魔術猜牌-由再生訊息延伸推展猜中比值之研究
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「經由巧妙的猜牌方法保證可以猜中不少於 26 張花色」，並得出「當總張數趨近於無窮大時，保證可以猜中不少於 81 . 07 ％的牌，並且證出若僅使用獨立的訊息猜牌，無論任何猜牌方法皆無法猜中多於 87 . 37 ％的牌」 · 其中一個猜中多於 80 ％的例子是‘「當總張數等於 23006 張時，保證可以猜中不少於 1 8405 張牌（18405/23006 > 4/5）」 ·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 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 this study, we can prove that through mathematical method, we can assure 26 or more cards can be correctly guessed. Furthermore, when the total amount of cards is close to infinity, we can assure 81.07% or more of the cards can be correctly guessed, and prove that if the cards are guessed from independent information, no more than 87.37% of the cards will be correctly guessed by any guessing methods. One of the examples, which 80% of the cards are correctly guessed, is that when the amount of the cards is 23006, 18405 or more of the cards can be correctly guessed. （18405/23006 > 4/5）
六圓定理
在實驗中學2007 年校內科展，參展作品《三角形中的切圓》的研究中，研究三角形內的切圓時，發現連續切圓的圓心與拋物線的軌跡有關。於是去查資料，在偶然的情況下，翻閱《平面幾何中的小花》時，接觸了「六圓定理」。因為覺得這問題非常有趣，於是便著手證明（見報告內文）。 又發現，當移動六個圓中的起始圓時，總是在某種情況下，六個圓會重合成三個圓。繼續研究其重合的狀況，發現了馬爾法蒂問題（Malfatti's Problem）的一種代數解法。 當我試著推廣六圓定理至多邊形時，發現奇數邊的多邊形似乎也有如六圓定理般圓循環的狀況，於是著手證明，但目前尚未證明成功。而偶數邊的多邊形則無類似的結果。 ;In 2007 National Experimental High School Science Exhibition, one of the exhibit works, "Inscribed Circles in Triangles", shows that the centers of the consecutive inscribed circles has something to do with the parabola's trajectory. To learn more about inscribed circles and parabolas, I referred to literature. By accident, I am faced with the problem on six circles theorem, in the book The Small Flower of Plane Geometry(平面幾何中的小花). Out of my interest in this problem, I tried to prove it. The other results are as follows: With the initial circle of six circles moved, in certain circumstances, the six circles merge into three. Further in studying this coincidence leads to an algebraic method to solve the Malfatti's Problem. Applying six circles theorem to the odd-number-sided polygons exists the same characteristic. It indicates that the inscribed circles will form a cycle. However, it hasn’t been successfully proven. The even-number-sided polygons show no similar results.
心手相連的正方形
正方形兩條對角線的交點（即中心點）距四頂點等長，也與四邊等距。如果將正方形的頂點比擬成它的「手」，兩對角線的交點當成它的「心」，則兩個正方形頂點間、中心點間、或頂點與中心點間的線段相連（或重合），就如同「手」或「心」彼此相連。本文即探索當多個正方形間「心手相連」時，衍生圖形間的面積關係。而四個正方形中某幾個頂點相接（邊未重疊），恰圍出兩個三角形的圖形則是本內容討論圖形的主體架構，我們以此架構向外作出「層出不窮」的正方形，再配合中心點連接成四邊形，將推導出這些四邊形與基準正方形（Reference Square）間的面積關係。In a square, the lengths from the intersection point （center point） of two diagonal lines to the four apexes are the same, and so are they from that point to the four sides. If the apexes are “hands” and the intersection point of two diagonal lines is the “heart” of a square, the connection or overlap of two squares’ apexes and apexes, center point and center point, or apexes and center points is just like the connection of hands with hearts. In this article, hence, we are to explore the relation in area of derivative graphs formed by several squares connected “heart in hand.” When some apexes of four squares are overlain without sides overlapped, two triangles are created. And that’s the theme we are going to discuss. Furthermore, we extend the operation to infinitely overlain squares and frame out quadrangles referring to the center points of some squares. Then, the relation in areas of these overlapped squares and the Reference Square would be deduced.
N 元二次不定方程式的整數解探討
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.
死亡巧克力—切切割割好計謀
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.
共點圓、共圓點
我的研究是利用一些特殊的手法來探討所有情況皆會產生共點圓或共圓點。在一個由四條直線(無平行線組、無共點)所構成的圖形中，可以找到四個三角形及它們的外接圓。我知道它會共點，在此稱其為限制點。且若再添加一條直線，則可以任意的取出四條直線，分別找出它的限制點，而這些限制點又會共圓，吾稱其為限制圓。我欲證明此種情況會不斷延續下去。即是六條線時又會有限制點，七條線時又會有限制圓…。在本研究中，我利用了數學歸納法、特殊的編號方法以及「方向角」來做出此證明。由於固定的線組對應至固定的限制點或限制圓，希望能向找出其性質的方向發展。In my study, I use some skills to discuss all the situations which satisfy following conditions. The result is that concurrent circles or concyclic points will be found in every situation. In a graph consisting of four lines, conforming to conditions that any three lines won’t be parallel or intersect at one point, I can find out four triangles and their circumscribed circles. I know these circumscribed circles will be concurrent and I call the point at which all the circles meet “restricted point”. If another line is additionally added in the graph, I can discover that restricted points determined by any four lines in the graph will be concyclic. I call the circle “restricted circle”. What I want to prove is that the above situation will go on. In other words, restricted points will exist when I have six lines, and restricted circles will exist when I have seven lines and so on. In my study, I used Principal of Mathematical Induction, special ways of numbering points and circles, and “orientated angle” to prove my hypothesis. Because of particular line groups corresponding with particular restricted points or restricted circles, the further work I want to attain is to find the relation of them.
步步為營
Two soldiers walk on a checkerboard. They can only walk one step once a time and two directions, front and left, are decided randomly. The gunshot is the column and row where a soldier is located, and one will die if he enters the gunshot area of the other. To treat the probability of winning, we first study the cases of 1×n, 2×n, 3×n, and 4×n rectangles iterately. Then we establish a general form of the probability of winning in a general n×k rectangle by using recurrence technique and generating function, respectively. Finally, we extend to the general n×m×k cuboid case to obtain the first soldier’s probability of winning.在一個長方形的棋盤中，兩士兵行走，每一次只走一步，而且上和左兩個方向是隨機的，射程範圍是所在的此行和此列，而進入他人射程範圍則死亡。探討其獲勝機率，從1×n 、2×n、3×n、4×n 矩形的情形逐步研究，並分別運用遞迴式的技巧及生成函數，導出 n×k 矩形中先走士兵獲勝機率的一般式。更進一步地，我們也獲得了n×m×k 立體空間先走士兵的獲勝機率。