兵庫医科大学 2020年度 数学 Q2
(1) ω 2 = ω − 1 \omega^2=\omega-1 ω 2 = ω − 1
原点中心に角 π / 3 \pi/3 π /3 ω = cos π 3 + i sin π 3 = 1 2 + 3 2 i . \displaystyle
\omega=\cos\frac\pi3+i\sin\frac\pi3=\frac12+\frac{\sqrt3}{2}i. ω = cos 3 π + i sin 3 π = 2 1 + 2 3 i . ω 2 = cos 2 π 3 + i sin 2 π 3 = − 1 2 + 3 2 i = ω − 1. \displaystyle
\omega^2=\cos\frac{2\pi}3+i\sin\frac{2\pi}3
=-\frac12+\frac{\sqrt3}{2}i=\omega-1. ω 2 = cos 3 2 π + i sin 3 2 π = − 2 1 + 2 3 i = ω − 1.
以後、正三角形の外側頂点を表すとき、η \eta η ω \omega ω ω ˉ \bar\omega ω ˉ A B C ABC A B C η \eta η A B C ABC A B C η = ω ˉ \eta=\bar\omega η = ω ˉ η = ω \eta=\omega η = ω ∣ η ∣ = 1 , η 2 = η − 1 , η 3 = − 1 |\eta|=1,\qquad \eta^2=\eta-1,\qquad \eta^3=-1 ∣ η ∣ = 1 , η 2 = η − 1 , η 3 = − 1
(2) △ D E F \triangle DEF △ D E F
外側正三角形の頂点は、辺の一端を中心に辺ベクトルを η \eta η d = β + η ( γ − β ) , e = γ + η ( α − γ ) , f = α + η ( β − α ) . d=\beta+\eta(\gamma-\beta),\quad
e=\gamma+\eta(\alpha-\gamma),\quad
f=\alpha+\eta(\beta-\alpha). d = β + η ( γ − β ) , e = γ + η ( α − γ ) , f = α + η ( β − α ) . d + e + f = α + β + γ . d+e+f=\alpha+\beta+\gamma. d + e + f = α + β + γ . △ D E F \triangle DEF △ D E F α + β + γ 3 . \displaystyle
\boxed{\frac{\alpha+\beta+\gamma}{3}}. 3 α + β + γ .
(3) A D = B E = C F AD=BE=CF A D = B E = C F
p = β − α p=\beta-\alpha p = β − α q = γ − β q=\gamma-\beta q = γ − β X = d − α = p + η q X=d-\alpha=p+\eta q X = d − α = p + η q α − γ = − p − q \alpha-\gamma=-p-q α − γ = − p − q e − β = q + η ( − p − q ) = − η p + ( 1 − η ) q = − η p − η 2 q = − η ( p + η q ) = − η X , \begin{aligned}
e-\beta&=q+\eta(-p-q)\\
&=-\eta p+(1-\eta)q\\
&=-\eta p-\eta^2q=-\eta(p+\eta q)=-\eta X,
\end{aligned} e − β = q + η ( − p − q ) = − η p + ( 1 − η ) q = − η p − η 2 q = − η ( p + η q ) = − η X , η 2 = η − 1 \eta^2=\eta-1 η 2 = η − 1 η 3 = − 1 \eta^3=-1 η 3 = − 1 f − γ = ( − p − q ) + η p = ( η − 1 ) p − q = η 2 p − q = η 2 ( p + η q ) = η 2 X . \begin{aligned}
f-\gamma&=(-p-q)+\eta p\\
&=(\eta-1)p-q=\eta^2p-q=\eta^2(p+\eta q)=\eta^2X.
\end{aligned} f − γ = ( − p − q ) + η p = ( η − 1 ) p − q = η 2 p − q = η 2 ( p + η q ) = η 2 X . ∣ η ∣ = 1 |\eta|=1 ∣ η ∣ = 1 A D = ∣ d − α ∣ = ∣ X ∣ , B E = ∣ e − β ∣ = ∣ − η X ∣ = ∣ X ∣ , C F = ∣ f − γ ∣ = ∣ η 2 X ∣ = ∣ X ∣ . AD=|d-\alpha|=|X|,
\quad BE=|e-\beta|=|-\eta X|=|X|,
\quad CF=|f-\gamma|=|\eta^2X|=|X|. A D = ∣ d − α ∣ = ∣ X ∣ , B E = ∣ e − β ∣ = ∣ − η X ∣ = ∣ X ∣ , C F = ∣ f − γ ∣ = ∣ η 2 X ∣ = ∣ X ∣. A D = B E = C F . \boxed{AD=BE=CF}. A D = B E = C F .
(4) △ M N P \triangle MNP △ M N P
G = ( α + β + γ ) / 3 G=(\alpha+\beta+\gamma)/3 G = ( α + β + γ ) /3 △ A B C \triangle ABC △ A B C 3 ( m − G ) = d − α = X , 3 ( n − G ) = e − β = − η X , 3 ( p − G ) = f − γ = η 2 X . 3(m-G)=d-\alpha=X,\quad
3(n-G)=e-\beta=-\eta X,\quad
3(p-G)=f-\gamma=\eta^2X. 3 ( m − G ) = d − α = X , 3 ( n − G ) = e − β = − η X , 3 ( p − G ) = f − γ = η 2 X . m = G + X 3 , n = G − η X 3 , p = G + η 2 X 3 . \displaystyle
m=G+\frac X3,\qquad
n=G-\frac{\eta X}{3},\qquad
p=G+\frac{\eta^2X}{3}. m = G + 3 X , n = G − 3 η X , p = G + 3 η 2 X . η = e i π / 3 \eta=e^{i\pi/3} η = e iπ /3 e − i π / 3 e^{-i\pi/3} e − iπ /3 ∣ 1 + η ∣ = 3 , ∣ η + η 2 ∣ = ∣ η ∣ ∣ 1 + η ∣ = 3 , ∣ 1 − η 2 ∣ = 3 . |1+\eta|=\sqrt3,
\quad |\eta+\eta^2|=|\eta||1+\eta|=\sqrt3,
\quad |1-\eta^2|=\sqrt3. ∣1 + η ∣ = 3 , ∣ η + η 2 ∣ = ∣ η ∣∣1 + η ∣ = 3 , ∣1 − η 2 ∣ = 3 . ∣ n − m ∣ = ∣ 1 + η ∣ ∣ X ∣ 3 = ∣ η + η 2 ∣ ∣ X ∣ 3 = ∣ 1 − η 2 ∣ ∣ X ∣ 3 = ∣ X ∣ 3 . \displaystyle
|n-m|=\frac{|1+\eta||X|}{3}
=\frac{|\eta+\eta^2||X|}{3}
=\frac{|1-\eta^2||X|}{3}
=\frac{|X|}{\sqrt3}. ∣ n − m ∣ = 3 ∣1 + η ∣∣ X ∣ = 3 ∣ η + η 2 ∣∣ X ∣ = 3 ∣1 − η 2 ∣∣ X ∣ = 3 ∣ X ∣ . A , D A,D A , D B C BC B C X = d − α ≠ 0 X=d-\alpha\ne0 X = d − α = 0 △ M N P \triangle MNP △ M N P
**使用解法:** 回転を複素数の積で表し、重心を頂点の複素数の平均として計算する。距離は複素数の絶対値と単位回転の保存性で比較する。
**issues:** なし。
