Explanation:
Okay, so I need to solve this recurrence relation problem. Let me read it again to make sure I understand.

We have a function f defined for integers n ≥ 1, with the recurrence relation:

f(n + 1) = (-1)^{n + 1} * n - 2f(n)

And it's given that f(1) = f(1986). The task is to compute the sum f(1) + f(2) + ... + f(1985).

Hmm, okay. So we have a recurrence that relates f(n+1) to f(n), and the function is periodic in some sense because f(1) = f(1986). The period is 1985, but maybe there's a shorter period? Not sure yet. The problem is asking for the sum of f(1) through f(1985). So maybe we can find a pattern or express the sum in terms of itself?

First, let me write down the recurrence again to make it clearer:

For each n ≥ 1, f(n + 1) = (-1)^{n + 1} * n - 2f(n)

And f(1) = f(1986). So the sequence is periodic with period 1985? Wait, not necessarily. It's only that f(1) equals f(1986), which would make the sequence repeat every 1985 terms, but since the recurrence is defined for n ≥ 1, if f(1) = f(1986), then perhaps we can consider the sequence as a cycle?

But maybe I should try to find a closed-form expression for f(n), or find a relation between the sum S = f(1) + f(2) + ... + f(1985). Since the sum is up to 1985, and f(1986) is equal to f(1). Maybe the sum can be related to itself through the recurrence?

Let me think. Let me denote S = f(1) + f(2) + ... + f(1985). Also, since f(1986) = f(1), we can consider that the sequence is cyclic with period 1985. But maybe I can use the recurrence relation to express terms in terms of previous ones and then sum them up?

Alternatively, perhaps I can express the sum S in terms of itself. Let me try that.

Let me first write down the recurrence for each n from 1 to 1985. For each n from 1 to 1985, f(n + 1) = (-1)^{n + 1} * n - 2f(n). But when n = 1985, we get f(1986) = (-1)^{1986} * 1985 - 2f(1985). But since f(1986) = f(1), this gives us an equation involving f(1) and f(1985). Maybe we can find another equation?

Alternatively, let me try to compute S = sum_{n=1}^{1985} f(n). Let me try to relate S with the recurrence.

Let me consider sum_{n=1}^{1985} f(n + 1) = sum_{n=1}^{1985} [ (-1)^{n + 1} * n - 2f(n) ].

But the left side is sum_{n=1}^{1985} f(n + 1) = sum_{k=2}^{1986} f(k) = [sum_{k=1}^{1986} f(k)] - f(1) = [S + f(1986)] - f(1) = [S + f(1)] - f(1) = S. Because f(1986) = f(1). Therefore, the left side is equal to S.

On the right side, we have sum_{n=1}^{1985} [ (-1)^{n + 1} * n - 2f(n) ] = sum_{n=1}^{1985} (-1)^{n + 1} * n - 2 sum_{n=1}^{1985} f(n) = [sum_{n=1}^{1985} (-1)^{n + 1} * n] - 2S.

Therefore, equating left and right sides:

S = [sum_{n=1}^{1985} (-1)^{n + 1} * n] - 2S.

Therefore, bringing the 2S to the left:

S + 2S = sum_{n=1}^{1985} (-1)^{n + 1} * n.

So 3S = sum_{n=1}^{1985} (-1)^{n + 1} * n.

Therefore, if I can compute that sum, then I can find S.

Alright, so the problem reduces to calculating this alternating sum. Let's denote this sum as T = sum_{n=1}^{1985} (-1)^{n + 1} * n.

Let me compute T. Let me note that (-1)^{n + 1} is equal to (-1)^{n - 1}. So T is an alternating sum of integers from 1 to 1985. Let me recall that for such sums, there's a formula.

Let me consider the sum S = a_1 - a_2 + a_3 - a_4 + ... + (-1)^{n+1} a_n, where a_i = i.

In our case, n is 1985. Let's compute T.

First, note that if the number of terms is odd, the last term is positive. Since 1985 is odd, the last term is +1985.

Let me think of grouping terms in pairs. Let me consider pairing the first two terms, the next two, etc.

For example, if we have an odd number of terms, say 2k + 1 terms, then we can pair the first 2k terms and have one term left.

So for T, with 1985 terms:

Number of pairs = floor(1985 / 2) = 992 pairs, and one remaining term.

Each pair is (1 - 2) + (3 - 4) + ... + (1983 - 1984) + 1985.

Each pair is (-1) + (-1) + ... + (-1). Wait, 1 - 2 = -1, 3 - 4 = -1, etc.

How many such pairs are there?

Since 1985 = 2*992 + 1. So there are 992 pairs and one term.

Therefore, the sum T is equal to (-1) * 992 + 1985.

Calculating that: -992 + 1985 = 1985 - 992 = 993.

Wait, let me check:

1985 - 992: 1985 - 900 = 1085, then subtract 92 more: 1085 - 92 = 993. Yes.

So T = 993. Therefore, 3S = 993, so S = 993 / 3 = 331. So the answer is 331?

Wait, that seems straightforward, but let me verify my steps again to be sure.

First, I used the recurrence relation to express sum_{n=1}^{1985} f(n + 1) which is equal to S. Then equated that to the other side which involved the alternating sum T minus 2S, leading to 3S = T. Then computed T as 993, so S = 331.

But let me check the computation of T again. Maybe I made a mistake there.

So T = sum_{n=1}^{m} (-1)^{n + 1} * n, where m = 1985.

If m is odd, then T = (number of pairs) * (-1) + last term.

Number of pairs is floor(m / 2) = floor(1985 / 2) = 992. Then the last term is m = 1985. Therefore, T = 992*(-1) + 1985 = 1985 - 992 = 993. That seems correct.

Alternatively, I can think of T as follows. Let me recall that for the sum of (-1)^{n+1} * n from n=1 to N, the formula is:

If N is even: (-1)^{N+1} * (N / 2)

Wait, maybe not. Let me recall:

Let me consider the sum S = 1 - 2 + 3 - 4 + ... + (-1)^{N+1} N.

There's a formula for this. Let me recall:

Let me write S = sum_{k=1}^N (-1)^{k+1} k.

If N is even, say N = 2m:

S = (1 - 2) + (3 - 4) + ... + (2m - 1 - 2m) = m*(-1) = -m.

If N is odd, N = 2m + 1:

S = [sum_{k=1}^{2m} (-1)^{k+1} k] + (-1)^{2m + 2}*(2m + 1) = -m + (2m + 1) = m + 1.

So, in general,

If N is even: S = -N/2

If N is odd: S = (N + 1)/2

So for N = 1985, which is odd, S = (1985 + 1)/2 = 1986 / 2 = 993. Which matches my previous calculation. Therefore, T = 993. Therefore, 3S = 993 => S = 331. Therefore, the answer is 331. So is that the answer? Let me check once more.

Wait, but let me check for small N. Let me take N=1: sum is 1. According to formula, (1 + 1)/2 = 1. Correct. N=3: 1 - 2 + 3 = 2. (3 + 1)/2 = 2. Correct. N=5: 1 - 2 + 3 - 4 + 5 = 3. (5 + 1)/2 = 3. Correct. So formula holds. Thus, T = 993. Therefore, S = 993 / 3 = 331. So answer is 331?

But let me check again the step where I equated the sum_{n=1}^{1985} f(n + 1) to S. Let me re-examine that.

Original sum: sum_{n=1}^{1985} f(n + 1) = sum_{k=2}^{1986} f(k) = [sum_{k=1}^{1986} f(k)] - f(1). But sum_{k=1}^{1986} f(k) = sum_{k=1}^{1985} f(k) + f(1986) = S + f(1). Therefore, sum_{k=2}^{1986} f(k) = S + f(1) - f(1) = S. Therefore, left side is S. So yes, that step is correct.

Then the right side is T - 2S, so S = T - 2S => 3S = T. Therefore, yes, 3S = 993 => S = 331.

Therefore, the answer is 331. But wait, let me check with a smaller case. Let me take a small period where I can compute manually.

Suppose instead of 1985, let's take N=1. Wait, but N=1 would require f(1 + 1) = f(2) = (-1)^{2} *1 - 2f(1). But f(2) = f(1) because f(1) = f(1 + N), but if N=1, then f(1) = f(2). So let's see:

For N=1, the sum would be f(1). Let's check.

Given f(2) = (-1)^{2}*1 - 2f(1) => f(2) = 1 - 2f(1). But f(2) = f(1). Therefore:

f(1) = 1 - 2f(1) => 3f(1) = 1 => f(1) = 1/3. Then the sum is 1/3. Let's check with formula:

If N=1 (which is odd), then T = (1 + 1)/2 = 1. Then 3S = 1 => S = 1/3. Which matches. So the formula works here.

Another test case: N=3. Let me try N=3. So f(1) = f(4). Compute sum f(1) + f(2) + f(3).

First, let me write the recurrence:

For n=1: f(2) = (-1)^{2} * 1 - 2f(1) = 1 - 2f(1)

For n=2: f(3) = (-1)^{3} * 2 - 2f(2) = -2 - 2f(2)

For n=3: f(4) = (-1)^{4} * 3 - 2f(3) = 3 - 2f(3)

But since f(4) = f(1), so:

f(1) = 3 - 2f(3)

Now, let's express everything in terms of f(1).

From n=1: f(2) = 1 - 2f(1)

From n=2: f(3) = -2 - 2f(2) = -2 - 2*(1 - 2f(1)) = -2 - 2 + 4f(1) = -4 + 4f(1)

From n=3: f(1) = 3 - 2f(3) = 3 - 2*(-4 + 4f(1)) = 3 + 8 - 8f(1) = 11 - 8f(1)

Thus, f(1) + 8f(1) = 11 => 9f(1) = 11 => f(1) = 11/9

Then sum S = f(1) + f(2) + f(3) = f(1) + [1 - 2f(1)] + [-4 + 4f(1)] = f(1) + 1 - 2f(1) - 4 + 4f(1) = (1 - 4) + (f(1) - 2f(1) + 4f(1)) = -3 + 3f(1) = -3 + 3*(11/9) = -3 + 11/3 = (-9/3 + 11/3) = 2/3.

Now, according to our formula, T = (N + 1)/2 where N = 3. (3 + 1)/2 = 2. Then 3S = 2 => S = 2/3. Which matches! Perfect. So our formula works here.

Another check. Let me take N=2. Wait, but N must be such that f(1) = f(N + 1). Wait, if N=2, then f(1) = f(3). Let me see:

For N=2, compute sum f(1) + f(2). Let me compute using recurrence.

But perhaps the formula would have T = sum_{n=1}^2 (-1)^{n + 1} n = 1 - 2 = -1. Then according to formula 3S = T => S = -1/3. Let me check manually.

For N=2, the recurrence:

For n=1: f(2) = (-1)^{2} * 1 - 2f(1) = 1 - 2f(1)

For n=2: f(3) = (-1)^{3} * 2 - 2f(2) = -2 - 2f(2). But since f(3) = f(1), we have:

f(1) = -2 - 2f(2). Substitute f(2) from above: f(1) = -2 - 2*(1 - 2f(1)) = -2 - 2 + 4f(1) = -4 + 4f(1). Then f(1) - 4f(1) = -4 => -3f(1) = -4 => f(1) = 4/3. Then f(2) = 1 - 2*(4/3) = 1 - 8/3 = -5/3. Then sum S = f(1) + f(2) = 4/3 -5/3 = -1/3, which matches 3S = -1. So yes, the formula works here as well.

Therefore, our formula seems solid. Therefore, for N=1985, the answer is 331.

Wait, but let me check one more thing. Let me check the general case. Suppose the recurrence is f(n + 1) + 2f(n) = (-1)^{n + 1} n. Then, if we consider summing both sides from n=1 to N, where f(N + 1) = f(1). Then:

sum_{n=1}^N [f(n + 1) + 2f(n)] = sum_{n=1}^N (-1)^{n + 1} n.

Left side: sum_{n=1}^N f(n + 1) + 2 sum_{n=1}^N f(n) = [sum_{k=2}^{N + 1} f(k)] + 2S = [S + f(N + 1) - f(1)] + 2S = [S + f(1) - f(1)] + 2S = S + 2S = 3S.

Therefore, 3S = sum_{n=1}^N (-1)^{n + 1} n. Which is exactly what we had. Therefore, regardless of N, if we have f(N + 1) = f(1), then 3S = T where T is the alternating sum. Therefore, for N=1985, T = 993, so S=331. Therefore, answer is 331.

But let me check with another N. Let's take N=3 again. T = 1 - 2 + 3 = 2. Then 3S = 2 => S = 2/3, which was correct. So formula works.

Therefore, confident that the answer is 331. Therefore, the answer is \boxed{331}.

**Final Answer**
\boxed{331}

We are given a recurrence relation for the function $ f(n) $, defined for $ n \geq 1 $, by:

$$
f(n + 1) = (-1)^{n + 1} n - 2f(n)
$$

with a periodic condition:

$$
f(1) = f(1986)
$$

We are to compute the sum:

$$
f(1) + f(2) + \dots + f(1985)
$$

---

### Step 1: Define the sum $ S $

Let:
$$
S = \sum_{n=1}^{1985} f(n)
$$

From the recurrence, we can rewrite the relation for each $ n $ from 1 to 1985 as:
$$
f(n + 1) = (-1)^{n + 1} n - 2f(n)
$$

Now, summing both sides from $ n = 1 $ to $ 1985 $:

$$
\sum_{n=1}^{1985} f(n + 1) = \sum_{n=1}^{1985} [(-1)^{n+1} n - 2f(n)]
$$

The left-hand side becomes:
$$
\sum_{n=1}^{1985} f(n + 1) = \sum_{k=2}^{1986} f(k) = \left( \sum_{k=1}^{1986} f(k) \right) - f(1)
$$

Because $ f(1986) = f(1) $, we have:
$$
\sum_{k=1}^{1986} f(k) = S + f(1)
$$

Therefore:
$$
\sum_{k=2}^{1986} f(k) = S + f(1) - f(1) = S
$$

Thus, the equation becomes:
$$
S = \sum_{n=1}^{1985} [(-1)^{n+1} n - 2f(n)] = \sum_{n=1}^{1985} (-1)^{n+1} n - 2 \sum_{n=1}^{1985} f(n)
$$

Hence:
$$
S = T - 2S
\quad \Rightarrow \quad 3S = T
$$

---

### Step 2: Evaluate the alternating sum $ T $

Let:
$$
T = \sum_{n=1}^{1985} (-1)^{n+1} n
$$

This is an alternating sum of integers from 1 to 1985. There is a known formula for such alternating sums:

- If $ N $ is even:
  $$
  \sum_{n=1}^N (-1)^{n+1} n = -\frac{N}{2}
  $$
- If $ N $ is odd:
  $$
  \sum_{n=1}^N (-1)^{n+1} n = \frac{N+1}{2}
  $$

Since 1985 is odd, we use the second formula:
$$
T = \frac{1985 + 1}{2} = \frac{1986}{2} = 993
$$

---

### Step 3: Solve for the sum $ S $

From earlier, we had:
$$
3S = T = 993 \quad \Rightarrow \quad S = \frac{993}{3} = 331
$$

---

### Final Answer

$$
\boxed{331}
$$

[Assistant]
Explanation:
Okay, so I need to solve this recurrence relation problem. Let me read it again to make sure I understand.

We have a function f defined for integers n ≥ 1, with the recurrence relation:

f(n + 1) = (-1)^{n + 1} * n - 2f(n)

And it's given that f(1) = f(1986). The task is to compute the sum f(1) + f(2) + ... + f(1985).

Hmm, okay. So we have a recurrence that relates f(n+1) to f(n), and the function is periodic in some sense because f(1) = f(1986). The period is 1985, but maybe there's a shorter period? Not sure yet. The problem is asking for the sum of f(1) through f(1985). So maybe we can find a pattern or express the sum in terms of itself?

First, let me write down the recurrence again to make it clearer:

For each n ≥ 1, f(n + 1) = (-1)^{n + 1} * n - 2f(n)

And f(1) = f(1986). So the sequence is periodic with period 1985? Wait, not necessarily. It's only that f(1) equals f(1986), which would make the sequence repeat every 1985 terms, but since the recurrence is defined for n ≥ 1, if f(1) = f(1986), then perhaps we can consider the sequence as a cycle?

But maybe I should try to find a closed-form expression for f(n), or find a relation between the sum S = f(1) + f(2) + ... + f(1985). Since the sum is up to 1985, and f(1986) is equal to f(1). Maybe the sum can be related to itself through the recurrence?

Let me think. Let me denote S = f(1) + f(2) + ... + f(1985). Also, since f(1986) = f(1), we can consider that the sequence is cyclic with period 1985. But maybe I can use the recurrence relation to express terms in terms of previous ones and then sum them up?

Alternatively, perhaps I can express the sum S in terms of itself. Let me try that.

Let me first write down the recurrence for each n from 1 to 1985. For each n from 1 to 1985, f(n + 1) = (-1)^{n + 1} * n - 2f(n). But when n = 1985, we get f(1986) = (-1)^{1986} * 1985 - 2f(1985). But since f(1986) = f(1), this gives us an equation involving f(1) and f(1985). Maybe we can find another equation?

Alternatively, let me try to compute S = sum_{n=1}^{1985} f(n). Let me try to relate S with the recurrence.

Let me consider sum_{n=1}^{1985} f(n + 1) = sum_{n=1}^{1985} [ (-1)^{n + 1} * n - 2f(n) ].

But the left side is sum_{n=1}^{1985} f(n + 1) = sum_{k=2}^{1986} f(k) = [sum_{k=1}^{1986} f(k)] - f(1) = [S + f(1986)] - f(1) = [S + f(1)] - f(1) = S. Because f(1986) = f(1). Therefore, the left side is equal to S.

On the right side, we have sum_{n=1}^{1985} [ (-1)^{n + 1} * n - 2f(n) ] = sum_{n=1}^{1985} (-1)^{n + 1} * n - 2 sum_{n=1}^{1985} f(n) = [sum_{n=1}^{1985} (-1)^{n + 1} * n] - 2S.

Therefore, equating left and right sides:

S = [sum_{n=1}^{1985} (-1)^{n + 1} * n] - 2S.

Therefore, bringing the 2S to the left:

S + 2S = sum_{n=1}^{1985} (-1)^{n + 1} * n.

So 3S = sum_{n=1}^{1985} (-1)^{n + 1} * n.

Therefore, if I can compute that sum, then I can find S.

Alright, so the problem reduces to calculating this alternating sum. Let's denote this sum as T = sum_{n=1}^{1985} (-1)^{n + 1} * n.

Let me compute T. Let me note that (-1)^{n + 1} is equal to (-1)^{n - 1}. So T is an alternating sum of integers from 1 to 1985. Let me recall that for such sums, there's a formula.

Let me consider the sum S = a_1 - a_2 + a_3 - a_4 + ... + (-1)^{n+1} a_n, where a_i = i.

In our case, n is 1985. Let's compute T.

First, note that if the number of terms is odd, the last term is positive. Since 1985 is odd, the last term is +1985.

Let me think of grouping terms in pairs. Let me consider pairing the first two terms, the next two, etc.

For example, if we have an odd number of terms, say 2k + 1 terms, then we can pair the first 2k terms and have one term left.

So for T, with 1985 terms:

Number of pairs = floor(1985 / 2) = 992 pairs, and one remaining term.

Each pair is (1 - 2) + (3 - 4) + ... + (1983 - 1984) + 1985.

Each pair is (-1) + (-1) + ... + (-1). Wait, 1 - 2 = -1, 3 - 4 = -1, etc.

How many such pairs are there?

Since 1985 = 2*992 + 1. So there are 992 pairs and one term.

Therefore, the sum T is equal to (-1) * 992 + 1985.

Calculating that: -992 + 1985 = 1985 - 992 = 993.

Wait, let me check:

1985 - 992: 1985 - 900 = 1085, then subtract 92 more: 1085 - 92 = 993. Yes.

So T = 993. Therefore, 3S = 993, so S = 993 / 3 = 331. So the answer is 331?

Wait, that seems straightforward, but let me verify my steps again to be sure.

First, I used the recurrence relation to express sum_{n=1}^{1985} f(n + 1) which is equal to S. Then equated that to the other side which involved the alternating sum T minus 2S, leading to 3S = T. Then computed T as 993, so S = 331.

But let me check the computation of T again. Maybe I made a mistake there.

So T = sum_{n=1}^{m} (-1)^{n + 1} * n, where m = 1985.

If m is odd, then T = (number of pairs) * (-1) + last term.

Number of pairs is floor(m / 2) = floor(1985 / 2) = 992. Then the last term is m = 1985. Therefore, T = 992*(-1) + 1985 = 1985 - 992 = 993. That seems correct.

Alternatively, I can think of T as follows. Let me recall that for the sum of (-1)^{n+1} * n from n=1 to N, the formula is:

If N is even: (-1)^{N+1} * (N / 2)

Wait, maybe not. Let me recall:

Let me consider the sum S = 1 - 2 + 3 - 4 + ... + (-1)^{N+1} N.

There's a formula for this. Let me recall:

Let me write S = sum_{k=1}^N (-1)^{k+1} k.

If N is even, say N = 2m:

S = (1 - 2) + (3 - 4) + ... + (2m - 1 - 2m) = m*(-1) = -m.

If N is odd, N = 2m + 1:

S = [sum_{k=1}^{2m} (-1)^{k+1} k] + (-1)^{2m + 2}*(2m + 1) = -m + (2m + 1) = m + 1.

So, in general,

If N is even: S = -N/2

If N is odd: S = (N + 1)/2

So for N = 1985, which is odd, S = (1985 + 1)/2 = 1986 / 2 = 993. Which matches my previous calculation. Therefore, T = 993. Therefore, 3S = 993 => S = 331. Therefore, the answer is 331. So is that the answer? Let me check once more.

Wait, but let me check for small N. Let me take N=1: sum is 1. According to formula, (1 + 1)/2 = 1. Correct. N=3: 1 - 2 + 3 = 2. (3 + 1)/2 = 2. Correct. N=5: 1 - 2 + 3 - 4 + 5 = 3. (5 + 1)/2 = 3. Correct. So formula holds. Thus, T = 993. Therefore, S = 993 / 3 = 331. So answer is 331?

But let me check again the step where I equated the sum_{n=1}^{1985} f(n + 1) to S. Let me re-examine that.

Original sum: sum_{n=1}^{1985} f(n + 1) = sum_{k=2}^{1986} f(k) = [sum_{k=1}^{1986} f(k)] - f(1). But sum_{k=1}^{1986} f(k) = sum_{k=1}^{1985} f(k) + f(1986) = S + f(1). Therefore, sum_{k=2}^{1986} f(k) = S + f(1) - f(1) = S. Therefore, left side is S. So yes, that step is correct.

Then the right side is T - 2S, so S = T - 2S => 3S = T. Therefore, yes, 3S = 993 => S = 331.

Therefore, the answer is 331. But wait, let me check with a smaller case. Let me take a small period where I can compute manually.

Suppose instead of 1985, let's take N=1. Wait, but N=1 would require f(1 + 1) = f(2) = (-1)^{2} *1 - 2f(1). But f(2) = f(1) because f(1) = f(1 + N), but if N=1, then f(1) = f(2). So let's see:

For N=1, the sum would be f(1). Let's check.

Given f(2) = (-1)^{2}*1 - 2f(1) => f(2) = 1 - 2f(1). But f(2) = f(1). Therefore:

f(1) = 1 - 2f(1) => 3f(1) = 1 => f(1) = 1/3. Then the sum is 1/3. Let's check with formula:

If N=1 (which is odd), then T = (1 + 1)/2 = 1. Then 3S = 1 => S = 1/3. Which matches. So the formula works here.

Another test case: N=3. Let me try N=3. So f(1) = f(4). Compute sum f(1) + f(2) + f(3).

First, let me write the recurrence:

For n=1: f(2) = (-1)^{2} * 1 - 2f(1) = 1 - 2f(1)

For n=2: f(3) = (-1)^{3} * 2 - 2f(2) = -2 - 2f(2)

For n=3: f(4) = (-1)^{4} * 3 - 2f(3) = 3 - 2f(3)

But since f(4) = f(1), so:

f(1) = 3 - 2f(3)

Now, let's express everything in terms of f(1).

From n=1: f(2) = 1 - 2f(1)

From n=2: f(3) = -2 - 2f(2) = -2 - 2*(1 - 2f(1)) = -2 - 2 + 4f(1) = -4 + 4f(1)

From n=3: f(1) = 3 - 2f(3) = 3 - 2*(-4 + 4f(1)) = 3 + 8 - 8f(1) = 11 - 8f(1)

Thus, f(1) + 8f(1) = 11 => 9f(1) = 11 => f(1) = 11/9

Then sum S = f(1) + f(2) + f(3) = f(1) + [1 - 2f(1)] + [-4 + 4f(1)] = f(1) + 1 - 2f(1) - 4 + 4f(1) = (1 - 4) + (f(1) - 2f(1) + 4f(1)) = -3 + 3f(1) = -3 + 3*(11/9) = -3 + 11/3 = (-9/3 + 11/3) = 2/3.

Now, according to our formula, T = (N + 1)/2 where N = 3. (3 + 1)/2 = 2. Then 3S = 2 => S = 2/3. Which matches! Perfect. So our formula works here.

Another check. Let me take N=2. Wait, but N must be such that f(1) = f(N + 1). Wait, if N=2, then f(1) = f(3). Let me see:

For N=2, compute sum f(1) + f(2). Let me compute using recurrence.

But perhaps the formula would have T = sum_{n=1}^2 (-1)^{n + 1} n = 1 - 2 = -1. Then according to formula 3S = T => S = -1/3. Let me check manually.

For N=2, the recurrence:

For n=1: f(2) = (-1)^{2} * 1 - 2f(1) = 1 - 2f(1)

For n=2: f(3) = (-1)^{3} * 2 - 2f(2) = -2 - 2f(2). But since f(3) = f(1), we have:

f(1) = -2 - 2f(2). Substitute f(2) from above: f(1) = -2 - 2*(1 - 2f(1)) = -2 - 2 + 4f(1) = -4 + 4f(1). Then f(1) - 4f(1) = -4 => -3f(1) = -4 => f(1) = 4/3. Then f(2) = 1 - 2*(4/3) = 1 - 8/3 = -5/3. Then sum S = f(1) + f(2) = 4/3 -5/3 = -1/3, which matches 3S = -1. So yes, the formula works here as well.

Therefore, our formula seems solid. Therefore, for N=1985, the answer is 331.

Wait, but let me check one more thing. Let me check the general case. Suppose the recurrence is f(n + 1) + 2f(n) = (-1)^{n + 1} n. Then, if we consider summing both sides from n=1 to N, where f(N + 1) = f(1). Then:

sum_{n=1}^N [f(n + 1) + 2f(n)] = sum_{n=1}^N (-1)^{n + 1} n.

Left side: sum_{n=1}^N f(n + 1) + 2 sum_{n=1}^N f(n) = [sum_{k=2}^{N + 1} f(k)] + 2S = [S + f(N + 1) - f(1)] + 2S = [S + f(1) - f(1)] + 2S = S + 2S = 3S.

Therefore, 3S = sum_{n=1}^N (-1)^{n + 1} n. Which is exactly what we had. Therefore, regardless of N, if we have f(N + 1) = f(1), then 3S = T where T is the alternating sum. Therefore, for N=1985, T = 993, so S=331. Therefore, answer is 331.

But let me check with another N. Let's take N=3 again. T = 1 - 2 + 3 = 2. Then 3S = 2 => S = 2/3, which was correct. So formula works.

Therefore, confident that the answer is 331. Therefore, the answer is \boxed{331}.

**Final Answer**
\boxed{331}

We are given a recurrence relation for the function $ f(n) $, defined for $ n \geq 1 $, by:

$$
f(n + 1) = (-1)^{n + 1} n - 2f(n)
$$

with a periodic condition:

$$
f(1) = f(1986)
$$

We are to compute the sum:

$$
f(1) + f(2) + \dots + f(1985)
$$

---

### Step 1: Define the sum $ S $

Let:
$$
S = \sum_{n=1}^{1985} f(n)
$$

From the recurrence, we can rewrite the relation for each $ n $ from 1 to 1985 as:
$$
f(n + 1) = (-1)^{n + 1} n - 2f(n)
$$

Now, summing both sides from $ n = 1 $ to $ 1985 $:

$$
\sum_{n=1}^{1985} f(n + 1) = \sum_{n=1}^{1985} [(-1)^{n+1} n - 2f(n)]
$$

The left-hand side becomes:
$$
\sum_{n=1}^{1985} f(n + 1) = \sum_{k=2}^{1986} f(k) = \left( \sum_{k=1}^{1986} f(k) \right) - f(1)
$$

Because $ f(1986) = f(1) $, we have:
$$
\sum_{k=1}^{1986} f(k) = S + f(1)
$$

Therefore:
$$
\sum_{k=2}^{1986} f(k) = S + f(1) - f(1) = S
$$

Thus, the equation becomes:
$$
S = \sum_{n=1}^{1985} [(-1)^{n+1} n - 2f(n)] = \sum_{n=1}^{1985} (-1)^{n+1} n - 2 \sum_{n=1}^{1985} f(n)
$$

Hence:
$$
S = T - 2S
\quad \Rightarrow \quad 3S = T
$$

---

### Step 2: Evaluate the alternating sum $ T $

Let:
$$
T = \sum_{n=1}^{1985} (-1)^{n+1} n
$$

This is an alternating sum of integers from 1 to 1985. There is a known formula for such alternating sums:

- If $ N $ is even:
  $$
  \sum_{n=1}^N (-1)^{n+1} n = -\frac{N}{2}
  $$
- If $ N $ is odd:
  $$
  \sum_{n=1}^N (-1)^{n+1} n = \frac{N+1}{2}
  $$

Since 1985 is odd, we use the second formula:
$$
T = \frac{1985 + 1}{2} = \frac{1986}{2} = 993
$$

---

### Step 3: Solve for the sum $ S $

From earlier, we had:
$$
3S = T = 993 \quad \Rightarrow \quad S = \frac{993}{3} = 331
$$

---

### Final Answer

$$
\boxed{331}
$$

Solution:
We can list the equations
\begin{align*}
f(2) &= 1 - 2f(1), \\
f(3) &= -2 - 2f(2), \\
f(4) &= 3 - 2f(3), \\
f(5) &= -4 - 2f(4), \\
&\dots, \\
f(1985) &= -1984 - 2f(1984), \\
f(1986) &= 1985 - 2f(1985).
\end{align*}Adding these equations, we get
\[f(2) + f(3) + \dots + f(1986) = (1 - 2 + 3 - 4 + \dots + 1983 - 1984 + 1985) - 2f(1) - 2f(2) - \dots - 2f(1985).\]To find $1 - 2 + 3 - 4 + \dots + 1983 - 1984 + 1985,$ we can pair the terms
\begin{align*}
1 - 2 + 3 - 4 + \dots + 1983 - 1984 + 1985 &= (1 - 2) + (3 - 4) + \dots + (1983 - 1984) + 1985 \\
&= (-1) + (-1) + \dots + (-1) + 1985 \\
&= -\frac{1984}{2} + 1985 \\
&= 993.
\end{align*}Hence,
\[f(2) + f(3) + \dots + f(1986) = 993 - 2f(1) - 2f(2) - \dots - 2f(1985).\]Then
\[2f(1) + 3f(2) + 3f(3) + \dots + 3f(1985) + f(1986) = 993.\]Since $f(1986) = f(1),$
\[3f(1) + 3f(2) + 3f(3) + \dots + 3f(1985) = 993.\]Therefore, $f(1) + f(2) + f(3) + \dots + f(1985) = \boxed{331}.$


Explanation:
Okay, so I need to find the area of an isosceles right triangle where the altitude to the hypotenuse is 4√2. Hmm, let me think about how to approach this.

First, let me recall what an isosceles right triangle is. It's a triangle with two legs of equal length and a right angle between them. The hypotenuse is the side opposite the right angle, and it's longer than either of the legs.

Since it's a right triangle, the area can be calculated as (base * height)/2. But in this case, they're giving me the length of the altitude to the hypotenuse, not the legs. So maybe I need to relate the altitude to the hypotenuse with the area?

Wait, actually, the altitude to the hypotenuse in a right triangle... I remember there's a formula or some relationship here. Let me think. In any right triangle, the altitude to the hypotenuse relates the legs and the hypotenuse. Also, the area can be expressed in two ways: once using the legs and once using the hypotenuse and the altitude.

Let me write down what I know:

Let the legs of the triangle be of length 'a'. Since it's an isosceles right triangle, both legs are equal. The hypotenuse would then be a√2 by the Pythagorean theorem.

The area of the triangle can be calculated as (a * a)/2 = a²/2.

Now, the altitude to the hypotenuse. Let me denote the altitude as h. In any right triangle, the altitude to the hypotenuse is equal to (product of the legs)/hypotenuse. Wait, yes, because area is also (hypotenuse * altitude)/2. So:

Area = (hypotenuse * h)/2

But we also have Area = (a²)/2. Therefore:

(a²)/2 = (hypotenuse * h)/2

Multiply both sides by 2:

a² = hypotenuse * h

We know that hypotenuse is a√2, so substituting:

a² = (a√2) * h

Then, divide both sides by a:

a = √2 * h

So, h = a / √2

But wait, they gave us the altitude h is 4√2. So plugging in:

4√2 = a / √2

Multiply both sides by √2:

4√2 * √2 = a

Calculate that:

√2 * √2 is 2, so 4 * 2 = 8. Therefore, a = 8.

So the legs are each 8 units long. Therefore, the area is (8 * 8)/2 = 64/2 = 32.

Wait, hold on. But let me check again. If the legs are 8, hypotenuse is 8√2. Then the altitude h is given by (a²)/hypotenuse = (64)/ (8√2) = 8 / √2 = 4√2. Which matches the given altitude. So that checks out. Therefore, the area is indeed 32.

But let me think if there's another way to approach this problem, maybe using properties of isosceles right triangles.

Alternatively, maybe I can let the hypotenuse be length c, and the altitude h to the hypotenuse is given. Then, since the area is (c * h)/2. But also, in an isosceles right triangle, the legs are c / √2 each? Wait, no. Wait, hypotenuse is a√2 if the legs are length a. So if hypotenuse is c, then each leg is c / √2. Therefore, the area would be (c / √2)^2 / 2 = (c² / 2)/2 = c² / 4.

Alternatively, area is also (c * h)/2. Therefore, c² / 4 = (c * h)/2 => divide both sides by c: c / 4 = h / 2 => c = 2h.

So hypotenuse is twice the altitude. Since h is 4√2, hypotenuse is 8√2. Then area is (c * h)/2 = (8√2 * 4√2)/2. Let me calculate that:

Multiply numerator: 8√2 * 4√2 = 32 * (√2 * √2) = 32 * 2 = 64. Then divide by 2: 64 / 2 = 32. Same result. So area is 32.

So that seems consistent. Therefore, the area is 32.

But let me just make sure I didn't make any wrong assumptions here. Is the formula for the altitude to the hypotenuse in a right triangle always equal to (product of legs)/hypotenuse? Let me recall.

Yes, in any right triangle, the altitude to the hypotenuse is equal to (leg1 * leg2)/hypotenuse. Because area is (leg1 * leg2)/2, and also (hypotenuse * altitude)/2, so equate them: (leg1 * leg2)/2 = (hypotenuse * altitude)/2 => leg1 * leg2 = hypotenuse * altitude => altitude = (leg1 * leg2)/hypotenuse. So that formula is correct.

In our case, since it's an isosceles right triangle, leg1 = leg2 = a. So altitude h = (a * a)/hypotenuse. Since hypotenuse is a√2, then h = a²/(a√2) = a / √2. Which is the same as before. Therefore, solving for a gives a = h√2. Then with h = 4√2, a = 4√2 * √2 = 4 * 2 = 8. Then area is (a²)/2 = 64 / 2 = 32.

So all methods lead to the area being 32.

Alternatively, maybe I can think of the triangle and the altitude. In an isosceles right triangle, the altitude to the hypotenuse also acts as a median and angle bisector? Wait, in an isosceles triangle, the altitude from the right angle to the hypotenuse would divide the triangle into two smaller congruent triangles. Wait, but in this case, the triangle is already isosceles, so the altitude to the hypotenuse would split it into two congruent isosceles right triangles?

Wait, maybe not. Let me visualize. The original triangle has legs of length a, hypotenuse c = a√2. The altitude h divides the hypotenuse into two segments. In a right triangle, the altitude to the hypotenuse divides it into two smaller similar triangles, each similar to the original triangle.

But since the original triangle is isosceles, those two smaller triangles would also be congruent. So each of them would be an isosceles right triangle? Wait, if the original triangle is isosceles right, then when you drop the altitude to hypotenuse, which splits the hypotenuse into two equal parts? Wait, in an isosceles triangle, the altitude from the apex to the base is also a median and angle bisector. But is that the case here?

Wait, in our original triangle, the altitude is from the right angle to the hypotenuse. Wait, no. Wait, the original triangle has a right angle, so the altitude to the hypotenuse would be from the opposite vertex. Wait, but all three vertices are... Wait, in a right triangle, the hypotenuse is opposite the right angle. So the altitude to the hypotenuse would be from the right angle vertex to the hypotenuse. Wait, but that is the same as the altitude. Wait, no, in a right triangle, the altitude from the right angle to the hypotenuse is actually the same as the height, which is given here as 4√2. Wait, but in that case, in an isosceles right triangle, does that altitude split the hypotenuse into two equal parts?

Wait, if we have an isosceles right triangle, and we drop an altitude from the right angle to the hypotenuse, then since the triangle is symmetric, this altitude should split the hypotenuse into two equal segments. Therefore, each segment is c/2.

So, we can think of this altitude forming two smaller isosceles right triangles? Wait, no, because if we split the hypotenuse into two equal parts, each of length c/2, then the altitude h is given as 4√2. In the original triangle, the legs are a, hypotenuse c = a√2, and the altitude h is related by h = a / √2, as we had before.

Alternatively, maybe using Pythagoras on one of the smaller triangles. Let me try that. If we split the original triangle into two smaller triangles by the altitude h. Each of those smaller triangles would have:

- One leg: h, which is 4√2.

- Another leg: c/2, which is (a√2)/2.

- And the hypotenuse is the original leg of the big triangle, which is a.

Wait, because the altitude divides the original triangle into two smaller right triangles. Each of these smaller triangles has legs h and c/2, and hypotenuse equal to the original legs of the big triangle, which is a.

So, let's write Pythagoras for one of the smaller triangles:

h² + (c/2)² = a²

We know that c = a√2, so substituting:

h² + ( (a√2)/2 )² = a²

Calculate that:

h² + ( (a² * 2) / 4 ) = a²

Simplify:

h² + (a² / 2 ) = a²

Subtract a² / 2:

h² = a² - a² / 2 = a² / 2

Therefore, h = a / √2, which is the same as before. So this again gives h = a / √2. Then with h = 4√2, solving for a gives a = 4√2 * √2 = 4 * 2 = 8. Then the area is (a²)/2 = 64 / 2 = 32. So again, same answer.

Therefore, regardless of the method used, the area comes out to be 32.

But just to make sure there isn't any alternative interpretation of the question. The problem says "the altitude to the hypotenuse has length 4√2". So that altitude is the one from the right angle vertex to the hypotenuse. In a right triangle, this altitude is also related to the legs.

Alternatively, could there be another altitude? But in any triangle, there are three altitudes. However, in the context of the problem, since it's an isosceles right triangle, the altitude to the hypotenuse is a specific one. The other altitudes would be the legs themselves. For example, if we take a leg as the base, then the altitude would be the other leg. So the altitude to a leg is the other leg. But since the problem mentions the altitude to the hypotenuse, which is different.

So no, the problem is referring to the altitude from the right angle to the hypotenuse, which we have already considered.

Another way to think: maybe using coordinate geometry. Let me try that. Let me place the triangle on the coordinate plane. Let's say the right angle is at the origin (0,0), and the two legs are along the x-axis and y-axis. Then, since it's an isosceles right triangle, the legs are equal. Let me say the vertices are at (0,0), (a, 0), and (0, a). Then the hypotenuse is between (a, 0) and (0, a). The equation of the hypotenuse is x + y = a.

Now, the altitude from the right angle (0,0) to the hypotenuse. Wait, but the right angle is already on the hypotenuse? Wait no. Wait, hypotenuse is opposite the right angle. So in this case, hypotenuse is between (a,0) and (0,a), while the right angle is at (0,0). Therefore, the altitude from the right angle to the hypotenuse is the same as the length of the line segment from (0,0) to the hypotenuse. Wait, but (0,0) is not on the hypotenuse. Wait, the hypotenuse connects (a,0) and (0,a). So the right angle is at (0,0), which is not on the hypotenuse. So the altitude from (0,0) to the hypotenuse is a line segment from (0,0) to the hypotenuse, perpendicular to it. Wait, but in this case, isn't that the same as the altitude we already considered?

Wait, let me calculate this. Let me compute the distance from the point (0,0) to the line x + y = a. The formula for the distance from a point (x0, y0) to the line Ax + By + C = 0 is |Ax0 + By0 + C| / sqrt(A² + B²).

First, let me write the equation of the hypotenuse in standard form. The hypotenuse is between (a, 0) and (0, a), so its equation is x + y = a. Bringing all terms to one side: x + y - a = 0. Therefore, A = 1, B = 1, C = -a.

Then the distance from (0, 0) to this line is |1*0 + 1*0 - a| / sqrt(1 + 1) = | -a | / sqrt(2) = a / sqrt(2). Which is the same as we had before. Therefore, this distance is the altitude h = a / sqrt(2). Given h = 4√2, so solving for a:

a = h * sqrt(2) = 4√2 * sqrt(2) = 4 * 2 = 8. Then area is (a²)/2 = 64 / 2 = 32. Same result.

Therefore, regardless of the method—whether using area relations, coordinate geometry, Pythagorean theorem in the smaller triangles—all approaches lead to the area being 32.

Just to ensure there's no trick in the question, let me check if maybe the problem is referring to an altitude to the hypotenuse from one of the other vertices. Wait, but in that case, the altitude would be one of the legs. For example, if we take one of the legs as the base, then the altitude to that base would be the other leg. But since the triangle is isosceles, both legs are equal, so the altitude to either leg would be the other leg, which is length a. But the given altitude is 4√2. If that were the case, then a = 4√2, and area would be (a²)/2 = (16 * 2)/2 = 16. However, this would be different. But the problem specifically says "the altitude to the hypotenuse", so that's not the case. So that's probably not the case.

Alternatively, if someone misunderstood and thought that the altitude to the hypotenuse is one of the legs... but no, the altitude to the hypotenuse is a separate segment.

So, given that, I think my answer is solid. The area is 32.

Just to recap:

- Given: isosceles right triangle, altitude to hypotenuse is 4√2.

- Found that altitude h = a / √2, so a = h√2 = 4√2 * √2 = 8.

- Therefore, legs are 8 each, area is (8*8)/2 = 32.

Everything checks out. Multiple methods confirm this.

I think that's thorough enough. I can't see any mistakes in the reasoning.

**Final Answer**
The area of the triangle is \boxed{32}.

To determine the area of the isosceles right triangle, we are given that the **altitude to the hypotenuse** is $ 4\sqrt{2} $. Let’s analyze the situation and solve it methodically.

---

### Step 1: Understanding the Triangle

An **isosceles right triangle** has:

- Two equal legs, say of length $ a $,
- A hypotenuse of length $ a\sqrt{2} $ (by the Pythagorean Theorem),
- A right angle between the two legs.

The **altitude to the hypotenuse** is a line segment from the right angle vertex to the hypotenuse, perpendicular to it. This altitude divides the triangle into two smaller right triangles.

---

### Step 2: Use of Area Formulas

We can use two distinct expressions for the area of the triangle:

1. **Using the legs:**
   $$
   \text{Area} = \frac{1}{2} \cdot a \cdot a = \frac{a^2}{2}
   $$

2. **Using the hypotenuse and the altitude to it:**
   $$
   \text{Area} = \frac{1}{2} \cdot \text{hypotenuse} \cdot h = \frac{1}{2} \cdot a\sqrt{2} \cdot h
   $$

Setting the two expressions equal:
$$
\frac{a^2}{2} = \frac{1}{2} \cdot a\sqrt{2} \cdot h
$$

Multiply both sides by 2:
$$
a^2 = a\sqrt{2} \cdot h
$$

Divide both sides by $ a $ (assuming $ a \ne 0 $):
$$
a = \sqrt{2} \cdot h
$$

Substitute the given value of $ h = 4\sqrt{2} $:
$$
a = \sqrt{2} \cdot 4\sqrt{2} = 4 \cdot 2 = 8
$$

---

### Step 3: Calculate the Area

Now that we know each leg is of length $ a = 8 $, the area of the triangle is:
$$
\text{Area} = \frac{1}{2} \cdot a \cdot a = \frac{1}{2} \cdot 8 \cdot 8 = \frac{64}{2} = 32
$$

---

### Final Answer

$$
\boxed{32}
$$

[Assistant]
Explanation:
Okay, so I need to find the area of an isosceles right triangle where the altitude to the hypotenuse is 4√2. Hmm, let me think about how to approach this.

First, let me recall what an isosceles right triangle is. It's a triangle with two legs of equal length and a right angle between them. The hypotenuse is the side opposite the right angle, and it's longer than either of the legs.

Since it's a right triangle, the area can be calculated as (base * height)/2. But in this case, they're giving me the length of the altitude to the hypotenuse, not the legs. So maybe I need to relate the altitude to the hypotenuse with the area?

Wait, actually, the altitude to the hypotenuse in a right triangle... I remember there's a formula or some relationship here. Let me think. In any right triangle, the altitude to the hypotenuse relates the legs and the hypotenuse. Also, the area can be expressed in two ways: once using the legs and once using the hypotenuse and the altitude.

Let me write down what I know:

Let the legs of the triangle be of length 'a'. Since it's an isosceles right triangle, both legs are equal. The hypotenuse would then be a√2 by the Pythagorean theorem.

The area of the triangle can be calculated as (a * a)/2 = a²/2.

Now, the altitude to the hypotenuse. Let me denote the altitude as h. In any right triangle, the altitude to the hypotenuse is equal to (product of the legs)/hypotenuse. Wait, yes, because area is also (hypotenuse * altitude)/2. So:

Area = (hypotenuse * h)/2

But we also have Area = (a²)/2. Therefore:

(a²)/2 = (hypotenuse * h)/2

Multiply both sides by 2:

a² = hypotenuse * h

We know that hypotenuse is a√2, so substituting:

a² = (a√2) * h

Then, divide both sides by a:

a = √2 * h

So, h = a / √2

But wait, they gave us the altitude h is 4√2. So plugging in:

4√2 = a / √2

Multiply both sides by √2:

4√2 * √2 = a

Calculate that:

√2 * √2 is 2, so 4 * 2 = 8. Therefore, a = 8.

So the legs are each 8 units long. Therefore, the area is (8 * 8)/2 = 64/2 = 32.

Wait, hold on. But let me check again. If the legs are 8, hypotenuse is 8√2. Then the altitude h is given by (a²)/hypotenuse = (64)/ (8√2) = 8 / √2 = 4√2. Which matches the given altitude. So that checks out. Therefore, the area is indeed 32.

But let me think if there's another way to approach this problem, maybe using properties of isosceles right triangles.

Alternatively, maybe I can let the hypotenuse be length c, and the altitude h to the hypotenuse is given. Then, since the area is (c * h)/2. But also, in an isosceles right triangle, the legs are c / √2 each? Wait, no. Wait, hypotenuse is a√2 if the legs are length a. So if hypotenuse is c, then each leg is c / √2. Therefore, the area would be (c / √2)^2 / 2 = (c² / 2)/2 = c² / 4.

Alternatively, area is also (c * h)/2. Therefore, c² / 4 = (c * h)/2 => divide both sides by c: c / 4 = h / 2 => c = 2h.

So hypotenuse is twice the altitude. Since h is 4√2, hypotenuse is 8√2. Then area is (c * h)/2 = (8√2 * 4√2)/2. Let me calculate that:

Multiply numerator: 8√2 * 4√2 = 32 * (√2 * √2) = 32 * 2 = 64. Then divide by 2: 64 / 2 = 32. Same result. So area is 32.

So that seems consistent. Therefore, the area is 32.

But let me just make sure I didn't make any wrong assumptions here. Is the formula for the altitude to the hypotenuse in a right triangle always equal to (product of legs)/hypotenuse? Let me recall.

Yes, in any right triangle, the altitude to the hypotenuse is equal to (leg1 * leg2)/hypotenuse. Because area is (leg1 * leg2)/2, and also (hypotenuse * altitude)/2, so equate them: (leg1 * leg2)/2 = (hypotenuse * altitude)/2 => leg1 * leg2 = hypotenuse * altitude => altitude = (leg1 * leg2)/hypotenuse. So that formula is correct.

In our case, since it's an isosceles right triangle, leg1 = leg2 = a. So altitude h = (a * a)/hypotenuse. Since hypotenuse is a√2, then h = a²/(a√2) = a / √2. Which is the same as before. Therefore, solving for a gives a = h√2. Then with h = 4√2, a = 4√2 * √2 = 4 * 2 = 8. Then area is (a²)/2 = 64 / 2 = 32.

So all methods lead to the area being 32.

Alternatively, maybe I can think of the triangle and the altitude. In an isosceles right triangle, the altitude to the hypotenuse also acts as a median and angle bisector? Wait, in an isosceles triangle, the altitude from the right angle to the hypotenuse would divide the triangle into two smaller congruent triangles. Wait, but in this case, the triangle is already isosceles, so the altitude to the hypotenuse would split it into two congruent isosceles right triangles?

Wait, maybe not. Let me visualize. The original triangle has legs of length a, hypotenuse c = a√2. The altitude h divides the hypotenuse into two segments. In a right triangle, the altitude to the hypotenuse divides it into two smaller similar triangles, each similar to the original triangle.

But since the original triangle is isosceles, those two smaller triangles would also be congruent. So each of them would be an isosceles right triangle? Wait, if the original triangle is isosceles right, then when you drop the altitude to hypotenuse, which splits the hypotenuse into two equal parts? Wait, in an isosceles triangle, the altitude from the apex to the base is also a median and angle bisector. But is that the case here?

Wait, in our original triangle, the altitude is from the right angle to the hypotenuse. Wait, no. Wait, the original triangle has a right angle, so the altitude to the hypotenuse would be from the opposite vertex. Wait, but all three vertices are... Wait, in a right triangle, the hypotenuse is opposite the right angle. So the altitude to the hypotenuse would be from the right angle vertex to the hypotenuse. Wait, but that is the same as the altitude. Wait, no, in a right triangle, the altitude from the right angle to the hypotenuse is actually the same as the height, which is given here as 4√2. Wait, but in that case, in an isosceles right triangle, does that altitude split the hypotenuse into two equal parts?

Wait, if we have an isosceles right triangle, and we drop an altitude from the right angle to the hypotenuse, then since the triangle is symmetric, this altitude should split the hypotenuse into two equal segments. Therefore, each segment is c/2.

So, we can think of this altitude forming two smaller isosceles right triangles? Wait, no, because if we split the hypotenuse into two equal parts, each of length c/2, then the altitude h is given as 4√2. In the original triangle, the legs are a, hypotenuse c = a√2, and the altitude h is related by h = a / √2, as we had before.

Alternatively, maybe using Pythagoras on one of the smaller triangles. Let me try that. If we split the original triangle into two smaller triangles by the altitude h. Each of those smaller triangles would have:

- One leg: h, which is 4√2.

- Another leg: c/2, which is (a√2)/2.

- And the hypotenuse is the original leg of the big triangle, which is a.

Wait, because the altitude divides the original triangle into two smaller right triangles. Each of these smaller triangles has legs h and c/2, and hypotenuse equal to the original legs of the big triangle, which is a.

So, let's write Pythagoras for one of the smaller triangles:

h² + (c/2)² = a²

We know that c = a√2, so substituting:

h² + ( (a√2)/2 )² = a²

Calculate that:

h² + ( (a² * 2) / 4 ) = a²

Simplify:

h² + (a² / 2 ) = a²

Subtract a² / 2:

h² = a² - a² / 2 = a² / 2

Therefore, h = a / √2, which is the same as before. So this again gives h = a / √2. Then with h = 4√2, solving for a gives a = 4√2 * √2 = 4 * 2 = 8. Then the area is (a²)/2 = 64 / 2 = 32. So again, same answer.

Therefore, regardless of the method used, the area comes out to be 32.

But just to make sure there isn't any alternative interpretation of the question. The problem says "the altitude to the hypotenuse has length 4√2". So that altitude is the one from the right angle vertex to the hypotenuse. In a right triangle, this altitude is also related to the legs.

Alternatively, could there be another altitude? But in any triangle, there are three altitudes. However, in the context of the problem, since it's an isosceles right triangle, the altitude to the hypotenuse is a specific one. The other altitudes would be the legs themselves. For example, if we take a leg as the base, then the altitude would be the other leg. So the altitude to a leg is the other leg. But since the problem mentions the altitude to the hypotenuse, which is different.

So no, the problem is referring to the altitude from the right angle to the hypotenuse, which we have already considered.

Another way to think: maybe using coordinate geometry. Let me try that. Let me place the triangle on the coordinate plane. Let's say the right angle is at the origin (0,0), and the two legs are along the x-axis and y-axis. Then, since it's an isosceles right triangle, the legs are equal. Let me say the vertices are at (0,0), (a, 0), and (0, a). Then the hypotenuse is between (a, 0) and (0, a). The equation of the hypotenuse is x + y = a.

Now, the altitude from the right angle (0,0) to the hypotenuse. Wait, but the right angle is already on the hypotenuse? Wait no. Wait, hypotenuse is opposite the right angle. So in this case, hypotenuse is between (a,0) and (0,a), while the right angle is at (0,0). Therefore, the altitude from the right angle to the hypotenuse is the same as the length of the line segment from (0,0) to the hypotenuse. Wait, but (0,0) is not on the hypotenuse. Wait, the hypotenuse connects (a,0) and (0,a). So the right angle is at (0,0), which is not on the hypotenuse. So the altitude from (0,0) to the hypotenuse is a line segment from (0,0) to the hypotenuse, perpendicular to it. Wait, but in this case, isn't that the same as the altitude we already considered?

Wait, let me calculate this. Let me compute the distance from the point (0,0) to the line x + y = a. The formula for the distance from a point (x0, y0) to the line Ax + By + C = 0 is |Ax0 + By0 + C| / sqrt(A² + B²).

First, let me write the equation of the hypotenuse in standard form. The hypotenuse is between (a, 0) and (0, a), so its equation is x + y = a. Bringing all terms to one side: x + y - a = 0. Therefore, A = 1, B = 1, C = -a.

Then the distance from (0, 0) to this line is |1*0 + 1*0 - a| / sqrt(1 + 1) = | -a | / sqrt(2) = a / sqrt(2). Which is the same as we had before. Therefore, this distance is the altitude h = a / sqrt(2). Given h = 4√2, so solving for a:

a = h * sqrt(2) = 4√2 * sqrt(2) = 4 * 2 = 8. Then area is (a²)/2 = 64 / 2 = 32. Same result.

Therefore, regardless of the method—whether using area relations, coordinate geometry, Pythagorean theorem in the smaller triangles—all approaches lead to the area being 32.

Just to ensure there's no trick in the question, let me check if maybe the problem is referring to an altitude to the hypotenuse from one of the other vertices. Wait, but in that case, the altitude would be one of the legs. For example, if we take one of the legs as the base, then the altitude to that base would be the other leg. But since the triangle is isosceles, both legs are equal, so the altitude to either leg would be the other leg, which is length a. But the given altitude is 4√2. If that were the case, then a = 4√2, and area would be (a²)/2 = (16 * 2)/2 = 16. However, this would be different. But the problem specifically says "the altitude to the hypotenuse", so that's not the case. So that's probably not the case.

Alternatively, if someone misunderstood and thought that the altitude to the hypotenuse is one of the legs... but no, the altitude to the hypotenuse is a separate segment.

So, given that, I think my answer is solid. The area is 32.

Just to recap:

- Given: isosceles right triangle, altitude to hypotenuse is 4√2.

- Found that altitude h = a / √2, so a = h√2 = 4√2 * √2 = 8.

- Therefore, legs are 8 each, area is (8*8)/2 = 32.

Everything checks out. Multiple methods confirm this.

I think that's thorough enough. I can't see any mistakes in the reasoning.

**Final Answer**
The area of the triangle is \boxed{32}.

To determine the area of the isosceles right triangle, we are given that the **altitude to the hypotenuse** is $ 4\sqrt{2} $. Let’s analyze the situation and solve it methodically.

---

### Step 1: Understanding the Triangle

An **isosceles right triangle** has:

- Two equal legs, say of length $ a $,
- A hypotenuse of length $ a\sqrt{2} $ (by the Pythagorean Theorem),
- A right angle between the two legs.

The **altitude to the hypotenuse** is a line segment from the right angle vertex to the hypotenuse, perpendicular to it. This altitude divides the triangle into two smaller right triangles.

---

### Step 2: Use of Area Formulas

We can use two distinct expressions for the area of the triangle:

1. **Using the legs:**
   $$
   \text{Area} = \frac{1}{2} \cdot a \cdot a = \frac{a^2}{2}
   $$

2. **Using the hypotenuse and the altitude to it:**
   $$
   \text{Area} = \frac{1}{2} \cdot \text{hypotenuse} \cdot h = \frac{1}{2} \cdot a\sqrt{2} \cdot h
   $$

Setting the two expressions equal:
$$
\frac{a^2}{2} = \frac{1}{2} \cdot a\sqrt{2} \cdot h
$$

Multiply both sides by 2:
$$
a^2 = a\sqrt{2} \cdot h
$$

Divide both sides by $ a $ (assuming $ a \ne 0 $):
$$
a = \sqrt{2} \cdot h
$$

Substitute the given value of $ h = 4\sqrt{2} $:
$$
a = \sqrt{2} \cdot 4\sqrt{2} = 4 \cdot 2 = 8
$$

---

### Step 3: Calculate the Area

Now that we know each leg is of length $ a = 8 $, the area of the triangle is:
$$
\text{Area} = \frac{1}{2} \cdot a \cdot a = \frac{1}{2} \cdot 8 \cdot 8 = \frac{64}{2} = 32
$$

---

### Final Answer

$$
\boxed{32}
$$

Solution:
In isosceles right triangle $\triangle ABC$ below, $\overline{AD}$ is the altitude to the hypotenuse.

[asy]
import olympiad;
unitsize(0.8inch);
pair A,B,C,D;
A = (0,1);
B= (1,0);
C = -B;
D = (0,0);
draw(A--B--C--A,linewidth(1));
draw(A--D,linewidth(0.8));
draw(rightanglemark(C,A,B,s=5));
draw(rightanglemark(C,D,A,s=5));
label("$A$",A,N);
label("$B$",B,S);
label("$C$",C,S);
label("$D$",D,S);
[/asy]

Because $\triangle ABC$ is an isosceles right triangle, $\angle ABC = 45^\circ$.  Since $\angle ADB = 90^\circ$, we know that $\angle DAB = 45^\circ$, so $\triangle ABD$ is also a 45-45-90 triangle.  Similarly, $\triangle ACD$ is a 45-45-90 triangle.  Therefore, $DB=DC = DA = 4\sqrt{2}$, so $BC = BD+DC = 8\sqrt{2}$, and  \[[ABC] = \frac{(AD)(BC)}{2} = \frac{(4\sqrt{2})(8\sqrt{2})}{2} = \boxed{32}.\]


Explanation:
Okay, so I need to find the domain of the function f(x) = (2x - 7)/sqrt(x² - 5x + 6). Let me think about how to approach this.

First, I remember that the domain of a function consists of all the real numbers x for which the function is defined. Since this function has a square root in the denominator, I need to make sure that the expression inside the square root is positive. Also, the denominator can't be zero because division by zero is undefined. So, there are two main conditions here: the expression under the square root must be positive, and the denominator itself must not be zero.

Let me write down the denominator: sqrt(x² - 5x + 6). The square root requires that the inside is non-negative, right? But since it's in the denominator, it can't be zero. Therefore, x² - 5x + 6 must be greater than zero.

So my first step is to solve the inequality x² - 5x + 6 > 0. Let me factor the quadratic. Let me try factoring x² - 5x + 6. Looking for two numbers that multiply to 6 and add up to -5. Hmm, -2 and -3? Let me check: (-2)*(-3) = 6, and -2 + (-3) = -5. Yes, that works. So the quadratic factors as (x - 2)(x - 3).

So the inequality becomes (x - 2)(x - 3) > 0. To solve this inequality, I can use a sign chart or test intervals. The critical points are x = 2 and x = 3. These divide the number line into three intervals:

1. x < 2
2. 2 < x < 3
3. x > 3

I need to test each interval to see where the product is positive.

For the first interval, let's pick x = 1. Plugging into (x - 2)(x - 3): (1 - 2)(1 - 3) = (-1)(-2) = 2, which is positive. So the first interval is part of the solution.

For the second interval, let's pick x = 2.5. Then (2.5 - 2)(2.5 - 3) = (0.5)(-0.5) = -0.25, which is negative. So this interval is not part of the solution.

For the third interval, let's pick x = 4. Then (4 - 2)(4 - 3) = (2)(1) = 2, which is positive. So this interval is part of the solution.

Therefore, the solution to the inequality x² - 5x + 6 > 0 is x < 2 or x > 3.

But wait, let me make sure. Since the quadratic opens upwards (coefficient of x² is positive), the parabola is above the x-axis outside the roots. So yes, that matches with the intervals x < 2 and x > 3.

So the denominator is defined and non-zero in those intervals. Therefore, the domain of f(x) is all real numbers x such that x < 2 or x > 3.

But wait, is there anything else I need to consider? The numerator is 2x - 7, which is a linear function and defined for all real numbers. So the only restrictions come from the denominator.

Therefore, combining these, the domain should be (-∞, 2) ∪ (3, ∞).

Let me check if there's a possibility that the denominator could be zero in these intervals. But we already excluded the points where the denominator is zero (x=2 and x=3), so the intervals are open, not closed.

Wait, just to confirm, plugging x=2 into the denominator: sqrt(2² -5*2 +6) = sqrt(4 -10 +6) = sqrt(0) = 0. Similarly for x=3: sqrt(9 -15 +6) = sqrt(0) = 0. So those points are excluded. Therefore, the domain is indeed all real numbers except x=2 and x=3, but since the expression inside the square root must be positive, it's only x < 2 or x > 3.

Therefore, the domain is (-∞, 2) union (3, ∞).

Let me just verify with an example. Let's take x=1, which is in the interval x < 2. Then denominator is sqrt(1 -5 +6) = sqrt(2), which is fine. The numerator is 2*1 -7 = -5, so the function is defined.

Take x=4, which is in x > 3. Denominator sqrt(16 -20 +6) = sqrt(2), same as before. Numerator is 8 -7 = 1, so defined.

What about x=2.5? Well, that's between 2 and 3, which we said is excluded. Let's check denominator: sqrt(6.25 -12.5 +6) = sqrt(-0.25). Wait, that's sqrt of negative number, which is not real. So that's why x between 2 and 3 is excluded.

Therefore, the domain is indeed all real numbers except between 2 and 3, inclusive. So written in interval notation, (-∞, 2) ∪ (3, ∞).

Therefore, I think that is the answer.

**Final Answer**
The domain of the function is \boxed{(-\infty, 2) \cup (3, \infty)}.

To determine the **domain** of the real-valued function
$$
f(x) = \frac{2x - 7}{\sqrt{x^2 - 5x + 6}},
$$
we analyze the conditions under which the function is defined. A function is defined if and only if the **denominator is not zero** and the **expression inside the square root is positive** (since square roots of negative numbers are not real).

---

### Step 1: Analyze the denominator

The denominator is
$$
\sqrt{x^2 - 5x + 6}.
$$
For the square root to be defined (i.e., produce a real number), the expression inside must be **non-negative**:
$$
x^2 - 5x + 6 \geq 0.
$$
However, since the square root appears in the **denominator**, the expression inside must be **strictly positive** to avoid division by zero. Hence, we need:
$$
x^2 - 5x + 6 > 0.
$$

---

### Step 2: Solve the inequality

Factor the quadratic:
$$
x^2 - 5x + 6 = (x - 2)(x - 3).
$$
So the inequality becomes:
$$
(x - 2)(x - 3) > 0.
$$
We solve this by identifying the **critical points** $ x = 2 $ and $ x = 3 $, and testing intervals around them.

- For $ x < 2 $, say $ x = 1 $:
  $$
  (1 - 2)(1 - 3) = (-1)(-2) = 2 > 0.
  $$
- For $ 2 < x < 3 $, say $ x = 2.5 $:
  $$
  (2.5 - 2)(2.5 - 3) = (0.5)(-0.5) = -0.25 < 0.
  $$
- For $ x > 3 $, say $ x = 4 $:
  $$
  (4 - 2)(4 - 3) = (2)(1) = 2 > 0.
  $$

Thus, the inequality is satisfied when $ x < 2 $ or $ x > 3 $.

---

### Step 3: Exclude points where the denominator is zero

The expression inside the square root is zero when $ x^2 - 5x + 6 = 0 $, i.e., when $ x = 2 $ or $ x = 3 $. At these points, the denominator becomes zero, which makes the function **undefined**. So these values must be **excluded** from the domain.

---

### Step 4: Write the domain

The domain includes all real numbers **excluding** the interval $ [2, 3] $, and in particular, $ x = 2 $ and $ x = 3 $, since the expression inside the square root becomes zero there. So the domain is:

$$
(-\infty, 2) \cup (3, \infty)
$$

---

### Final Answer

$$
\boxed{(-\infty, 2) \cup (3, \infty)}
$$

[Assistant]
Explanation:
Okay, so I need to find the domain of the function f(x) = (2x - 7)/sqrt(x² - 5x + 6). Let me think about how to approach this.

First, I remember that the domain of a function consists of all the real numbers x for which the function is defined. Since this function has a square root in the denominator, I need to make sure that the expression inside the square root is positive. Also, the denominator can't be zero because division by zero is undefined. So, there are two main conditions here: the expression under the square root must be positive, and the denominator itself must not be zero.

Let me write down the denominator: sqrt(x² - 5x + 6). The square root requires that the inside is non-negative, right? But since it's in the denominator, it can't be zero. Therefore, x² - 5x + 6 must be greater than zero.

So my first step is to solve the inequality x² - 5x + 6 > 0. Let me factor the quadratic. Let me try factoring x² - 5x + 6. Looking for two numbers that multiply to 6 and add up to -5. Hmm, -2 and -3? Let me check: (-2)*(-3) = 6, and -2 + (-3) = -5. Yes, that works. So the quadratic factors as (x - 2)(x - 3).

So the inequality becomes (x - 2)(x - 3) > 0. To solve this inequality, I can use a sign chart or test intervals. The critical points are x = 2 and x = 3. These divide the number line into three intervals:

1. x < 2
2. 2 < x < 3
3. x > 3

I need to test each interval to see where the product is positive.

For the first interval, let's pick x = 1. Plugging into (x - 2)(x - 3): (1 - 2)(1 - 3) = (-1)(-2) = 2, which is positive. So the first interval is part of the solution.

For the second interval, let's pick x = 2.5. Then (2.5 - 2)(2.5 - 3) = (0.5)(-0.5) = -0.25, which is negative. So this interval is not part of the solution.

For the third interval, let's pick x = 4. Then (4 - 2)(4 - 3) = (2)(1) = 2, which is positive. So this interval is part of the solution.

Therefore, the solution to the inequality x² - 5x + 6 > 0 is x < 2 or x > 3.

But wait, let me make sure. Since the quadratic opens upwards (coefficient of x² is positive), the parabola is above the x-axis outside the roots. So yes, that matches with the intervals x < 2 and x > 3.

So the denominator is defined and non-zero in those intervals. Therefore, the domain of f(x) is all real numbers x such that x < 2 or x > 3.

But wait, is there anything else I need to consider? The numerator is 2x - 7, which is a linear function and defined for all real numbers. So the only restrictions come from the denominator.

Therefore, combining these, the domain should be (-∞, 2) ∪ (3, ∞).

Let me check if there's a possibility that the denominator could be zero in these intervals. But we already excluded the points where the denominator is zero (x=2 and x=3), so the intervals are open, not closed.

Wait, just to confirm, plugging x=2 into the denominator: sqrt(2² -5*2 +6) = sqrt(4 -10 +6) = sqrt(0) = 0. Similarly for x=3: sqrt(9 -15 +6) = sqrt(0) = 0. So those points are excluded. Therefore, the domain is indeed all real numbers except x=2 and x=3, but since the expression inside the square root must be positive, it's only x < 2 or x > 3.

Therefore, the domain is (-∞, 2) union (3, ∞).

Let me just verify with an example. Let's take x=1, which is in the interval x < 2. Then denominator is sqrt(1 -5 +6) = sqrt(2), which is fine. The numerator is 2*1 -7 = -5, so the function is defined.

Take x=4, which is in x > 3. Denominator sqrt(16 -20 +6) = sqrt(2), same as before. Numerator is 8 -7 = 1, so defined.

What about x=2.5? Well, that's between 2 and 3, which we said is excluded. Let's check denominator: sqrt(6.25 -12.5 +6) = sqrt(-0.25). Wait, that's sqrt of negative number, which is not real. So that's why x between 2 and 3 is excluded.

Therefore, the domain is indeed all real numbers except between 2 and 3, inclusive. So written in interval notation, (-∞, 2) ∪ (3, ∞).

Therefore, I think that is the answer.

**Final Answer**
The domain of the function is \boxed{(-\infty, 2) \cup (3, \infty)}.

To determine the **domain** of the real-valued function
$$
f(x) = \frac{2x - 7}{\sqrt{x^2 - 5x + 6}},
$$
we analyze the conditions under which the function is defined. A function is defined if and only if the **denominator is not zero** and the **expression inside the square root is positive** (since square roots of negative numbers are not real).

---

### Step 1: Analyze the denominator

The denominator is
$$
\sqrt{x^2 - 5x + 6}.
$$
For the square root to be defined (i.e., produce a real number), the expression inside must be **non-negative**:
$$
x^2 - 5x + 6 \geq 0.
$$
However, since the square root appears in the **denominator**, the expression inside must be **strictly positive** to avoid division by zero. Hence, we need:
$$
x^2 - 5x + 6 > 0.
$$

---

### Step 2: Solve the inequality

Factor the quadratic:
$$
x^2 - 5x + 6 = (x - 2)(x - 3).
$$
So the inequality becomes:
$$
(x - 2)(x - 3) > 0.
$$
We solve this by identifying the **critical points** $ x = 2 $ and $ x = 3 $, and testing intervals around them.

- For $ x < 2 $, say $ x = 1 $:
  $$
  (1 - 2)(1 - 3) = (-1)(-2) = 2 > 0.
  $$
- For $ 2 < x < 3 $, say $ x = 2.5 $:
  $$
  (2.5 - 2)(2.5 - 3) = (0.5)(-0.5) = -0.25 < 0.
  $$
- For $ x > 3 $, say $ x = 4 $:
  $$
  (4 - 2)(4 - 3) = (2)(1) = 2 > 0.
  $$

Thus, the inequality is satisfied when $ x < 2 $ or $ x > 3 $.

---

### Step 3: Exclude points where the denominator is zero

The expression inside the square root is zero when $ x^2 - 5x + 6 = 0 $, i.e., when $ x = 2 $ or $ x = 3 $. At these points, the denominator becomes zero, which makes the function **undefined**. So these values must be **excluded** from the domain.

---

### Step 4: Write the domain

The domain includes all real numbers **excluding** the interval $ [2, 3] $, and in particular, $ x = 2 $ and $ x = 3 $, since the expression inside the square root becomes zero there. So the domain is:

$$
(-\infty, 2) \cup (3, \infty)
$$

---

### Final Answer

$$
\boxed{(-\infty, 2) \cup (3, \infty)}
$$

Solution: The function is defined when the value inside the square root is positive, i.e. we must have $x^2-5x+6>0$.  Factoring, we get $(x-3)(x-2)>0$. So either both factors in the left hand side are negative or they are both positive. They are both negative when $x<2$. They are both positive when $x>3$. So the domain of $f(x)$ is $x<2 \text{ or } x>3$, or $x \in \boxed{(-\infty, 2) \cup (3, \infty)}$ in interval notation.