2. Find the greatest value of the parameter a for which the equation: $(|x-2|+2 a)^{2}-3(|x-2|+2 a)+4 a(3-4 a)=0$ has three solutions. In the answer, specify the greatest of them.
Solution:

Let $|\mathrm{x}-2|+2 \mathrm{a}=\mathrm{t}$, then

$$
t^{2}-3 t+4 a(3-4 a)=0
$$

$t^{2}-2 \cdot \frac{3}{2} t+\frac{4}{9}-\frac{4}{9}+12 a-16 a^{2}=0$

$\left(\mathrm{t}-\frac{3}{2}\right)^{2}-\left(\frac{9}{4}-12 \mathrm{a}+16 a^{2}\right)=0$;

$\left(t-\frac{3}{2}\right)^{2}-\left(4 a-\frac{3}{2}\right)^{2}=0 ;\left(t-\frac{3}{2}+4 a-\frac{3}{2}\right) \cdot\left(t-\frac{3}{2}-4 a+\frac{3}{2}\right)=0 ;$

$(t+4 a-3) \cdot(t-4 a)=0$

We get $(|x-2|+2 a+4 a-3) \cdot(|x-2|+2 a-4 a)=0$

$\left[\begin{array}{l}|\mathrm{x}-2|+6 \mathrm{a}-3=0 \\ |\mathrm{x}-2|-2 \mathrm{a}=0 ;\end{array} \quad\left[\begin{array}{l}a=-\frac{1}{6} \cdot|x-2|+\frac{1}{2} \\ a=\frac{1}{2} \cdot|x-2|\end{array}\right.\right.$

Online (correspondence) stage of the "Step into the Future" School Students Olympiad in the subject of Mathematics

Let's construct the graphs of the equations

$a=-\frac{1}{6}|x-2|+\frac{1}{2}$ and

$\mathrm{a}=\frac{1}{2}|\mathrm{x}-2|$ in the XOA system

The equation has three solutions when

$\mathrm{a}=0, \mathrm{a}=0.5$

Answer: 0.5

![](https://cdn.mathpix.com/cropped/2024_05_06_79c0420da3c6e388f719g-2.jpg?height=1071&width=1539&top_left_y=938&top_left_x=287)

## №3: Triangles.

A circle is inscribed in a right triangle $\mathrm{ABC}$, touching its sides at points P, Q, and R. Find the area of triangle $\mathrm{PQR}$, if the lengths of the legs of triangle $\mathrm{ABC}$ are 3 and 4.

Answer: 1.2

## №4: Properties of numbers.