2024 AMC 12A Problems/Problem 11

Problem

In regular tetrahedron $ABCD$ , points $E$ and $F$ lie on segments $\overline{AB}$ and $\overline{AC}$ , respectively, such that $BE = CF = 3$ . If $EF = 8$ , what is the area of $\triangle DEF$ ?

$\textbf{(A)}~32\qquad\textbf{(B)}~35\qquad\textbf{(C)}~36\qquad\textbf{(D)}~42\qquad\textbf{(E)}~48$

Solution

Note that $\triangle AEF$ is an equilateral triangle. Since $EF = 8$ , $AE = AF = 8$ as well. Therefore, the side length of the tetrahedron is $AB = 8 + 3 = 11$ . Using $\angle ABD = 60^{\circ}$ and applying the Law of Cosines on $\triangle BDE$ gives $DE^{2} = 11^{2} + 3^{2} - 2 \cdot 11 \cdot 3 \cdot \cos(60^{\circ}) = 121 + 9 - 33 = 97.$ By symmetry, $DE = DF$ , so we also have $DF^{2} = 97$ . Let $X$ be the foot of the altitude from $D$ in $\triangle DEF$ . Because $\triangle DEF$ is isosceles, $X$ is the midpoint of $\overline{EF}$ and $EX = \tfrac{EF}{2} = \tfrac{8}{2} = 4$ . By the Pythagorean theorem, $DH = \sqrt{97 - 4^{2}} = \sqrt{81} = 9$ , and the area of $\triangle DEF$ is $\tfrac{1}{2} \cdot 9 \cdot 8 = \boxed{\textbf{(C)}~36}$ .

2024 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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
All AMC 12 Problems and Solutions

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2024 AMC 12A Problems/Problem 11

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