Equilateral Triangle with X Inside

First, let us rotate $\triangle AXC$ [about $C$ by $60^\circ$ ] to $\triangle BYC$ , with $Y$ being the new position of $X$ . Now, $\angle YCX = \angle YCB + \angle BCX = \angle ACX + \angle BCX = 60^\circ$ . Again, $CX = CY = 15$ (after rotation). As $\triangle CXY$ isosceles and $\angle YCX = 60^\circ$ , $\triangle CXY$ is equilateral, so $XY = 15$ . Now if we use cosine law on $\angle BYX$ , we get $\cos BYX = \frac {8^2 + 15^2 - 13^2}{2 \cdot 8 \cdot 15} = \frac 1 2$ so, $\angle BYX = 60^\circ$ and $\angle BYC = 60^\circ + 60^\circ = 120 ^ \circ$ . So, as rotated angle, $\angle AXC = 120 ^\circ$ . Now if we use cosine law on $\triangle AXC$ , we get $AC ^2 = AX^2 + CX^2 - 2 AX \cdot CX \cdot \cos 120^\circ = 8^2+15^2 + 2 \cdot 8 \cdot 15 \cdot (-\frac {1}2) = 409$ . So, $N^2 = 409$ , which is our desired answer.

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Rotate AXB 60 degrees clockwise around B

Rotate CXA 60 degrees clockwise around A

Rotate BXC 60 degrees clockwise around C

AXC is congruent to ADB, so <XAC=<DAB. Thus, <CAB=<DAX=60. Since <DAX=60 and AD=AX, <ADX=<XDA=(180-60)/2=60,and DX=8.

We could argue similarly that FX=15

By the cosine law, cosAXF=(8^2+15^2-13^2)/(2 8 15)=1/2. Thus, AXF=60

<FXC=60, so <AXC=120. cos120=-1/2, so through the cosine law, -1/2=(8^2+15^2-n^2)/(2 8 15). Thus, -120=289-n^2, and n^2=409, which is what we are looking for.

N^2=409.

Rotating triangle APC 60 degrees towards side C gets you point P'. You know PP' is also 8 from the fact that AP' and AP are 8 and P'AP is 60 degrees. Use law of cosines on triangle P'PC toobtain angle P'PB. Add 60 degrees to that to obtain angle APC. Finally use law of cosine again on triangle APC to obtain N^2=409.

Rotate triangle AXB 60 degree about the centre A such that AB maps onto AC and X maps onto Y. $\angle XAY=\angle XAC+\angle YAC=\angle XAC+\angle XAB=60$ degrees. Note that AX=AY=8, therefore triangle AXY is equilateral and XY=8. Since YC=XB=13, by Cosine rule, $13^2=15^2+8^2-2(15)(8)(\cos \angle YXC)$ , which gives $\angle YXC=60$ degrees. So $\angle AXC =60+60=120$ . By Cosine rule again, $N^2=AC^2=8^2+15^2-2(8)(15)(\cos 120)=409$ .

{Latex edits - Calvin]

We rotate point X 60 degrees clockwise about A to get point A', and so we have <A'AX=60. Since line segment AA' is a rotation of line segment AX by 60 degrees also, AA'=8. Since triangle AA'X is isosceles, <AA'X=<AXA'. Since the two angles sum to 180-60=120, each angle is 60 degrees, and so triangle AA'X is equilateral.

Similarly, we rotate X around B and C by 60 degrees to get B' and C', respectively. Note that by the same reason, triangles BB'X and CC'X are also equilateral.

Also, note that the area of the hexagon AA'BB'CC' is twice the area of triangle ABC, since we have rotated pieces of triangle ABC to get the hexagon (for example, triangles AA'B and AXC are congruent because they both have 8, 15, and N as side lengths). Lastly, we can see that the hexagon is composed of three equilateral triangles (AA'X, BB'X, CC'X) and three congruent triangles with side lengths 8, 13, and 15 (A'XB, B'XC, C'XA). Thus using Heron's formula and the formula for the area of an equilateral triangle, we get

$[ABC]=\frac {1}{2}[AA'BB'CC']=\frac {1}{2}([AA'X]+[BB'X]+[CC'X]+3[A'XB])$

$=\frac {1}{2}(8^2\sqrt{3}/4+13^2\sqrt{3}/4+15^2\sqrt{3}/4+3 \times \sqrt{(18)(18-8)(18-13)(18-15)})$

$=\frac {1}{2}(458\sqrt{3}/4+3 \times 30\sqrt{3})$

$=409\sqrt{3}/4.$

Since [ABC] also equals $N^2\sqrt{3}/4$ , we have

$N^2\sqrt{3}/4=409\sqrt{3}/4.$

Therefore, $N^2=409$ .

if we rotate ABC in 60 degree centered at B,we get the image BCC'(A coincides with C)and X' is the image of X.so,.XC=15,XX'=13(BXX' will be equilateral)and CX=8.using cosine rule on angle XX'C,we get,cos(x'xc)=11/13.again using cosine rule on BXC(angleBXC=60+x'xc), we get N^2=409

Consider Cartesian coordinate system,taking origin at B & x-axis along BC. So $B:(0,0)$ , $C:(N,0)$ & $A:(N/2,\sqrt{3}N/2)$ (Height of equilateral triangle is $\sqrt{3}/2 \times side$ ). Now consider a circle of radius 13 with centre B, one of radius 15 with centre C & another of radius 13 with centre A. We get their equations: $x^2 + y^2 = 13^2$ ; $(x-N)^2 + y^2 = 15^2$ ; $(x-N/2)^2 + (y-\sqrt{3}N/2)^2 = 8^2$ . Solving these 3 equations for $N^2$ , we get $N^2=409$ .

60+60=120 8^2+15^2+8*15=409

Label the vertices in a clockwise manner. Let $Y$ be the rotation of $X$ by $60^\circ$ clockwise around A. Then $YC = XB = 13$ , $AXY$ is an isosceles triangle with $\angle XAY = 60^\circ$ , hence is equilateral and $XY=8$ . By cosine rule, $\cos \angle YXC = \frac {8^2 + 15^2 - 13^2}{2 \cdot 8 \cdot 15 }= \frac {1}{2}$ , so $\angle YXC = 60^\circ$ . As such, $\angle AXC = \angle AXY + \angle YXC = 60^\circ + 60^\circ = 120^\circ$ , which gives $AC^2 = 8^2 + 15^2 -2 \cdot 8 \cdot 15 \cdot \cos 120^\circ = 409$ .

A general formula is according to Gardner 1977, pp. 56-57 and 63 :

Given the distances of a point from the three corners of an equilateral triangle, a, b, and c, the length of a side s is given by: $3(a^4+b^4+c^4+s^4)=(a^2+b^2+c^2+s^2)^2$

in our case: $3(8^4 + 13^4 + 15^4 + N^4) = (8^2 + 13^2 + 15^2 + N^2)^2$

let $x = N^2$ :

$3 (x^2 + 83282) = (x + 458)^2$

this yields : $x = 49$ or $x = 409$

The first solution is too small to have the point $X$ inside $ABC$ , so that $N^2 = 409$

NB: $N^2 = 49$ is possible with point $X$ outside $ABC$

area of equilateral triangle = sqrt{3}N^{2}/4

By Herons Formula sqrt{s(s-a)(s-b)(s-c)} = area of triangle

Sum of area of 3 traingles = total area of triangle

(1/4)sqrt{(28+N)(28-N)(N+2)(N-2)} + (1/4)sqrt{(21+N)(21-N)(N+5)(N-5)} +(1/4)sqrt{(23+N)(23-N)(N+7)(N-7)} = sqrt{3}N^{2}/4

sqrt{(7-N^2)(N}