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2023 AMC 12B Problem 17

Problem 17 of 25IntermediateAlgebraGeometry

Triangle ABCABC has side lengths in arithmetic progression, and the smallest side has length 6.6. If the triangle has an angle of 120∘,120^\circ, what is the area of ABC?ABC?

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Solution

Let the sides be 6, 6+d, 6+2d.6,\ 6+d,\ 6+2d. The 120∘120^\circ angle faces the longest side, so (6+2d)2=62+(6+d)2(6+2d)^2=6^2+(6+d)^2 −2⋅6⋅(6+d)cos⁡120∘.-2\cdot 6\cdot(6+d)\cos 120^\circ. Using cos⁡120∘=−12\cos 120^\circ=-\tfrac12 gives 3d2+6d−72=0,3d^2+6d-72=0, so d=4d=4 and the sides are 6,10,14.6,10,14. The area is 12⋅6⋅10⋅sin⁡120∘\tfrac12\cdot 6\cdot 10\cdot\sin 120^\circ =30⋅32=30\cdot\tfrac{\sqrt3}{2} =153.=15\sqrt3. Thus, the correct answer is E.
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Tagged: arithmetic sequence · law of cosines · triangle area

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