# Area of triangle inscribed in a circle with a specific position

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.

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 ...

• 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. – Donald Splutterwit Apr 30 at 20:38
• Alright, but how can I use it ...$R=20+x=16+y=13+z$ ? – Khosrotash Apr 30 at 20:58
• 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. – user Apr 30 at 22:05
• @user:Can you turn a light to the numerical solution? – Khosrotash Apr 30 at 22:06
• @user: Can you start answer, but not full solution? – 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

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

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