數學

在這篇研究報告中,我用了三種觀點來推廣幾何中的反演變換,首先,把反演變換視為是一種圓內與圓外的一種1-1且onto的映射,第一種推廣,是將變換中心移到視圓心以外的圓內的地方,馬上我們得到一個結論「反演半徑會隨著動點而改變」,接著,我們實驗了一下反演變換用有的一些性質,保角性,保圓性,…等在這個變換視中是否依然存在;接著我們用第二種方法來推廣反演變換,我們將邊界的形狀由圓視改成別的形狀(如三角形,四邊形…等等),然後也試試看在這種變換之下是否還擁視有反演變換的一些性質;第三種推廣,則是在研究的過程中,我發現了一種新的幾視變換,承接第一種推廣,我們將原先為定點的變換中心改為動點,將原先的動點改為定點,做出來的一種新變換。In the study, a new geometric Inversive transformation through three points is discovered. Here is the main result：（1）The first, onto cycle of inside and outside can be proved under invasive transformation. It is changed moving the center from center of cycle, we can get a new ” Inversive radius can be changed by moving drop. (2) We hope to find the answer to this problem by experiment, it is exist with the inversive properties. (3) A new geometric transformation is discovered, a fixed drop can be changed moving drop, then the first moving drop shifted the fixed drop. This leads to a new construction if the new transformation.

Sicherman 已經找出與兩顆六面的正常骰子有相同機率分布的Sicherman 骰子，並進一步獲得與三顆六面的正常骰子有相同機率分布的骰子必為一對Sicherman 骰子與一顆六面的正常骰子之結果，我們試圖由已知的Sicherman 六面骰子的處理方法出發，透過對割圓多項式的分析來累積足夠的相關資料，以處理由兩顆四面骰子至兩顆三十面骰子，處理由三顆四面骰子至三顆三十面骰子的各種Sicherman 骰子的答案，來探索兩顆與三顆的n 面Sicherman骰子存在的充要條件與求法，並進一步將所得之結果分類，得到 ”有相同標準分解式的類型的數n，會具有相同組數的Sicherman 骰子”之猜測結果與特殊情形下的證明。 Sicherman has found out the Sicherman dice which have the same probability distribution as the normal two six-sides dice. Furthermore , he also found out a pair of Sicherman dice and a normal six-sides dice has the same result as 3 normal six-sides dice . We try to begin with the given algorithm of six-sides Sicherman dice , through the analysis of Cyclotomic Polynomials to accumulate sufficient related information then to come up with the solution from discussion of 2 four-sides dice to 2 thirty-sides dice , from 3 four-sides dice to 3 thirty-sides dice to explore the existence of necessary and sufficient condition and solution of 2 n-sides Sicherman dice and 3- sides Sicherman dice , and even to classify the results to come to a conclusion of the guessing results and proofs under special cases about “the numbers n which have the same Canonical Prime Factorization will have the same numbers of n-sides Sicherman dice.”

We inferred the original Feynman triangle theorem from equilateral triangle into common triangle and from same ratios of dividing points into different ones. The conclusions are below： 由原始費曼三角形原理中的正三角形等比例分點問題，推廣至一般三角形的分點等比例及不相同之比例，得到如下之結論：

本研究所探討的主題是一道分割棋盤的題目：\r 給定一個黑白相間的mxn棋盤(設m為縱向邊，n為橫向邊)，將其分割成若干個面積大小都不等的矩形區域，且每個矩形的黑格和白格個數相等。設f(m,n)為符合上述條件的最多分割矩形的個數，則f(m,n)是多少？\r 在研究過程中，我們首先藉由對各棋盤進行分類，利用不等式的運算找出其中的關係，求得當m=n以及mn=k(k+1) 時(其中K為正整數)的f(m,n)值，並構造出其分割方法。而k(k+1)

k(2αβ ,α2 ? β2,α2 + β2 )是大家熟悉畢氏定理的通式解，且一般書籍的証明大都採用代數的手法證明。以國中生而言，上述的代數方對國中生來說不夠直接且較無推展的實用性。因此幾何觀點出發發展另一種思考方式，利用角平線的性質給予畢氏定理比例解另一種全新的詮釋，並賦予比例解中的參數α 、β 在幾何的意義。在推理的過程中，我們得到一個相當有用的對應關係：一個有理數對應到一個直角三角形、兩個有理數對應到海倫三角形，再將此對應關係運用到各種幾何圖形上面，即可證明出他們所對應的通式解。最後我的興趣鎖定在海倫三角形、完美海倫多邊形與超完美海倫多邊形上的做圖方法上，善用我們所發展的對應關係，上述的問題皆可迎刃而解。k(2αβ ,α2 ? β2,α2 + β2 ) is a popular formula in Pythagoras Theory, often proved in algebra approach among books. Nevertheless, in light of junior high students, the aforementioned algebra method is neither direct nor practical. Hence, a different thinking method is derived from geometry perspective, using the straight line concept to reinterpret Pythagoras Theory and define the geometric meanings of α andβ . In the process of logical development, a useful correlation emerges: a rational number correlates with a straight-angled triangle, and two rational numbers correlate with Heron Triangle. This correlation can be applied to all kinds of geometrical diagrams to prove the correlated homogenous solution. Ultimately, my interest lies in the diagram methods of Heron Triangle, Perfect Heron Polygon, and Super Perfect Heron Polygon in order to apply our developed correlations to solve the above mentioned problems.

滿足之M點是否為重心之探索

滿足之M 點，我們稱之為Pi(i=1…n)的均值點。當n=3，M 恰為△P1P2P3 的重心 (G); n=4 時，M 亦為三角錐P1P2P3P4 的重心！因此不免引人遐思：滿足之M 點是否皆為其重心？ 我們藉由電腦幾何作圖軟體GSP 協助觀察，掌握了圖形變化間之不變性，再配合向量解析及推理，得以發現均值點、多邊形的重心、以至多面體的重心、及平行多邊形的一般性作法。附帶又發現：任意相鄰三頂點即可決定一平行n 邊形。並進而證實：平行四邊形為四邊形M＝G 的充要條件。但當n≧5 時，平行n 邊形只是n 邊形M＝G 的充分非必要條件！一般而言，具有對稱中心O 的n 個點所構成的圖形必可使M 與G 重合於O 點上。 The point M satisfying is called “the mean point of Pi(i=1…n)”. As n=3, M is the center of gravity (G) of the △P1P2P3. If n=4, then M is also the center of gravity of the triangular pyramid P1P2P3P4. Therefore, I began to wonder if the following assumption stands: The point M that satisfies is always a center of gravity. By using the computer software GSP (The Geometer’s Sketchpad) to observe figures. It is found that when a figure is changing there is still constancy. Furthermore, supported by the analysis based on vectors, general constructions can be established concerning the mean point, the center of gravity of polygon, the center of gravity of polyhedron, and the parallel polygon. Also, I find that any three neighboring vertexes decide a parallel polygon. And thus it is verified that the parallelogram is the sufficient and necessary condition for quadrilateral M=G. As n≧5, the parallel n-sides shape is the sufficient, not necessary condition, for n-sides shape M=G. In general, a central figure of n points having the center of symmetry O can make M and G meet on O.

傑克船長和他的海盜們掠奪到許多箱珍寶，每箱含有數量不等的金幣及鑽石。船員們深怕傑克船長又出什麼陰謀，一致同意讓船長任選一半的箱子拿走。\r 當然，傑克不知道金幣和鑽石的價格比，為了保證可以得到一半的利益，傑克希望他拿到的金幣和鑽石都各占一半。傑克的願望會實現嗎？\r 這個問題已用高等數學證明其解，而本研究利用初等數學的方法，除證明傑克需取的最少箱子數外，同時也能更快速的算出取法。\r 類似的結果可應用至分配災區物資等情形，或任何無法轉移、獨立的資源，如各式專長的人才、多功能的機械等，期望可對更有效的分配做出貢獻。

我們從日常生活中的酒瓶填塞問題，延伸出圓在相守條件下產生的衍生圖面\r 積極值及相守圓排法等問題，研究過程中我們發現以下的結果：\r 一、 我們可以利用較少個相守圓的排法，以繁殖或增加的方式排出較多個相守圓\r 的衍生圖。\r 二、 當相守圓數量為2個、4個及6個時，相守圓以對角線排列，且衍生圖為正方形\r 時，面積會產生最大值。而當相守圓數量為3 個及5 個時，衍生圖面積最大\r 值會出現在相守圓排成波浪形時，但其夾角並非特殊角。\r 三、當相守圓數量為2個至6個時，相守圓以直線排列，衍生圖面積會產生最小值。\r 四、透過不同層數、個數的研究得知：當相守圓的數量為11個或14個以上時，存\r 在正三角形排法小於直線排法的情形 。而相守圓個數在10個以下時，直線排\r 法面積都叫正三角形排法小。\r 五、我們可以用質單元分割的方式討論更多圓數時，相守圓間的排列方式，使得\r 生多元間的排列變為數字間的加法排列，可以大大降低討論情形的複雜度。\r 六、我們可以用密度的概念，搭配質單元分割的方式，得到不同圓數時，以不同\r 衍生多元搭配波浪形法形成之衍生圖密度最小值。\r 七、我們可以算幾及柯西不等式來驗證特定類型的衍生圖面積極值，也可以列出\r 面積函數來分析不同類型的衍生圖面積極值。

最後留下數字會是多少？該問題在台灣的全國中小學科學展覽出現多次。而資訊界演算法大師Donlad E. Knuth 在其著作The Art of Programing，CONCRETE MATHEMATICS （具體數學），針對該數列作詳細的說明；但是，不論是歷屆全國中小學科學展覽或是大師著作，對於該問題，都只是談及殺1 留β或是殺α留1。本研究利用獨創α分類、n 及k 分類、d 函數、b 函數及循環、n 及y 分類、碎形數列和演變關係，將約瑟夫問題探討範圍提升至殺α(個數)留β(個數)，直到剩下最後1 個數時就不能再殺了，遊戲終止，倒數第k 個留下的自然數是多少？同時，本研究在殺α(個數)留β(個數)下，指定自然數y 為酋長，酋長不能被殺，殺到酋長時遊戲停止，求剩下的自然數有幾個？會發生什麼情形？The Josephus problem refers to what will be remaining when arranging n natural numbers in a circle and starting killing one and leaving the next one alive. The problem has been on display for many times in Taiwan National Primary and High School Science Exhibitions (as shown in Table 1). And, the information algorithm master, Donald E. Knuth has elaborated on the array in his works The Art of Programming, CONCRETE MATHEMATICS. However, both the past science exhibitions and the master’s works are limited to discussions on cases of killing 1 leaving β or killing α and leaving 1. This research employs uniquely created α classification, n and k classifications, d function, b function and loop theory to extend the Josephus problem scope to killing α leaving β to find out what the remaining natural number is by No. k counted recursively. Meanwhile, this research designates natural number y as the chieftain, which can never be killed. The game is over when the chieftain is to be killed. The problem is to work out how many natural numbers are remaining. And what happened?