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数学・算数関連を気ままに

2020年度　センター試験　数学I　第3問

センター試験

(1)
f:id:coco_math_1801:20200126185019j:plain:w300
余弦定理より,
\cos{\angle{ABC}} = \dfrac{AB^{2}+BC^{2}-CA^{2}}{2\cdot AB\cdot BC} = \dfrac{25+36-21}{2\cdot 5\cdot 6} = \dfrac{40}{60} = \dfrac{2}{3}

\sin{\angle{ABC}} = \sqrt{1-\left( \dfrac{2}{3} \right) ^{2}} = \sqrt{1 - \dfrac{4}{9}} = \dfrac{\sqrt{5}}{3}

\triangle{ABC} = \dfrac{1}{2} \cdot AB\cdot BC \cdot \sin{\angle{ABC}} = \dfrac{1}{2} \cdot 5 \cdot 6 \cdot \dfrac{\sqrt{5}}{3} = 5\sqrt{5}

→ア=2, イ=3, ウ=5, エ=3, オ=5, カ=5


(2)
f:id:coco_math_1801:20200126190049j:plain:w300
(i)
\cos{\angle{DIG}} = \dfrac{3}{5}より,
IG: GD: DI = 3: 4: 5
\therefore DI = GD \dfrac{5}{4} = 10
IG = GD \dfrac{3}{4} = 6
IF =GF - IG = 8-6 = 2

\tan{\angle{FIH}} = 2より,
IF:FH:HI = 1:2:\sqrt{5}
\therefore HI = IF \cdot sqrt{5} = 2\sqrt{5}

→キ=3, ク=1


(ii)
DH = \sqrt{HE^{2}+DE^{2}} =\sqrt{4^{2}+8^{2}} = 4\sqrt{5}

HF: FI: IH =2:1:\sqrt{5}
DE:EH:HD = 8:4:4\sqrt{5} =2:1:\sqrt{5}
DG:GI:ID = 8:6:10=4:3:5
DH:HI:ID =4\sqrt{5}:2\sqrt{5}:10 = 2:1:\sqrt{5}
\therefore \triangle{HFI}と相似なものは,
\triangle{DEH}と \triangle{DHI}
また, \tan{\angle{DIG}}=\dfrac{8}{6} =\dfrac{4}{3}
\tan{\angle{DIH}} = \dfrac{4\sqrt{5}}{2\sqrt{5}} =2
\therefore \tan{\angle{DIG}} < \tan{\angle{DIH}}より,
\angle{DIG} < \angle{DIH}

→ケ=1, コ=3


(iii)
f:id:coco_math_1801:20200126192634j:plain:w300
\angle{DHI} = 90^{\circ}より, DIは外接円の直径
\therefore外接円の半径は,
\dfrac{1}{2} \cdot DI = 5

\triangle{DHI}=\dfrac{1}{2} \cdot DH \cdot HI = \dfrac{1}{2} \cdot 4\sqrt{5} \cdot 2\sqrt{5} = 20
内接円の半径を rとおくと,
\dfrac{1}{2}\left(DH+HI+ID \right)r =20
\Leftrightarrow r = \dfrac{2\cdot 20}{4\sqrt{5}+2\sqrt{5}+10}=\dfrac{40}{2\sqrt{5}(3+\sqrt{5})}=\dfrac{4\sqrt{5}(3-\sqrt{5})}{4} =3\sqrt{5}-5

→サ=5, シ=3, ス=5, セ=5


(3)
f:id:coco_math_1801:20200126195923j:plain:w300
DJ = \sqrt{\left(4\sqrt{5} \right)^{2} + 8^{2}} = 12

f:id:coco_math_1801:20200126195031j:plain:w300
IJ =\sqrt{8^2+\left(2\sqrt{5}\right)^2} = 2\sqrt{21}

f:id:coco_math_1801:20200126200139j:plain:w300
\triangle{IDJ}は \triangle{ABC}を2倍に拡大したものだから,
\triangle{IDJ} = 2^{2} \cdot \triangle{ABC} = 4\cdot 5\sqrt{5}=20\sqrt{5}

四面体 JDHIの体積を Vとおくと,
V = \dfrac{1}{3}\triangle{DHI} \cdot HJ = \dfrac{1}{3}\triangle{IDJ}\cdot HK
\Leftrightarrow HK = HJ \cdot \dfrac{\triangle{DHI}}{\triangle{IDJ}}=8\cdot \dfrac{20}{20\sqrt{5}}=\dfrac{8\sqrt{5}}{5}

→ソタ=20, チ=5, ツ=8, テ=5, ト=5