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| − | ==Problem==
| + | #redirect[[2024 AMC 10A Problems/Problem 18]] |
| − | 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}}
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