## Task B-1.5.

In a right-angled triangle $A B C$, with a right angle at vertex $C$, the length of the hypotenuse is 12. Squares $A B D E$ and $A C G F$ are constructed outward on sides $\overline{A B}$ and $\overline{A C}$. If points $D, E$, $F$ and $G$ lie on the same circle, calculate the perimeter of triangle $A B C$.
## Solution.

Let $a=\overline{B C}, b=\overline{A C}$, and $c=\overline{A B}$.

The segments $\overline{D E}$ and $\overline{F G}$ are chords of the circle, so their perpendicular bisectors intersect at the center of the given circle. The perpendicular bisector of $\overline{D E}$ intersects the hypotenuse $\overline{A B}$ at point $S$. The perpendicular bisector of $\overline{F G}$ is parallel to the side $\overline{B C}$ of triangle $A B C$ and intersects the leg $\overline{A C}$ at its midpoint $R$.

Then, by Thales' theorem on proportional segments (or the similarity of triangles $A S R$ and $A B C$), the hypotenuse is intersected at its midpoint, i.e., at point $S$.

(This conclusion can also be derived from the fact that the perpendicular bisectors of the sides $\overline{D E}$ and $\overline{F G}$ are also the perpendicular bisectors of the sides $\overline{A B}$ and $\overline{A C}$, and the perpendicular bisectors of the sides in a right triangle intersect at the midpoint of the hypotenuse.)

We conclude that the center of the circle is at the midpoint of the hypotenuse and that $|R S|=\frac{a}{2}$. Then $|S E|=|S G|=r$. Applying the Pythagorean theorem to triangle $S A E$ gives us

$$
\begin{aligned}
& |S A|^{2}+|A E|^{2}=|S E|^{2} \\
& \left(\frac{c}{2}\right)^{2}+c^{2}=r^{2} \\
& r^{2}=36+144=180
\end{aligned}
$$

In triangle $G Q S$, we have $|S Q|^{2}+|Q G|^{2}=|G S|^{2}$, or

$$
\begin{aligned}
& \left(\frac{a}{2}+b\right)^{2}+\left(\frac{b}{2}\right)^{2}=r^{2}=180 \\
& \frac{a^{2}}{4}+a b+b^{2}+\frac{b^{2}}{4}=180 \\
& a b+b^{2}=144
\end{aligned}
$$

Since $a^{2}+b^{2}=144$ (Pythagorean theorem in triangle $A B C$), it follows that $a b-a^{2}=0$, or $a(b-a)=0$.

This is only possible if $a=b$. Then $2 a^{2}=144$ and $a=6 \sqrt{2}$.

The perimeter of triangle $A B C$ is $o=a+b+c=2 a+c=12 \sqrt{2}+12$.

## NATIONAL MATHEMATICS COMPETITION

2nd grade - high school - B variant

Poreč, March 29, 2019.