Verified answer

Draw a line AB intersecting NR at A, intersecting LQ at B,

and AB is in parallel with RQ.

The ratio of areas of equilateral triangles can be transformed

into ratio of the altitudes of equilateral triangles.

(area of triangle PAB)/(area of triangle PRQ)=((6-2)^2)/(6^2)=16/36.

(area of triangle LONP)/(area of triangle PAB)=

(area of triangle PAB-area of triangle ONA-area of triangle OLB)

/(area of triangle PAB)=(4^2-(1/2)*3^2-(1/2)*1^2)/4^2=11/16

So a/b=(11/16)*(16/36)=11/36

I got a*b = ab = 396

Similarity considerations allow us the dimensions on the diagram to be:

|OL| = 1, |OM| = 2, |ON| = 3

It's well-known (and easy to prove) that for every internal point O the sum

|OL| + |OM| + |ON| = 6 is equal to the length of the altitude of the equilateral triangle PQR, so its side length is 4√3, hence

Area(PQR) = 12√3

Let's connect O with P and consider the angles

λ = angle(OPL) and ν = angle(OPN), obviously

(*) λ + ν = π/3 and sin λ : sin ν = |OL| : |ON| = 1 : 3

Now Area(LONP) = Area(OPL) + Area(OPN) =

= |OL| * |PL| / 2 + |ON| * |PN| / 2 =

= (cot λ + 9 cot ν)/2

Further sin v / sin λ = sin(π/3 - λ) / sin λ =

= (√3/2) * cot λ - 1/2 = 3, so we get

cot λ = 7/√3 and then, using (*),

cot ν = cot(π/3 - λ) = 5√3/9

Now the irreducible fraction

a / b = Area(LONP) / Area(PQR) =

= (1/2)( cot λ + 9 cot ν) / (12√3) =

= (11√3/3) / (12√3) = 11 / 36, so a = 11, b = 36

P.S.(After having read Philo's answer, the end of the 2nd paragraph - last 2 text lines):

"√3 / 7 = sin α / cos α = tan α

since 1/PL = tan α, PL = √3/7 and area POL = √3/14"

Well, tan α = √3/7 indeed (λ in my notations above), but

since 1/PL = tan α, I get

PL = 1/ tan α = 7/√3 instead of √3/7, or I'm wrong?!?

P.S.2.(Plagiarizing Lester L) If O is an internal point for the equilateral triangle PQR, then the 3 lines through O, parallel to the sides PQ, QR, RP decompose PQR into 3 parallelograms and 3 triangles - let the latter altitudes are in ratio l : m : n /l, m, n > 0/, then easily can be seen, their areas A_{l}, A_{m}, A_{n} satisfy:

√A_{l} : √A_{m} : √A_{n} = l : m : n and

√A_{l} + √A_{m} + √A_{n} = √Area(PQR)

Hence the ratios of the roots of areas of the small triangles to the area of the big one are

l/(l + m + n), m/(l + m + n) and n/(l + m + n)

Then the required ratio (2 adjacent sides of the quadrilateral in ratio l : n) is

(l + n)²/(l + m + n)² - (1/2)l²/(l + m + n)² - (1/2)n²/(l + m + n)² =

=(2ln + l²/2 + n²/2) / (l + m + n)²,

the latter for l = 1, m = 2, n = 3 yields

a/b = (6 + 1/2 + 9/2)/36 = 11/36, a + b =47.

Lester L doesn't cease astonishing me with inventive and inspired solutions! Congratulations, Lester - reading Yours, I didn't like mine first!

I agree with Duke that if 1,2,3 are the lengths of OL, OM, and ON, then area(PQR) = 12√3. And the key is splitting the angle at P and finding areas of triangles POL and PON. After that we differ enough to get different answers.

Let angle OPL be α and length PO be x.

sin α = 1/x and sin (60-α) = 3/x, so

sin(60-α) = 3 sin α

sin 60 cos α - cos 60 sin α = 3 sin α

½√3 cos α - ½ sin α = 3 sin α

√3 cos α - sin α = 6 sin α

√3 cos α = 7 sin α

√3 / 7 = sin α / cos α = tan α

since 1/PL = tan α, PL = √3/7 and area POL = √3/14

for triangle PON,

tan(60-α) = [tan 60 - tan α] / [1 + tan 60 • tan α]

tan(60-α) = [ √3 - √3/7] / [ 1 + √3 • √3/7]

tan(60-α) = [ 6√3 / 7] / [ 1 + 3/7]

tan(60-α) = [ 6√3 / 7] / [ 10/7]

tan(60-α) = 3√3 / 5

so 3/PN = (3√3)/5, 1/PN = √3/5, PN = 5/√3

then area PON = ½•3•5/√3 = 15/(2√3) = (5√3)/2

finally, areas POL + PON =

√3 / 14 + 5√3 / 2 =

√3 / 14 + 35√3 / 14 =

36√3 / 14

that makes the ratio of the shaded area to the area of the triangle

[ 36√3 / 14] / [12√3] = 3/14

so a = 3, b = 14, and ab = 42

which is the meaning of life, the universe, and everything, so we're told.