變形的橢圓—從距離及距離和談起
給定一平面E，A為平面上一點。取r>0，則我們知道到其距離為定值的點形成一圓，而A為此圓圓心。如果把A改成一平面圖形，則到其距離為定值的點形成的集合會是什麼樣子？類似地，給定平面上兩焦點F1及F2在平面上，則到其距離和為定值的點形成橢圓。同樣的，若把F1及F2改成平面圖形，其圖形會是什麼樣子？藉著GSP的輔助，到目前為止，我們得到了以下的結果： \r 1. 給定一平面E及此平面上的一個凸多邊形, 我們描繪出在此平面上到此凸多邊形之距離為定值的點所形成的圖形。\r 2. 設F1和F2分別為平面E上之點或線段或多邊形（未必是凸多邊形），我們利用包絡線描繪出所有滿足d(P,F1)+d(P,F2)=k（k夠大）的點所形成的圖形。 \r 3. 設C1,C2為平面E上之兩圓，我們討論所有滿足 d(P,C1)+d(P,C2)=k\r （k夠大）的點形成的圖形並討論其性質。 \r 4. 設L1和L2分別為平面E上之兩線段，我們討論所有滿足d(P,L1)+d(P,L2)=k（k夠大）的點形成的圖形並討論其性質。 \r 5. 設A為平面E上之一點，Γ為平面上一凸多邊形，我們討論所有滿足d(P,A)+D(P,Γ)=k（k夠大）的點形成的集合並討論其特性。 \r 6. 藉由和圓作比較，我們研究了變形圓的光學性質；而對變形橢圓也做類似的討論。\r Let E be a plane and A a fixed point on E. Given , it is known that all of the points on E with distance to 0r>rA form a circle and the point A is called the center of this circle. What is the corresponding graph if we replace the point A with a set (for example,a segament or a polygon) contained in FE? Similarly, what is the case when we modify the two focuses and in the definition of an ellcpse to sets and (or example,two segments or two polygons) contained in 1F2F1F2FE ? Taking advantages of GSP and analytic geomety, we research related situations and so far we have obtained the following results:\r 1. Let Γ?E be a segment, a convex polygon or a circle , etc. and r>0 be fixed. We sketch the graph of points on E with distance r to Γ and study properties of such graphs.\r 2. Let F1 and F2 be singletons, line segments , polygons(may not be convex), or circles,etc., on E Taking advantage of envelopes, we sketch the graph of those points P on E satisfying d(P,F1)=k(K>0 is large enough).\r 3. Let C1 and C2 be circles on 1C2CE. We sketch the graph of the points P on E that satisfiy d(P,C1)6d(P,C2)=k (k>0 is large enough) and study properties of this graph.\r 4. Let L1 and L2 be two line segments on E and be a large enough constant. We sketch the graph of points P on E that satisfy d(P,L1)+d(P,L2)=k(k >0is large enough) and research properties of this graph. 0k>\r 5. Let A?E and be a convex polygon on ΓE. We sketch the graph of points on E that satisfy d(P,L1)+d(P,L2)=k(k>0 is large enough) and research properties of this graph.\r 6.We compare the optical properties of metamorphic circles with circles and we deal with metamorphic ellipses similiarly.
A New 3-Dimensional Model for the Periodic Table of Codons
a. Purpose of research- Since the discovery of genetic codes and the dogma of 64 codons coding for 21 amino acids, scientists worldwide have been interested to know the reason(s) behind this unique number ratio (64:21). This ratio indicates certain form of inefficiency in the replication of amino acids. Such inefficiency can be explained through symmetries in the condons coding for the same amino acids. In the light of that, my project looks for patterns in the properties of amino acids and symmetries in the codons combinations. Using these analysis findings, I invented a three dimensional periodic table for the codons and amino acids that has a points to layers ratio of 64:21. b. Procedures- To get started with the project, I searched for relevant information in books and the Internet. After locating the relevant materials, I began my analysis by looking for non-random patterns in the correlation between codons and the respective amino acids they code for. At the same time, I try to look for symmetries in the codon distributions and suggest new and innovative models for a periodic table of codon combinations. I have come out with mainly a new model, with its own unique ideas and concepts behind it. Finally, I will try to match a property of the amino acids to the positions of the codons such that the table shows a gradual change in property of the amino acids, together with the symmetries. This will effectively explain the unique codons to amino acids ratio and lead to discovery of possible amino acids. c. Data- This research is primarily conducted based on the conventional 2D periodic table and no experimental data is collected. After much analysis, I have come up with the 3-sided triangular pyramid model. This model is inspired by the ratio of 64 codons coding for 21 amino acids, which can be easily approximated to 3:1. It is made up of a triangular pyramid that is three-faced, with the bottom side unutilized. As a triangular structure, each layer has dimensions in the multiples of 3. Layer 1 consists of 1 point, layer 2 with 3 points, layer 3 with 6 points and so on… until layer 7 with 18 points, having a total of 64 points. This 64 points to 21 layers ratio is consistent with the codons to amino acids ratio! d. Conclusions- The unique 64:21 ratio suggest certain form of inefficiency in the replication of amino acids. This may be explained through symmetries in condons coding for the same amino acids. A general 3:1 ratio can be approximated and this suggests a high possibility for the existence of a three-sided symmetry in codon combinations. Thus, this idea of a three-sided symmetry gives rise to my 3-sided triangular pyramid model. This new model of a 3-dimensional periodic table for codon combinations would be useful in explaining such a unique 64:21 ratio and serves to provide a basis for better understanding of the relationship between codons and amino acids. This new model may also lead to the discovery of currently unknown amino acids.
總站該設在哪裡？—另類費馬點的研究
The definition of "Fermat Point" is that a dot, which lies in a triangle, has the minimum distance to the three apexes. In other words, "Fermat Point" has the minimum distance to three dots which are not on the same line. In the broad sense, then, in a N polygon, a dot which has the minimum distance to the N apexes could be named "Fermat Point." But what if we link up the N apexes and find out that they cannot make a convex polygon? The above is what we wish to fully discuss. Our inspiration comes from a paper on"Fermat Point." It just describes N convex polygon, so we think of putting the case to naturally polygon. The case may be that it is a concave polygon or part of the apexes which lies on the same line. We would not base our study on the conventional methods. Moreover, strictly defined, the repeated line segment will not be taken into account. That is, if the "Fermat Point" drops on the line with more than two dots on it, we just count the\r line segments except for the shorter line segments which were originally included in other studies. According to the theorem, our conclusions are as follows: 1. If N points lie on the same line segment, then the "Fermat Point"can be any point on the line segment. 2. If (N-1) points are on the same line segment, then the "Fermat Point" is on the point which two lines join together. One is that the line segment, and the other is the one which passes the remaining point and\r perpendicular to the first line segment. 3. Now there are (M+N) points. Among them, M points will make a M jog-polygon. The others all drop in the polygon. As the diagram shown beneath, we know that the "Fermat Point" drops on the point which two lines join together. The two lines must pass as many points as possible. 所謂的「費馬點」是指三角形內到三頂點距離和最小的點。換言之，「費馬點」就是到平面上不共線三點距離和最小的點。因此，我們可定義，廣義的「費馬點」即是n 多邊形內到各頂點距離和最小的點，亦即到平面上不共線n 點距離和最小的點，但若平面上n 點不能恰為n 多邊形的頂點呢？這就是我們所要討論的。由於我們的靈感來自一份關於「費馬點」的科展作品，所以我們想到，當平面上n 點不能恰為n 凸多邊形的頂點，甚或其中有一部分的點共線時，將不能以n邊形的方法來探討，但我們可以將之化為m 邊形內（n-m）個點來討論。而更重要的是，我\r 們增加了另一個限制，重複的線段將不被我們列入計算。亦即當所求點落在某一多點共線的線段上時，我們只計算該線段的總長，而不計其中重複的較短線段。根據這個原則，我們試行證明平面上三點、四點、五點及六點的可能情況，期望能從中找出足以推廣至平面上n 點的一般性。結果雖不完美，但我們總算差強人意的歸納出了下列結論：1.若n 點共線段，所求點可為所共線段上任一點。2.若（n-1）點共線段，則由該不共線點引一線與共線段垂直，其交點即為所求。3.若（n+m）個點中有m 個點為一m 多邊形的頂點，另外n 個點落在該m 多邊形內，則由兩個外頂點引直線盡可能通過最多點，該兩直線的交點即為所求。
探討「避開矩形框」的配置方法與推廣
一、若Mn×n(s)表示在n×n 的正方形棋盤中，排列s 顆棋子在方格內，且每一方格最多只能排1子，其中s 顆棋子的配置需滿足兩個條件：1. 並無任意4 子可以形成矩形框的4 個頂點。（此矩形框的邊需與棋盤的邊平行）2. 在沒有棋子的方格中，無法再加入棋子。二、若Vn×n×n(a1,……,an) 表示在n×n×n 的正方體棋盤中，每層的棋子個數分別為a1,……,an，且s= a1+……+an，其中s 顆棋子的配置需滿足兩個條件：1. 並無任意8 子可以形成長方體的8 個頂點。（此長方體的邊需與立體棋盤的邊平行）2. 在沒有棋子的方格中，無法再加入棋子。本研究即在Mn×n(s)與Vn×n×n(a1,……,an) , s= a1+……+an 中探討s 的最小值、最大值及變化情形，並分析其配置方法。之後推廣至長方形Mn×m(s)及長方體Vn×m×k(a1,……,ak) , s= a1+……+ak。最後根據其研究結果設計一個「避開矩形框棋」，並加以分析出致勝的策略。一．If Mn×n(s) indicates in the n×n square chessboard, we put s chesses to line in the square and each square only can put one chess. Then the station of s chesses must satisfy the following two conditions：1. No any 4 chesses can form the tops of the rectangular frame ( The sides of rectangular frame must be parallel to the sides of chessboard )2. If there’s no chess in the square, we can’t add any chess. 二．If Mn×n×n(a1,……,an) indicates in the n×n×n square chessboard, the chess number in each layer are a1,……,an and s= a1+……+an. The station of s chesses must satisfy the following two conditions： 1. No any 8 chesses can form eight tops of the rectangular cube ( The sides of rectangular cube must be parallel to the sides of cubic chessboard ) 2. If there’s no chess in the square, we can’t add any chess. This research try to explore the minimum, maximum and variation of s which in Mn×n(s) and Mn×n×n(a1,……,an), s= a1+……+an, and analyze its station. Then we will extend the research to rectangle Mn×m(s) and rectangular cube Vn×m×k(a1,……,ak), s= a1+……+ak. Finally, according to the result of research we wish can design one “avert rectangular frame chess“ and analyze the strategies to triumph.
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.
四面體體積平分面的包絡方程探討
剛開始考慮平分物件時，我們從二維的多邊形部分著手，後來發現已經有人做過相關研究，並且得到類似的結論。這個部份顯現出面積平分線與其包絡曲線間的密切關係。我們將其中的方法和結果加以歸納、改善，為了更全面地研究，我們推導出一般性的包絡方程。之後當我們推廣到三維領域時，發現四面體體積平分面與之前的結論有些相似之處，平分的情況卻也更複雜，我們將推導的結果用電腦軟體呈現出來，以便更深入地了解它。最後嘗試了相當抽象的高維積平分，結果仍具有工整的對稱性，讓我們充分領略了數學之美！When considering bisecting a subject, at first we focused our attention on 2-D case, polygons. But afterwards, we found there were already some similar studies conducted by other students, which indicated the close relation between the area-bisecting lines of a polygon and their envelope. We rearranged their methods and results, and then made further improvement. Moreover, in order to study the bisecting problem entirely, we derived the general envelope equation. Then when extending the generalization to the 3-D case, we came to the conclusion that tetrahedrons’ volume-bisecting planes is similar to that in 2-D, but the circumstances are more complex. We tried to show our result with the aid of software, hoping to understand it fully. Finally, we tried to do the case in higher dimension, which is very abstract, and the result was clear-cut symmetrical. During the studying process, we had seen “the beauty of mathematics.”