【数学A】例題3.2.4：方べきの定理（One More）★

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% 例題A3.2.4：方べきの定理（One More）★
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次の図において，$x$の値を求めよ．ただし，(2)では，$\mathrm{O}$を円の中心とし，$\mathrm{PT}$を点$\mathrm{T}$における接線とする．

(1)

(2)

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% 例題A3.2.4の解答（One More）★
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(1) $\mathrm{PB}=7+x,\mathrm{PD}=6+8=14$であるから，方べきの定理より，$\mathrm{PA}\cdot\mathrm{PB}=\mathrm{PC}\cdot\mathrm{PD}$

したがって，

\[
7\cdot(7+x)=6\cdot 14
\]

よって，$x=5$

(2) 方べきの定理より，$\mathrm{PA}\cdot\mathrm{PB}=\mathrm{PT}^2$

したがって，$2\cdot 8=\mathrm{PT}^2$より，$\mathrm{PT}^2=16$

$\triangle\mathrm{PTB}$は直角三角形であるから，三平方の定理より，

\[
\mathrm{PB}^2=\mathrm{PT}^2+\mathrm{BT}^2
\]

ゆえに，$8^2=16+(2x)^2$より，$x^2=12$

よって，$x>0$より，$x=2\sqrt{3}$

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% 問題A3.2.4：方べきの定理（One More）★
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次の図において，$x$の値を求めよ．ただし，(2)では，$\mathrm{O}$を円の中心とし，$\mathrm{PT}$を点$\mathrm{T}$における接線とする．

(1)

(2)

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% 問題A3.2.4の解答（One More）★
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(1) COの延長と円との交点をDとすると，
\[
\mathrm{PD}=2x+2
\]

方べきの定理より，$\mathrm{PA}\cdot\mathrm{PB}=\mathrm{PC}\cdot\mathrm{PD}$

したがって，$4\cdot 4=2\cdot(2x+2)$

よって，$x=3$

(2) 方べきの定理より，$\mathrm{PA}\cdot\mathrm{PB}=\mathrm{PT}^2$したがって，$2\sqrt{2}\cdot 4\sqrt{2}=\mathrm{PT}^2$より，$\mathrm{PT}^2=16$

$\triangle\mathrm{PTB}$は直角三角形であるから，三平方の定理より，

\[
\mathrm{PB}^2=\mathrm{PT}^2+\mathrm{BT}^2
\]

ゆえに，$(4\sqrt{2})^2=16+(2x)^2$より，$x^2=4$

よって，$x>0$より，$x=2$

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% 例題A3.2.4の側注：方べきの定理（One More）★
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% 側注1
$\mathrm{PB}=2+6=8$

% 側注2
$\triangle\mathrm{PTB}$は$1:2:\sqrt{3}$の直角三角形であることを利用して，$x$の値を求めてもよい．

% 側注3
$\mathrm{BT}=2\mathrm{OT}=2x$

% 側注4
\begin{align*}
&\mathrm{PB}=\mathrm{PA}+\mathrm{AB},\\
&\mathrm{PD}=\mathrm{PC}+\mathrm{CD}
\end{align*}

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% 問題A3.2.4の側注：方べきの定理（One More）★
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% 側注1
半径は，$\mathrm{OP}+\mathrm{PC}=x+2$であり，$\mathrm{PD}=\mathrm{PO}+\mathrm{OD}$より，$\mathrm{PD}=2x+2$

% 側注2
$\mathrm{PB}=2\sqrt{2}+2\sqrt{2}=4\sqrt{2}$

% 側注3
$\triangle\mathrm{PTB}$は$1:1:\sqrt{2}$の直角三角形であることを利用して，$x$の値を求めてもよい．

% 側注4
$\mathrm{BT}=2\mathrm{OT}=2x$

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% 例題A3.2.4の考え方・補足など：方べきの定理（One More）★
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方べきの定理を用いて，$x$の値を求める．

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% 方べきの定理
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(1) ，$(1)^\prime$の図において，

\[
{\mathrm{PA}\cdot\mathrm{PB}=\mathrm{PC}\cdot\mathrm{PD}}
\]

また，(2)の図において，

\[
{\mathrm{PA}\cdot\mathrm{PB}=\mathrm{PT}^2}
\]

(1)

$(1)^\prime$

(2)