Find Area From Versines

Carnot's Theorem tells us that $(R - v_A) + (R - v_B) + (R - v_C) \; = \; R + r$ , where $R$ is the outradius and $r$ the inradius, and hence we deduce that $2R - r \; = \; 49$ Standard results tell us that $v_A = R(1 - \cos A) = \frac{2R(s-b)(s-c)}{bc}$ , with similar results for $v_B,v_C$ , and hence $v_Av_Bv_C \; = \; \frac{8R^3(s-a)^2(s-b)^2(s-c)^2}{a^2b^2c^2} \; =\; \frac{8R^3\Delta^4}{s^2a^2b^2c^2} \; = \; \frac{8R^3\Delta^4}{16s^2R^2\Delta^2} \; = \; \frac{R\Delta^2}{2s^2} \; = \; \tfrac12Rr^2$ where $\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \frac{abc}{4R} = rs$ is the area of the triangle and $s$ its semiperimeter, so that $Rr^2 \; = \; 8320$ Thus we deduce that $R(2R-49)^2 = 8320$ , so that $R$ is one of $\tfrac{65}{2}$ , $\tfrac14(33 + \sqrt{65})$ and $\tfrac14(33-\sqrt{65})$ . Neither the second nor the third of these is possible, since they both imply that $R < 13 = v_C$ . Thus we deduce that $R = \tfrac{65}{2}$ and therefore $r=16$ .

It is now easy to calculate that $\cos A = \tfrac{33}{65}$ , $\cos B = \tfrac{5}{13}$ and $\cos C = \tfrac35$ , so that $a = 56$ , $b=60$ , $c = 52$ and hence $s = 84$ , so that $\Delta = rs = \boxed{1344}$ .

Extending the three versines will meet at the circumcenter $O$ of $\triangle ABC$ and $\triangle ABC$ will be split into three isosceles triangles. Let the circumradius be $r$ and the angles of the isosceles triangles be $\alpha$ , $\beta$ , and $\gamma$ as shown in the figure. Then $\sin \alpha = \dfrac {r-13}r$ , $\sin \beta = \dfrac {r-16}r$ , and $\sin \gamma = \dfrac {r-20}r$ and

$\begin{aligned} 2 (\alpha + \beta + \gamma) & = 180^\circ \\ \alpha + \beta + \gamma & = 90^\circ \\ \alpha + \beta & = 90^\circ - \gamma \\ \cos (\alpha + \beta) & = \cos (90^\circ - \gamma) = \sin \gamma \\ \cos \alpha \cos \beta - \sin \alpha \sin \beta & = \sin \gamma \\ \cos \alpha \cos \beta & = \sin \alpha \sin \beta + \sin \gamma \\ \frac {\sqrt{2ar-a^2}}r \cdot \frac {\sqrt{2br-b^2}}r & = \frac {r-a}r \cdot \frac {r-b}r + \frac {r-c}r & \small \blue{\text{where }a = 13, b=16, c=20} \\ \sqrt{ab(2r-a)(2r-b)} & = 2r^2 - (a+b+c)r + ab & \small \blue{\text{Squaring both sides}} \end{aligned}$

$\begin{aligned} 4abr^2 - 2ab(a+b)r + a^2b^2 & = 4r^4 - 4(a+b+c)r^3 + ((a+b+c)^2 + 4ab)r^2 -2ab(a+b+c)r + a^2b^2 \end{aligned}$

$\begin{aligned} \implies 4r^3 - 4(a+b+c)r^2 + (a+b+c)^2r -2abc & = 0 \\ 4r^3 - 196r^2 + 2401r - 8320 & = 0 \\ (2r-65)(2r^2 - 33r + 128) & = 0 \end{aligned}$

$\implies r = \begin{cases} \dfrac {65}2 = 32.5 \\ \dfrac {33+\sqrt{65}}4 \approx 10.266 \\ \dfrac {33-\sqrt{65}}4 \approx 6.234 \end{cases} \implies$ the acceptable $r=32.5$ and the area of $\triangle ABC$ is

$\begin{aligned} A_\triangle & = r^2(\sin \alpha \cos \alpha + \sin \beta \cos \beta + \sin \gamma \cos \gamma) \\ & = r^2 \left(\frac {(r-13)\sqrt{13(2r-13)}}{r^2} + \frac {(r-16)\sqrt{16(2r-16)}}{r^2} + \frac {(r-20)\sqrt{20(2r-20)}}{r^2} \right) \\ & = (r-13)\sqrt{13(2r-13)} + (r-16)\sqrt{16(2r-16)} + (r-20)\sqrt{20(2r-20)} \\ & = 19.5 \cdot 26 + 16.5 \cdot 28 + 12.5 \cdot 30 \\ & = \boxed{1344} \end{aligned}$

->From the cosine law we know that : c²=a²+b²-2ab.cos(C)

->From the product of chords in a circle we know that in this case

a²=64(2r-16)

b²=80(2r-20)

c²=52(2r-13)

->We also know that the radius r can be expressed as : r= $\frac{c}{2sin(C)}$

->From this equality we easily get that Cos²(C)= 1-sin²(C)=1- $\frac{13(2r-13)}{r²}$

We use the law of cosine knowing the values of the a,b,c and cos(c) in terms of r :

->52(2r-13)=64(2r-16)+80(2r-20)-2ab.cos(C)

we sort the values and square both sides

->(92r-974)²=a²b².cos²(C)

Finally we get this equation :

->(92r-974)²=64(2r-16)80(2r-20) (1- $\frac{13(2r-13)}{r²}$ )

We find several values for r, the valid one being r= $\frac{65}{2}$

After Solving for r, we solve for a, b and c a=56 b=60 c=52

We use this formula to find the area : Area= $\frac{abc}{4r}$ = 1344

Solve

a^2=4x(2r-x), b^2=4y(2r-y), c^2=4z(2r-z), t=[a(r-x)+b(r-y)+c(r-z)]/2, t=(abc)/(4r),

Here, a, b,c are sides of triangle, x=13, y=16 and z=20. “t” is area of the triangle.

Solve by WolframAlpha to get

a=52, b=56, c=60, r=65/2, t=1344

Answer=1344

R = (C^2 + 4 M^2)/(8 M) where M =20 ,16 &13 R= radius of the circle C = chord (sides of the triangle)

(a^2 + 4 * 16^2) / (8 * 16) = (b^2 + 4 * 20^2) / (8 * 20) = (c^2 + 4 * 13^2) / (8 * 13) solve for a^2 ,b^2 & c^2 implies a=56 , b = 60 and c = 52 and R = 32.5 Area = 1344