14. $A B C D$ is a square with side length 9 . Let $P$ be a point on $A B$ such that $A P: P B=7: 2$. Using $C$ as centre and $C B$ as radius, a quarter circle is drawn inside the square. The tangent from $P$ meets the circle at $E$ and $A D$ at $Q$. The segments $C E$ and $D B$ meet at $K$, while $A K$ and $P Q$ meet at $M$. Find the length of $A M$.
$A B C D$ 是一個邊長爲 9 的正方形。設 $P$ 爲 $A B$ 上的一點, 使得 $A P: P B=7: 2 \circ$ 以 $C$ 緺圓心 、 $C B$ 爲牛徑在正方形內作一個四分一圓, 從 $P$ 點到圓的切線與圓相交於 $E$, 與 $A D$ 相交於 $Q$ 。 $C E$ 和 $D B$ 交於 $K$, 而 $A K$ 和 $P Q$ 則交於 $M$ 。求 $A M$ 的長度。
14. $\frac{85}{22}$
14. By tangent properties, we have $P E=P B=2$ and we may let $Q D=Q E=x$. Then $Q A=9-x$ and $P Q=x+2$. Applying Pythagoras' Theorem in $\triangle A P Q$, we have $(9-x)^{2}+7^{2}=(x+2)^{2}$, which gives $x=\frac{63}{11}$.
Next observe that $C E P B$ is a cyclic quadrilateral, so we have $\angle M P A=\angle K C B=\angle K A B$, which means $M P=M A$. Likewise we have $M Q=M A$ and hence $A M=\frac{1}{2} P Q=\frac{1}{2}\left(\frac{63}{11}+2\right)=\frac{85}{22}$.