The lengths of two conjugate diameters of an ellipse are $2 r$ and $2 r^{\prime}$, and the angle between them is $\varphi$. Express the lengths of the axes of the ellipse and the angles between the given diameters and the axes using these. Example: $r=1, r^{\prime}=\sqrt{2}, \varphi=45^{\circ}$.[^0]


[^0]:    ${ }^{1}$ See problem 1416, K. M. L. 32 (1966)208. o.
I. Solution. I. We follow the steps of the Rytz construction as cited in problem 1416. Let the center of the ellipse be $O$, the two conjugate semi-diameters be $O C=r, O D=r^{\prime}$, where $r^{\prime} \geq r$, and the acute angle between them be $C O D \varangle=\psi$. (Indeed, $\psi \neq 0^{\circ}$, and the case $\psi=90^{\circ}$ is unnecessary to examine, as $r$ and $r^{\prime}$ are the semi-axes in this case.)

![](https://cdn.mathpix.com/cropped/2024_05_02_35e7ef813bce8d6fd414g-1.jpg?height=637&width=675&top_left_y=305&top_left_x=714)

Rotate $D$ around $O$ so that in its new position $P$, $C O P \varangle=\psi+90^{\circ}$. Let the midpoint of $C P$ be $Q$, and let the circle centered at $Q$ with radius $Q O$ intersect the ray $Q C$ at $R$ and the ray $Q P$ at $S$. Then, as shown in the cited problem, the semi-axes of the ellipse are $C S=a, C R=S P=b$, so $C P=a+b, R S=2 Q O=a-b$. Applying the cosine rule,

$$
(a+b)^{2}=C P^{2}=O C^{2}+O P^{2}-2 \cdot O C \cdot O P \cdot \cos \left(90^{\circ}+\psi\right)=r^{2}+r^{\prime 2}+2 r r^{\prime} \sin \psi
$$

Furthermore, the reflection of $O$ over $Q$ to $T$ completes the triangles $P O C$ and $S O R$ into a parallelogram, and the latter is a right triangle; thus $R S$ is equal to the diagonal of the parallelogram $O C T P$:

$$
(a-b)^{2}=R S^{2}=O T^{2}=2\left(O C^{2}+O P^{2}\right)-C P^{2}=r^{2}+r^{\prime 2}-2 r r^{\prime} \sin \psi
$$

Therefore,

$$
a, b=\frac{1}{2}\left(\sqrt{r^{2}+r^{\prime 2}+2 r r^{\prime} \sin \psi} \pm \sqrt{r^{2}+r^{\prime 2}-2 r r^{\prime} \sin \psi}\right)
$$

Moreover, the direction of the major axis is $O R$, so

$$
C O R \varangle=C O Q \varangle-R O Q \varangle=C O Q \varangle-90^{\circ}+O Q C \varangle / 2,
$$

since $O Q R$ is an isosceles triangle. For this purpose, based on (1) and (2),

$$
\begin{gathered}
\cos C O Q \varangle=\frac{O C^{2}+O Q^{2}-C Q^{2}}{2 O C \cdot O Q}=\frac{r^{2}-r r^{\prime} \sin \psi}{2 r \cdot O Q}= \\
=\frac{r-r^{\prime} \sin \psi}{\sqrt{r^{2}+r^{\prime 2}-2 r r^{\prime} \sin \psi}}, \\
\cos O Q C \varangle=\frac{Q O^{2}+Q C^{2}-O C^{2}}{2 \cdot Q O \cdot Q C}=\frac{r^{\prime 2}-r^{2}}{\sqrt{\left(r^{2}+r^{\prime 2}\right)^{2}-4 r^{2} r^{\prime 2} \sin ^{2} \psi}}
\end{gathered}
$$

From the comparison of (1) and (2), we see that $R S < C P$, so $R$ and $S$ always lie on the segment $C P$. Thus, $R$ is in the sector $C O D$, because rotating the angle $P O S$ by a right angle, its sides fall on $O D$ and $O R$, as $S O R$ is a right angle by Thales' theorem.

II. In the given numerical system,

$$
\begin{gathered}
C P^{2}=5, \quad R S^{2}=1, \quad a=(\sqrt{5}+1) / 2, \quad b=(\sqrt{5}-1) / 2, \\
C O Q \varangle=90^{\circ}, \quad \cos O Q C \varangle=1 / \sqrt{5}, \quad O Q C \varangle=63.43^{\circ}, \quad C O R \varangle=31.72^{\circ} .
\end{gathered}
$$

András Wagner (Pécs, Zipernovszky K. Gépip. Techn., III. o. t.) Imre Gallatz (Pannonhalma, Bencés g. III. o. t.)

![](https://cdn.mathpix.com/cropped/2024_05_02_35e7ef813bce8d6fd414g-2.jpg?height=583&width=507&top_left_y=126&top_left_x=798)

II. Solution. We place our ellipse in the standard position in a Cartesian coordinate system, so its equation is $b^{2} x^{2}+a^{2} y^{2}=a^{2} b^{2}$, and we take the half of the conjugate diameters $O C=r$ and $O D=r^{\prime}$ such that a $+45^{\circ}$ rotation of $O C$ takes it to the direction of $O D$, and in the process, it crosses the positive $X$-axis. Let $D^{*}$ be the point on the main circle $x^{2}+y^{2}=a^{2}$ of the ellipse from which, using the concentric circles with radii $a$ and $b$, we obtain $D$, and let the parameter angle $X O D^{*}$ be $t$; then the coordinates of $D$ - as seen in problem 1416 - are $(a \cos t, b \sin t)$. Furthermore, for the point $C^{*}$ that generates $C$, $X O C^{*} \varangle=t-90^{\circ}$, because $O C^{*}$ and $O D^{*}$ are conjugate in the main circle and thus perpendicular, so the coordinates of $C$ are $(a \sin t, -b \cos t)$. From these, we have

$$
O C^{2}=a^{2} \sin ^{2} t+b^{2} \cos ^{2} t=r^{2}, \quad O D^{2}=a^{2} \cos ^{2} t+b^{2} \sin ^{2} t=r^{\prime 2}
$$

and by adding them,

$$
a^{2}+b^{2}=r^{2}+r^{\prime 2}
$$

Furthermore, denoting the rotation angles $X O C$ and $X O D$ by $\psi_{1}$ and $\psi_{2}$ respectively,

$$
\begin{array}{ll}
\sin \psi_{1}=-\frac{b}{r} \cos t, & \cos \psi_{1}=\frac{a}{r} \sin t \\
\sin \psi_{2}=\frac{b}{r^{\prime}} \sin t, & \cos \psi_{2}=\frac{a}{r^{\prime}} \cos t
\end{array}
$$

and thus

$$
\sin \psi=\sin \left(\psi_{2}-\psi_{1}\right)=\frac{a b}{r r^{\prime}}\left(\sin ^{2} t+\cos ^{2} t\right)=\frac{a b}{r r^{\prime}}
$$

from which

$$
a b=r r^{\prime} \sin \psi
$$

Now, adding twice (6) to (4) and then subtracting $(a+b)^{2}$ and $(a-b)^{2}$ from it, we obtain $(a+b)^{2}$ and $(a-b)^{2}$, and from these, by taking the square root, we get the results (3) of the first solution. After this, the direction of the major axis can be determined based on any of the expressions in (5).

István Bodor (Veszprém, Lovassy L. g. III. o. t.) Ferenc Gegesy (Budapest, Móricz Zs. g. II. o. t.)