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Difference between revisions of "2024 AMC 12A Problems/Problem 11"

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==Problem==
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In regular tetrahedron <math>ABCD</math>, points <math>E</math> and <math>F</math> lie on segments <math>\overline{AB}</math> and <math>\overline{AC}</math>, respectively, such that <math>BE = CF = 3</math>. If <math>EF = 8</math>, what is the area of <math>\triangle DEF</math>?
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<math>\textbf{(A)}~32\qquad\textbf{(B)}~35\qquad\textbf{(C)}~36\qquad\textbf{(D)}~42\qquad\textbf{(E)}~48</math>
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==Solution==
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Note that <math>\triangle AEF</math> is an equilateral triangle. Since <math>EF = 8</math>, <math>AE = AF = 8</math> as well. Therefore, the side length of the tetrahedron is <math>AB = 8 + 3 = 11</math>. Using <math>\angle ABD = 60^{\circ}</math> and applying the Law of Cosines on <math>\triangle BDE</math> gives <cmath>DE^{2} = 11^{2} + 3^{2} - 2 \cdot 11 \cdot 3 \cdot \cos(60^{\circ}) = 121 + 9 - 33 = 97.</cmath> By symmetry, <math>DE = DF</math>, so we also have <math>DF^{2} = 97</math>. Let <math>X</math> be the foot of the altitude from <math>D</math> in <math>\triangle DEF</math>. Because <math>\triangle DEF</math> is isosceles, <math>X</math> is the midpoint of <math>\overline{EF}</math> and <math>EX = \tfrac{EF}{2} = \tfrac{8}{2} = 4</math>. By the Pythagorean theorem, <math>DH = \sqrt{97 - 4^{2}} = \sqrt{81} = 9</math>, and the area of <math>\triangle DEF</math> is <math>\tfrac{1}{2} \cdot 9 \cdot 8 = \boxed{\textbf{(C)}~36}</math>.
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==See also==
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{{AMC12 box|year=2024|ab=A|num-b=10|num-a=12}}
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{{MAA Notice}}

Revision as of 21:42, 20 March 2025

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 $\tfrac{1}{2} \cdot 9 \cdot 8 = \boxed{\textbf{(C)}~36}$.

See also

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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