The Tiny Triangle

Imgur

Refer to the diagram.

To find $[MNO]$ , I will try to find the area of other shapes and subtract them from each other to get $MNO$ . In this case, I notice: $[MNO]=[NGF]-[OMGF]$

Extending $GF$ and $CH$ to meet at $P$ :

$[OMGF]=[OFP]-[PGM]$

Thus $[MNO]=[NGF]-[OFP]+[PGM]$

We see that: $[NGF]=\dfrac{1}{2}\cdot(1)\cdot (\sqrt{3})=\dfrac{\sqrt{3}}{2}$

$FP=1+\dfrac{1}{2}\cdot\sqrt{3}\cdot 2=1+\sqrt{3}$

$\therefore OF \dfrac{1+\sqrt{3}}{\sqrt{3}}=1+\dfrac{\sqrt{3}}{3}$

$\therefore [OFP]=\dfrac{1}{2}(1+\sqrt{3})\left(1+\dfrac{\sqrt{3}}{3}\right)=\dfrac{2\sqrt{3}}{3}+1$

$PG=\sqrt{3}\implies [PGM]=\dfrac{3\sqrt{3}}{4}$

Thus $[MNO]=\dfrac{\sqrt{3}}{2}-\left(\dfrac{2\sqrt{3}}{3}+1\right)+\dfrac{3\sqrt{3}}{4}=\dfrac{7\sqrt{3}-12}{12}=\dfrac{\sqrt{147}-12}{12}$

and so our answer is $147+12+12=\boxed{171}$

$MN=ON-OM=(2-\sqrt{3})$ $\frac{MN}{2}=\frac{2-\sqrt{3}}{2}$ $OS=\frac{MN}{2}tan(30^{0})=\frac{MN}{2} \times \frac{1}{\sqrt{3}}$ $[MNO]=\frac{MN*OS}{2}=\frac{1}{2} \times MN(MN/2) \times (1/ \sqrt{3})$ $=\frac{MN^2}{4\sqrt{3}}=\frac{7-4\sqrt{3}}{4\sqrt{3}}$ $[MNO]=\frac{7\sqrt{3}-12}{12}=\frac{\sqrt{49*3}-12}{12}$ $(a+b+c)=\boxed{171}$

I always like to solve most of your problems by the hard way :D, this time I used analytic geometry and complex numbers. Rotate the dodecahedron $30°$ to the left, with the objective to make that $D$ and $J$ be in the x-axis. Now, we know that the side length is $1$ , so the radius, with a little of trigonometry, will be $r=2\cos(15°)$ . Notice that the vertices of the figure, in the Argand diagram are: $V=2\cos(15°) \times cis(30°k)$ from $k=0$ to $11$ .

So, $F=2\cos(15°) \times cis(60°)$ , $G=2\cos(15°) \times cis(90°)$ , $H=2\cos(15°) \times cis(120°)$ , $A=2\cos(15°) \times cis(270°)$ , $B=2\cos(15°) \times cis(300°)$ and $C=2\cos(15°) \times cis(330°)$

In the cartesian plane, they are: $F=\left(\dfrac{\sqrt{6}+\sqrt{2}}{4},\dfrac{\sqrt{6}+3\sqrt{2}}{4}\right)$ $G=\left(0,\dfrac{\sqrt{6}+\sqrt{2}}{2}\right)$ $H=\left(-\dfrac{\sqrt{6}+\sqrt{2}}{4},\dfrac{\sqrt{6}+3\sqrt{2}}{4}\right)$ $A=\left(0,-\dfrac{\sqrt{6}+\sqrt{2}}{2}\right)$ $B=\left(\dfrac{\sqrt{6}+\sqrt{2}}{4},-\dfrac{\sqrt{6}+3\sqrt{2}}{4}\right)$ $C=\left(\dfrac{\sqrt{6}+3\sqrt{2}}{4},-\dfrac{\sqrt{6}+\sqrt{2}}{4}\right)$

Let's find the equation of the line that passes through: $\overline{AF} \Rightarrow y=(2+\sqrt{3})x-\dfrac{\sqrt{6}+\sqrt{2}}{2}$ $\overline{BG} \Rightarrow y=-(2+\sqrt{3})x+\dfrac{\sqrt{6}+\sqrt{2}}{2}$ $\overline{CH} \Rightarrow y=-x+\dfrac{\sqrt{2}}{2}$

Their intersection points are the vertices of the triangle, and they are: $O=\left(\dfrac{\sqrt{6}-\sqrt{2}}{2},0\right)$ $N=\left(\dfrac{\sqrt{6}+3\sqrt{2}}{12},\dfrac{3\sqrt{2}-\sqrt{6}}{12}\right)$ $M=\left(\dfrac{3\sqrt{2}-\sqrt{6}}{4},\dfrac{\sqrt{6}-\sqrt{2}}{4}\right)$

And finally, find the area of the triangle using determinants: $A_T=\dfrac{7\sqrt{3}-12}{12}$

Hence, $a=147$ , $b=12$ , $c=12$ and $a+b+c=\boxed{171}$

It can be shown that $\angle ABG=90^{ \circ}$ and $\angle FAB= 60^{\circ}$ , therefore, the tiny $\triangle MNO$ is a $30^{\circ }-120^{ \circ}-30^{\circ }$ isosceles triangle. If $|MN|=d$ then the height of the tiny $\triangle MNO$ is $\frac{d}{2}tan30^{\circ}$ and its area is given by:

$A=\frac{1}{2}d(\frac{d}{2}tan30^{\circ})=\frac {d^{2}}{4\sqrt{3}}$

To get $d$ , we use $|BG|-|BN|-|MG|=|MN|=d$ . From $\triangle ABN$ ( $30^{\circ }-60^{ \circ}-90^{\circ }$ ), we find that $|BN|=\sqrt{3}$ ; similarly, $|MG|=\sqrt{3}$ .

Let the center of the dodecahedron be $O'$ , then $\triangle O'AB$ is a $75^{\circ}-(30^{\circ}-(75^{\circ}$ isosceles triangle. If $|BG|=H$ , then:

$\frac{H}{2}=\frac{1}{2}tan75^{\circ}\Rightarrow H=tan75^{\circ}=2+\sqrt{3}$

$\Rightarrow d = |BG|-|BN|-|MG| = H - 2\sqrt{3} = 2+\sqrt{3}-2\sqrt{3} = 2-\sqrt{3}$

$\Rightarrow A=\frac {d^{2}}{4\sqrt{3}}=\frac{(2-\sqrt{3})^{2}}{4\sqrt{3}}=\frac{7-4\sqrt{3}}{4\sqrt{3}}=\frac{7\sqrt{3}-12}{12}=\frac{\sqrt{147}-12}{12}$

Therefore, $a = 147$ and $b=c=12$ , and $a+b+c=\boxed{171}$

Let st. line POQR through O meet CF at P, NM at Q and AH at R.
COF,.NOM,.AOH are (isosceles) SIMILAR triangles with respective altitudes ,OP, OQ, OP.
RQ = 1, RO = RQ + QO = 1 + QO...>>> QO = RO - 1....RP = RO + OP.

Ratio of area of similar triangles = square of ratio of respective sides, or altitude......ect.

This is a dodecahedron hence external angles are 30, its sides are 1.
By simple trigo we find out the base lengths, PQ, and PR, as under.
CF = 1+sqrt(3).....AH = 2+sqrt(3)....PQ = sqrt(3)/2....RP = 1+sqrt(3)/2 =(1/2)(2+sqrt(3))

By ratios of similar triangles and ratio rules we find altitudes OR and OQ.
RO/AH=OP/CF=(RO+OP)/(AH+CF)=RP/(AH+CF)...>RO=AH RP/(AH+CF).
RO =( 2+sqrt(3))
(1/2)(2+sqrt(3))./.( 2 + sqrt(3) + 1 + sqrt(3) )
RO=(1/2) ( (2+sqrt(3)) )^2./.( 3+2 sqrt(3) ) ...................................................(1)
..... = 1 + 1/(2* ( 3+2 sqrt(3) ) = 1 + QO......QO = (1/2)/( 3+2 sqrt(3) ).....(2)

Areas NOM./.AOH = ( QO./.RO )^2 = ( (2)/( 1) ) ^2 = 1./.( (2+sqrt(3)) )^4
Areas AOH = (1/2)AH RO =(1/2) * ( 2+sqrt(3 ) ) * (1/2) * ( (2+sqrt(3)) )^2./.( 3+2 sqrt(3) ) )
..= (1/4) * ( 2 + sqrt(3 ) )^3./..( 3+2 sqrt(3) ) )
Areas NOM = (1/4) * ( 2+sqrt(3) )^3./..( 3+2 sqrt(3) ) ) * 1./.( (2+sqrt(3))^4
..= (1/4)./.( ( 2+sqrt(3) ) * ( 3+2 * sqrt(3)) ) = (- 12 + 7 * sqrt(3))/12 =
= (- 12+7 * sqrt(3) )./.12
= (- 12+sqrt(147) )./.12 = ( - b + sqrt(a) ) / c...>>>a+b+c= 171.

I have not given details of multiplication and division of surds.