【数学A】例題3.2.11：2次方程式の解と作図（One More）★★

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% 例題A3.2.11：2次方程式の解と作図（One More）★★
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長さ$a,b$の線分が与えられたとき，2次方程式$x^2\pm ax-b^2=0$の正の解を長さとする線分を作図せよ．

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% 例題A3.2.11の解答（One More）★★
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長さ$a$の線分$\mathrm{AB}$を直径とする円$\mathrm{O}$をかく．

[2]円O上の点$\mathrm{B}$を通る線分$\mathrm{AB}$の垂線を引き，この直線上で$\mathrm{BC}=b$となる点$\mathrm{C}$をとる．

[3]点$\mathrm{C}$と線分$\mathrm{AB}$の中点である$\mathrm{O}$を結ぶ直線と円$\mathrm{O}$との交点を，右の図のように$\mathrm{D}$，$\mathrm{E}$とすると，線分$\mathrm{CD}$，$\mathrm{CE}$が求める線分である．

$x^2+ax-b^2=0$より，$x(x+a)=b^2$であり，$x^2-ax-b^2=0$より，$x(x-a)=b^2$である．

また，$\mathrm{CB}$は円に接するから，方べきの定理より，$\mathrm{CD}\cdot\mathrm{CE}=\mathrm{CB}^2=b^2$

$\mathrm{CD}=x$とすると，$\mathrm{CE}=x+a$より，$x(x+a)=b^2$

$\mathrm{CE}=x$とすると，$\mathrm{CD}=x-a$より，$x(x-a)=b^2$

よって，線分CD，CEは2次方程式$x^2\pm ax-b^2=0$の正の解を長さとする線分である．

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% 問題A3.2.11：2次方程式の解と作図（One More）★★
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長さ$1$の線分が与えられたとき，2次方程式$x^2+4x-1=0$の正の解を長さとする線分を作図せよ．

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% 問題A3.2.11の解答（One More）★★
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$x^2+4x-1=0$より，$x(x+4)=1^2$

[1]長さ$4$の線分$\mathrm{AB}$を直径とする円$\mathrm{O}$をかく．

[2]円O上の点$\mathrm{B}$を通る線分$\mathrm{AB}$の垂線を引き，この直線上で$\mathrm{BC}=1$となる点$\mathrm{C}$をとる．

[3]点$\mathrm{C}$と$\mathrm{AB}$の中点である$\mathrm{O}$を結ぶ直線と円$\mathrm{O}$との交点を，右の図のように$\mathrm{D}$，$\mathrm{E}$とすると，線分$\mathrm{CD}$が求める線分である．

$\mathrm{CB}$は円に接するから，方べきの定理より，$\mathrm{CD}\cdot\mathrm{CE}=\mathrm{CB}^2=1^2$

$\mathrm{CD}=x$とすると，$\mathrm{CE}=x+4$より，$x(x+4)=1^2$

よって，線分CDは2次方程式$x^2+4x-1=0$の正の解を長さとする線分である．

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% 問題A3.2.11の側注：2次方程式の解と作図（One More）★★
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% 側注1
作図によって得られた図形が条件を満たすことを確認する．

% 側注2
$x=-2+\sqrt{5}$

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% 例題A3.2.11の考え方・補足など：2次方程式の解と作図（One More）★★
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% 考え方
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2次方程式$x^2\pm ax-b^2=0$を変形すると，$x(x\pm a)=b^2$となる．そこで，方べきの定理$\mathrm{PA}\cdot\mathrm{PB}=\mathrm{PT}^2$を利用することを考える．