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设 ABCD 为凸四边形, AC 交 BD 于 P.
△ABP,△BCP,△CDP,△DAP的内心为 $I_1, I_2, I_3, I_4 $.
求证: $I_1, I_2, I_3, I_4 $ 四点共圆当且仅当 ABCD 有内切圆.
双心四边形的八个三角形内心共圆
证明:
先证明必要性. 当 $I_1, I_2, I_3, I_4$ 四点共圆时,
$$\tag1
PI_1 \cdot P I_3=P I_2 \cdot PI_4
$$
设 $P A=x, P B=y, P C=z, P D=w$.
$A B=a, \quad B C=b, \quad C D=c, \quad D A=d . \quad \angle A P B=\alpha,\quad P I_1=\dfrac{r_1}{\sin \frac{\alpha}{2}}$ ($r_1$ 为 $\odot I_1$ 的半径)
从而可知$(1) \Leftrightarrow$\[\tag2\frac{r_1 r_3}{\sin ^2 \frac{\alpha}{2}}=\frac{r_2 r_4}{\cos ^2 \frac{\alpha}{2}} \Leftrightarrow \frac{\frac{1}{r_1}+\frac{1}{r_3}}{\frac{1}{r_2}+\frac{1}{r_4}}=\frac{\left(r_1+r_3\right)(1+\cos \alpha)}{\left(r_2+r_4\right)(1-\cos \alpha)} \]\begin{aligned}
r_1&=\frac{x y \sin \alpha}{x+y+a}\\r_3&=\frac{w z \sin \alpha}{w+z+c}\\
1+\cos \alpha&=\frac{(x+y+z)(x+y-z)}{x y}=\frac{(w+z-c)(w+z-c)}{w z} \\
1-\cos \alpha&=\frac{(y+z-b)(y+z+b)}{y z}=\frac{(x+w-d)(x+w+d)}{x w} \end{aligned}故$(2) \Leftrightarrow$
\[\tag3 \frac{\frac{a}{x y}+\frac{c}{w z}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{w}}{\frac{b}{y z}+\frac{d}{x w}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{w}}=\frac{r_1(1+\cos \alpha)+r_3(1+\cos \alpha)}{r_2(1-\cos \alpha)+r_4(1-\cos \alpha)}
=\frac{x+y+z+w-a-c}{x+y+z+w-b-d}\]设 $a+c \geq b+d$, 则 $x+y+z+w-a-c \leq x+y+z+w-b-d$
因为 $(a+c)^2 \geq(b+d)^2, a^2+c^2-b^2-d^2+2(a c-b d) \geq 0$
故\begin{multline*}x^2+y^2-2 x y \cos \alpha+w^2+z^2-2 z w \cos \alpha\\
-\left(x^2+w^2+2 x w \cos \alpha+y^2+z^2+2 y z \cos \alpha\right)+2(a c-b d) \geq 0
\end{multline*}$$
\begin{aligned}
& \Leftrightarrow \frac{2(a c-b d)}{x y z w} \geq \frac{2 \cos \alpha}{x y z w}(x y+z w+x w+y z) \\
& \left(\cos \alpha=\frac{x^2+y^2-a^2}{2 x y}=\frac{w^2+z^2-c^2}{2 w z}=-\frac{y^2+z^2-b^2}{2 y z}=-\frac{x^2+w^2-d^2}{2 x w}\right) \\
& \Leftrightarrow \frac{2(a c-b d)}{x y z w} \geq-\left(\frac{a^2}{x^2 y^2}+\frac{c^2}{z^2 w^2}\right)+\frac{b^2}{y^2 z^2}+\frac{d^2}{x^2 w^2} \\
& \Leftrightarrow\left(\frac{a}{x y}+\frac{c}{z w}\right)^2 \geq\left(\frac{b}{y z}+\frac{d}{x w}\right)^2 \Leftrightarrow \frac{a}{x y}+\frac{c}{z w} \geq \frac{b}{y z}+\frac{d}{x w}
\end{aligned}
$$
由 (3) 式, 有:
$$
1 \leq \frac{\frac{a}{y z}+\frac{c}{z w}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{w}}{\frac{b}{y z}+\frac{d}{x w}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{w}}=\frac{x+y+z+w-a-c}{x+y+z+w-b-d} \leq 1
$$
∴ 它们均等于 1. ∴ $a + c = b + d$. 必要性证毕.
充分性. 由上述证明知
\begin{aligned}a+c=b+d &\Leftrightarrow\left(r_{1}+r_{3}\right)(1+\cos \alpha)=\left(r_{2}+r_{4}\right)(1-\cos \alpha)\\&⇔\frac{1}{r_{1}}+\frac{1}{r_{3}}=\frac{1}{r_{2}}+\frac{1}{r_{4}}\end{aligned}从而(2)成立.
得出 $I_1 I_2 I_3 I_4 $ 四点共圆,证毕. |
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