✔ 最佳答案
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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：
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