In the right-angled triangle $ABC$, the point $M$ moves from $A$ towards $C$ on the leg $CA$. The line passing through $M$ and perpendicular to $BM$ intersects the hypotenuse $AB$ at point $N$, and its projection on $CA$ is point $P$. How close can $P$ get to $C$? Construct the closest position of this point!
I. solution. Let us phrase the question in this reverse formulation: can the point $P$ reach any point $P^{*}$ on the segment $A C$? We prepare the answer by reversing the procedure that produces $P$. Let the perpendicular to $A C$ at $P^{*}$ intersect $A B$ at $N^{*}$. $P$ can reach $P^{*}$ if and only if there is a point $M^{*}$ on $A C$ from which the segment $N^{*} B$ is seen at a right angle; in other words: if the Thales circle $k^{*}$ with diameter $N^{*} B$ has a common point with $A C$. We will only deal with such $P^{*}$ from now on.

![](https://cdn.mathpix.com/cropped/2024_05_02_38e6059eff1a001a7ed7g-1.jpg?height=710&width=959&top_left_y=388&top_left_x=564)

We will show that if $k^{*}$ has two common points with $A C$: $M^{*}$ and $M^{* *}$ such that $A M^{*}B M_{1}$. Then, for the projections of the two diameters on $A C$, $C P_{1}<C P^{*}$, as we stated.

![](https://cdn.mathpix.com/cropped/2024_05_02_38e6059eff1a001a7ed7g-1.jpg?height=493&width=528&top_left_y=1618&top_left_x=777)

Figure 2

According to these, the best approximation of $C$ by $P^{*}$ - which cannot be improved - occurs only if $k^{*}$ has exactly one common point with $A C$, if they touch at $M^{*}$. In this case, the ray $M^{*} O^{*}$ is perpendicular to $A C$ and parallel to $B C$, and the triangle $M^{*} B O^{*}$ is isosceles (Figure 2):

$$
A B M^{*} C \varangle=B M^{*} O^{*} \varangle=M^{*} B C \varangle,
$$

for the point $M^{*}$ producing the desired $P^{*}$, $M^{*} B$ bisects the angle $A B C$. In the construction of $P^{*}$, $N^{*}$ can be omitted, $M^{*} O^{*} \| C B$ and based on $B N^{*}=2 \cdot B O^{*}$, $P^{*}$ is the reflection of $C$ over $M^{*}$.

According to the angle bisector theorem, denoting the sides of triangle $A B C$ in the usual way,

$$
C P^{*}=2 C M^{*}=\frac{2 a b}{a+c}
$$

is the minimum distance of $C P$.

II. solution. From the definition of $P$, it follows that in the desired position, $N$ also comes closest to $B$, it is enough to determine the point $M$ leading to the minimal $B N$. Let us characterize its position by the angle $A B M \varangle=x$. Then $M B C \varangle=\beta-x$, and using a known identity (Figure 3):

$$
B N=\frac{B M}{\cos x}=\frac{B C}{\cos (\beta-x) \cos x}=\frac{2 a}{\cos \beta+\cos (\beta-2 x)} \geq \frac{2 a}{\cos \beta+1}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_38e6059eff1a001a7ed7g-2.jpg?height=442&width=528&top_left_y=522&top_left_x=777)

Equality can indeed occur, specifically when $\cos (\beta-2 x)=1$, i.e., $x=\beta / 2$.

We can proceed further according to the first solution.