拖線溜點
原題目是環球城市盃中，一個圖論的問題。而題目提供了一個證明，是證 明此種連線都是偶數的圖形，一定會在三的倍數邊形成立。在經過一番思考過 後，我們希望能將原本的偶數連線性質加以驗證，並確定奇存在性。此時，我們 也不禁聯想到：奇數是否也有所特別的性質。因此，我們也向奇數連線做研究。 就在平面得到了部分結論的同時，我們想到這個問題是否可以推廣至三維 空間。然而在推至三維空間的過程中，我們又聯想到，另一種平面：球面。在球 面上放點，能否也找到一些不同的性質。因此，我們分別從平面、球面、立體圖 下手。 基本上，探討平面和立體問題的方法，是以土法煉鋼的方式來求出結果。 然而這種圖論的問題，不可能嘗試到無限多點的情形。因此，我們是著找出一個 關鍵的key，那就是結合性質和外接合性質。以這兩種方法，我們可以將一個簡 單的基本圖形，推向無限多點和無限多邊的情況。 接下來，還有討論一些特殊狀況，例如： deg v＝3n+1，探討其結果。 最後得到的結論是： 1、平面偶圖成立的條件為：此多邊形為三倍數邊形， 而且除了內 部一、二、四點以外， 其他點數都可以成為偶圖。 2、平面奇圖成立的條件為：奇數邊形的情形下，除了三點以外，其 他的內部奇數點的都可以成為奇圖。偶數情形下， 除了四 點以外， 其他的內部奇數點的都可以成為偶圖。 3、三角形平面圖，d eg n 皆為m 成立的條件：2＜ m＜ 6( m? N ) 4、三角形內外任意點d eg 皆為3n ( n ? N )的成立條件： 三角形內部4 x+1 個點( x ? N )。 5、三角形內外任意點d eg 皆為3n+1 ( n ? N )的成立條件： 三角形內部3 x 個點( x ? N )。 6、立體偶圖n 頂點(n＞4)面體的成立條件為： 內部點數為5m+ n- 3、5m+ n- 1、5m+ n、5m+ n +1、5m+ n +3。(m 為大於或等於零的整數) 7、立體奇圖四面體的成立條件為： 內部點數為偶數皆存在。 The original problem is a question of Graph Theory in IMTOT ,which provides\r a proof that proving the figure which its linking-line number is even ,should also be\r contented in the triple-sides figure. After profound consideration ,we try to make sure\r the existence of the properties the we mentioned above. Meanwhile ,it also occurs to\r us that whether the properties would be contented ,in the figure which its linking-line\r number is odd. So we make our way to it. Additionally ,three-dimensional and\r spherical figures are part of our research as well.\r Basically ,we discuss the problem in two-dimensional and three-dimensional\r aspects with the simplest method .However ,it is impossible to discuss the problem in\r unlimited dots .Hence , we are going to find a “key” to solve this problem .As a\r result ,we can find a simple basic-picture , and expand to infinite-multiple lateral\r pictures.\r Next step ,we also discussed some special situations , for example: for each\r point v , deg v=3n+1.\r At last the conclusion is following:\r 1、The conditions of linking-line number is even: triple-sides. And the amount of\r points inside the figure is without 1,2,and 3.\r 2、The conditions of linking-line number is odd: In the odd-sides figure , all number\r of the points inside the figure can be content without 3 point. In the even-sides\r figure , all number of the points inside the figure can be content without 4 point.\r 3、In a triangle , each point’s deg is the same number m: 2
會逆轉的石頭－RattleBack 逆旋現象
具有長短兩軸的對稱性剛體，譬如半橢球，在施加適當質量配重後，給予其Z 軸方向的 初角速度，將會逐漸上下震盪、搖擺，而後改變原先的角速度方向，朝相反的方向逆轉回來。 此現象稱為「逆旋現象」，本研究分析關於配重、質心距離與表面摩擦係數等因素的影響，利 用光點投影方式判讀其軌跡，並建立數學模型以驗證物理原理。研究發現，動摩擦垂直長軸 的分力所造成的力矩，將會導致逆旋的發生，並且，對本研究的剛體模型而言，加以20g 配 重，與長軸夾45°距質心5.5cm 且動摩擦系數為0.1227 時，逆轉角度有最大值。當長短兩軸 的震盪頻率相近時，甚至會發生兩軸能量的耦合現象，而有不只一次的逆旋。 The phenomenon was found for rigid bodies with long and short symmetrical axes, such as half-ellipsoids. We add extra masses ( Δm ) to a stone of the shape of half-ellipsoid and then give it an initial angular velocity ( 0 ω ) to make it spin. While the stone spins, the two symmetric axes start oscillating up and down. Then, remarkably, the spin of the stone slows down while the amplitude of oscillation increases until the direction of angular velocity reverses. This process is called “Rattle Back Spin-Reversed Phenomenon.” Our study involves the effect of several parameters on the above phenomenon, such as the shape of the stone, the added mass, and the friction between the stone and the table surface. We try to build a physical model to explain the spin-reversed phenomenon which can be recorded by tracing the motion of the stone with optical means. According to our study, the shape of the stone and the distribution of the added mass played crucial roles. The torque generated by kinetic friction along the long axis also helps the spin to reverse. Furthermore, in our experiment, we record a maximum reverse angle with following settings: two sets of 10 grams of mass are symmetrically added to the flat side of half-ellipsoid, in the distance of 5.5 cm to the C.M. of stone and an included 45 degrees angle with a long axis. And the most surprising finding is that when the oscillation frequencies of two symmetric axes are close to each other, the spin of the stone can reverse twice.
密碼鎖
一個有3 個旋鈕，每個位置的號碼數分別是a、b、c 的密碼鎖，如果有兩個位置的數字正確就能打開，最少需要猜多少次才能保證打開這個鎖。在本論文中，我們將密碼鎖三個位置的號碼數分成：a＝b＝c＝n、a＝b＜c，a＝b＞c 和a＞b＞c 四個部份來討論。前兩部份的研究已經找到最少次數開鎖的方法 ，後兩部份則是給了一個演算法可求出開鎖次數的上界。If a combination lock with three rotate wheels can be opened when two wheels are adjusted to the correct numbers, then how many guesses does one need to make before he or she can actually open this lock? Let us say a , b and c respectively represents the numbers that should show on each wheel. In this paper, we divide the numbers into shown on the three wheels, and they are a = b = c = n , a = b c and a = b < c . The research on the first two combinations has already given us the method we can use to open he lock with the least number of trials. On the other hand, the latter two offer us an algorithm that can be uses to obtain the upper bound of tries needed to open the lock.
大鳳蝶衣櫥裡的祕密-大鳳蝶母蝶的型態研究
Swallowtail species, Papilio memnon heronus Fruhstorfer, demonstrates sexual dimorphism with four forms in females: melanism with tails, melanism without tails, albinism with tails, and albinism without tails. This study is to understand whether environment or genetics causes the differences. Besides, we hope to discover any possible link between characteristics of larvae and the morphology of adults. We understand it is the genetic factor that controls the expression of different morphologies. The two characteristics (body color and existence of tails) are controlled by two separate genes located on different chromosomes, and the inheritance pattern is consistent with that of Mendel’s law of inheritance. As for the observation in larva stage, we fail to notice any characteristics can indicate the future morphology of a particular larva. 本研究主要是在探討為何大鳳蝶雌蝶會有四種不同的形態：1.有尾白化2.有尾黑化3. 無尾白化4.無尾黑化，而造成此形態差異的原因又是什麼？探討造成此形態差異的成因分 為兩種：1.環境因子造成的2.遺傳因子造成的，我們也就是從這兩大方向去做實驗，對於環 境因子，我們的實驗設計是列出可能影響的因子（例：光照、溫度），而我們第一個探討的 因子就是光照，然而，後來發現有尾黑化的雌蝶，其子代在有光照及無光照的環境下皆有 無尾黑化、無尾白化、有尾黑化的雌蝶出現，而且可以從實驗很明顯的看出有尾的雌蝶， 生下無尾的雌蝶為多數而非定值，故我們可以確定造成此形態差異的因子不是環境，所以 我們又趕快從遺傳因子去做探討是何種遺傳類型造成的，實驗結果確定影響翅色及尾突的 遺傳皆為孟德爾遺傳。
永不妥協
本文籍由一套數學遊戲的必勝方法及其背後潛藏的數學原理，來作為研究目標。透過研究德國數學家E.Sperner 提出的方法所延伸的數學遊戲，來解決潘建強、邵慰慈兩位教授留下來沒有證完的遊戲結果[1]，並將遊戲增廣至三維空間的探討且得到如下的結論：
一、平面棋盤
(1)不可換色，先下者恆勝，其最快獲勝方法，為依所下位置的三角形衍生子圖周界走。
(2)可換色，獲勝規則由棋盤的總頂點數決定，若棋盤的總頂點數為奇數，先下者獲勝；若棋盤的總頂點數為偶數，則後下者獲勝。
二、空間棋盤
(3) 不可換色，先下者恆勝，而最佳下法，則是下在大四面體本身內部的某一點，且其最快獲勝方法為，依正四面體稜邊所下位置走。
This study is mainly about an invincible method of a mathematical game and its theory from which it is derived. We want to solve the problems left by Professor Poon, K.K and Professor Shiu，W.C. and meanwhile extend it into three dimensions through the method brought up by E. Sperner[1].
On two dimensional case, the first player will win the game forever on condition that these two players can't change their chesses colors at will. And the fastest way to win will be just putting the chesses that along the baby triangle boundaries. If both players can change their chesses colors randomly, count the chesses number before starting the game. It is calculated that if the number of the total chesses is odd, the first player will win the game in normal and logical circumstances. On the contrary, if the number of total chesses is even, the latter will win.
On three dimensional case, the first player will definitely win the game without allowing changing chesses colors. And the best strategy is putting chesses in the inner of the big tetrahedron; what’s more, going along the edge of the tetrahedron will be shortest way to win the game.