卡諾定理 (內切圓、外接圓)

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设ABC为三角形，O为其外心。则O到ABC各边的距离之和为

${\displaystyle {\begin{aligned}&{}\qquad DG+DH+DF\\&{}=|DG|+|DH|-|DF|\\&{}=R+r\end{aligned}}}$

${\displaystyle OO_{A}+OO_{B}+OO_{C}=R+r}$，

其中r为内切圆半径，R为外接圆半径。这个定理叫做卡诺定理（法語：Théorème de Carnot），以拉扎尔·卡诺為名。

引理

在${\displaystyle \triangle ABC}$ 中，${\displaystyle R}$ 為${\displaystyle \triangle ABC}$ 之外接圓半徑，且${\displaystyle r}$ 為${\displaystyle \triangle ABC}$ 之內切圓半徑，則

${\displaystyle r=4R\sin({\frac {A}{2}})\sin({\frac {B}{2}})\sin({\frac {C}{2}})}$ 

證明

假設${\displaystyle \triangle ABC}$ 為銳角三角形，${\displaystyle D}$ 為${\displaystyle \triangle ABC}$ 之外接圓圓心，${\displaystyle D}$ 至${\displaystyle \triangle ABC}$ 三邊之距離分別為${\displaystyle {\overline {DG}}}$ 、${\displaystyle {\overline {DH}}}$ 、${\displaystyle {\overline {DF}}}$ ，其中${\displaystyle {\overline {DG}}}$ 為${\displaystyle D}$ 至${\displaystyle {\overline {AB}}}$ 之距離，${\displaystyle {\overline {DH}}}$ 為${\displaystyle D}$ 至${\displaystyle {\overline {BC}}}$ 之距離，${\displaystyle {\overline {DF}}}$ 為${\displaystyle D}$ 至${\displaystyle {\overline {AC}}}$ 之距離。連接${\displaystyle D}$ 與${\displaystyle B}$ ，在${\displaystyle \triangle HDB}$ 中，根據三角形外心性質，可以得到

${\displaystyle {\overline {DB}}=R}$ 

${\displaystyle \angle {HDB}=\angle {A}}$ 

所以，可以得到${\displaystyle {\overline {DH}}}$ 的表示式，

${\displaystyle {\overline {DH}}=R\cos(A)}$ 

同理，亦可得到${\displaystyle {\overline {DG}}}$ 和${\displaystyle {\overline {DF}}}$ 的表示式，

${\displaystyle {\overline {DG}}=R\cos(C)}$ 

${\displaystyle {\overline {DF}}=R\cos(B)}$ 

因此，

${\displaystyle {\overline {DG}}+{\overline {DH}}+{\overline {DF}}\,}$ 

${\displaystyle =R(\cos(A)+\cos(B)+\cos(C))\,}$ 

${\displaystyle =R(2\cos({\frac {A+B}{2}})\cos({\frac {A-B}{2}})+1-2\sin ^{2}({\frac {C}{2}}))\,}$ 

${\displaystyle =R(2\cos({\frac {\pi -C}{2}})\cos({\frac {A-B}{2}})+1-2\sin({\frac {\pi -(A+B)}{2}})\sin({\frac {C}{2}}))\,}$ 

${\displaystyle =R(2\sin({\frac {C}{2}})\cos({\frac {A-B}{2}})+1-2\cos({\frac {(A+B)}{2}})\sin({\frac {C}{2}}))\,}$ 

${\displaystyle =R(2\sin({\frac {C}{2}})(\cos({\frac {A-B}{2}})-\cos({\frac {(A+B)}{2}}))+1)\,}$ 

${\displaystyle =R(4\sin({\frac {A}{2}})\sin({\frac {B}{2}})\sin({\frac {C}{2}})+1)\,}$ 

${\displaystyle =4R\sin({\frac {A}{2}})\sin({\frac {B}{2}})\sin({\frac {C}{2}})+R\,}$ 

根據引理，即可得證，

${\displaystyle {\overline {DG}}+{\overline {DH}}+{\overline {DF}}=R+r}$ 

此外，若${\displaystyle \triangle ABC}$ 為鈍角三角形，且${\displaystyle \angle {B}}$ 大於${\displaystyle 90}$ 度，其餘符號假設均與上面相同，則可以得到，

${\displaystyle {\overline {DH}}=R\cos(A)\,}$ 

${\displaystyle {\overline {DF}}=R\cos(\pi -B)=-R\cos(B)\,}$ 

${\displaystyle {\overline {DG}}=R\cos(C)\,}$ 

所以，

${\displaystyle {\overline {DG}}+{\overline {DH}}-{\overline {DF}}\,}$ 

${\displaystyle =R(\cos(A)+\cos(B)+\cos(C))\,}$ 

${\displaystyle =R+r\,}$ 

故得證卡諾定理。

參考資料

• Perrier, Frédéric. Carnot's Theorem in Trigonometric Disguise. The Mathematical Gazette. 2007-03, 91 (520): 115-117 [2023-05-15]. （原始内容于2020-10-20）.