areageometrypublishingtrianglestrigonometry
I created this trigonometric formula for finding the area of oblique triangles: $$R=\frac{(a^2+b^2-c^2)^2\sec C\tan C}{8ab}$$ where $a, b$ and $c$ are sides of the triangle, angle $C$ is the angle opposite to side $c$, and $R$ is the area.
Does this formula deserve being in textbooks or contests? And where can I publish this or find out if someone already discovered the formula?
Also see this post Have I found a formula for the area of a triangle?
A simple derivation exploits the cosine theorem. We have $\Delta=\frac{1}{2}ab\sin C$, hence $$ 4\Delta^2 = a^2 b^2 \sin^2 C = a^2 b^2 (1-\cos C)(1+\cos C).\tag{1}$$ On the other hand, $ 2ab\cos C = a^2+b^2-c^2$, hence $$ 2ab(1+\cos C) = (a+b)^2-c^2 = (a+b+c)(a+b-c), \tag{2}$$ $$ 2ab(1-\cos C) = c^2-(a-b)^2 = (a-b+c)(-a+b+c),\tag{3}$$ and by multiplying $(2)$ and $(3)$ and exploiting $(1)$ $$ 16\Delta^2 = (a+b+c)(-a+b+c)(a-b+c)(a+b-c)\tag{4} $$ which is equivalent to $$ \Delta = \sqrt{s(s-a)(s-b)(s-c)}\tag{5} $$ as wanted.
Alternative derivation: by considering the circumcenter $O$ and its distances from the sides we have $$2\Delta = R\sum_{cyc}a\cos A,\qquad 16\Delta^2 = \sum_{cyc}a^2\cdot 2bc(\cos A)=\sum_{cyc}a^2(b^2+c^2-a^2)\tag{6}$$ through $4R\Delta=abc$ and the cosine theorem. By rearranging the RHS of $(6)$ $$ 16\Delta^2 = (a^2+b^2+c^2)^2-2(a^4+b^4+c^4)\tag{7} $$ immediately follows.
This is from an article I remember reading in an AMS journal.
Let $D$ be the point of intersection of the internal bisectors of $\angle A$, $\angle B$, and $\angle C$. Let $\alpha$, $\beta$, and $\gamma$ be the measures of the corresponding bisected angles. Note $$\alpha + \beta + \gamma = \frac 12 \pi.$$ Let $r$ be the common distance from the point $D$ to the sides $\overline{AB}$, $\overline{BC}$, and $\overline{CA}$. Define $$s = \frac 12(a+b+c).$$ Note that the feet of the perpendiculars drawn from point $D$ to the sides $\overline{AB}$, $\overline{BC}$, and $\overline{CA}$ create segments with the indicated lengths. Let $u,v,$ and $w$ be the respective lengths of the segments drawn from point $D$ to points $A$, $B$, and $C$. Then
\begin{align} (s-a)+ir = ue^{i \alpha}\\ (s-b)+ir = ve^{i \beta}\\ (s-c)+ir = we^{i \gamma}\\ \end{align}
So $(s-a+ir)(s-b+ir)(s-c+ir)=uvwe^{\frac 12 i\pi}=iuvw$. Hence the real part of $(s-a+ir)(s-b+ir)(s-c+ir)$ is equal to $0$.
We compute \begin{align} \Re[(s-a+ir)(s-b+ir)(s-c+ir)] &= 0 \\ (s-a)(s-b)(s-c)-r^2[(s-a)+(s-b)+(s-c)] &= 0 \\ sr^2 &= (s-a)(s-b)(s-c) \\ r &= \sqrt{\frac{(s-a)(s-b)(s-c)}{s}} \end{align}
Hence the area of $\triangle ABC$ is
$$ \frac 12r(a+b+c) = rs = \sqrt{s(s-a)(s-b)(s-c)} $$
Best Answer
I can do you one better. $$ A=\frac{1}{2}ab \sin C. $$ This identity is equivalent to yours after a use of the Law of Cosines.