Let the height dropped from one vertex of a triangle be $m$, and the segment of this height from the vertex to the foot of the perpendicular be $m_{1}$. Calculate the value of the product $m m_{1}$ from the sides of the triangle.
I. solution: In his article "On the Pedal Triangle" (see the March 1956 issue), Jenő Berkes proves that \( m_{1}=2 r \cos \alpha \), where \( r \) is the radius of the circumcircle of the triangle. If we consider that

\[
r=\frac{a b c}{4 t}
\]

where \( t \) is the area of the triangle, and by the cosine rule,

\[
\cos \alpha=\frac{b^{2}+c^{2}-a^{2}}{2 b c}
\]

and \( m=\frac{2 t}{a} \) (where \( a \) is the side corresponding to \( m \)), then

\[
m m_{1}=\frac{2 t}{a} \cdot \frac{a b c}{2 t} \cdot \frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{b^{2}+c^{2}-a^{2}}{2}
\]

Remark: If \( \alpha \) is an obtuse angle, then either \( m_{1} \) is taken as negative, or the absolute value of the right-hand side is considered.

Ferenc Kolonits (Bp., VIII., Piarist g. I. o. t.)

II. solution: The first figure shows an acute-angled triangle at \( A \), and the second figure shows an obtuse-angled triangle at \( A \), both with the labeling.

![](https://cdn.mathpix.com/cropped/2024_05_02_b2b16de483092d912ab1g-1.jpg?height=737&width=886&top_left_y=1011&top_left_x=598)

1. figure

In both figures,

\[
A B_{1} M \triangle \sim A A_{1} C \triangle
\]

because they are right-angled triangles in which the acute angle at \( A \) is common or vertically opposite.

Thus,

\[
m_{1}: b_{1}=b: m
\]

from which

\[
m m_{1}=b b_{1}
\]

But \( b_{1}=c \cos \alpha \), and thus (considering both the acute and obtuse cases for \( \alpha \))

\[
m m_{1}=b c \cos \alpha=\frac{\left|b^{2}+c^{2}-a^{2}\right|}{2}
\]

If \( \alpha \) is a right angle, then \( M=A \), or \( m_{1}=0 \), and thus \( m m_{1}=0 \), as the above formula also shows \( \left(b^{2}+c^{2}-a^{2}=0\right) \).

Péter Vámos (Bp., II., Than Károly chem. t. II. o. t.)

III. solution: Consider the circumcircles of the cyclic quadrilaterals \( M B_{1} C A_{1} \) (in the second figure \( M B_{1} A_{1} C \)), \( B C_{1} B_{1} C \) (in the second figure \( B B_{1} C_{1} C \)), etc., in both figures.

![](https://cdn.mathpix.com/cropped/2024_05_02_b2b16de483092d912ab1g-2.jpg?height=794&width=903&top_left_y=145&top_left_x=600)

For the secants starting from vertex \( A \),

\[
m \cdot m_{1}=b \cdot b_{1}=c \cdot c_{1}
\]

Similarly, in the first figure, for the secant starting from \( C \),

\[
b\left(b-b_{1}\right)=a\left(a-a_{1}\right)
\]

and for the secant starting from \( B \),

\[
c\left(c-c_{1}\right)=a \cdot a_{1}
\]

The sum of (2) and (3) is

\[
b^{2}+c^{2}-\left(b b_{1}+c c_{1}\right)=a^{2}
\]

or, considering (1),

\[
b^{2}+c^{2}-2 m \cdot m_{1}=a^{2},
\]

from which

\[
m m_{1}=\frac{b^{2}+c^{2}-a^{2}}{2}
\]

In the second figure, similarly,

\[
\begin{gathered}
b\left(b+b_{1}\right)=\left(a-a_{1}\right) a \\
c\left(c+c_{1}\right)=a_{1} a
\end{gathered}
\]

and thus

\[
b^{2}+c^{2}+\left(b b_{1}+c c_{1}\right)=a^{2}
\]

from which, considering (1),

\[
m m_{1}=\frac{a^{2}-b^{2}-c^{2}}{2}
\]