Mayan Pyramid
| Time Limit: 1000MS |
|
Memory Limit: 65536K |
| Total Submissions: 850 |
|
Accepted: 84 |
Description
When a king dies, Mayan people build a pyramid as a monument. The pyramid is a tetrahedron. So the bottom of the pyramid is a triangle. One side of the triangle represents the year of the kings death, another represents the month, and the last represents the day. The height of the pyramid represents the king's age.
In the following picture, we see a Mayan pyramid. We have already known the coordinates of A, B and C. And we know the height of P. Q is the projection of P. Because of the Mayan religion, they want PD2+PE2+PF2 to be as small as possible.
Input
The input consists of one or more test cases ended with EOF. Each line contains one test case. Each test case contains 7 non-negative float numbers meaning the coordinates of A, B, C and the height of P. All numbers are less than 1000.
Output
For each test case, output the coordinates of Q in a single line. Round all numbers to the second digit after the decimal point.
Sample Input
0 0 2 0 1 1.73205 1
Sample Output
1.00 0.58
題目要求使PD2+PE2+PF2最小的點(diǎn)P在底面上的投影Q的坐標(biāo)。因?yàn)楦咭欢耍跃娃D(zhuǎn)化為求使QD2+QE2+QF2最小的點(diǎn)Q，也就是三角形點(diǎn)一點(diǎn)到三邊的距離的平方和最小。
這個(gè)點(diǎn)是重心的等角共軛點(diǎn)，求等角共軛點(diǎn)的方法：先求出重心關(guān)于三邊的對(duì)稱點(diǎn)，由這三個(gè)對(duì)稱點(diǎn)組成的三角形的外心即為所求。
方法就是這樣。至于怎么實(shí)現(xiàn)就是簡單的事情了..計(jì)算幾何還是很繁瑣的啊....
posted on 2008-12-29 11:07
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