2021 AMC 10B Problem 20

Let $F$ be the interior point joined to $A$, $B$, $C$ and $G$ the interior point joined to $A$, $E$, $D$. The eleven segments are the five sides plus $AF, BF, CF, AG, EG, DG$, all of length $2$.

Triangles $ABF$ and $BCF$ have all sides equal to $2$, so both are equilateral. Therefore $\angle ABC = 60^\circ + 60^\circ = 120^\circ$, and $ABCF$ is a rhombus with side $2$ and a $60^\circ$ angle at $A$; its long diagonal is $AC = 2\cdot 2\cos 30^\circ = 2\sqrt{3}$. By the same reasoning on the right, $\angle AED = 120^\circ$ and $AD = 2\sqrt{3}$.

Now split the pentagon by the diagonals from $A$:

• $[ABC] = \frac12 \cdot 2 \cdot 2 \sin 120^\circ = \sqrt{3}$, and likewise $[ADE] = \sqrt{3}$.
• Triangle $ACD$ is isosceles with $AC = AD = 2\sqrt{3}$ and base $CD = 2$. Its altitude from $A$ is $\sqrt{12 - 1} = \sqrt{11}$, so $[ACD] = \frac12 \cdot 2 \cdot \sqrt{11} = \sqrt{11}$.

Total area: $2\sqrt{3} + \sqrt{11} = \sqrt{12} + \sqrt{11}$, so $m + n = 12 + 11 = 23$.

The answer is $\boxed{\textbf{(D)}\ 23}$.

Many equal segments means equilateral triangles hiding in the figure; find them first, because they hand you angles ($60^\circ$, $120^\circ$) and lengths ($2\sqrt{3}$ diagonals). Once the key diagonals $AC$ and $AD$ are known, the pentagon is a fan of three triangles from one vertex, each computable with basic tools. The final form $\sqrt{m} + \sqrt{n}$ tells you to absorb the coefficient: $2\sqrt{3} = \sqrt{12}$.

• Assuming the pentagon is regular, or adding four equilateral triangles plus a 'middle region' without noticing that segments FC and GD cross.
• Leaving the answer as $2\sqrt{3} + \sqrt{11}$ and reading $m + n$ as $3 + 11 = 14$ or $2 + 3 + 11$.