AMC 12B 2024 (Problem 23)A right pyramid has regular octagon ABCDEFGHABCDEFGHABCDEFGH with side length 111 as its base and apex VVV. Segments AV‾\overline{AV}AV and DV‾\overline{DV}DV are perpendicular. What is the square of the height of the pyramid?(A) 1\text{(A)}\;1(A)1(B) 1+22\text{(B)}\;\frac{1+\sqrt{2}}{2}(B)21+2(C) 2\text{(C)}\;\sqrt{2}(C)2(D) 32\text{(D)}\;\frac{3}{2}(D)23(E) 2+23\text{(E)}\;\frac{2+\sqrt{2}}{3}(E)32+2Related TopicsCoreRegular PolygonsCheck AnswerYour answer:ABCDECheckHints (4)Hint 1Hint 2By Regular Polygons,AE2=4+22=(2r)2AE^2=4+2\sqrt{2}=(2r)^2AE2=4+22=(2r)2AD2=(2+1)2=3+22AD^2=(\sqrt{2}+1)^2=3+2\sqrt{2}AD2=(2+1)2=3+22Hint 3AV2+DV2=AD2AV^2+DV^2=AD^2AV2+DV2=AD2AV2=r2+h2AV^2=r^2+h^2AV2=r2+h2DV2=h2+r2DV^2=h^2+r^2DV2=h2+r2Therefore, by Hint 2,2(h2+r2)=AD2=3+222(h^2+r^2)=AD^2=3+2\sqrt{2}2(h2+r2)=AD2=3+22Hint 4By Hint 2,AE2=4+22=(2r)2AE^2=4+2\sqrt{2}=(2r)^2AE2=4+22=(2r)2r2=1+22r^2=1+\frac{\sqrt{2}}{2}r2=1+22By Hint 3,2(h2+1+22)=3+222\left(h^2+1+\frac{\sqrt{2}}{2}\right)=3+2\sqrt{2}2(h2+1+22)=3+222h2+2+2=3+222h^2+2+\sqrt{2}=3+2\sqrt{2}2h2+2+2=3+222h2=1+22h^2=1+\sqrt{2}2h2=1+2h2=1+22h^2=\frac{1+\sqrt{2}}{2}h2=21+2Final Answer(B)1+22(B) \tfrac{1+\sqrt{2}}{2}(B)21+2Related Problems (1)AMC 10B/12B 2023 (Problem 25)
