B1. Let points $A$ and $B$ and a circle $\mathcal{K}$ with diameter $A B$ be given. On the segment $A B$, choose a point $T$ different from $A$ and $B$. The perpendicular to the segment $A B$ through point $T$ intersects the circle $\mathcal{K}$ at points $M$ and $N$. Let $|A T|=x,|T B|=y$ and $|T N|=z$. Calculate the value of the expression

$$
\frac{\log _{y} z+\log _{x} z}{\log _{x} z \log _{y} z}
$$
B1.

![](https://cdn.mathpix.com/cropped/2024_06_07_9b375d7a0e2f6834f525g-17.jpg?height=574&width=606&top_left_y=621&top_left_x=271)

By Thales' theorem about the angle in a semicircle, the triangle $ABN$ is a right triangle with a right angle at $N$. By the altitude theorem in a right triangle, we have $z^2 = xy$. The given expression can be written as the sum of two fractions

$$
\frac{\log_{y} z + \log_{x} z}{\log_{x} z \log_{y} z} = \frac{1}{\log_{x} z} + \frac{1}{\log_{y} z}
$$

and then we use the change of base formula for logarithms $\log_{a} b = \frac{\log b}{\log a}$ to get

$$
\frac{1}{\log_{x} z} + \frac{1}{\log_{y} z} = \frac{1}{\frac{\log z}{\log x}} + \frac{1}{\frac{\log z}{\log y}} = \frac{\log x}{\log z} + \frac{\log y}{\log z} = \log_{z} x + \log_{z} y
$$

We then use the formula for the sum of logarithms and the relationship $z^2 = xy$ to get

$$
\log_{z} x + \log_{z} y = \log_{z}(x \cdot y) = \log_{z} z^2 = 2
$$

Thus,

$$
\frac{\log_{y} z + \log_{x} z}{\log_{x} z \log_{y} z} = 2
$$

Identifying that the triangle $ABN$ is a right triangle with a right angle at $N$ ..... 2 points
Writing or using the relationship $z^2 = xy$. ..... 4 points
Writing the expression $\frac{\log_{y} z + \log_{x} z}{\log_{x} z \log_{y} z}$ as $\frac{1}{\log_{x} z} + \frac{1}{\log_{y} z}$ ..... 3 points
Calculating $\frac{1}{\log_{x} z} + \frac{1}{\log_{y} z} = \log_{z} x + \log_{z} y$ ..... 5 points
Writing or using the relationship $\log_{z} x + \log_{z} y = \log_{z}(x \cdot y)$ ..... 3 points
Calculating $\log_{z}(x \cdot y) = \log_{z} z^2$ ..... 2 points
Calculating $\log_{z} z^2 = 2$ ..... 1 point