# 06-05 FRI

心血来潮，写写Latex。

![image.png](https://i.loli.net/2020/06/05/i2qd3JvwWk56tAP.png)

**16 题.**

![draft.png](https://i.loli.net/2020/06/06/p7ZsaYPMdiOu1WB.png)

发现自己真的是东亚做题家，兴趣爱好就只有打游戏和刷题，社交一塌糊涂。成就感仅仅来自于做题和打游戏吧，嗯。

## **附**

> Github的markdown解析是通过SunDown库实现的。这个库的宗旨就是*"Standards compliant, fast, secure* markdown processing library in C"。

\
由于GitHub的Markdown解释器不支持LATEX，遂上传图片，源码如下：

```
设线段  $ AB $  长度为  $ a $   ，易得：

$ AC = \sqrt{2} $  ，  $ BM = \frac{\sqrt{5}}{3}a $ ，   $ DM = BD - BM = 5 - \frac{\sqrt{5}}{3}a $ ，  $ \sin{\angle{ABD}} = \frac{\sqrt{5}}{5} $ 



观察并计算得：

$\sin{\angle{AMD}} = \sin{\angle{BMC}} = \sin{(\angle{BAC} + \angle{ABD})} = \sin{\angle{ABD}}\sin{\angle{BAC}} + \cos{\angle{ABD}}\cos{\angle{BAC}}$ 

​                    $ = \frac{\sqrt{2}}{2}(\frac{\sqrt{5}}{5} + \frac{2\sqrt{5}}{5})$

​                    $ = \frac{3\sqrt{10}}{10}$ 



在$ \bigtriangleup{ACD}$ 作 $AC$ 边的高，长度为 $h$ ，则：

 $ h = DM\sin{\angle{AMD}} = \frac{3\sqrt{10}}{10}(5 - \frac{\sqrt{5}}{3}a)$ 



由此可得：

$S_{ \bigtriangleup{ACD}} = \frac{AC \cdot h}{2} = \frac{1}{2}\cdot\sqrt{2}a\cdot\frac{3\sqrt{10}}{10}(5 - \frac{\sqrt{5}}{3}a)$ 

​                            $= \frac{9}{10} \times \frac{\sqrt{5}}{3}a(5 -\frac{\sqrt{5}}{3}a)$ 



由算术-几何平均值不等式可得：

$S_{ \bigtriangleup{ACD}} \leq \frac{9}{10} \times \frac{(\frac{\sqrt{5}}{3}a + 5 - \frac{\sqrt{5}}{3}a)^{2}}{4} = \frac{45}{8}$    

 (当且仅当  $\frac{\sqrt{5}}{3}a = 5 - \frac{\sqrt{5}}{3}a$  ，即  $ a = \frac{3\sqrt{5}}{2}$  时取相等)



综上，当  $AB = \frac{3\sqrt{5}}{2}$  时， $S_{ \bigtriangleup{ACD}}$  面积最大为 $\frac{45}{8}$
```


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