Triangle $ABC$ inscribed in a circle. Versine (green)are drawn from the midpoints of the sides of the triangle perpendicular to them. They have lengths as shown in this figure. enter image description here

find the area of $\Delta ABC$
honestly, I got stuck on this problem and I was far from geometry for years. Please show me a clue or guide me to get over on this problem. Thanks in advance.
I just find out the lines(green) must cross at one point, because of the middle and perpendicularly. but not go more ...

  • $\begingroup$ Hint: The perpendicular bisetors of a triangle concur at the circumcenter. In other words ... extend your green lines to meet at the center of the circle. $\endgroup$ – Donald Splutterwit Apr 30 at 20:38
  • $\begingroup$ Alright, but how can I use it ...$R=20+x=16+y=13+z$ ? $\endgroup$ – Khosrotash Apr 30 at 20:58
  • $\begingroup$ It is easy to find the value of $R$ numerically and then prove it to be the exact result, but I wonder if it is posssible to find the value analytically. $\endgroup$ – user Apr 30 at 22:05
  • $\begingroup$ @user:Can you turn a light to the numerical solution? $\endgroup$ – Khosrotash Apr 30 at 22:06
  • $\begingroup$ @user: Can you start answer, but not full solution? $\endgroup$ – Khosrotash Apr 30 at 22:16

Let $\alpha, \beta, \gamma = 16,20,13$ be the heights of circular segment opposite to vertices $A,B,C$.

Let $R$ be the circumradius. It is easy to see $$\alpha = R(1-\cos A),\quad \beta = R(1-\cos B) \quad\text{ and }\quad \gamma = R(1-\cos C)$$

Notice for any $3$ angles $A,B,C$ which sums to $180^\circ$, we have the trigonometry identity:

$$1 - \cos^2A - \cos^2 B - \cos^2 C - 2\cos A \cos B \cos C = 0\\ \iff 2(1-\cos A)(1-\cos B)(1-\cos C) = (1- \cos A - \cos B - \cos C)^2$$ In terms of $\alpha,\beta,\gamma$, this leads to

$$2\frac{\alpha\beta\gamma}{R^3} = \left(\frac{\alpha+\beta+\gamma}{R} - 2\right)^2 \quad\iff\quad (2R - (\alpha+\beta+\gamma))^2 R - 2\alpha\beta\gamma = 0 $$ Plug back the values of $\alpha,\beta,\gamma$, this becomes

$$R(2R - 49)^2 - 8320 = (2R-65)(2R^2 - 33R + 128) = 0$$

This cubic equations has $3$ real roots. However, the two roots from the quadratic factor is too small (both $\le 20$). This leaves us with $R = \frac{65}{2}$.

Apply intersecting chord theorem to side $BC$ and its perpendicular bisector, we find $$\left(\frac{a}{2}\right)^2 = \alpha(2R - \alpha) \implies a = 2\sqrt{\alpha(2R-\alpha)} = 2\sqrt{16(65-16)} = 56$$ By a similar argument, we find $$b = 2\sqrt{20(65-20)} = 60\quad\text{ and }\quad c = 2\sqrt{13(65-13)} = 52$$

By Euler's formula between a triangle's area, circumradius and sides, the desired area equals to

$$ \verb/Area/(ABC) = \frac{abc}{4R} = \frac{56\cdot 60 \cdot 52}{2\cdot 65} = 1344$$

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Hint: One can easily check that $R=\dfrac{65}2$ is the only solution of the equation $$ \arccos\left(1-\frac{13}R\right)+\arccos\left(1-\frac{16}R\right)+\arccos\left(1-\frac{20}R\right)=\pi. $$

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