.平面座標上長方形沙發旋轉問題之解的存在性
這篇報告要探討下列的「轉沙發的問題」是否有解？有一個長方形的沙發，如圖一，若要求每次只能以「四個頂點逆時針或順時針連續旋轉90度」的方式轉動，請問當長寬具備何種關係時，沙發經數次轉動後，剛好可以「轉」到相鄰的位置，如圖一，而且沙發坐人的正面方向仍保持不變呢？
我們把原問題看成「平面座標上長方形旋轉的數學問題」，再利用「平面座標、三角函數、複數、複數的極式表示及向量」等數學工具，導出符合題目要求的方程式，最後證出當長與寬的比值為正實數時，有下列的結果：
1.當長與寬比值為無理數時，此問題無解。
2.當長與寬比值是最簡分數時，若分子為奇數，此問題無解。
3.當長與寬比值是最簡分數時，若分子為偶數，分母為奇數，此問題有解。
4.在有解的情況下，我們可以找出特定轉法的最小值。
5.當長與寬比值是最簡分數時，若分子為偶數，分母為奇數，沙發可轉至A點座標為(αp,0) 的位置，其中 α∈Z，且沙發坐人的正面方向保持不變。
6.當長與寬比值是最簡分數時，若分子為奇數，分母為偶數，沙發可轉至A點座標為(0,βq) 的位置，其中 β∈Z，且沙發坐人的正面方向保持不變。
7當的長與寬比值為正實數時，可將沙發轉至A點的座標為(2αp + 2βq,2γp + 2qω)的位置，其中 α,β,γ,ω∈Z，且沙發坐人的正面方向保持不變。 In this paper we discuss the solution of rotating sofa problem as follows : The condition is : Merely allow to rotate the sofa several times by rotating 90 degrees clockwise or counterclockwise around the vertex. (maybe A, B, C, or D in Fig. 1) The question is : What’s the relationship between the length and the width of the sofa, if we request the sofa translated next to the original position with direction unchanged. (as shown in Fig. 1 with A’B’C’D’).
We take this problem as a mathematical one of rotating a rectangle in plane coordinates. Then we derive the desired equations by using the tools of plane coordinates, trigonometric functions, complex number, polar form of complex number, and vector. Finally, we prove that：
1. When the ratio of length and width is irrational, the problem has no solution.
2. When the length of sofa is odd in the ratio of length and width, the problem has no solution.
3. When the ratio of length and width is even, the problem has solutions.
4. When the solutions exist , we can find the minimum of the number of rotations.
5. When the ratio of length and width is an irreducible fraction, which has the even numerator and the odd denominator, the sofa can be rotated to the coordinate (αp,0)(α∈Z)which is the new position of A and keep the original position with direction unchanged.
6. When the ratio of length and width is an irreducible fraction, which has the odd numerator and the even denominator, the sofa can be rotated to the coordinate (0,βq)(β∈Z) which is the new position of A and keep the original position with direction unchanged.
7. When the ratio of length and width is a real positive number, the sofa can be rotated to the coordinate (2αp + 2βq,2γp + 2qω)(α,β,γ,ω∈Z)which is the new position of A and keep the original position with direction unchanged.
都是氣泡惹的禍
在物理馬戲團這本書中提到：「當你泡即溶咖啡或攪拌奶精的時候，用湯匙輕敲杯壁看看，添加奶精後攪拌時，敲擊的聲音與添加前明顯不同，為什麼？」這本書的解答是：「當粉末溶解的時候，藏在粉末裡的空氣就會跑出來。因為空氣裡的音速低於水裡的音速，在空氣與水混合的環境裡，音速也比在水裡低。當水裡不斷有空氣混進去時，這個容器的共振頻率和它裡面的音速有關，所以也會降低。因此你會聽到較低的音調，直到空氣全部跑光。」我們利用指向性麥克風以電腦錄音後以Adobe Audition 軟體分析聲波頻率，覺得這個說法有點問題。例鹽水溶液音速較水高，敲擊時的音調卻較水低。由敲擊一黏於裝水水盆中之空杯，與敲擊杯內裝同一水位之水之杯子，頻率非常接近。告訴我們影響頻率的是靠近杯壁一層有效質量。因鹽水溶液密度較高有效質量較大，所以頻率較低。以密度的觀念檢視裝有溶液之杯子被敲後的頻率是對的。但對杯中有懸浮物就不然，例如流體中含有氣泡，則混合體之密度必定變低，有效質量變小頻率應變高。但實驗發現含有氣泡時頻率是變低的。可見氣泡還有其他的影響力高於密度對音調的影響。 流體的振動應是會壓縮到氣泡，氣泡與流體間之力學交互作用為何會使頻率下降，正是我們要找出的。 The Flying Circus of Physics has a question “As you stir instant cream or instant coffee into a cup of water, tap the side with your spoon. The pitch of the tapping changes radically as the powder is added and during the stirring. Why?” The answer is, “The air trapped in the powder is released as the powder dissolves. Since the speed of sound is lower in air than that in water, the speed of sound in the air-water mixture is lower than that in pure water. During that period while the air escapes the container, the resonant frequencies of the water, which depend directly on the speed of sound, will also be lower. Hence, you hear a lower tone until the air escapes. “We then tap the coffee cup and generate an audible tone. The signal picked up by the microphone . The same signal is also studied using Adobe Audition, a waveform processing and analyzing software. We find the assumption is wrong, the speed of sound is higher in sugar solution than that in water, but we hear a lower tone. An effective layer of fluid adjacent to the glass wall is set into motion when we gently rub the rim of the wineglass. The thickness is about the same whether the fluid is inside or outside the glass. This explains why the frequency drops when the liquid is added to the system. When the density of the sugar solution is higher, the mass of the effective layer is higher. But what the presence of the bubbles and the theoretical explanations must NOT rely on are: Use effective density argument: One should not just use a change in the main density to try to explain why the frequency is lower. I would think that the bubbles are compressed a little bit by the vibrational motion of the glass communicated to them through the fluid. But how do the bubbles interact with the fluid under this setting? This is what we need to work out.
盡可能擁擠
給定一個有n個頂點的簡單圖 G，將頂點標號為1,2,…n；考慮 任意相鄰的兩頂點標號和中最大值的最小值，稱此極值發生時的標號為圖G的擁擠標號。在這個研究中，我們得出方格表、m×n×l長方體、環狀圖、圓柱圖及樹圖的擁擠標號和其極值的通式，並討論相關的問題。 Given a simplicial graph G of n vertices, label the vertices with 1,2,…n. Consider the minimum of the maximum of the sum of any two close vertices’ labeling, and we call the labeling that has this extremum the “crowded labeling”. In this study, we found out the “crowded labeling” and the common equation of the grid, m×n×l cuboid, cycle, cylinder, and trees. And discuss the correlative questions.
長方體中切割正立方體之研究
我們這個作品是先由在長方形中切割出正方形的研究著手，先研究出在平面中，在一個邊長為任意正整數的長方形中，如何找到在其中切割出正方形，但正方形的邊長為最大，而且正方形的個數為最少的方法和規則。 緊接著，我們更進一步想研究這個問題在長方體中的研究：在長方體三邊長a、b、c（a、b、c均為正整數）中，如何在其中切割出正立方體，每次切割出邊長為最大的正立方體，而且正方體的個數為最少的方法和規則。 This study began with investigation of how to segment squares from a rectangle. We studied from a rectangle, with random positive integer sides, trying to figure out the methods and regulations to segments squares with the longest side length but the fewest number of squares within. Moreover, we took further step to examine a cuboid. We found out the methods and regulations to segment cubes with longest side length but fewest number of cubes from a cuboid with sides a, b, and c(a ,b ,c are positive integers).