「Codeforces 1164B」Triangle Area

Description

已知 $S_{\bigtriangleup ABC} = 16$，$M,N,P$ 分别在边 $AB,BC,CA$ 上，且 $\frac{AM}{MB} = \frac{BN}{NC} = \frac{CP}{PA} = \frac{1}{3}$，求 $S_{\bigtriangleup MNP}$ 。

Source

[Codeforces]1164B

Solution

提交答案 题。

先画出图：

连结 $CM$ 。

$$\begin{aligned} \because &\frac{AM}{MB} = \frac{1}{3}\\[2ex] \therefore &\frac{MB}{AB} = \frac{S_{\bigtriangleup MBC}}{S_{\bigtriangleup ABC}} = \frac{3}{4}\\[2ex] &S_{\bigtriangleup MBC} = 16 \times \frac{3}{4} = 12\\[2ex] \because &\frac{BN}{NC} = \frac{1}{3}\\[2ex] \therefore &\frac{BN}{BC} = \frac{S_{\bigtriangleup MBN}}{S_{\bigtriangleup MBC}} = \frac{1}{4}\\[2ex] &S_{\bigtriangleup MBN} = 12 \times \frac{1}{4} = 3\\[2ex] 同理可以求出，&S_{\bigtriangleup AMP} = S_{\bigtriangleup PNC} = 3\\[2ex] \therefore &S_{\bigtriangleup MNP} = S_{\bigtriangleup ABC} - S_{\bigtriangleup MBN} - S_{\bigtriangleup AMP} - S_{\bigtriangleup PNC} = 16 - 3 \times 3 = 7 \end{aligned}$$

Answer

7