Triangle ABC has ∠A=40∘, ∠B=60∘, ∠C=80∘. Points M,N trisect the side BC and points P,Q trisect the side AC. The lines AM,AN,BP,BQ intersect at the points S,T,U,V as shown in the figure below, dividing the triangle into 9 regions. Determine the smallest possible value of [ABC]+[STUV] such that both [ABC] and [STUV] are positive integers.

To determine the smallest possible value of [ABC]+[STUV], we first need to find the area of triangle ABC and the area of the region STUV.

To find the area of triangle ABC, we can use the formula:

[ABC] = 0.5 * |AB| * |AC| * sin(∠BAC)

First, let's calculate the lengths of each side of triangle ABC. Since points M and N trisect side BC, we can assume BC = 3x and MC = CN = x. Similarly, since points P and Q trisect side AC, we can assume AC = 3y and AP = PQ = QC = y.

Using the law of sines, we can find the lengths of AB and AC:

sin(∠ABC) / BC = sin(∠ACB) / AB

sin(60∘) / 3x = sin(80∘) / AB

AB = (3x * sin(60∘)) / sin(80∘)

Similarly, using the law of sines again, we can find the length of AC:

sin(∠ACB) / BC = sin(∠ABC) / AC

sin(80∘) / 3x = sin(40∘) / AC

AC = (3x * sin(40∘)) / sin(80∘)

Now, we can calculate the area of triangle ABC:

[ABC] = 0.5 * AB * AC * sin(∠BAC)

Substituting the values for AB and AC, we get:

[ABC] = 0.5 * ( (3x * sin(60∘)) / sin(80∘) ) * ( (3x * sin(40∘)) / sin(80∘) ) * sin(40∘)

Simplifying further, we have:

[ABC] = (9x^2 * sin(60∘) * sin(40∘) * sin(40∘)) / (2 * sin(80∘) * sin(80∘))

Now, let's find the area of the region STUV.

Since S, T, U, and V are formed by the intersection of lines AM, AN, BP, and BQ, they divide the triangle ABC into 9 smaller triangles. We can find the area of each smaller triangle and sum them up to get the total area of STUV.

To find the area of each smaller triangle, we can use the formula:

[XYZ] = 0.5 * |XY| * |XZ| * sin(∠YXZ)

Now, let's analyze the triangle formed by points A, M, and S. Since points M and N trisect side BC, we know that MS = x and AS = 2x (since AM = 3x). Furthermore, since angle AMS is common with angle ABC and AM/BC = 3x/3x = 1, we can conclude that triangle AMS is similar to triangle ABC. Therefore, if [ABC] = K, then [AMS] = (x^2/K) * [ABC].

Similarly, we can analyze the triangles formed by points A, N, and T, and points C, P, and U. Using the same logic, we can determine that the areas of triangles ANT and CPU are both (x^2/K) * [ABC].

Lastly, let's analyze the triangle formed by points B, Q, and V. Similar to the previous cases, we can deduce that triangle BQV is similar to triangle ABC. Therefore, its area will also be (x^2/K) * [ABC].

Now, we have four triangles with areas that are all proportional to [ABC] with the same constant of proportionality (x^2/K). Therefore, we can calculate the area of STUV as:

[STUV] = 4 * (x^2/K) * [ABC]

Now, we want to find the smallest possible values of [ABC] and [STUV] such that both are positive integers.

Let's analyze the expression for [ABC]:

[ABC] = (9x^2 * sin(60∘) * sin(40∘) * sin(40∘)) / (2 * sin(80∘) * sin(80∘))

In order for [ABC] to be an integer, we need the numerator to be divisible by both 2 * sin(80∘) and sin(80∘). Since these are irrational numbers, it won't be possible to find an exact value for x that satisfies this condition. However, by substituting values for x, we can find the closest possible values of [ABC] that are positive integers.

Similarly, to find the smallest possible value of [STUV], we need (x^2/K) * [ABC] to be an integer. By adjusting the value of x, we can find the closest possible values of [STUV] that are positive integers.

Therefore, by experimenting with different values of x and finding the corresponding values of [ABC] and [STUV], we can determine the smallest possible value of [ABC] + [STUV] such that both [ABC] and [STUV] are positive integers.