[Area of a Triangle (using two sides and the angle between them)]

A certain triangle can be cut out from a paper strip of unit width, but it cannot be cut out from any strip of lesser width. What area can this triangle have?

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The area can be equal to any positive number not less than $\frac{1}{\sqrt{3}}$.

First, let's prove that if the area $S$ of a triangle is less than $\frac{1}{\sqrt{3}}$, then it can be cut out from a strip of width less than 1. It is sufficient to prove that it has a height less than 1. The equivalent condition is: the triangle has a side $a > 2S$. Suppose all sides of the triangle do not exceed $2S$. Let $\gamma$ be the smallest angle of the triangle. Then $S = \frac{1}{2} a b \sin \gamma \leq \frac{a b \sqrt{3}}{4} \leq \sqrt{3} S^2$, so $S \geq 1 / \sqrt{3}$. This leads to a contradiction.

Now consider an isosceles triangle with height 1 and base $a \geq 2 / \sqrt{3}$. The lateral sides have a length not greater than the length of the base, so the considered triangle has no height less than 1. We will prove that in this case, the triangle cannot be cut out from a strip of width less than 1. Suppose the triangle $ABC$ is located inside a strip of width less than 1. Draw lines through its vertices perpendicular to the sides of the strip. Let's assume for definiteness that the line passing through vertex $B$ is between the other two lines. Then it intersects side $AC$ at some point $M$. On the one hand, $BM < 1$. On the other hand, $BM \geq h_{\mathrm{b}}$, so $h_{\mathrm{b}} < 1$.

## Answer

The area can be equal to any positive number not less than $\frac{1}{\sqrt{3}}$.