Duplicate area

Claim: The largest area that is enclosed by $n$ sides of length 1 is obtained in a regular polygon. This is a "well-known" fact that follows from the Isoperimetric Inequality . It will not be proven here.

Recall that for a regular polygon with $n$ sides of length 1, the height of an individual triangle is $\frac{1}{2} \cot \frac{ \pi } { n}$ , and the area is

$\frac{1}{4} n \cot \frac{ \pi } { n }.$

The goal is to double the area from 4 to 8.

1. If we didn't use any additional matchsticks, then we would have $n = 1 0$ . This gives us a maximum area of $\frac{1}{4} \times 10 \cot \frac{ \pi }{ 10} \approx 7.69$ , which isn't sufficient.
2. If we used 1 additional matchstick, then we would have $n = 11$ . This gives us a maximum area of $\frac{1}{4} \times 11 \cot \frac{ \pi}{11} \approx 9.365$ . Hence, we can squish this shape slightly, in order to get an area of 8.

This establishes that

1. With 0 matchsticks, the goal cannot be ahcieved (so 1 is a lower bound).
2. With 1 matchstick, the goal can be acheived.

Hence, the minimum is 1.

Make a 3 x 3 square, and turn two matches to exclude one corner square. This requires 4 x 3 = 12 matches, $\boxed{\ 2\ }$ more than the current number of matches.