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标题: 角度 [打印本页]

作者: 烟 时间: 2005-1-11 17:20
标题: 角度

已知角AOB等于40度，P是角内一点，角POA＝30度。M、N分别是OA和OB上的一点，要使▲PMN的周长最小，角MPN应是几度？

作者: cn 时间: 2005-1-11 18:16
标题: 0度

0度

作者: fzy 时间: 2005-1-12 18:28
标题: 100°

We assume P is a fixed point and we adjust M and N to achieve the minimum. Otherwise they all shrink to O.

When the perimeter reaches the minimum, we must have 角NMO = 角NMA and 角MNO = 角MNB. Because 角NMO + 角MNO = 180° - 角AOB, we have 角NMP + 角MNP = 360° - 2 * (角NMO + 角MNO) = 2 * 角AOB. Therefore 角MPN = 180° - 2 * 角AOB. In this particular case, it is 100°

This is a very nice problem with a very nice result.

BTW Mr or Ms 烟, I would like to post this problem to another forum if I have your permission. I will, of course, credit it to you and this forum.

作者: 呵呵 时间: 2005-1-12 20:55
标题: 80度

80度

作者: 呵呵 时间: 2005-1-12 22:46
标题: Sorry, 再加20度

Sorry, 再加20度

作者: husonghu 时间: 2005-1-14 21:19
标题: Good answer, but there is a typo.....

"When the perimeter reaches the minimum, we must have 角NMO = 角AMP (not NMA) and 角MNO = 角BNP (not MNB)",
I guess, hehe.

作者: fzy 时间: 2005-1-15 20:22
标题: 回复：Good answer, but there is a typo.....

Yes, thanks. You know what I mean.

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