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 here. To do that, 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, so 56-55=1. Yes! So 7*8 ≡ 1 mod 11. Therefore, the inverse is 8.

Therefore, multiplying 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, so 40-33=7. Therefore, k ≡ 7 mod 11. So 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 here. Again, find the inverse of 12 mod 13. Since 12 ≡ -1 mod 13, so inverse of -1 mod 13 is -1, because (-1)*(-1) = 1. Therefore, inverse of 12 mod 13 is 12.

Therefore, multiplying 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, we need to 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, divided by 7: 1440 / 7. Let me compute 7*205 = 1435. Then 1440 - 1435 = 5. So remainder 5. Good.

Divided by 11: 1440 / 11. Let me compute 11*130 = 1430. 1440 - 1430 = 10. So remainder 10. Good.

Divided by 13: 1440 / 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 solution after 1440 would be 1440 + 1001 = 2441, which is more than 2010. Therefore, 1440 is indeed the largest one less than 2010. But wait, 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 there's a mistake in my calculation. Let me check the step where I solved for m. Let me go back through the steps again.

Starting with x ≡ 5 mod 7 and x ≡ 10 mod 11. I set x = 7k + 5. Then 7k + 5 ≡ 10 mod 11 => 7k ≡ 5 mod 11. Then inverse of 7 mod 11 is 8, so k ≡ 5*8 = 40 ≡ 7 mod 11. So k = 11m + 7. Then x = 7*(11m + 7) + 5 = 77m + 49 + 5 = 77m + 54. Then moving to the third congruence: x ≡ 10 mod 13. So 77m + 54 ≡ 10 mod 13. Then 77 mod 13 is 12, 54 mod 13 is 2. So 12m + 2 ≡ 10 mod 13 => 12m ≡ 8 mod 13. Then inverse of 12 mod 13 is 12, since 12*12 = 144 ≡ 144 - 11*13 = 144 - 143 = 1 mod 13. Therefore, m ≡ 8*12 mod 13. 8*12 is 96. 96 divided by 13: 13*7=91, 96-91=5. So m ≡ 5 mod 13. Therefore, m = 13n + 5. Therefore, x = 77*(13n + 5) + 54 = 1001n + 385 + 54 = 1001n + 439. So that seems correct.

Therefore, the solutions are 439, 1440, 2441, etc. So yes, 1440 is the one less than 2010. Therefore, the answer is 1440. Wait, but let me check if there is a larger number less than 2010. For example, maybe 1440 + 1001 = 2441 is too big, but maybe there is another solution between 1440 and 2010? But since the modulus is 1001, the next solution after 1440 is 1440 + 1001 = 2441, which is more than 2010. Therefore, 1440 is indeed the largest one less than 2010.

But let me check if there is a mistake in the calculation of 1001n + 439. Let me check for n=1: 1001*1 + 439 = 1440. Correct. For n=0: 439. For n=2: 2441. So yes, 1440 is the correct one. Therefore, the answer is 1440. Wait, 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 and less than 2010? Well, 1440 + 1001 = 2441, which is over 2010. So no. Therefore, 1440 is the answer.

But let me check another way. Let me check if 2010 - 1440 = 570. So, is there a number between 1440 and 2010 that satisfies the conditions? For example, maybe 1440 + k, where k is between 1 and 570. But since the solutions are spaced 1001 apart, there are no solutions in between. Therefore, 1440 is the only one. Therefore, the answer is 1440.

But just to be thorough, let me check if there is a mistake in my initial solving. Let me try solving the congruences again.

First, x ≡ 5 mod 7, x ≡ 10 mod 11, x ≡ 10 mod 13.

Alternatively, since the last two congruences are x ≡ 10 mod 11 and x ≡ 10 mod 13. Since 11 and 13 are coprime, by CRT, x ≡ 10 mod (11*13) = 143. Therefore, x ≡ 10 mod 143. Therefore, x can be written as 143k + 10. Then, we can combine this with the first congruence x ≡ 5 mod 7.

So, substituting x = 143k + 10 into x ≡ 5 mod 7:

143k + 10 ≡ 5 mod 7.

Compute 143 mod 7. Let me divide 143 by 7. 7*20 = 140, so 143 - 140 = 3. Therefore, 143 ≡ 3 mod 7. Therefore:

3k + 10 ≡ 5 mod 7 => 3k ≡ 5 - 10 mod 7 => 3k ≡ -5 mod 7 => -5 mod 7 is 2, since -5 +7=2. Therefore, 3k ≡ 2 mod 7.

Now, solving for k: find inverse of 3 mod 7. 3*5=15≡1 mod7, so inverse is 5. Therefore, k ≡ 2*5 mod7 => k≡10 mod7 => k≡3 mod7. Therefore, k=7m + 3 for some integer m. Therefore, substituting back into x=143k +10:

x=143*(7m +3) +10= 1001m + 429 +10=1001m + 439. Which is the same result as before. Therefore, x=1001m +439. Therefore, same conclusion. Therefore, the solutions are indeed 439, 1440, 2441,... So, 1440 is the answer. Therefore, I think my answer is correct.

But just to be 100% sure, let me check if there is a larger number less than 2010. Let me compute 2010 - 439 = 1571. So, 1571 divided by 1001 is approximately 1.569. So, the next solution after 439 is 439 + 1001 = 1440, then 1440 + 1001 = 2441. Therefore, 1440 is indeed the largest less than 2010. Therefore, the answer is 1440.

But just to check, let me check if 1440 is less than 2010. Yes, 1440 is less than 2010. And the next one is 2441, which is more than 2010. Therefore, 1440 is the correct answer.

Alternatively, maybe I can check if there is another solution by considering that maybe the modulus is different? Wait, but we have already combined the congruences properly. So, since 7, 11, 13 are coprime, the combined modulus is 1001, so the solutions are spaced 1001 apart. Therefore, there are no other solutions between 1440 and 2010. Therefore, 1440 is the answer.

Therefore, I think the answer is 1440. Wait, but let me check once again with another method. Let me try to find numbers less than 2010 that satisfy x ≡ 10 mod 11 and x ≡ 10 mod 13. Since x ≡ 10 mod 11 and x ≡ 10 mod 13, then x ≡ 10 mod 143. Therefore, numbers congruent to 10 mod 143. So, starting from 10, 153, 296, 439, 582, 725, 868, 1011, 1154, 1297, 1440, 1583, 1726, 1869, 2012,... So, up to 2010. So, the numbers are 10, 153, 296, 439, 582, 725, 868, 1011, 1154, 1297, 1440, 1583, 1726, 1869, 2012. So, the last one less than 2010 is 1869. Wait, but 1869 + 143 = 2012, which is over. So, 1869 is the last one. Wait, but according to our previous calculation, the solutions are 1001n + 439. Let me check if 1869 is in that sequence. Let me compute 1869 - 439 = 1430. 1430 divided by 1001 is 1.428... So, 1001*1 + 439 = 1440, then 1001*2 + 439 = 2441. So, 1869 is not in that sequence. Wait, so there is a contradiction here. Which means that my previous conclusion is wrong?

Wait, this is confusing. Let me check. If x ≡ 10 mod 143, then x can be written as 143k + 10. Then combining with x ≡ 5 mod 7 gives us x = 1001m + 439. But according to the list above, numbers congruent to 10 mod 143 include 10, 153, 296, 439, 582, 725, 868, 1011, 1154, 1297, 1440, 1583, 1726, 1869, 2012. So, for example, 1440 is in that list. Let me check if 1440 satisfies x ≡ 5 mod 7. 1440 divided by 7: 7*205 = 1435, 1440 - 1435 = 5. Yes, remainder 5. So 1440 is in the solution set. Similarly, 1869: let me check if 1869 ≡ 5 mod 7. 1869 divided by 7: 7*267 = 1869. Wait, 7*267 is 1869? Let me check 7*200=1400, 7*60=420, 7*7=49. So 1400 + 420 = 1820, plus 49 is 1869. Therefore, 1869 is divisible by 7, so remainder 0. But we need remainder 5. Therefore, 1869 does not satisfy the first condition. Therefore, even though it satisfies the last two conditions, it doesn't satisfy the first one. Therefore, my previous list of numbers congruent to 10 mod 143 includes numbers that don't satisfy the first condition. Therefore, the correct solutions are only those numbers which are congruent to 10 mod 143 and also congruent to 5 mod 7. Which we found to be x = 1001m + 439. Therefore, in the list above, which numbers are congruent to 5 mod 7?

Let me check:

Take 10: 10 mod 7 is 3. Not 5.

153: 153 /7: 7*21=147, 153-147=6. Not 5.

296: 296 /7: 7*42=294, 296-294=2. Not 5.

439: 439 /7: 7*62=434, 439-434=5. Yes, remainder 5. So 439 is good.

582: 582 /7: 7*83=581, 582-581=1. Not 5.

725: 725 /7: 7*103=721, 725-721=4. Not 5.

868: 868 /7: 7*124=868. Remainder 0. Not 5.

1011: 1011 /7: 7*144=1008, 1011-1008=3. Not 5.

1154: 1154 /7: 7*164=1148, 1154-1148=6. Not 5.

1297: 1297 /7: 7*185=1295, 1297-1295=2. Not 5.

1440: 1440 /7: 7*205=1435, 1440-1435=5. Yes, remainder 5.

1583: 1583 /7: 7*226=1582, 1583-1582=1. Not 5.

1726: 1726 /7: 7*246=1722, 1726-1722=4. Not 5.

1869: 1869 /7=267, remainder 0. Not 5.

2012: 2012 /7: 7*287=2009, 2012-2009=3. Not 5.

So, only 439 and 1440 in that list satisfy x ≡ 5 mod 7. Therefore, the solutions are indeed 439, 1440, 2441,... So, 1440 is the correct answer. Therefore, my initial conclusion was correct. Therefore, the answer is 1440.

But just to make sure, let me check if there is another way to approach this problem. For example, maybe using the Chinese Remainder Theorem step by step.

Alternatively, since x ≡ 10 mod 11 and x ≡ 10 mod 13, as we saw, x ≡ 10 mod 143. Then, combining with x ≡ 5 mod 7. So, solving x ≡ 10 mod 143 and x ≡ 5 mod 7. Let me write x = 143k + 10. Then, 143k + 10 ≡ 5 mod 7. As before, 143 ≡ 3 mod 7, so 3k + 10 ≡ 5 mod 7 => 3k ≡ -5 mod 7 => 3k ≡ 2 mod 7. Then, k ≡ 2*3^{-1} mod 7. Since 3^{-1} mod 7 is 5, as 3*5=15≡1 mod7. Therefore, k ≡ 2*5=10≡3 mod7. Therefore, k=7m + 3. Therefore, x=143*(7m +3)+10=1001m + 429 +10=1001m + 439. Same result. Therefore, same conclusion. Therefore, 1440 is the answer.

Therefore, I think I can be confident that 1440 is the correct answer. Therefore, the largest integer less than 2010 satisfying all three conditions is 1440.

**Final Answer**
\boxed{1440}

To solve the problem, we are given the following system of congruences:

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

We are to find the **largest integer less than 2010** that satisfies all three conditions.

---

### Step 1: Combine the Last Two Congruences

We first combine the congruences:

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

Since 11 and 13 are coprime, by the **Chinese Remainder Theorem**, we can combine these into a single congruence:

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

So now we have:

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

---

### Step 2: Combine the Two Congruences

Let’s write $ x = 143k + 10 $, and substitute into the first congruence:

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

We compute $ 143 \mod 7 $:

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

So the equation becomes:

$$
3k + 10 \equiv 5 \pmod{7}
\Rightarrow 3k \equiv -5 \equiv 2 \pmod{7}
$$

Now, we solve for $ k $:

- Find the modular inverse of 3 modulo 7. Since $ 3 \cdot 5 = 15 \equiv 1 \pmod{7} $, the inverse is 5.
- Multiply both sides by 5:

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

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

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

---

### Step 3: Find the Largest $ 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:

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

---

### Step 4: Verify the Solution

Let’s check that $ x = 1440 $ satisfies all three original congruences:

- $ 1440 \div 7 = 205 $ remainder $ 5 $ → ✅
- $ 1440 \div 11 = 130 $ remainder $ 10 $ → ✅
- $ 1440 \div 13 = 110 $ remainder $ 10 $ → ✅

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 here. To do that, 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, so 56-55=1. Yes! So 7*8 ≡ 1 mod 11. Therefore, the inverse is 8.

Therefore, multiplying 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, so 40-33=7. Therefore, k ≡ 7 mod 11. So 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 here. Again, find the inverse of 12 mod 13. Since 12 ≡ -1 mod 13, so inverse of -1 mod 13 is -1, because (-1)*(-1) = 1. Therefore, inverse of 12 mod 13 is 12.

Therefore, multiplying 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, we need to 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, divided by 7: 1440 / 7. Let me compute 7*205 = 1435. Then 1440 - 1435 = 5. So remainder 5. Good.

Divided by 11: 1440 / 11. Let me compute 11*130 = 1430. 1440 - 1430 = 10. So remainder 10. Good.

Divided by 13: 1440 / 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 solution after 1440 would be 1440 + 1001 = 2441, which is more than 2010. Therefore, 1440 is indeed the largest one less than 2010. But wait, 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 there's a mistake in my calculation. Let me check the step where I solved for m. Let me go back through the steps again.

Starting with x ≡ 5 mod 7 and x ≡ 10 mod 11. I set x = 7k + 5. Then 7k + 5 ≡ 10 mod 11 => 7k ≡ 5 mod 11. Then inverse of 7 mod 11 is 8, so k ≡ 5*8 = 40 ≡ 7 mod 11. So k = 11m + 7. Then x = 7*(11m + 7) + 5 = 77m + 49 + 5 = 77m + 54. Then moving to the third congruence: x ≡ 10 mod 13. So 77m + 54 ≡ 10 mod 13. Then 77 mod 13 is 12, 54 mod 13 is 2. So 12m + 2 ≡ 10 mod 13 => 12m ≡ 8 mod 13. Then inverse of 12 mod 13 is 12, since 12*12 = 144 ≡ 144 - 11*13 = 144 - 143 = 1 mod 13. Therefore, m ≡ 8*12 mod 13. 8*12 is 96. 96 divided by 13: 13*7=91, 96-91=5. So m ≡ 5 mod 13. Therefore, m = 13n + 5. Therefore, x = 77*(13n + 5) + 54 = 1001n + 385 + 54 = 1001n + 439. So that seems correct.

Therefore, the solutions are 439, 1440, 2441, etc. So yes, 1440 is the one less than 2010. Therefore, the answer is 1440. Wait, but let me check if there is a larger number less than 2010. For example, maybe 1440 + 1001 = 2441 is too big, but maybe there is another solution between 1440 and 2010? But since the modulus is 1001, the next solution after 1440 is 1440 + 1001 = 2441, which is more than 2010. Therefore, 1440 is indeed the largest one less than 2010.

But let me check if there is a mistake in the calculation of 1001n + 439. Let me check for n=1: 1001*1 + 439 = 1440. Correct. For n=0: 439. For n=2: 2441. So yes, 1440 is the correct one. Therefore, the answer is 1440. Wait, 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 and less than 2010? Well, 1440 + 1001 = 2441, which is over 2010. So no. Therefore, 1440 is the answer.

But let me check another way. Let me check if 2010 - 1440 = 570. So, is there a number between 1440 and 2010 that satisfies the conditions? For example, maybe 1440 + k, where k is between 1 and 570. But since the solutions are spaced 1001 apart, there are no solutions in between. Therefore, 1440 is the only one. Therefore, the answer is 1440.

But just to be thorough, let me check if there is a mistake in my initial solving. Let me try solving the congruences again.

First, x ≡ 5 mod 7, x ≡ 10 mod 11, x ≡ 10 mod 13.

Alternatively, since the last two congruences are x ≡ 10 mod 11 and x ≡ 10 mod 13. Since 11 and 13 are coprime, by CRT, x ≡ 10 mod (11*13) = 143. Therefore, x ≡ 10 mod 143. Therefore, x can be written as 143k + 10. Then, we can combine this with the first congruence x ≡ 5 mod 7.

So, substituting x = 143k + 10 into x ≡ 5 mod 7:

143k + 10 ≡ 5 mod 7.

Compute 143 mod 7. Let me divide 143 by 7. 7*20 = 140, so 143 - 140 = 3. Therefore, 143 ≡ 3 mod 7. Therefore:

3k + 10 ≡ 5 mod 7 => 3k ≡ 5 - 10 mod 7 => 3k ≡ -5 mod 7 => -5 mod 7 is 2, since -5 +7=2. Therefore, 3k ≡ 2 mod 7.

Now, solving for k: find inverse of 3 mod 7. 3*5=15≡1 mod7, so inverse is 5. Therefore, k ≡ 2*5 mod7 => k≡10 mod7 => k≡3 mod7. Therefore, k=7m + 3 for some integer m. Therefore, substituting back into x=143k +10:

x=143*(7m +3) +10= 1001m + 429 +10=1001m + 439. Which is the same result as before. Therefore, x=1001m +439. Therefore, same conclusion. Therefore, the solutions are indeed 439, 1440, 2441,... So, 1440 is the answer. Therefore, I think my answer is correct.

But just to be 100% sure, let me check if there is a larger number less than 2010. Let me compute 2010 - 439 = 1571. So, 1571 divided by 1001 is approximately 1.569. So, the next solution after 439 is 439 + 1001 = 1440, then 1440 + 1001 = 2441. Therefore, 1440 is indeed the largest less than 2010. Therefore, the answer is 1440.

But just to check, let me check if 1440 is less than 2010. Yes, 1440 is less than 2010. And the next one is 2441, which is more than 2010. Therefore, 1440 is the correct answer.

Alternatively, maybe I can check if there is another solution by considering that maybe the modulus is different? Wait, but we have already combined the congruences properly. So, since 7, 11, 13 are coprime, the combined modulus is 1001, so the solutions are spaced 1001 apart. Therefore, there are no other solutions between 1440 and 2010. Therefore, 1440 is the answer.

Therefore, I think the answer is 1440. Wait, but let me check once again with another method. Let me try to find numbers less than 2010 that satisfy x ≡ 10 mod 11 and x ≡ 10 mod 13. Since x ≡ 10 mod 11 and x ≡ 10 mod 13, then x ≡ 10 mod 143. Therefore, numbers congruent to 10 mod 143. So, starting from 10, 153, 296, 439, 582, 725, 868, 1011, 1154, 1297, 1440, 1583, 1726, 1869, 2012,... So, up to 2010. So, the numbers are 10, 153, 296, 439, 582, 725, 868, 1011, 1154, 1297, 1440, 1583, 1726, 1869, 2012. So, the last one less than 2010 is 1869. Wait, but 1869 + 143 = 2012, which is over. So, 1869 is the last one. Wait, but according to our previous calculation, the solutions are 1001n + 439. Let me check if 1869 is in that sequence. Let me compute 1869 - 439 = 1430. 1430 divided by 1001 is 1.428... So, 1001*1 + 439 = 1440, then 1001*2 + 439 = 2441. So, 1869 is not in that sequence. Wait, so there is a contradiction here. Which means that my previous conclusion is wrong?

Wait, this is confusing. Let me check. If x ≡ 10 mod 143, then x can be written as 143k + 10. Then combining with x ≡ 5 mod 7 gives us x = 1001m + 439. But according to the list above, numbers congruent to 10 mod 143 include 10, 153, 296, 439, 582, 725, 868, 1011, 1154, 1297, 1440, 1583, 1726, 1869, 2012. So, for example, 1440 is in that list. Let me check if 1440 satisfies x ≡ 5 mod 7. 1440 divided by 7: 7*205 = 1435, 1440 - 1435 = 5. Yes, remainder 5. So 1440 is in the solution set. Similarly, 1869: let me check if 1869 ≡ 5 mod 7. 1869 divided by 7: 7*267 = 1869. Wait, 7*267 is 1869? Let me check 7*200=1400, 7*60=420, 7*7=49. So 1400 + 420 = 1820, plus 49 is 1869. Therefore, 1869 is divisible by 7, so remainder 0. But we need remainder 5. Therefore, 1869 does not satisfy the first condition. Therefore, even though it satisfies the last two conditions, it doesn't satisfy the first one. Therefore, my previous list of numbers congruent to 10 mod 143 includes numbers that don't satisfy the first condition. Therefore, the correct solutions are only those numbers which are congruent to 10 mod 143 and also congruent to 5 mod 7. Which we found to be x = 1001m + 439. Therefore, in the list above, which numbers are congruent to 5 mod 7?

Let me check:

Take 10: 10 mod 7 is 3. Not 5.

153: 153 /7: 7*21=147, 153-147=6. Not 5.

296: 296 /7: 7*42=294, 296-294=2. Not 5.

439: 439 /7: 7*62=434, 439-434=5. Yes, remainder 5. So 439 is good.

582: 582 /7: 7*83=581, 582-581=1. Not 5.

725: 725 /7: 7*103=721, 725-721=4. Not 5.

868: 868 /7: 7*124=868. Remainder 0. Not 5.

1011: 1011 /7: 7*144=1008, 1011-1008=3. Not 5.

1154: 1154 /7: 7*164=1148, 1154-1148=6. Not 5.

1297: 1297 /7: 7*185=1295, 1297-1295=2. Not 5.

1440: 1440 /7: 7*205=1435, 1440-1435=5. Yes, remainder 5.

1583: 1583 /7: 7*226=1582, 1583-1582=1. Not 5.

1726: 1726 /7: 7*246=1722, 1726-1722=4. Not 5.

1869: 1869 /7=267, remainder 0. Not 5.

2012: 2012 /7: 7*287=2009, 2012-2009=3. Not 5.

So, only 439 and 1440 in that list satisfy x ≡ 5 mod 7. Therefore, the solutions are indeed 439, 1440, 2441,... So, 1440 is the correct answer. Therefore, my initial conclusion was correct. Therefore, the answer is 1440.

But just to make sure, let me check if there is another way to approach this problem. For example, maybe using the Chinese Remainder Theorem step by step.

Alternatively, since x ≡ 10 mod 11 and x ≡ 10 mod 13, as we saw, x ≡ 10 mod 143. Then, combining with x ≡ 5 mod 7. So, solving x ≡ 10 mod 143 and x ≡ 5 mod 7. Let me write x = 143k + 10. Then, 143k + 10 ≡ 5 mod 7. As before, 143 ≡ 3 mod 7, so 3k + 10 ≡ 5 mod 7 => 3k ≡ -5 mod 7 => 3k ≡ 2 mod 7. Then, k ≡ 2*3^{-1} mod 7. Since 3^{-1} mod 7 is 5, as 3*5=15≡1 mod7. Therefore, k ≡ 2*5=10≡3 mod7. Therefore, k=7m + 3. Therefore, x=143*(7m +3)+10=1001m + 429 +10=1001m + 439. Same result. Therefore, same conclusion. Therefore, 1440 is the answer.

Therefore, I think I can be confident that 1440 is the correct answer. Therefore, the largest integer less than 2010 satisfying all three conditions is 1440.

**Final Answer**
\boxed{1440}

To solve the problem, we are given the following system of congruences:

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

We are to find the **largest integer less than 2010** that satisfies all three conditions.

---

### Step 1: Combine the Last Two Congruences

We first combine the congruences:

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

Since 11 and 13 are coprime, by the **Chinese Remainder Theorem**, we can combine these into a single congruence:

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

So now we have:

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

---

### Step 2: Combine the Two Congruences

Let’s write $ x = 143k + 10 $, and substitute into the first congruence:

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

We compute $ 143 \mod 7 $:

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

So the equation becomes:

$$
3k + 10 \equiv 5 \pmod{7}
\Rightarrow 3k \equiv -5 \equiv 2 \pmod{7}
$$

Now, we solve for $ k $:

- Find the modular inverse of 3 modulo 7. Since $ 3 \cdot 5 = 15 \equiv 1 \pmod{7} $, the inverse is 5.
- Multiply both sides by 5:

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

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

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

---

### Step 3: Find the Largest $ 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:

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

---

### Step 4: Verify the Solution

Let’s check that $ x = 1440 $ satisfies all three original congruences:

- $ 1440 \div 7 = 205 $ remainder $ 5 $ → ✅
- $ 1440 \div 11 = 130 $ remainder $ 10 $ → ✅
- $ 1440 \div 13 = 110 $ remainder $ 10 $ → ✅

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 √x, then that should mean y is equal to some constant divided by √x. So the formula would be y = k / √x. Yeah, that makes sense. So instead of dividing by x, we divide by the square root of x.

Alright, so given that when x = 24, y = 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:

15 = k / √24

I need to solve for k. To do that, I can multiply both sides by √24.

k = 15 * √24

Hmm, let me compute that. But maybe I can simplify √24 first. √24 is the same as √(4*6) which is √4 * √6 = 2√6. So √24 = 2√6. Therefore, k = 15 * 2√6 = 30√6.

So the constant of proportionality k is 30√6. Therefore, the equation that relates y and x is y = 30√6 / √x.

Now, the question asks for the value of x when y = 3. So I need to substitute y = 3 into the equation and solve for x. Let me write that down:

3 = 30√6 / √x

Okay, so I need to solve for x here. 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:

√x = (30√6) / 3

Simplify the right side:

√x = 10√6

Now, to solve for x, I need to square both sides of the equation.

(√x)^2 = (10√6)^2

Which simplifies to:

x = 100 * (√6)^2

Since (√6)^2 is just 6, this becomes:

x = 100 * 6 = 600

Wait, so x is 600? 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. Then with x=24, y=15, so k = 15 * √24. √24 is 2√6, so k = 15*2√6 = 30√6. 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. Seems correct.

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

So y = 30√6 / √600. Let me compute √600. √600 is √(100*6) = 10√6. Therefore, y = 30√6 / (10√6) = (30/10)*(√6/√6) = 3*1 = 3. Perfect, that matches the given y=3. So x=600 is correct.

Wait, but let me think again. Maybe there's another way to approach this problem? For example, using proportions? Since it's inverse variation, maybe I can set up a proportion.

If y varies inversely as √x, then y1 * √x1 = y2 * √x2. Wait, is that right? Let me recall. For inverse variation, if y = k / x, then y1 * x1 = y2 * x2. But here, since it's y = k / √x, then multiplying both sides by √x gives y * √x = k. Therefore, for two different points, y1 * √x1 = y2 * √x2.

So maybe I can use that formula directly. Let me try that.

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

So according to the formula: y1 * √x1 = y2 * √x2

Plugging in the numbers:

15 * √24 = 3 * √x2

Then, solving for √x2:

√x2 = (15 * √24) / 3 = 5 * √24

Then, square both sides:

x2 = (5 * √24)^2 = 25 * (√24)^2 = 25 * 24 = 600

Same answer! So that confirms it. So x is 600.

Therefore, I think the answer is 600.

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.

Another way to think about it: if x increases, then √x increases, so y decreases, which is inverse. So that makes sense.

So, given that, the steps are correct.

Alternatively, maybe I can think in terms of exponents. Since √x is x^(1/2), so inverse variation would be y = k * x^(-1/2). Then, using the given values to find k, then solving for x when y=3.

Let me try that approach.

y = k * x^(-1/2)

So when x=24, y=15:

15 = k * (24)^(-1/2)

Which is the same as 15 = k / √24, which is the same equation as before. So k = 15√24 = 30√6.

Then, when y=3:

3 = 30√6 * x^(-1/2)

Which is 3 = 30√6 / √x, same as before. So solving gives x=600.

So all approaches lead to the same answer.

Therefore, I think I can be confident that x=600 is the correct answer.

Just to recap:

1. Recognize inverse variation with √x: y = k / √x
2. Use given values to find k: k = y * √x
3. Substitute k into the equation and solve for x when y=3
4. Verified through multiple methods: algebraic manipulation, proportion, exponent rules
5. Checked by plugging back into original equation to ensure consistency

No steps seem to have errors. Therefore, 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} $, which means:

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

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} $:

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

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:

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

Set $ y = 3 $:

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

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 √x, then that should mean y is equal to some constant divided by √x. So the formula would be y = k / √x. Yeah, that makes sense. So instead of dividing by x, we divide by the square root of x.

Alright, so given that when x = 24, y = 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:

15 = k / √24

I need to solve for k. To do that, I can multiply both sides by √24.

k = 15 * √24

Hmm, let me compute that. But maybe I can simplify √24 first. √24 is the same as √(4*6) which is √4 * √6 = 2√6. So √24 = 2√6. Therefore, k = 15 * 2√6 = 30√6.

So the constant of proportionality k is 30√6. Therefore, the equation that relates y and x is y = 30√6 / √x.

Now, the question asks for the value of x when y = 3. So I need to substitute y = 3 into the equation and solve for x. Let me write that down:

3 = 30√6 / √x

Okay, so I need to solve for x here. 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:

√x = (30√6) / 3

Simplify the right side:

√x = 10√6

Now, to solve for x, I need to square both sides of the equation.

(√x)^2 = (10√6)^2

Which simplifies to:

x = 100 * (√6)^2

Since (√6)^2 is just 6, this becomes:

x = 100 * 6 = 600

Wait, so x is 600? 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. Then with x=24, y=15, so k = 15 * √24. √24 is 2√6, so k = 15*2√6 = 30√6. 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. Seems correct.

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

So y = 30√6 / √600. Let me compute √600. √600 is √(100*6) = 10√6. Therefore, y = 30√6 / (10√6) = (30/10)*(√6/√6) = 3*1 = 3. Perfect, that matches the given y=3. So x=600 is correct.

Wait, but let me think again. Maybe there's another way to approach this problem? For example, using proportions? Since it's inverse variation, maybe I can set up a proportion.

If y varies inversely as √x, then y1 * √x1 = y2 * √x2. Wait, is that right? Let me recall. For inverse variation, if y = k / x, then y1 * x1 = y2 * x2. But here, since it's y = k / √x, then multiplying both sides by √x gives y * √x = k. Therefore, for two different points, y1 * √x1 = y2 * √x2.

So maybe I can use that formula directly. Let me try that.

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

So according to the formula: y1 * √x1 = y2 * √x2

Plugging in the numbers:

15 * √24 = 3 * √x2

Then, solving for √x2:

√x2 = (15 * √24) / 3 = 5 * √24

Then, square both sides:

x2 = (5 * √24)^2 = 25 * (√24)^2 = 25 * 24 = 600

Same answer! So that confirms it. So x is 600.

Therefore, I think the answer is 600.

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.

Another way to think about it: if x increases, then √x increases, so y decreases, which is inverse. So that makes sense.

So, given that, the steps are correct.

Alternatively, maybe I can think in terms of exponents. Since √x is x^(1/2), so inverse variation would be y = k * x^(-1/2). Then, using the given values to find k, then solving for x when y=3.

Let me try that approach.

y = k * x^(-1/2)

So when x=24, y=15:

15 = k * (24)^(-1/2)

Which is the same as 15 = k / √24, which is the same equation as before. So k = 15√24 = 30√6.

Then, when y=3:

3 = 30√6 * x^(-1/2)

Which is 3 = 30√6 / √x, same as before. So solving gives x=600.

So all approaches lead to the same answer.

Therefore, I think I can be confident that x=600 is the correct answer.

Just to recap:

1. Recognize inverse variation with √x: y = k / √x
2. Use given values to find k: k = y * √x
3. Substitute k into the equation and solve for x when y=3
4. Verified through multiple methods: algebraic manipulation, proportion, exponent rules
5. Checked by plugging back into original equation to ensure consistency

No steps seem to have errors. Therefore, 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} $, which means:

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

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} $:

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

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:

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

Set $ y = 3 $:

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

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|² + |B|² + |C|² = -2(A·B + B·C + C·A). Therefore, substituting back into the previous equation:

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

Wait, that seems recursive. Let me write:

Let me denote S = |A|² + |B|² + |C|². Then from above:

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

But also, since A + B + C = 0, then (A + B + C)·(A + B + C) = 0 => S + 2(A·B + B·C + C·A) = 0 => A·B + B·C + C·A = -S/2

Therefore, substituting back into the previous equation:

a² + b² + c² = 2S - 2(-S/2) = 2S + S = 3S => S = (a² + b² + c²)/3

Therefore, |A|² + |B|² + |C|² = (a² + b² + c²)/3. Therefore, GA² + GB² + GC² = (a² + b² + c²)/3. 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. So 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 states that the distance between the orthocenter H and centroid G is HG = 2GO, where O is the circumcenter. Therefore, GH = 2GO. Therefore, GH² = 4GO². Therefore, if I can find GO², then I can compute GH².

Alternatively, there is a formula for the distance between H and G. Let me recall. In terms of the sides and circumradius. Alternatively, there is a relation involving the circumradius R. Let me recall that in any triangle, the distance between the orthocenter and centroid can be expressed as:

GH² = 9R² - (a² + b² + c²)

Wait, is that correct? Wait, I need to verify. Alternatively, there is a formula for OH², the distance between orthocenter and circumcenter, which is OH² = 9R² - (a² + b² + c²). Wait, no, that's not. Wait, actually, the formula for OH² is OH² = 9R² - (a² + b² + c²). Wait, let me check.

Wait, actually, Euler's formula states that OH² = 9R² - (a² + b² + c²). Wait, no, Euler's formula is OH² = 9R² - (a² + b² + c²)? Wait, no, Euler's formula is OH² = 9R² - (a² + b² + c²)? Wait, no, actually, Euler's formula is OH² = 9R² - (a² + b² + c²)? Wait, I need to recall correctly.

Wait, Euler's theorem states that OH² = 9R² - (a² + b² + c²). Wait, no, that's not. Wait, actually, the formula is 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. Okay, that works. So that formula seems correct. Therefore, OH² = 9R² - (a² + b² + c²). Therefore, OH² = 9R² - (a² + b² + c²). Therefore, since GH = 2GO, and O lies on Euler line, so HO = HG + GO? Wait, no. Wait, the Euler line has H, G, O colinear with HG = 2GO. Therefore, the distance from H to O is HG + GO? Wait, if G is between H and O, then HO = HG + GO. But since HG = 2GO, then HO = 2GO + GO = 3GO. Therefore, HO = 3GO. Therefore, GO = HO / 3. Therefore, GH = 2GO = 2*(HO / 3) = (2/3)HO. Therefore, GH² = (4/9)HO². Therefore, since HO² = 9R² - (a² + b² + c²), then GH² = (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:

AF² + BF² + CF² = (3/4)GH² + (a² + b² + c²)/3 = (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. Suppose ABC is equilateral with side length a. 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 3AG². In an equilateral triangle, AG is the distance from centroid to vertex, which is (2/3) height. The height is (√3/2)a, so AG = (2/3)(√3/2)a = (√3/3)a. Therefore, AG² = (1/3)a². Therefore, 3AG² = a². On the other hand, R for an equilateral triangle is a / √3, so 3R² = 3*(a² / 3) = a². So it matches. Therefore, in this case, the formula gives 3R² = a², which is correct. Therefore, seems like the answer is 3R². But wait, the problem says "Express AF² + BF² + CF² in terms of the side lengths a, b, c and circumradius R". But according to this, it's just 3R², independent of a, b, c? But in the case of an equilateral triangle, it's correct. Let me check another case.

Take a right-angled triangle. Let me take a 3-4-5 triangle. Let me compute R. For a right-angled triangle, circumradius R is half the hypotenuse. So hypotenuse is 5, so R = 5/2. Let me compute AF² + BF² + CF² where F is the midpoint of GH.

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, hypotenuse AB is 5. Let me compute 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 (1, 4/3).

Orthocenter H of a right-angled triangle is at the vertex of the right angle, which is C (0, 0). Wait, in a right-angled triangle, the orthocenter is at the right-angled vertex. Therefore, H is (0, 0). Therefore, GH is the segment from G (1, 4/3) to H (0, 0). The midpoint F of GH is ((1 + 0)/2, (4/3 + 0)/2) = (0.5, 2/3).

Now, compute AF² + BF² + CF². Let me compute coordinates of A (0,4), B (3,0), C (0,0), F (0.5, 2/3).

Compute AF²: distance from A (0,4) to F (0.5, 2/3):

Δx = 0.5 - 0 = 0.5, Δy = 2/3 - 4 = 2/3 - 12/3 = -10/3. Therefore, AF² = (0.5)^2 + (-10/3)^2 = 0.25 + 100/9 = 0.25 + 11.111... ≈ 11.361...

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

Δx = 0.5 - 3 = -2.5, Δy = 2/3 - 0 = 2/3. Therefore, BF² = (-2.5)^2 + (2/3)^2 = 6.25 + 4/9 ≈ 6.25 + 0.444... ≈ 6.694...

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

Δx = 0.5 - 0 = 0.5, Δy = 2/3 - 0 = 2/3. Therefore, CF² = 0.25 + 4/9 ≈ 0.25 + 0.444... ≈ 0.694...

Adding them up: AF² + BF² + CF² ≈ 11.361 + 6.694 + 0.694 ≈ 18.749...

Now, compute 3R². R is 5/2, so 3*(25/4) = 75/4 = 18.75. Which matches the approximate value. Therefore, the formula gives 3R², which is exactly 18.75 in this case. Therefore, this seems correct. Therefore, the answer is 3R². Therefore, AF² + BF² + CF² = 3R². Therefore, the answer is \boxed{3R^2}.

But wait, the problem says "in terms of the side lengths a, b, c and circumradius R". But according to this, it's only in terms of R. However, in the case of the equilateral triangle, 3R² is equal to a², but in general, is there a way to express 3R² in terms of a, b, c? Wait, but the problem says "in terms of the side lengths a, b, c and circumradius R". So maybe they just want it expressed as 3R², since it's already in terms of R. However, maybe there is another expression? Wait, but in the problem statement, they might expect an expression involving a, b, c and R. However, according to our derivation, it's 3R² regardless of a, b, c. But in the case of the right-angled triangle, 3R² is 3*(5/2)^2 = 75/4, which is equal to AF² + BF² + CF². However, if we wanted to write it in terms of a, b, c, we could use the formula that relates R to a, b, c. For example, R = (a b c)/(4Δ), where Δ is the area. But since the problem allows expressing in terms of R, maybe 3R² is acceptable. However, let me check again the steps to make sure I didn't make a mistake.

Earlier, I had AF² + BF² + CF² = 3R². But let me check again the steps:

We had AF² + BF² + CF² = 3GF² + (a² + b² + c²)/3. Then, GF = (1/2)GH, so GF² = (1/4)GH². Then, GH² = 4R² - (4/9)(a² + b² + c²). Therefore, substituting:

AF² + BF² + CF² = 3*(1/4)*(4R² - (4/9)(a² + b² + c²)) + (a² + b² + c²)/3

= 3*(R² - (1/9)(a² + b² + c²)) + (a² + b² + c²)/3

= 3R² - (1/3)(a² + b² + c²) + (1/3)(a² + b² + c²)

= 3R².

Yes, that's correct. Therefore, the terms involving a² + b² + c² cancel out, leaving 3R². Therefore, regardless of the triangle, AF² + BF² + CF² equals 3R². Therefore, the answer is 3R². Therefore, the problem says "in terms of the side lengths a, b, c and circumradius R", but since the answer is purely in terms of R, maybe that's acceptable. However, maybe there's a different approach where it's expressed in terms of a, b, c and R? But according to the derivation, it's 3R². 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. For the 3-4-5 triangle, 3R² = 3*(25/4) = 75/4, which matched the numerical calculation. Therefore, seems correct.

But let me think again. The problem says "Express AF² + BF² + CF² in terms of the side lengths a, b, c and circumradius R". So maybe they want an expression that includes both a, b, c and R? But according to our derivation, it's 3R². However, since R can be expressed in terms of a, b, c, but the problem allows using R as a variable. Therefore, maybe 3R² is the answer. However, let me check if there is a mistake in the assumption that GH² = 4R² - (4/9)(a² + b² + c²). Let me verify this formula.

Earlier, we had OH² = 9R² - (a² + b² + c²). Then, since GH = 2GO and HO = 3GO, so GH = (2/3)HO. Therefore, GH² = (4/9)HO² = (4/9)(9R² - (a² + b² + c²)) = 4R² - (4/9)(a² + b² + c²). Therefore, that is correct. Therefore, the rest of the steps are correct. Therefore, the answer is indeed 3R². Therefore, the answer is \boxed{3R^2}.

But let me check once more with 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 triangle with sides 2, 2, 3. Wait, check if it's valid. The triangle inequality: 2 + 2 > 3? 4 > 3, yes. Let me compute R for this triangle. First, compute area. Using Heron's formula: s = (2 + 2 + 3)/2 = 7/2. Area Δ = √[s(s - a)(s - b)(s - c)] = √[(7/2)(7/2 - 2)(7/2 - 2)(7/2 - 3)] = √[(7/2)(3/2)(3/2)(1/2)] = √[(7 * 3 * 3 * 1) / 16] = √(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 for this triangle. Let me assign 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 BC = 3. Then, point A is somewhere above the x-axis. Let me compute coordinates of A. Since AB = AC = 2, coordinates of A are (0, h). Then, distance from A to B is √[(1.5)^2 + h²] = 2. Therefore, (2.25) + h² = 4 => h² = 1.75 => h = √(7)/2 ≈ 1.322. Therefore, coordinates of A: (0, √7/2), B: (-1.5, 0), C: (1.5, 0).

Compute centroid G: average of coordinates. G_x = (0 - 1.5 + 1.5)/3 = 0/3 = 0. G_y = (√7/2 + 0 + 0)/3 = √7/6. Therefore, G is (0, √7/6).

Orthocenter H. In an isosceles triangle, the orthocenter lies along the altitude from A to BC. Let me compute coordinates of H. In an isosceles triangle, the orthocenter, centroid, and circumcenter lie along the same line. Let me compute H. Since the triangle is isosceles with AB = AC, the altitude from A is also the median and the angle bisector. Let me find the coordinates of H. Let me recall that in any triangle, the orthocenter can be found by intersecting altitudes. Since we already have one altitude as the y-axis (from A to BC). Let me find another altitude, say from B to AC.

First, find equation of side AC. Points A (0, √7/2) and C (1.5, 0). The slope of AC is (0 - √7/2)/(1.5 - 0) = (-√7/2)/1.5 = (-√7)/3. Therefore, the altitude from B to AC is perpendicular to AC. Therefore, its slope is the negative reciprocal: 3/√7. Therefore, equation of altitude from B (-1.5, 0) is y - 0 = (3/√7)(x + 1.5). Now, find intersection of this altitude with the altitude from A (the y-axis x=0). Substitute x=0 into the equation:

y = (3/√7)(0 + 1.5) = (3/√7)(3/2) = 9/(2√7). Therefore, coordinates of H are (0, 9/(2√7)).

Wait, but centroid G is at (0, √7/6). Let me check if this makes sense. In an isosceles triangle, the centroid divides the median in ratio 2:1. The median from A to BC is from (0, √7/2) to (0, 0). The centroid is located 1/3 from the base. Therefore, the y-coordinate of centroid is (2/3)*0 + (1/3)*(√7/2) = √7/6, which matches. The orthocenter is at (0, 9/(2√7)). Let me compute 9/(2√7) ≈ 9/(2*2.6458) ≈ 9/5.2916 ≈ 1.70. The y-coordinate of A is √7/2 ≈ 1.322. Wait, but the orthocenter is above A? That can't be. Wait, in an acute triangle, the orthocenter lies inside the triangle. Wait, but in this triangle with sides 2, 2, 3, is it acute?

Let me check. The triangle with sides 2, 2, 3. The largest side is 3. Check if 3² < 2² + 2²? 9 < 4 + 4? 9 < 8? No, 9 > 8. Therefore, it's an obtuse triangle. Therefore, the orthocenter lies outside the triangle. Therefore, the coordinates of H being (0, 9/(2√7)) ≈ (0, 1.70) while the y-coordinate of A is √7/2 ≈ 1.322. Therefore, H is above A, outside the triangle. That makes sense for an obtuse triangle. Therefore, coordinates of H are (0, 9/(2√7)).

Therefore, midpoint F of GH. Coordinates of G: (0, √7/6), coordinates of H: (0, 9/(2√7)). Therefore, midpoint F has coordinates (0, (√7/6 + 9/(2√7))/2). Let me compute that. First, let me rationalize denominators:

√7/6 remains as is. 9/(2√7) = 9√7/(2*7) = 9√7/14.

Therefore, adding √7/6 + 9√7/14. Let me find common denominator, which is 42.

√7/6 = 7√7/42, 9√7/14 = 27√7/42. Therefore, total is 34√7/42 = 17√7/21. Then, divide by 2: 17√7/42. Therefore, coordinates of F are (0, 17√7/42).

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

First, coordinates:

A: (0, √7/2)

F: (0, 17√7/42)

Therefore, AF² is the square of the distance between A and F. Since they share the same x-coordinate, it's just the difference in y-coordinates squared.

Δy = 17√7/42 - √7/2 = 17√7/42 - 21√7/42 = (-4√7)/42 = (-2√7)/21. Therefore, AF² = [(-2√7)/21]^2 = (4*7)/441 = 28/441 = 4/63.

Next, compute BF². Coordinates of B: (-1.5, 0) = (-3/2, 0). Coordinates of F: (0, 17√7/42). Therefore, Δx = 0 - (-3/2) = 3/2, Δy = 17√7/42 - 0 = 17√7/42. Therefore, BF² = (3/2)^2 + (17√7/42)^2.

Compute (3/2)^2 = 9/4.

(17√7/42)^2 = (289 * 7) / (42^2) = 2023 / 1764.

Therefore, BF² = 9/4 + 2023/1764. Convert 9/4 to 3969/1764. Therefore, total BF² = (3969 + 2023)/1764 = 6, 3969 + 2023: 3969 + 2000 = 5969, +23 = 5992. Therefore, BF² = 5992 / 1764. Simplify: divide numerator and denominator by 4: 1498 / 441. Let me check if this reduces further. 1498 ÷ 2 = 749, 441 ÷ 2 = 220.5, no. So BF² = 1498/441.

Similarly, compute CF². Coordinates of C: (1.5, 0) = (3/2, 0). Coordinates of F: (0, 17√7/42). Therefore, Δx = 0 - 3/2 = -3/2, Δy = same as BF. Therefore, CF² is same as BF², which is 1498/441.

Therefore, AF² + BF² + CF² = 4/63 + 1498/441 + 1498/441.

Convert 4/63 to denominator 441: 4/63 = 28/441. Therefore, total is 28/441 + 1498/441 + 1498/441 = (28 + 1498 + 1498)/441 = (28 + 2996)/441 = 3024 / 441. Simplify: divide numerator and denominator by 21: 3024 ÷ 21 = 144, 441 ÷ 21 = 21. Therefore, 144/21 = 48/7 ≈ 6.857, which is equal to 3R². Since earlier we had R = 4/√7, so 3R² = 3*(16/7) = 48/7. Therefore, matches. Therefore, this confirms that AF² + BF² + CF² = 3R². Therefore, the answer is indeed 3R².

Therefore, despite the problem mentioning side lengths a, b, c, the answer simplifies to 3R², which is independent of the side lengths. 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 sum $ 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

A key identity in triangle geometry is:

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

This identity holds for any point $ P $ in the plane of triangle $ ABC $, and $ G $ is the centroid. If we apply this identity with $ P = F $, we get:

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

---

### Step 2: Express $ GF^2 $ in Terms of $ GH $

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:

$$
AF^2 + BF^2 + CF^2 = \frac{3}{4}GH^2 + GA^2 + GB^2 + GC^2
$$

---

### Step 3: 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:

$$
GA^2 + GB^2 + GC^2 = \frac{1}{3}(a^2 + b^2 + c^2)
$$

---

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

To find $ GH^2 $, we use the fact that the centroid $ G $, orthocenter $ H $, and circumcenter $ O $ lie on the Euler line, and the distance between $ G $ and $ H $ is related to the distance between $ O $ and $ H $ by:

$$
GH = \frac{2}{3}OH
$$

Thus:

$$
GH^2 = \frac{4}{9}OH^2
$$

From Euler's formula, we know:

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

So:

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

---

### Step 5: Substitute into the Expression

Now substitute $ GH^2 $ and $ GA^2 + GB^2 + GC^2 $ into the expression:

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

$$
= \frac{3}{4} \left(4R^2 - \frac{4}{9}(a^2 + b^2 + c^2)\right) + \frac{1}{3}(a^2 + b^2 + c^2)
$$

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

$$
= 3R^2
$$

---

### Final Result

The sum $ AF^2 + BF^2 + CF^2 $ simplifies to:

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

This result is independent of the side lengths $ a, b, c $, and only depends on the circumradius $ R $ of triangle $ ABC $.

[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|² + |B|² + |C|² = -2(A·B + B·C + C·A). Therefore, substituting back into the previous equation:

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

Wait, that seems recursive. Let me write:

Let me denote S = |A|² + |B|² + |C|². Then from above:

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

But also, since A + B + C = 0, then (A + B + C)·(A + B + C) = 0 => S + 2(A·B + B·C + C·A) = 0 => A·B + B·C + C·A = -S/2

Therefore, substituting back into the previous equation:

a² + b² + c² = 2S - 2(-S/2) = 2S + S = 3S => S = (a² + b² + c²)/3

Therefore, |A|² + |B|² + |C|² = (a² + b² + c²)/3. Therefore, GA² + GB² + GC² = (a² + b² + c²)/3. 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. So 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 states that the distance between the orthocenter H and centroid G is HG = 2GO, where O is the circumcenter. Therefore, GH = 2GO. Therefore, GH² = 4GO². Therefore, if I can find GO², then I can compute GH².

Alternatively, there is a formula for the distance between H and G. Let me recall. In terms of the sides and circumradius. Alternatively, there is a relation involving the circumradius R. Let me recall that in any triangle, the distance between the orthocenter and centroid can be expressed as:

GH² = 9R² - (a² + b² + c²)

Wait, is that correct? Wait, I need to verify. Alternatively, there is a formula for OH², the distance between orthocenter and circumcenter, which is OH² = 9R² - (a² + b² + c²). Wait, no, that's not. Wait, actually, the formula for OH² is OH² = 9R² - (a² + b² + c²). Wait, let me check.

Wait, actually, Euler's formula states that OH² = 9R² - (a² + b² + c²). Wait, no, Euler's formula is OH² = 9R² - (a² + b² + c²)? Wait, no, Euler's formula is OH² = 9R² - (a² + b² + c²)? Wait, no, actually, Euler's formula is OH² = 9R² - (a² + b² + c²)? Wait, I need to recall correctly.

Wait, Euler's theorem states that OH² = 9R² - (a² + b² + c²). Wait, no, that's not. Wait, actually, the formula is 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. Okay, that works. So that formula seems correct. Therefore, OH² = 9R² - (a² + b² + c²). Therefore, OH² = 9R² - (a² + b² + c²). Therefore, since GH = 2GO, and O lies on Euler line, so HO = HG + GO? Wait, no. Wait, the Euler line has H, G, O colinear with HG = 2GO. Therefore, the distance from H to O is HG + GO? Wait, if G is between H and O, then HO = HG + GO. But since HG = 2GO, then HO = 2GO + GO = 3GO. Therefore, HO = 3GO. Therefore, GO = HO / 3. Therefore, GH = 2GO = 2*(HO / 3) = (2/3)HO. Therefore, GH² = (4/9)HO². Therefore, since HO² = 9R² - (a² + b² + c²), then GH² = (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:

AF² + BF² + CF² = (3/4)GH² + (a² + b² + c²)/3 = (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. Suppose ABC is equilateral with side length a. 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 3AG². In an equilateral triangle, AG is the distance from centroid to vertex, which is (2/3) height. The height is (√3/2)a, so AG = (2/3)(√3/2)a = (√3/3)a. Therefore, AG² = (1/3)a². Therefore, 3AG² = a². On the other hand, R for an equilateral triangle is a / √3, so 3R² = 3*(a² / 3) = a². So it matches. Therefore, in this case, the formula gives 3R² = a², which is correct. Therefore, seems like the answer is 3R². But wait, the problem says "Express AF² + BF² + CF² in terms of the side lengths a, b, c and circumradius R". But according to this, it's just 3R², independent of a, b, c? But in the case of an equilateral triangle, it's correct. Let me check another case.

Take a right-angled triangle. Let me take a 3-4-5 triangle. Let me compute R. For a right-angled triangle, circumradius R is half the hypotenuse. So hypotenuse is 5, so R = 5/2. Let me compute AF² + BF² + CF² where F is the midpoint of GH.

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, hypotenuse AB is 5. Let me compute 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 (1, 4/3).

Orthocenter H of a right-angled triangle is at the vertex of the right angle, which is C (0, 0). Wait, in a right-angled triangle, the orthocenter is at the right-angled vertex. Therefore, H is (0, 0). Therefore, GH is the segment from G (1, 4/3) to H (0, 0). The midpoint F of GH is ((1 + 0)/2, (4/3 + 0)/2) = (0.5, 2/3).

Now, compute AF² + BF² + CF². Let me compute coordinates of A (0,4), B (3,0), C (0,0), F (0.5, 2/3).

Compute AF²: distance from A (0,4) to F (0.5, 2/3):

Δx = 0.5 - 0 = 0.5, Δy = 2/3 - 4 = 2/3 - 12/3 = -10/3. Therefore, AF² = (0.5)^2 + (-10/3)^2 = 0.25 + 100/9 = 0.25 + 11.111... ≈ 11.361...

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

Δx = 0.5 - 3 = -2.5, Δy = 2/3 - 0 = 2/3. Therefore, BF² = (-2.5)^2 + (2/3)^2 = 6.25 + 4/9 ≈ 6.25 + 0.444... ≈ 6.694...

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

Δx = 0.5 - 0 = 0.5, Δy = 2/3 - 0 = 2/3. Therefore, CF² = 0.25 + 4/9 ≈ 0.25 + 0.444... ≈ 0.694...

Adding them up: AF² + BF² + CF² ≈ 11.361 + 6.694 + 0.694 ≈ 18.749...

Now, compute 3R². R is 5/2, so 3*(25/4) = 75/4 = 18.75. Which matches the approximate value. Therefore, the formula gives 3R², which is exactly 18.75 in this case. Therefore, this seems correct. Therefore, the answer is 3R². Therefore, AF² + BF² + CF² = 3R². Therefore, the answer is \boxed{3R^2}.

But wait, the problem says "in terms of the side lengths a, b, c and circumradius R". But according to this, it's only in terms of R. However, in the case of the equilateral triangle, 3R² is equal to a², but in general, is there a way to express 3R² in terms of a, b, c? Wait, but the problem says "in terms of the side lengths a, b, c and circumradius R". So maybe they just want it expressed as 3R², since it's already in terms of R. However, maybe there is another expression? Wait, but in the problem statement, they might expect an expression involving a, b, c and R. However, according to our derivation, it's 3R² regardless of a, b, c. But in the case of the right-angled triangle, 3R² is 3*(5/2)^2 = 75/4, which is equal to AF² + BF² + CF². However, if we wanted to write it in terms of a, b, c, we could use the formula that relates R to a, b, c. For example, R = (a b c)/(4Δ), where Δ is the area. But since the problem allows expressing in terms of R, maybe 3R² is acceptable. However, let me check again the steps to make sure I didn't make a mistake.

Earlier, I had AF² + BF² + CF² = 3R². But let me check again the steps:

We had AF² + BF² + CF² = 3GF² + (a² + b² + c²)/3. Then, GF = (1/2)GH, so GF² = (1/4)GH². Then, GH² = 4R² - (4/9)(a² + b² + c²). Therefore, substituting:

AF² + BF² + CF² = 3*(1/4)*(4R² - (4/9)(a² + b² + c²)) + (a² + b² + c²)/3

= 3*(R² - (1/9)(a² + b² + c²)) + (a² + b² + c²)/3

= 3R² - (1/3)(a² + b² + c²) + (1/3)(a² + b² + c²)

= 3R².

Yes, that's correct. Therefore, the terms involving a² + b² + c² cancel out, leaving 3R². Therefore, regardless of the triangle, AF² + BF² + CF² equals 3R². Therefore, the answer is 3R². Therefore, the problem says "in terms of the side lengths a, b, c and circumradius R", but since the answer is purely in terms of R, maybe that's acceptable. However, maybe there's a different approach where it's expressed in terms of a, b, c and R? But according to the derivation, it's 3R². 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. For the 3-4-5 triangle, 3R² = 3*(25/4) = 75/4, which matched the numerical calculation. Therefore, seems correct.

But let me think again. The problem says "Express AF² + BF² + CF² in terms of the side lengths a, b, c and circumradius R". So maybe they want an expression that includes both a, b, c and R? But according to our derivation, it's 3R². However, since R can be expressed in terms of a, b, c, but the problem allows using R as a variable. Therefore, maybe 3R² is the answer. However, let me check if there is a mistake in the assumption that GH² = 4R² - (4/9)(a² + b² + c²). Let me verify this formula.

Earlier, we had OH² = 9R² - (a² + b² + c²). Then, since GH = 2GO and HO = 3GO, so GH = (2/3)HO. Therefore, GH² = (4/9)HO² = (4/9)(9R² - (a² + b² + c²)) = 4R² - (4/9)(a² + b² + c²). Therefore, that is correct. Therefore, the rest of the steps are correct. Therefore, the answer is indeed 3R². Therefore, the answer is \boxed{3R^2}.

But let me check once more with 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 triangle with sides 2, 2, 3. Wait, check if it's valid. The triangle inequality: 2 + 2 > 3? 4 > 3, yes. Let me compute R for this triangle. First, compute area. Using Heron's formula: s = (2 + 2 + 3)/2 = 7/2. Area Δ = √[s(s - a)(s - b)(s - c)] = √[(7/2)(7/2 - 2)(7/2 - 2)(7/2 - 3)] = √[(7/2)(3/2)(3/2)(1/2)] = √[(7 * 3 * 3 * 1) / 16] = √(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 for this triangle. Let me assign 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 BC = 3. Then, point A is somewhere above the x-axis. Let me compute coordinates of A. Since AB = AC = 2, coordinates of A are (0, h). Then, distance from A to B is √[(1.5)^2 + h²] = 2. Therefore, (2.25) + h² = 4 => h² = 1.75 => h = √(7)/2 ≈ 1.322. Therefore, coordinates of A: (0, √7/2), B: (-1.5, 0), C: (1.5, 0).

Compute centroid G: average of coordinates. G_x = (0 - 1.5 + 1.5)/3 = 0/3 = 0. G_y = (√7/2 + 0 + 0)/3 = √7/6. Therefore, G is (0, √7/6).

Orthocenter H. In an isosceles triangle, the orthocenter lies along the altitude from A to BC. Let me compute coordinates of H. In an isosceles triangle, the orthocenter, centroid, and circumcenter lie along the same line. Let me compute H. Since the triangle is isosceles with AB = AC, the altitude from A is also the median and the angle bisector. Let me find the coordinates of H. Let me recall that in any triangle, the orthocenter can be found by intersecting altitudes. Since we already have one altitude as the y-axis (from A to BC). Let me find another altitude, say from B to AC.

First, find equation of side AC. Points A (0, √7/2) and C (1.5, 0). The slope of AC is (0 - √7/2)/(1.5 - 0) = (-√7/2)/1.5 = (-√7)/3. Therefore, the altitude from B to AC is perpendicular to AC. Therefore, its slope is the negative reciprocal: 3/√7. Therefore, equation of altitude from B (-1.5, 0) is y - 0 = (3/√7)(x + 1.5). Now, find intersection of this altitude with the altitude from A (the y-axis x=0). Substitute x=0 into the equation:

y = (3/√7)(0 + 1.5) = (3/√7)(3/2) = 9/(2√7). Therefore, coordinates of H are (0, 9/(2√7)).

Wait, but centroid G is at (0, √7/6). Let me check if this makes sense. In an isosceles triangle, the centroid divides the median in ratio 2:1. The median from A to BC is from (0, √7/2) to (0, 0). The centroid is located 1/3 from the base. Therefore, the y-coordinate of centroid is (2/3)*0 + (1/3)*(√7/2) = √7/6, which matches. The orthocenter is at (0, 9/(2√7)). Let me compute 9/(2√7) ≈ 9/(2*2.6458) ≈ 9/5.2916 ≈ 1.70. The y-coordinate of A is √7/2 ≈ 1.322. Wait, but the orthocenter is above A? That can't be. Wait, in an acute triangle, the orthocenter lies inside the triangle. Wait, but in this triangle with sides 2, 2, 3, is it acute?

Let me check. The triangle with sides 2, 2, 3. The largest side is 3. Check if 3² < 2² + 2²? 9 < 4 + 4? 9 < 8? No, 9 > 8. Therefore, it's an obtuse triangle. Therefore, the orthocenter lies outside the triangle. Therefore, the coordinates of H being (0, 9/(2√7)) ≈ (0, 1.70) while the y-coordinate of A is √7/2 ≈ 1.322. Therefore, H is above A, outside the triangle. That makes sense for an obtuse triangle. Therefore, coordinates of H are (0, 9/(2√7)).

Therefore, midpoint F of GH. Coordinates of G: (0, √7/6), coordinates of H: (0, 9/(2√7)). Therefore, midpoint F has coordinates (0, (√7/6 + 9/(2√7))/2). Let me compute that. First, let me rationalize denominators:

√7/6 remains as is. 9/(2√7) = 9√7/(2*7) = 9√7/14.

Therefore, adding √7/6 + 9√7/14. Let me find common denominator, which is 42.

√7/6 = 7√7/42, 9√7/14 = 27√7/42. Therefore, total is 34√7/42 = 17√7/21. Then, divide by 2: 17√7/42. Therefore, coordinates of F are (0, 17√7/42).

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

First, coordinates:

A: (0, √7/2)

F: (0, 17√7/42)

Therefore, AF² is the square of the distance between A and F. Since they share the same x-coordinate, it's just the difference in y-coordinates squared.

Δy = 17√7/42 - √7/2 = 17√7/42 - 21√7/42 = (-4√7)/42 = (-2√7)/21. Therefore, AF² = [(-2√7)/21]^2 = (4*7)/441 = 28/441 = 4/63.

Next, compute BF². Coordinates of B: (-1.5, 0) = (-3/2, 0). Coordinates of F: (0, 17√7/42). Therefore, Δx = 0 - (-3/2) = 3/2, Δy = 17√7/42 - 0 = 17√7/42. Therefore, BF² = (3/2)^2 + (17√7/42)^2.

Compute (3/2)^2 = 9/4.

(17√7/42)^2 = (289 * 7) / (42^2) = 2023 / 1764.

Therefore, BF² = 9/4 + 2023/1764. Convert 9/4 to 3969/1764. Therefore, total BF² = (3969 + 2023)/1764 = 6, 3969 + 2023: 3969 + 2000 = 5969, +23 = 5992. Therefore, BF² = 5992 / 1764. Simplify: divide numerator and denominator by 4: 1498 / 441. Let me check if this reduces further. 1498 ÷ 2 = 749, 441 ÷ 2 = 220.5, no. So BF² = 1498/441.

Similarly, compute CF². Coordinates of C: (1.5, 0) = (3/2, 0). Coordinates of F: (0, 17√7/42). Therefore, Δx = 0 - 3/2 = -3/2, Δy = same as BF. Therefore, CF² is same as BF², which is 1498/441.

Therefore, AF² + BF² + CF² = 4/63 + 1498/441 + 1498/441.

Convert 4/63 to denominator 441: 4/63 = 28/441. Therefore, total is 28/441 + 1498/441 + 1498/441 = (28 + 1498 + 1498)/441 = (28 + 2996)/441 = 3024 / 441. Simplify: divide numerator and denominator by 21: 3024 ÷ 21 = 144, 441 ÷ 21 = 21. Therefore, 144/21 = 48/7 ≈ 6.857, which is equal to 3R². Since earlier we had R = 4/√7, so 3R² = 3*(16/7) = 48/7. Therefore, matches. Therefore, this confirms that AF² + BF² + CF² = 3R². Therefore, the answer is indeed 3R².

Therefore, despite the problem mentioning side lengths a, b, c, the answer simplifies to 3R², which is independent of the side lengths. 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 sum $ 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

A key identity in triangle geometry is:

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

This identity holds for any point $ P $ in the plane of triangle $ ABC $, and $ G $ is the centroid. If we apply this identity with $ P = F $, we get:

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

---

### Step 2: Express $ GF^2 $ in Terms of $ GH $

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:

$$
AF^2 + BF^2 + CF^2 = \frac{3}{4}GH^2 + GA^2 + GB^2 + GC^2
$$

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### Step 3: 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:

$$
GA^2 + GB^2 + GC^2 = \frac{1}{3}(a^2 + b^2 + c^2)
$$

---

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

To find $ GH^2 $, we use the fact that the centroid $ G $, orthocenter $ H $, and circumcenter $ O $ lie on the Euler line, and the distance between $ G $ and $ H $ is related to the distance between $ O $ and $ H $ by:

$$
GH = \frac{2}{3}OH
$$

Thus:

$$
GH^2 = \frac{4}{9}OH^2
$$

From Euler's formula, we know:

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

So:

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

---

### Step 5: Substitute into the Expression

Now substitute $ GH^2 $ and $ GA^2 + GB^2 + GC^2 $ into the expression:

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

$$
= \frac{3}{4} \left(4R^2 - \frac{4}{9}(a^2 + b^2 + c^2)\right) + \frac{1}{3}(a^2 + b^2 + c^2)
$$

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

$$
= 3R^2
$$

---

### Final Result

The sum $ AF^2 + BF^2 + CF^2 $ simplifies to:

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

This result is independent of the side lengths $ a, b, c $, and only depends on the circumradius $ R $ of triangle $ ABC $.

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, $AF^2 + BF^2 + CF^2 = \boxed{3R^2}.$