5. Given is $\triangle A B C, \measuredangle C=90^{\circ}$. Squares $A E D C$ and $C F G B$ are constructed externally on its sides. Segment $E B$ intersects segments $A C$ and $A G$ at points $H$ and $I$, respectively. Segment $A G$ intersects segment $B C$ at point $J$. If the area of $\triangle A I B$ is 2017, find the area of quadrilateral $H C J I$.

## SOLUTIONS TO THE MUNICIPAL COMPETITION IN MATHEMATICS FOR HIGH SCHOOL STUDENTS, 21. 1. 2017.

## First Grade - A category
5. Let $B C=a, A C=b, C J=x$ and $H C=y$. It is obvious that $\triangle A F G \sim \triangle A C J$, so we get $\frac{a+b}{a}=\frac{b}{x}$, and from this we calculate $x=\frac{a b}{a+b}$. Similarly, $\triangle D B E \sim \triangle C B H$, so we get $\frac{a+b}{b}=\frac{a}{y}$, and from this we calculate $y=\frac{a b}{a+b}$. Now observe:

$$
\begin{aligned}
\frac{a b}{2} & =P(\triangle A C B) \\
& =P(\triangle A I B)+P(\triangle A C J)+P(\triangle B C H)-P(H C J I) \\
& =P(\triangle A I B)+\frac{b \frac{a b}{a+b}}{2}+\frac{\frac{a b}{a+b} a}{2}-P(H C J I) \\
& =P(\triangle A I B)+(b+a) \frac{\frac{a b}{a+b}}{2}-P(H C J I) \\
& =P(\triangle A I B)+\frac{a b}{2}-P(H C J I)
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_06_04_f254ba4f25ce7f8d9df1g-64.jpg?height=404&width=406&top_left_y=728&top_left_x=1133)

Dr 2017 4B 5

and from here it follows that $P(H C J I)=P(\triangle A I B)=2017$.

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