\section*{Task 4 - 011124}

Three segments of different lengths \(a, b\), and \(c\) are to originate from a point \(M\) and lie in a plane such that their endpoints \(A, B\), and \(C\) lie on a straight line in this order and \(A B = B C\).

Perform the construction and justify it!

Let \(a > c\). Provide the conditions for \(b\) under which the task is solvable!
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\section*{I. Analysis:}

Considering triangle \(A M C\), \(M B\) is a median because \(A B = B C\). Therefore, we need to construct a triangle from two sides and the included median. To do this, we extend \(A M C\) to a parallelogram \(A M C N\), in which the diagonals \(A C\) and \(M N\) bisect each other.

The sub-triangle \(A M N\) can thus be constructed from the given segments.

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1680.jpg?height=440&width=488&top_left_y=2127&top_left_x=790)

II. Construction Description:

We construct triangle \(A M N\) from the side lengths \(a, c\), and \(2b\) using the SSS (Side-Side-Side) congruence criterion. The midpoint of segment \(M N\) is then \(B\), and doubling \(A B\) gives point \(C\).

III. Proof:

According to the above construction, \(A M N\) is a triangle where \(A M = a\), \(A N = c\), and \(M B = B N = b\), and \(A B\) is a median. Since \(C\) is obtained by doubling \(A B\), the congruence \(\triangle A B N \cong \triangle C B M\) (SAS) holds, meaning \(A N = M C = c\). The points \(A, B\), and \(C\) thus have the required distances from \(M\).

IV. Construction:

Triangle \(A M N\) exists if and only if the triangle inequalities are satisfied: \(a + c > 2b\) and \(c + 2b > a\). This leads to the required conditions for the length \(b\):

\[
\frac{1}{2}(a - c) < b < \frac{1}{2}(a + c)
\]