數學

在1940 年代，Bouwkamp 提出一系列有關如何將矩形切割成若干個正方形的研究報告，但是如何找出正方形個數最少的方法仍是長久以來懸而未決的問題。在本研究報告中，首先引進「四角切割」的方法，並結合輾轉相除法的概念，來研究矩形的切割問題。我們的方法能大幅度降低正方形的個數，也適合做為此問題的上界函數。有關如何在長方體中切割出正立方體的組合，我們也將輾轉相除法的概念延伸到三維空間，進而建立所切割出最少個正立體數的一個上界模式。此外，藉由四角切割概念的延伸，我們也發現這個上界亦可再予修正。In 1940’s, Bouwkamp proposed the study of dissecting squares from rectangles. Among the study, the problem of the least number of dissected squares has been open for decades. In this project, we first propose a corner dissection method, associated with the famous Euclidean algorithm. By reducing nearly three fourths of the number dissected by the primitive Euclidian algorithm, our method indeed establish a suitable upper bound of the minimal number of dissected squares from the given rectangles Meanwhile, the Euclidean algorithm has also been considered to dissect the cubes from cuboids. We analyze the fundamental properties of the method and establish a prototype of upper bound function for the minimal number of dissected cubes. Moreover, the method of corner dissection has also been implemented for some cuboids, which also exhibits the acceptable improvement being a suitable upper bound.

在升學主義越來越興盛的社會中，考試成績成為人人關心的重點，這\r 次研究就是藉由數理資優班同學的各學期在校成績和指定考科成\r 績，透過迴歸分析，找出各學期成績與指考成績之間的關係，並利用\r 圖表來解釋各種科目在各學期的課程，在高中三年所學的重要性，在\r 藉由此結果，希望能對目前老師的教育重點及學生學習方式能有所幫\r 助，亦可了解學生在高中求學過程中，哪些階段對指考成績較有正面\r 影響，進而強化該學習階段，以有助在指定考科時能充分發揮所學。\r \r In a society that emphasize on degrees, examination scores become the\r spotlight, and the ultimate goal for a high school student who had worked\r so hard for three years is to achieve high scores in the J.C.E.E. In the\r three years of high school, each subject has different topics each semester,\r but which semester has the most decisive effect on the J.C.E.E. score?\r This research is to study the effect of each semester on the J.C.E.E. by\r analyzing the grades of a science and math talented class in Senior High\r School using Regression analysis to find out the connections between\r term grades and the J.C.E.E. Then finding out which term grades had the\r most decisive effect in each subject. By using the result, we hope it can\r help teachers in their teaching and students in their learning. Also, it can\r provide the information about which stage in high school has positive\r effects on J.C.E.E. grades, therefore enabling students to emphasize on\r that stage in order to perform well on the J.C.E.E.

「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.

看到環球城市數學競賽2003年春季賽國中組試題中，一題有關方格遊戲的問題： \r \r 在一塊9 × 9的正方形方格紙板中，最多可以挑選幾個小方格， \r 使得沿著這些小方格的二條對角線割開後，原正方形方格紙板 \r 不會分裂為二片或二片以上(即沒有小片紙板會從原正方形紙板 \r 中“掉下來”)？\r \r 原題目雖然只有一種圖形解，但我們發覺在其他方格紙板中，圖形解不一，在對幾個圖形分析和研究過後，發覺“似乎”有其特定作圖法，而且可挑選的小方格數也頗有發展的地方，令我們覺得相當有趣，而且此題目和之前看過方格類的問題不大一樣，因此，決定以此問題當作科展主題，加以延伸、研究，自我挑戰。

給定正n邊形，於內部區域新增對角線，使得對角線不交叉且內部區域皆為三角形，則將此圖形稱為正n邊形的一個『三角剖分』。考慮正n邊形的所有三角剖分，已知其數量為卡特蘭數Catalan(n-2)。在所有三角剖分的情形中，考慮旋轉與翻轉，將同構的情形視為相同，則將所有不同構的三角剖分總數記為Dn。本文對於正n邊形的不同構三角剖分進行研究，以三種面向進行探討，首先我們以不同構三角剖分的對稱性分類，發現其和化學式CnHn+2的同分異構物有相關性；再者，以相鄰三角形的最大數量進行分類，當相鄰三角形的最大數量為n-2、n-3、n-4與n-5時，得出不同構三角剖分的計算通式；最後，以三個相鄰頂點組成的外圍三角形進行分類，將『恰包含兩個外圍三角形的不同構三角剖分』、『圖論中的毛毛蟲圖（Caterpillar）』以及『化學結構相關的Losanitsch’s triangle』進行深入探討。

壹、研究目的：培養建構式思考方式，提高解決問題的能力。 貳、研究過程：一先查數學辭典，確定F.S.之定義。二以文字敘述替代數字敘述F.S.，並分析歸納規律性。三將發表過的有關關係式，挑選適合以代數分析研究者，研究採逆命題角度處理，共有下列七種關係式採論之。 壹、Motivation and Purpose In this study, we expect to know something about Fibonacci Sequence (F.S.) that we can understand and enjoy as a high school student. 貳、Procedure 一.Make sure the definition of F.S. 二.Use algebra instead of numerical to state F.S. 三.Select the related formulas and discuss by fundamental algebra. We get 7 types as follows

Within the framework of the new educational model for mathematics based on constructivism, results are presented of the design, application, and evaluation processes of a series of didactic sequences aimed at developing the student’s abilities for problem solving as part of the geometry curriculum for technological preparatory schools, using the Cabri-Geometre II software. In this case, subjects of study were ten newly enrolled students from CETis 18 preparatory school in Mexicali, Baja California, Mexico. The theoretical basis for this work is the constructivist approach, mainly emphasizing Mashbits views (1997) regarding problem solving. This didactic proposal was longitudinally applied in a quasi-experimental qualitative design under the following analysis categories: problem solving skills and the impact of Cabri- Geometre II in geometry learning. Recognizing the potentiality this research can have with the proper follow-up, it is intended to include it in the preparatory school curricula. For this purpose, teachers should be trained to focus their work on learning instead of on teaching. As a result of this, designing educational programs will require for teachers to become more knowledgeable not only in discipline, but in the use of computer technology, the teaching process, learning, and the students themselves. The final objective of this project is to instill educators to play this new role. As a final point, conclusions on various psychological, pedagogical, and technological aspects are given placing emphasis on the creation of learning situations with their appropriate theoretical support. Using the Cabri-Geometre II as a resource, these situations will provide geometry teaching with a more dynamic and interesting concept applicable to real-life situations.

本文中我們探討一個有趣的數列。這個數列有一個非常特殊的性質：將數列相鄰兩項的前項當分子，後項當分母，所產生的分數數列，恰好會出現所有的正有理數。 這個特殊的性質表示，可以將正有理數按照這個方式作排序，這個排序將完全不同於常見的正有理數排序的方法。 (1). 在正有理數的排序的結構中，我們做出許多有關於此數列的定理。 (2). 用數學歸納法證明此分數數列涵蓋所有正有理數，且每一正有理數只出現過一次。 (3). 將數列分割後，利用試算表製成數列規則表，並整理出快速的方法將數列表達出來。 (4). 將an 數列排成“樹＂的模式，可更快速的把正有理數寫下來。 (5). 最後，設計出搜尋正有理數的演算法，解決在分數數列中第n個正有理數會是多少；以及正有理數會出現在數列中第幾項的問題。 Let’s discuss an interesting sequence. There is a very special quality in it. In this sequence, choose two numbers, which are close to each other, and suppose the first number as “member” while the second one as “denominator.” Then we can get a fraction sequence that includes all of the positive rational numbers! According to this special quality, we can arrange positive rational numbers by the following method. Then we can get a brand-new way of the arrangements. (1). We can find many theorems about this sequence according to this special arrangement of the positive rational numbers. (2). We can prove the rule that this fraction sequence includes all of the positive rational numbers by mathematical induction. Furthermore, every positive rational number appears only once. (3). After dividing this sequence into several parts, we can get a sequence rule list by using trial balance and find a faster method to express the sequence. (4). Arrange the an sequence by the tree model. By this way, we can get all of the positive rational numbers much faster. (5). Finally, we can develop the operation method to solve the questions that what position would one positive rational number be in the sequence and what is the first, second, third or nth positive rational number of the sequence.

單淘汰賽是一種失敗一次即遭淘汰的賽制；在此假定每位選手都有一相對應的能力數值，本文主要探討在均高的單淘汰賽程表之下，若賽程安排完全依照種子安排原則(亦即最強的選手對最弱的選手、次強隊次弱….)，則對於能力越強的選手越有保障，直觀上而言能力最強的選手應有最大的奪冠機率，探討此種賽程安排是否滿足能力較強的選手有較大的勝率？因發現在某些特殊的選手能力數值分佈之下會發生次強選手勝率大於最強選手的情況，令A、B代表最強與次強選手，P(A)、P(B)代表A、B奪冠的機率，故擬定P(B)/ P(A)為參考依據，尋求P(B)/ P(A)的最大值發生處作為最極端的狀況。發現四位選手的情況下，P(B) / P(A)最大值 = 1；八位選手的情況下，P(B) / P(A)最大值=(196+98) / 343=1.0938，當選手數為2n時，P(B)/ P(A)最大值隨n的增加而遞增。 Knockout Tournament is a highly competitive system in which any player losing a game can no longer play in the tournament. Here we suppose that every player has a numerical value that corresponds to his ability. We consider a totally-seeded knockout tournament with 2n players where in the first round, the strongest player matches the weakest player, the second strongest player matches the second weakest player, and so on. We examine whether a stronger player always has a greater probability of winning the tournament. The answer is in the affirmative for n = 2. For a tournament with eight players(n = 3), the situation is much more complicated. In certain cases, the second strongest player has the greatest probability of winning the tournament. Specifically, let A and B denote the strongest and second strongest players, P(A) and P(B) their respective probability of winning the tournament. We find that the maximum value of P(B)/P(A)equals (196+98) / 343 = 1.0938. For n > 3, we have not obtained the maximum value of P(B) / P(A) . However, it can be readily seen that the maximum value of P(B) / P(A) is non-decreasing as n increases.