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 FormulasHints (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 (3)MathCounts 2026 Chapter Sprint Round (Problem 28)AMC 10B/12B 2024 (Problem 14/9)AMC 12A 2024 (Problem 20)
