Explanation:
Okay, so I need to find the largest integer less than 2010 that satisfies three conditions: when divided by 7, it leaves a remainder of 5; when divided by 11, it leaves a remainder of 10; and when divided by 13, it also leaves a remainder of 10. Hmm, let me think about how to approach this.

First, I remember that these types of problems are usually solved using the Chinese Remainder Theorem (CRT). CRT allows us to solve systems of congruences with pairwise coprime moduli. Let me check if the moduli here are coprime. The moduli are 7, 11, and 13. Since 7, 11, and 13 are all prime numbers, they are pairwise coprime. That means CRT applies here, and there should be a unique solution modulo 7*11*13. Let me compute that product first.

Calculating 7*11 is 77, and then 77*13. Let me do that step by step. 70*13 is 910, and 7*13 is 91, so adding them together gives 910 + 91 = 1001. So the moduli are 7, 11, 13, and their product is 1001. Therefore, the solutions will repeat every 1001 numbers. So, if I can find the smallest positive solution, then I can add multiples of 1001 to it to find other solutions. Since we need the largest one less than 2010, I can find how many times 1001 fits into 2010 and then adjust accordingly.

But first, let me set up the system of congruences:

Let the number be x. Then:

x ≡ 5 mod 7

x ≡ 10 mod 11

x ≡ 10 mod 13

So, I need to solve for x in these congruences. Let me start by solving the first two congruences and then combine the result with the third one.

Starting with x ≡ 5 mod 7 and x ≡ 10 mod 11.

Let me write x as 7k + 5 for some integer k. Then substitute this into the second equation:

7k + 5 ≡ 10 mod 11

Subtract 5 from both sides:

7k ≡ 5 mod 11

Now, I need to solve for k. To do this, I need the multiplicative inverse of 7 modulo 11. Let me find that. The inverse of 7 mod 11 is a number m such that 7m ≡ 1 mod 11.

Trying m=8: 7*8=56. 56 divided by 11 is 5*11=55, remainder 1. Yes! So 7*8 ≡ 1 mod 11. Therefore, the inverse is 8.

Multiply both sides of 7k ≡ 5 mod 11 by 8:

k ≡ 5*8 mod 11 => k ≡ 40 mod 11. 40 divided by 11 is 3*11=33, remainder 7. So k ≡ 7 mod 11. Therefore, k can be written as 11m + 7 for some integer m.

Therefore, substituting back into x = 7k + 5:

x = 7*(11m + 7) + 5 = 77m + 49 + 5 = 77m + 54.

So, the solutions to the first two congruences are x ≡ 54 mod 77. Now, we need to incorporate the third congruence: x ≡ 10 mod 13.

So, we have x ≡ 54 mod 77 and x ≡ 10 mod 13. Let me write x as 77m + 54 and substitute into the third congruence:

77m + 54 ≡ 10 mod 13.

First, let me compute 77 mod 13 and 54 mod 13 to simplify.

Calculating 13*5 = 65, so 77 - 65 = 12. Therefore, 77 ≡ 12 mod 13.

Similarly, 54 divided by 13: 13*4 = 52, so 54 - 52 = 2. Therefore, 54 ≡ 2 mod 13.

Therefore, substituting back:

12m + 2 ≡ 10 mod 13

Subtract 2 from both sides:

12m ≡ 8 mod 13

Now, need to solve for m. Let me find the inverse of 12 mod 13. Since 12 ≡ -1 mod 13, so inverse of -1 is -1, because (-1)*(-1) = 1 mod 13. Therefore, inverse of 12 mod 13 is 12.

Multiply both sides by 12:

m ≡ 8*12 mod 13

Calculate 8*12 = 96. Now divide 96 by 13: 13*7 = 91, so 96 - 91 = 5. Therefore, m ≡ 5 mod 13. Therefore, m can be written as 13n + 5 for some integer n.

Therefore, substituting back into x = 77m + 54:

x = 77*(13n + 5) + 54 = 77*13n + 385 + 54 = 1001n + 439.

Therefore, the solutions to all three congruences are x ≡ 439 mod 1001. Therefore, the general solution is x = 1001n + 439, where n is an integer.

Now, we need the largest integer less than 2010. So, let's find the maximum n such that 1001n + 439 < 2010.

Let me solve for n:

1001n + 439 < 2010

Subtract 439:

1001n < 2010 - 439 = 1571

Therefore, n < 1571 / 1001 ≈ 1.569...

Since n must be an integer, the maximum possible n is 1. Therefore, the largest x less than 2010 is when n=1: x = 1001*1 + 439 = 1001 + 439 = 1440.

Wait, but let me check if there is a larger n? Wait, n=1 gives 1440, n=2 would be 1001*2 + 439 = 2002 + 439 = 2441, which is way above 2010. So n=1 is the maximum. Therefore, 1440 is the answer?

Wait, but let me check if 1440 satisfies all the original conditions. Let me verify:

First, check remainder when divided by 7: 1440 divided by 7. Let me compute 7*200 = 1400. 1440 - 1400 = 40. 40 divided by 7 is 5*7=35, remainder 5. So 1440 ≡ 5 mod 7. Good.

Next, check remainder when divided by 11: 1440 divided by 11. Let me compute 11*130 = 1430. 1440 - 1430 = 10. So remainder 10. Perfect.

Third, remainder when divided by 13: 1440 divided by 13. Let me compute 13*110 = 1430. 1440 - 1430 = 10. So remainder 10. Perfect. So 1440 satisfies all three conditions. But wait, is there a larger number less than 2010? Let me check if there is another solution between 1440 and 2010. Since the solutions are spaced 1001 apart, the next one after 1440 would be 1440 + 1001 = 2441, which is more than 2010. Therefore, 1440 is indeed the largest one less than 2010. Wait, but let me check if there is another solution between 1440 and 2010. For example, maybe there is another number in that range that satisfies the congruences? But according to our solution, the solutions are x = 1001n + 439. So for n=1, 1440; n=0, 439; n=2, 2441. So between 1440 and 2010, there's no other solution. Therefore, 1440 is the answer. But wait, let me check if 1440 is indeed less than 2010. Yes, 1440 is less than 2010. But wait, maybe there is a larger number? Let me think again.

Wait, perhaps I made a mistake in solving the congruences? Let me check again.

Starting with x ≡ 5 mod 7, x ≡ 10 mod 11, x ≡ 10 mod 13.

First step: solving x ≡ 5 mod 7 and x ≡ 10 mod 11. Let me check my steps again.

x = 7k + 5. Then 7k + 5 ≡ 10 mod 11 => 7k ≡ 5 mod 11. Then inverse of 7 mod 11 is 8, since 7*8=56≡1 mod11. Then k ≡ 5*8=40≡7 mod11. So k=11m +7. Then x=7*(11m +7)+5=77m +49 +5=77m +54. So x≡54 mod77. Then combining with x≡10 mod13.

So x=77m +54. Then 77m +54 ≡10 mod13. 77 mod13: 13*5=65, 77-65=12, so 77≡12 mod13. 54 mod13: 13*4=52, 54-52=2, so 54≡2 mod13. Therefore, 12m +2 ≡10 mod13 => 12m≡8 mod13. Then inverse of 12 mod13 is 12, since 12*12=144≡144-11*13=144-143=1 mod13. Therefore, m≡8*12=96≡96-7*13=96-91=5 mod13. Therefore, m=13n +5. Therefore, x=77*(13n +5)+54= 77*13n + 385 +54= 1001n + 439. So that's correct. Therefore, solutions are 439, 1440, 2441,... So yes, 1440 is the one before 2441. So 1440 is the answer. But wait, let me check if there is another solution between 1440 and 2010. For example, maybe 1440 + 1001 = 2441 is too big, but maybe there is another solution? Wait, no, because the modulus is 1001, so the solutions are spaced exactly 1001 apart. Therefore, between 0 and 1001, there is 439; between 1001 and 2002, there is 1440; and then next is 2441. Therefore, 1440 is indeed the largest less than 2010. Therefore, the answer is 1440. Wait, but let me check if 1440 is indeed less than 2010. Yes, 1440 is 1440, and 2010 is larger. So 1440 is the answer. But wait, let me check if there is a larger number less than 2010. For example, maybe 1440 + something?

Wait, suppose I made a mistake in the calculation of the modulus? Let me check again. The modulus is 7*11*13=1001. So the solutions are 439 + 1001n. So 439 + 1001*1=1440, 439 + 1001*2=2441. So yes, 1440 is the largest less than 2010. Therefore, 1440 is the answer. But let me check if there is another number between 1440 and 2010 that satisfies the conditions. For example, maybe there is a number that is congruent to 439 mod 1001 but larger than 1440 but less than 2010? Well, 1440 + 1001 = 2441, which is more than 2010. Therefore, no. Therefore, 1440 is the answer. Wait, but let me check if 1440 is indeed the correct solution. Let me check again:

Divide 1440 by 7: 7*205 = 1435, so 1440 - 1435 = 5. So remainder 5. Correct.

Divide 1440 by 11: 11*130 = 1430, 1440 - 1430 = 10. Correct.

Divide 1440 by 13: 13*110 = 1430, 1440 - 1430 = 10. Correct. So all conditions are satisfied. Therefore, 1440 is correct. But wait, the problem says "the largest integer less than 2010". Is there a possibility that there is another number between 1440 and 2010 that satisfies the conditions? For example, maybe there is a number that is congruent to 439 mod 1001 but larger than 1440? Well, the next one is 1440 + 1001 = 2441, which is more than 2010. Therefore, no. Therefore, 1440 is the answer. Wait, but let me check if there is another solution that I didn't consider. For example, maybe there is a different way to solve the congruences?

Alternatively, maybe I can check numbers near 2010. Let me think. Let me check if 2010 - 1001 = 1009. Is 1009 a solution? Let me check. 1009 divided by 7: 7*144=1008, so remainder 1. Not 5. So no. How about 2010 - 1001*1 = 1009, which we saw. Then 2010 - 1001*0 = 2010. Let me check 2010. But 2010 is not less than 2010, so it's excluded. So the previous one is 1009, which doesn't work. Then maybe check numbers between 1440 and 2010. For example, 1440 + 1001 = 2441, which is too big. So maybe there is no other solution. Therefore, 1440 is the answer. Wait, but let me check another way. Let me think about solving the congruences again.

Alternatively, since x ≡ 10 mod 11 and x ≡ 10 mod 13. So x ≡ 10 mod lcm(11,13). Since 11 and 13 are coprime, lcm(11,13)=143. Therefore, x ≡ 10 mod 143. Therefore, x can be written as 143k + 10. Then, we also have x ≡ 5 mod 7. Therefore, substituting x = 143k + 10 into x ≡ 5 mod 7:

143k + 10 ≡ 5 mod 7. Let me compute 143 mod 7. 7*20=140, so 143-140=3. Therefore, 143 ≡ 3 mod7. Therefore, 3k + 10 ≡ 5 mod7. Then 3k ≡ 5 -10 mod7 => 3k ≡ -5 mod7 => -5 mod7 is 2, since -5 +7=2. Therefore, 3k ≡ 2 mod7. Solve for k. Multiply both sides by inverse of 3 mod7. The inverse of 3 mod7 is 5, since 3*5=15≡1 mod7. Therefore, k ≡ 2*5=10≡3 mod7. Therefore, k=7m +3. Therefore, x=143*(7m +3) +10=143*7m + 429 +10=1001m + 439. Which is the same result as before. Therefore, x=1001m +439. Therefore, same conclusion. Therefore, the solutions are 439, 1440, 2441,... So 1440 is the answer. Therefore, I think that is correct.

But just to be thorough, let me check if there is a larger number less than 2010. Let me compute 1001*1 + 439 = 1440. Then 1001*2 + 439 = 2441. Since 2441 is more than 2010, so no. Therefore, 1440 is the answer. Therefore, I think the answer is 1440. Wait, but let me check if there is another solution between 1440 and 2010. For example, maybe there is a number that is congruent to 439 mod 1001 but not necessarily 1001n + 439? No, because the modulus is 1001, so all solutions are of the form 1001n + 439. Therefore, there are no other solutions. Therefore, 1440 is the answer.

But just to be 100% sure, let me check if there is a mistake in the calculation of 1001*1 + 439. 1001 + 439: 1000 + 400 = 1400, 1 + 39 = 40, so total 1400 + 40 = 1440. Correct. And 1440 is less than 2010. Therefore, yes. Therefore, I think the answer is 1440.

But wait, let me check if there is a larger number less than 2010. For example, suppose I take x = 1440 + 1001 = 2441, which is over. But maybe there is another number between 1440 and 2010 that satisfies the congruences? For example, maybe there is a number that is congruent to 439 mod 1001 but not in the form 1001n + 439? No, because modulus is 1001, so all solutions are in that form. Therefore, no. Therefore, 1440 is the answer.

Alternatively, maybe I can check numbers near 2010. Let me try 2010 - 10 = 2000. Let me check if 2000 satisfies the conditions. 2000 divided by 7: 7*285=1995, remainder 5. So 2000 ≡ 5 mod7. Good. Then check mod11: 2000 divided by 11. Let me compute 11*181= 1991. 2000 - 1991=9. So remainder 9. But we need remainder 10. Not good. Next, 2000 - 1 = 1999. Check 1999 mod11. 1999 - 1991=8. Not 10. 1999 - 8=1991. Not helpful. How about 2000 - 10=1990. Check 1990 mod11. 11*180=1980. 1990 - 1980=10. So remainder 10. Good. Then check mod13. 1990 divided by 13. Let me compute 13*153= 1989. 1990 - 1989=1. So remainder 1. Not 10. Not good. So 1990 is not a solution. How about 1990 + 11=2001. Check 2001 mod7: 2001 - 2002= -1, so 2001 ≡ -1 mod7, which is 6 mod7. Not 5. Not good. How about 1990 + 13=2003. Check mod7: 2003 - 2002=1, so 1 mod7. Not 5. Not helpful. Maybe this approach is too tedious. Alternatively, since we know the solutions are 1001n + 439, and the next one after 1440 is 2441, which is over 2010, so 1440 is the answer. Therefore, I think I can be confident that 1440 is the correct answer.

But just to check another number, let me check 1440 + 13 = 1453. Check if it satisfies the conditions. 1453 mod7: 1440 is 5 mod7, so 1453 is 1440 +13. 13 mod7 is 6. So 5 +6=11≡4 mod7. Not 5. Not good. Similarly, adding 11: 1440 +11=1451. 1451 mod7: 1440 +11=1451. 1440≡5, 11≡4 mod7, so 5+4=9≡2 mod7. Not 5. Not good. So adding 11 or 13 doesn't help. Therefore, no other numbers between 1440 and 2010. Therefore, 1440 is the answer.

Therefore, I think the answer is 1440. Wait, but let me check once more if there is a mistake in the calculation of the initial solution. Let me check solving the congruences again.

Starting with x ≡ 5 mod7, x ≡10 mod11, x≡10 mod13.

First, solving x ≡10 mod11 and x≡10 mod13. Since 11 and 13 are coprime, x ≡10 mod(11*13)=143. Therefore, x=143k +10. Then, substitute into x≡5 mod7:

143k +10 ≡5 mod7. As before, 143 mod7 is 3, so 3k +10 ≡5 mod7. 10 mod7 is 3, so 3k +3 ≡5 mod7 => 3k ≡2 mod7. Multiply by inverse of 3 mod7, which is 5: k≡10≡3 mod7. Therefore, k=7m +3. Therefore, x=143*(7m +3)+10=1001m + 429 +10=1001m +439. So same result. Therefore, correct. Therefore, x=1001m +439. Therefore, the answer is 1440. Therefore, I think that is correct.

Therefore, I think the answer is 1440. Therefore, I can write that as the final answer.

**Final Answer**
\boxed{1440}

To solve the problem, we are looking for the **largest integer less than 2010** that satisfies the following system of congruences:

$$
\begin{cases}
x \equiv 5 \pmod{7} \\
x \equiv 10 \pmod{11} \\
x \equiv 10 \pmod{13}
\end{cases}
$$

---

### Step 1: Combine the congruences

We begin by combining the last two congruences:

$$
x \equiv 10 \pmod{11} \quad \text{and} \quad x \equiv 10 \pmod{13}
$$

Since 11 and 13 are coprime, we can apply the **Chinese Remainder Theorem (CRT)** to combine them:

$$
x \equiv 10 \pmod{143}
$$

So now we have:

$$
x \equiv 5 \pmod{7} \quad \text{and} \quad x \equiv 10 \pmod{143}
$$

---

### Step 2: Solve the combined system

Let’s express $ x $ in terms of the second congruence:

$$
x = 143k + 10
$$

Substitute into the first congruence:

$$
143k + 10 \equiv 5 \pmod{7}
$$

Now compute $ 143 \mod 7 $:

$$
143 \div 7 = 20 \text{ remainder } 3 \Rightarrow 143 \equiv 3 \pmod{7}
$$

So the equation becomes:

$$
3k + 10 \equiv 5 \pmod{7}
$$

Subtract 10 from both sides:

$$
3k \equiv -5 \equiv 2 \pmod{7}
$$

Now solve for $ k $:

We need the **multiplicative inverse** of 3 modulo 7. Since $ 3 \times 5 = 15 \equiv 1 \pmod{7} $, the inverse is 5.

Multiply both sides by 5:

$$
k \equiv 2 \times 5 = 10 \equiv 3 \pmod{7}
$$

So $ k = 7m + 3 $, and substituting back:

$$
x = 143k + 10 = 143(7m + 3) + 10 = 1001m + 429 + 10 = 1001m + 439
$$

---

### Step 3: Find the largest such $ x < 2010 $

We now have the general solution:

$$
x = 1001m + 439
$$

We want the **largest** such $ x $ less than 2010:

$$
1001m + 439 < 2010
\Rightarrow 1001m < 1571
\Rightarrow m < \frac{1571}{1001} \approx 1.569
$$

So the largest integer $ m $ is $ m = 1 $, which gives:

$$
x = 1001 \times 1 + 439 = 1440
$$

---

### Step 4: Verify the solution

Check that $ x = 1440 $ satisfies all three original congruences:

- $ 1440 \div 7 = 205 $ remainder $ 5 $ → $ 1440 \equiv 5 \pmod{7} $
- $ 1440 \div 11 = 130 $ remainder $ 10 $ → $ 1440 \equiv 10 \pmod{11} $
- $ 1440 \div 13 = 110 $ remainder $ 10 $ → $ 1440 \equiv 10 \pmod{13} $

All conditions are satisfied.

---

### Final Answer

$$
\boxed{1440}
$$

[Assistant]
Explanation:
Okay, so I need to find the largest integer less than 2010 that satisfies three conditions: when divided by 7, it leaves a remainder of 5; when divided by 11, it leaves a remainder of 10; and when divided by 13, it also leaves a remainder of 10. Hmm, let me think about how to approach this.

First, I remember that these types of problems are usually solved using the Chinese Remainder Theorem (CRT). CRT allows us to solve systems of congruences with pairwise coprime moduli. Let me check if the moduli here are coprime. The moduli are 7, 11, and 13. Since 7, 11, and 13 are all prime numbers, they are pairwise coprime. That means CRT applies here, and there should be a unique solution modulo 7*11*13. Let me compute that product first.

Calculating 7*11 is 77, and then 77*13. Let me do that step by step. 70*13 is 910, and 7*13 is 91, so adding them together gives 910 + 91 = 1001. So the moduli are 7, 11, 13, and their product is 1001. Therefore, the solutions will repeat every 1001 numbers. So, if I can find the smallest positive solution, then I can add multiples of 1001 to it to find other solutions. Since we need the largest one less than 2010, I can find how many times 1001 fits into 2010 and then adjust accordingly.

But first, let me set up the system of congruences:

Let the number be x. Then:

x ≡ 5 mod 7

x ≡ 10 mod 11

x ≡ 10 mod 13

So, I need to solve for x in these congruences. Let me start by solving the first two congruences and then combine the result with the third one.

Starting with x ≡ 5 mod 7 and x ≡ 10 mod 11.

Let me write x as 7k + 5 for some integer k. Then substitute this into the second equation:

7k + 5 ≡ 10 mod 11

Subtract 5 from both sides:

7k ≡ 5 mod 11

Now, I need to solve for k. To do this, I need the multiplicative inverse of 7 modulo 11. Let me find that. The inverse of 7 mod 11 is a number m such that 7m ≡ 1 mod 11.

Trying m=8: 7*8=56. 56 divided by 11 is 5*11=55, remainder 1. Yes! So 7*8 ≡ 1 mod 11. Therefore, the inverse is 8.

Multiply both sides of 7k ≡ 5 mod 11 by 8:

k ≡ 5*8 mod 11 => k ≡ 40 mod 11. 40 divided by 11 is 3*11=33, remainder 7. So k ≡ 7 mod 11. Therefore, k can be written as 11m + 7 for some integer m.

Therefore, substituting back into x = 7k + 5:

x = 7*(11m + 7) + 5 = 77m + 49 + 5 = 77m + 54.

So, the solutions to the first two congruences are x ≡ 54 mod 77. Now, we need to incorporate the third congruence: x ≡ 10 mod 13.

So, we have x ≡ 54 mod 77 and x ≡ 10 mod 13. Let me write x as 77m + 54 and substitute into the third congruence:

77m + 54 ≡ 10 mod 13.

First, let me compute 77 mod 13 and 54 mod 13 to simplify.

Calculating 13*5 = 65, so 77 - 65 = 12. Therefore, 77 ≡ 12 mod 13.

Similarly, 54 divided by 13: 13*4 = 52, so 54 - 52 = 2. Therefore, 54 ≡ 2 mod 13.

Therefore, substituting back:

12m + 2 ≡ 10 mod 13

Subtract 2 from both sides:

12m ≡ 8 mod 13

Now, need to solve for m. Let me find the inverse of 12 mod 13. Since 12 ≡ -1 mod 13, so inverse of -1 is -1, because (-1)*(-1) = 1 mod 13. Therefore, inverse of 12 mod 13 is 12.

Multiply both sides by 12:

m ≡ 8*12 mod 13

Calculate 8*12 = 96. Now divide 96 by 13: 13*7 = 91, so 96 - 91 = 5. Therefore, m ≡ 5 mod 13. Therefore, m can be written as 13n + 5 for some integer n.

Therefore, substituting back into x = 77m + 54:

x = 77*(13n + 5) + 54 = 77*13n + 385 + 54 = 1001n + 439.

Therefore, the solutions to all three congruences are x ≡ 439 mod 1001. Therefore, the general solution is x = 1001n + 439, where n is an integer.

Now, we need the largest integer less than 2010. So, let's find the maximum n such that 1001n + 439 < 2010.

Let me solve for n:

1001n + 439 < 2010

Subtract 439:

1001n < 2010 - 439 = 1571

Therefore, n < 1571 / 1001 ≈ 1.569...

Since n must be an integer, the maximum possible n is 1. Therefore, the largest x less than 2010 is when n=1: x = 1001*1 + 439 = 1001 + 439 = 1440.

Wait, but let me check if there is a larger n? Wait, n=1 gives 1440, n=2 would be 1001*2 + 439 = 2002 + 439 = 2441, which is way above 2010. So n=1 is the maximum. Therefore, 1440 is the answer?

Wait, but let me check if 1440 satisfies all the original conditions. Let me verify:

First, check remainder when divided by 7: 1440 divided by 7. Let me compute 7*200 = 1400. 1440 - 1400 = 40. 40 divided by 7 is 5*7=35, remainder 5. So 1440 ≡ 5 mod 7. Good.

Next, check remainder when divided by 11: 1440 divided by 11. Let me compute 11*130 = 1430. 1440 - 1430 = 10. So remainder 10. Perfect.

Third, remainder when divided by 13: 1440 divided by 13. Let me compute 13*110 = 1430. 1440 - 1430 = 10. So remainder 10. Perfect. So 1440 satisfies all three conditions. But wait, is there a larger number less than 2010? Let me check if there is another solution between 1440 and 2010. Since the solutions are spaced 1001 apart, the next one after 1440 would be 1440 + 1001 = 2441, which is more than 2010. Therefore, 1440 is indeed the largest one less than 2010. Wait, but let me check if there is another solution between 1440 and 2010. For example, maybe there is another number in that range that satisfies the congruences? But according to our solution, the solutions are x = 1001n + 439. So for n=1, 1440; n=0, 439; n=2, 2441. So between 1440 and 2010, there's no other solution. Therefore, 1440 is the answer. But wait, let me check if 1440 is indeed less than 2010. Yes, 1440 is less than 2010. But wait, maybe there is a larger number? Let me think again.

Wait, perhaps I made a mistake in solving the congruences? Let me check again.

Starting with x ≡ 5 mod 7, x ≡ 10 mod 11, x ≡ 10 mod 13.

First step: solving x ≡ 5 mod 7 and x ≡ 10 mod 11. Let me check my steps again.

x = 7k + 5. Then 7k + 5 ≡ 10 mod 11 => 7k ≡ 5 mod 11. Then inverse of 7 mod 11 is 8, since 7*8=56≡1 mod11. Then k ≡ 5*8=40≡7 mod11. So k=11m +7. Then x=7*(11m +7)+5=77m +49 +5=77m +54. So x≡54 mod77. Then combining with x≡10 mod13.

So x=77m +54. Then 77m +54 ≡10 mod13. 77 mod13: 13*5=65, 77-65=12, so 77≡12 mod13. 54 mod13: 13*4=52, 54-52=2, so 54≡2 mod13. Therefore, 12m +2 ≡10 mod13 => 12m≡8 mod13. Then inverse of 12 mod13 is 12, since 12*12=144≡144-11*13=144-143=1 mod13. Therefore, m≡8*12=96≡96-7*13=96-91=5 mod13. Therefore, m=13n +5. Therefore, x=77*(13n +5)+54= 77*13n + 385 +54= 1001n + 439. So that's correct. Therefore, solutions are 439, 1440, 2441,... So yes, 1440 is the one before 2441. So 1440 is the answer. But wait, let me check if there is another solution between 1440 and 2010. For example, maybe 1440 + 1001 = 2441 is too big, but maybe there is another solution? Wait, no, because the modulus is 1001, so the solutions are spaced exactly 1001 apart. Therefore, between 0 and 1001, there is 439; between 1001 and 2002, there is 1440; and then next is 2441. Therefore, 1440 is indeed the largest less than 2010. Therefore, the answer is 1440. Wait, but let me check if 1440 is indeed less than 2010. Yes, 1440 is 1440, and 2010 is larger. So 1440 is the answer. But wait, let me check if there is a larger number less than 2010. For example, maybe 1440 + something?

Wait, suppose I made a mistake in the calculation of the modulus? Let me check again. The modulus is 7*11*13=1001. So the solutions are 439 + 1001n. So 439 + 1001*1=1440, 439 + 1001*2=2441. So yes, 1440 is the largest less than 2010. Therefore, 1440 is the answer. But let me check if there is another number between 1440 and 2010 that satisfies the conditions. For example, maybe there is a number that is congruent to 439 mod 1001 but larger than 1440 but less than 2010? Well, 1440 + 1001 = 2441, which is more than 2010. Therefore, no. Therefore, 1440 is the answer. Wait, but let me check if 1440 is indeed the correct solution. Let me check again:

Divide 1440 by 7: 7*205 = 1435, so 1440 - 1435 = 5. So remainder 5. Correct.

Divide 1440 by 11: 11*130 = 1430, 1440 - 1430 = 10. Correct.

Divide 1440 by 13: 13*110 = 1430, 1440 - 1430 = 10. Correct. So all conditions are satisfied. Therefore, 1440 is correct. But wait, the problem says "the largest integer less than 2010". Is there a possibility that there is another number between 1440 and 2010 that satisfies the conditions? For example, maybe there is a number that is congruent to 439 mod 1001 but larger than 1440? Well, the next one is 1440 + 1001 = 2441, which is more than 2010. Therefore, no. Therefore, 1440 is the answer. Wait, but let me check if there is another solution that I didn't consider. For example, maybe there is a different way to solve the congruences?

Alternatively, maybe I can check numbers near 2010. Let me think. Let me check if 2010 - 1001 = 1009. Is 1009 a solution? Let me check. 1009 divided by 7: 7*144=1008, so remainder 1. Not 5. So no. How about 2010 - 1001*1 = 1009, which we saw. Then 2010 - 1001*0 = 2010. Let me check 2010. But 2010 is not less than 2010, so it's excluded. So the previous one is 1009, which doesn't work. Then maybe check numbers between 1440 and 2010. For example, 1440 + 1001 = 2441, which is too big. So maybe there is no other solution. Therefore, 1440 is the answer. Wait, but let me check another way. Let me think about solving the congruences again.

Alternatively, since x ≡ 10 mod 11 and x ≡ 10 mod 13. So x ≡ 10 mod lcm(11,13). Since 11 and 13 are coprime, lcm(11,13)=143. Therefore, x ≡ 10 mod 143. Therefore, x can be written as 143k + 10. Then, we also have x ≡ 5 mod 7. Therefore, substituting x = 143k + 10 into x ≡ 5 mod 7:

143k + 10 ≡ 5 mod 7. Let me compute 143 mod 7. 7*20=140, so 143-140=3. Therefore, 143 ≡ 3 mod7. Therefore, 3k + 10 ≡ 5 mod7. Then 3k ≡ 5 -10 mod7 => 3k ≡ -5 mod7 => -5 mod7 is 2, since -5 +7=2. Therefore, 3k ≡ 2 mod7. Solve for k. Multiply both sides by inverse of 3 mod7. The inverse of 3 mod7 is 5, since 3*5=15≡1 mod7. Therefore, k ≡ 2*5=10≡3 mod7. Therefore, k=7m +3. Therefore, x=143*(7m +3) +10=143*7m + 429 +10=1001m + 439. Which is the same result as before. Therefore, x=1001m +439. Therefore, same conclusion. Therefore, the solutions are 439, 1440, 2441,... So 1440 is the answer. Therefore, I think that is correct.

But just to be thorough, let me check if there is a larger number less than 2010. Let me compute 1001*1 + 439 = 1440. Then 1001*2 + 439 = 2441. Since 2441 is more than 2010, so no. Therefore, 1440 is the answer. Therefore, I think the answer is 1440. Wait, but let me check if there is another solution between 1440 and 2010. For example, maybe there is a number that is congruent to 439 mod 1001 but not necessarily 1001n + 439? No, because the modulus is 1001, so all solutions are of the form 1001n + 439. Therefore, there are no other solutions. Therefore, 1440 is the answer.

But just to be 100% sure, let me check if there is a mistake in the calculation of 1001*1 + 439. 1001 + 439: 1000 + 400 = 1400, 1 + 39 = 40, so total 1400 + 40 = 1440. Correct. And 1440 is less than 2010. Therefore, yes. Therefore, I think the answer is 1440.

But wait, let me check if there is a larger number less than 2010. For example, suppose I take x = 1440 + 1001 = 2441, which is over. But maybe there is another number between 1440 and 2010 that satisfies the congruences? For example, maybe there is a number that is congruent to 439 mod 1001 but not in the form 1001n + 439? No, because modulus is 1001, so all solutions are in that form. Therefore, no. Therefore, 1440 is the answer.

Alternatively, maybe I can check numbers near 2010. Let me try 2010 - 10 = 2000. Let me check if 2000 satisfies the conditions. 2000 divided by 7: 7*285=1995, remainder 5. So 2000 ≡ 5 mod7. Good. Then check mod11: 2000 divided by 11. Let me compute 11*181= 1991. 2000 - 1991=9. So remainder 9. But we need remainder 10. Not good. Next, 2000 - 1 = 1999. Check 1999 mod11. 1999 - 1991=8. Not 10. 1999 - 8=1991. Not helpful. How about 2000 - 10=1990. Check 1990 mod11. 11*180=1980. 1990 - 1980=10. So remainder 10. Good. Then check mod13. 1990 divided by 13. Let me compute 13*153= 1989. 1990 - 1989=1. So remainder 1. Not 10. Not good. So 1990 is not a solution. How about 1990 + 11=2001. Check 2001 mod7: 2001 - 2002= -1, so 2001 ≡ -1 mod7, which is 6 mod7. Not 5. Not good. How about 1990 + 13=2003. Check mod7: 2003 - 2002=1, so 1 mod7. Not 5. Not helpful. Maybe this approach is too tedious. Alternatively, since we know the solutions are 1001n + 439, and the next one after 1440 is 2441, which is over 2010, so 1440 is the answer. Therefore, I think I can be confident that 1440 is the correct answer.

But just to check another number, let me check 1440 + 13 = 1453. Check if it satisfies the conditions. 1453 mod7: 1440 is 5 mod7, so 1453 is 1440 +13. 13 mod7 is 6. So 5 +6=11≡4 mod7. Not 5. Not good. Similarly, adding 11: 1440 +11=1451. 1451 mod7: 1440 +11=1451. 1440≡5, 11≡4 mod7, so 5+4=9≡2 mod7. Not 5. Not good. So adding 11 or 13 doesn't help. Therefore, no other numbers between 1440 and 2010. Therefore, 1440 is the answer.

Therefore, I think the answer is 1440. Wait, but let me check once more if there is a mistake in the calculation of the initial solution. Let me check solving the congruences again.

Starting with x ≡ 5 mod7, x ≡10 mod11, x≡10 mod13.

First, solving x ≡10 mod11 and x≡10 mod13. Since 11 and 13 are coprime, x ≡10 mod(11*13)=143. Therefore, x=143k +10. Then, substitute into x≡5 mod7:

143k +10 ≡5 mod7. As before, 143 mod7 is 3, so 3k +10 ≡5 mod7. 10 mod7 is 3, so 3k +3 ≡5 mod7 => 3k ≡2 mod7. Multiply by inverse of 3 mod7, which is 5: k≡10≡3 mod7. Therefore, k=7m +3. Therefore, x=143*(7m +3)+10=1001m + 429 +10=1001m +439. So same result. Therefore, correct. Therefore, x=1001m +439. Therefore, the answer is 1440. Therefore, I think that is correct.

Therefore, I think the answer is 1440. Therefore, I can write that as the final answer.

**Final Answer**
\boxed{1440}

To solve the problem, we are looking for the **largest integer less than 2010** that satisfies the following system of congruences:

$$
\begin{cases}
x \equiv 5 \pmod{7} \\
x \equiv 10 \pmod{11} \\
x \equiv 10 \pmod{13}
\end{cases}
$$

---

### Step 1: Combine the congruences

We begin by combining the last two congruences:

$$
x \equiv 10 \pmod{11} \quad \text{and} \quad x \equiv 10 \pmod{13}
$$

Since 11 and 13 are coprime, we can apply the **Chinese Remainder Theorem (CRT)** to combine them:

$$
x \equiv 10 \pmod{143}
$$

So now we have:

$$
x \equiv 5 \pmod{7} \quad \text{and} \quad x \equiv 10 \pmod{143}
$$

---

### Step 2: Solve the combined system

Let’s express $ x $ in terms of the second congruence:

$$
x = 143k + 10
$$

Substitute into the first congruence:

$$
143k + 10 \equiv 5 \pmod{7}
$$

Now compute $ 143 \mod 7 $:

$$
143 \div 7 = 20 \text{ remainder } 3 \Rightarrow 143 \equiv 3 \pmod{7}
$$

So the equation becomes:

$$
3k + 10 \equiv 5 \pmod{7}
$$

Subtract 10 from both sides:

$$
3k \equiv -5 \equiv 2 \pmod{7}
$$

Now solve for $ k $:

We need the **multiplicative inverse** of 3 modulo 7. Since $ 3 \times 5 = 15 \equiv 1 \pmod{7} $, the inverse is 5.

Multiply both sides by 5:

$$
k \equiv 2 \times 5 = 10 \equiv 3 \pmod{7}
$$

So $ k = 7m + 3 $, and substituting back:

$$
x = 143k + 10 = 143(7m + 3) + 10 = 1001m + 429 + 10 = 1001m + 439
$$

---

### Step 3: Find the largest such $ x < 2010 $

We now have the general solution:

$$
x = 1001m + 439
$$

We want the **largest** such $ x $ less than 2010:

$$
1001m + 439 < 2010
\Rightarrow 1001m < 1571
\Rightarrow m < \frac{1571}{1001} \approx 1.569
$$

So the largest integer $ m $ is $ m = 1 $, which gives:

$$
x = 1001 \times 1 + 439 = 1440
$$

---

### Step 4: Verify the solution

Check that $ x = 1440 $ satisfies all three original congruences:

- $ 1440 \div 7 = 205 $ remainder $ 5 $ → $ 1440 \equiv 5 \pmod{7} $
- $ 1440 \div 11 = 130 $ remainder $ 10 $ → $ 1440 \equiv 10 \pmod{11} $
- $ 1440 \div 13 = 110 $ remainder $ 10 $ → $ 1440 \equiv 10 \pmod{13} $

All conditions are satisfied.

---

### Final Answer

$$
\boxed{1440}
$$

Solution:
We want a remainder of $10$ when divided by both $11$ and $13$. The least common multiple of $11$ and $13$ is $143$. We add $10$ to the number such that the remainder would be $10$ when divided by $11$ and $13$ so we get $143+10=153$. However, that does not give a remainder of $5$ when divided by $7$, so we add more $143$s until we get a value that works. We get that $153+143+143=439$ gives a remainder of $5$ when divided by $7$.

Since we want the largest integer less than 2010, we keep adding the least common multiple of $7$, $11$, and $13$ until we go over. The least common multiple is $7 \cdot 11 \cdot 13 =1001$. We add it to $439$ to get $1440$, adding it again would give a value greater than $2010$, so our answer is $\boxed{1440}$.


Explanation:
Okay, so I need to solve this problem where y varies inversely as the square root of x. When x is 24, y is 15. Then I need to find x when y is 3. Hmm, let me think about inverse variation first.

I remember that when something varies inversely, it means that as one quantity increases, the other decreases. The general formula for inverse variation is y = k/x, where k is the constant of proportionality. But wait, in this case, it's not just x, it's the square root of x. So maybe the formula is different? Let me check.

If y varies inversely as the square root of x, that should mean y is equal to some constant divided by the square root of x. So, mathematically, that would be y = k / √x. Yeah, that makes sense. So instead of just x in the denominator, it's √x.

Alright, so the problem gives me a specific case where x is 24 and y is 15. I can use this to find the constant k. Let me plug those values into the equation.

So substituting x = 24 and y = 15 into y = k / √x, I get:

15 = k / √24

Now I need to solve for k. To do that, I can multiply both sides of the equation by √24. Let me write that down:

k = 15 * √24

Hmm, okay. Let me compute √24. I know that 24 can be broken down into 4*6, and since √4 is 2, that simplifies to 2√6. So √24 is 2√6. Therefore, substituting back:

k = 15 * 2√6 = 30√6

So the constant of proportionality k is 30√6.

Now, the question asks for the value of x when y is 3. So I need to use the same formula y = k / √x, but now with y = 3 and k = 30√6. Let me set up the equation:

3 = (30√6) / √x

Now I need to solve for x. Let me rearrange this equation. First, multiply both sides by √x to get rid of the denominator:

3 * √x = 30√6

Then, divide both sides by 3 to isolate √x:

√x = (30√6) / 3

Simplifying the right side:

√x = 10√6

Now, to solve for x, I need to square both sides of the equation. Let me do that:

(√x)^2 = (10√6)^2

Which simplifies to:

x = 10^2 * (√6)^2

Calculating each part:

10^2 is 100, and (√6)^2 is 6. So:

x = 100 * 6 = 600

Wait, so x is 600 when y is 3? Let me check my steps again to make sure I didn't make a mistake.

Starting from the beginning: inverse variation with √x. So y = k / √x. Given x=24, y=15. So k = y * √x = 15 * √24. √24 is 2√6, so 15*2√6 is 30√6. That seems right.

Then, when y=3, 3 = 30√6 / √x. Multiply both sides by √x: 3√x = 30√6. Divide by 3: √x = 10√6. Square both sides: x = 100*6 = 600. Yeah, that seems consistent.

Let me verify by plugging x=600 back into the original equation to see if y=3.

Compute y = k / √x = 30√6 / √600. Let me simplify √600. 600 is 100*6, so √600 = √(100*6) = 10√6. Therefore, y = 30√6 / 10√6. The √6 cancels out, and 30/10 is 3. So y=3. Perfect, that checks out.

So my answer is x=600. Wait, but let me think again if there's another way to approach this problem, maybe using proportions?

Since y varies inversely as √x, then y1 * √x1 = y2 * √x2. Is that right? Because if y = k / √x, then k = y * √x. So if k is constant, then y1 * √x1 = y2 * √x2. Let me try that approach.

Given y1 = 15, x1 = 24, y2 = 3, and we need to find x2.

So:

15 * √24 = 3 * √x2

Which is the same equation as before. Then solving for √x2:

√x2 = (15 * √24) / 3 = 5 * √24

Wait, hold on, earlier I had √x = 10√6. Let me check if 5 * √24 equals 10√6.

√24 is 2√6, so 5 * 2√6 = 10√6. Yes, that's correct. Then squaring both sides:

x2 = (10√6)^2 = 100 * 6 = 600. Same result. So that's another way to think about it, using the proportionality.

So both methods lead me to x=600. Therefore, I think that's the correct answer.

But just to make sure, let me think if there's any chance I messed up the inverse variation. Sometimes, people confuse direct and inverse variation. Let me recall: if y varies inversely as x, then y = k/x. If it varies inversely as √x, then y = k / √x. So that's correct.

Alternatively, if it had said y varies inversely as x squared, then it would be y = k / x². But here it's square root, so inverse variation with square root. So the formula is correct.

Another check: when x increases, y should decrease. Since when x went from 24 to 600, which is a significant increase, y went from 15 to 3, which is a decrease. That makes sense for inverse variation.

Alternatively, if I think about the relationship, if x is multiplied by a factor, then y is divided by the square root of that factor. Let me see. Suppose x is multiplied by some factor m, so new x is m*x. Then y would be k / √(m*x) = k / (√m * √x) = (k / √x) / √m = y / √m. So y is divided by √m.

So in our case, from x=24 to x=600, the factor m is 600 / 24 = 25. So √m is 5. Therefore, y should be divided by 5. Original y was 15, so 15 / 5 = 3. Which matches the given y=3. Perfect! That's another way to see it.

So that's three different methods: using the formula, using proportionality, and using the factor of change. All leading to x=600.

Therefore, I'm confident that the answer is 600.

**Final Answer**
The value of $ x $ when $ y = 3 $ is \boxed{600}.

To solve the problem, we begin by understanding the nature of the relationship between $ y $ and $ x $. The problem states that $ y $ varies inversely as $ \sqrt{x} $. This means that:

$$
y = \frac{k}{\sqrt{x}}
$$

where $ k $ is the constant of proportionality.

---

### Step 1: Find the constant $ k $

We are given that when $ x = 24 $, $ y = 15 $. Substituting into the equation:

$$
15 = \frac{k}{\sqrt{24}}
$$

Solving for $ k $:

$$
k = 15 \cdot \sqrt{24}
$$

We simplify $ \sqrt{24} $ as:

$$
\sqrt{24} = \sqrt{4 \cdot 6} = 2\sqrt{6}
$$

So:

$$
k = 15 \cdot 2\sqrt{6} = 30\sqrt{6}
$$

---

### Step 2: Use the constant to find $ x $ when $ y = 3 $

Now that we know $ k = 30\sqrt{6} $, we substitute into the original equation:

$$
3 = \frac{30\sqrt{6}}{\sqrt{x}}
$$

Multiply both sides by $ \sqrt{x} $:

$$
3\sqrt{x} = 30\sqrt{6}
$$

Divide both sides by 3:

$$
\sqrt{x} = 10\sqrt{6}
$$

Now square both sides to solve for $ x $:

$$
x = (10\sqrt{6})^2 = 100 \cdot 6 = 600
$$

---

### Final Answer

$$
\boxed{600}
$$

[Assistant]
Explanation:
Okay, so I need to solve this problem where y varies inversely as the square root of x. When x is 24, y is 15. Then I need to find x when y is 3. Hmm, let me think about inverse variation first.

I remember that when something varies inversely, it means that as one quantity increases, the other decreases. The general formula for inverse variation is y = k/x, where k is the constant of proportionality. But wait, in this case, it's not just x, it's the square root of x. So maybe the formula is different? Let me check.

If y varies inversely as the square root of x, that should mean y is equal to some constant divided by the square root of x. So, mathematically, that would be y = k / √x. Yeah, that makes sense. So instead of just x in the denominator, it's √x.

Alright, so the problem gives me a specific case where x is 24 and y is 15. I can use this to find the constant k. Let me plug those values into the equation.

So substituting x = 24 and y = 15 into y = k / √x, I get:

15 = k / √24

Now I need to solve for k. To do that, I can multiply both sides of the equation by √24. Let me write that down:

k = 15 * √24

Hmm, okay. Let me compute √24. I know that 24 can be broken down into 4*6, and since √4 is 2, that simplifies to 2√6. So √24 is 2√6. Therefore, substituting back:

k = 15 * 2√6 = 30√6

So the constant of proportionality k is 30√6.

Now, the question asks for the value of x when y is 3. So I need to use the same formula y = k / √x, but now with y = 3 and k = 30√6. Let me set up the equation:

3 = (30√6) / √x

Now I need to solve for x. Let me rearrange this equation. First, multiply both sides by √x to get rid of the denominator:

3 * √x = 30√6

Then, divide both sides by 3 to isolate √x:

√x = (30√6) / 3

Simplifying the right side:

√x = 10√6

Now, to solve for x, I need to square both sides of the equation. Let me do that:

(√x)^2 = (10√6)^2

Which simplifies to:

x = 10^2 * (√6)^2

Calculating each part:

10^2 is 100, and (√6)^2 is 6. So:

x = 100 * 6 = 600

Wait, so x is 600 when y is 3? Let me check my steps again to make sure I didn't make a mistake.

Starting from the beginning: inverse variation with √x. So y = k / √x. Given x=24, y=15. So k = y * √x = 15 * √24. √24 is 2√6, so 15*2√6 is 30√6. That seems right.

Then, when y=3, 3 = 30√6 / √x. Multiply both sides by √x: 3√x = 30√6. Divide by 3: √x = 10√6. Square both sides: x = 100*6 = 600. Yeah, that seems consistent.

Let me verify by plugging x=600 back into the original equation to see if y=3.

Compute y = k / √x = 30√6 / √600. Let me simplify √600. 600 is 100*6, so √600 = √(100*6) = 10√6. Therefore, y = 30√6 / 10√6. The √6 cancels out, and 30/10 is 3. So y=3. Perfect, that checks out.

So my answer is x=600. Wait, but let me think again if there's another way to approach this problem, maybe using proportions?

Since y varies inversely as √x, then y1 * √x1 = y2 * √x2. Is that right? Because if y = k / √x, then k = y * √x. So if k is constant, then y1 * √x1 = y2 * √x2. Let me try that approach.

Given y1 = 15, x1 = 24, y2 = 3, and we need to find x2.

So:

15 * √24 = 3 * √x2

Which is the same equation as before. Then solving for √x2:

√x2 = (15 * √24) / 3 = 5 * √24

Wait, hold on, earlier I had √x = 10√6. Let me check if 5 * √24 equals 10√6.

√24 is 2√6, so 5 * 2√6 = 10√6. Yes, that's correct. Then squaring both sides:

x2 = (10√6)^2 = 100 * 6 = 600. Same result. So that's another way to think about it, using the proportionality.

So both methods lead me to x=600. Therefore, I think that's the correct answer.

But just to make sure, let me think if there's any chance I messed up the inverse variation. Sometimes, people confuse direct and inverse variation. Let me recall: if y varies inversely as x, then y = k/x. If it varies inversely as √x, then y = k / √x. So that's correct.

Alternatively, if it had said y varies inversely as x squared, then it would be y = k / x². But here it's square root, so inverse variation with square root. So the formula is correct.

Another check: when x increases, y should decrease. Since when x went from 24 to 600, which is a significant increase, y went from 15 to 3, which is a decrease. That makes sense for inverse variation.

Alternatively, if I think about the relationship, if x is multiplied by a factor, then y is divided by the square root of that factor. Let me see. Suppose x is multiplied by some factor m, so new x is m*x. Then y would be k / √(m*x) = k / (√m * √x) = (k / √x) / √m = y / √m. So y is divided by √m.

So in our case, from x=24 to x=600, the factor m is 600 / 24 = 25. So √m is 5. Therefore, y should be divided by 5. Original y was 15, so 15 / 5 = 3. Which matches the given y=3. Perfect! That's another way to see it.

So that's three different methods: using the formula, using proportionality, and using the factor of change. All leading to x=600.

Therefore, I'm confident that the answer is 600.

**Final Answer**
The value of $ x $ when $ y = 3 $ is \boxed{600}.

To solve the problem, we begin by understanding the nature of the relationship between $ y $ and $ x $. The problem states that $ y $ varies inversely as $ \sqrt{x} $. This means that:

$$
y = \frac{k}{\sqrt{x}}
$$

where $ k $ is the constant of proportionality.

---

### Step 1: Find the constant $ k $

We are given that when $ x = 24 $, $ y = 15 $. Substituting into the equation:

$$
15 = \frac{k}{\sqrt{24}}
$$

Solving for $ k $:

$$
k = 15 \cdot \sqrt{24}
$$

We simplify $ \sqrt{24} $ as:

$$
\sqrt{24} = \sqrt{4 \cdot 6} = 2\sqrt{6}
$$

So:

$$
k = 15 \cdot 2\sqrt{6} = 30\sqrt{6}
$$

---

### Step 2: Use the constant to find $ x $ when $ y = 3 $

Now that we know $ k = 30\sqrt{6} $, we substitute into the original equation:

$$
3 = \frac{30\sqrt{6}}{\sqrt{x}}
$$

Multiply both sides by $ \sqrt{x} $:

$$
3\sqrt{x} = 30\sqrt{6}
$$

Divide both sides by 3:

$$
\sqrt{x} = 10\sqrt{6}
$$

Now square both sides to solve for $ x $:

$$
x = (10\sqrt{6})^2 = 100 \cdot 6 = 600
$$

---

### Final Answer

$$
\boxed{600}
$$

Solution:
Since $y$ and $\sqrt{x}$ are inversely proportional, this means that $y\sqrt{x}=k$ for some constant $k$.  Substituting the given values, when $x=24$ and $y=15$, we find that $15\sqrt{24}=30\sqrt{6}=k$.  Therefore, when $y=3$, we can solve for $x$: \begin{align*}
3\cdot\sqrt{x}&=30\sqrt{6}\\
\Rightarrow\qquad (\sqrt{x})^2&=(10\sqrt{6})^2\\
\Rightarrow\qquad x&=100\cdot6\\
&=\boxed{600}
\end{align*}


Explanation:
Okay, so I need to find AF² + BF² + CF² where F is the midpoint of GH, with G being the centroid and H the orthocenter of triangle ABC. The answer should be in terms of the side lengths a, b, c and the circumradius R. Hmm, let me start by recalling some properties of centroids and orthocenters.

First, the centroid G divides the medians in a 2:1 ratio. The orthocenter H is the intersection point of the altitudes. I remember there's a relationship between the centroid, orthocenter, and other centers like the circumcenter O. In fact, Euler's line states that G, H, and O are colinear, and HG = 2GO. Maybe that will come into play here.

Since F is the midpoint of GH, then GF = FH = (1/2)GH. But I need to relate this to AF² + BF² + CF². Maybe coordinate geometry could help here? Let me try setting up coordinates for triangle ABC. But before jumping into coordinates, perhaps there are vector approaches or formulas that can relate these points?

Alternatively, I remember that for any point P in the plane of triangle ABC, the formula for the sum of squared distances from P to the vertices is given by:

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

Wait, is that right? Let me recall. There's a formula that relates the sum of squared distances from a point to the vertices with the centroid. Yes, I think it's:

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

But I need to verify this. Let me think. If P is the centroid G, then PA² + PB² + PC² = 3PG² + ... but PG would be zero, so PA² + PB² + PC² = GA² + GB² + GC². Wait, but that seems not quite. Wait, actually, the formula is:

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

Wait, maybe I need to recall the formula properly. Let me think of coordinates. Let me take coordinates where G is the origin. Then, the coordinates of A, B, C satisfy (A + B + C)/3 = 0, so A + B + C = 0. Then, for any point P, PA² + PB² + PC² = |P - A|² + |P - B|² + |P - C|². Expanding each term:

|P - A|² = |P|² - 2P·A + |A|²

Similarly for others. Adding them up:

3|P|² - 2P·(A + B + C) + |A|² + |B|² + |C|²

But since A + B + C = 0, this simplifies to 3|P|² + |A|² + |B|² + |C|². Therefore, if G is the origin, then PG is just |P|, so PA² + PB² + PC² = 3PG² + |A|² + |B|² + |C|². Wait, but if G is the origin, then PG is the distance from P to G, which is |P - G|, but since G is the origin, it's just |P|. So yes, the formula is PA² + PB² + PC² = 3PG² + GA² + GB² + GC²? Wait, no. Wait, GA² is |A - G|². Since G is the origin, GA² is |A|². Similarly for others. Therefore, PA² + PB² + PC² = 3PG² + GA² + GB² + GC². So that formula is correct.

Therefore, if I can express AF² + BF² + CF² in terms of PG² where P is F, then I can use this formula. So if I can find the position vector of F relative to G, then compute PG², which would be GF² since G is the origin? Wait, no. Wait, if I take G as the origin, then F is some point, and AF² + BF² + CF² would be 3GF² + GA² + GB² + GC². Wait, but in this case, if I take G as the origin, then the formula is PA² + PB² + PC² = 3PG² + GA² + GB² + GC². Therefore, if P is F, then AF² + BF² + CF² = 3GF² + GA² + GB² + GC². Therefore, if I can compute GF² and GA² + GB² + GC², then I can find the desired sum.

So first, let me compute GA² + GB² + GC². For that, since G is centroid, there is a formula. In general, for centroid, GA² + GB² + GC² = (1/3)(a² + b² + c²). Wait, is that correct? Wait, let me recall. Let me think of coordinates again. Let me take coordinates where centroid is at origin. Then, A + B + C = 0. Then, GA² = |A|², GB² = |B|², GC² = |C|². Then, GA² + GB² + GC² = |A|² + |B|² + |C|². But how does that relate to the sides?

Alternatively, using the formula for the centroid. Let me recall that in any triangle, the sum of the squares of the medians is related to the sides. But maybe I need another approach. Alternatively, using vectors. Let me consider vectors from centroid.

Alternatively, since in coordinate system with centroid at origin, A + B + C = 0. Then, the square of the sides can be expressed as |B - C|² = a², etc. Let me compute |A|² + |B|² + |C|². Let me compute |A - B|² = |A - B|² = |A|² + |B|² - 2A·B. Similarly for others. So:

a² = |B - C|² = |B|² + |C|² - 2B·C

b² = |C - A|² = |C|² + |A|² - 2C·A

c² = |A - B|² = |A|² + |B|² - 2A·B

Adding all three equations:

a² + b² + c² = 2(|A|² + |B|² + |C|²) - 2(A·B + B·C + C·A)

Therefore, |A|² + |B|² + |C|² = [a² + b² + c² + 2(A·B + B·C + C·A)] / 2

But since A + B + C = 0, then (A + B + C)·(A + B + C) = 0 => |A|² + |B|² + |C|² + 2(A·B + B·C + C·A) = 0. Therefore, A·B + B·C + C·A = - (|A|² + |B|² + |C|²)/2. Substituting back into the previous equation:

|A|² + |B|² + |C|² = [a² + b² + c² + 2*(- (|A|² + |B|² + |C|²)/2)] / 2

Simplify:

|A|² + |B|² + |C|² = [a² + b² + c² - (|A|² + |B|² + |C|²)] / 2

Multiply both sides by 2:

2(|A|² + |B|² + |C|²) = a² + b² + c² - (|A|² + |B|² + |C|²)

Bring terms together:

2(|A|² + |B|² + |C|²) + (|A|² + |B|² + |C|²) = a² + b² + c²

So 3(|A|² + |B|² + |C|²) = a² + b² + c² => |A|² + |B|² + |C|² = (a² + b² + c²)/3

Therefore, GA² + GB² + GC² = (a² + b² + c²)/3. So that's settled. Therefore, going back to the formula:

AF² + BF² + CF² = 3GF² + (a² + b² + c²)/3

So now, I need to compute GF². Since F is the midpoint of GH, GF is half of GH. Therefore, GF = (1/2)GH, so GF² = (1/4)GH². Therefore, AF² + BF² + CF² = 3*(1/4)GH² + (a² + b² + c²)/3 = (3/4)GH² + (a² + b² + c²)/3.

Therefore, I need to compute GH². So now, the problem reduces to finding GH² in terms of a, b, c, and R.

So how to compute GH? There is a formula for the distance between centroid and orthocenter. Let me recall. In Euler's line, the distance between centroid G and orthocenter H is given by GH = 3GO, where O is the circumcenter. Wait, Euler's line states that HG = 2GO, so GH = 2GO? Wait, let me check.

Euler's line: The centroid G divides the segment joining the orthocenter H and the circumcenter O in the ratio HG : GO = 2 : 1. So HG = 2GO. Therefore, GH = 2GO. Therefore, GH² = 4GO². Therefore, if I can find GO², then I can compute GH².

Alternatively, there is a formula for GO². Let me recall. There is a relation between the distances from the centroid to the circumcenter and other elements. Also, there is a formula involving the sides and the circumradius. Let me recall that in any triangle, the distance between the centroid and circumcenter is given by:

GO² = R² - (a² + b² + c²)/9

Wait, is that correct? Wait, I need to verify. Alternatively, there is a formula:

OH² = 9R² - (a² + b² + c²)

Where O is circumcenter and H is orthocenter. Wait, that's the formula for OH². Then, since GH = 2GO, then GH² = 4GO². But how to relate GO² to OH²?

Since O, G, H are colinear with HG = 2GO, so OH = OG + GH? Wait, depending on the direction. Let me think. If G is between O and H, then OH = OG + GH. But since HG = 2GO, then if we let OG = x, then GH = 2x, so OH = OG + GH = x + 2x = 3x. Therefore, OH = 3x, so x = OH / 3. Therefore, OG = OH / 3, so GO² = (OH²)/9. Therefore, GH² = 4GO² = 4*(OH²)/9 = (4/9)OH². Therefore, if I can find OH², then I can compute GH².

But there is a formula for OH². Yes, Euler's formula states that OH² = 9R² - (a² + b² + c²). Wait, let me check. Wait, actually, Euler's theorem states that OH² = 9R² - (a² + b² + c²). Wait, no, I think it's OH² = 9R² - (a² + b² + c²). Wait, let me check with an equilateral triangle. In an equilateral triangle, O and H coincide, so OH = 0. Then, 9R² - (a² + b² + c²) should be zero. Let me check. In an equilateral triangle, a = b = c, and R = a / √3. Therefore, 9R² = 9*(a² / 3) = 3a². Then, a² + b² + c² = 3a². Therefore, 9R² - (a² + b² + c²) = 3a² - 3a² = 0. Correct. So that formula seems correct. Therefore, OH² = 9R² - (a² + b² + c²). Therefore, GH² = (4/9)OH² = (4/9)(9R² - (a² + b² + c²)) = 4R² - (4/9)(a² + b² + c²). Therefore, GH² = 4R² - (4/9)(a² + b² + c²).

Therefore, going back to AF² + BF² + CF² = (3/4)GH² + (a² + b² + c²)/3. Let me substitute GH²:

AF² + BF² + CF² = (3/4)[4R² - (4/9)(a² + b² + c²)] + (a² + b² + c²)/3

Let me compute this step by step. First, expand the first term:

(3/4)*4R² = 3R²

(3/4)*(-4/9)(a² + b² + c²) = - (3/4)*(4/9)(a² + b² + c²) = - (1/3)(a² + b² + c²)

Therefore, AF² + BF² + CF² = 3R² - (1/3)(a² + b² + c²) + (1/3)(a² + b² + c²) = 3R².

Wait, that's interesting! The terms involving (a² + b² + c²) cancel out, leaving just 3R². Therefore, AF² + BF² + CF² = 3R². Is that possible? Let me check with an equilateral triangle. If ABC is equilateral, then centroid and orthocenter coincide, so G = H. Therefore, F is the midpoint of GH, which is the same as G. Therefore, AF² + BF² + CF² would be GA² + GB² + GC². In an equilateral triangle, GA = GB = GC. Let me compute that. Let’s say the side length is a. The centroid divides the median in ratio 2:1. The length of the median in an equilateral triangle is (sqrt(3)/2)a. Therefore, GA = (2/3)*(sqrt(3)/2)a = (sqrt(3)/3)a. Therefore, GA² = (1/3)a². Therefore, AF² + BF² + CF² = 3*(1/3)a² = a². On the other hand, the circumradius R of an equilateral triangle is (a)/sqrt(3). Therefore, 3R² = 3*(a²/3) = a². So it matches. Therefore, in this case, the formula gives 3R², which equals a². So that checks out. Therefore, maybe the answer is 3R²? But the problem says to express it in terms of a, b, c and R. Wait, but according to this, it's just 3R². However, the problem says "in terms of the side lengths a, b, c and circumradius R". But if it's 3R², then it's already in terms of R. However, maybe there's a mistake here? Let me check another case.

Take a right-angled triangle. Let me take a 3-4-5 triangle. Let me compute AF² + BF² + CF² and check if it equals 3R². Let me recall that in a right-angled triangle, the circumradius R is half the hypotenuse. So for a 3-4-5 triangle, hypotenuse is 5, so R = 5/2. Therefore, 3R² = 3*(25/4) = 75/4 = 18.75.

Now, let me compute AF² + BF² + CF². First, let me find coordinates. Let me place the triangle with right angle at C, so coordinates: Let me set C at (0,0), B at (3,0), A at (0,4). Then, the triangle has sides: AB = 5, BC = 3, AC = 4. Let me find centroid G. The centroid is the average of the coordinates: G_x = (0 + 3 + 0)/3 = 1, G_y = (0 + 0 + 4)/3 = 4/3. So G is at (1, 4/3).

Orthocenter H of a right-angled triangle is at the vertex of the right angle, which is C (0,0). Wait, is that correct? In a right-angled triangle, the orthocenter is at the right-angled vertex. Yes, because the altitudes are the legs themselves. Therefore, H is at (0,0).

Therefore, F is the midpoint of GH. G is (1, 4/3), H is (0,0). Therefore, midpoint F has coordinates ((1 + 0)/2, (4/3 + 0)/2) = (0.5, 2/3).

Now, compute AF² + BF² + CF². Let me compute each term:

Point A is (0,4). Point F is (0.5, 2/3).

AF² = (0 - 0.5)^2 + (4 - 2/3)^2 = (-0.5)^2 + (10/3)^2 = 0.25 + 100/9 = 0.25 + 11.111... ≈ 11.361...

Similarly, BF²: Point B is (3,0). So BF² = (3 - 0.5)^2 + (0 - 2/3)^2 = (2.5)^2 + (-2/3)^2 = 6.25 + 4/9 ≈ 6.25 + 0.444... ≈ 6.694...

CF²: Point C is (0,0). CF² = (0 - 0.5)^2 + (0 - 2/3)^2 = 0.25 + 4/9 ≈ 0.25 + 0.444... ≈ 0.694...

Adding them up: 11.361 + 6.694 + 0.694 ≈ 18.749... which is approximately 18.75, which is 75/4. Which is equal to 3R², since R = 5/2, 3*(25/4) = 75/4. So that checks out. Therefore, the formula seems correct. Therefore, AF² + BF² + CF² = 3R². Therefore, the answer is 3R². But the problem says "in terms of the side lengths a, b, c and circumradius R". However, 3R² is already in terms of R. But maybe there's a way to express it in terms of a, b, c and R? But since the problem allows expressing in terms of a, b, c and R, and 3R² is already in terms of R, but maybe they want it expressed in terms of a, b, c and R? However, since in the case of the equilateral triangle, 3R² equals a², but in general, it's 3R². However, maybe there's a different expression? Wait, but according to the derivation, it cancels out. Let me check again.

Wait, in the derivation, we had AF² + BF² + CF² = 3R². But the problem says "in terms of the side lengths a, b, c and circumradius R". So maybe they want an expression that includes a, b, c and R? But according to the derivation, it's 3R². However, in the case of the 3-4-5 triangle, 3R² is 3*(25/4) = 75/4, which is equal to AF² + BF² + CF². However, if we wanted to write it in terms of a, b, c and R, but since R is already given, maybe 3R² is acceptable. However, maybe there's a mistake in my derivation?

Wait, let me check again. Let me go through the steps again.

Starting with AF² + BF² + CF² = 3GF² + (a² + b² + c²)/3. Then GH² = 4R² - (4/9)(a² + b² + c²). Then GF² = GH² / 4. Therefore, AF² + BF² + CF² = 3*(GH²)/4 + (a² + b² + c²)/3 = 3/4*(4R² - 4/9(a² + b² + c²)) + (a² + b² + c²)/3 = 3R² - (1/3)(a² + b² + c²) + (1/3)(a² + b² + c²) = 3R². So yes, the terms cancel. Therefore, regardless of a, b, c, the answer is 3R². Therefore, the answer is 3R². Therefore, maybe the problem is designed to have this answer. However, the problem says "in terms of the side lengths a, b, c and circumradius R". But since 3R² is already in terms of R, but maybe they want it expressed in terms of a, b, c and R? But unless there's a different expression. Wait, but maybe there's a formula that relates R with a, b, c? For example, in terms of the sides, R = (a b c)/(4Δ), where Δ is the area. But unless they want it expressed in terms of a, b, c and R, but since R is already given, maybe 3R² is acceptable. However, maybe the problem expects an expression that includes a, b, c? But according to the derivation, it's 3R². However, let me check another triangle.

Take an equilateral triangle with side length a. Then, R = a / √3. Therefore, 3R² = 3*(a² / 3) = a². Which matches with the earlier calculation. So that's correct. For the 3-4-5 triangle, 3R² = 3*(25/4) = 75/4, which matches the calculation. Therefore, seems correct. Therefore, maybe the answer is simply 3R². However, the problem says "in terms of the side lengths a, b, c and circumradius R". So maybe they want it expressed as 3R², which is in terms of R. But maybe there's a different way? Wait, but if you want to write it in terms of a, b, c and R, but since R is already given, maybe they just want 3R². However, maybe there's a mistake in my derivation? Let me check again.

Wait, let me think again. The formula AF² + BF² + CF² = 3GF² + (a² + b² + c²)/3. Then GH² = 4R² - (4/9)(a² + b² + c²). Therefore, GF² = GH² / 4 = [4R² - (4/9)(a² + b² + c²)] / 4 = R² - (1/9)(a² + b² + c²). Therefore, AF² + BF² + CF² = 3[R² - (1/9)(a² + b² + c²)] + (a² + b² + c²)/3 = 3R² - (1/3)(a² + b² + c²) + (1/3)(a² + b² + c²) = 3R². So yes, that's correct. Therefore, regardless of the triangle, AF² + BF² + CF² = 3R². Therefore, the answer is 3R². Therefore, maybe the problem is designed to have this answer. However, the problem says "in terms of the side lengths a, b, c and circumradius R". But since R is already given, maybe they just want 3R². However, maybe there's a different approach where it's expressed in terms of a, b, c and R? For example, if we use the formula that relates R with the sides, but since R is already given, maybe it's redundant. Alternatively, maybe the problem expects an answer in terms of a, b, c and R, but since 3R² is already in terms of R, and the problem allows that, then 3R² is the answer.

Alternatively, maybe I made a mistake in assuming that GH² = 4R² - (4/9)(a² + b² + c²). Let me check that again. Earlier, I used Euler's formula OH² = 9R² - (a² + b² + c²). Then, since GH = 2GO, and OH = 3GO, so GO = OH / 3. Therefore, GH = 2*(OH / 3) = 2OH / 3. Therefore, GH² = (4/9)OH² = (4/9)(9R² - (a² + b² + c²)) = 4R² - (4/9)(a² + b² + c²). Therefore, that is correct. Therefore, the rest follows. Therefore, the answer is 3R². Therefore, maybe the answer is simply 3R². Therefore, I think that is the answer.

But let me check another triangle. Let me take an isosceles triangle with sides 2, 2, 2. Wait, that's equilateral. Let me take a different triangle. Let me take a triangle with sides 2, 2, 3. Wait, is that possible? Let me check. The triangle inequality: 2 + 2 > 3? 4 > 3, yes. 2 + 3 > 2, yes. So okay. Let me compute R for this triangle. First, compute the area. Using Heron's formula: s = (2 + 2 + 3)/2 = 7/2. Area Δ = sqrt[s(s - a)(s - b)(s - c)] = sqrt[(7/2)(7/2 - 2)(7/2 - 2)(7/2 - 3)] = sqrt[(7/2)(3/2)(3/2)(1/2)] = sqrt[(7*3*3*1)/(16)] = sqrt(63/16) = (3√7)/4. Then, R = (a b c)/(4Δ) = (2*2*3)/(4*(3√7)/4) = 12 / (3√7) = 4 / √7. Therefore, 3R² = 3*(16 / 7) = 48/7 ≈ 6.857.

Now, let me compute AF² + BF² + CF². Let me find coordinates. Let me place the triangle with base BC = 3, and AB = AC = 2. Let me place coordinates: Let me put point B at (-1.5, 0), point C at (1.5, 0), so that BC = 3. Then, point A is somewhere above the x-axis. Let me compute coordinates of A. Since AB = AC = 2, the triangle is isoceles with AB = AC = 2. Let me find coordinates of A. Let me denote A as (0, h). Then, distance from A to B is sqrt[(0 - (-1.5))² + (h - 0)^2] = sqrt[(1.5)^2 + h²] = 2. Therefore, (1.5)^2 + h² = 4 => 2.25 + h² = 4 => h² = 1.75 => h = sqrt(7)/2 ≈ 1.322. Therefore, coordinates of A are (0, sqrt(7)/2). Therefore, coordinates:

A: (0, sqrt(7)/2)

B: (-1.5, 0)

C: (1.5, 0)

Now, centroid G is the average of coordinates:

G_x = (0 + (-1.5) + 1.5)/3 = 0/3 = 0

G_y = (sqrt(7)/2 + 0 + 0)/3 = sqrt(7)/6

Therefore, G is at (0, sqrt(7)/6)

Orthocenter H. In an isoceles triangle, the orthocenter lies along the altitude from A, which is the y-axis. Let me compute coordinates of H. In an isoceles triangle, the orthocenter, centroid, and circumcenter lie along the same line. Let me compute H. Alternatively, since we know coordinates of A, B, C, we can compute equations of altitudes.

Altitude from A is the y-axis. Altitude from B: Let me compute. The altitude from B to AC. First, find equation of AC. Points A(0, sqrt(7)/2) and C(1.5, 0). The slope of AC is (0 - sqrt(7)/2)/(1.5 - 0) = (-sqrt(7)/2)/1.5 = -sqrt(7)/3. Therefore, the altitude from B is perpendicular to AC, so its slope is reciprocal and opposite: 3/sqrt(7). Therefore, equation of altitude from B: passes through B(-1.5, 0), slope 3/sqrt(7). Let me write the equation:

y - 0 = (3/sqrt(7))(x + 1.5)

Similarly, the orthocenter H lies at the intersection of this altitude and the altitude from A (the y-axis). Therefore, substitute x = 0 into the equation:

y = (3/sqrt(7))(0 + 1.5) = (3/sqrt(7))*(3/2) = 9/(2 sqrt(7)) = (9 sqrt(7))/14

Therefore, coordinates of H are (0, 9√7 / 14). Therefore, midpoint F of GH: G is (0, sqrt(7)/6) and H is (0, 9√7 / 14). Therefore, midpoint F has coordinates:

x-coordinate: 0

y-coordinate: [sqrt(7)/6 + 9√7 /14]/2. Let me compute that:

First, common denominator for 6 and 14 is 42.

sqrt(7)/6 = 7 sqrt(7)/42

9√7 /14 = 27 sqrt(7)/42

Adding them: 7 sqrt(7)/42 + 27 sqrt(7)/42 = 34 sqrt(7)/42 = 17 sqrt(7)/21

Divide by 2: 17 sqrt(7)/42

Therefore, coordinates of F are (0, 17√7 / 42)

Now, compute AF² + BF² + CF².

First, compute AF²: distance from A(0, sqrt(7)/2) to F(0, 17√7 / 42):

Difference in y-coordinates: sqrt(7)/2 - 17√7 /42 = (21√7 / 42 - 17√7 /42) = 4√7 /42 = 2√7 /21

Therefore, AF² = (2√7 /21)^2 = 4*7 / 441 = 28 / 441 = 4 / 63 ≈ 0.0635

Wait, that seems small. Let me check again.

Wait, sqrt(7)/2 is equal to 21√7 / 42? Wait, sqrt(7)/2 = (sqrt(7)/2)*(21/21) = 21 sqrt(7)/42? Wait, no. Wait, sqrt(7)/2 is equal to (sqrt(7)/2)*(21/21) = 21 sqrt(7)/42? Wait, no. Wait, sqrt(7)/2 = (sqrt(7)/2)*(21/21) = (21 sqrt(7))/42? Wait, no. Wait, 21/21 is 1. So sqrt(7)/2 = sqrt(7)/2. To write it over 42, multiply numerator and denominator by 21: sqrt(7)/2 = (sqrt(7)*21)/42. Therefore, yes, sqrt(7)/2 = 21 sqrt(7)/42. Similarly, 17 sqrt(7)/42 is as is. Therefore, difference is 21 sqrt(7)/42 - 17 sqrt(7)/42 = 4 sqrt(7)/42 = 2 sqrt(7)/21. Therefore, AF² = (2 sqrt(7)/21)^2 = 4*7 / 441 = 28 / 441 = 4 / 63 ≈ 0.0635. Okay.

Now compute BF²: distance from B(-1.5, 0) to F(0, 17√7 /42). Let me compute coordinates:

Δx = 0 - (-1.5) = 1.5 = 3/2

Δy = 17√7 /42 - 0 = 17√7 /42

Therefore, BF² = (3/2)^2 + (17√7 /42)^2 = 9/4 + (289 * 7)/ (42^2)

Compute 289 * 7 = 2023

42^2 = 1764

Therefore, BF² = 9/4 + 2023 / 1764

Convert 9/4 to denominator 1764: 9/4 = (9 * 441)/1764 = 3969 / 1764

Therefore, BF² = (3969 + 2023)/1764 = 6, 3969 + 2023 = let me compute:

3969 + 2000 = 5969, plus 23 = 5992. Therefore, BF² = 5992 / 1764. Simplify:

Divide numerator and denominator by 4: 5992 / 4 = 1498, 1764 / 4 = 441. So 1498 / 441 ≈ 3.396...

Similarly, compute CF²: distance from C(1.5, 0) to F(0, 17√7 /42). It's symmetric to BF², so CF² = BF² = 5992 / 1764.

Therefore, AF² + BF² + CF² = 4/63 + 2*(5992 / 1764). Let me compute this.

First, convert 4/63 to denominator 1764: 4/63 = (4 * 28)/1764 = 112 / 1764

Then, 2*(5992 / 1764) = 11984 / 1764

Therefore, total is 112 + 11984 = 12096 / 1764

Simplify 12096 / 1764. Let me divide numerator and denominator by 12: 12096 / 12 = 1008, 1764 / 12 = 147. So 1008 / 147. Divide numerator and denominator by 21: 1008 / 21 = 48, 147 / 21 = 7. Therefore, 48 / 7 ≈ 6.857. Which is equal to 48/7. Now, 3R² was 48/7. Since earlier, R = 4 / sqrt(7), so 3R² = 3*(16 / 7) = 48/7. Correct. Therefore, this also checks out. Therefore, the formula holds. Therefore, the answer is indeed 3R². Therefore, the answer is \boxed{3R^2}.

**Final Answer**
\boxed{3R^2}

To solve the problem, we are given a triangle $ ABC $ with centroid $ G $, orthocenter $ H $, and a point $ F $ defined as the midpoint of segment $ \overline{GH} $. We are to compute the expression:

$$
AF^2 + BF^2 + CF^2
$$

in terms of the side lengths $ a, b, c $ and the circumradius $ R $ of triangle $ ABC $.

---

### Step 1: Use of a Known Vector Identity

There is a well-known identity in triangle geometry that relates the sum of squared distances from any point $ P $ to the vertices of triangle $ ABC $ to the centroid $ G $:

$$
PA^2 + PB^2 + PC^2 = 3PG^2 + GA^2 + GB^2 + GC^2
$$

Applying this identity with $ P = F $, we get:

$$
AF^2 + BF^2 + CF^2 = 3GF^2 + GA^2 + GB^2 + GC^2
$$

---

### Step 2: Compute $ GA^2 + GB^2 + GC^2 $

It is a known result that for the centroid $ G $ of triangle $ ABC $, the sum of the squares of the distances from $ G $ to the vertices is:

$$
GA^2 + GB^2 + GC^2 = \frac{a^2 + b^2 + c^2}{3}
$$

---

### Step 3: Compute $ GF^2 $

Since $ F $ is the midpoint of $ \overline{GH} $, we have:

$$
GF = \frac{1}{2}GH \quad \Rightarrow \quad GF^2 = \frac{1}{4}GH^2
$$

So, the expression becomes:

$$
AF^2 + BF^2 + CF^2 = 3 \cdot \frac{1}{4}GH^2 + \frac{a^2 + b^2 + c^2}{3} = \frac{3}{4}GH^2 + \frac{a^2 + b^2 + c^2}{3}
$$

---

### Step 4: Compute $ GH^2 $

Using Euler's line and the known relation between the centroid $ G $, orthocenter $ H $, and circumcenter $ O $, we have:

$$
GH = 2GO \quad \Rightarrow \quad GH^2 = 4GO^2
$$

Also, from Euler's theorem, the square of the distance between the orthocenter and the circumcenter is:

$$
OH^2 = 9R^2 - (a^2 + b^2 + c^2)
$$

Since $ G $ lies on the Euler line between $ O $ and $ H $, and $ OG = \frac{1}{3}OH $, we get:

$$
GO^2 = \frac{1}{9}OH^2 = \frac{1}{9}(9R^2 - (a^2 + b^2 + c^2)) = R^2 - \frac{1}{9}(a^2 + b^2 + c^2)
$$

Thus:

$$
GH^2 = 4GO^2 = 4\left(R^2 - \frac{1}{9}(a^2 + b^2 + c^2)\right) = 4R^2 - \frac{4}{9}(a^2 + b^2 + c^2)
$$

---

### Step 5: Final Substitution

Now substitute $ GH^2 $ into the expression for $ AF^2 + BF^2 + CF^2 $:

$$
AF^2 + BF^2 + CF^2 = \frac{3}{4} \left(4R^2 - \frac{4}{9}(a^2 + b^2 + c^2)\right) + \frac{a^2 + b^2 + c^2}{3}
$$

Simplify:

$$
= 3R^2 - \frac{1}{3}(a^2 + b^2 + c^2) + \frac{1}{3}(a^2 + b^2 + c^2) = 3R^2
$$

---

### Final Answer

$$
\boxed{3R^2}
$$

[Assistant]
Explanation:
Okay, so I need to find AF² + BF² + CF² where F is the midpoint of GH, with G being the centroid and H the orthocenter of triangle ABC. The answer should be in terms of the side lengths a, b, c and the circumradius R. Hmm, let me start by recalling some properties of centroids and orthocenters.

First, the centroid G divides the medians in a 2:1 ratio. The orthocenter H is the intersection point of the altitudes. I remember there's a relationship between the centroid, orthocenter, and other centers like the circumcenter O. In fact, Euler's line states that G, H, and O are colinear, and HG = 2GO. Maybe that will come into play here.

Since F is the midpoint of GH, then GF = FH = (1/2)GH. But I need to relate this to AF² + BF² + CF². Maybe coordinate geometry could help here? Let me try setting up coordinates for triangle ABC. But before jumping into coordinates, perhaps there are vector approaches or formulas that can relate these points?

Alternatively, I remember that for any point P in the plane of triangle ABC, the formula for the sum of squared distances from P to the vertices is given by:

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

Wait, is that right? Let me recall. There's a formula that relates the sum of squared distances from a point to the vertices with the centroid. Yes, I think it's:

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

But I need to verify this. Let me think. If P is the centroid G, then PA² + PB² + PC² = 3PG² + ... but PG would be zero, so PA² + PB² + PC² = GA² + GB² + GC². Wait, but that seems not quite. Wait, actually, the formula is:

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

Wait, maybe I need to recall the formula properly. Let me think of coordinates. Let me take coordinates where G is the origin. Then, the coordinates of A, B, C satisfy (A + B + C)/3 = 0, so A + B + C = 0. Then, for any point P, PA² + PB² + PC² = |P - A|² + |P - B|² + |P - C|². Expanding each term:

|P - A|² = |P|² - 2P·A + |A|²

Similarly for others. Adding them up:

3|P|² - 2P·(A + B + C) + |A|² + |B|² + |C|²

But since A + B + C = 0, this simplifies to 3|P|² + |A|² + |B|² + |C|². Therefore, if G is the origin, then PG is just |P|, so PA² + PB² + PC² = 3PG² + |A|² + |B|² + |C|². Wait, but if G is the origin, then PG is the distance from P to G, which is |P - G|, but since G is the origin, it's just |P|. So yes, the formula is PA² + PB² + PC² = 3PG² + GA² + GB² + GC²? Wait, no. Wait, GA² is |A - G|². Since G is the origin, GA² is |A|². Similarly for others. Therefore, PA² + PB² + PC² = 3PG² + GA² + GB² + GC². So that formula is correct.

Therefore, if I can express AF² + BF² + CF² in terms of PG² where P is F, then I can use this formula. So if I can find the position vector of F relative to G, then compute PG², which would be GF² since G is the origin? Wait, no. Wait, if I take G as the origin, then F is some point, and AF² + BF² + CF² would be 3GF² + GA² + GB² + GC². Wait, but in this case, if I take G as the origin, then the formula is PA² + PB² + PC² = 3PG² + GA² + GB² + GC². Therefore, if P is F, then AF² + BF² + CF² = 3GF² + GA² + GB² + GC². Therefore, if I can compute GF² and GA² + GB² + GC², then I can find the desired sum.

So first, let me compute GA² + GB² + GC². For that, since G is centroid, there is a formula. In general, for centroid, GA² + GB² + GC² = (1/3)(a² + b² + c²). Wait, is that correct? Wait, let me recall. Let me think of coordinates again. Let me take coordinates where centroid is at origin. Then, A + B + C = 0. Then, GA² = |A|², GB² = |B|², GC² = |C|². Then, GA² + GB² + GC² = |A|² + |B|² + |C|². But how does that relate to the sides?

Alternatively, using the formula for the centroid. Let me recall that in any triangle, the sum of the squares of the medians is related to the sides. But maybe I need another approach. Alternatively, using vectors. Let me consider vectors from centroid.

Alternatively, since in coordinate system with centroid at origin, A + B + C = 0. Then, the square of the sides can be expressed as |B - C|² = a², etc. Let me compute |A|² + |B|² + |C|². Let me compute |A - B|² = |A - B|² = |A|² + |B|² - 2A·B. Similarly for others. So:

a² = |B - C|² = |B|² + |C|² - 2B·C

b² = |C - A|² = |C|² + |A|² - 2C·A

c² = |A - B|² = |A|² + |B|² - 2A·B

Adding all three equations:

a² + b² + c² = 2(|A|² + |B|² + |C|²) - 2(A·B + B·C + C·A)

Therefore, |A|² + |B|² + |C|² = [a² + b² + c² + 2(A·B + B·C + C·A)] / 2

But since A + B + C = 0, then (A + B + C)·(A + B + C) = 0 => |A|² + |B|² + |C|² + 2(A·B + B·C + C·A) = 0. Therefore, A·B + B·C + C·A = - (|A|² + |B|² + |C|²)/2. Substituting back into the previous equation:

|A|² + |B|² + |C|² = [a² + b² + c² + 2*(- (|A|² + |B|² + |C|²)/2)] / 2

Simplify:

|A|² + |B|² + |C|² = [a² + b² + c² - (|A|² + |B|² + |C|²)] / 2

Multiply both sides by 2:

2(|A|² + |B|² + |C|²) = a² + b² + c² - (|A|² + |B|² + |C|²)

Bring terms together:

2(|A|² + |B|² + |C|²) + (|A|² + |B|² + |C|²) = a² + b² + c²

So 3(|A|² + |B|² + |C|²) = a² + b² + c² => |A|² + |B|² + |C|² = (a² + b² + c²)/3

Therefore, GA² + GB² + GC² = (a² + b² + c²)/3. So that's settled. Therefore, going back to the formula:

AF² + BF² + CF² = 3GF² + (a² + b² + c²)/3

So now, I need to compute GF². Since F is the midpoint of GH, GF is half of GH. Therefore, GF = (1/2)GH, so GF² = (1/4)GH². Therefore, AF² + BF² + CF² = 3*(1/4)GH² + (a² + b² + c²)/3 = (3/4)GH² + (a² + b² + c²)/3.

Therefore, I need to compute GH². So now, the problem reduces to finding GH² in terms of a, b, c, and R.

So how to compute GH? There is a formula for the distance between centroid and orthocenter. Let me recall. In Euler's line, the distance between centroid G and orthocenter H is given by GH = 3GO, where O is the circumcenter. Wait, Euler's line states that HG = 2GO, so GH = 2GO? Wait, let me check.

Euler's line: The centroid G divides the segment joining the orthocenter H and the circumcenter O in the ratio HG : GO = 2 : 1. So HG = 2GO. Therefore, GH = 2GO. Therefore, GH² = 4GO². Therefore, if I can find GO², then I can compute GH².

Alternatively, there is a formula for GO². Let me recall. There is a relation between the distances from the centroid to the circumcenter and other elements. Also, there is a formula involving the sides and the circumradius. Let me recall that in any triangle, the distance between the centroid and circumcenter is given by:

GO² = R² - (a² + b² + c²)/9

Wait, is that correct? Wait, I need to verify. Alternatively, there is a formula:

OH² = 9R² - (a² + b² + c²)

Where O is circumcenter and H is orthocenter. Wait, that's the formula for OH². Then, since GH = 2GO, then GH² = 4GO². But how to relate GO² to OH²?

Since O, G, H are colinear with HG = 2GO, so OH = OG + GH? Wait, depending on the direction. Let me think. If G is between O and H, then OH = OG + GH. But since HG = 2GO, then if we let OG = x, then GH = 2x, so OH = OG + GH = x + 2x = 3x. Therefore, OH = 3x, so x = OH / 3. Therefore, OG = OH / 3, so GO² = (OH²)/9. Therefore, GH² = 4GO² = 4*(OH²)/9 = (4/9)OH². Therefore, if I can find OH², then I can compute GH².

But there is a formula for OH². Yes, Euler's formula states that OH² = 9R² - (a² + b² + c²). Wait, let me check. Wait, actually, Euler's theorem states that OH² = 9R² - (a² + b² + c²). Wait, no, I think it's OH² = 9R² - (a² + b² + c²). Wait, let me check with an equilateral triangle. In an equilateral triangle, O and H coincide, so OH = 0. Then, 9R² - (a² + b² + c²) should be zero. Let me check. In an equilateral triangle, a = b = c, and R = a / √3. Therefore, 9R² = 9*(a² / 3) = 3a². Then, a² + b² + c² = 3a². Therefore, 9R² - (a² + b² + c²) = 3a² - 3a² = 0. Correct. So that formula seems correct. Therefore, OH² = 9R² - (a² + b² + c²). Therefore, GH² = (4/9)OH² = (4/9)(9R² - (a² + b² + c²)) = 4R² - (4/9)(a² + b² + c²). Therefore, GH² = 4R² - (4/9)(a² + b² + c²).

Therefore, going back to AF² + BF² + CF² = (3/4)GH² + (a² + b² + c²)/3. Let me substitute GH²:

AF² + BF² + CF² = (3/4)[4R² - (4/9)(a² + b² + c²)] + (a² + b² + c²)/3

Let me compute this step by step. First, expand the first term:

(3/4)*4R² = 3R²

(3/4)*(-4/9)(a² + b² + c²) = - (3/4)*(4/9)(a² + b² + c²) = - (1/3)(a² + b² + c²)

Therefore, AF² + BF² + CF² = 3R² - (1/3)(a² + b² + c²) + (1/3)(a² + b² + c²) = 3R².

Wait, that's interesting! The terms involving (a² + b² + c²) cancel out, leaving just 3R². Therefore, AF² + BF² + CF² = 3R². Is that possible? Let me check with an equilateral triangle. If ABC is equilateral, then centroid and orthocenter coincide, so G = H. Therefore, F is the midpoint of GH, which is the same as G. Therefore, AF² + BF² + CF² would be GA² + GB² + GC². In an equilateral triangle, GA = GB = GC. Let me compute that. Let’s say the side length is a. The centroid divides the median in ratio 2:1. The length of the median in an equilateral triangle is (sqrt(3)/2)a. Therefore, GA = (2/3)*(sqrt(3)/2)a = (sqrt(3)/3)a. Therefore, GA² = (1/3)a². Therefore, AF² + BF² + CF² = 3*(1/3)a² = a². On the other hand, the circumradius R of an equilateral triangle is (a)/sqrt(3). Therefore, 3R² = 3*(a²/3) = a². So it matches. Therefore, in this case, the formula gives 3R², which equals a². So that checks out. Therefore, maybe the answer is 3R²? But the problem says to express it in terms of a, b, c and R. Wait, but according to this, it's just 3R². However, the problem says "in terms of the side lengths a, b, c and circumradius R". But if it's 3R², then it's already in terms of R. However, maybe there's a mistake here? Let me check another case.

Take a right-angled triangle. Let me take a 3-4-5 triangle. Let me compute AF² + BF² + CF² and check if it equals 3R². Let me recall that in a right-angled triangle, the circumradius R is half the hypotenuse. So for a 3-4-5 triangle, hypotenuse is 5, so R = 5/2. Therefore, 3R² = 3*(25/4) = 75/4 = 18.75.

Now, let me compute AF² + BF² + CF². First, let me find coordinates. Let me place the triangle with right angle at C, so coordinates: Let me set C at (0,0), B at (3,0), A at (0,4). Then, the triangle has sides: AB = 5, BC = 3, AC = 4. Let me find centroid G. The centroid is the average of the coordinates: G_x = (0 + 3 + 0)/3 = 1, G_y = (0 + 0 + 4)/3 = 4/3. So G is at (1, 4/3).

Orthocenter H of a right-angled triangle is at the vertex of the right angle, which is C (0,0). Wait, is that correct? In a right-angled triangle, the orthocenter is at the right-angled vertex. Yes, because the altitudes are the legs themselves. Therefore, H is at (0,0).

Therefore, F is the midpoint of GH. G is (1, 4/3), H is (0,0). Therefore, midpoint F has coordinates ((1 + 0)/2, (4/3 + 0)/2) = (0.5, 2/3).

Now, compute AF² + BF² + CF². Let me compute each term:

Point A is (0,4). Point F is (0.5, 2/3).

AF² = (0 - 0.5)^2 + (4 - 2/3)^2 = (-0.5)^2 + (10/3)^2 = 0.25 + 100/9 = 0.25 + 11.111... ≈ 11.361...

Similarly, BF²: Point B is (3,0). So BF² = (3 - 0.5)^2 + (0 - 2/3)^2 = (2.5)^2 + (-2/3)^2 = 6.25 + 4/9 ≈ 6.25 + 0.444... ≈ 6.694...

CF²: Point C is (0,0). CF² = (0 - 0.5)^2 + (0 - 2/3)^2 = 0.25 + 4/9 ≈ 0.25 + 0.444... ≈ 0.694...

Adding them up: 11.361 + 6.694 + 0.694 ≈ 18.749... which is approximately 18.75, which is 75/4. Which is equal to 3R², since R = 5/2, 3*(25/4) = 75/4. So that checks out. Therefore, the formula seems correct. Therefore, AF² + BF² + CF² = 3R². Therefore, the answer is 3R². But the problem says "in terms of the side lengths a, b, c and circumradius R". However, 3R² is already in terms of R. But maybe there's a way to express it in terms of a, b, c and R? But since the problem allows expressing in terms of a, b, c and R, and 3R² is already in terms of R, but maybe they want it expressed in terms of a, b, c and R? However, since in the case of the equilateral triangle, 3R² equals a², but in general, it's 3R². However, maybe there's a different expression? Wait, but according to the derivation, it cancels out. Let me check again.

Wait, in the derivation, we had AF² + BF² + CF² = 3R². But the problem says "in terms of the side lengths a, b, c and circumradius R". So maybe they want an expression that includes a, b, c and R? But according to the derivation, it's 3R². However, in the case of the 3-4-5 triangle, 3R² is 3*(25/4) = 75/4, which is equal to AF² + BF² + CF². However, if we wanted to write it in terms of a, b, c and R, but since R is already given, maybe 3R² is acceptable. However, maybe there's a mistake in my derivation?

Wait, let me check again. Let me go through the steps again.

Starting with AF² + BF² + CF² = 3GF² + (a² + b² + c²)/3. Then GH² = 4R² - (4/9)(a² + b² + c²). Then GF² = GH² / 4. Therefore, AF² + BF² + CF² = 3*(GH²)/4 + (a² + b² + c²)/3 = 3/4*(4R² - 4/9(a² + b² + c²)) + (a² + b² + c²)/3 = 3R² - (1/3)(a² + b² + c²) + (1/3)(a² + b² + c²) = 3R². So yes, the terms cancel. Therefore, regardless of a, b, c, the answer is 3R². Therefore, the answer is 3R². Therefore, maybe the problem is designed to have this answer. However, the problem says "in terms of the side lengths a, b, c and circumradius R". But since 3R² is already in terms of R, but maybe they want it expressed in terms of a, b, c and R? But unless there's a different expression. Wait, but maybe there's a formula that relates R with a, b, c? For example, in terms of the sides, R = (a b c)/(4Δ), where Δ is the area. But unless they want it expressed in terms of a, b, c and R, but since R is already given, maybe 3R² is acceptable. However, maybe the problem expects an expression that includes a, b, c? But according to the derivation, it's 3R². However, let me check another triangle.

Take an equilateral triangle with side length a. Then, R = a / √3. Therefore, 3R² = 3*(a² / 3) = a². Which matches with the earlier calculation. So that's correct. For the 3-4-5 triangle, 3R² = 3*(25/4) = 75/4, which matches the calculation. Therefore, seems correct. Therefore, maybe the answer is simply 3R². However, the problem says "in terms of the side lengths a, b, c and circumradius R". So maybe they want it expressed as 3R², which is in terms of R. But maybe there's a different way? Wait, but if you want to write it in terms of a, b, c and R, but since R is already given, maybe they just want 3R². However, maybe there's a mistake in my derivation? Let me check again.

Wait, let me think again. The formula AF² + BF² + CF² = 3GF² + (a² + b² + c²)/3. Then GH² = 4R² - (4/9)(a² + b² + c²). Therefore, GF² = GH² / 4 = [4R² - (4/9)(a² + b² + c²)] / 4 = R² - (1/9)(a² + b² + c²). Therefore, AF² + BF² + CF² = 3[R² - (1/9)(a² + b² + c²)] + (a² + b² + c²)/3 = 3R² - (1/3)(a² + b² + c²) + (1/3)(a² + b² + c²) = 3R². So yes, that's correct. Therefore, regardless of the triangle, AF² + BF² + CF² = 3R². Therefore, the answer is 3R². Therefore, maybe the problem is designed to have this answer. However, the problem says "in terms of the side lengths a, b, c and circumradius R". But since R is already given, maybe they just want 3R². However, maybe there's a different approach where it's expressed in terms of a, b, c and R? For example, if we use the formula that relates R with the sides, but since R is already given, maybe it's redundant. Alternatively, maybe the problem expects an answer in terms of a, b, c and R, but since 3R² is already in terms of R, and the problem allows that, then 3R² is the answer.

Alternatively, maybe I made a mistake in assuming that GH² = 4R² - (4/9)(a² + b² + c²). Let me check that again. Earlier, I used Euler's formula OH² = 9R² - (a² + b² + c²). Then, since GH = 2GO, and OH = 3GO, so GO = OH / 3. Therefore, GH = 2*(OH / 3) = 2OH / 3. Therefore, GH² = (4/9)OH² = (4/9)(9R² - (a² + b² + c²)) = 4R² - (4/9)(a² + b² + c²). Therefore, that is correct. Therefore, the rest follows. Therefore, the answer is 3R². Therefore, maybe the answer is simply 3R². Therefore, I think that is the answer.

But let me check another triangle. Let me take an isosceles triangle with sides 2, 2, 2. Wait, that's equilateral. Let me take a different triangle. Let me take a triangle with sides 2, 2, 3. Wait, is that possible? Let me check. The triangle inequality: 2 + 2 > 3? 4 > 3, yes. 2 + 3 > 2, yes. So okay. Let me compute R for this triangle. First, compute the area. Using Heron's formula: s = (2 + 2 + 3)/2 = 7/2. Area Δ = sqrt[s(s - a)(s - b)(s - c)] = sqrt[(7/2)(7/2 - 2)(7/2 - 2)(7/2 - 3)] = sqrt[(7/2)(3/2)(3/2)(1/2)] = sqrt[(7*3*3*1)/(16)] = sqrt(63/16) = (3√7)/4. Then, R = (a b c)/(4Δ) = (2*2*3)/(4*(3√7)/4) = 12 / (3√7) = 4 / √7. Therefore, 3R² = 3*(16 / 7) = 48/7 ≈ 6.857.

Now, let me compute AF² + BF² + CF². Let me find coordinates. Let me place the triangle with base BC = 3, and AB = AC = 2. Let me place coordinates: Let me put point B at (-1.5, 0), point C at (1.5, 0), so that BC = 3. Then, point A is somewhere above the x-axis. Let me compute coordinates of A. Since AB = AC = 2, the triangle is isoceles with AB = AC = 2. Let me find coordinates of A. Let me denote A as (0, h). Then, distance from A to B is sqrt[(0 - (-1.5))² + (h - 0)^2] = sqrt[(1.5)^2 + h²] = 2. Therefore, (1.5)^2 + h² = 4 => 2.25 + h² = 4 => h² = 1.75 => h = sqrt(7)/2 ≈ 1.322. Therefore, coordinates of A are (0, sqrt(7)/2). Therefore, coordinates:

A: (0, sqrt(7)/2)

B: (-1.5, 0)

C: (1.5, 0)

Now, centroid G is the average of coordinates:

G_x = (0 + (-1.5) + 1.5)/3 = 0/3 = 0

G_y = (sqrt(7)/2 + 0 + 0)/3 = sqrt(7)/6

Therefore, G is at (0, sqrt(7)/6)

Orthocenter H. In an isoceles triangle, the orthocenter lies along the altitude from A, which is the y-axis. Let me compute coordinates of H. In an isoceles triangle, the orthocenter, centroid, and circumcenter lie along the same line. Let me compute H. Alternatively, since we know coordinates of A, B, C, we can compute equations of altitudes.

Altitude from A is the y-axis. Altitude from B: Let me compute. The altitude from B to AC. First, find equation of AC. Points A(0, sqrt(7)/2) and C(1.5, 0). The slope of AC is (0 - sqrt(7)/2)/(1.5 - 0) = (-sqrt(7)/2)/1.5 = -sqrt(7)/3. Therefore, the altitude from B is perpendicular to AC, so its slope is reciprocal and opposite: 3/sqrt(7). Therefore, equation of altitude from B: passes through B(-1.5, 0), slope 3/sqrt(7). Let me write the equation:

y - 0 = (3/sqrt(7))(x + 1.5)

Similarly, the orthocenter H lies at the intersection of this altitude and the altitude from A (the y-axis). Therefore, substitute x = 0 into the equation:

y = (3/sqrt(7))(0 + 1.5) = (3/sqrt(7))*(3/2) = 9/(2 sqrt(7)) = (9 sqrt(7))/14

Therefore, coordinates of H are (0, 9√7 / 14). Therefore, midpoint F of GH: G is (0, sqrt(7)/6) and H is (0, 9√7 / 14). Therefore, midpoint F has coordinates:

x-coordinate: 0

y-coordinate: [sqrt(7)/6 + 9√7 /14]/2. Let me compute that:

First, common denominator for 6 and 14 is 42.

sqrt(7)/6 = 7 sqrt(7)/42

9√7 /14 = 27 sqrt(7)/42

Adding them: 7 sqrt(7)/42 + 27 sqrt(7)/42 = 34 sqrt(7)/42 = 17 sqrt(7)/21

Divide by 2: 17 sqrt(7)/42

Therefore, coordinates of F are (0, 17√7 / 42)

Now, compute AF² + BF² + CF².

First, compute AF²: distance from A(0, sqrt(7)/2) to F(0, 17√7 / 42):

Difference in y-coordinates: sqrt(7)/2 - 17√7 /42 = (21√7 / 42 - 17√7 /42) = 4√7 /42 = 2√7 /21

Therefore, AF² = (2√7 /21)^2 = 4*7 / 441 = 28 / 441 = 4 / 63 ≈ 0.0635

Wait, that seems small. Let me check again.

Wait, sqrt(7)/2 is equal to 21√7 / 42? Wait, sqrt(7)/2 = (sqrt(7)/2)*(21/21) = 21 sqrt(7)/42? Wait, no. Wait, sqrt(7)/2 is equal to (sqrt(7)/2)*(21/21) = 21 sqrt(7)/42? Wait, no. Wait, sqrt(7)/2 = (sqrt(7)/2)*(21/21) = (21 sqrt(7))/42? Wait, no. Wait, 21/21 is 1. So sqrt(7)/2 = sqrt(7)/2. To write it over 42, multiply numerator and denominator by 21: sqrt(7)/2 = (sqrt(7)*21)/42. Therefore, yes, sqrt(7)/2 = 21 sqrt(7)/42. Similarly, 17 sqrt(7)/42 is as is. Therefore, difference is 21 sqrt(7)/42 - 17 sqrt(7)/42 = 4 sqrt(7)/42 = 2 sqrt(7)/21. Therefore, AF² = (2 sqrt(7)/21)^2 = 4*7 / 441 = 28 / 441 = 4 / 63 ≈ 0.0635. Okay.

Now compute BF²: distance from B(-1.5, 0) to F(0, 17√7 /42). Let me compute coordinates:

Δx = 0 - (-1.5) = 1.5 = 3/2

Δy = 17√7 /42 - 0 = 17√7 /42

Therefore, BF² = (3/2)^2 + (17√7 /42)^2 = 9/4 + (289 * 7)/ (42^2)

Compute 289 * 7 = 2023

42^2 = 1764

Therefore, BF² = 9/4 + 2023 / 1764

Convert 9/4 to denominator 1764: 9/4 = (9 * 441)/1764 = 3969 / 1764

Therefore, BF² = (3969 + 2023)/1764 = 6, 3969 + 2023 = let me compute:

3969 + 2000 = 5969, plus 23 = 5992. Therefore, BF² = 5992 / 1764. Simplify:

Divide numerator and denominator by 4: 5992 / 4 = 1498, 1764 / 4 = 441. So 1498 / 441 ≈ 3.396...

Similarly, compute CF²: distance from C(1.5, 0) to F(0, 17√7 /42). It's symmetric to BF², so CF² = BF² = 5992 / 1764.

Therefore, AF² + BF² + CF² = 4/63 + 2*(5992 / 1764). Let me compute this.

First, convert 4/63 to denominator 1764: 4/63 = (4 * 28)/1764 = 112 / 1764

Then, 2*(5992 / 1764) = 11984 / 1764

Therefore, total is 112 + 11984 = 12096 / 1764

Simplify 12096 / 1764. Let me divide numerator and denominator by 12: 12096 / 12 = 1008, 1764 / 12 = 147. So 1008 / 147. Divide numerator and denominator by 21: 1008 / 21 = 48, 147 / 21 = 7. Therefore, 48 / 7 ≈ 6.857. Which is equal to 48/7. Now, 3R² was 48/7. Since earlier, R = 4 / sqrt(7), so 3R² = 3*(16 / 7) = 48/7. Correct. Therefore, this also checks out. Therefore, the formula holds. Therefore, the answer is indeed 3R². Therefore, the answer is \boxed{3R^2}.

**Final Answer**
\boxed{3R^2}

To solve the problem, we are given a triangle $ ABC $ with centroid $ G $, orthocenter $ H $, and a point $ F $ defined as the midpoint of segment $ \overline{GH} $. We are to compute the expression:

$$
AF^2 + BF^2 + CF^2
$$

in terms of the side lengths $ a, b, c $ and the circumradius $ R $ of triangle $ ABC $.

---

### Step 1: Use of a Known Vector Identity

There is a well-known identity in triangle geometry that relates the sum of squared distances from any point $ P $ to the vertices of triangle $ ABC $ to the centroid $ G $:

$$
PA^2 + PB^2 + PC^2 = 3PG^2 + GA^2 + GB^2 + GC^2
$$

Applying this identity with $ P = F $, we get:

$$
AF^2 + BF^2 + CF^2 = 3GF^2 + GA^2 + GB^2 + GC^2
$$

---

### Step 2: Compute $ GA^2 + GB^2 + GC^2 $

It is a known result that for the centroid $ G $ of triangle $ ABC $, the sum of the squares of the distances from $ G $ to the vertices is:

$$
GA^2 + GB^2 + GC^2 = \frac{a^2 + b^2 + c^2}{3}
$$

---

### Step 3: Compute $ GF^2 $

Since $ F $ is the midpoint of $ \overline{GH} $, we have:

$$
GF = \frac{1}{2}GH \quad \Rightarrow \quad GF^2 = \frac{1}{4}GH^2
$$

So, the expression becomes:

$$
AF^2 + BF^2 + CF^2 = 3 \cdot \frac{1}{4}GH^2 + \frac{a^2 + b^2 + c^2}{3} = \frac{3}{4}GH^2 + \frac{a^2 + b^2 + c^2}{3}
$$

---

### Step 4: Compute $ GH^2 $

Using Euler's line and the known relation between the centroid $ G $, orthocenter $ H $, and circumcenter $ O $, we have:

$$
GH = 2GO \quad \Rightarrow \quad GH^2 = 4GO^2
$$

Also, from Euler's theorem, the square of the distance between the orthocenter and the circumcenter is:

$$
OH^2 = 9R^2 - (a^2 + b^2 + c^2)
$$

Since $ G $ lies on the Euler line between $ O $ and $ H $, and $ OG = \frac{1}{3}OH $, we get:

$$
GO^2 = \frac{1}{9}OH^2 = \frac{1}{9}(9R^2 - (a^2 + b^2 + c^2)) = R^2 - \frac{1}{9}(a^2 + b^2 + c^2)
$$

Thus:

$$
GH^2 = 4GO^2 = 4\left(R^2 - \frac{1}{9}(a^2 + b^2 + c^2)\right) = 4R^2 - \frac{4}{9}(a^2 + b^2 + c^2)
$$

---

### Step 5: Final Substitution

Now substitute $ GH^2 $ into the expression for $ AF^2 + BF^2 + CF^2 $:

$$
AF^2 + BF^2 + CF^2 = \frac{3}{4} \left(4R^2 - \frac{4}{9}(a^2 + b^2 + c^2)\right) + \frac{a^2 + b^2 + c^2}{3}
$$

Simplify:

$$
= 3R^2 - \frac{1}{3}(a^2 + b^2 + c^2) + \frac{1}{3}(a^2 + b^2 + c^2) = 3R^2
$$

---

### Final Answer

$$
\boxed{3R^2}
$$

Solution:
Let the circumcenter $O$ of triangle $ABC$ be the origin.  Then
\[\overrightarrow{G} = \frac{\overrightarrow{A} + \overrightarrow{B} + \overrightarrow{C}}{3}\]and $\overrightarrow{H} = \overrightarrow{A} + \overrightarrow{B} + \overrightarrow{C},$ so
\[\overrightarrow{F} = \frac{2}{3} (\overrightarrow{A} + \overrightarrow{B} + \overrightarrow{C}).\]Then
\begin{align*}
AF^2 &= \|\overrightarrow{A} - \overrightarrow{F}\|^2 \\
&= \left\| \overrightarrow{A} - \frac{2}{3} (\overrightarrow{A} + \overrightarrow{B} + \overrightarrow{C}) \right\|^2 \\
&= \left\| \frac{1}{3} \overrightarrow{A} - \frac{2}{3} \overrightarrow{B} - \frac{2}{3} \overrightarrow{C} \right\|^2 \\
&= \frac{1}{9} \|\overrightarrow{A} - 2 \overrightarrow{B} - 2 \overrightarrow{C}\|^2 \\
&= \frac{1}{9} (\overrightarrow{A} - 2 \overrightarrow{B} - 2 \overrightarrow{C}) \cdot (\overrightarrow{A} - 2 \overrightarrow{B} - 2 \overrightarrow{C}) \\
&= \frac{1}{9} (\overrightarrow{A} \cdot \overrightarrow{A} + 4 \overrightarrow{B} \cdot \overrightarrow{B} + 4 \overrightarrow{C} \cdot \overrightarrow{C}  - 4 \overrightarrow{A} \cdot \overrightarrow{B} - 4 \overrightarrow{A} \cdot \overrightarrow{C} + 8 \overrightarrow{B} \cdot \overrightarrow{C}) \\
&= \frac{1}{9} (9R^2 - 4 \overrightarrow{A} \cdot \overrightarrow{B} - 4 \overrightarrow{A} \cdot \overrightarrow{C} + 8 \overrightarrow{B} \cdot \overrightarrow{C}).
\end{align*}Similarly,
\begin{align*}
BF^2 &= \frac{1}{9} (9R^2 - 4 \overrightarrow{A} \cdot \overrightarrow{B} + 8 \overrightarrow{A} \cdot \overrightarrow{C} - 4 \overrightarrow{B} \cdot \overrightarrow{C}), \\
CF^2 &= \frac{1}{9} (9R^2 + 8 \overrightarrow{A} \cdot \overrightarrow{B} - 4 \overrightarrow{A} \cdot \overrightarrow{C} - 4 \overrightarrow{B} \cdot \overrightarrow{C}).
\end{align*}Thus, $AF^2 + BF^2 + CF^2 = \boxed{3R^2}.$