|
|
Author |
hbghlyj
Posted 2021-5-13 23:38
下面是一种笨重的计算方法
由引理得$e=\frac{2bc^2\left(a^2+b^2-c^2\right)}{\left(a+b+c\right)\left(-a+b+c\right)\left(a-b+c\right)\left(a+b-c\right)},f=\frac{2b^2c\left(a^2-b^2+c^2\right)}{\left(a+b+c\right)\left(-a+b+c\right)\left(a-b+c\right)\left(a+b-c\right)}.$
由塞瓦定理,$\frac{BD}{DC}=\frac{e\left(c-f\right)}{f\left(b-e\right)}.$对△CDE用余弦定理,$DE^2=\left(b-e\right)^2\left[\left(\frac{af}{bf+ce-2ef}\right)^2-\frac{2af\cos{C}}{bf+ce-2ef}+1\right]$,同理,$DF^2=\left(c-f\right)^2\left[\left(\frac{ae}{bf+ce-2ef}\right)^2-\frac{2ae\cos{B}}{bf+ce-2ef}+1\right]$.对△AEF用余弦定理,$EF^2=\frac{4b^2c^2G}{\left(a+b+c\right)^2\left(-a+b+c\right)^2\left(a-b+c\right)^2\left(a+b-c\right)^2}$,其中
$G=a^6-3a^2b^4+2b^6+6a^2b^2c^2-2b^4c^2-3a^2c^4-2b^2c^4+2c^6$
对△DEF用中线长公式,$OD^2=\frac{1}{2}\left(ED^2+FD^2\right)-\frac{1}{4}EF^2$.全部代入得$OD^2=\frac{U}{\left(a+b+c\right)\left(-a+b+c\right)\left(a-b+c\right)\left(a+b-c\right)V^2}$,其中
$U=a^{12}-2a^{10}b^2+5a^4b^8-6a^2b^{10}+2b^{12}-2a^{10}c^2+4a^8b^2c^2-12a^4b^6c^2+18a^2b^8c^2-8b^{10}c^2+14a^4b^4c^4-12a^2b^6c^4+14b^8c^4-12a^4b^2c^6-12a^2b^4c^6-16b^6c^6+5a^4c^8+18a^2b^2c^8+14b^4c^8-6a^2c^{10}-8b^2c^{10}+2c^{12}$
$V=a^4b^2-2a^2b^4+b^6+a^4c^2+4a^2b^2c^2-b^4c^2-2a^2c^4-b^2c^4+c^6$
对△ACD用余弦定理,$MT^2=AD^2=\frac{b^2c^2W}{\left(a^4b^2-2a^2b^4+b^6+a^4c^2+4a^2b^2c^2-b^4c^2-2a^2c^4-b^2c^4+c^6\right)^2}$,其中$
W=-a^{10}+2a^8b^2-2a^6b^4+4a^4b^6-5a^2b^8+2b^{10}+2a^8c^2+4a^6b^2c^2-4a^4b^4c^2+20a^2b^6c^2-6b^8c^2-2a^6c^4-4a^4b^2c^4-30a^2b^4c^4+4b^6c^4+4a^4c^6+20a^2b^2c^6+4b^4c^6-5a^2c^8-6b^2c^8+2c^{10}$
设∠OMQ=α,则$\tan{\alpha}=\frac{EF}{2R}⇒OR=R\cos{2\alpha}=R\cdot\frac{4R^2-EF^2}{4R^2+EF^2}.$
设RN中点为K,对△ATK用Stewart定理,$OT^2=\frac{2R}{3R+OR}\left(\frac{R-OR}{2}\right)^2+\frac{R+OR}{3R+OR}MT^2-\frac{R\left(R+OR\right)}{2}.$
全部代入得$OT^2=\frac{U}{\left(a+b+c\right)\left(-a+b+c\right)\left(a-b+c\right)\left(a+b-c\right)V^2}$,故OD=OT. |
|
