Equilateral triangle point

My solution was this.

First moves $\triangle APC$ together $\triangle APB$ as image shows, clearly, $\triangle AP'P$ is an equilateral triangle and $\triangle PP'B$ is an rectangle triangle so $\angle BPA = 60° + 90° = 150°$ .

By law of cosines, $AB^2 = 5^2 + 12^2 - 2\times5\times 12 \times cos 150° = 169 + 60\sqrt{3}$ .

Therefore $\triangle ABC$ is

$\frac{\sqrt{3}}{4} \times (169 + 60\sqrt{3})=45+\frac{169\sqrt{3}}{4} \approx 118.179$

Rotate $\triangle ABC$ with a magnitude of $60^\circ$ in a counter-clockwise direction making point $B$ as the pivot point. $\triangle BPP'$ is an equilateral $\triangle$ and $\triangle 5-12-13$ is a right triangle. Then, $\cos \angle P'PC=\dfrac{5}{12}$ , such that $\angle P'PC \approx 22.62$ . By cosine law (or cosine rule or law of cosines) on $\triangle BPC$ , we have $(BC)^2=12^2+13^2-2(12)(13)[\cos (22.62+60)] \approx 272.924$ . Since $\triangle ABC$ is equilateral, the area is $\dfrac{1}{2}(BC)^2(\sin 60)=\dfrac{1}{2}(272.924)(\sin 60) \approx \boxed{118.179}$

Note:

I used the formula $\boxed{A=\dfrac{1}{2}ab~\sin C}$ for the area of the triangle.

I solved this by using analytic geometry.

Let $s$ be the side length of $\triangle ABC$ . Place the triangle on a Cartesian coordinate system such that $A = (0,0), B = \left(\dfrac{\sqrt{3}{s}}{2}, -\dfrac{s}{2}\right), C = \left(\dfrac{\sqrt{3}{s}}{2}, \dfrac{s}{2}\right)$ and $P = (a,b)$ . Using the lengths of $PA, PB, PC$ and the formula for the distance between 2 points, we get the following system of equations:

$\left\{ \begin{array}{ll} (1) : & a^2 + b^2 = 5^2 = 25 \\ (2) : & \left(\frac{\sqrt{3}}{2}s - a\right)^2 + \left(b+\frac{s}{2}\right)^2 = 13^2 = 169\\ (3) : & \left(\frac{\sqrt{3}}{2}s - a\right)^2 + \left(b-\frac{s}{2}\right)^2 = 12^2 = 144 \end{array} \right.$

Subtracting $(3)$ from $(2)$ gives us $2bs = 25$ . $(4)$

Now using $(1)$ and $(4)$ we get $a^2 s^2 + b^2 s^2 = 25s^2 \rightarrow a^2 s^2 = 25s^2 - \frac{625}{4}$ . $(5)$

Next let's expand equation $(2)$ : $\frac{3}{4}s^2 - \sqrt{3}sa+a^2 + b^2 + bs + \frac{s^2}{4} = 169$ Simplify using $(1)$ and $(4)$ : $s^2 - \frac{263}{2} = \sqrt{3}sa$ Now square the equation and rewrite using $(5)$ : $s^4 - 263s^2 + \frac{69169}{4} = 3a^2 s^2 = 75s^2 - \frac{1875}{4}$ $s^4 - 338s^2 + 17761 = 0$ $(s^2 - 169)^2 = 10800$ $s^2 = 169 \pm 60\sqrt{3}$

Hence the area of the triangle is $\dfrac{\sqrt{3}}{4} s^2 = \dfrac{169\sqrt{3}}{4} \pm 45$ .

The minus option would create a triangle that is too small to fit point $P$ inside it. Therefore the area is $\dfrac{169\sqrt{3}}{4} + 45 \approx \boxed{118.79}$

It is easy to realize that three triangles PCR, PAQ, BQR are isosceles triangle with three angles are 30, 30, 60 degrees. So applying heron theory to get: $S(ABC)=\sqrt{(\frac{\sqrt{3}}{2})^{4} \times (13+5-12) \times (13-5+12) \times (13+5+12) \times (13+5+12)} +$ $\frac{1}{2} \times \frac{\sqrt{3}}{2} \times (13^2+5^2+12^2)=\boxed{118.179}$

Rotate APB about A until AB coincide with AC. Let the new location of P be denoted by Q. So triangle APQ is equilateral. Moreover, PQC is a 5-12-13 right triangle. Therefore AQC = AQP+PQC = 60+90 = 150. By law of cosines,

AC^2

= AQ^2+QC^2-2 AQ QC*cos(AQC)

= 25+144-60*sqrt(3)

= 169+60*sqrt(3).

It follows that the area of triangle ABC is AC^2 sqrt(3)/4 = 45+169 sqrt(3)/4 ~ 118.18.

We first rotate $\triangle APC$ so that $C$ goes to $B$ . Name the image of point $Q$ . Rotate triangles $CPB$ and $BPA$ in a similar fashion, naming the images of point $P$ $R$ and $S$ respectively. Notice that the area of hexagon $ASCQBR$ is twice the area of triangle $ABC$ because each of the triangles $APC$ , $CPB$ , and $BPA$ occur twice in $ASCQBR$ . So we will calculate the area of the hexagon and divide by $2$ . Draw the lines $PQ$ , $PR$ , and $PS$ and observe the resulting triangles. The angles $PCQ$ , $PBR$ , and $PAS$ are $60$ degrees because triangle $ABC$ is equilateral and our construction preserves angles. Therefore triangles $QPC$ , $RPB$ , and $PSA$ are equilateral with side lengths 13,12,5 respectively. These will have total area $\frac{5^2\sqrt{3}}{4}+\frac{12^2\sqrt{3}}{4}+\frac{13^2\sqrt{3}}{4}=\frac{169\sqrt{3}}{2}$ The remaining triangles are 5-12-13 triangles, so they will have total area $\frac{1}{2}\cdot5\cdot12+\frac{1}{2}\cdot5\cdot12+\frac{1}{2}\cdot5\cdot12=90$ So the total area of the hexagon is $[ASCQBR]=90+\frac{169\sqrt{3}}{2}$ Which means $[ABC]=\frac{1}{2}[ASCQBR]=45+\frac{169\sqrt{3}}{4}\approx 118.18$

a b c De, Prithwijit, "Curious properties of the circumcircle and incircle of an equilateral triangle," Mathematical Spectrum 41(1), 2008-2009, 32-35. (For any point P in the plane, with distances p, q, and t from the vertices A, B, and C respectively,[17]) http://en.wikipedia.org/wiki/Equilateral triangle#cite note-De-17

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My solution's a bit more rigorous (basically, it involves solving a $3x3$ system of the Law of Cosines). Let angle $\angle{APB} = x, \angle{BPC} = y, \angle{APC} = 360-x-y$ degrees. Let the side of the equilateral $\triangle{ABC}$ equal $s$ . One now now obtains the system:

$s^2 = 5^2 + 12^2 - 2(5)(12) \cdot \cos(x),$

$s^2 = 12^2 + 13^2 - 2(12)(13) \cdot \cos(y),$

$s^2 = 5^2 + 13^2 - 2(5)(13) \cdot \cos(360-x-y).$

I solved the system for $x = 150$ degrees, which gives the area of $\triangle{ABC}$ as $\frac{\sqrt{3}}{4} \cdot s^2 = \frac{\sqrt{3}}{4} \cdot [5^2 + 12^2 - 120*(-\frac{\sqrt{3}}{2})] = 169 \cdot \frac{\sqrt{3}}{4} + 45 = \boxed{118.18}$ square units.