2022-M2-12

$\begin{array}{cl}
& \av{AD} \\
= & \dfrac{b\av{AB}+c\av{AC}}{b+c} \\
= & \dfrac{b}{b+c}(\av{OB}-\av{OA})+\dfrac{c}{b+c}(\av{OC}-\av{OA}) \\
= & \dfrac{-(b+c)}{b+c}\av{OA}+\dfrac{b}{b+c}\av{OB}+\dfrac{c}{b+c}\av{OC} \\
= & -\av{OA}+\dfrac{b}{b+c}\av{OB}+\dfrac{c}{b+c}\av{OC}
\end{array}$

$\begin{array}{cl}
& \av{AJ} \\
= & \av{OJ}-\av{OA} \\
= & \dfrac{a}{a+b+c}\av{OA}+\dfrac{b}{a+b+c}\av{OB}+\dfrac{c}{a+b+c}\av{OC}-\av{OA} \\
= & \dfrac{-(b+c)}{a+b+c}\av{OA}+\dfrac{b}{a+b+c}\av{OB}+\dfrac{c}{a+b+c}\av{OC} \\
= & \dfrac{b+c}{a+b+c}\left(-\av{OA}+\dfrac{b}{b+c}\av{OB}+\dfrac{c}{b+c}\av{OC} \right) \\
= & \dfrac{b+c}{a+b+c}\av{AD}
\end{array}$

Since $a$, $b$ and $c$ are positive real numbers, then we have $0 < \dfrac{b+c}{a+b+c} < 1$.

Hence, $\av{AJ}=t\av{AD}$ for $t\in(0,1)$.

Therefore, $J$ lies on $AD$.

By comparing the given condition of (a) and (a)(ii), we have $AE:EC=c:a$. Then by the same method of (a)(i), we have

$\begin{array}{rcl}
\av{BE} & = & \dfrac{a}{a+c}\av{OA}-\av{OB}+\dfrac{c}{a+c}\av{OC}
\end{array}$

Therefore, we have

$\begin{array}{cl}
& \av{BJ} \\
= & \av{OJ}-\av{OB} \\
= & \dfrac{a}{a+b+c}\av{OA}+\dfrac{b}{a+b+c}\av{OB}+\dfrac{c}{a+b+c}\av{OC}-\av{OB} \\
= & \dfrac{a}{a+b+c}\av{OA}+\dfrac{-(a+c)}{a+b+c}\av{OB}+\dfrac{c}{a+b+c}\av{OC} \\
= & \dfrac{a+c}{a+b+c}\left(\dfrac{a}{a+c}\av{OA}-\av{OB}+\dfrac{c}{a+c}\av{OC}\right) \\
= & \dfrac{a+c}{a+b+c}\av{BE}
\end{array}$

Since $a$, $b$ and $c$ are positive real numbers, then we have $0 < \dfrac{a+c}{a+b+c} < 1$.

Hence, $\av{BJ}=t\av{BE}$ for $t\in(0,1)$.

Therefore, $J$ lies on $BE$.

Hence, $AD$ and $BE$ intersect at $J$.

In (b), $I$ is the incentre of $\Delta ABC$. Hence, we have

$\begin{array}{rcl}
\av{OI} & = & \dfrac{a}{a+b+c}\av{OA}+\dfrac{b}{a+b+c}\av{OB}+\dfrac{c}{a+b+c}\av{OC} \ \ldots \unicode{x2460}
\end{array}$

$\begin{array}{cl}
& a \\
= & BC \\
= & |\av{OC}-\av{OB}| \\
= & | -3\bv{j}+\bv{k}-40\bv{i}+3\bv{j}-\bv{k}| \\
= & |-40\bv{i}| \\
= & 40
\end{array}$

$\begin{array}{cl}
& b \\
= & AC \\
= & |\av{OC}-\av{OA}| \\
= & |-3\bv{j}+\bv{k}-35\bv{i}-9\bv{j}-\bv{k}| \\
= & |-35\bv{i}-12\bv{j}| \\
= & \sqrt{(-35)^2+(-12)^2} \\
= & 37
\end{array}$

$\begin{array}{cl}
& c \\
= & AB \\
= & |\av{OB}-\av{OA}| \\
= & |40\bv{i}-3\bv{j}+\bv{k}-35\bv{i}-9\bv{j}-\bv{k}| \\
= & |5\bv{i}-12\bv{j}| \\
= & \sqrt{(5)^2+(-12)^2} \\
= & 13
\end{array}$

Hence from $\unicode{x2460}$, we have

$\begin{array}{cl}
& \av{OI} \\
= & \dfrac{40}{40+37+13}(35\bv{i}+9\bv{j}+\bv{k})+\dfrac{37}{40+37+13}(40\bv{i}-3\bv{j}+\bv{k})+\dfrac{13}{40+37+13}(-3\bv{j}+\bv{k}) \\
= & 32\bv{i}+\dfrac{7}{3}\bv{j}+\bv{k}
\end{array}$

Consider $\Delta AFI$,

$\begin{array}{cl}
\sin \angle FAI & = & \dfrac{FI}{AI} \\
FI & = & AI\sin \angle FAI \ \ldots \unicode{x2461}
\end{array}$

Also,

$\begin{array}{cl}
& \av{AI}\times \av{AB} \\
= & (\av{OI}-\av{OA})\times \av{AB} \\
= & \left(32\bv{i}+\dfrac{7}{3}\bv{j}+\bv{k}-35\bv{i}-9\bv{j}-\bv{k}\right)\times (5\bv{i}-12\bv{j}) \\
= & \left(-32\bv{i}-\dfrac{20}{3}\bv{j}\right)\times (5\bv{i}-12\bv{j}) \\
= & \begin{vmatrix} \bv{i} & \bv{j} & \bv{k} \\ -3 & \dfrac{-20}{3} & 0 \\ 5 & -12 & 0 \end{vmatrix} \\
= & \bv{k} \begin{vmatrix} -3 & \dfrac{-20}{3} \\ 5 & -12 \end{vmatrix} \\
= & \dfrac{208}{3}\bv{k}
\end{array}$

On the other hand,

$\begin{array}{rcl}
|\av{AI}\times \av{AB}| & = & (AI)(AB)\sin\angle BAI \\
\left|\dfrac{208}{3}\bv{k}\right| & = & 13(AI)\sin\angle FAI \\
AI\sin \angle FAI & = & \dfrac{16}{3} \ \ldots \unicode{x2462}
\end{array}$

By $\unicode{x2461}$ and $\unicode{x2462}$, we have $FI=\dfrac{16}{3}$.

Therefore, the radius of the inscribed circle is $\dfrac{16}{3}$.