What are the angles of the isosceles triangle in which the sum of the heights corresponding to the base and the leg is equal to half the base?
I. solution. In the triangle $ABC$, let $CA=CB$, the foot of the perpendicular from $A$ to the line $BC$ be $D$, and the foot of the perpendicular from $C$ to the line $AB$ be $E$. Let $AE=1$ and $CE=x$. From the similarity of the right-angled triangles $ABD$ and $CBE$, which share an acute angle,

$$
AD=\frac{AB}{BC} \cdot CE=\frac{2x}{\sqrt{1+x^{2}}}
$$

thus, according to the assumption,

$$
x+\frac{2x}{\sqrt{1+x^{2}}}=1
$$

from which it is clearly $00$.)

Karsai Kornélia (Szeged, Radnóti M. g. I. o. t.)

Remarks. 1. Writing equation (3) as $\sin 2\alpha(\sin 2\alpha+1)=1$, or equivalently $\sin 2\alpha : 1 = 1 : (\sin 2\alpha+1)$, we see that $\sin 2\alpha$ is the smaller part of the segment $\sin 2\alpha+1$ obtained by the golden section (division according to continuous proportion). Based on this, the shape of the triangle $ABC$ can be constructed: at the endpoint $G$ of the segment $FG$, we measure the segment $GH = FG / 2$ on the perpendicular, draw a circle with radius $GH$ around $H$, which intersects the line $FH$ at points $J$ and $K$, $FJ3 \sin \alpha$, so $\sin \alpha0.3333, \alpha>18.43^{\circ}$. These two limits confine $\alpha$ to within about a degree of accuracy.

For $\alpha=19.47^{\circ}$, $2 \sin \alpha+\operatorname{tg} \alpha=0.6666+0.3535=1.0201$, so we need to try a smaller angle, for which the left side is $0.0201-$ smaller. The tangent function changes faster than the sine function, so we need to get at least $0.0201: 3=0.0067$ decrease in $\operatorname{tg} \alpha$:

$$
\operatorname{tg} \alpha0.3333-0.0067=0.3266, \alpha>19.06^{\circ}$.

From here, $\alpha=19.1^{\circ}$ is already with an error less than $0.1^{\circ}$.

Máté Zoltán (Bonyhád, Petófi S. g. IV. o. t.)

3. We can obtain an approximate value for the angle from the observation that the graph of the function $y=2 \sin x+\operatorname{tg} x$ in the interval $0 \leq x \leq \pi / 6$ approximates the function $y=3 x$ (with $x$ in radians) with an error less than $1/2 \%$. Therefore, a good approximate value is $3 x=1$-bó $x=1/3=0.3333$ radians $\approx 19.1^{\circ}$. Testing the values $\alpha=19.0^{\circ}, 19.1^{\circ}$, and $19.2^{\circ}$ in (5), the left side values are respectively $0.9955, 1.0007$, and $1.0060$, the middle one being closest to the required value of 1, so with an accuracy of $0.1^{\circ}$, $\alpha=19.1^{\circ}$.

[^0]:    ${ }^{1}$ The point $E$ should be added in the figure.

    ${ }^{2}$ For example, as seen in problem 1176, K. M. L. 26 (1963/1) p. 56.