Let's construct a right-angled triangle where the longer leg is equal to the geometric mean of the other two sides. Furthermore, we draw the altitude from the right angle vertex in the original triangle, and in the two resulting smaller triangles, and in the four resulting smaller triangles. Find all pairs of segments in the figure whose geometric mean is equal to one of the segments in the figure.
I. Let the hypotenuse $AB$ of the triangle $ABC$ be of unit length, and the smaller acute angle $\angle BAC = \alpha$. Thus, the legs are $AC = \cos \alpha$ and $BC = \sin \alpha$, which we will abbreviate as $c$ and $s$ respectively. According to the requirement, $c^2 = 1 \cdot s$, and by the Pythagorean theorem,

$$
s^2 + s - 1 = 0, \quad s = \frac{\sqrt{5} - 1}{2} \quad (\approx 0.6180)
$$

Accordingly, the triangle can be obtained by extending the known (so-called Ptolemy-Dürer) construction of the side of a regular decagon (and pentagon) inscribed in a unit circle by one step. The center of the circle $k$ is $B$, the side of the decagon is $BC$, and the perpendicular raised at $C$ intersects $k$ at $A$ (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_4213117147a3cc86acc3g-1.jpg?height=384&width=556&top_left_y=529&top_left_x=768)

Figure 1

II. Let the foot of the altitude from the right angle vertex in the successive triangles be denoted as follows (Figure 2):

![](https://cdn.mathpix.com/cropped/2024_05_02_4213117147a3cc86acc3g-1.jpg?height=637&width=699&top_left_y=1134&top_left_x=694)

triangle: $ABC, \quad ACD, \quad BCD, \quad ADE, \quad CDE, \quad BDF, \quad CDF$, foot: $\quad D, \quad E, \quad F, \quad G, \quad H, \quad J, \quad K$.

The resulting triangles are similar, and each leg becomes the hypotenuse in a further subdivided triangle, so most segments can be expressed in terms of $s$ and $c$, or according to the requirement, solely in terms of $c$:

$$
\begin{aligned}
& DC = AC \sin \alpha = BC \cos \alpha = c s = c^3, \\
& DA = AC \cos \alpha = c^2; \quad DB = BC \sin \alpha = s^2 = c^4 \\
& ED = CF = CD \cos \alpha = c^4, \quad FD = CE = CD \sin \alpha = c^3 s = c^5 \\
& EA = c^3, \quad FB = c^6 \\
& GE = DH = c^5, \quad GD = EH = c^6, \quad GA = c^4, \quad HC = c^7 \\
& JF = DK = c^7, \quad JD = FK = c^6, \quad JB = c^8, \quad KC = c^5;
\end{aligned}
$$

Finally, the segments that are not sides of any of the above triangles:

$$
\begin{gathered}
AJ = AB - JB = 1 - c^8 = 1 - s^4 = (1 - s^2)(1 + s^2) = c^2 \cdot \sqrt{5} s = \sqrt{5} c^4 \\
BG = AB - AG = 1 - c^4 = 1 - s^2 = c^2 \\
GJ = GD + DJ = 2 c^6 \\
HK = CK - CH = c^5 (1 - c^2) = c^5 s^2 = c^9
\end{gathered}
$$

Thus, out of the 28 segments drawn between the 10 points, 26 have a length equal to a non-negative integer power of $c$. We can say that each of these segments has an exponent of $c$, and conversely, each exponent from $0, 1, \ldots, 9$ is associated with at least one segment of the diagram. Grouping by exponents:

$$
\begin{aligned}
c^0 & = AB & c^5 = FD, CE, EG, DH, KC \\
c & = AC & c^6 = BF, DG, EH, JD, FK \\
c^2 & = BC, AD, BG & c^7 = CH, JF, DK \\
c^3 & = CD, AE & c^8 = JB \\
c^4 & = BD, ED, CF, AG & c^9 = HK
\end{aligned}
$$

The remaining 2 segments are multiples of a power of $c$:

$$
AJ = \sqrt{5} c^4, \quad GJ = 2 c^6
$$

Thus, a geometric mean can be found for every pair of segments whose exponents are of equal parity (and of course different), for example, $CD$ and $AE$ are the geometric means of $AB$ and $BF$. Conversely, except for the first and last terms of the found geometric sequence ($AB$ and $HK$), a pair of different segments can be found for every segment that is the geometric mean of those segments, for example, for $JB$, one can choose $HK$, $CH$, $JF$, or $DK$.

The segments (1) cannot appear in such a triplet, because the constant factors $\sqrt{5}$ and 2 are not integer powers of $c$, nor is their product, and these exponents are not of the form $n + 1/2$, where $n$ is an integer: $\sqrt{5} = c^{-3.345}$, $2 = c^{-2.881}$, $2\sqrt{5} = c^{-6.225}$.

Finally, we determine the number of such segment pairs. A segment of length $c^5$ is the geometric mean of every possible $(c^4, c^6)$, $(c^3, c^7)$, $(c^2, c^8)$, and $(c, c^9)$ pair, and there are $4 \cdot 5 + 2 \cdot 3 + 3 \cdot 1 + 1 \cdot 1 = 30$ ways to choose these pairs. (If we also consider which of the 5 segments of length $c^5$ is chosen as the geometric mean, then the number of such triplets is $5 \cdot 30 = 150$.) Similarly,

$$
c, \quad c^2, \quad c^3, \quad c^4, \quad c^6, \quad c^7, \quad c^8
$$

are the geometric means of

$$
3, \quad 6, \quad 22, \quad 29, \quad 21, \quad 10, \quad 3
$$

(different length) segment pairs, so the total number of pairs is 124. (We have ignored the relationships between segments of equal length.)

Szendrei Ágnes (Szeged, Ságvári E. Gyak. Gimn., III. o. t.)