2010-II-26

Ans: C
Let $AM=x\text{ cm}$. Consider $\mathrm{\Delta }ABC$, we have

$\begin{array}{rcl}AC& =& \sqrt{8^2+6^2}\\ & =& 10\text{ cm}\end{array}$

Then, $MC=(10-x)\text{ cm}$.

Consider $\mathrm{\Delta }AMD$, we have

$\begin{array}{rcl}A{D}^2& =& A{M}^2+D{M}^2\\ D{M}^2& =& 6^2–x^2\\ & =& (36-x^2){\text{ cm}}^2\dots \text{①}\end{array}$

Consider $\mathrm{\Delta }CMD$, we have

$\begin{array}{rcl}C{D}^2& =& D{M}^2+C{M}^2\\ D{M}^2& =& 8^2–(10-x{)}^2\\ & =& 64–100+20x-x^2\\ & =& (-x^2+20x–36){\text{ cm}}^2\dots \text{②}\end{array}$

By solving $\text{①}$ and $\text{②}$, we have

$\begin{array}{rcl}36-x^2& =& -x^2+20x–36\\ 20x& =& 72\\ x& =& {\displaystyle \frac{18}{5}}\end{array}$

Therefore, we have

$\begin{array}{rcl}AM:MC& =& {\displaystyle \frac{18}{5}}:10–{\displaystyle \frac{18}{5}}\\ & =& {\displaystyle \frac{18}{5}}:{\displaystyle \frac{32}{5}}\\ & =& 9:16\end{array}$