【数学I】例題4.2.6：三角形の成立条件（One More）★★★

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% 例題I4.2.6：三角形の成立条件（One More）★★★
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$\triangle\mathrm{ABC}$において，$\mathrm{AB}=x,\mathrm{AC}=x+1,\mathrm{BC}=x+2$のとき，次の問いに答えよ．

(1) $x$のとり得る値の範囲を求めよ．

(2) $\triangle\mathrm{ABC}$が鈍角三角形となる$x$の値の範囲を求めよ．

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% 例題I4.2.6の解答（One More）★★★
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(1) $x<x+1<x+2$であるから，三角形が成立する条件より，
\[
x+2<x+(x+1)
\]

よって，$x$のとり得る値の範囲は，$x>1\cdots(\mathrm{i})$

(2) 辺$\mathrm{BC}$が最大の辺であるから，鈍角三角形となる条件は$A>90^{\circ}$，すなわち，$\cos A<0$である．

余弦定理より，

\[
\cos A=\frac{x^2+(x+1)^2-(x+2)^2}{2x(x+1)}<0
\]

したがって，$x^2+(x+1)^2-(x+2)^2<0$

整理すると，$x^2-2x-3<0$

ゆえに，$(x-3)(x+1)<0$

これを解くと，$-1<x<3\cdots(\mathrm{ii})$

よって，(i)，(ii)より，求める$x$の値の範囲は，$1<x<3$

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% 問題I4.2.6：三角形の成立条件（One More）★★★
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3辺の長さが$2,3,x$である三角形について，次の問いに答えよ．

(1) $x$のとり得る値の範囲を求めよ．

(2) この三角形が鈍角三角形となるような$x$の値の範囲を求めよ．

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(1) 三角形が成立する条件より，$|3-2|<x<3+2$

よって，$x$のとり得る値の範囲は，$1<x<5\cdots(\mathrm{i})$

(2)

（ア）$1<x<3$のとき

最大の辺の長さは3であるから，その対角が$90^\circ$より大きいとき，鈍角三角形となる．

したがって，$3^2>2^2+x^2$

整理すると，$x^2-5<0$

ゆえに，$(x+\sqrt{5})(x-\sqrt{5})<0$

これを解くと，$-\sqrt{5}<x<\sqrt{5}\cdots(\mathrm{ii})$

(i)，(ii)の共通範囲を求めると，${1<x<\sqrt{5}}$

（イ）$3\leqq x<5$のとき

最大の辺の長さは$x$であるから，その対角が$90^\circ$より大きいとき，鈍角三角形となる．

したがって，$x^2>2^2+3^2$

整理すると，$x^2-13>0$

ゆえに，$(x+\sqrt{13})(x-\sqrt{13})>0$

これを解くと，$x<-\sqrt{13},\sqrt{13}<x\cdots(\mathrm{iii})$

(i)，(iii)の共通範囲を求めると，$\sqrt{13}<x<5$

よって，（ア），（イ）より，求める$x$の値の範囲は，$1<x<\sqrt{5},\sqrt{13}<x<5$

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% 例題I4.2.6の側注：三角形の成立条件（One More）★★★
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% 側注1
最大の辺の長さは，他の2辺の長さの和より小さい．

% 側注2
三角形の成立する条件が成り立つならば，3辺の長さが正である条件は考えなくてよい．

% 側注3
鈍角になることができるのは，最大の角のみである．

% 側注4
(i)より，分母が正であるから，分子は負である．

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% 側注1
最大の辺になることができる辺は，$3$と$x$の2通りあるので注意すること．

% 側注2
三角形の成立条件$|b-c|<a<b+c$を利用する．$|x-3|<2<x+3$または$|2-x|<3<2+x$を解いてもよいが，計算に手間が掛かる．

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% 例題I4.2.6の考え方・補足など：三角形の成立条件（One More）★★★
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(1)は，例えば，3辺の長さが$\frac{1}{2},\frac{3}{2},\frac{5}{2}$であるとき，右の図のようになり三角形を作れない（辺の長さが足りない）．三角形ができるためには，$a+b>c$が成り立たなければならない．なお，解答では余弦定理を用いたが，$A\text{が鈍角}\Longleftrightarrow b^2+c^2<a^2$を用いても同じ結果が得られる．

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$A$が鋭角$\Leftrightarrow\cos A>0\Leftrightarrow b^2+c^2>a^2$，$A$が直角$\Leftrightarrow\cos A=0\Leftrightarrow b^2+c^2=a^2$，$A$が鈍角$\Leftrightarrow\cos A<0\Leftrightarrow b^2+c^2<a^2$

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\begin{align*}
a,b,c\text{を3辺の長さとする三角形が成立する}&\Longleftrightarrow a<b+c\text{かつ}b<c+a\text{かつ}c<a+b\\
&\Longleftrightarrow|b-c|<a<b+c\\
&\Longleftrightarrow\text{(最大の辺の長さ)}<\text{(他の2辺の長さの和)}
\end{align*}

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(1)では3辺の長さが正である条件を考えなかったが，一般に，$a<b+c\cdots(\mathrm{i}),b<c+a\cdots(\mathrm{ii}),c<a+b$が成り立つとき，$a>0,b>0,c>0$も成り立つ．

例えば，(i)と(ii)の辺々を足し合わせると，$a+b<a+b+2c$

よって，$0<2c$，すなわち，$c>0$が成り立つ．同様に，$a>0,b>0$も成り立つことがわかる．