A plane intersects the lateral edges of a regular triangular prism at points $A, B, C$. What is the length of the side $d$ of the base triangle if $A B=c, B C=a, C A=b ?$

How is $d$ related to a specific value of $d$ determined in the solution of problem 1527. I?[^0]


[^0]:    ${ }^{1}$ K. M. L. 36 (1968) 99.
I. Let the orthogonal projection of the intersection point $A$ onto the side containing $B$ be $A_{1}$, and similarly, the projections of $B$ and $C$ onto the sides containing $C$ and $A$ be $B_{1}$ and $C_{1}$, respectively.

![](https://cdn.mathpix.com/cropped/2024_05_02_8b7ed9e82150b8be1618g-1.jpg?height=683&width=596&top_left_y=266&top_left_x=754)

Thus, $A A_{1}=B B_{1}=C C_{1}=d$, and the height differences of the vertices of the triangle $A B C$ (considering the base triangle of the prism on a horizontal plane) from the right triangles $A B A_{1}$, $B C B_{1}$, and $C A C_{1}$ are respectively

$$
A_{1} B=\sqrt{c^{2}-d^{2}}, \quad B_{1} C=\sqrt{a^{2}-d^{2}}, \quad C_{1} A=\sqrt{b^{2}-d^{2}}
$$

The lengths $a, b, c$ of the segments play an equivalent role in the problem, so we can assume their magnitude relationship is:

$$
a \leqq b \leqq c,
$$

therefore

$$
A_{1} B \geqq C_{1} A \geqq B_{1} C
$$

and from this

$$
\begin{gathered}
A_{1} B=C_{1} A+B_{1} C \\
\sqrt{c^{2}-d^{2}}=\sqrt{b^{2}-d^{2}}+\sqrt{a^{2}-d^{2}}
\end{gathered}
$$

since the largest height difference is equal to the sum of the other two, meaning we ascend the same amount in two steps on the broken line $A C B$ as on the straight line segment connecting its endpoints. Squaring, rearranging, squaring again, and finally reducing the equation to zero:

$$
\begin{gathered}
c^{2}-d^{2}=a^{2}+b^{2}-2 d^{2}+2 \sqrt{\left(a^{2}-d^{2}\right)\left(b^{2}-d^{2}\right)} \\
d^{2}+c^{2}-a^{2}-b^{2}=2 \sqrt{\left(a^{2}-d^{2}\right)\left(b^{2}-d^{2}\right)} \\
d^{4}+2\left(c^{2}-a^{2}-b^{2}\right) d^{2}+\left(c^{2}-a^{2}-b^{2}\right)^{2}=4 a^{2} b^{2}-4\left(a^{2}+b^{2}\right) d^{2}+4 d^{4}, \\
3 d^{4}-2\left(a^{2}+b^{2}+c^{2}\right) d^{2}+\left\{4 a^{2} b^{2}-\left(c^{2}-a^{2}-b^{2}\right)^{2}\right\}=0
\end{gathered}
$$

which is a quadratic equation in $d^{2}$.

The discriminant's $\frac{1}{4}$ part can be transformed as follows:

$$
\begin{aligned}
D & =\left(a^{2}+b^{2}+c^{2}\right)^{2}-12 a^{2} b^{2}+3\left(c^{2}-a^{2}-b^{2}\right)^{2}= \\
& =4\left(a^{4}+b^{4}+c^{4}-a^{2} b^{2}-a^{2} c^{2}-b^{2} c^{2}\right)= \\
& =\left(a^{2}+c^{2}-2 b^{2}\right)^{2}+3\left(c^{2}-a^{2}\right)^{2} \geqq 0
\end{aligned}
$$

thus the values obtained for $d^{2}$ are real (in the last step, we applied one of the transformations from the cited solution in reverse).

On the other hand, it is easy to show by applying the difference of squares formula $U^{2}-V^{2}=(U-V)(U+V)$ three times that the term of the equation not containing $d$ is equal to four times the square of the area of triangle $A B C$, and thus is a positive number. The coefficient of $d^{2}$, however, is negative (and the coefficient of $d^{4}$ is positive), so the product and sum of the two (real) roots of $d^{2}$ are also positive, meaning both are positive.

The roots of (2) are equal only if $D=0$, and this only holds if both terms of (3) are 0, i.e., $c^{2}=a^{2}$, and thus from the first term $a^{2}=b^{2}$, or if $a=b=c$. In this case, as can be easily seen, $d$ is also equal to them.

However, if $a, b$, and $c$ are not all equal, then $D>0$, and the larger value obtained for $d^{2}$ is greater than $a^{2}$, and thus, according to (1), cannot be used in our problem. Therefore, the side length of the equilateral triangle base of the prism is uniquely

$$
d^{2}=\frac{1}{3}\left\{a^{2}+b^{2}+c^{2}-2 \sqrt{a^{4}+b^{4}+c^{4}-a^{2} b^{2}-a^{2} c^{2}-b^{2} c^{2}}\right\}
$$

II. In problem 1527, the sides of the base triangle of the prism were $a, b, c$, and the side $d$ of the equilateral (oblique) section of the prism obtained was (4) with only a difference in the sign of the square root. According to this, the value of $d$ - although obtained from slightly different relationships - is numerically equal to the larger root of (2).

Draschitz Rudolf (Budapest, Landler J. Gépip. Techn.)

Nagy Zsigmond (Budapest, Kaffa M. Gimn.)