Let $A B C$ be a right triangle with $\angle B A C=90^{\circ}$ and $I$ the point of intersection of its angle bisectors. A line through $I$ intersects the sides $A B$ and $A C$ at $P$ and $Q$, respectively. The distance from $I$ to the side $B C$ is $1 \mathrm{~cm}$.

a) Find the value of $P M \cdot N Q$.

b) Determine the minimum possible value for the area of triangle $A P Q$.

Hint: If $x$ and $y$ are two non-negative real numbers, then $x+y \geq 2 \sqrt{x y}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_12f29789e10e6749ebb9g-16.jpg?height=423&width=864&top_left_y=2036&top_left_x=488)
Solution

a) If $\angle A P Q=\alpha$, it follows that $\angle M I P=90^{\circ}-\alpha$ and

$$
\angle N I Q=180^{\circ}-\angle M I N-\angle M I P=\alpha
$$

Therefore, triangles $I M P$ and $N I Q$ are similar, and thus

$$
\frac{I M}{N Q}=\frac{P M}{I N} \Rightarrow P M \cdot N Q=I M \cdot I N=1
$$

b) Let $x=M P$ and $y=N Q$. Since $A M I N$ is a square with a side length of $1 \mathrm{~cm}$, the area of triangle $A P Q$ is given by

$$
\begin{aligned}
\frac{A P \cdot A Q}{2} & =\frac{(1+x)(1+y)}{2} \\
& =\frac{1+x y+x+y}{2}
\end{aligned}
$$

From the previous item, $x y=1$. Additionally, by the inequality presented in the hint, $x+y \geq 2 \sqrt{1}=2$. Thus, the minimum area is $\frac{1+1+2}{2}=2$ and can be achieved by setting $A P=A Q=2$. Note: The inequality presented as a hint is a special case of the arithmetic mean-geometric mean inequality $(M A \geq M G)$. To verify this case, from $(\sqrt{x}-\sqrt{y})^{2} \geq 0$, it follows that

$$
x-2 \sqrt{x y}+y \geq 0 \Rightarrow x+y \geq 2 \sqrt{x y}
$$