We inscribe triangles in a quarter ellipse, one vertex of which is the endpoint of the minor axis, the other is the focus, and the third is a point on the ellipse. Which triangle will have the maximum area, if the first two points are fixed, and the third one runs along the quarter ellipse?
I. Solution: Let the semi-major axis of the centrally positioned ellipse be denoted by $a$, the semi-minor axis by $b$, the distance of the focus from the center by $c$, and the coordinates of a point on the quarter-ellipse by $x, y$. The parametric equation system of the ellipse is

$$
x=a \cos t, \quad y=b \sin t
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_2b87032ec7a082bac19ag-1.jpg?height=444&width=577&top_left_y=350&top_left_x=752)

Figure 1

From Figure 1, we can see that the area of the triangle in question is obtained by subtracting the area of a right triangle, whose one leg is the semi-minor axis and whose opposite vertex is $F$, from the area of a trapezoid with parallel sides $b$ and $y$ and height $x$. Additionally, we subtract or add (depending on whether $x>c$ or $x<c$) the area of a right triangle, whose one vertex is also $F$ and whose opposite side is the $y$ ordinate. In both cases, the formula

$$
T=\frac{1}{2}[(b+y) x-b c-(x-c) y]=\frac{1}{2}(b x+c y-b c)=\frac{b}{2}(a \cos t+c \sin t-c)
$$

gives the area.

The area is a function of the parameter $t$ and is maximal when the expression $a \cos t+c \sin t$ is maximal. According to the solution of problem 827, part III (Középiskolai Mat. Lapok XVI. (1958.) 1. issue, p. 20), this occurs when $a \sin t=c \cos t$, i.e., (since $a \sin t$ and $c \cos t$ are non-negative for a quarter-ellipse) when

$$
a^{2} \sin ^{2} t=c^{2} \cos ^{2} t=c^{2}-c^{2} \sin ^{2} t
$$

From the obtained equality,

$$
\sin ^{2} t=\frac{c^{2}}{a^{2}+c^{2}}, \quad \text { or } \quad \sin t=\frac{c}{\sqrt{a^{2}+c^{2}}}
$$

and thus

$$
\cos t=\frac{a}{\sqrt{a^{2}+c^{2}}}
$$

(In both cases, it is sufficient to take the positive value of the square root.)

The coordinates of the point on the ellipse, based on the parametric equation system, are

$$
x=\frac{a^{2}}{\sqrt{a^{2}+c^{2}}}, \quad y=\frac{b c}{\sqrt{a^{2}+c^{2}}}
$$

With this, we have determined the missing vertex of the triangle with the maximum area.

From the expression for $T$, the maximum area is:

$$
T_{\max }=\frac{b}{2}\left(\frac{a^{2}+c^{2}}{\sqrt{a^{2}+c^{2}}}-c\right)
$$

This completes our solution.

Remark: If the angle $t$ corresponding to the maximum area is constructed based on the relationship $\frac{\sin t}{\cos t}=\operatorname{tg} t=\frac{c}{a}$, the missing vertex of the triangle with the maximum area can be easily constructed using the affinity with the main circle.

II. Solution: Let one of the foci of the centrally positioned ellipse be $F$, and the endpoint of the semi-minor axis be $B$ (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_2b87032ec7a082bac19ag-2.jpg?height=507&width=672&top_left_y=126&top_left_x=713)

Figure 2

The base $B F$ of the triangles to be drawn is common, so the one with the largest height will have the maximum area.

It is obvious that among the points of the quarter-ellipse, the point $E$ where the tangent parallel to $B F$ touches the ellipse is the farthest from the segment $B F$. Therefore, the triangle with the maximum area among the triangles that can be drawn is the $E B F$ triangle.

Kálmán Győry (Ózd, József A. g. III. o. t.)