問題の解答
検索用コード（LaTeX）
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% 例題A3.1.4：三角形の外心・内心の角の大きさ（One More）★
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次の図において，$\triangle\mathrm{ABC}$の外心をO，内心をIとするとき，角$x,y$を求めよ．
(1)
(2)
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% 例題A3.1.4の解答（One More）★
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(1) Oは$\triangle\mathrm{ABC}$の外心であるから，
\[
\angle\mathrm{OAB}=\angle\mathrm{OBA}=30^{\circ}
\]
したがって，$\angle\mathrm{OAC}=40^{\circ}$
よって，$x=\angle\mathrm{OAC}=40^{\circ}$
また，$\mathrm{OB}=\mathrm{OC}$であるから，$\angle\mathrm{OBC}=\angle\mathrm{OCB}=y$
したがって，$30^{\circ}+70^{\circ}+40^{\circ}+2y=180^{\circ}$
よって，$y=20^{\circ}$
(2) Iは$\triangle\mathrm{ABC}$の内心であるから，
\[
\angle\mathrm{IBA}=\angle\mathrm{IBC}=35^{\circ},\angle\mathrm{IAB}=\angle\mathrm{IAC}=y
\]
$\triangle\mathrm{ABC}$において，$2y+2\times 35^{\circ}+60^{\circ}=180^{\circ}$
よって，$y=25^{\circ}$
また，$\triangle\mathrm{IAB}$において，$x=180^{\circ}-(\angle\mathrm{IAB}+\angle\mathrm{IBA})=120^{\circ}$
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 問題A3.1.4：三角形の外心・内心の角の大きさ（One More）★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
次の図において，$\triangle\mathrm{ABC}$の外心をO，内心をIとするとき，角$x,y$を求めよ．
(1)
(2)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 問題A3.1.4の解答（One More）★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
(1) Oは$\triangle\mathrm{ABC}$の外心であるから，
\[
\angle\mathrm{OAC}=\angle\mathrm{OCA}=40^{\circ}
\]
したがって，$\angle\mathrm{OAB}=20^{\circ}$
よって，$x=\angle\mathrm{OBA}=20^{\circ}$
$\triangle\mathrm{ABC}$において，$60^\circ+(20^\circ+\angle\mathrm{OBC})+(40^\circ+\angle\mathrm{OCB})=180^\circ$
したがって，$\angle\mathrm{OBC}+\angle\mathrm{OCB}=60^\circ$
よって，$y=180^{\circ}-(\angle\mathrm{OBC}+\angle\mathrm{OCB})=120^{\circ}$
(2) Iは$\triangle\mathrm{ABC}$の内心であるから，
\[
\angle\mathrm{ICA}=\angle\mathrm{ICB}=35^{\circ},\angle\mathrm{IAB}=\angle\mathrm{IAC}=y
\]
$\triangle\mathrm{ABC}$において，$2y+2\times 35^{\circ}+70^{\circ}=180^{\circ}$
よって，$y=20^{\circ}$
また，$\triangle\mathrm{IAC}$において，$x=180^{\circ}-(\angle\mathrm{IAC}+\angle\mathrm{ICA})=125^{\circ}$
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