A three-sided right prism has base edges of $3,4,5 \mathrm{~cm}$. A plane section of the prism is an equilateral triangle. What is the angle between the cutting plane and the base plane?

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I. Solution. In a triangular prism, the planes that intersect the lateral edges and the planes that cut off one vertex intersect the prism to form a triangle. First, let's find a plane of the first type, for a prism with arbitrary base edges $a, b, c$ where $a \leq b \leq c$. Let the side of the equilateral triangle cut off by the plane be $d$. Thus, the height differences of the endpoints of the intersection lines on the lateral faces of width $a, b, c$ are respectively

$$
\sqrt{d^{2}-a^{2}}=x, \quad \sqrt{d^{2}-b^{2}}=y, \quad \sqrt{d^{2}-c^{2}}=z
$$

and these satisfy $x \geq y \geq z(\geq 0)$.

![](https://cdn.mathpix.com/cropped/2024_05_02_d6ec3cc9a666b7747c61g-1.jpg?height=773&width=1242&top_left_y=508&top_left_x=428)

Therefore (1st figure):

$$
x=y+z
$$

since the height difference between the lower and upper vertices of $H$ is the largest, and this is the same as the height we gain when we go up to the upper vertex from the intermediate height vertex through two other sides of $H$. According to (1),

$$
a^{2}+x^{2}=b^{2}+y^{2}=c^{2}+z^{2} \quad\left(=d^{2}\right)
$$

From the extreme expressions, according to (2),

$$
a^{2}+(y+z)^{2}=c^{2}+z^{2}, \quad z=\frac{c^{2}-a^{2}}{2 y}-\frac{y}{2}
$$

(we can assume that $y \neq 0$, because otherwise $z=0$ and from (2) $x=0$, which only follows if $a=b=c$, and in this case the sought plane must be parallel to the base). Then from (3), after the usual rearrangement steps, we get a quadratic equation for $y^{2}$:

$$
\begin{aligned}
& b^{2}+y^{2}=c^{2}+\frac{\left(c^{2}-a^{2}\right)^{2}}{4 y^{2}}-\frac{c^{2}-a^{2}}{2}+\frac{y^{2}}{4} \\
& 3 y^{4}-2\left(a^{2}+c^{2}-2 b^{2}\right) y^{2}-\left(c^{2}-a^{2}\right)^{2}=0
\end{aligned}
$$

The product of the two roots for $y^{2}$ is $-\left(c^{2}-a^{2}\right)^{2} / 30,5=\cos 60^{\circ}\right)$, so the base of the isosceles triangle cannot be $d$, as it is smaller than $f$, which is smaller than $d$.

Then the plane $S$ cuts off congruent right triangles from two lateral faces, and according to the above considerations, the triangle cut off from the base is also congruent to the other two. Thus, $S$ cuts off the same segment $f$ from all three edges, the symmetry axis of the triangle cut off from the base is $f / \sqrt{2}$, so for the angle $\varphi^{\prime}$

$$
\operatorname{tg} \varphi^{\prime}=\frac{f}{f / \sqrt{2}}=\sqrt{2}, \quad \varphi^{\prime}=54,74^{\circ}
$$

The cutting plane cannot have a segment on both the base and the top face of the prism, because then the intersection would have at least 4 vertices. Therefore, there is no other possibility.

Andor Csaba (Budapest, Berzsenyi D. g. III. o. t.)

Remark. The French mathematician S. Lhuilier (1750-1840) proved that among the plane sections of a triangular prism, every possible triangular shape can occur.

II. Solution (only for the case of a plane intersecting the lateral edges). Consider the circumcircle $k$ of $H$. The projection of $k$ onto the base plane is an ellipse circumscribing the base triangle. It would be enough to know the semi-axes of this ellipse - $a-t$ and $b$, with which $\cos \varphi=b / a$, because $a$ is precisely the radius of $k$, i.e., the projection of the radius parallel to the base plane, and $b$ - the half of the shortest diameter of the ellipse - is the projection of the radius that makes the sought angle $\varphi$ with the base plane.

$a$ and $b$ can be calculated from the lengths $r, r^{\prime}$ and the angle $\psi$ between a pair of conjugate semi-diameters of the ellipse using the method of problem 1502. ${ }^{2}$ The value triplet is given by any pair of perpendicular radii of $k$. Take one of them to be the radius corresponding to a vertex of $H$; its projection is the $2/3$ part of the median from the corresponding vertex of the base triangle. Indeed, the center of $k$ is the centroid of $H$, and this property is preserved in the projection (because the ratio of the projections of segments on a straight line is equal to the ratio of the original segments, so the projection of the midpoint of a side bisects the projection of the corresponding side, and the projection of the point dividing the median in the ratio 1:2 divides the projection of the median in the same ratio). The projection of the radius perpendicular to the mentioned radius falls on the line passing through the centroid of the base triangle and parallel to the opposite side of the selected vertex, because the projections of parallel lines are parallel. The mentioned radius is $\sqrt{3}$ times the part inside $H$. Therefore, the corresponding semi-axis on the base is $\sqrt{3}$ times the segment from the centroid to the midpoint of the opposite side of the triangle, or the $1 / \sqrt{3}$ part of the side, since the segment in the triangle is the $1 / 3$ part of the side by simple calculation.[^0]![](https://cdn.mathpix.com/cropped/2024_05_02_d6ec3cc9a666b7747c61g-3.jpg?height=526&width=896&top_left_y=122&top_left_x=602)

3rd figure

Focusing on the numerical example, let's start from the vertex of $H$ whose projection is the vertex of the right angle of the base triangle; thus, the median is half the hypotenuse, and the two conjugate semi-axes of the ellipse are $r=c / 3=5 / 3, r^{\prime}=5 / \sqrt{3}$, and the acute angle $\psi$ between them is twice the smaller acute angle of the base triangle: $\sin \psi / 2=3 / 5, \cos \psi / 2=4 / 5$, so $\sin \psi=24 / 25$.

Now, using the formulas (3) of the cited problem,

$$
\begin{gathered}
\sqrt{r^{2}+r^{\prime 2}+2 r r^{\prime} \sin \psi}=\frac{2}{3} \sqrt{25+12 \sqrt{3}} \\
\sqrt{r^{2}+r^{\prime 2}-2 r r^{\prime} \sin \psi}=\frac{2}{3} \sqrt{25-12 \sqrt{3}}, \text { and thus } \\
\cos \varphi=\frac{b}{a}=\frac{\sqrt{25+12 \sqrt{3}}-\sqrt{25-12 \sqrt{3}}}{\sqrt{25+12 \sqrt{3}}+\sqrt{25-12 \sqrt{3}}}= \\
=\frac{1}{24 \sqrt{3}}(\sqrt{25+12 \sqrt{3}}-\sqrt{25-12 \sqrt{3}})^{2}=\frac{25-\sqrt{193}}{12 \sqrt{3}}
\end{gathered}
$$

which we also found in the first solution.

Remark. The connection of the centroid and the endpoint of the segment on it to $H$ is essentially similar to how we connected additional points constructible in the projection to the regular octagon in the 1st solution and the remarks attached to problem 1424.3.[^1]


[^0]:    ${ }^{1}$ See, for example, H. Dörrie: Triumph of Mathematics, 100 Famous Problems from Two Millennia of Mathematical Culture, Gondolat Publishing, Budapest, $1965,324.0$

    ${ }^{2}$ K. M. L. $35(1967) 206$. о.

[^1]:    ${ }^{3}$ K. M. L. 32 (1966) 212. o.