Challenge yourself part 3

On the first step we will use the formula to search for the circumradius of the triangle ABC, however it's AC who is being asked:

$R = \frac{AB\times BC\times AC}{4[ABC]}$

where $[ABC]$ denotes the area of triangle ABC.

$12=\frac{24AC}{\sqrt{(AB+BC+AC)(AB+BC-AC)(AC+AB-BC)(AC-AB+BC)}}$

$\sqrt{(10+AC)(10-AC)(AC+2)(AC-2)}=2AC$

$(100-AC^2)(AC^2-4)=4AC^2$

$(AC^2-100)(AC^2-4)+4AC^2=0$

Assume $AC^2=x$ . Expanding the LHS, we get:

$x^2-100x+400=0$

$x_{1, 2}=\frac{100\pm \sqrt{100^2-4\times 1\times 400}}{2}$

Simplifying it, we get:

$x_{1, 2}=50\pm 10\sqrt{21}=35+15 \pm 2\sqrt{525}=(\sqrt{35}\pm\sqrt{15})^2$

$AC_{1, 2}=\sqrt{35}\pm\sqrt{15}\approx 5.91608\pm 3.87298$

Then, the possible values of AC are $\boxed{9.78906}$ and $\boxed{2.04310}$ . However, in this problem, the chosen one is $\boxed{9.78906}$ .

Let the center of the circle be O. Let angle AOC = 2\theta ,then angle ACB=\theta.Using _ cosine rule_ ,we have \cos\theta = \frac {4^{2} + AC^{2} _ 36}{8\timesAC}. \cos\2 theta = \frac {144+144-36}{288}.We get \cos\2 theta=\frac{7}{8}. 2\cos^{2}\theta -1 =\frac{7}{8}. We get \cos\theta=\frac {15^{1/2} }{4} .We get a _ quadratic_ in AC.Solving we get AC= 9.789

we can also use cosine rule.!!! that will also be a good solution.!!!