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  • » 2025.02
ヘロンの公式
ヘロンの公式 をmathで作ってみた。
・三角形ABCを考える A ⃗ + B ⃗ = C ⃗ ・余弦定理より B ⃗ 2 = C ⃗ 2 − 2 C ⃗ ⋅ A ⃗ ⋅ + A ⃗ 2 C ⃗ ⋅ A ⃗ = B ⃗ 2 − C ⃗ 2 − A ⃗ 2 2 ・垂線のベクトル P ⃗ を求める C ⃗ ⋅ P ⃗ = C ⃗ ⋅ ( t C ⃗ − A ⃗ ) = 0 t = C ⃗ ⋅ A ⃗ C ⃗ 2 ・ P ⃗ を求める P ⃗ 2 = C ⃗ ⋅ A ⃗ C ⃗ 2 C ⃗ − A ⃗ 2 = C ⃗ ⋅ A ⃗ C ⃗ 2 C ⃗ 2 − 2 ( C ⃗ ⋅ A ⃗ ) 2 C ⃗ 2 + A ⃗ 2 = A ⃗ 2 − ( C ⃗ ⋅ A ⃗ ) 2 C ⃗ 2 = ( C ⃗ A ⃗ ) 2 − ( C ⃗ ⋅ A ⃗ ) 2 C ⃗ 2 = ( C ⃗ A ⃗ − C ⃗ ⋅ A ⃗ ) ( C ⃗ A ⃗ + C ⃗ ⋅ A ⃗ ) C ⃗ 2 = C ⃗ A ⃗ − B ⃗ 2 − C ⃗ 2 − A ⃗ 2 2 C ⃗ A ⃗ + B ⃗ 2 − C ⃗ 2 − A ⃗ 2 2 C ⃗ 2 = 2 C ⃗ A ⃗ − B ⃗ 2 + C ⃗ 2 + A ⃗ 2 2 C ⃗ A ⃗ + B ⃗ 2 − C ⃗ 2 − A ⃗ 2 4 C ⃗ 2 = ( C ⃗ + A ⃗ ) 2 − B ⃗ 2 B ⃗ 2 − ( C ⃗ − A ⃗ ) 2 4 C ⃗ 2 = ( C ⃗ + A ⃗ ) − B ⃗ ( C ⃗ + A ⃗ ) + B ⃗ B ⃗ − ( C ⃗ − A ⃗ ) B ⃗ + ( C ⃗ − A ⃗ ) 4 C ⃗ 2 = 4 s − b s s − c s − a c 2 よって三角形の面積は 1 2 C ⃗ P ⃗ = 1 2 c 4 s − b s s − c s − a c 2 = s s − b s − c s − a "・三角形ABCを考える" newline widevec{A}+widevec{B}=widevec{C} newline "・余弦定理より" newline {abs{widevec{B}}}^2={abs{widevec{C}}}^2-2 widevec{C}cdot widevec{A} cdot+{abs{widevec{A}}}^2 newline {vec{C}cdot vec{A}}={{{abs{vec{B}}}^2-{abs{vec{C}}}^2-{abs{vec{A}}}^2}}over{2} newline "・垂線のベクトル" vec{P} "を求める" newline vec{C}cdot vec{P} = widevec{C}cdot (t widevec{C}-widevec{A})=0 newline t={{vec{C}cdot vec{A}}over{{abs{vec{C}}}^2}} newline "・" {abs{vec{P}}} "を求める" newline {abs{vec{P}}}^2={abs{{{vec{C}cdot vec{A}}over{{abs{vec{C}}}^2}}vec{C}-vec{A}}}^2={abs{{{vec{C}cdot vec{A}}over{{abs{vec{C}}}^2}}vec{C}}}^2-2{{(vec{C}cdot vec{A})}^2 over{{abs{vec{C}}}^2}}+{abs{vec{A}}}^2 newline {}={{abs{vec{A}}}^2-{(vec{C}cdot vec{A})}^2 over{{abs{vec{C}}}^2}} newline {}={{({abs{vec{C}}}{abs{vec{A}}})^2-(vec{C}cdot vec{A})}^2 over{{abs{vec{C}}}^2}} newline {}={{({abs{vec{C}}}{abs{vec{A}}}-vec{C}cdot vec{A})({abs{vec{C}}}{abs{vec{A}}}+vec{C}cdot vec{A})} over{{abs{vec{C}}}^2}} newline {}={{left({abs{vec{C}}}{abs{vec{A}}}-{{{abs{vec{B}}}^2-{abs{vec{C}}}^2-{abs{vec{A}}}^2}}over{2} right)left({abs{vec{C}}}{abs{vec{A}}}+{{{abs{vec{B}}}^2-{abs{vec{C}}}^2-{abs{vec{A}}}^2}}over{2} right)} over{{abs{vec{C}}}^2}} newline {}={{left({2{abs{vec{C}}}{abs{vec{A}}}-{{abs{vec{B}}}^2+{abs{vec{C}}}^2+{abs{vec{A}}}^2}} right)left({2{abs{vec{C}}}{abs{vec{A}}}+{{abs{vec{B}}}^2-{abs{vec{C}}}^2-{abs{vec{A}}}^2}} right)} over {4{{abs{vec{C}}}^2}}} newline {}={{left(({abs{vec{C}}}+{abs{vec{A}}})^2-{{abs{vec{B}}}^2} right)left({{{abs{vec{B}}}^2-({abs{vec{C}}}-{abs{vec{A}}})^2}} right)} over {4{{abs{vec{C}}}^2}}} newline {}={{left(({abs{vec{C}}}+{abs{vec{A}}})-{{abs{vec{B}}}} right)left(({abs{vec{C}}}+{abs{vec{A}}})+{{abs{vec{B}}}} right)left({{{abs{vec{B}}}-({abs{vec{C}}}-{abs{vec{A}}})}} right)left({{{abs{vec{B}}}+({abs{vec{C}}}-{abs{vec{A}}})}} right)} over {4{{abs{vec{C}}}^2}}} newline {}={{4 left(s-b right)left(s right)left(s-c right)left(s-a right)} over {{c^2}}} newline newline "よって三角形の面積は" newline {{1} over {2}}{abs{vec{C}}}{abs{vec{P}}}={{1} over {2}}c{sqrt{{{4 left(s-b right)left(s right)left(s-c right)left(s-a right)} over {{c^2}}}}} newline {}={sqrt{{{left(s right)left(s-b right)left(s-c right)left(s-a right)}}}}

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【2013/01/27 22:05 】 | 論理 | 有り難いご意見(0)
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