【数学A】例題3.3.2：三垂線の定理（One More）★★★

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% 例題A3.3.2：三垂線の定理（One More）★★★
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$l$を平面$\alpha$上の直線，$\mathrm{P}$を平面$\alpha$上にない点，$\mathrm{A}$を直線$l$上の点，$\mathrm{O}$を$l$上にない平面$\alpha$上の点とするとき，

$\mathrm{PA}\perp l,\mathrm{OA}\perp l,\mathrm{PO}\perp\mathrm{OA}$ならば，$\mathrm{PO}\perp\alpha$

が成り立つことを示せ．

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% 例題A3.3.2の解答（One More）★★★
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$\mathrm{PA}\perp l,\mathrm{OA}\perp l$より，$l$は平面$\mathrm{AOP}$上の交わる2直線$\mathrm{PA}$，$\mathrm{OA}$に垂直であるから，$l\perp$平面$\mathrm{AOP}$

$\mathrm{PO}$は平面$\mathrm{AOP}$上にあるから，

\[
\mathrm{PO}\perp l\cdots(\mathrm{i})
\]

また，仮定より，$\mathrm{PO}\perp\mathrm{OA}\cdots(\mathrm{ii})$

よって，(i)，(ii)より，直線$\mathrm{PO}$は平面$\alpha$上の交わる2直線$l$，$\mathrm{OA}$と垂直であるから，$\mathrm{PO}\perp\alpha\blacksquare$

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% 例題A3.3.2の別解
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直線$l$上に，$\mathrm{A}$と異なる点$\mathrm{B}$をとる．

三平方の定理より，

\begin{align*}
&\mathrm{PA}^2+\mathrm{AB}^2=\mathrm{PB}^2\cdots(\mathrm{i}),\\
&\mathrm{OA}^2+\mathrm{AB}^2=\mathrm{OB}^2\cdots(\mathrm{ii}),\\
&\mathrm{PO}^2+\mathrm{OA}^2=\mathrm{PA}^2\cdots(\mathrm{iii})
\end{align*}

(i)〜(iii)より，

\[
\mathrm{PO}^2+\mathrm{OB}^2=\mathrm{PB}^2
\]

三平方の定理の逆より，$\angle\mathrm{POB}=90^{\circ}$，すなわち，$\mathrm{PO}\perp\mathrm{OB}\cdots(\mathrm{iv})$

また，仮定より，$\mathrm{PO}\perp\mathrm{OA}\cdots(\mathrm{v})$

よって，(iv)，(v)より，直線$\mathrm{PO}$は平面$\alpha$上の交わる2直線$\mathrm{OB}$，$\mathrm{OA}$と垂直であるから，$\mathrm{PO}\perp\alpha\blacksquare$

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% 問題A3.3.2：三垂線の定理（One More）★★★
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$l$を平面$\alpha$上の直線，$\mathrm{P}$を平面$\alpha$上にない点，$\mathrm{A}$を直線$l$上の点，$\mathrm{O}$を$l$上にない平面$\alpha$上の点とするとき，次のことを示せ．

(1) $\mathrm{PO}\perp\alpha,\mathrm{PA}\perp l$ならば$\mathrm{OA}\perp l$

(2) $\mathrm{PO}\perp\alpha,\mathrm{OA}\perp l$ならば$\mathrm{PA}\perp l$

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(1) $\mathrm{PO}\perp\alpha$より，$\mathrm{PO}\perp l\cdots(\mathrm{i})$

仮定より，$\mathrm{PA}\perp l\cdots(\mathrm{ii})$

(i)，(ii)より，$l$は平面$\mathrm{AOP}$上の交わる2直線$\mathrm{PA},\mathrm{PO}$に垂直であるから，$l\perp\text{平面}\mathrm{AOP}$

よって，$\mathrm{OA}$は平面$\mathrm{AOP}$上にあるから，$\mathrm{OA}\perp l\blacksquare$

(2) $\mathrm{PO}\perp\alpha$より，$\mathrm{PO}\perp l\cdots(\mathrm{i})$

仮定より，$\mathrm{OA}\perp l\cdots(\mathrm{ii})$

(i)，(ii)より，$l$は平面$\mathrm{AOP}$上の交わる2直線$\mathrm{PO},\mathrm{OA}$に垂直であるから，$l\perp\text{平面}\mathrm{AOP}$

よって，$\mathrm{PA}$は平面$\mathrm{AOP}$上にあるから，$\mathrm{PA}\perp l\blacksquare$

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% 例題A3.3.2の側注：三垂線の定理（One More）★★★
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% 側注1
直線PA，OAは点Aで交わる．

% 側注2
$\triangle\mathrm{PAB}$，$\triangle\mathrm{OAB}$，$\triangle\mathrm{AOP}$は直角三角形である．

% 側注3
(ii)より，
\[
\mathrm{AB}^2=\mathrm{OB}^2-\mathrm{OA}^2
\]
これと(iii)を(i)に代入すると，

\[
\mathrm{PO}^2+\mathrm{OB}^2=\mathrm{PB}^2
\]
が得られる．

% 側注4
直線$l,$OAは点Aで交わる．

% 側注5
三垂線の定理のうちの1つが示された．

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% 問題A3.3.2の側注：三垂線の定理（One More）★★★
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% 側注1
直線PA，POは点Pで交わる．

% 側注2
直線PO，OAは点Oで交わる．

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% 例題A3.3.2の考え方・補足など：三垂線の定理（One More）★★★
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% 考え方
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この例題と下の問題は，三垂線の定理といわれる定理を証明している．$\mathrm{PO}\perp\alpha$を示すためには，直線$\mathrm{PO}$が平面$\alpha$上の交わる2直線に垂直であることを示せばよい．そこで，直線$l$に注目し，$\mathrm{PO}\perp l$を示すことから考えるとよい．