Okay, so I need to find an expression for AF² + BF² + CF² in terms of the side lengths a, b, c and the circumradius R of triangle ABC. Let me start by recalling some properties of centroids, orthocenters, and midpoints in triangles. 

First, the centroid G of a triangle divides the medians in a 2:1 ratio. The orthocenter H is the intersection point of the altitudes. The midpoint F of GH is given, so maybe there's a relationship between G, H, and F that can be useful here. 

I remember that in a triangle, the centroid, orthocenter, circumcenter, and the center of the nine-point circle are colinear on a line called the Euler line. The nine-point circle has a radius equal to half the circumradius. Wait, so the nine-point circle passes through the midpoints of the sides, the feet of the altitudes, and the midpoints of the segments from each vertex to the orthocenter. Since F is the midpoint of GH, which are both points on the Euler line, maybe F lies on the nine-point circle? Let me check that.

The nine-point circle's center is the midpoint between the orthocenter H and the circumcenter O. But in this case, F is the midpoint of GH. Hmm, unless G and O have a particular relationship. Wait, G divides the Euler line such that OG : GH = 1:2. So G is located one-third of the way from the circumcenter O to the orthocenter H. If F is the midpoint of GH, then F is the midpoint between G and H. Let's see, if G is one-third from O to H, then the midpoint between G and H would be at (2/3) of the way from O to H. But the nine-point center is the midpoint of O and H, so it's at (1/2) of OH. So F is not the nine-point center unless (2/3) of OH is equal to (1/2) of OH. Which would only be true if OH is zero, which is only possible if the triangle is equilateral. So F is not the nine-point center in general. So perhaps this line of thinking isn't helpful.

Alternatively, maybe coordinate geometry could help here. Let me consider placing the triangle in a coordinate system to simplify calculations. Let me place the centroid G at the origin (0, 0) for simplicity. Then, the coordinates of the centroid are the average of the coordinates of the vertices. If I let A be (x1, y1), B be (x2, y2), C be (x3, y3), then the centroid G is ((x1 + x2 + x3)/3, (y1 + y2 + y3)/3). If I set G at (0, 0), then x1 + x2 + x3 = 0 and y1 + y2 + y3 = 0.

Now, the orthocenter H. The coordinates of the orthocenter in terms of the coordinates of the vertices can be a bit complicated, but maybe there's a formula or relationship we can use. Alternatively, since we know the centroid divides the line joining the orthocenter and circumcenter in a 2:1 ratio, maybe we can relate H and O (the circumcenter) through G. Wait, the Euler line states that OG : GH = 1:2. So if G is the centroid, then the vector from O to G is 1/3 of the vector from O to H. So vector OG = (1/3) vector OH. Therefore, vector OH = 3 vector OG. But if G is at the origin, then vector OG is just the vector from O to G, which is G - O. Wait, if G is the centroid at (0,0), then the vector from O to G is (-O_x, -O_y). Hmm, maybe this is getting too complicated.

Alternatively, maybe I can use properties of vectors. Let me denote vectors from the centroid G as the origin. Then, the position vectors of A, B, C are vectors a, b, c respectively, such that a + b + c = 0 (since G is the average of the position vectors and is at the origin). 

The orthocenter H in terms of vectors a, b, c. Wait, there is a formula for the orthocenter in terms of the position vectors. If we have a triangle with centroid at the origin, then the orthocenter H can be expressed as a + b + c. But wait, if a + b + c = 0, then H would be at the origin, which can't be right. Wait, maybe I'm confusing something. Let me check.

Wait, if I take the centroid as the origin, then the position vectors of A, B, C are a, b, c with a + b + c = 0. The formula for the orthocenter in terms of these vectors is H = a + b + c. But since a + b + c = 0, then H would be at the origin. But that's not correct unless the triangle is equilateral? Hmm, maybe my approach here is flawed. Let me check another resource.

Wait, perhaps better to recall that in barycentric coordinates, the orthocenter can be expressed in terms of the triangle's side lengths and angles. Alternatively, maybe use coordinate geometry with specific coordinates.

Alternatively, maybe use properties of the centroid and orthocenter. Let me recall that in a triangle, the centroid divides the Euler line in the ratio OG : GH = 1:2. So if G is the centroid, then the distance from O to G is one-third of the distance from O to H. But since F is the midpoint of GH, then the distance from G to F is half of GH. So GF = (1/2)GH. Since OG : GH = 1:2, then GH = 2 OG. Therefore, GF = (1/2)(2 OG) = OG. So F is located at a distance OG from G in the direction opposite to O. Wait, that might not be correct. Let me think again.

If OG : GH = 1:2, and F is the midpoint of GH, then GF = FH = (1/2)GH. Since GH = 2 OG, then GF = (1/2)(2 OG) = OG. So F is located at a distance OG from G, but in the direction from G to H. Since G is the centroid, and O is the circumradius, maybe this gives a relationship between F and O. But I'm not sure if this helps directly.

Alternatively, maybe using coordinate geometry. Let me try to assign coordinates to the triangle to simplify calculations. Let me place the centroid G at (0, 0). Let me choose coordinates such that the triangle is in a plane with centroid at the origin. Let me denote the coordinates of A, B, C as (x1, y1), (x2, y2), (x3, y3). Then, since G is the centroid, (x1 + x2 + x3)/3 = 0 and (y1 + y2 + y3)/3 = 0. Therefore, x1 + x2 + x3 = 0 and y1 + y2 + y3 = 0.

Now, the centroid G is at (0, 0). The orthocenter H. To find H, maybe we can use the formula for the orthocenter in terms of coordinates. The coordinates of the orthocenter can be calculated using the formula involving the slopes of the sides and the altitudes. However, this might be too involved. Alternatively, if we can relate the coordinates of H to the coordinates of A, B, C in some way.

Alternatively, if we can use vector approaches. Let me denote vectors with boldface letters. Let me let vectors a, b, c be the position vectors of points A, B, C respectively. Then, since G is the centroid at the origin, we have a + b + c = 0.

The orthocenter H can be expressed in terms of the vectors a, b, c. There is a formula for the orthocenter in terms of these vectors. Wait, perhaps if I recall that in vector terms, the orthocenter H is given by H = a + b + c - 2O, where O is the circumcenter. Wait, is that correct? Wait, maybe not. Let me think.

Alternatively, if we consider that in the Euler line, the orthocenter H, centroid G, and circumcenter O are colinear with OG : GH = 1:2. So if G is at the origin, then O is at some point, say, O = (p, q), and then H would be at (3p, 3q). Wait, because OG : GH = 1:2, so the vector from O to G is (-p, -q), and the vector from G to H is (3p, 3q). Wait, if G is the centroid at (0,0), then vector OG is from O to G, which is G - O = (0 - p, 0 - q) = (-p, -q). The vector GH is H - G = H - (0,0) = H. So OG : GH = 1:2 implies that (-p, -q) : H = 1:2. Therefore, H = 2(-p, -q) = (-2p, -2q). Wait, that seems off. Let me check.

Wait, the ratio is OG : GH = 1:2. So the distance from O to G is one-third the distance from O to H. So if O is at (p, q), then G is at (0,0), so OG = sqrt(p² + q²). Then GH = 2 OG, so H is located at a distance 2 OG from G in the direction away from O. Wait, maybe in vector terms, H = G + 2(G - O). Since G is the centroid at (0,0), then H = 0 + 2(0 - O) = -2O. Therefore, H = -2O. So if O is at (p, q), then H is at (-2p, -2q). That seems more consistent.

Therefore, if I can find the coordinates of O, then H is simply -2 times the coordinates of O. But how do I find O in terms of the coordinates of A, B, C?

The circumcenter O is the intersection of the perpendicular bisectors of the sides of the triangle. In coordinate terms, given points A(x1, y1), B(x2, y2), C(x3, y3), the circumcenter O can be found by solving the perpendicular bisector equations. However, this might be tedious. Alternatively, there is a formula for the circumradius R in terms of the side lengths and coordinates, but that might not be helpful here.

Wait, maybe there's a different approach. Since we need AF² + BF² + CF², where F is the midpoint of GH. If we can express F in terms of the coordinates of A, B, C, then we can compute AF² + BF² + CF².

Given that G is the centroid at (0,0), and H is at (-2O) as per the previous deduction, then the midpoint F of GH would be (G + H)/2 = (0 + (-2O))/2 = -O. Therefore, F is located at -O, which is the reflection of O over the origin. Therefore, F is the reflection of the circumcenter O through the centroid G (which is at the origin).

Therefore, F is the point such that F = -O. Therefore, if O is at (p, q), then F is at (-p, -q). 

Therefore, since F is the reflection of O over G, and G is at the origin, F is simply the negative of the coordinates of O. Therefore, if we can express O in terms of the coordinates of A, B, C, then F is known.

Alternatively, perhaps we can express AF² + BF² + CF² in terms of coordinates. Let me denote F as (h, k). Then AF² + BF² + CF² = (x1 - h)² + (y1 - k)² + (x2 - h)² + (y2 - k)² + (x3 - h)² + (y3 - k)². 

But since G is the centroid at (0,0), we have x1 + x2 + x3 = 0 and y1 + y2 + y3 = 0. So expanding AF² + BF² + CF²:

= (x1² - 2x1h + h² + y1² - 2y1k + k²) + (x2² - 2x2h + h² + y2² - 2y2k + k²) + (x3² - 2x3h + h² + y3² - 2y3k + k²)

= (x1² + x2² + x3²) + (y1² + y2² + y3²) - 2h(x1 + x2 + x3) - 2k(y1 + y2 + y3) + 3h² + 3k²

But since x1 + x2 + x3 = 0 and y1 + y2 + y3 = 0, the terms involving h and k simplify, so we have:

= (x1² + x2² + x3²) + (y1² + y2² + y3²) + 3h² + 3k²

Therefore, AF² + BF² + CF² = (x1² + x2² + x3² + y1² + y2² + y3²) + 3(h² + k²)

Now, h and k are the coordinates of F, which is the midpoint of GH. But since G is at (0,0) and H is at (-2O), then F is at (-O). Therefore, h = -p and k = -q, where O is at (p, q). Therefore, h² + k² = p² + q².

But O is the circumcenter, so the coordinates of O can be expressed in terms of the coordinates of A, B, C. However, this might be complicated. Alternatively, we can relate h² + k² to other known quantities.

Wait, since O is the circumradius center, the distance from O to any vertex is R. Therefore, OA = OB = OC = R. Therefore, p² + (y1 - q)^2 = R², similarly for the other vertices. Wait, no. If O is at (p, q), then OA² = (x1 - p)^2 + (y1 - q)^2 = R². Similarly for OB and OC. But since F is at (-p, -q), then h² + k² = p² + q². So OA² = (x1 - p)^2 + (y1 - q)^2 = x1² - 2x1p + p² + y1² - 2y1q + q² = (x1² + y1²) - 2p x1 - 2q y1 + (p² + q²) = R².

Similarly, for OB²: (x2 - p)^2 + (y2 - q)^2 = R², and OC²: (x3 - p)^2 + (y3 - q)^2 = R².

Therefore, summing up OA² + OB² + OC²:

= [ (x1² + y1²) - 2p x1 - 2q y1 + (p² + q²) ] + [ (x2² + y2²) - 2p x2 - 2q y2 + (p² + q²) ] + [ (x3² + y3²) - 2p x3 - 2q y3 + (p² + q²) ]

= (x1² + x2² + x3² + y1² + y2² + y3²) - 2p(x1 + x2 + x3) - 2q(y1 + y2 + y3) + 3(p² + q²)

But since x1 + x2 + x3 = 0 and y1 + y2 + y3 = 0, this simplifies to:

= (x1² + x2² + x3² + y1² + y2² + y3²) + 3(p² + q²)

But OA² + OB² + OC² = 3R², since each is equal to R². Therefore:

3R² = (x1² + x2² + x3² + y1² + y2² + y3²) + 3(p² + q²)

Thus, (x1² + x2² + x3² + y1² + y2² + y3²) = 3R² - 3(p² + q²)

Therefore, substituting back into the expression for AF² + BF² + CF²:

AF² + BF² + CF² = (x1² + x2² + x3² + y1² + y2² + y3²) + 3(h² + k²) = [3R² - 3(p² + q²)] + 3(p² + q²) = 3R²

Wait, that's interesting. So AF² + BF² + CF² = 3R²? But that seems too simple. Wait, let me check.

Wait, we have AF² + BF² + CF² = (x1² + x2² + x3² + y1² + y2² + y3²) + 3(h² + k²)

But h² + k² = p² + q². Therefore, AF² + BF² + CF² = (x1² + x2² + x3² + y1² + y2² + y3²) + 3(p² + q²)

But we also found that (x1² + x2² + x3² + y1² + y2² + y3²) = 3R² - 3(p² + q²). Therefore, substituting:

AF² + BF² + CF² = [3R² - 3(p² + q²)] + 3(p² + q²) = 3R²

Therefore, the sum AF² + BF² + CF² is equal to 3R². Wait, but that seems counterintuitive. The result is independent of the side lengths a, b, c? But the problem statement says to express it in terms of the side lengths a, b, c and the circumradius R. So according to this calculation, AF² + BF² + CF² = 3R², regardless of the side lengths. But that might not be correct. Let me verify.

Wait, maybe I made a mistake in the steps. Let's go through again.

Starting with AF² + BF² + CF², expressed in terms of coordinates:

= (x1 - h)² + (y1 - k)² + (x2 - h)² + (y2 - k)² + (x3 - h)² + (y3 - k)²

Expanding each term:

= x1² - 2x1h + h² + y1² - 2y1k + k² + x2² - 2x2h + h² + y2² - 2y2k + k² + x3² - 2x3h + h² + y3² - 2y3k + k²

Combine like terms:

= (x1² + x2² + x3² + y1² + y2² + y3²) + 3h² + 3k² - 2h(x1 + x2 + x3) - 2k(y1 + y2 + y3)

Since G is the centroid at (0,0), x1 + x2 + x3 = 0 and y1 + y2 + y3 = 0. Therefore, the terms involving h and k vanish:

= (x1² + x2² + x3² + y1² + y2² + y3²) + 3h² + 3k²

Now, F is the midpoint of GH. Since G is at (0,0) and H is at (-2O), then F is at (-O). Therefore, h = -p and k = -q, so h² + k² = p² + q². Therefore:

AF² + BF² + CF² = (x1² + x2² + x3² + y1² + y2² + y3²) + 3(p² + q²)

But we also have that OA² = (x1 - p)^2 + (y1 - q)^2 = R², similarly for OB and OC. Summing these:

OA² + OB² + OC² = [ (x1 - p)^2 + (y1 - q)^2 ] + [ (x2 - p)^2 + (y2 - q)^2 ] + [ (x3 - p)^2 + (y3 - q)^2 ] = 3R²

Expanding each OA²:

= (x1² - 2x1p + p² + y1² - 2y1q + q²) + (x2² - 2x2p + p² + y2² - 2y2q + q²) + (x3² - 2x3p + p² + y3² - 2y3q + q²)

= (x1² + x2² + x3² + y1² + y2² + y3²) - 2p(x1 + x2 + x3) - 2q(y1 + y2 + y3) + 3(p² + q²)

Since x1 + x2 + x3 = 0 and y1 + y2 + y3 = 0, this simplifies to:

= (x1² + x2² + x3² + y1² + y2² + y3²) + 3(p² + q²) = 3R²

Therefore, (x1² + x2² + x3² + y1² + y3²) + 3(p² + q²) = 3R²

Hence, substituting back into AF² + BF² + CF²:

AF² + BF² + CF² = [3R² - 3(p² + q²)] + 3(p² + q²) = 3R²

So according to this calculation, AF² + BF² + CF² = 3R². But this seems to ignore the side lengths a, b, c. However, the problem statement asks to express it in terms of a, b, c and R. Therefore, maybe there's a mistake here.

Wait, perhaps I made an incorrect assumption in the coordinate system. Let me check again. I placed G at (0,0), but does this affect the generality of the result? Since the expression is supposed to be in terms of side lengths and R, which are independent of the coordinate system, maybe the result is indeed 3R². But that seems strange.

Alternatively, maybe there's a different approach. Let me recall the formula for the sum of squares of distances from a point to the vertices of a triangle. There is a formula that says for any point P in the plane of triangle ABC, PA² + PB² + PC² = GA² + GB² + GC² + 3PG². This is known as Leibniz's formula. 

In this case, F is a specific point. Let's see if we can apply this formula here. Let me recall that formula:

PA² + PB² + PC² = 3PG² + GA² + GB² + GC²

Where G is the centroid. Therefore, in our case, if we take P = F, then:

AF² + BF² + CF² = 3FG² + GA² + GB² + GC²

But since F is the midpoint of GH, and G is the centroid, we can relate FG to GH. Since F is the midpoint of GH, FG = (1/2)GH. But GH is the distance between G and H. 

Wait, in the Euler line, we know that GH = 2 OG, since OG : GH = 1:2. Therefore, FG = (1/2)GH = OG. But the distance from centroid G to circumcenter O is OG. Therefore, FG = OG. Therefore, FG² = OG².

Also, GA, GB, GC are the distances from centroid to the vertices. There is a formula that says GA² + GB² + GC² = (a² + b² + c²)/3. Wait, is that correct?

Wait, actually, the centroid divides the medians in a 2:1 ratio, and the distance from the centroid to a vertex can be expressed in terms of the side lengths. The formula for the distance from centroid G to vertex A is GA² = (2b² + 2c² - a²)/9. Similarly for GB² and GC².

Therefore, GA² + GB² + GC² = [(2b² + 2c² - a²) + (2a² + 2c² - b²) + (2a² + 2b² - c²)] / 9

Simplify numerator: 2b² + 2c² - a² + 2a² + 2c² - b² + 2a² + 2b² - c²

= (2b² - b² + 2b²) + (2c² + 2c² - c²) + (2a² + 2a² - a²)

= (3b²) + (3c²) + (3a²) = 3a² + 3b² + 3c²

Therefore, GA² + GB² + GC² = (3a² + 3b² + 3c²)/9 = (a² + b² + c²)/3

Therefore, going back to Leibniz's formula:

AF² + BF² + CF² = 3FG² + GA² + GB² + GC² = 3FG² + (a² + b² + c²)/3

But FG is the distance from centroid G to point F. Since F is the midpoint of GH, and G is the centroid, FG = (1/2)GH. But GH is the distance between centroid G and orthocenter H. From Euler line, GH = 2 OG. Therefore, FG = OG. 

But how can we express OG in terms of side lengths? There is a formula for the distance between centroid and circumcenter: OG² = R² - (a² + b² + c²)/9. Wait, let me verify.

Yes, the formula for the distance between centroid G and circumcenter O is OG² = R² - (a² + b² + c²)/9. Therefore, FG² = OG² = R² - (a² + b² + c²)/9

Therefore, substituting back into Leibniz's formula:

AF² + BF² + CF² = 3[ R² - (a² + b² + c²)/9 ] + (a² + b² + c²)/3

Simplify:

= 3R² - (a² + b² + c²)/3 + (a² + b² + c²)/3

= 3R²

Therefore, AF² + BF² + CF² = 3R²

But this contradicts the original problem's request to express it in terms of a, b, c and R. However, according to this calculation, AF² + BF² + CF² is equal to 3R², which is independent of the side lengths a, b, c. Therefore, the answer is 3R².

But wait, this seems to contradict the problem statement which says "Express AF² + BF² + CF² in terms of the side lengths a, b, c and circumradius R". However, according to the calculations, the result is only dependent on R. Therefore, perhaps the answer is 3R².

But let me check with a specific example. Let's take an equilateral triangle with side length a. The centroid G coincides with the circumcenter O and orthocenter H. Therefore, F is the midpoint of GH, which is the same as G (since G and H coincide). Therefore, F = G = O. Therefore, AF = AG. In an equilateral triangle, the distance from centroid to vertex is (2/3) the height, which is (2/3)(√3/2 a) = √3/3 a. Therefore, AF = √3/3 a. Therefore, AF² + BF² + CF² = 3*( (√3/3 a)^2 ) = 3*( (3/9 a² ) ) = 3*(1/3 a² ) = a². 

But according to our previous result, AF² + BF² + CF² = 3R². In an equilateral triangle, the circumradius R is a/√3. Therefore, 3R² = 3*(a²/3) = a², which matches the specific case. Therefore, in this case, the result is correct. 

Another test case: consider a right-angled triangle. Let's take a right-angled triangle with legs of length 3 and 4, hypotenuse 5. Then the circumradius R is half the hypotenuse, so R = 5/2.

Let me compute AF² + BF² + CF². 

First, find the centroid G. Coordinates: let’s place the triangle with right angle at the origin, vertices at (0,0), (3,0), (0,4). Centroid G is at ((0 + 3 + 0)/3, (0 + 0 + 4)/3) = (1, 4/3).

Orthocenter H in a right-angled triangle is at the right-angled vertex, so H is at (0,0). Therefore, midpoint F of GH is ((1 + 0)/2, (4/3 + 0)/2) = (0.5, 2/3). 

Therefore, AF² + BF² + CF² is the sum of squares of distances from F to each vertex.

Compute AF²: distance from (0.5, 2/3) to (0,0):

= (0.5 - 0)^2 + (2/3 - 0)^2 = 0.25 + 4/9 = 9/36 + 16/36 = 25/36

BF²: distance from (0.5, 2/3) to (3,0):

= (3 - 0.5)^2 + (0 - 2/3)^2 = (2.5)^2 + ( -2/3)^2 = 6.25 + 4/9 = 55.25/9 = 221/36

CF²: distance from (0.5, 2/3) to (0,4):

= (0 - 0.5)^2 + (4 - 2/3)^2 = 0.25 + (10/3)^2 = 0.25 + 100/9 = (9/36 + 400/36) = 409/36

Total sum: 25/36 + 221/36 + 409/36 = (25 + 221 + 409)/36 = 655/36 ≈ 18.194...

But according to our formula, AF² + BF² + CF² should be 3R² = 3*(5/2)^2 = 3*(25/4) = 75/4 = 18.75. However, in our calculation, it's 655/36 ≈ 18.194..., which is not equal to 75/4. Therefore, there's a discrepancy here. 

This suggests that the previous result is incorrect. Therefore, my earlier approach must have an error. 

Wait, perhaps the mistake is in assuming that AF² + BF² + CF² = 3R². In the equilateral triangle case, it worked, but in the right-angled triangle, it doesn't. Therefore, the result must be different.

Therefore, my mistake was perhaps in the Leibniz formula. Let me re-examine that.

Leibniz's formula states that for any point P, PA² + PB² + PC² = 3PG² + GA² + GB² + GC². Therefore, if we take P = F, then:

AF² + BF² + CF² = 3FG² + GA² + GB² + GC²

In the right-angled triangle example, let's compute each term.

First, compute FG². F is the midpoint of GH, which in the right-angled triangle is the midpoint between centroid G (1, 4/3) and orthocenter H (0,0). So F is (0.5, 2/3). Therefore, distance FG is the distance between G (1, 4/3) and F (0.5, 2/3):

FG² = (1 - 0.5)^2 + (4/3 - 2/3)^2 = (0.5)^2 + (2/3)^2 = 0.25 + 4/9 = 9/36 + 16/36 = 25/36

GA²: distance from G (1, 4/3) to A (0,0):

GA² = (1)^2 + (4/3)^2 = 1 + 16/9 = 25/9

GB²: distance from G (1, 4/3) to B (3,0):

GB² = (3 - 1)^2 + (0 - 4/3)^2 = 4 + 16/9 = 52/9

GC²: distance from G (1, 4/3) to C (0,4):

GC² = (1 - 0)^2 + (4/3 - 4)^2 = 1 + (-8/3)^2 = 1 + 64/9 = 73/9

Therefore, GA² + GB² + GC² = 25/9 + 52/9 + 73/9 = (25 + 52 + 73)/9 = 150/9 = 50/3

Therefore, according to Leibniz's formula:

AF² + BF² + CF² = 3FG² + GA² + GB² + GC² = 3*(25/36) + 50/3 = 75/36 + 600/36 = 675/36 = 75/4 = 18.75

Which matches 3R², since R = 5/2, so 3*(25/4) = 75/4. But when we computed the actual distances, we got 655/36 ≈ 18.194..., which is different. Therefore, there's a contradiction. Therefore, my mistake is in the calculation of AF² + BF² + CF². 

Wait, perhaps I miscalculated the distances. Let me recalculate AF², BF², CF² in the right-angled triangle.

Coordinates:

F = (0.5, 2/3)

A = (0,0)

B = (3,0)

C = (0,4)

Compute AF²:

(0.5 - 0)^2 + (2/3 - 0)^2 = 0.25 + 4/9 = 25/36 ≈ 0.694

BF²:

(0.5 - 3)^2 + (2/3 - 0)^2 = (-2.5)^2 + (2/3)^2 = 6.25 + 4/9 ≈ 6.25 + 0.444 ≈ 6.694

CF²:

(0.5 - 0)^2 + (2/3 - 4)^2 = 0.25 + (-10/3)^2 = 0.25 + 100/9 ≈ 0.25 + 11.111 ≈ 11.361

Sum: 25/36 + 6.25 + 100/9 ≈ 0.694 + 6.694 + 11.361 ≈ 18.749

Which is approximately 18.75, which is 75/4. Therefore, the actual calculation gives 75/4, which matches 3R². Therefore, my mistake was in the earlier calculation where I thought I got 655/36. Wait, 655/36 is approximately 18.194, but the correct sum is 75/4 = 18.75. Therefore, the correct result is indeed 3R².

Therefore, in the right-angled triangle, AF² + BF² + CF² = 3R², which matches the formula. Therefore, the initial result is correct. Therefore, in both test cases, the formula holds. Therefore, the answer is 3R².

But the problem statement says "Express AF² + BF² + CF² in terms of the side lengths a, b, c and circumradius R". However, according to our calculations, the result is 3R², which does not involve the side lengths a, b, c. Therefore, the answer is simply 3R².

But this seems counterintuitive. Let me check if there's another approach that takes into account the side lengths. Let me recall that in the Leibniz formula, we have:

PA² + PB² + PC² = 3PG² + GA² + GB² + GC²

In our case, P = F, so AF² + BF² + CF² = 3FG² + GA² + GB² + GC²

We found that FG = OG, and OG² = R² - (a² + b² + c²)/9

Therefore, FG² = R² - (a² + b² + c²)/9

GA² + GB² + GC² = (a² + b² + c²)/3

Therefore, AF² + BF² + CF² = 3[ R² - (a² + b² + c²)/9 ] + (a² + b² + c²)/3

= 3R² - (a² + b² + c²)/3 + (a² + b² + c²)/3

= 3R²

Therefore, regardless of the side lengths, AF² + BF² + CF² = 3R². Therefore, the answer is 3R².

Therefore, the boxed answer is \boxed{3R^2}.
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