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[几何] 圆条件下线段长内接三角形的面积

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isee posted 2020-9-21 14:23 |Read mode
Last edited by isee 2020-9-21 15:39看到 MindYourDecisions 的 Impossible Viral Problem ,觉得有点意思,个人暂无结果,,,就仅转题来了,如图。


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我把[原作者的过程看完了,原来图中的选项全是错的,近似的结果为17.19。
area.jpg

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色k posted 2020-9-21 14:32
cosA=(R-2)/R
cosB=(R-3)/R
cosC=(R-1)/R
代入cos²A+cos²B+cos²C+2cosAcosBcosC=1

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original poster isee posted 2020-9-21 14:54
回复 2# 色k


看上去次数偏高,可能化简之后是三次,求的面积,可能可以整体求出

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色k posted 2020-9-21 14:59
回复 3# isee

你开挂算算,我爪机ing…

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original poster isee posted 2020-9-21 15:02
Last edited by isee 2020-9-21 15:11用$1=\sin^2 B+\cos^2 B=\left(\frac {2\sqrt{6r-9}}{2r}\right)^2+\left(\frac{r-3}{r}\right)^2$可能会好算一点点。。。。

===

擦恒成立!



先扔这吧,等你来化简。。。

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kuing posted 2020-9-21 15:21
开挂算了,R没简单解,面积也不能整体求出,此题数据不凑不能撸

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original poster isee posted 2020-9-21 15:39
回复 6# kuing


哈哈哈,果然是题目没截全啊。。主楼已经修改。。。。看看就好。。

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kuing posted 2020-9-21 15:41
将 123 换成一般的 xyz,则 2# 代入后化简为
\[4R^3-4(x+y+z)R^2+(x+y+z)^2R-2xyz=0,\quad(1)\]而由面积公式 `S=2R^2\sin A\sin B\sin C` 两边平方
\[S^2=4R^4(1-\cos^2A)(1-\cos^2B)(1-\cos^2C),\]同样代入 2# 的,化简为
\[4xyz(2R-x)(2R-y)(2R-z)-S^2R^2=0,\quad(2)\]如果由式 (1)、(2) 消 `R`,则会得出一串很长的式子,这里就不写出来了,还是代回 123,就是
\[\led
R^3-6R^2+9R-3&=0,\\
192R^3-(S^2+576)R^2+528R-144&=0,
\endled\]消 `R` 得
\[S^6-288S^4-2304S^2+36864=0.\]

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kuing posted 2020-9-21 15:47
咦,我发现 R 虽然没有简单根式解,却有非常简单的三角解:
\[R=2+2\cos20^\circ=4\cos^210^\circ,\]!!!难道还有啥没发现嘀东西……

于是 `S` 也同样可以写成这样的三角解……

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kuing posted 2020-9-21 16:07
擦,才发现楼主加的链接里面也解了这个方程,白写鸟……

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zhcosin posted 2020-9-22 19:22
回复 10# kuing

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kuing posted 2020-9-22 19:28
回复 11# zhcosin

好久没见!

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