婆罗摩笈多公式

四边形给定四条边长 ${\displaystyle a,b,c,d}$, 则它是圆内接四边形时面积最大. 记半周长为 ${\displaystyle s}$, 则最大面积为 $$\sqrt{(s-a)(s-b)(s-c)(s-d)}.$$

^[1] 记四边形为 ${\displaystyle ABCD}$, 且 ${\displaystyle AB=a}$, ${\displaystyle BC=b}$, ${\displaystyle CD=c}$, ${\displaystyle DA=d}$. 则四边形的面积为 $$S=\frac{1}{2}ab\sin B+\frac{1}{2}cd\sin D,$$ 则 $$\begin{array}{rl}{S}^2& =\frac{1}{4}(ad\sin A{)}^2+\frac{1}{4}(bc\sin C{)}^2+\frac{1}{2}abcd\sin A\sin C\\ & =\frac{1}{4}(a^2{d}^2+b^2{c}^2)+\frac{1}{4}(ad\cos A{)}^2+\frac{1}{4}(bc\cos C{)}^2+\frac{1}{2}abcd\sin A\sin C\\ & =\frac{1}{4}(ad+bc{)}^2-\frac{1}{4}(2abcd)+\frac{1}{4}(a^2{d}^2{\cos }^{2}A)+\frac{1}{4}(b^2{c}^2{\cos }^{2}C)+\frac{1}{2}abcd\sin A\sin C.\end{array}$$ 记对角线 ${\displaystyle BD=f}$, 则分别在 ${\displaystyle \mathrm{\Delta }ABD}$, ${\displaystyle \mathrm{\Delta }BCD}$ 中: $$\begin{array}{rl}& f^2=a^2+d^2-2ad\cos A=b^2+c^2-2bc\cos C\\ \Rightarrow & a^2+d^2-b^2-c^2=2ad\cos A-2bc\cos C.\end{array}$$ 等式两边平方得 $$(a^2+d^2-b^2-c^2{)}^2=(2ad\cos A{)}^2+(2bc\cos C{)}^2-8abcd\cos A\cos C.$$ 代入 ${\displaystyle S^2}$ 有$$\begin{array}{rl}{S}^2=& \frac{1}{4}(ad+bc{)}^2-\frac{1}{16}(a^2+d^2-b^2-c^2{)}^2-\frac{1}{4}(2abcd)\\ & -\frac{1}{2}abcd\cos A\cos C+\frac{1}{2}abcd\sin A\sin C\\ =& \frac{1}{4}(ad+bc{)}^2-\frac{1}{16}(a^2+d^2-b^2-c^2{)}^2-\frac{1}{2}abcd\\ & -\frac{1}{2}abcd\cos (A+C)\\ =& \frac{1}{4}(ad+bc{)}^2-\frac{1}{16}(a^2+d^2-b^2-c^2{)}^2-abcd{\cos }^2\frac{A+C}{2}\\ =& \frac{1}{16}(2ad+2bc+a^2+d^2-b^2-c^2)(2ad+2bc-a^2-d^2+b^2+c^2)\\ & -abcd{\cos }^2\frac{A+C}{2}\\ =& \frac{1}{16}((a+d{)}^2-(b-c{)}^2)((b+c{)}^2-(a-d{)}^2)-abcd{\cos }^2\frac{A+C}{2}\\ =& (s-a)(s-b)(s-c)(s-d)-abcd{\cos }^2\frac{A+C}{2}.\end{array}$$ 所以 $$S=\sqrt{(s-a)(s-b)(s-c)(s-d)-abcd{\cos }^2\frac{A+C}{2}}.$$ 它的最小值是 ${\displaystyle \sqrt{(s-a)(s-b)(s-c)(s-d)}}$, 取等条件是 ${\displaystyle A,C}$ 互补, 也即四边形 ${\displaystyle ABCD}$ 是圆内接四边形.