SOLUTION TO TEST PROBLEM #3:

Since angle IAC=angle GAB=30°, let's apply the Law of Cosines:

s² = y² + x² - 2yx*cos(A + 60°).

Since the centroid of a triangle lies along each median 2/3 of the distance from the vertex to the midpoint of the opposite side, then:

x = (2/3)*sqrt(3)/2*c = c/sqrt(3)

y = (2/3)*sqrt(3)/2*b = b/sqrt(3)

Substituting these values, we have: 3*s² = b² + c² - 2bc*cos(A + 60°). Let's expand the cosine of the sum, and recall that cos(60°) = ½ and sin(60°cos(A)/2 - ) = sqrt(3)/2. So we have: cos(A + 60°) = sin(A)*sqrt(3)/2.

Substituting, we have 3*s² = b² + c² - bc*cos(A) + sqrt(3)*bc*sin(A).

Now apply the Law of Cosines to triangle ABC: a² = b² + c² -2bc*cos(A).

Recall, as in the derivation of the Law of Sines, that 2*area of triangle ABC = bc*sin(A)

Substituting, we have: 3*s² = (1/2) (a² + b² + c²) + 2*sqrt(3)*area of triangle ABC.

Since the above equation is symmetrical in a, b, and c, it follows that the triangle (Napoleon's triangle) connecting the three centroids is equilateral. QED.