有關三角形的問題 @ 飛航模式

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有關三角形的問題

發問:

有一直角三角形 ABC 內有另一三角形 AXC. 如在 AXC 的三角形內任意取一點(x,y), 之後將三角形 AXC 擴展到與三角形 ABC 一樣大小. 請問用什麼方法或公式去求得剛才在三角形 AXC 上的點(x,y) 在擴展後的新座標? 更新: 有圖. 但我唔知點貼將圖上發問區.. 可唔可以教下我. thanks. 更新 2: http://img186.imageshack.us/img186/2179/question1.jpg

最佳解答:

圖片參考：http://img510.imageshack.us/img510/5298/dddt.png http://img510.imageshack.us/img510/5298/dddt.png Consider a general point D = (p,q) within ABC, AD1 = h - q; AB = h; BC = w By similar triangles, D1D2 = w(h - q) / h ... (1) DD2 = D1D2 - D1D = w(h - q) / h - p = [w(h - q) - hp] / h ... (2) When we consider X(a,b) as D, by sub (a,b) into (p, q) XX2 = [w(h - b) - ha] / h X1X2 = w(h - b) / h When X is dragged horizontally to X1, XX2 will be expanded to X1X2 The magnification factor is X1X2 / XX2 = w(h - b) / [w(h - b) - ha] Consider the triangles AXX2 and AEE2, the expansion of EE2 to E1E2 has the same magnification factor due to similar triangles Consider the triangles CXX2 and CFF2, the expansion of FF2 to F1F2 has the same magnification factor due to similar triangles Now consider a point M(x,y) within the triangle, its horizontal distance from the hypotenuse is given by sub (x,y) into (2), i.e. [w(h - y) - hx] / h and M1M2 is w(h - y) / h After expansion, the distance from the hypotenuse will increase to {[w(h - y) - hx] / h}{w(h - b) / [w(h - b) - ha]} The new x-ordinate will be w(h - y) / h - {[w(h - y) - hx] / h}{w(h - b) / [w(h - b) - ha]} which is simplified to w(hx - bx - ah + ay)/[w(h - b) - ha] ... (3) We can verify this equation, by sub A, X and C into it, i.e. For A, new x-ordinate = w(- ah + ah)/[w(h - b) - ha] = 0 no change For X, new x-ordinate = w(ha - ba - ah + ab)/[w(h - b) - ha] = 0 move to side of the triangle For C, new x-ordinate = w(hw - bw - ah)/[w(h - b) - ha] = w no change Now by symmetry, the translation in the y-axis can easily obtained by interchanging x and y related entities in equation (3), i.e. new y-ordinate is h(wy - ay - bw + bx)/[h(w - a) - wb] ... (4) 2009-09-05 11:06:38 補充： Upload 到 http://img215.imageshack.us/ 再提供條link

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