Among right-angled triangles with the same hypotenuse, which one has the maximum perimeter?
I. solution: Our question can also be phrased as: in which of the triangles in question is the sum of the two legs maximal? To produce this sum, we measure on the extension of the leg $AC$ of a right-angled triangle $ABC$ standing on the constant hypotenuse $AB$, beyond $C$, a segment $CB'$ equal to $CB$ and examine the variation of $AB' = AC + CB$ (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_b9d07dba772010915cc1g-1.jpg?height=656&width=677&top_left_y=339&top_left_x=713)

Due to the isosceles right-angled triangle $CBB'$, $\angle AB'B = 45^\circ$, so $B'$ lies on the $45^\circ$ angle subtended by $AB$. The angle subtended by $AB$ from the center $O$ of this arc is $2 \cdot 45^\circ = 90^\circ$, so $O$ is the intersection of the perpendicular bisector of $AB$ with the Thales semicircle of diameter $AB$. $AB'$ is a chord of a part of the angle subtended arc, and it is maximal when $B'$ coincides with the point $B^*$ opposite to $A$ on the arc, i.e., $AB^*$ passes through $O$. Then the corresponding $C^*$ is the foot of the perpendicular from $B$ to $AB^*$, which is $O$, and thus the sought triangle with the maximal perimeter is the isosceles right-angled triangle $ABO$.

Géza Beke (Hatvan, Bajza J. g. III. o. t.)

Remark: Similarly, it can be shown that among the triangles with a given base and an arbitrary given angle opposite to it, the one with the maximal perimeter is isosceles.

II. solution: According to a known relationship, in a right-angled triangle, with the usual notation: $a + b = c + 2\rho$, where $\rho$ is the radius of the inscribed circle. Thus, $a + b + c = 2c + 2\rho$ is maximal when this radius is maximal.

Furthermore, we know (from the 1st grade textbook) that the center $O$ of the inscribed circle lies on the arc from which the segment $AB$ subtends an angle of $135^\circ$ (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_b9d07dba772010915cc1g-1.jpg?height=439&width=656&top_left_y=1640&top_left_x=713)

Clearly, $\rho$ is maximal when $O$ is at the midpoint of this arc, on the perpendicular bisector of $AB$; in this case, due to symmetry, the triangle is isosceles.

Sándor Gaál (Veszprém, Lovassy L. g. III. o. t.)

III. solution: $a + b + c$ is maximal when $a + b$ is; this is when $(a + b)^2$ is; this, according to the transformation

$$
(a + b)^2 = a^2 + b^2 + 2ab = c^2 + 2ab = c^2 + 4t = c^2 + 2cm_c
$$

is when $ab$, or $t$, $cm_c$, or $m_c$ is maximal. Already now, the maximum value of $m_c$ in the Thales semicircle is $\frac{c}{2}$, and with this, the triangle is isosceles, and the perimeter is $c(1 + \sqrt{2})$.

Csaba Máthé (Győr, Révay M. g. II. o. t.)

IV. solution: Related to the third solution, the following inequality holds for $ab$:

$$
ab = \sqrt{a^2 b^2} \leq \frac{a^2 + b^2}{2} = \frac{c^2}{2}
$$

$ab$ also takes its maximum value, and from $ab = \frac{c^2}{2}$, we get $a = b$.

Balázs Nagy (Eger, Dobó I. g. IV. o. t.)

V. solution: The function $y = a + b + c = c(1 + \sin \alpha + \cos \alpha)$ has its maximum where the function $z = \sin \alpha + \cos \alpha$ does, or - since here $0^\circ < \alpha < 90^\circ$ and both terms are positive - where $v = (\sin \alpha + \cos \alpha)^2 = 1 + \sin 2\alpha$ does. This follows when $2\alpha = 90^\circ$, then $\alpha = 45^\circ = \beta$.

Lajos Ducza (Székesfehérvár, József A. g. IV. o. t.)

VI. solution: The maximum of the function $z$ can also be obtained as follows:

$$
\sin \alpha + \cos \alpha = \sin \alpha + \sin (90^\circ - \alpha) = 2 \sin \frac{\alpha + 90^\circ - \alpha}{2} \cos \frac{\alpha - (90^\circ - \alpha)}{2} = \sqrt{2} \cos (\alpha - 45^\circ)
$$

is maximal when $\cos (\alpha - 45^\circ) = 1$, $\alpha - 45^\circ = 0^\circ$, $\alpha = 45^\circ = \beta$.

László Tusnády (Sátoraljaújhely, Kossuth L. g. IV. o. t.)