2024 AMC 10A Problems/Problem 22

Problem

Let $\mathcal K$ be the kite formed by joining two right triangles with legs $1$ and $\sqrt3$ along a common hypotenuse. Eight copies of $\mathcal K$ are used to form the polygon shown below. What is the area of triangle $\Delta ABC$ ?

$\textbf{(A) }2+3\sqrt3\qquad\textbf{(B) }\dfrac92\sqrt3\qquad\textbf{(C) }\dfrac{10+8\sqrt3}{3}\qquad\textbf{(D) }8\qquad\textbf{(E) }5\sqrt3$

Solution 1

Let $\mathcal K$ be quadrilateral $MNOP$ . Drawing line $MO$ splits the triangle into $\Delta MNO$ . Drawing the altitude from $N$ to point $Q$ on line $MO$ , we know $NQ$ is $\frac{\sqrt{3}}{2}$ , $MQ$ is $\frac{3}{2}$ , and $QO$ is $\frac{1}{2}$ .

Due to the many similarities present, we can find that $AB$ is $4(MQ)$ , and the height of $\Delta ABC$ is $NQ+MN$

$AB$ is $4(\frac{3}{2})=6$ and the height of $\Delta ABC$ is $\sqrt3+\frac{\sqrt{3}}{2}=\frac{3\sqrt{3}}{2}$ .

Solving for the area of $\Delta ABC$ gives $6\cdot\frac{3\sqrt{3}}{2}\cdot\frac{1}{2}$ which is $\boxed{\textbf{(B) }\dfrac92\sqrt3}$

~9897 (latex beginner here)

~i_am_suk_at_math(very minor latex edits)

Solution 2

Let's start by looking at kite $\mathcal K$ . We can quickly deduce based off of the side lengths that the kite can be split into two $30-60-90$ triangles. Going back to the triangle $\triangle ABC$ , focus on side $AB$ . There are $4$ kites, they are all either reflected over the line $AB$ or a line perpendicular to $AB$ , meaning the length of $AB$ can be split up into 4 equal parts.

Pick out the bottom-left kite, and we can observe that the kite and the triangle formed by the intersection of the kite and $\Delta ABC$ share a $60$ degree angle. (this was deduced from the $30-60-90$ triangles in the kite) The line AB and the right side of the kite are perpendicular, forming a $90^{\circ}$ angle. Because that is also a $30-60-90$ triangle with a hypotenuse of $\sqrt3$ , so we find the length of AB to be $4*3/2$ , which is $6$ .

Then, we can drop an altitude from $C$ to $AB$ . We know that will be equivalent to the sum of the longer side of the kite and the shorter side of the triangle formed by the intersection of the kite and $\Delta ABC$ . (Look at the line formed on the left of $C$ that drops down to $AB$ if you are confused) We already have those values from the $30-60-90$ triangles, so we can just plug it into the triangle area formula, $bh/2$ . We get $6\cdot\dfrac{\sqrt3+\frac{\sqrt3}{2}}{2}\rightarrow3\cdot(\sqrt3+\dfrac{\sqrt3}{2})\rightarrow3\cdot\dfrac{\sqrt3}{2}\rightarrow\boxed{\textbf{(B) }\dfrac92\sqrt3}$

~YTH (Need help with Latex and formatting)

~WIP (Header)

~Tacos_are_yummy_1 ( $\LaTeX$ & Formatting)

Solution 3

(latexing a WIP) ~mathboy282 ~Thesmartgreekmathdude

Solution 4 (wip)

Let the point of intersection of $AB$ and the kite with $A$ as vertex be $D$ .

Let the left kite with $C$ as a vertex touch the kite with $A$ as vertex at point $E$ .

$\triangle ADE$ is a $30-60-90$ so $AD = \frac{3}{2}$ and $DE = \frac{\sqrt3}{2}$ .

So, $AB = 4AD = 6$ and $CD=CE+DE= \frac{3\sqrt3}{2}$ , and the area is $\frac12\cdot AB \cdot CD = \boxed{ \textbf{(B) } \frac{9\sqrt3}{2} }.$

~Mintylemon66

Video Solution by Innovative Minds

https://www.youtube.com/watch?v=bhC58BB3kJA

~i_am_suk_at_math_2

2024 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
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