\section*{Problem 3 - 051243}

Among all segments \(M N\) that divide the triangle \(\triangle A B C\) into two regions of equal area, the number and the length of all those that are as short as possible are to be determined.
}

We first consider a segment \(M N\), whose endpoint \(M\) lies on the ray from \(A\) through \(B\) and whose endpoint \(N\) lies on the ray from \(A\) through \(C\), and for which the area of triangle \(A M N\) is half the area of triangle \(A B C\).

The segment \(M N\) may also partially lie outside the given triangle. If we set \(A M=x\) and \(A N=y\), then according to a known area formula, the area of triangle \(A M N\) is exactly half the area of triangle \(A B C\) if

\[
\frac{1}{2} x y \sin \alpha=\frac{1}{2} \cdot \frac{1}{2} b c \sin \alpha
\]

thus \(2 x y=b c\) holds.

If \(d_{a}=M N\), then by the Law of Cosines applied to \(\triangle A M N\),

\[
\begin{aligned}
d_{a}^{2} & =x^{2}+y^{2}-2 x y \cos \alpha=(x-y)^{2}+2 x y(1-\cos \alpha) \\
& =(x-y)^{2}+b c(1-\cos \alpha)
\end{aligned}
\]

On the other hand, by the Law of Cosines applied to triangle \(A B C\),

\[
\begin{aligned}
a^{2}=b^{2}+c^{2}-2 b c \cos \alpha & =(b-c)^{2}+2 b c(1-\cos \alpha) \quad \text { so } \\
1-\cos \alpha & =\frac{a^{2}-(b-c)^{2}}{2 b c}
\end{aligned}
\]

Consequently,

\[
d_{a}^{2}=(x-y)^{2}+\frac{1}{2}\left[a^{2}-(b-c)^{2}\right]=(x-y)^{2}+2(s-b)(s-c)
\]

with \(2 s=a+b+c\). Using the triangle area formula \(I_{\triangle A B C}=\sqrt{s(s-a)(s-b)(s-c)}\), we thus get

\[
d_{a}^{2}=(x-y)^{2}+\frac{2 I^{2}}{s(s-a)}
\]

By cyclic permutation of (2), we get

\[
d_{b}^{2}=\left(x^{\prime}-y^{\prime}\right)^{2}+\frac{2 I^{2}}{s(s-b)} ; \quad d_{c}^{2}=\left(x^{\prime \prime}-y^{\prime \prime}\right)^{2}+\frac{2 I^{2}}{s(s-c)}
\]

if \(x^{\prime}, y^{\prime}, x^{\prime \prime}, y^{\prime \prime}\) are the corresponding segments for a segment whose endpoints lie on the rays from \(B\) and \(C\) respectively, and \(d_{b}\) and \(d_{c}\) are the lengths of these segments.

We assume w.l.o.g. that \(a \leq b \leq c\). Then

\[
\frac{2 I^{2}}{s(s-a)} \leq \frac{2 I^{2}}{s(s-b)} \leq \frac{2 I^{2}}{s(s-c)}
\]

and the first or second equality holds if and only if \(a=b\) or \(b=c\). From this and from (2) and (3), it follows that each of the lengths \(d_{a}, d_{b}, d_{c}\) is not less than

\[
\frac{\sqrt{2} I}{\sqrt{s(s-a)}}
\]

To show that

\[
d=\frac{\sqrt{2} I}{\sqrt{s(s-a)}}=\sqrt{2(s-b)(s-c)}
\]

is the desired minimum, it suffices to show that the segment \(M N\) formed for \(x=y\), whose length \(d_{a}=d\), does not enter the exterior of the triangle.

In the case \(x=y\), from (1) we get

\[
x=y=\sqrt{\frac{b c}{2}}
\]

Since \(a \leq b \leq c\) is assumed, due to the triangle inequality \(c<a+b \leq 2 b\) and therefore

\[
\sqrt{\frac{b c}{2}}<\sqrt{b^{2}}=b \leq c
\]

so that each of the points \(M\) and \(N\) lies on one of the sides of the triangle.

If \(a<b\), there is exactly one shortest among these segments, namely the one for \(x=y\).

If \(a=b<c\), there are exactly two shortest among these segments, namely the ones for \(x=y\) and \(x^{\prime}=y^{\prime}\).

If \(a=b=c\), there are correspondingly three shortest among these segments, the ones for \(x=y, x^{\prime}=y^{\prime}\) and \(x^{\prime \prime}=y^{\prime \prime}\).

\section*{Adapted from \([5]\)}