Kshitij's circumcircle

It's very helpful to notice that we can solve for $\cos A$ , $\cos B$ , and $\cos C$ using the Law of Cosines, and similarly for $\sin A$ , $\sin B$ , and $\sin C$ using the Extended Law of Sines in ways that will reduce one side of the equation and relate to the circumcircle at the same time.

First, note that, by the Law of Cosines, $BC^2 = AC^2 + AB^2 - 2AC \cdot AB \cos A,$ And therefore, $\cos A = \frac{AC^2 + AB^2 - BC^2}{2AC \cdot AB}.$ Similarly, we can find that $\cos B = \frac{AB^2 + BC^2 - AC^2}{2AB \cdot BC} \\ \cos C = \frac{BC^2 + AC^2 - AB^2}{2BC \cdot AC}.$ Now note that by the Extended Law of Sines, $\frac{BC}{\sin A} = \frac{AC}{\sin B} = \frac{AB}{\sin C} = 2R$ where $R$ is the radius of the circumcircle of $\triangle ABC$ . Therefore, $\sin A = \frac{BC}{2R} \\ \sin B = \frac{AC}{2R} \\ \sin C = \frac{AB}{2R}.$

If we substitute these values in the original equation, we find that $\begin{aligned} AB^2 + BC^2 + AC^2 &= \frac{AC^2 + AB^2 - BC^2}{2AC \cdot AB} \cdot \frac{AC}{2R} \cdot \frac{AB}{2R} + \frac{BC}{2R} \cdot \frac{AB^2 + BC^2 - AC^2}{2AB \cdot BC} \cdot \frac{AB}{2R} + \frac{BC}{2R} \cdot \frac{AC}{2R} \cdot \frac{BC^2 + AC^2 - AB^2}{2BC \cdot AC} \\ &= \frac{AB^2 + BC^2 + AC^2}{8R^2} \end{aligned}$ which cancels quite nicely with the lefthand side of the equation: $\begin{aligned} 1 &= \frac{1}{8R^2} \\ R^2 &= \frac{1}{8} \\ \pi R^2 &= \frac{1 \cdot \pi}{8} \end{aligned}$ Therefore, $a + b = \boxed{9}$ .

using cosine law and sine law together,
a = BC, b = CA, c = AB and R = circumradius.

b^2 + c^2 - a^2 = 2bc cosA = 8R^2 sinB sinC cosC

c^2 + a^2 - b^2 = 2ac cosB = 8R^2 sinA sinC cosB

a^2 + b^2 - c^2 = 2ab cosC = 8R^2 sinA sinB cosC

adding all these and comparing with the given equation,

8R^2 = 1

so area of circumcircle is pi/8.

hence a = 1 and b = 8.

a + b = 9.

The initial equation is equivalent to: $2(AB^2+BC^2+AC^2)=2\cdot \cos A \sin B \sin C+\\ 2\cdot \sin A \cos B \sin C+ 2\cdot \sin A\sin B\cos C=\\ \sin A(\sin B\cos C+\sin C\cos B)+\\ \sin B(\sin A\cos C+\sin C\cos A)+\\ \sin C(\sin A\cos B+\sin B\cos A)=\\ \sin A\sin (B+C)+\sin B\sin(A+C)+\sin C\sin (A+B)=\\ \sin^2A+\sin^2B+\sin^2C\quad (*)$

Since that $A+B+C=180^{\circ}$ and $\sin(x)=\sin(180^{\circ}-x)$

Now, by the Sine Rule we have: $4R^2=\frac{AB^2}{\sin^2 C}=\frac{AC}{\sin^2 B}=\frac{BC^2}{\sin^2 A}=\\ \frac{AB^2+BC^2+AC^2}{\sin^2C+\sin^2A+\sin^2B}=\frac{1}{2}$

by $(*)$ .

Thus $R^2=\frac{1}{8}$ and then the area of the circumcircle of $\triangle ABC$ is $\frac{\pi}{8}=\frac{a\pi}{b}$ , where $a=1, b=8, \text{mdc}(a, b)=1$ .

Then $a+b=9$ .

We just use sin(A+B) identity for each of the 3 pair of angles. This can be done by multiplying the given condition by 2. We take sin(C) and its corresponding sin of other angles to use the sum identity on the right hand side of the condition. We also sine rule to write each side in terms of the sine of the opposite angle and circumradius. This will give us, after a lot if computations, 8R^2=1 where R is is the circumradius. So we can simply calculate the area and we have 9 as the answer to this problem.

Assume it is a right triangle with side x,x, and √2x...hence x²+x²+2x²=0.5+0.5, gives x=0.5...hence r=1/(2√2), area=pi/8..hence a=1 and b=8, a+b=9

$c^{2} + b^{2} + a^{2} = \cos A \sin B \sin C + \cos B \sin A \sin C + \cos C \sin A \sin B$

From this, one can easily see that switching the letters around with each other doesn't change the equation at all. Therefore, A, B and C must all be identical. Therefore, triangle ABC is equilateral.

Letting s be the side length of this triangle, we can easily discover $3s^{2} = 3\cos 60 (\sin 60)^{2}$

$s =\sqrt{3/8}$

By symmetry arguments, one can show that the circumcenter and any two points on the triangle form a 30-30-120 triangle, and of this triangle we know 1 side length. Therefore, you can use cosine law to show that the circumradius is $\sqrt\frac{1}{8}$

From the formula for a circle, we know that $\frac{a}{b} = \frac{1}{8}$

So, a+b = 9 .

Using cosine rule: a^2 = b^2 + c^2 - 2bc cosA Writing similar equations for involving cosB and cosC and adding these equations 2bc cosA + 2ab cosC + 2ac cosB = a^2 + b^2 + c^2

Using given condition 2bc cosA + 2ab cosC + 2ac*cosB = cosAsinBsinC+sinAcosBsinC+sinAsinBcosC

Putting a = 2RsinA, b= 2RsinB, c=2RsinC in above equation, we get R^2 = 1/8