Circumcenter Cevians Take 2

If $R$ is the outradius, define $u = AA_1-R$ and $v = BB_1-R$ .

Set up a coordinate system with the triangle outcentre at the origin, and with $B$ having coordinates $(-R,0)$ . Then $B_1$ has coordinates $(v,0)$ , and $A$ and $C$ both lie on a straight line through $B_1$ . Without loss of generality, we can assume that $A$ lies above the $x$ -axis, and so that $\angle AOB_1 = \theta$ lies between $0$ and $\pi$ . Then $A$ and $C$ are the points of intersection of the circle and line $x^2 + y^2 \; = \; R^2 \hspace{2cm} y \; = \; \tan\theta(x-u)$ and hence we have coordinates $\begin{aligned} A&\;\;\big(v\sin^2\theta + \cos\theta\sqrt{R^2 - v^2\sin^2\theta},-v\sin\theta\cos\theta + \sin\theta\sqrt{R^2 - v^2\sin^2\theta}\big) \\ C&\;\;\big(v\sin^2\theta - \cos\theta\sqrt{R^2 - v^2\sin^2\theta},-v\sin\theta\cos\theta - \sin\theta\sqrt{R^2 - v^2\sin^2\theta}\big) \end{aligned}$ and $A_1$ will have coordinates $A_1\;\; \big(-\tfrac{u}{R}\big(v\sin^2\theta + \cos\theta\sqrt{R^2 - v^2\sin^2\theta}\big),-\tfrac{u}{R}\big(-v\sin\theta\cos\theta + \sin\theta\sqrt{R^2 - v^2\sin^2\theta}\big)\big)$ and from this it follows that $B$ , $A_1$ and $C$ will be collinear precisely when $R(R+u)v \cos\theta + (R^2 - Ru - 2uv)\sqrt{R^2 - v^2\sin^2\theta} \; = \; 0 \hspace{2cm} (\star)$ It is a fairly standard result that $\frac{1}{AA_1} + \frac{1}{BB_1} + \frac{1}{CC_1} \; = \; \frac{2}{R}$ where $R$ is the outradius. With $AA_1 = 26\tfrac{26}{279}$ , $BB_1 = 24\tfrac{16}{21}$ and $CC_1 = 22\tfrac{224}{323}$ we deduce that $R = \tfrac{65}{4}$ . Putting $t = \tan\tfrac12\theta$ , the equation $(\star)$ squares to give $0 \; = \; 3969t^4 - 8962 t^2 + 3969 \; = \; (7t-9)(7t+9)(9t-7)(9t+7)$ Since $0 < \theta < \pi$ , we have $t = \tfrac79$ or $\tfrac97$ . It turns out that only $t = \tfrac79$ is a proper solution of $(\star)$ --- the other solution is introduced by the squaring process, and so we deduce that we have coordinates $A \; \big(\tfrac{595}{52}, \tfrac{150}{13}\big) \hspace{1cm} B \; \big(-\tfrac{65}{4},0\big) \hspace{1cm} C \; \big(\tfrac{91}{20}, -\tfrac{78}{5}\big)$ which makes the area of the triangle equal to $\boxed{336}$ .

I use

1.Circumcenter.html

2. ExactTrilinearCoordinates.html

and I find formula for length for the cevians, and find lengths sides of triangle by Mathcad. And after I use HeronsFormula .

Let $r = AO = BO = CO$ . Since an inscribed angle is half of a central angle that subtends the same arc, $\angle BOC = 2A$ , so $\angle BOC' = \angle COB' = 180° - 2A$ (from the straight line), and $\angle CBO = \angle BCO = 90° - A$ (from the isosceles triangle $\triangle BOC$ ). Similarly, $\angle AOC' = \angle COA' = 180° - 2B$ and $\angle AOB' = \angle BOA' = 180° - 2C$ , and $\angle ACO = \angle CAO = 90° - B$ and $\angle ABO = \angle BAO = 90° - C$ . Using the law of sines on $\triangle BOA'$ gives $OA' = \frac{r \cos A}{\cos(B - C)}$ . Similarly, $OB' = \frac{r \cos B}{\cos(A - C)}$ and $OC' = \frac{r \cos C}{\cos(A - B)}$ . Therefore, $AA' = AO + OA' = r + \frac{r\cos A}{\cos(B - C)} = r(1 + \frac{\cos A}{\sin B \sin C + \cos B \cos C})$ . Similarly, $BB' = r(1 + \frac{\cos B}{\sin A \sin C + \cos A \cos C})$ and $CC' = r(1 + \frac{\cos C}{\sin A \sin B + \cos A \cos B})$ .

Using the theorem $\frac{1}{AA'} + \frac{1}{BB'} + \frac{1}{CC'} = \frac{2}{r}$ (see theorem and proof here ) and the given values for $AA'$ , $BB'$ , and $CC'$ , $r$ solves to $r = \frac{65}{4}$ .

Letting $x = \cos A$ , $y = \cos B$ , then $\sin A = \sqrt{1 - x^2}$ and $\sin B = \sqrt{1 - y^2}$ , and $\cos C = \cos(180° - (A + B)) = -\cos(A + B) = \sin A \sin B - \cos A \cos B = \sqrt{(1 - x^2)(1 - y^2)} - xy$ and $\sin C = \sin(180° - (A + B)) = \sin(A + B) = \sin A \cos B + \cos A \sin B = y\sqrt{1 - x^2} + x\sqrt{1 - y^2}$ .

Substituting these values in the $AA'$ and $BB'$ equations gives:

$AA' = \frac{7280}{279} = \frac{65}{4}\bigg(1 + \frac{x}{\sqrt{1 - y^2}(y\sqrt{1 - x^2} + x\sqrt{1 - y^2}) + y(\sqrt{(1 - x^2)(1 - y^2)} - xy)}\bigg)$

$BB' = \frac{520}{21} = \frac{65}{4}\bigg(1 + \frac{y}{\sqrt{1 - x^2}(y\sqrt{1 - x^2} + x\sqrt{1 - y^2}) + x(\sqrt{(1 - x^2)(1 - y^2)} - xy)}\bigg)$

which has two solutions $(x, y) = (-\frac{3}{5}, -\frac{33}{65})$ and $(x, y) = (\frac{3}{5}, \frac{33}{65})$ . However, $(x, y) = (-\frac{3}{5}, -\frac{33}{65})$ gives two obtuse angles for $\angle A$ and $\angle B$ , which is not possible, so $(x, y) = (\frac{3}{5}, \frac{33}{65})$ .

Using these values $x = \frac{3}{5}$ and $y = \frac{33}{65}$ gives $\sin A = \sqrt{1 - x^2} = \sqrt{1 - (\frac{3}{5})^2} = \frac{4}{5}$ , $\sin B = \sqrt{1 - y^2} = \sqrt{1 - (\frac{33}{65})^2} = \frac{56}{65}$ , and $\sin C = y\sqrt{1 - x^2} + x\sqrt(1 - y^2) = \frac{33}{65}\sqrt{1 - (\frac{3}{5})^2} + \frac{3}{5}\sqrt(1 - (\frac{33}{65})^2) = \frac{12}{13}$ .

Therefore, the area of $\triangle ABC = 2r^2 \sin A \sin B \sin C = 2 \cdot (\frac{65}{4})^2 \cdot \frac{4}{5} \cdot \frac{56}{65} \cdot \frac{12}{13} = \boxed{336}$ .

I don't have a solution, but just for the record, the sides of the triangle are $BC = 26$ , $AC = 28$ , and $AB = 30$ .