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\begin{document}
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\twocolumn[
\begin{flushleft}
\underline{
{\Large ヘロンの公式の証明}\hspace{5mm}
{\color{cyan!70!blue}\bf 内接円と傍接円を利用した幾何学的な証明}\hspace{60mm} \quad 
} 
\end{flushleft}
]


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


$\triangle ABC$に内接円をかき、その半径を$r$とする。
図１のように、接点でそれぞれの辺を分割すると、
$a=y+z$、$b=x+z$、$c=x+y$と表すことができる。

\definecolor{ccqqqq}{named}{magenta}
\definecolor{cyan}{named}{cyan}
\definecolor{ududff}{named}{black}
\definecolor{xdxdff}{named}{black}
\definecolor{uuuuuu}{named}{black}


\begin{figure}[h]
\caption{内接円の接点で辺を分割}
\begin{tikzpicture}[scale=0.8]
\draw [line width=1pt] (6,4)-- (0,0);
\draw [line width=1pt] (0,0)-- (9,0);
\draw [line width=1pt] (9,0)-- (6,4);
\draw [line width=0.8pt] (5.60555127546399,1.6972243622680052) circle (1.6972243622680052cm);
\draw [line width=1pt,color=ccqqqq] (5.60555127546399,1.6972243622680052)-- (5.6055512754639905,0);
\draw [line width=1pt,color=ccqqqq] (5.60555127546399,1.6972243622680052)-- (6.9633307652783945,2.715558979628808);
\draw [line width=1pt,color=ccqqqq] (5.60555127546399,1.6972243622680052)-- (4.664100666485255,3.1094004443235033);
\draw [line width=1pt,dash pattern=on 4pt off 4pt,color=cyan] (5.60555127546399,1.6972243622680052)-- (0,0);
\draw [line width=1pt,dash pattern=on 4pt off 4pt,color=cyan] (6,4)-- (5.60555127546399,1.6972243622680052);
\draw [line width=1pt,dash pattern=on 4pt off 4pt,color=cyan] (5.60555127546399,1.6972243622680052)-- (9,0);
\draw [shift={(4.485558958652373,-0.22833843797856013)},line width=1pt,dotted]  plot[domain=1.2268665404912604:3.090731320193668,variable=\t]({1*4.491367009475517*cos(\t r)+0*4.491367009475517*sin(\t r)},{0*4.491367009475517*cos(\t r)+1*4.491367009475517*sin(\t r)});
\draw [shift={(6.833333333333333,1.5)},line width=1pt,dotted]  plot[domain=-0.6055446636049702:1.892546881191539,variable=\t]({1*2.635231383473649*cos(\t r)+0*2.635231383473649*sin(\t r)},{0*2.635231383473649*cos(\t r)+1*2.635231383473649*sin(\t r)});
\draw [shift={(4.5,3.375)},line width=1pt,dotted]  plot[domain=3.7850937623830774:5.639684198386302,variable=\t]({1*5.625*cos(\t r)+0*5.625*sin(\t r)},{0*5.625*cos(\t r)+1*5.625*sin(\t r)});
\draw [shift={(5.377582129970967,-3.013597472930759)},line width=1pt,dotted,color=ccqqqq]  plot[domain=1.6867980470837152:2.6307998136012127,variable=\t]({1*6.16442692335936*cos(\t r)+0*6.16442692335936*sin(\t r)},{0*6.16442692335936*cos(\t r)+1*6.16442692335936*sin(\t r)});
\draw [shift={(5.876915696959399,2.737402176586593)},line width=1pt,dotted,color=ccqqqq]  plot[domain=1.4736184242900654:2.8439794363948616,variable=\t]({1*1.2685830715184803*cos(\t r)+0*1.2685830715184803*sin(\t r)},{0*1.2685830715184803*cos(\t r)+1*1.2685830715184803*sin(\t r)});
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\draw [shift={(6.265087886909069,0.0703463680168079)},line width=1pt,dotted,color=ccqqqq]  plot[domain=-0.025715948997707017:1.312718166584276,variable=\t]({1*2.7358166747471695*cos(\t r)+0*2.7358166747471695*sin(\t r)},{0*2.7358166747471695*cos(\t r)+1*2.7358166747471695*sin(\t r)});
\draw [shift={(7.302775637731995,2.2415560151056995)},line width=1pt,dotted,color=ccqqqq]  plot[domain=4.064318983644818:5.360458977124561,variable=\t]({1*2.8116087751912744*cos(\t r)+0*2.8116087751912744*sin(\t r)},{0*2.8116087751912744*cos(\t r)+1*2.8116087751912744*sin(\t r)});
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\draw [fill=uuuuuu] (0,0) node[left]{B};
\draw [fill=xdxdff] (9,0) node[right]{C};
\draw [fill=ududff] (6,4) node[above]{A};
\draw [fill=uuuuuu] (5.60555127546399,1.6972243622680052) circle (2pt)node[left]{$\rm O_1$};
\draw [fill=uuuuuu] (5.6055512754639905,0) node[below]{L};
\draw [fill=uuuuuu] (4.664100666485255,3.1094004443235033) node[above left]{N};
\draw [fill=uuuuuu] (6.9633307652783945,2.715558979628808) node[above right]{M};
\draw [fill=ududff] (1.88,3.43) node[fill=white]{$c$};
\draw [fill=ududff] (9,3)  node[fill=white]{$b$};
\draw [fill=ududff] (4.5,-2.25)  node[fill=white]{$a$};
\draw [fill=ccqqqq] (1.98,2.13) node[fill=white]{\color{ccqqqq}$y$};
\draw [fill=ccqqqq] (5.14,3.9) node[fill=white]{\color{ccqqqq}$x$};
\draw [fill=ccqqqq] (6.78,3.59) node[fill=white]{\color{ccqqqq}$x$};
\draw [fill=ccqqqq] (8.44,1.73) node[fill=white]{\color{ccqqqq}$z$};
\draw [fill=ccqqqq] (7.32,-0.57) node[fill=white]{\color{ccqqqq}$z$};
\draw [fill=ccqqqq] (2.7,-0.77) node[fill=white]{\color{ccqqqq}$y$};
\draw [fill=uuuuuu] (5.6055512754639905,0.9) node[left]{\color{ccqqqq}$r$};
\end{tikzpicture}
\end{figure}

$\triangle ABC$の面積は、
$$S=r(x+y+z)\qquad \cdots\cdots(1)$$
と表すことができる。

\begin{figure}[h]
\caption{$\rm \triangle ABC$の面積}
\begin{tikzpicture}[scale=0.7]
\draw [line width=1pt] (0,0)-- (9,0);
\draw [line width=1pt,color=ccqqqq] (5.60555127546399,1.6972243622680052)-- (5.6055512754639905,0);
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\draw [shift={(5.6055512754639905,0)},line width=0.5pt]  plot[domain=1.5707963267948972:2.8475913518160096,variable=\t]({1*1.6972243622680052*cos(\t r)+0*1.6972243622680052*sin(\t r)},{0*1.6972243622680052*cos(\t r)+1*1.6972243622680052*sin(\t r)});
\draw [shift={(5.60555127546399,1.6972243622680052)},line width=0.5pt]  plot[domain=4.71238898038469:5.81953769817878,variable=\t]({1*1.6972243622680052*cos(\t r)+0*1.6972243622680052*sin(\t r)},{0*1.6972243622680052*cos(\t r)+1*1.6972243622680052*sin(\t r)});
\draw [shift={(9,0)},line width=0.5pt]  plot[domain=1.5707963267948957:2.677945044588987,variable=\t]({1*1.6972243622680052*cos(\t r)+0*1.6972243622680052*sin(\t r)},{0*1.6972243622680052*cos(\t r)+1*1.6972243622680052*sin(\t r)});
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\draw [fill=ududff] (-1.941175475998729,0.8377699691019566) node[fill=white]{\color{ccqqqq}$r$};
\draw [fill=ududff] (-0.8703946395737411,-0.41977496669947467) node[fill=white]{\color{ccqqqq}$x$};
\draw [fill=ududff] (2.8648873479552863,-0.49448060645005476) node[fill=white]{\color{ccqqqq}$y$};
\draw [fill=ududff] (7.372127612906979,-0.4571277865747647) node[fill=white]{\color{ccqqqq}$z$};
\end{tikzpicture}
\end{figure}

図3のように傍接円をかく。傍心を$O_2$、その半径を$R$とする。
$O_2$から三辺へ垂線を下し、その足をX,Y,Zとする。



\begin{figure}[h]
\caption{$\rm \triangle ABC$と傍接円$\rm O_2$}
\begin{tikzpicture}[scale=0.6]
\draw [line width=1pt] (6,4)-- (0,0);
\draw [line width=1pt] (0,0)-- (9,0);
\draw [line width=1pt] (9,0)-- (6,4);
\draw [line width=0.5pt] (5.60555127546399,1.6972243622680052) circle (1.6972243622680052cm);
\draw [shift={(5.377582129970967,-3.013597472930759)},line width=1pt,dotted,color=ccqqqq]  plot[domain=1.6867980470837152:2.6307998136012127,variable=\t]({1*6.16442692335936*cos(\t r)+0*6.16442692335936*sin(\t r)},{0*6.16442692335936*cos(\t r)+1*6.16442692335936*sin(\t r)});
\draw [shift={(5.876915696959399,2.737402176586593)},line width=1pt,dotted,color=ccqqqq]  plot[domain=1.4736184242900654:2.8439794363948616,variable=\t]({1*1.2685830715184803*cos(\t r)+0*1.2685830715184803*sin(\t r)},{0*1.2685830715184803*cos(\t r)+1*1.2685830715184803*sin(\t r)});
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\draw [fill=uuuuuu] (0,0) node[left]{B};
\draw [fill=xdxdff] (9,0) node[below]{C};
\draw [fill=ududff] (6,4) node[above]{A};
\draw [fill=uuuuuu] (5.60555127546399,1.6972243622680052) circle (2pt)node[above]{$\rm O_1$};
\draw [fill=uuuuuu] (5.6055512754639905,0) node[below]{L};
\draw [fill=uuuuuu] (4.664100666485255,3.1094004443235033) circle (2pt);
\draw [fill=uuuuuu] (6.9633307652783945,2.715558979628808) circle (2pt);
\draw [fill=ududff] (1.98,2.13)node[fill=white]{\color{ccqqqq}$y$};
\draw [fill=ududff] (5.0,3.9) node[fill=white]{\color{ccqqqq}$x$};
\draw [fill=ududff] (6.0,3.1) node[fill=white]{\color{ccqqqq}$x$};
\draw [fill=ududff] (7.54,1.01) node[fill=white]{\color{ccqqqq}$z$};
\draw [fill=ududff] (7.32,-0.57) node[fill=white]{\color{ccqqqq}$z$};
\draw [fill=ududff] (2.7,-0.77) node[fill=white]{\color{ccqqqq}$y$};
\draw [fill=xdxdff] (13,0);
\draw [fill=xdxdff] (10.421538461538463,6.947692307692307) ;
\draw [fill=uuuuuu] (10.605551275463986,3.2111025509279774) circle (2pt)node[right]{$\rm O_2$};
\draw [fill=uuuuuu] (10.605551275463986,1.6) node[right]{\color{ccqqqq}$R$};
\draw [fill=uuuuuu] (10.605551275463986,0)  node[below]{X};
\draw [fill=uuuuuu] (8.036669234721606,1.2844410203711918) node[above]{W};
\draw [fill=uuuuuu] (8.824352060364905,5.882901373576602) node[above]{Y};
\end{tikzpicture}
\end{figure}

$\rm AY=AW, \quad CX=CW$
となるので$\rm BX+BY$は$\rm \triangle ABC$の周の長さと等しくなる。
また$\rm BY=BX$であるので、以下の式が成り立つ。
$$
\begin{cases}
AY=z \\
CX=x
\end{cases}
\qquad \cdots\cdots(2)
$$

\begin{figure}[h]
\caption{内接円の半径$t$と傍接円の半径$R$の関係(1)}
\begin{tikzpicture}[scale=0.6]
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%\draw [fill=uuuuuu] (5.6055512754639905,0) circle (2pt);
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%\draw [fill=uuuuuu] (6.9633307652783945,2.715558979628808) circle (2pt);
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\draw [fill=ududff] (2.7,-0.77) node[fill=white]{\color{ccqqqq}$y$};
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\draw [fill=uuuuuu] (5.6055512754639905,0.9) node[left]{\color{ccqqqq}$r$};

\end{tikzpicture}
\end{figure}




図4で、相似な三角形$\triangle O_2 BX$と$\triangle O_1 BL$であるから,

$$
(x+y+z):R=y:r
$$

$$
r(x+y+z)=yR \qquad \cdots\cdots(3)
$$

\definecolor{qqqqff}{named}{cyan}
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\begin{figure}[h]
\caption{内接円の半径$t$と傍接円の半径$R$の関係(2)}
\begin{tikzpicture}[scale=0.6]
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\draw [fill=uuuuuu] (0,0) node[left]{B};
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\end{tikzpicture}
\end{figure}




図5で$CO_1$と$CO_2$は角の二等分線なので、$O_1CO_2=90^\circ$。
したがって、$\triangle O_1LC$と$\triangle XCO_2$は相似となる。
$$
z:r=R:x
$$

$$
rR=xz \qquad \cdots\cdots(4)
$$


さて、(3)式と(4)式を辺々掛けると
$$
r^2R(x+y+z)=xyzR
$$
$R$で割り、両辺に$(x+y+z)$を掛けると、
$$
r^2(x+y+z)^2=(x+y+z)xyz
$$
(1)式により、左辺は$S^2$となるから、
$$
S^2=(x+y+z)xyz\qquad \cdots\cdots(5)
$$

ここで、$s=x+y+z$とすると、図１より、
$$
s=a+x,\quad s=b+y, \quad s=c+z
$$

$$
x=s-a,\quad y=s-b,\quad z=s-c
$$
であるから、 式(5)は以下のように書き直すことができる。

$$
S^2=s(s-a)(s-b)(s-c)
$$

$S,s,a,b,c$は正であるので、

$$
S=\sqrt{s(s-a)(s-b)(s-c)}
$$




\end{document}
%\hfill\underline{\fbox{\scriptsize 答}\quad \hspace{20mm}}\\


