三角形外心

realnumber

设O是$ΔABC$的外心，$\vv{AO}=x\vv{AB}+y\vv{AC}$,且$\abs{\vv{AB}}=4,\abs{\vv{AC}}=6,2x+\frac{1}{2}y=1 $,
则$\vv{AB}$·$\vv{AC}$．


我是这样解$\vv{AO}=x\vv{AB}+y\vv{AC}$两边分别与$\vv{AB},\vv{AC}$数量积下，得到三元二次方程组，似乎不好解．
有什么好办法？

realnumber

$\vv{AB}$·$\vv{AC}＝s$
那么三个方程是8=16x+ys,18=36y+xs,2x+0.5y=1．而第１，３个方程正好可以把x,y同时消去，得到s=4.
看来数据是凑好的．如果换成$3x+4y=1$,也许该老老实实地解了吧．

kuing

\(\newcommand\led[1]{\left\{\begin{aligned}
#1
\end{aligned}\right.}\)

\begin{align*}
\led{
\vv{AB}\cdot\bigl(2\vv{AO}-\vv{AB}\bigr)&=0, \\
\vv{AC}\cdot\bigl(2\vv{AO}-\vv{AC}\bigr)&=0,
}
&\riff
\led{
\vv{AB}\cdot\bigl((2x-1)\vv{AB}+2y\vv{AC}\bigr)&=0, \\
\vv{AC}\cdot\bigl(2x\vv{AB}+(2y-1)\vv{AC}\bigr)&=0,
}\\
&\riff
\led{
16(2x-1)+2y\vv{AB}\cdot\vv{AC}&=0, \\
2x\vv{AB}\cdot\vv{AC}+36(2y-1)&=0, \\
2x+\frac12y&=1,
}\\
&\riff\cdots
\end{align*}

kuing

看来是一样的意思

kuing

$\vv{AO}=x\vv{AB}+y\vv{AC}$, $\vv{AB}^2=c^2$, $\vv{AC}^2=b^2$, $s=\vv{AB}\cdot\vv{AC}$, $px+qy=1$，则
\begin{align*}
\led{
\vv{AB}\cdot\bigl(2\vv{AO}-\vv{AB}\bigr)&=0, \\
\vv{AC}\cdot\bigl(2\vv{AO}-\vv{AC}\bigr)&=0,
}
&\riff
\led{
\vv{AB}\cdot\bigl((2x-1)\vv{AB}+2y\vv{AC}\bigr)&=0, \\
\vv{AC}\cdot\bigl(2x\vv{AB}+(2y-1)\vv{AC}\bigr)&=0,
}\\
&\riff
\led{
c^2(2x-1)+2sy&=0, \\
2sx+b^2(2y-1)&=0,
}\\
&\riff
\led{
x&=\frac{b^2(c^2-s)}{2b^2c^2-2s^2},\\
y&=\frac{c^2(b^2-s)}{2b^2c^2-2s^2},
}
\end{align*}
故
\[pb^2(c^2-s)+qc^2(b^2-s)=2b^2c^2-2s^2,\]
即
\[2s^2-(pb^2+qc^2)s+(p+q-2)b^2c^2=0,\]
然后就可以凑数得到好数据……

hongxian

回复 1# realnumber


    这种题通常用三点共线吧！不知答案是不是4?

其妙

这儿有一道类似题吧？？blog.sina.com.cn/s/blog_54df069f0101hpws.html

hongxian

取$AB$中点$D$,$\vv{AC}$延长一倍到$E$
$\vv{AO}=2x \cdot\frac12\vv{AB}+\frac12y\cdot 2\vv{AC}$,$2x+\frac12y=1$
所以$E$、$O$、$D$三点共线，易得$\vv{AC}$在$\vv{AB}$上投影为1