Let's calculate the hypotenuse of a right-angled triangle, if the sum of the legs is $p$, and the sum of the two segments connecting the right-angle vertex to the points that trisect the hypotenuse is $q$.

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Let $C_1$ and $C_2$ be the points that trisect the hypotenuse $AB$ of the right triangle $ABC$, such that $AC_1 = C_1C_2 = C_2B$. The projections of these points on the leg $AC$ are $B_1$ and $B_2$, and $BC = a$, $CA = b$, $AB = c$. It is evident that $AB_1 = B_1B_2 = B_2C = b/3$, $C_1B_1 = a/3$, and $C_2B_2 = 2a/3$. Thus, from the right triangles $CC_1B_1$ and $CC_2B_2$,

\[
CC_1 + CC_2 = \frac{1}{3} \sqrt{a^2 + 4b^2} + \frac{1}{3} \sqrt{4a^2 + b^2} = q
\]

![](https://cdn.mathpix.com/cropped/2024_05_02_ff70643169d5e75ff0b9g-1.jpg?height=439&width=631&top_left_y=496&top_left_x=736)

Removing the square roots by squaring,

\[
5a^2 + 5b^2 + 2 \sqrt{(a^2 + 4b^2)(4a^2 + b^2)} = 9q^2
\]

Rearranging and squaring again,

\[
\begin{gathered}
16(a^4 + b^4) + 68a^2b^2 = 81q^4 - 90q^2(a^2 + b^2) + 25(a^4 + b^4) + 50a^2b^2 \\
a^4 + b^4 - 2a^2b^2 - 10q^2(a^2 + b^2) + 9q^4 = 0
\end{gathered}
\]

The first three terms are the square of $a^2 - b^2$, which can be rewritten using

\[
a + b = p
\]

as follows:

\[
\begin{gathered}
a^4 + b^4 - 2a^2b^2 = (a^2 - b^2)^2 = (a + b)^2(a - b)^2 = \\
= p^2(a^2 + b^2 - 2ab) = p^2[(a + b)^2 - 4ab] = p^4 - 4p^2ab
\end{gathered}
\]

Substituting this into (2) and replacing $a^2 + b^2$ with $p^2 - 2ab$ from (3), we get a linear equation in $ab$:

\[
p^4 - 4p^2ab + 20q^2ab - 10p^2q^2 + 9q^4 = 0
\]

Thus,

\[
ab = \frac{p^4 - 10p^2q^2 + 9q^4}{4(p^2 - 5q^2)} = \frac{(p^2 - q^2)(p^2 - 9q^2)}{4(p^2 - 5q^2)}
\]

and the hypotenuse $c$ from (3) is

\[
\begin{gathered}
c^2 = a^2 + b^2 = (a + b)^2 - 2ab = p^2 - \frac{(p^2 - q^2)(p^2 - 9q^2)}{2(p^2 - 5q^2)} = \\
= \frac{2p^4 - 10q^2p^2 - p^4 + 10p^2q^2 - 9q^4}{2(p^2 - 5q^2)} \\
c^2 = \frac{p^4 - 9q^4}{2(p^2 - 5q^2)}
\end{gathered}
\]

If this value is positive, its positive square root can be the length of the hypotenuse. However, since we arrived at (2) through two squarings, it is possible that this is satisfied by values of $a$ and $b$ that do not satisfy (1). We must therefore check for which values of $p$ and $q$ there exist positive real $a$ and $b$ that satisfy (3) and (4), and whether (1) holds for them.

From (3) and (4), $a$ and $b$ are the roots of the equation

\[
t^2 - pt + \frac{(p^2 - q^2)(p^2 - 9q^2)}{4(p^2 - 5q^2)} = 0
\]

This has (real) roots if and only if its discriminant is non-negative:

\[
p^2 - \frac{(p^2 - q^2)(p^2 - 9q^2)}{(p^2 - 5q^2)} = \frac{q^2(5p^2 - 9q^2)}{p^2 - 5q^2} \geq 0
\]

The following can be determined about the ratio of $p$ and $q$: From the right triangles $CC_1B_1$ and $CC_2B_2$,

\[
CC_1C B_1 = \frac{2b}{3}, \quad CC_1 > C_1B_1 = \frac{a}{3}, \quad CC_2 > \frac{b}{3}, \quad CC_2 > \frac{2a}{3}
\]

Adding these,

\[
2q > a + b = p
\]

If these hold, the denominator in (7) is negative according to (9), so the numerator cannot be positive:

\[
5p^2 - 9q^2 \leq 0, \quad p \leq q \sqrt{9/5} = q \sqrt{1.8}
\]

This provides a smaller upper bound for the ratio $p/q$ than (9).

Under these conditions, the values obtained for $a$ and $b$ are positive if their product, the right-hand side of (4), is positive. This is true because the first factor in the numerator is positive according to (8), and the second factor and the denominator are negative according to (10). In this case, the value obtained for $c^2$, as the sum of the squares of two positive values, is positive (which can also be seen directly from the expression obtained). Therefore, we get a solution if

\[
q < p < q \sqrt{1.8}
\]

(We see that it would not have been sufficient to require the positivity of the expression for $c^2$. From this, $p < q \sqrt{3}$ would follow according to (8), and it would be possible for $c^2$ to be positive while $a$ and $b$ are not real.)

Tibor Pernek (Budapest, Apáczai Csere J. Gyak. G. IV. o. t.)

Remark. We can also arrive at the result (5) using coordinate geometry. Let the $X$-axis be the line $AB$, and the origin be the midpoint of the segment $AB$, so the abscissas of $A, C_1, C_2, B$ are $-c/2, -c/6, c/6, c/2$, respectively. We get an equation for the desired $c$ by noting that the right angle vertex $C$ lies on three lines: the circle of radius $c/2$ centered at the origin (Thales' circle), the ellipse with foci $C_1$ and $C_2$ and major axis $q$, and the ellipse with foci $A$ and $B$ and major axis $p$. Thus, the square of half the minor axis of the ellipses is

\[
\left(\frac{q}{2}\right)^2 - \left(\frac{c}{6}\right)^2, \quad \text{and} \quad \left(\frac{p}{2}\right)^2 - \left(\frac{c}{2}\right)^2
\]

The equations of the ellipses are

\[
\frac{9q^2 - c^2}{36} x^2 + \frac{q^2}{4} y^2 = \frac{q^2(9q^2 - c^2)}{144}, \quad \text{and} \quad \frac{p^2 - c^2}{4} x^2 + \frac{p^2}{4} y^2 = \frac{p^2(p^2 - c^2)}{16}
\]

Solving for $x^2$ and $y^2$ and substituting into the equation of the Thales' circle

\[
x^2 + y^2 = \frac{c^2}{4}
\]

we get a (deficient) quadratic equation for $c^2$, whose non-zero root is identical to (5).