2024 AMC 10A Problem 22

Gluing two $1$–$\sqrt3$–$2$ triangles along the hypotenuse gives a kite with a $60^\circ$ corner (where the two $\sqrt3$ sides meet), a $120^\circ$ corner (where the two unit sides meet), and two right angles. Its long diagonal is the hypotenuse, length $2$, bisecting the $60^\circ$ and $120^\circ$ corners.

Let $P$ be the polygon vertex directly below $C$, where the thick edge from $A$ meets the vertical thick edge. Four kites meet at $P$ with their $60^\circ$ corners.

Length of $AB$. Let $N$ be the point where the boundary touches $AB$ between $A$ and $B$. The two kites below $P$ have right-angle corners $A$ and $N$, so $PA = PN = \sqrt3$ and $\angle APN = 120^\circ$. An isosceles triangle with legs $\sqrt3$ and apex $120^\circ$ has base $3$ and height $\tfrac{\sqrt3}{2}$, so $AN = 3$. At the lowest right vertex $V$, two kites meet at $60^\circ$ corners with right-angle corners $N$ and $B$, so $NB = 3$ likewise. Both bases are horizontal (perpendicular to vertical bisectors), so $AB = 6$.

Height of $C$. $C$ is the $120^\circ$ corner of the kite whose $60^\circ$ corner is $P$, so $PC = 2$ is that kite's long diagonal, bisecting the $60^\circ$ angle between the vertical edge at $P$ and the edge rising at $30^\circ$. Hence $PC$ makes $60^\circ$ with the horizontal, $C$ is $2\sin 60^\circ = \sqrt3$ above $P$, and $P$ is $\tfrac{\sqrt3}{2}$ above $AB$: total height $\tfrac{3\sqrt3}{2}$.

$$[ABC] = \tfrac12\cdot 6\cdot\tfrac{3\sqrt3}{2} = \tfrac{9\sqrt3}{2}.$$

The answer is $\boxed{\textbf{(B)}\ \frac{9}{2}\sqrt3}$.

The whole figure lives on a $30^\circ$–$60^\circ$ grid: every kite edge is $1$ or $\sqrt3$ and every angle is a multiple of $30^\circ$, so lengths along $AB$ and heights above it come from $30$-$60$-$90$ pieces. Identify the special corners ($60^\circ$ where $\sqrt3$ sides meet, $120^\circ$ where unit sides meet) and the base $AB$ decomposes into two $120^\circ$ isosceles bases.

Coordinates with $P$ at the origin and $AB$ horizontal: $A = (-\tfrac32, -\tfrac{\sqrt3}{2})$, $N = (\tfrac32, -\tfrac{\sqrt3}{2})$, $V = (3, -\sqrt3)$, $B = (\tfrac92, -\tfrac{\sqrt3}{2})$, $C = (1, \sqrt3)$. Then $AB = 6$ and $C$ is $\sqrt3 + \tfrac{\sqrt3}{2}$ above line $AB$, giving $\tfrac{9\sqrt3}{2}$.

• Misreading the kite: its angles are 60°, 90°, 120°, 90° (the two √3 sides meet at 60°, the two unit sides at 120°), and its long diagonal is 2.
• Assuming $C$ is above the midpoint of $AB$ or that $ABC$ is equilateral; only the base and the height are needed, and $C$ is offset toward $A$.