3. The condition that the area of triangle $ABC$ is to equal $\frac{1}{8}$ of the area $S$ of the square with side $AB$ means that the height of triangle $ABC$ to side $AB$ has a length of $\frac{1}{4}|AB|$, so point $C$ must lie on one of the two lines parallel to line $AB$ and at a distance of $\frac{1}{4}|AB|$ from line $AB$.

The condition that the sum of the areas of the squares with sides $AC$ and $BC$ is to equal the area of the square with side $AB$ means, according to the Pythagorean theorem for triangle $ABC$, that this triangle is a right triangle with hypotenuse $AB$, so point $C$ must lie on the circle with center at the midpoint of hypotenuse $AB$ and radius $\frac{1}{2}|AB|$.

The construction of point $C$ is therefore simple. Both mentioned parallel lines will intersect the circle above the diameter $AB$ at four points (Fig. 1). Since this is a positional problem, the problem has four solutions.

![](https://cdn.mathpix.com/cropped/2024_04_17_1eb44fe4e07a44a83a20g-3.jpg?height=411&width=500&top_left_y=1822&top_left_x=778)

Fig. 1

For a complete solution, 6 points are awarded, of which 2 points are for correctly determining the number of solutions. For identifying that point $C$ lies on the specified parallel lines to line $AB$, award 2 points. For proving that point $C$ lies on the Thales' circle above the diameter $AB$, award 2 points. For stating that the problem has two solutions, award 1 point; for stating that the problem has four solutions, award 2 points. If no number of solutions is given, do not award the 2 points designated for this purpose.
3. Given a segment $A B$. Construct a point $C$ such that the area of triangle $A B C$ equals $1 / 8$ of the area $S$ of a square with side $A B$, and the sum of the areas of squares with sides $A C$ and $B C$ equals $S$. How many solutions does the problem have for a given placement of segment $A B$ in the plane?

The school-based written part of the first round of category C takes place

on Tuesday, January 25, 2005

so that it starts in the morning and the contestants have 4 hours of clean time to solve the problems. Contestants can earn 6 points for each problem, and a successful solver is a student who scores 10 points or more. These details will be communicated to the students before the competition begins. Allowed aids are writing and drawing tools, school MF tables, and calculators without a graphical display.