OI⊥EF,且OI:EF只与∠A有关

hbghlyj

△ABC的内心为I,外心为O,点E,F分别在射线CA,BA上,使BF=CE=BC,求证
(1)OI⊥EF
(2)当A在⊙O上运动时,EF:OI不变

证明
(1)取AC中点Mb,取AC上的内切圆切点Tb,则$OE^2=OMb^2+EMb^2=R^2\cos^2{B}+\left(\frac{b}{2}-a\right)^2=R^2-ab+a^2,IE^2=ITb^2+ETb^2=r^2+\left(\frac{a-b+c}{2}\right)^2$⇒$OE^2-IE^2=R^2-r^2+\frac{3a^2-b^2-c^2-2a\left(b+c\right)+2bc}{4}$⇒$OE^2-IE^2=OF^2-IF^2⇒OI⊥EF$
(2)对△AEF用余弦定理,\begin{aligned}
\frac{EF^2}{OI^2}=&4\frac{\sin^2{A}\left(3-2\cos{A}\right)+2\sin{A}\left(\sin{B}+\sin{C}\right)\left(\cos{A}-1\right)}{3-2\left(\cos{A}+\cos{B}+\cos{C}\right)}\\
=&4\frac{\sin^2{A}\left(3-2\cos{A}\right)-8\sin{A}\cos{\frac{A}{2}}\cos{\frac{B-C}{2}}\sin^2{\frac{A}{2}}}{3-2\left(\cos{A}+\cos{B}+\cos{C}\right)}\\
=&4\frac{\sin^2{A}\left(3-2\cos{A}\right)-4\sin^2{A}\cos{\frac{B-C}{2}}\cos{\frac{B+C}{2}}}{3-2\left(\cos{A}+\cos{B}+\cos{C}\right)}\\
=&4\frac{\sin^2{A}\left(3-2\cos{A}\right)-2\sin^2{A}\left(\cos{B}+\cos{C}\right)}{3-2\left(\cos{A}+\cos{B}+\cos{C}\right)}\\
=&4\sin^2{A}
\end{aligned}
⇒$\frac{EF}{OI}=2\sin{A}.$

kuing

\begin{align*}
EF^2&=(b-a)^2+(c-a)^2-2(b-a)(c-a)\cos A\\
&=(b-a)^2+(c-a)^2-(b-a)(c-a)\frac{b^2+c^2-a^2}{bc}\\
&=a^2\cdot\frac{(a+b+c)^3-4(a+b+c)(ab+bc+ca)+9abc}{abc},
\end{align*}代 `a+b+c=2s`, `ab+bc+ca=s^2+4Rr+r^2`, `abc=4sRr` 化简得
\[EF^2=a^2\cdot\frac{R-2r}R=a^2\cdot\frac{IO^2}{R^2}=4\sin^2A\cdot IO^2.\]