在generalized Petersen graph P(n,5)中的hyper Hamiltonian
Generalized Petersen graph P(n,k)，定義為n 為不小於2 的整數以及1≤ k ≤ n−1，有頂點{ u0, u1, . . . , un−1, v0 , v1 , . . . , vn−1 }，及路徑{ uiui+1 , uivi , vivi+k：1≤ i ≤ n−1 }。在 [2] 中，我們可以知道P(n,5) 是Hamiltonian 等價於當n≠11。
在這一篇報告中，我們證明當generalized Petersen graph P(n,5) 是hyper Hamiltonian（一種Hamiltonian graph 再去掉任何一點後，仍然是Hamiltonian graph）的充要條件是n 為不等於11 的奇數且n ≥ 7。
The generalized Petersen graph P(n,k), n ≥ 2 and 1≤ k ≤ n−1, has vertex-set { u0, u1, . . . , un−1, v0 , v1 , . . . , vn−1 } and edge-set { uiui+1 , uivi , vivi+k：1≤ i ≤ n−1 with subscripts reduced modulo n}. And we can know that P(n,5) is Hamiltonian if and only if n≠11 from [2].In this paper it is proved that generalized Petersen graph P(n,5) is Hyper Hamiltonian (A Hamiltonian graph can still be a Hamiltonian graph when any one of the nodes fault) if and only if n is odd and n≠11.
蜘蛛數
We understood the definition and meaning of spider number by reading〝Wonders of Numbers〞. It interested us so much. So, we took further step to study the situation of extreme value when the gap sometimes lie on the line and sometimes on the circle or even on both. That is to say, we explored the relation between spider number and the gap when the spider number is maximum or minimum. New research for the application of spider number involves several directions. First, we design a new game called〝Stepping Land Mine〞with the rule of spider number. Give you a net with several hidden gaps, trying to find the right positions of gaps. Second is the further result for a different type of net about regular n-polygon. Third is a tactic for a net with destroying of the strategy points. In this situation, the gaps amount on the circle and on the line are fixed. At the same time, consider the situation of circles and lines designing the tactic of placing the gaps to attain the maximum of the destructive effect. 在本文中我們定義一個蜘蛛網上的蜘蛛數，若在蜘蛛網中加入缺口後，會影響蜘蛛數的大小。我們探討蜘蛛網上的缺口，該如何分配才能夠得到蜘蛛數的極值(最大值及最小值)。先觀察一直線和圓上缺口如何放置蜘蛛數有極值，再探討許多條直線及圓上的情況，進而推展至許多同心圓及通過圓心的許多條放射線的缺口，該如何放置，蜘蛛數才會有極值發生。
「圖形板」的圖形軌跡之探討及其延伸
Starting from the problem in AMC competition of Australia, we try to find out the locus and its length when a point in a regular polygon rolls in a circle. The result is that the locus has a wonderful and regular cycle.Next, we discuss the regularity of the cycle when a regular polygon（n sides） rolls in another regular polygon. Furthermore,we discuss the the equation of the locus by changing the radius and the angle of rolling. we find out the argument function of the locus of a point inside when a a regular polygon（n sides）rolls in another regular polygon （m sides）： ， Aj is the summits of the regular polygon（m sides）, Bjcorresponds Aj when a point inside the regular polygon (n sides) rolls, ) And then, we do some moving simulation with some computer math software, such as Cabri Geometry、Mupad, etc. We discuss the regularity of the locus and its equation of a point inside when some special cycloids, like asteroids, cardioids, etc, roll in a certain condition. Moreover, with the result of research 2, we create the “plate＂ and apply for a patent on it. We hope to study math by playing games.
從澳洲AMC 競賽題出發，嘗試探討一正n 邊形中的一點在單位圓內滾動軌跡及其軌跡長度，發現該軌跡均會產生奇妙的循環規律。
接下來，推廣探討正n 邊形在其他正多邊形中滾動時循環的規律，並利用旋轉半徑及角度之間的變化深入探討其滾動軌跡方程式，發現正n 邊形繞正m 邊形滾動時其內部一點軌跡參數式為，其中， Aj 為 正m 邊形之各頂點、Bj 為正n 邊形中內部一點旋轉時對應 Aj 之點，。
進一步想嘗試使用數學電腦軟體如：Cabri Geometry、Mupad 等對以上研究去做一些動態模擬，並再探討一些特殊擺線如：星狀線、心臟線…等，在條件下相切滾動時，圖中某一點的軌跡規律性及其方程式。另外，應用研究二中的結果，創造出寓數學於遊戲的「圖形板」，並申請了新型專利。
二次函數上正三角形建構之研究及探討
在拋物線上置掛正三角形看似簡單，其實不然。本篇文章研究在二次函數的各種不同情況下，可做正三角形的分佈以及其個數。
1. 在一條拋物線上時，最多只能作正三角形。
4. 在三條對稱軸相等的拋物線和共頂點開口大小不同之拋物線上，本篇文章證明一定能找出正三角形落在它們之上。但由於最多有四個分界點，要解四次方乘組過於繁複，於是本篇文章對分界點作了一些估計，找出了分界點的極限值。
5. 本篇文章證明了對於給定的正n 邊形，存在一1 元n-1 次方程式可以通過它所有頂點。
Building a regular triangle on a parabolic curve looks easy . In fact , it doesn’t . This Article researches regular triangles distributions and its numbers in different conditions.
1. On one parabolic curve can only build regular triangles , squares and other regular polygons can’t be built.
4. For three parabolic curves which has same symmetrical axis or three concurrent parabolic curves, we prove that it can build at least one regular triangle on them .But because it can have at most 4 boundary points, to solve quartic equation is to complicated. So we do some estimation of boundary points, and find out some limits.
5. This Article prove that for given regular polygons , there exists a one dimension n-1 orders equation can pass all its apexes.
Bezier曲線與蚶線間之關聯性的探討與推廣
在這篇報告中，我們以貝斯曲線的做圖原理建立出一種新的曲線－環狀貝斯曲線，進而得到不少有趣的結果。我們發現有名的古典曲線－蚶線，也是屬於二次環狀貝斯曲線。軌跡方程式為：，此時，係數恰符合二項式定理。之後我們推廣至n次環狀貝斯曲線的軌跡方程式：，也符合二項式定理。
在複數平面上，給定z0、z1、z2三點，我們定義出一個二次變換 ，若，，可映射成蚶線的圖形；若z∈實數，則可映射成拋物線。利用此結果類推我們找到一個複數平面上由 z0、z1、...、zn 所決定的n次變換將以原點為圓心的單位圓，映射成n次環狀Bezier曲線。
In this essay, we use the method of forming a Bezier Curve to establish a new curve, circular Bezier Curve, and find a lot of interesting results. We discover the famous classical curve ＂limacon＂, which belongs to the Quadratic Circular Bezier Curve. The locus of Quadratic Circular Bezier Curve is, where. Its coefficients match the binomial theorem. Then we apply it to the locus of nth-circular Bezier Curve:, and it also matches the binomial theorem.On the complex plane, we define a quadratic transformation corresponding to three points—z0,z1 and z2 as .If , where , a limacon is mapped. If z is a real number, a parabola is mapped. With this result, we will find a nth transformation defined by z0、z1、...、zn on the complex plane. It will form a nth-circular Bezier Curve with unit circle centering on the origin.
M&m Sequences 之研究
本專題的目的是研究以任意實數 a1 、 a2 、 a3 為起始的M&m Sequences 之穩定性質。我們主要關心的問題是：(1) 是否任給定三數a1 、 a2 、 a3 為起始的M&m 數列皆會穩定？(2) 若上述的M&m 數列穩定，則其穩定的長度與a1 、 a2 、 a3的關係為何？(3) 其穩定的值與a1 、 a2 、 a3的關係為何？我們研究的主要步驟及結果如下︰1. 當1 2 3 a 1) 為起始的M&m 數列。3. 我們證明了下列性質：(1) 若M&m 數列中前n 項所成數列的中位數為n m ，則下式成立： (2) 當存在 k > 4 , k ? N ，使得 ?1 ?2 = k k m m 成立時，則此數列穩定，且穩定長度p 滿足：min{ | 4 } ?1 ?2 = > = k k p k k 且m m ，其中p 必為奇數。(3) { n m }為單調遞增且, 5 1 ? ? ? a m n n n4. 如果x ? 41.625，則{?x,1, x}為起始的M&m 數列，其對應的數列有相同的大小次序且此M&m 數列會穩定，穩定值為41.625，且穩定長度為73。5. 我們觀察發現：如果x 1). 3. We prove the following properties: (1) If the median of the former n numbers of the M&m sequence is n m , we obtain (2) There exist k > 4 , k ? N such that ?1 ?2 = k k m m , then the sequence is stable and the stable length min{ | 4 }?1 ?2 = > = k k p k k and m m , where p must be an odd number. (3) { n m } is monotone increasing and , 5 1 ? ? ? a m n n n . 4. Suppose x ? 41.625, then the all M&m Sequences beginning with –x , 1 , x are the same, and the sequences will be stable, the stable value is 41.625 and the stable length is 73. 5. By the computer experiments, we observe that if x is any positive real number less than 41.625, the M&m Sequence starting with –x, 1, x, will be also stable but does not appear to follow any clearly discernible pattern of behavior. However, the stable lengths are much variant and exist some unknown relation with point format of x. Moreover, we have the following properties: (1)If x is a node, then the stable value is x and the stable length equals to the index of median of the node + 2； (2)Near the branch of 41.625, the stable length is almost a constant except at the edge area，the stable length of (-x,1,x) as x around branch 1 is chaos； (3)If x near the node (K= 3, 5, 7, …, 67, 69), then the stable length is l(K)+K?1 where the positive integral l(K) is determined by Prop1 (see Table 6 and 7).
正N 邊形光圈之路徑追蹤
本研究是[對於正n 邊形A1A2…An邊上一點P(含頂點)，想像自定點P 朝鄰邊發出一條光線，若依逆(順)時針方向依序與每邊皆碰撞一次，經一圈而可回到P 點，則此路徑稱為「光圈」。過程試著追蹤在正n 邊形內能形成光圈的光線行進路徑及其相關問題。
本研究令，且以逆時針得光圈來討論：
1.根據[光的反射原理]，探討光圈之存在性，發現除定點P 在正2m 邊形或正三角形的頂點外，其餘皆有光圈。
2.將可形成光圈的路徑圖展開成[直線路徑圖]來探討。
3.由[直線路徑圖]，觀察到形成光圈的光線行進路徑，可能存在下列情況： (1)不通過正n 邊形的頂點，且產生路徑循環與不循環問題。 (2)通過正n 邊形的頂點。
4.發現正2m 邊形光圈皆為[完美光圈]。
5.發現正2m+1 邊形光圈之路徑與有理數、無理數之特質有關。即當s 值為有理數時，路徑會循環；當s 值為無理數時，路徑不循環。
The research is about [on Point P (including the angles) on the side of regular polygons A1、A2…An , imagine the light goes from Point P to the closest side, then bumps each side sequentially counterclockwise. After going a circle, it’s back to Point P. The track is called “the circle of light.” I try to trace the light track of the circle of light and other correlative questions.]
In this research, we suppose,and we discuss the circle of light according counterclockwise direction:1.According to the light reflective principles, we discuss whether the circle of light exists or not. And then we discover that the circle of light really exists except when Point P is on the angles of regular triangle or regular 2m polygons. 2.Spread out the circle of light’s track to [rectilinear track.] 3.By [the picture of rectilinear track], observing there are two kinds of the circle of light’s track: (1)If the light doesn’t go through the angles of regular polygons, it can be a circulative track or a non-circulative track. (2)When the light goes through the angles, it stops. 4.We discover that all the circles of light in regular 2m polygons are [the perfect circles of light.] 5.We discover the circle of light’s track is correlative with rational numbers and irrantional numbers. When s is a rational number, the track is circulative, if s is a irrantional number, the track is not circulative.
數列生成遞迴
這個題目是源自2003年的TRML思考賽的題目，原題目並不難，它只有用到簡單的排列方法，主要是討論 an 、bn 兩種數字的排列，其中 an 為滿足下列所有條件之N位數A的個數。
I. A中每一個數字為1或2
II. A中至少有相鄰的兩數字是1
而 bn 表示滿足下列所有條件的N位數B的個數
I. B中每一個數字為0或1
II. B中至少有相鄰的兩數字是1
以及探討an 、bn 與費氏數列cn之關係，其中 cn = cn-1 + cn-2 ，n≧3 ,c1=1, c2=2 。
其中 an 如果改成考慮為一數列，其值不變；而 bn 如果改為數列，那麼就不需要考慮0不能為首位數字的情況。如此，讓人聯想到一個用生成函數解的題目「一個N項數列，其中每一項只能是0或1或2，其中0和2永不能相鄰，求這個數列個數的一般式。」，因此，我們嘗試將這個題目改變它的要求繼續做下去，發現其中有某些規則，例如：不只是原來的11相鄰，甚至是排列其它種方式，都可能從其遞迴式看出它排列的意義，甚至這種排列數是可以用遞迴式求出來的。這提供了我們另一種求數字排列的方法，也是我們覺得有趣的地方。
在過程中我們初步得到以下結論：
This solution is according to power contest of 2003 TRML. It is composed of two number arrangements, an , bn .
First, suppose an is the total number conforming to the following rules.
I. Each number is 1 or 2 in A.
II. There is a couple of (11) in A at least.
Then, suppose bn is the total number conforming to the following conditions.
I. Each number is 0 or 1 in B.
II. There is a couple of (11) in B at least.
Furthermore , we give the thought to the relation among an ， bn ，and cn (Fibonacci Sequence).
By the way, if an is changed to a sequence, and the result is the same. But if bn is to arrange number, we have to give thought to the fact that the first number can’t be zero. If it is a sequence, we don’t have to consider it.
The problem belongs to combinatorics. After we do this problem, we find not only original question but also other permutation can be understood by its formula. The problem provides us with other means to solve permutation and combination question. Then, we get the conclusion as follows：