全國中小學科展

數學

單淘汰制賽程分析

本研究報告針對單淘汰制賽程中存在的迷思,提出方法並加以討論。單淘汰制賽程規則為每場比賽皆有勝負(即沒有和局),負方即失去奪得冠軍的機會,且不得出現於另一場賽程。全文分別對影響選手勝率的因素──選手實力與賽程安排,提出方法與概念討論。\r 第一部分討論選手實力變化對勝率的影響。利用「假想選手」的概念,討論選手於各場賽程中,所有可能遇上的對手所造成的威脅。並透過「威脅門檻」判斷對手實力的變化對自己勝率的利弊。藉由「勝率一般式」計算選手於賽程中奪冠的機率,並以各種角度觀看賽程,判斷個體與群體的實力。\r 第二部分討論位置安排對勝率造成的影響。由「勝率實力比」討論賽程安排對選手的公平性,並定義「賽程表現率」討論選手因賽程安排對勝率所造成的影響。\r 文中並以實際數據範例,希望閱讀以後的你(妳),能認識並喜歡上淘汰賽的世界。

畢氏定理演繹的正三角形分割研究

畢氏定理(a²+b²=c²)歷經25世紀,發現了數百種的幾何論證法;而畢氏定理演繹出的正三角形 ( (/4) a²+(/4) b²=(/4) c² )幾何分割研究,卻一直沒有人研究。因此,承襲著之前處理幾何問題的經驗,決定挑戰畢氏定理演譯的正三角形分割研究。本文研究兩正三角形,經切割後拼成另一大正三角形;期間以GSP及AutoCAD繪製分析幾何圖形,並建立了4種分割模式,得到了3段式「最佳分割模式」及準「通用分割模式」,提供這方面問題一個可應用於所有條件之完善解決方案。本研究成果豐碩,補足了相關領域的空檔,且可製成益智又富挑戰性之拼圖系列,不管用做教具或遊戲,對建立意至己和相關資料有莫大貢獻! Twenty five centuries after its discovery, hundreds of proofs have been given for the Pythagorean Theorem (a²+b²=c²). But, research about regular triangle dissection extending from Pythagorean Theorem has always been lacking. So, based on previous experience with geometric dissection problems, I have decided to do a research on regular triangle dissection extending from Pythagorean theorem. This research dissects two regular triangles and assembles them into a large regular triangle. Using GSP and AutoCAD to draw and analyze geometric shapes, four dissection models and nine dissection methods are constructed. The extreme values under all conditions are also discussed, as are the best and generic dissection models. There is a Three-section type “best dissection model” and a semi “generic dissection model.” offering a perfect solution to this kind of problem that can be used under all conditions. This study yields numerous results as well as filling in blanks in similar fields. It can also be made into challenging jigsaw puzzles for educational or entertainment purposes.

傑克船長的心機'

作者受考題( International Mathematics Tournament of the Towns, Senior A-Level Paper, Fall 2009, No. 7 )?發而展開此研究。經搜尋文獻發現,這系列命題可追溯自 Scientific American ( Feb 1979 )中 Martin Gardner 的文章 The Rotating Table 。\r 命題之操作原在正方形桌上執行,後被Ted Lewis & Stephen Willard 推廣至正多邊形(1980),又再被Richard Ehrenborg & Chris M. Skinner推廣至任意置換群(1995)。\r 本文從均勻多面體的情形出發,藉諸自創的證明方式,重新詮釋上述論文之結果,給出較簡潔自然的證明。同時,作者改變命題裡的關鍵限制,發展全新的研究方向;並針對不同的情形(多邊形、多面體、置換群)分別求出各變數之上、下限。\r 本文使用到的技巧包括:群論、歸納法、組合設計。充分性之證明過程提供的演算法能應用於同步連絡管道,允許匿名用戶之間建立連線。

密碼鎖-拉丁超立方體的完美控制情形(Lucky Locks)

有個密碼鎖由D個旋鈕組成,每個旋鈕有N種不同的號碼,由於構造缺點若D個旋鈕中僅有1個號碼錯誤仍能打開密碼鎖,問最少嘗試多少組號碼才能保證一定能打開這個鎖?這個問題等同於在N元D維超立方中找一組點集,點集中的點各自向其D維度畫出延伸線,若超立方中的所有點都至少被1條延伸線所涵蓋,要求重複涵蓋的次數總和要最少。\r 43屆的科展中已經討論過3個旋鈕的情況,我們接著分析4個旋鈕的情況。在討論中發現D=4時並沒有如D=3時保證打開的最小次數公式,我們給出上下限的公式。但D=N+1且N≠6時卻很特別,恰可利用拉丁超立方挑出1組點集,其所有延伸線涵蓋的點都沒有重複,稱為完美控制,而保證打開鎖的最小次數是NN-1。

反正切函數,二階線性遞迴數列與疊在一起的方格紙

本文由三個結合 tan−1 與費波那契數列的等式及其所搭配的無字證明圖形出發, 做出和盧卡斯數列有關的圖形, 並由數學歸納法找出並證明 tan−1 與盧卡斯數列及一般二階線性遞迴數列的全新等式: This paper startes with three equations of tan−1 and the Fibonacci sequence combined with the diagrams used to prove the three equations without words. According to the principle of mathematical induction, we continued to find out the similar equations of the Lucas numbers and the second-order liner recursive sequences as follows.

約瑟夫數列(Josephus Series)

所謂約瑟夫數列,就是有n 個數排成一環狀,從頭開始,殺1(個數)留1(個數),求倒數第k 個留下的數會是多少?約瑟夫數列在台灣的全國中小學科學展覽出現多次(如下表)。全國科學展覽與本題類似的作品 資訊界演算法大師Donlad E. Knuth 在其著作The Art of Programing,CONCRETE MATHEMATICS,也針對該數列作詳細的說明。唯,不論是歷屆科學展覽或是大師的著作,對於該數列,都只是談及殺1 留β或是殺α留1。 筆者則在2005 年暑假,曾經提交於全國國小組比賽作品「老師無法解決的難題」討論到n 個人排成一圈經過殺α留β,最後留下來的情形。 本研究是將α、β、k 和n 作為變數,求:當有n 個數排成一環狀,從頭開始,殺α(個數) 留β(個數),則倒數第k 個留下的數會是多少? 需符合α、β、k、n 皆∈N,且n≧k 1.直觀觀察:發現在每一個循環中,當n 等差α時,Aα,β,n,k 則等差α+β、n- Aα,β,n,k 則等差β。 2.分類:將其分類為cα,n,使當中有規律可求。 3.循環觀察:發現每個循環的尾數n- Aα,β,n,k 都小於β。 4.循環尾數:設計公式求出每個循環節的尾數n、留下數Aα,β,n,k 及n-Aα,β,n,k 。 5.倒推:由與循環節中有等差的性質,則可以由循環節的尾數,推論出循環節中的任意一數。 Joseph Sequence is the problem that discussed the situation of eliminating1 and retaining1 in the circle formed by n people. Joseph Sequence has appeared a number of times in National Elementary School and Middle School Science Fair in Taiwan (as shown in the table below). Past national science fairs and researches on Joseph Sequence The publications,The Art of Programing,CONCRETE MATHEMATICS ,by the expert of mathematical calculation in the IT industry,Donlad E. Knuth,has provided detailed explanation on it. However, all of those only discussed eliminating 1 and retaining β or eliminating α and retaining 1. The researcher proposed “Problems unsolved by teachers” in the national competition, and discussed the situation of eliminating α and retaining β in the circle formed by n people. This study continued the summer project of 2005, and conducted research on the question of when is the last kth person eliminated in a circle formed by n people. In the paper, α, β, n and k were independent variables and the research process was as follows: 1. Direct observation: the series shows equal difference in each cycle. 2. Classification: to search the pattern of the series based on cα,n classification. 3. Use the end number of each cycle to obtain the pattern. 4. Reverse induction: use the equal difference of each cycle to induce when the kth person would be eliminated.

8x8 棋盤路徑解之一般化推廣

Abstract (一)、 In our study, we discuss a m×n chess and any beginning square p finding a directed path of chessman from p moving to an end square in which the chessman moves to adjacent squares including only three directions which are right move, up move and diagonal left down move. A m×n chess is ruled into m columns and n rows creating the number of (m×n) squares (二)、 A chess directed path moves from any beginning square to end square in a m×n chess and every other square is visited just once. In the view of the beginning squares, the chess paths are solvable paths in a mxn chess and the corresponding squares are solutions. (三)、 First, we find out that some beginning squares are located in a special area with no any solvable directed paths. We define the special area be no-solution area. (四)、 According the 3-color theorem, we determine more than two thirds of no-solution area. (五)、 Then, we derive properties of reversibility and symmetry in solvable paths. i.e. A solvable path exist another solvable path by reversibility and symmetry respectively. (六)、 Utilizing the generalization of no-solution area which is extended from the concept of no-solution area provides judgment for the next moves effectively. The judgment is defined as effective move principle. (七)、 Furthermore, using the other theorem called rules of shift Hamiltonian path gets augment solutions. (八)、 According to the effective move principle finding a number of solvable directed paths, use the reversibility and rules of shift Hamiltonian paths to get augment solutions. Finally, utilize symmetry to find out all solvable paths in the m×n chess. (一)、研究規則:在m×n 的格子中,任取一格A 當作「起點格」,在起點格上放一顆棋子,只能往「上」、往「右」、往「左下」的方向移動。(二)、定義:若棋子從「起點格」,按照上述規則能不重複的通過所有m×n 格子到達某一「終點格」,則對於「起點格」而言,此移動路徑稱為m×n 的「有解路徑」,其任4一「終點格」稱為「起點格」的「路徑解」。(三)、我們先研究出「基本無解區」。(四)、根據遊戲規則我們利用三種顏色將n × n 方格塗滿,並判斷出大部分的「無解起點格」。(五)、利用遊戲規則得到兩重要性質:(1)[可逆性性質] (2) [對稱性性質](六)、利用「廣義基本無解區」,當作我們[有效移動]的判斷,讓「有解路徑」快速的找出。(七)、利用本研究所稱的「平移哈式鏈」,得到[擴充解]。(八)、根據[有效移動]求出部分「路徑解」,再利用[可逆性性質]、 [擴充解] ,最後利用[對稱性性質]完成所有「路徑解」的尋找。

梯型的分割與調和數列

Refer to Figure 1. Suppose ABCD is a trapezoid and . Passing the intersection M of and we construct a parallel line intersecting and at E and F, respectively. We obtain that , and then we can generalize the result. 如(圖一),若ABCD為梯形且,過 和交點M 分別作平行線交、 於E 、F ,可得 的關係,再加以推廣。

擬-Lucas多項式的幾個性質

本篇文章從"將aⁿ+bⁿ分解成(a+b)及ab的非線性組合"出發,在同樣的遞迴精神下引進並定義擬-Lucas多項式 <Sn (X)>:

大富翁中的密秘—機率

記得小時候常玩大富翁,在走步數之前常要由擲骰子來決定。又在上國中之後學到了有關機率的問題,所以讓我想起此事,想藉由此次的科展對大富翁內的機率做一番徹底的研究與了解。