2024 AMC 10A Problem 22

Step 0: know the kite. Each half of the kite is a right triangle with legs $1$ and $\sqrt3$ and hypotenuse $2$: a $30$-$60$-$90$ triangle. Putting two together along the hypotenuse gives a kite with:

• a $60^\circ$ corner where the two $\sqrt3$ sides meet,
• a $120^\circ$ corner where the two sides of length $1$ meet,
• two $90^\circ$ corners,
• a long diagonal of length $2$ joining the $60^\circ$ and $120^\circ$ corners, which cuts both of those angles in half.

From the $60^\circ$ corner, the two $90^\circ$ corners are $\sqrt3$ away, and the $120^\circ$ corner is $2$ away.

Step 1: find the key point. Let $P$ be the corner of the outline where the thick edge from $A$ meets the vertical thick edge, below and to the left of $C$. Four kites meet at $P$, each with its $60^\circ$ corner there.

Step 2: the length of $AB$. Let $N$ be the point on segment $AB$ between $A$ and $B$ where the outline touches it.

1. The two kites just below $P$ have their $90^\circ$ corners at $A$ and at $N$. So $PA = PN = \sqrt3$, and the angle $APN$ is $60^\circ + 60^\circ = 120^\circ$.
2. Triangle $APN$ has two sides $\sqrt3$ with a $120^\circ$ angle between them. Cut it in half with a line from $P$ straight down to $AN$. Each half is a $30$-$60$-$90$ triangle with hypotenuse $\sqrt3$: its short side (the height) is $\frac{\sqrt3}{2}$ and its long side is $\frac{\sqrt3}{2}\cdot\sqrt3 = \frac32$. So $AN = 2 \cdot \frac32 = 3$, and $P$ is $\frac{\sqrt3}{2}$ above $AB$.
3. The same thing happens at the lowest corner on the right, below $AB$ between $N$ and $B$: two kites meet there with $60^\circ$ corners and have $90^\circ$ corners at $N$ and $B$. So $NB = 3$ as well.
4. So $AB = 3 + 3 = 6$.

Step 3: the height of $C$ above $AB$.

1. $C$ is the $120^\circ$ corner of the kite whose $60^\circ$ corner is at $P$ and which lies against the vertical edge. So $PC$ is that kite's long diagonal, with length $2$.
2. This kite's $60^\circ$ corner sits between the vertical edge (straight up) and an edge $60^\circ$ to the right of it. The diagonal cuts that corner in half, so $PC$ is $30^\circ$ from vertical, which is $60^\circ$ above horizontal.
3. So $P$, $C$ and the point level with $P$ below $C$ form a $30$-$60$-$90$ triangle with hypotenuse $2$. The side across from $60^\circ$ is $\sqrt3$, so $C$ is $\sqrt3$ above $P$.
4. Height of $C$ above $AB$: $\sqrt3 + \frac{\sqrt3}{2} = \frac{3\sqrt3}{2}$.

Step 4: the area.
$$\text{Area} = \tfrac12 \cdot 6 \cdot \tfrac{3\sqrt3}{2} = \tfrac{9\sqrt3}{2}.$$

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

Every edge of the figure is $1$ or $\sqrt3$ and every angle is a multiple of $30^\circ$, so all the distances you need come from $30$-$60$-$90$ triangles (sides in the ratio $1 : \sqrt3 : 2$). The work is in reading the picture: find where the kites' $60^\circ$ corners meet, and the base and height fall out from there.

Use coordinates, with $P$ at $(0, 0)$ and $AB$ horizontal. From the steps above, $A = \left(-\frac32, -\frac{\sqrt3}{2}\right)$, $N = \left(\frac32, -\frac{\sqrt3}{2}\right)$, $B = \left(\frac92, -\frac{\sqrt3}{2}\right)$, and $C = (1, \sqrt3)$ (it is $2$ units from $P$ in the direction $60^\circ$ above horizontal). The base $AB$ runs from $x = -\frac32$ to $x = \frac92$, so it is $6$ long. $C$ is $\sqrt3 + \frac{\sqrt3}{2} = \frac{3\sqrt3}{2}$ above the line $AB$. The area is $\frac12 \cdot 6 \cdot \frac{3\sqrt3}{2} = \frac{9\sqrt3}{2}$.