1014 至1014 +240000 之內的正整數時，猜想都是成立的，這已經超越了已知文獻的結果。This paper begins with the question: ‘‘Is 7/17 able to be the sum of two different Egyptian fractions?’’ to discuss the problem of Egyptian fractions. According to the complete division properties and the counter-evidence method, we get a back-check theorem which is about a true fraction can be the sum of two different Egyptian fractions (see theorem 1). Using the same method we obtain a new back-check theorem that is a fraction can be the sum of three or four different Egyptian fractions (thereom2, thereom3). Similarly, we can follow the same procedure to get the rule that a fraction can be the sum of five or six …or even more different Egyptian fractions. By the theorem1 and 2, we propose two programs written vie the Matlab software to examine that any true fraction can be the sum of two items and three items or not. Finally we focus on the Erdos-Straus Conjecture, which related about true fractions can be divided by three different Egyptian fractions. The conjecture is when the denominator is 4k, 4k+2, or 4k+3, the problem mentioned above can be solved. As for the denominator is 4k+1, then the conjecture also can be solved, as k equals to 3r+1 or 3r+2. Also, k being 3r and r is an odd number, the conjecture is satisfied. As for the case of r equals to even number, the problem has not been solved. But it is worth to mention here that we use Matlab software to examine the conjecture is agreeable as the denominator is between 1014to 1014+ 240000. This is beyond the results from the literatures.

This study was to explore the nature of two basic constitutes of the regular pentagon,With these two constitutes, the regular pentagon could be multiplied into any times. We used four multiplication methods (m2 = 2m1 + n1 、n2 = m1 + n1 、m2= k2m1 、n2= k2n1、a2 = a1 + 1、a2 = a1 + ) to show how the regular pentagon could enlarge and to verify that the enlarged regular pentagons derived from computer did exist. By integrating these four multiplication methods, we were able to arrange regular pentagon of any length of side, and evidenced the equation was ( If the side length of a regular pentagon is a form of m,n is the number of A,B respectively ) We further proved that the first multiplication method could be developed into a new modified method, which could divide a regular pentagon with a given side length into a combination of A and B. But only when the x and y of side length of a regular pentagon could be divided by a natural number, k, and made x/k into an item of the Fibonacci Sequence and y/k a successive item. When we tried to verify if any regular pentagon could be constituted by other smaller regular pentagons, we also found that it was un-dividable only if the length of pentagon side were ( the number of A, B were the 2n and 2n-1 item of Lucas Sequence). Otherwise, any regular pentagon might be able to be constituted by other smaller regular pentagons. 本研究是以正五邊形的兩個基本組成元素(B)作為討論對象，利用此二元素可以將正五邊形做任意倍數的放大。我們共使用4種繁殖法則(m2 = 2m1 + n1 、n2 = m1 + n1 、m2= k2m1 、n2= k2n1、a2 = a1 + 1、a2 = a1 + ) 來說明正五邊形的放大情形，並利用此4 種繁殖法驗證電腦運算出的放大圖形確實存在。利用這4 種繁殖法則的改良與整合，已達到能排出任意邊長之正五邊形的目標，並能計算並證明出其通式為。 (若正五邊形的邊長為形式,m、n代表、的個數) 更特別的是，我們能用第一繁殖法反推出一種方法，將給定邊長的正五邊形利用簡單的切割方式分成由A、B 組合成的形式，但只有正五邊形邊長之x、y 值可同除以任一自然數k 而使 x/k 為費波那契數列之一項且 y/k 為其後一項者才可以使用。 將此想法推廣至一個正五邊形能否由比他小的其他五邊形組合而成時，我們也發現當正五邊形之邊長為時（其A、B 個數為盧卡斯數列之第2n,2n-1 項），不可分解，否則應該皆可將一個正五邊形分解成比它小的其他五邊形組合（我們也可以利用這些質形檢驗出其他正五邊形是否也為質形）。但其分解形式，不只一種，而我們推測只用兩種較小的正五邊形就能達成，我們期待能找出一或多種分解方法，能將正五邊形分解成標準的分解形式。

一 Motivation and Purpose: In this study, we want to completely know about “The number abc…de, which times m/n, 1≦n≦m≦9?N can get ed… cba?”, and also expect to find out “The good rule within them”. 二 Procedure:Using method of enumeration, induction to collect sample of all and beginning from two digits to get information “good rule”. When get some useful idea, put them into the following research for the step easy go on, the method try and error is a very tiresome works, especially when we deal higher digits. till enough information is obtained, we solve problem and find new one, then likewise again research steps, just the basic science research ways, we are glad have the key of these problem. 三 Result and conclusion :Those number we named “converse No.” There are two groups: S=m+n=10 and 11 S=11, then Q=m/n=9/2,8/3,7/4,6/5=4.5,2.6,1.75,1.2 S=10, then Q=m/n=9/1,8/2,7/3,6/4=9,4,2.3,1.5 Each group have four type. When S=11,Q=7/4=1.75,if converse No.each digit is a multiple of 3, then can cancellation or extension of fraction to get another 3 or 4. Growth up rule: Converse No. = type factor x heritable factor x growth factor=rx hx g S=11,r=2~5,h=9, s=10, r=1~4, h=99 一 研究目的：盼能找出”顛倒一族”的族譜。二 研究過程：確定研究題目為ab…cde×m/n=edc...ba,0≦n≦m≦9?N 求ab… cde？以窮舉法收集觀察資料，歸納演繹尋求規律。1.先觀察兩位數，分析共有顛倒對36對。2.建立乘數Q=m/n一覽表，共有27個3.設計顛倒對大/小及其商一覽表，以利觀察、歸納獲得規律。4.接著觀察三位數，共有360對，綜合二、三位數規律，找出選擇式窮舉法：9之倍數法。5.再接著找出四位數，再綜合而知另有 全調法 重現法 半調法 GCD遺傳基因法等來繁衍高位數顛倒數。6.於是依諸法找得六位數資料，得知GCD遺傳基因法為繁衍通則，完成族譜建立模式。7.研究顛倒數位數與其個數間關係式，完成研究。研究結論：1.顛倒一族有兩大類：S=10與S=11 S=m+n。2.每一大類有四型： S=10中，Q =9/1,8/2,7/3,6/4(9,4,2.3,1.5)S=11中，Q =9/2,8/3,7/4,6/5(4.5,2.6,1.75,1.2)3.每一型均有一個顛倒數，除了S=11中，Q=7/4=1.75者可約、擴分而得3or4個。4.顛倒數原則上均為9之倍數，除了Q=7/4經約、擴分可能得非9倍數者。