MATHCOUNTS State Team 2025 (Problem 9)Five of the six edges of a tetrahedron each have length 333 inches, and the sixth edge has length 444 inches. What is the volume of the tetrahedron, in cubic inches? Express your answer in simplest radical form.Related TopicsCoreToolkit 82 — 3D Shapes: Surface Area and VolumeMajorToolkit 54 — Area Formulas and Important Geometry FormulasCheck AnswerYour answer:CheckHints (5)Hint 1MMM is the midpoint of BDBDBD. By symmetry, the altitude from AAA to plane BCDBCDBCD lies on line CMCMCM.Hint 2The height of the tetrahedron through AAA is the height of △AMC\triangle AMC△AMC through AAA.Hint 3AM=MC=332,AC=4,AE=?AM=MC=\frac{3\sqrt3}{2},\qquad AC=4,\qquad AE=?AM=MC=233,AC=4,AE=?Hint 4MF2=(332)2−22=274−4=114MF^2=\left(\frac{3\sqrt3}{2}\right)^2-2^2=\frac{27}{4}-4=\frac{11}{4}MF2=(233)2−22=427−4=411⟹MF=112\Longrightarrow MF=\frac{\sqrt{11}}{2}⟹MF=211AE⋅MC2=[AMC]=MF⋅AC2\frac{AE\cdot MC}{2}=[AMC]=\frac{MF\cdot AC}{2}2AE⋅MC=[AMC]=2MF⋅ACAE⋅332=112⋅4AE\cdot\frac{3\sqrt3}{2}=\frac{\sqrt{11}}{2}\cdot4AE⋅233=211⋅4⟹AE=41133\Longrightarrow AE=\frac{4\sqrt{11}}{3\sqrt3}⟹AE=33411Hint 5V=13⋅AE⋅[BCD]V=\frac13\cdot AE\cdot[BCD]V=31⋅AE⋅[BCD]=13⋅41133⋅3234=\frac13\cdot\frac{4\sqrt{11}}{3\sqrt3}\cdot\frac{3^2\sqrt3}{4}=31⋅33411⋅4323=11=\sqrt{11}=11Final Answer11\sqrt{11}11Related Problems (12)MathCounts 2026 Chapter Sprint Round (Problem 28)AMC 12A 2025 (Problem 10)AMC 12A 2025 (Problem 20)AMC 10B 2025 (Problem 6)AMC 10B 2025 (Problem 12)AMC 10B/12B 2025 (Problem 19/15)AMC 12B 2025 (Problem 21)AMC 8 2026 (Problem 11)View all related problems →
