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the-1111-geometry-problem.md

---
title: 一个几何问题与解法
createTime: 2025/11/14
categories:
    - study
tags:
    - maths
---

:::problem
已知 $\odot I, I_1$ 与平行线 $l, l_1$ 均相切，$l$ 上有两点 $B, C$，平面上有两点 $A, A_1$ 满足 $AB = AC$，且 $\odot I, I_1$ 分别是 $\triangle ABC, A_1BC$ 的内切圆。另有异于 $\odot I$ 的 $\odot O_1, O_2$ 与 $l, l_1$ 均相切且分别与 $AB, AC$ 相切。设 $\odot I_1$ 与 $l$ 切于 $D$，求证：

$$
\frac{\sin \angle A_1AB}{\sin \angle A_1AC} \cdot \frac{DB}{DC} = \frac{I_1O_1}{I_1O_2}
$$
:::

:::center
![图](the-1111-geometry-problem/1111problem.svg){width=800px}
:::

设 $\odot I, l$ 切于 $M$，直线 $AA_1$ 交 $l$ 与 $E$，过 $A$ 作 $l$ 的平行线分别交 $A_1B, A_1C$ 于 $F, G$。  
设 $\angle IBC = \angle ICB = \theta, \angle I_1BC = \alpha, \angle I_1CB = \beta$，  
$\cot\theta = u, \cot \alpha = v, \cot \beta = w$，则有

$$
1 = \frac{MD}{MD} = \left| \frac{u - v}{u - w} \right|
$$

于是

$$
\begin{align*}
\frac{\sin \angle A_1AB}{\sin \angle A_1AC} &= \frac{A_1B \sin(\angle AA_1B) / \cancel{AB}}{A_1C \sin(\angle AA_1C) / \cancel{AC}} \\
    &= \frac{EB \ \cancel{\sin \angle A_1EB}}{EC \ \cancel{\sin \angle A_1EC}} \\
    &= \frac{AF}{AG} \\
    &= \left| \dfrac{\cot 2\theta - \cot 2\alpha}{\cot 2\theta - \cot 2\beta} \right| \\
    &= \left|
        \frac{\frac{u^2 - 1}{2u} - \frac{v^2 - 1}{2v}}
        {\frac{u^2 - 1}{2u} - \frac{w^2 - 1}{2w}}
    \right| \\
    &= \left|
        \frac{\cancel{(u-v)}\ (uvw + w)}{\cancel{(u-w)}\ (uvw + v)}
    \right| \\
\end{align*}
$$

与此同时，另外一边：

$$
\begin{align*}
\frac{DC}{DB} \cdot \frac{I_1O_1}{I_2O_2} &= \left| \frac{\cot \beta}{\cot \alpha} \cdot \frac{\cot \alpha + \tan \theta}{\cot \beta + \tan \theta} \right| \\
&= \Biggl| \frac{w}{v} \cdot \frac{v + \frac{1}{u}}{w + \frac{1}{u}} \Biggr|
= \left| \frac{uvw + w}{uvw + v} \right|
\end{align*}
$$

:::right
$\square$
:::