How can we construct the vertex $C$ of a right-angled triangle to be erected over the given segment $AB$ as the hypotenuse, so that when a circle is drawn around $A$ with $AC$ as the radius and around $B$ with $BC$ as the radius, and the projection of $C$ onto $AB$ is denoted by $D$, the centroid of both triangles $ACD$ and $BCD$ is in the common part of the two circles?
$C$ can only be points on the Thales circle above the diameter $AB$. Let the centroid of triangle $ACD$ be $S$, and the centroid of triangle $BCD$ be $S_{1}$. The circle centered at $A$ with radius $AC$ always contains $S$, because it contains the triangle $ACD$, and the centroid is an interior point of the triangle. Similarly, $S_{1}$ is within the circle centered at $B$ with radius $BC$, so we only need to find the condition that the requirements $S_{1}A \leq CA$ and $SB \leq CB$ are simultaneously satisfied (if we include the boundary arcs in the common part of the circles).

![](https://cdn.mathpix.com/cropped/2024_05_02_6b1583ba07eae6468364g-1.jpg?height=433&width=786&top_left_y=388&top_left_x=651)

Let the midpoint of $AD$ be $F$, the projection of $S$ onto $AB$ be $S^{\prime}$, and choose the length unit to be $AB$, and denote $AD$ by $x$ (where $0 < x < 1$). Using the similarity of the right triangles $FSS^{\prime}$ and $FCD$ and the property of the centroid that it divides the median in a 2:1 ratio,

$$
\begin{gathered}
S^{\prime}D = 2FD / 3 = AD / 3 = x / 3, \text{ therefore } \\
S^{\prime}B = S^{\prime}D + DB = S^{\prime}D + 1 - x = 1 - 2x / 3
\end{gathered}
$$

Furthermore, by the mean proportional theorems in right triangles,

$$
SS^{\prime} = CD / 3 = \sqrt{AD \cdot BD} / 3 = \sqrt{x(1-x)} / 3 \quad \text{ and } \quad BC = \sqrt{AB \cdot DB} = \sqrt{1-x}
$$

With these, from the requirement $SB \leq CB$, squaring, applying the Pythagorean theorem, rearranging, and dividing by the positive $x$,

$$
\left(1 - \frac{2x}{3}\right)^2 + \frac{x(1-x)}{9} \leq 1 - x, \quad 3x^2 \leq 2x, \quad x \leq 2 / 3
$$

Similarly, with $BD = y (= 1 - x)$, from the requirement $S_{1}A \leq CA$, we get $y \leq 2 / 3$, or $x \geq 1 / 3$. According to these, both centroids are in the common part of the circles if $1 / 3 \leq x \leq 2 / 3$, that is, if $D$ is on the segment between the points $H_{1}$ and $H_{2}$ that trisect $AB$, including the endpoints of the segment, and $C$ is on the arc $I_{1}I_{2}$ above the segment $H_{1}H_{2}$. When $x = 2 / 3$, that is, $D = H_{2}$, from (1) we get $SB = CB$ (this case was seen in exercise 1033), while a simple calculation shows that $S_{1}A < CA$; when $x = 1 / 3$, that is, $D = H_{1}$, then $S_{1}A = CA$, and $SB < CB$, so only one of the centroids can be on the boundary of the common part of the two circles at the same time.

Katalin Szenes (Budapest, I. István g. II. o. t.)[^0]

[^0]:    ${ }^{1}$ on page 213 of this booklet (1966. December)