A question about an equilateral triangle Suppose that $\triangle ABC$ is an equilateral triangle. Let $D$ be a point inside the triangle so that $\overline{DA}=13$, $\overline{DB}=12$, and $\overline{DC}=5$. Find the length of $\overline{AB}$.
 A: 
Let s be the side of the equilateral triangle
Using cosine formula for the triangle
$$\frac{13^2 + 5^2 - s^2}{2.5.13} = \cos\alpha\tag1$$
$$\frac{12^2 + 5^2 - s^2}{2.5.12} = \cos\beta\tag2$$
$$\frac{12^2 + 13^2 - s^2}{2.13.12} = \cos\gamma\tag3$$
Now we have $\gamma + \alpha + \beta = \pi$
$$\cos\gamma = \cos(\pi - \alpha - \beta)$$
Knowing $\cos(\pi - \theta) = \cos(\theta)$
$$\Rightarrow \cos\gamma = \cos(\alpha + \beta)$$
$$\Rightarrow \cos\gamma = \cos\alpha\cos\beta - \sin\alpha\sin\beta$$
$$\Rightarrow (\cos\gamma - \cos\alpha\cos\beta)^2 = \sin^2\alpha\sin^2\beta$$
$$\Rightarrow (\cos\gamma - \cos\alpha\cos\beta)^2 = (1 - \cos^2\alpha)(1 - \cos^2\beta)$$
$$\Rightarrow \cos^2\alpha + \cos^2\beta + \cos^2\gamma - 2.\cos\alpha.\cos\beta.\cos\gamma -1 = 0\tag4$$
Applying $(1)$, $(2)$, $(3)$ to $(4)$ and simplifying we have 
$$s^4 - 338.s^2 + 17761 = 0$$
using Heron's formula we have
$$\Rightarrow s^2  = \pm 60 . \sqrt3 + 169$$
$$\Rightarrow s = \pm\sqrt{\pm 60 . \sqrt3 + 169}$$
as Area of a triangle is real and positive
$$s = \sqrt{60 . \sqrt3 + 169}$$
A: 
Consider an equilateral  triangle ABC and a Point P such that $\overline{AP} = 12$, $\overline{BP} = 13$ and $\overline{CP} = 5$.
Let $P'$, $P'_1$ and $P'_2$ be the reflection of the Point $P$ along  $\overline{BC}$, $\overline{AC}$ and $\overline{AB}$ 
Then we have
$${\triangle APB} \cong {\triangle AP'_2B}$$
$${\triangle APC} \cong {\triangle AP'_1B}$$
$${\triangle BPC} \cong {\triangle BP'C}$$
Thus we can safely say,
The area of the hexagon $AP'_1CP'BP'_2 = 2. {\triangle ABC}$ 
${\triangle AP'_2B}$, ${\triangle AP'_1B}$ and ${\triangle BP'C}$ are isosceles with an apex angle of $120^o$ and composed of $30^o$ and $60^o$ right angle triangle
Thus we have
$$\overline {P'_1P'_2} = 12\sqrt3$$
$$\overline {P'P'_2} = 13\sqrt3$$
$$\overline {P'_1P'} = 5\sqrt3$$
which implies $\angle P'P'_1P'_2 = 90^o$
So the area of the hexagon $AP'_1CP'BP'_2$
$$= \triangle AP'_1P'_2 + \triangle P'P'_1C + \triangle P'P'_2B + \triangle P'P'_1P'_2$$
$$= 12^2.\frac{\sqrt3}{4} + 5^2.\frac{\sqrt3}{4} + 13^2.\frac{\sqrt3}{4} + \frac{1}{2}(5\sqrt3)(12\sqrt3)$$
$$=169\frac{\sqrt3}{2} + 90$$
$${\triangle ABC} =169\frac{\sqrt3}{4} + 45$$
Now, given the sides of the triangle $\overline {AB}$ = $\overline {BC}$ = $\overline {CA}$ = $a$, then
$$\frac{\sqrt3}{4}a^2 = {\triangle ABC} =169\frac{\sqrt3}{4} + 45$$
$$\frac{\sqrt3}{4}a^2 =  169\frac{\sqrt3}{4} + 45$$
$$a^2 =  169 + 45\frac{4}{\sqrt3}$$
$$a = \sqrt{169 + 45\frac{4}{\sqrt3}}$$
Simplifying
$$a = \sqrt{169 + 60\sqrt3}$$
A: 
Consider an equilateral  triangle ABC and a Point P such that $\overline{AP} = 12$, $\overline{BP} = 13$ and $\overline{CP} = 5$.
Rotate $P$ clockwise along $A$ by $60^o$ to $B'$, along $B$ by $60^o$ to $C'$, along $C$ by $60^o$ to $A'$
We note that, 
$${\triangle APB} \cong ${\triangle ACB'}$$
$${\triangle APC} \cong ${\triangle A'CB}$$
$${\triangle BPC} \cong ${\triangle AC'B}$$
Thus we have, the area of the hexagon $AB'CA'BC'$
$$=2.{\triangle ABC}$$
We also observe that
${\triangle APB'}$, ${\triangle BPC'}$ and ${\triangle A'PC}$ are equilateral triangles
${\triangle PCB'} \cong {\triangle AC'P} \cong {\triangle A'BP} = RightAngle {\triangle}$
Thus we can say
$$2.{\triangle ABC} = [AB'CA'BC']$$
$$= {\triangle APB'} + {\triangle BPC'} + {\triangle A'PC} + 3.{\triangle PCB'}$$
$$= 12^2.\frac{\sqrt3}{4} + 13^2.\frac{\sqrt3}{4} + 5^2.\frac{\sqrt3}{4}  + 3.\frac{1}{2}(5)(12)$$
$$=169\frac{\sqrt3}{2} + 90$$
$${\triangle ABC} =169\frac{\sqrt3}{4} + 45$$
Now, given the sides of the triangle $\overline {AB}$ = $\overline {BC}$ = $\overline {CA}$ = $a$, then
$$\frac{\sqrt3}{4}a^2 = {\triangle ABC} =169\frac{\sqrt3}{4} + 45$$
$$\frac{\sqrt3}{4}a^2 =  169\frac{\sqrt3}{4} + 45$$
$$a^2 =  169 + 45\frac{4}{\sqrt3}$$
$$a = \sqrt{169 + 45\frac{4}{\sqrt3}}$$
Simplifying
$$a = \sqrt{169 + 60\sqrt3}$$
A: The Wikipedia article on equilateral triangles quotes the following theorem from "Curious Properties of the Circumcircle and Incircle of an Equilateral Triangle," by Prithwijit De (http://ms.appliedprobability.org/data/files/Abstracts%2041/41-1-7.pdf):
Theorem: Let $ABC$ be an equilateral triangle with side $s$, and let $P$ be a point in the plane of the triangle with distances $p$, $q$, and $r$ to $A$, $B$, and $C$, respectively. Then $$3(p^4+q^4+r^4+s^4) = (p^2+q^2+r^2+s^2)^2{\textrm.}$$
Solving $3(5^4+12^4+13^4+s^4) = (5^2+12^2+13^2+s^2)^2$ for $s$ gives $s=\sqrt{169\pm 60\sqrt3}$. Because $P$ is inside the triangle, $s>13$, which allows only $s=\sqrt{169+ 60\sqrt3}$, the same solution others have obtained.
A: If we are allowed to use Coordinate Geometry,

WLOG we can assume the coordinates of $A,B,C$ to be $(x,y),(-a,0),(a,0)$ respectively and $D$ to be $(h,k).$
As $\triangle  ABC$ is equilateral $(x-a)^2+(y-0)^2=(x+a)^2+(y-0)^2=(-a-a)^2+(0-0)^2$
From the first relation we find $4ax=0\implies x=0$ as $a\ne0$
From the second relation we find $y^2=3a^2\implies y=\sqrt3a$ assuming  $a>0$
So, $\overline{AB}=2a$
So, $\overline{AD}^2=(h-0)^2+(k-\sqrt3a)^2$ But $\overline{AD}=13$
$\implies (h-0)^2+(k-\sqrt3a)^2=13^2--->(1)$
Similarly, for $BD,CD$ respectively,
$(h+a)^2+k^2=12^2--->(2)$ and  $(h-a)^2+k^2=5^2--->(3)$
So here we have three equations with three unknowns $a,h,k$
One way to solve this could be  as follows :
From $(2)-(3),4ah=119\implies h=\frac{119}{4a}$
From $(2)-(1),2ah+2\sqrt3ak=2a^2-25\implies k=\frac{4a^2-169}{4\sqrt3a} $ (putting $ah=\frac{119}4$)
Putting the values of $h,k$ in terms of $a$ in $(1),$
$$\left(\frac{119}{4a}\right)^2+\left(\frac{4a^2-169}{4\sqrt3a}-\sqrt3a\right)^2=13^2$$
$$\implies 64a^4-32\cdot169a^2+169^2+3(119^2)=0$$
which is a Quadratic Equation in $a^2$ 
