We are given: A piecewise linear function defined by:
f(x) = { x if -1 ≤ x < 1, 2 - x if 1 ≤ x < 3 }
and f(x+4) = f(x) for all real x. So it's periodic with period 4. The graph depicts a sawtooth pattern (likely a triangle wave). The parabola x = 34 y^2 intersects the graph of f(x) at finitely many points. The sum of the y-coordinates of all these intersection points can be expressed in the form (a + b√c)/d, where a,b,c,d positive integers, a,b,d have gcd 1, c squarefree. Find a+b+c+d.

We need to compute the intersection points between the parabola x = 34 y^2 and the graph of the periodic function f(x). The parabola is a right-opening parabola, symmetric about x-axis, with vertex at (0,0). Since f(x) is defined for all x, and periodic with period 4, the graph consists of repeating "triangular" waves: from x=-1 to 1, f(x)=x (line slope 1), from 1 to 3, f(x)=2-x (slope -1), then repeats with period 4: so from x=3 to 5, f(x)=? Because f(x+4)=f(x), so for x in [3,5), we can write f(x)=f(x-4). Since x-4 in [-1,1) if x in [3,5)? Actually 3-4=-1, but careful: The original definition covers -1 ≤ x < 1 and 1 ≤ x < 3. So for any x, we subtract multiples of 4 to bring into [-1,3). But note that the interval [-1,3) has length 4, so it's exactly one period. So f is defined on [-1,3) as above, and then extended periodically. So for any x, let t = x - 4k where k is integer such that t ∈ [-1,3). Then f(x)=f(t). So the graph is a "triangle wave" with period 4, amplitude? At x=-1, f(-1)= -1? But note the definition: -1 ≤ x < 1 gives f(x)=x, so at x=-1, f(-1) = -1. At x=1, from first piece, f(1)=1? But note: at x=1, the first piece says x<1 so excludes 1; second piece says 1 ≤ x <3 gives f(1)=2-1=1. So f(1)=1. So the function is continuous: at x=1, both pieces give 1. At x=3, second piece gives f(3)=2-3=-1, but note the interval is 1≤x<3, so x=3 is excluded. However, periodicity gives f(3)=f(3-4)=f(-1)= -1? Actually f(3) would be defined by shifting: 3 is in [3,7) maybe? Since period 4, f(3) = f(3-4)=f(-1) = -1. And at x=-1, f(-1)=-1. So it's continuous at boundaries? At x=3, from left (1≤x<3) we approach -1, and from right (via periodicity) we have f(3)=f(-1)=-1, so continuous. So the function is a continuous triangle wave, with slope 1 on [-1,1] and slope -1 on [1,3], then repeats. The graph in the Asymptote code: they draw from -7 to 7 a sawtooth: blue lines: (-7,1) to (-5,-1) to (-3,1) to (-1,-1) to (1,1) to (3,-1) to (5,1) to (7,-1). So that matches: from x=-7 to -5, slope? Actually from (-7,1) to (-5,-1): that's slope (-2)/(2)= -1? Wait: change in y = -2, change in x = 2, slope -1. But from the definition, on intervals of length 2, the function goes from 1 to -1? Let's see: For period starting at -1? At x=-1, f=-1; at x=1, f=1; so it goes from -1 to 1 on [-1,1] (slope 1). Then from 1 to 3, it goes from 1 to -1 (slope -1). So the pattern: -1 at x=-1, 1 at x=1, -1 at x=3, 1 at x=5, etc. So the graph shows (-7,1) to (-5,-1) to (-3,1) to (-1,-1) to (1,1) to (3,-1) to (5,1) to (7,-1). So indeed the function values at odd integers (like -7, -5, -3, -1, 1, 3, 5, 7) alternate 1 and -1. So the function is symmetric: f(x+2) = -f(x)? Possibly. But anyway.

Now the parabola: x = 34 y^2. That's a parabola opening to the right, with vertex at (0,0). It is symmetric about the x-axis. Since x is a function of y, it's easier to think of y as independent variable. But we are intersecting with the graph of f(x), which is a function f: x -> y. So we need to find points (x, y) such that y = f(x) and also x = 34 y^2. So we have a system: y = f(x) and x = 34 y^2. So essentially we need to solve for (x, y) with y = f(x) and x = 34 y^2.

Since f is periodic, there will be infinitely many intersections? But the problem says "intersects the graph of f(x) at finitely many points". Because the parabola x = 34 y^2 is a curve that extends to infinite x as y grows. But f(x) is bounded between -1 and 1 (since its range is [-1,1] because it's triangle wave with amplitude 1). Indeed f(x) ∈ [-1,1] for all x. So y = f(x) is between -1 and 1. Then x = 34 y^2 ∈ [0, 34]. So the parabola only has points with x between 0 and 34 (since y^2 ≤ 1). So the intersection points are only for x in [0,34] and y ∈ [-1,1]. Since f(x) is periodic, there will be many x values where f(x) = y and x = 34y^2. But note that f(x) is defined for all x, but the parabola restricts x to 0..34. So we need to find all (x,y) such that y = f(x) and x = 34 y^2, with x in [0,34] (and y in [-1,1]). Since f is periodic with period 4, there may be multiple x values within [0,34] that map to the same y? Actually for a given y, the equation x = 34 y^2 gives a specific x. Then we need f(x) = y. So we need to check whether that specific x yields f(x) = y. So it's not like solving for y from f(x) and x = 34 y^2 simultaneously; we substitute x = 34 y^2 into f(x) = y. So we need to solve for y such that f(34 y^2) = y, with y ∈ [-1,1]. Then the intersection points are (34 y^2, y). And we need all such y that satisfy. Since f is periodic, there may be multiple solutions? But each y gives a unique x, so there is at most one intersection point for each y. But we need to solve the equation f(34 y^2) = y. Since f is not injective, there could be multiple y that satisfy? Actually for a given y, the equation is an equation in y. So we need to find all y such that f(34 y^2) = y. That's a single equation. So there might be several y values that satisfy. And the sum of the y-coordinates of all these intersection points is the sum of those y values.

Thus the problem reduces to solving: f(34 y^2) = y, where y ∈ ℝ, but since f outputs values in [-1,1], we need y ∈ [-1,1] as well. And also note that f is defined for all real x; we need to consider the periodic extension.

So we need to compute the set of y such that y = f(34 y^2). Because f is periodic with period 4, we can reduce x = 34 y^2 modulo 4 to the base interval [-1,3) (or any interval of length 4). However, careful: The definition f(x) = { x if -1 ≤ x < 1; 2-x if 1 ≤ x < 3 } and then f(x+4)=f(x). So for any x, we can write x = 4k + t, where k is integer (can be negative) and t ∈ [-1,3). Then f(x) = f(t) = t if t ∈ [-1,1) and = 2-t if t ∈ [1,3). But careful: At t=1, it belongs to second piece? Actually second piece includes 1, so f(1)=2-1=1, consistent with first piece's limit. So we can treat f(t) = t for t ∈ [-1,1] and = 2-t for t ∈ [1,3]? But t=1 included in both? Typically define piecewise: for -1 ≤ t ≤ 1, f(t)=t; for 1 ≤ t ≤ 3, f(t)=2-t; and at t=1, both give 1; at t=3, f(3)=2-3=-1? But t=3 is not in [1,3) but if we extend to 3? Since periodicity, f(3) = f(-1) = -1, but our base interval is [-1,3). Usually we take t ∈ [-1,3) and define f(t)=t for -1≤t<1, and f(t)=2-t for 1≤t<3. At t=3, it's not included, but we can use periodicity to treat t=3 as equivalent to -1. So better to treat the interval as [-1,3) with the definition, and for any x, let k = floor((x+1)/4)? Actually we want to map x to t = x - 4n where n is chosen so that t ∈ [-1,3). Typically, we can let n = floor((x+1)/4) or something.

We'll need to solve for y. Since y is between -1 and 1, 34 y^2 is between 0 and 34. So we need to evaluate f at that x. Since the period is 4, we can reduce x mod 4 to the interval [-1,3) (or maybe [0,4) but the definition uses [-1,3)). Let's define: Let t = (34 y^2) mod 4, but we need to bring it into [-1,3). Since modulo 4, numbers can be represented in [0,4) or [-1,3). It's easier to use the interval [-1,3) by subtracting appropriate multiples of 4. For any x, let k = floor((x+1)/4) such that t = x - 4k ∈ [-1,3). Then f(x) = f(t) = t if -1 ≤ t < 1; = 2 - t if 1 ≤ t < 3; and also treat t=3 as t=-1? Actually if t=3 exactly, then t=3 not in [-1,3), but we could define f(3) = f(-1) = -1. But since 34y^2 is a real number, it might equal 3 exactly? Possibly but unlikely. We'll handle if needed.

Thus we need to solve: for y ∈ [-1,1], let x = 34 y^2, compute t = x - 4k where k = floor((x+1)/4) (or equivalently, take x modulo 4 into [-1,3)), then f(x)= t if t ∈ [-1,1]; else f(x)=2-t if t∈[1,3). Then we require f(x)=y.

We need to find all y that satisfy. Since the parabola is symmetric, we might get symmetric solutions about y=0? Because if y is a solution, then -y? Let's check: f(34 (-y)^2) = f(34 y^2) = y? That would give f(34 y^2) = y. For -y, we need f(34 y^2) = -y. That would hold only if y=0 or if f(34 y^2)=0? Not generally. So not symmetric. However the parabola is symmetric in y, but f is not symmetric in x? Actually f(x) is odd? Let's check: f(x+2) = -f(x)? Because f(x+2) =? For x in [-1,1], f(x)=x, then x+2 in [1,3], f(x+2)=2-(x+2)= -x = -f(x). For x in [1,3], f(x)=2-x, then x+2 in [3,5], but using periodicity? f(x+2)=f(x+2-4)=f(x-2) =? For x in [1,3], x-2 in [-1,1], f(x-2)=x-2 = -(2-x)= -f(x). So indeed f(x+2) = -f(x). So the function is anti-symmetric with period 2. So f is an odd function with respect to shift by 2: f(x+2) = -f(x). This implies f(2) = -f(0)? f(0)=0, so f(2)=0. Also f(-x)? Not exactly. But note that f(x) is symmetric about the points (odd integers)? Actually it's a triangle wave: it's even about x=0? f(-x)? For x in [0,1], -x in [-1,0], f(-x) = -x, and f(x)=x, so f(-x) = -f(x). So f is odd on [-1,1]. For x in [1,2], -x in [-2,-1], but need periodicity: f(-x)=f(-x+4)=f(4-x)? Not straightforward. However the property f(x+2) = -f(x) is useful.

Now, we need to find y such that f(34 y^2) = y. Let's denote u = y. Then equation: f(34 u^2) = u. Since u ∈ [-1,1]. And note that 34 u^2 ∈ [0,34]. So we need to consider the values of f at those x.

Let’s try to solve by considering intervals of t = x mod 4. Since x = 34 u^2, we can compute the remainder when divided by 4. Let k = floor(x/4) maybe? But we need t ∈ [-1,3). Alternatively, we can define t = x - 4⌊(x+1)/4⌋. Since x≥0, we can also use t = x - 4⌊x/4⌋ gives t∈[0,4). But the definition uses [-1,3). So we can use t in [0,4) and then adjust: if t ∈ [0,1) then f = t? But the definition for -1≤t<1 gives f=t, so for t∈[0,1) it's the same; for t∈[1,3) gives f=2-t; for t∈[3,4) we need to subtract 4 to bring to [-1,0)? Since periodicity, f(t) = f(t-4). For t∈[3,4), t-4 ∈ [-1,0), then f(t)=t-4 (since t-4 in [-1,0) gives f = t-4). So we can also define f on [0,4) as: f(x) = x for 0≤x<1; = 2-x for 1≤x<3; = x-4 for 3≤x<4. That's consistent with periodic extension. Check: at x=3, f(3)=3-4=-1, good; at x=4, not included. So we can use t ∈ [0,4) as base. Then f(x) = f(t) where t = x mod 4 (taking remainder in [0,4)). So we'll use that.

Thus for x = 34 u^2, let t = 34 u^2 mod 4, i.e., t = 34u^2 - 4⌊34u^2/4⌋. Then f(x) = f(t) = t if 0≤t<1; = 2-t if 1≤t<3; = t-4 if 3≤t<4. (Note: at t=1, f=1; at t=3, f= -1; these are consistent.)

We need f(t)=u.

Now u ∈ [-1,1]. So we need to solve for u such that t = 34u^2 mod 4 gives f(t)=u.

Because 34u^2 is continuous and monotonic for u≥0? But u can be negative, but u^2 same. So it's symmetric in u. So we can consider u≥0 first, then include u<0? However f(t) is not symmetric in u because u appears as the right-hand side. But we can solve for u in [-1,1] directly.

Let’s denote v = u^2, so u = ±√v. But careful: u can be negative, so we have to consider both signs. However note that f(t) is a function of t, and t depends only on u^2, so for a given u^2, t is fixed. Then the equation becomes f(t) = u, which could have solutions for u positive or negative depending on f(t). Since f(t) ∈ [-1,1], it's possible that for some t, f(t) is positive, giving u positive; for some t, f(t) is negative, giving u negative. So we need to find all u in [-1,1] such that u = f(t) where t = 34 u^2 mod 4.

We can treat it as solving for u. Because u appears both sides, it's a fixed point type. Write u = f(34 u^2). Since f is periodic, we need to consider the various intervals for t.

We can break the possible values of t into three linear pieces: t ∈ [0,1): f(t)=t; t ∈ [1,3): f(t)=2-t; t ∈ [3,4): f(t)=t-4. Since 34u^2 ∈ [0,34], we will have t = 34u^2 mod 4. So t = 34u^2 - 4k for some integer k such that t ∈ [0,4). Then k = floor(34u^2/4). So k can be 0,1,...,8 because 34/4=8.5, so max floor = 8 when u^2=1 gives 34, floor(34/4)=8 (since 8*4=32, remainder 2). Actually 34/4=8.5, floor=8. So k ranges from 0 to 8 inclusive. So t = 34u^2 - 4k, with k = floor(34u^2/4). So we need to consider intervals of u that yield a specific k.

We can treat u ∈ [-1,1], but because u^2 is even, it's easier to consider u ≥ 0 first, and then handle negative u separately. However note that when u is negative, u^2 same as positive, so t is same, but u is negative. So if f(t) is positive, then u = f(t) would be positive, but u negative would not satisfy; if f(t) is negative, then u = f(t) would be negative, and a negative u could satisfy. So essentially, for a given u^2, there is a candidate u = f(t). That candidate is determined by u^2. Then we need to check whether u (the candidate) has the same square as u^2? Actually it's a self-consistency: u = f(34 u^2). So for a given u, we compute t from u, then f(t) gives some value; we need that value equals u. So it's an equation in u. Because t depends on u via u^2, and f(t) is then a number; we set equal to u. So we can solve for u.

We can treat u as unknown and consider the possible cases for t based on u. Since u ∈ [-1,1], u^2 ∈ [0,1], so 34u^2 ∈ [0,34]. So k = floor(34u^2/4) can be 0,1,...,8. But also note that u^2 small gives k=0; as u^2 increases, k increases. The boundaries where 34u^2/4 is an integer correspond to u^2 = 4k/34 = 2k/17. So u^2 = 2k/17 for k=0,...,8 gives boundaries? But 2k/17 for k=8 is 16/17 ≈ 0.941176; for k=8, 34u^2 = 32, so u^2=32/34=16/17≈0.941176. So that's within [0,1]. For k=9 would need u^2=36/34=18/17>1, not possible. So k ranges 0 to 8. But note: when 34u^2 = exactly 4k, then t=0? Actually t = 34u^2 - 4k = 0. Then t=0 belongs to [0,1) so f(0)=0. So that's fine. When 34u^2 = 4k+1? That gives t=1, which is boundary between [0,1) and [1,3); at t=1, f(1)=1 (both pieces). So okay. When 34u^2 = 4k+3, t=3, which is boundary between [1,3) and [3,4); at t=3, f(3)=? Using our [0,4) definition, for t=3, it's in [3,4) so f(3)=3-4=-1. Also from [1,3) piece, if we included 3, would be 2-3=-1. So consistent. So boundaries are okay.

Thus we can consider intervals for u^2 (or u) based on which k and which subinterval of t (0-1,1-3,3-4) we are in. Since u^2 is monotonic on u≥0, we can partition u≥0 into intervals determined by where 34u^2 lies relative to 4k, 4k+1, 4k+3 etc. But note that u can be negative; but u^2 same, so the intervals for u^2 are the same for positive and negative u. So the candidate f(t) will be the same for u and -u, because f(t) depends only on u^2. So if f(t) = some value v, then both u = v and u = -v would satisfy the equation? Not both; only the one that matches sign. So for a given u^2, f(t) yields a specific number w. Then the equation u = w holds if u = w. So the solutions are exactly those u for which u = f(34 u^2). So it's a fixed-point equation.

We can solve by considering the possible intervals for u (or u^2) and the corresponding expression for f(34 u^2) as a linear function of u^2. Then we set that equal to u and solve for u.

Let’s do that.

Define u ∈ ℝ, but we restrict to u ∈ [-1,1]. Let x = 34 u^2. Let k = floor(x/4). Write x = 4k + t, with t ∈ [0,4). Then f(x) = f(t) as defined:

If t ∈ [0,1): f(t)=t.
If t ∈ [1,3): f(t)=2-t.
If t ∈ [3,4): f(t)=t-4.

Now t = x - 4k = 34u^2 - 4k.

So f(x) =
Case A: 0 ≤ 34u^2 - 4k < 1 ⇒ f = 34u^2 - 4k.
Case B: 1 ≤ 34u^2 - 4k < 3 ⇒ f = 2 - (34u^2 - 4k) = 2 - 34u^2 + 4k.
Case C: 3 ≤ 34u^2 - 4k < 4 ⇒ f = (34u^2 - 4k) - 4 = 34u^2 - 4k - 4.

Now k = floor(34u^2/4). So for each u, the case is determined by the fractional part of x/4.

So the equation is:
u = f(x) =
either u = 34u^2 - 4k   (case A)
or u = 2 - 34u^2 + 4k   (case B)
or u = 34u^2 - 4k - 4   (case C).

We need to find u in [-1,1] that satisfy these with appropriate k and the interval condition for that case.

We can treat each possible k from 0 to 8 and solve the equation within the corresponding t-interval. Since u ∈ [-1,1], u^2 ≤1, so x ≤34. So k can be 0,...,8. For each k, the condition for case A is 0 ≤ 34u^2 - 4k < 1. That defines an interval for u^2: 4k/34 ≤ u^2 < (4k+1)/34. Similarly case B: 1 ≤ 34u^2 - 4k < 3 ⇒ (4k+1)/34 ≤ u^2 < (4k+3)/34. Case C: 3 ≤ 34u^2 - 4k < 4 ⇒ (4k+3)/34 ≤ u^2 < (4k+4)/34 = (4(k+1))/34. Note that u^2 max is 1, so for k=8, (4*8+3)/34 = (32+3)/34=35/34 >1, so case C for k=8 only part: (4*8+3)/34 = 35/34 >1, so only u^2 up to 1 is less than that bound? Actually the upper bound is (4k+4)/34 = (4k+4)/34. For k=8, (4*8+4)/34 = 36/34 ≈1.0588 >1, so the condition u^2<1.0588 is always true for u^2≤1. So the case C for k=8 would be u^2 ∈ [35/34,1) which is empty because 35/34≈1.0294 >1, so no u^2 in that range. So only cases A and B for k=8? Let's check: For k=8, x ∈ [32,36). But u^2 max is 1 gives x=34, so x ∈ [32,34]. So t = x-32 ∈ [0,2]. So t ∈ [0,2) which includes parts of case A (0≤t<1) and case B (1≤t<3). So indeed case C (t≥3) not relevant. So k=8 yields only cases A and B. For k=7: x ∈ [28,32); u^2 max 1 gives x=34, but k=7 only for x in [28,32), so for u^2 such that 34u^2∈[28,32). That is u^2 ∈ [28/34,32/34)=[14/17,16/17) ≈[0.8235,0.9412). And also possibly u^2 > 32/34? That would be k=8. So for k=7, u^2 ∈ [28/34,32/34). Within that, t ∈ [0,4) because x=4*7+t => t∈[0,4). Specifically t = x-28 ∈ [0,4). Then the intervals: case A for t∈[0,1) => u^2 ∈ [28/34,29/34); case B for t∈[1,3) => u^2 ∈ [29/34,31/34); case C for t∈[3,4) => u^2 ∈ [31/34,32/34). So all three cases possible for k=7. Similarly for lower k.

So we need to solve for u in each interval and each case.

Let’s do systematically.

We have u ∈ [-1,1]. For each u, u^2 determines k and case. We need to solve u = expression in terms of u^2. Since u appears also as u, we'll solve the equation.

Because u can be negative, we need to consider u as unknown, not u^2. For a given case and k, the equation becomes:

Case A: u = 34u^2 - 4k. (Equation A)
Case B: u = 2 - 34u^2 + 4k. (Equation B)
Case C: u = 34u^2 - 4k - 4. (Equation C)

These are quadratic equations in u. Let's rearrange:

Case A: 34u^2 - u - 4k = 0.
Case B: 34u^2 + u - (2+4k) = 0? Actually u = 2 - 34u^2 + 4k => bring terms: 34u^2 - u + ( -2 -4k?) Wait: u = 2 - 34u^2 + 4k => 34u^2 + u - (2+4k) = 0? Let's move: u = 2 - 34u^2 + 4k => bring all to one side: 34u^2 + u - 2 -4k = 0? Actually 34u^2 + u - 2 -4k = 0. Yes: 34u^2 + u - (2+4k)=0.
Case C: u = 34u^2 - 4k -4 => 34u^2 - u -4k -4 = 0 => 34u^2 - u - (4k+4)=0.

So each is a quadratic: 34u^2 - u - 4k =0 (A), 34u^2 + u - (4k+2)=0 (B), 34u^2 - u - (4k+4)=0 (C). Note that (C) is similar to (A) but with 4k+4 instead of 4k. (B) has +u.

We need to find real roots u in [-1,1] that also satisfy the interval condition for that case (i.e., that t = 34u^2 -4k lies in the appropriate range). Also note that k must equal floor(34u^2/4). But if we solve the quadratic for a given k, the resulting u might produce a different k (maybe not the one we assumed). So we need to verify consistency: for the candidate u, compute floor(34u^2/4) and ensure it equals k, and that t lies in the correct subinterval. So we need to solve each quadratic and then check the conditions.

Because u^2 appears, we might get two roots (positive and negative) from each quadratic. But note that u may be positive or negative; the equation involves u linearly. So for each case we solve quadratic.

Let's write them:

A: 34u^2 - u - 4k = 0 ⇒ u = [1 ± sqrt(1 + 4*34*4k)]/(2*34) = [1 ± sqrt(1 + 544k)]/68.
B: 34u^2 + u - (4k+2)=0 ⇒ u = [-1 ± sqrt(1 + 4*34*(4k+2))]/(2*34) = [-1 ± sqrt(1 + 544k + 272)]/68? Actually 4*34*(4k+2)=4*34*4k+4*34*2 = 544k + 272. So sqrt(544k+273)? Wait: 1 + 544k + 272 = 544k + 273. So u = [-1 ± sqrt(544k+273)]/68.
C: 34u^2 - u - (4k+4)=0 ⇒ u = [1 ± sqrt(1 + 4*34*(4k+4))]/68 = [1 ± sqrt(1 + 544k + 544)]/68 = [1 ± sqrt(544k+545)]/68.

We need u ∈ [-1,1]. Also note that the discriminant must be nonnegative, which it is for k≥0.

Now, k ranges from 0 to 8 (and maybe also negative? Since u^2≥0, k≥0). But also note that u can be negative; the ± choices give potentially negative u. For case A: u = (1 ± sqrt(D))/68. sqrt(D) >1 for k>=? For k=0, D=1 => sqrt=1, then u = (1±1)/68 => 0 and 2/68=1/34≈0.0294. Both nonnegative? Actually (1-1)/68=0, (1+1)/68=2/68=1/34. So both positive or zero. So case A yields nonnegative u (since 1 ± something). For k>0, sqrt(D)>1, so (1 - sqrt(D))/68 is negative? For k=1, D=1+544=545, sqrt≈23.336, then (1-23.336)/68 negative, magnitude about -22.336/68 ≈ -0.328. That's negative but maybe in [-1,1]. So case A can give negative u as well? Actually (1 - sqrt) negative. So case A yields two roots: one positive (1+sqrt)/68 >0, one negative (1-sqrt)/68 <0 (since sqrt>1). So both are possible. Case B: u = (-1 ± sqrt)/68. For k=0, D=273? Actually k=0: 544*0+273=273, sqrt≈16.522, then (-1+16.522)/68=15.522/68≈0.228; (-1-16.522)/68 negative large (-17.522/68≈ -0.258). So both could be in [-1,1] potentially. Case C: u = (1 ± sqrt)/68, similar to case A but with larger D.

So we need to consider all roots within [-1,1] that also satisfy the interval condition and k consistency.

Let's solve systematically. Since k ranges 0..8, we can compute approximate values and check conditions. But note: The sum of y-coordinates of all intersection points is asked as (a+b√c)/d. So there might be a finite number of solutions, and their sum can be expressed in that form. Possibly there are exactly two solutions? Or maybe four? The graph of f(x) is a sawtooth; the parabola x=34y^2 intersects it. The picture in the problem: they drew the parabola (red) as x=34y^2? Actually the Asymptote code draws: draw((0,0)..(1/4,sqrt(1/136))..(1/2,sqrt(1/68))..(0.75,sqrt(0.75/34))..(1, sqrt(1/34))..(2, sqrt(2/34))..(3, sqrt(3/34))..(4, sqrt(4/34))..(5, sqrt(5/34))..(6, sqrt(6/34))..(7, sqrt(7/34))..(8, sqrt(8/34)), red); and similarly the bottom half. So that's the parabola x=34y^2? Actually if x=34y^2, then y = ± sqrt(x/34). So points (x, sqrt(x/34)) for x from 0 to 8? They drew from x=0 to 8? Actually the red curve goes from (0,0) up to (8, sqrt(8/34)). That's the positive branch. And negative branch similarly. So the parabola is drawn only for x up to 8? But it extends further, but they only showed part. The blue sawtooth is drawn from -7 to 7. So there are intersections likely in the region where both exist. Since the parabola is symmetric, it will intersect the sawtooth at several points. The problem asks: "The parabola x = 34y^2 intersects the graph of f(x) at finitely many points." It says "finitely many points", probably because the parabola is bounded within x up to 34 (since y ∈ [-1,1] from f), so it's a finite number indeed. We need the sum of y-coordinates of all intersection points.

Given the sawtooth pattern, the intersections will occur where the parabola meets the linear segments of f. Since the parabola is symmetric about x-axis, and the sawtooth is symmetric? The sawtooth is odd about x=0? Actually f(-x) = -f(x)? Let's check: For x in [0,1], f(x)=x; for -x in [-1,0], f(-x) = -x = -f(x). For x in [1,2], f(x)=2-x; for -x in [-2,-1], using periodicity: f(-x)=f(-x+4)=f(4-x). For x=1.5, -1.5 => f(-1.5)=? -1.5+4=2.5, f(2.5)=2-2.5=-0.5, while f(1.5)=0.5, so f(-x) = -f(x)? Indeed 2.5 gives 2-2.5=-0.5, which is -f(1.5). So seems f is odd: f(-x) = -f(x) for all x? Check at x=2: f(2)=0, f(-2)=? -2+4=2, f(2)=0 => 0, so yes. So f is an odd function (since it's a triangle wave with symmetry). Indeed the sawtooth graph: from -7 to 7, it's odd symmetric about origin. The parabola x=34y^2 is symmetric about x-axis, i.e., if (x,y) is on parabola, (x,-y) also. So the intersection points will come in pairs symmetric about x-axis? Not necessarily, because f(x) is odd, so if (x,y) is an intersection (y = f(x)), then (x,-y) is on parabola but is it on f? For (x,-y) to be an intersection, we need -y = f(x). Since f is odd, f(x) = -f(-x)? Actually odd means f(-x) = -f(x). So if y = f(x), then -y = -f(x) = f(-x). So (x,-y) is not generally on f unless x = -x? That would require x=0. So intersections are not necessarily symmetric about x-axis. But note the parabola is symmetric about x-axis, so if (x,y) is intersection, then (x,-y) is also on parabola, but it may not be on f unless f(x) = -y, i.e., -y = f(x) ⇒ y = -f(x). That would be a different condition. So not symmetric. However the sawtooth is odd, so the graph of f has symmetry: f(-x) = -f(x). So the parabola might intersect at points that are symmetric in some way? Let's see later.

But we need to find all y such that y = f(34 y^2). That is the condition for intersection. Because for a given y, the parabola gives x = 34 y^2; then the point lies on the graph of f iff f(34 y^2)=y. So the y-coordinates of intersection points are exactly the solutions to that equation. So we need to find all y ∈ ℝ that satisfy, and then sum them. Since f is periodic and bounded, there will be finitely many solutions.

Let's solve the equation analytically by considering intervals for u = y.

We'll treat u as variable, and solve the quadratic equations for each case and k, then check consistency.

We need to be careful: For a given k and case, the equation yields potential u. But we also need that for that u, the value of t = 34u^2 - 4k satisfies the interval condition (e.g., for case A, 0 ≤ t < 1). And also that k = floor(34u^2/4). The latter is automatically satisfied if t ∈ [0,4) because k is chosen as floor(x/4). But if we solve for u from an equation that came from assuming a certain k and case, then we must verify that the computed u indeed yields t in that interval and that k indeed equals the floor. Since we used the relation t = 34u^2 - 4k, and the equation came from setting f(t)=u, we need to ensure t is in the appropriate region. So after solving, we need to check the condition on t.

We can solve each quadratic and then compute t and check.

Let's enumerate k from 0 to 8.

First, note that u ∈ [-1,1]. Since u^2 ≤1, 34u^2 ≤34. So k max = floor(34/4)=8. So k = 0,...,8.

We'll compute possible u roots from each quadratic for each k, then filter.

But note: For a given k, the quadratic might have two roots. However the interval condition might only allow one of them (or none) based on sign and magnitude of u? Also note that the interval condition involves u^2, so both positive and negative u have same u^2; thus if a root u is positive, its negative counterpart would have same u^2 but would not satisfy the same equation because equation is u = expression (which depends on u^2 only, not on sign of u). So for a given case and k, the quadratic equation is in u; it yields two roots that are generally not negatives of each other because the equation is not even in u (it has linear term). So the two roots are different numbers, possibly one positive and one negative. For case A: 34u^2 - u -4k=0. For a fixed k, the two roots are as given; they are not symmetric about 0. So we consider both.

Now let's compute exact expressions and then evaluate numeric approximations to check conditions. But we may need exact expressions for sum of y-coordinates in the form (a+b√c)/d. So we need to identify which solutions are valid and sum them exactly. Likely there are only a few solutions, maybe exactly two? Let's explore.

We'll go through k values systematically.

k = 0:
- Case A: condition: 0 ≤ 34u^2 < 1 => u^2 ∈ [0, 1/34) ≈ [0, 0.02941). Equation A: 34u^2 - u = 0 => 34u^2 = u => u(34u - 1)=0 => u=0 or u=1/34≈0.02941. Both are nonnegative. Check t = 34u^2 - 0 = 34u^2. For u=0: t=0, which is in [0,1) => ok. For u=1/34: u^2=1/34^2? Wait u=1/34, then u^2=1/1156≈0.000865, so t=34*(1/1156)=34/1156=1/34≈0.02941, which is <1, so ok. So both satisfy case A condition? But we also need that k=0 is indeed floor(34u^2/4). For u=1/34, 34u^2 = 34*(1/1156)=1/34≈0.02941, floor(0.02941/4)=0, yes. So both valid. So we have two solutions: u=0 and u=1/34. But u=0 is trivial? Check: y=0 gives x=0, f(0)=0, so (0,0) is intersection. u=1/34 gives y=1/34≈0.02941, x=34*(1/34)^2 = 34*(1/1156)=1/34≈0.02941? Wait x=34y^2 = 34*(1/1156)=1/34 ≈0.02941. So x is about 0.02941. f(0.02941)=0.02941 because 0≤x<1 => f(x)=x. So indeed f(x)=y. So that's also intersection.
- Case B for k=0: condition: 1 ≤ 34u^2 < 3 => u^2 ∈ [1/34, 3/34) ≈ [0.02941, 0.088235). Equation B: 34u^2 + u - 2 = 0 => u = [-1 ± sqrt(1+8*34? Actually compute discriminant: 1+4*34*2? Wait from earlier: D=544k+273, for k=0 => 273. sqrt(273)=√273. So u = (-1 ± √273)/68. √273 ≈ 16.522, so u ≈ (15.522)/68 ≈0.2283, and (-17.522)/68 ≈ -0.2577. Both are in [-1,1]. But we need u^2 to be in [1/34, 3/34). For u≈0.2283, u^2≈0.05214, which is between 0.02941 and 0.088235 => yes. For u≈ -0.2577, u^2≈0.0664, also within that range? 0.0664 is between 0.02941 and 0.088235 => yes. So both u satisfy the u^2 condition? However we also need to check that for these u, the resulting t = 34u^2 - 4k = 34u^2 lies in [1,3). Since u^2 is in that interval, t = 34u^2 will be in [1,3) exactly. So both satisfy the case B condition. Also k=0 because floor(34u^2/4)=0 since 34u^2<3 <4, so floor=0. So both are valid solutions. So we get two more: u ≈ 0.2283 and u ≈ -0.2577.
- Case C for k=0: condition: 3 ≤ 34u^2 < 4 => u^2 ∈ [3/34, 4/34)= [0.088235, 0.117647). Equation C: 34u^2 - u - 4 = 0 => u = [1 ± sqrt(1+544*0+544? Actually D=544k+545=545)]/68 = (1 ± √545)/68. √545≈23.336, so u ≈ (24.336)/68≈0.358, and (1-23.336)/68≈ -22.336/68≈ -0.3285. Check u^2: for u≈0.358, u^2≈0.1282, which is >0.117647? Actually 0.1282 > 0.117647, so not in [0.088235,0.117647). For u≈ -0.3285, u^2≈0.1079, which is between 0.088235 and 0.117647? 0.1079 is in that range? Yes, 0.1079 ∈ [0.088235,0.117647). So the negative root u≈ -0.3285 has u^2 in the interval, while positive root u≈0.358 has u^2 too high. So we need to check both: For u≈ -0.3285, u^2≈0.1079, then t=34u^2 ≈3.6686? Actually 34*0.1079 ≈ 3.6686, which is in [3,4) indeed. So that satisfies case C condition. Also k=0 because floor(34u^2/4)= floor(3.6686/4)=0? Wait 3.6686/4 = 0.91715, floor=0, yes. So valid. For u≈0.358, u^2≈0.1282, t=34*0.1282≈4.3588, which is not in [3,4) because it's ≥4. Actually 4.3588 is >=4, so t would be 4.3588 - 4*1? Wait careful: For k=0, we assumed k=0, but if 34u^2 is 4.3588, then floor(34u^2/4) would be 1, not 0. So that u would not correspond to k=0. So it fails the k consistency. So only negative root qualifies for k=0 case C. So we get one more solution: u ≈ -0.3285.

So for k=0, we have solutions: u = 0, 1/34, two from case B (both positive and negative), and one from case C (negative). That gives total: u=0, u=1/34, u≈0.2283, u≈ -0.2577, u≈ -0.3285. But note that u=0 is also from case A. So total 5 solutions for k=0? However we should verify that the positive root from case B (0.2283) is indeed distinct from 1/34 (0.0294) and 0. So that's fine. Also negative root from case B (-0.2577) distinct from negative from case C (-0.3285). So k=0 yields 5 solutions.

But wait, are there any other k values? For k=0, u^2 up to 1? Actually for k=0, the intervals cover u^2 from 0 to 4/34 ≈0.117647. So only small u^2. So these are the only solutions for u^2 less than 0.117647. But there are larger u^2 values that correspond to k=1,2,... up to 8. So we need to examine all k.

We should systematically go through k=0 to 8 and find all u satisfying the conditions. Since u is between -1 and 1, u^2 ≤1. For each k, the intervals for u^2 are as described. The total set of u^2 values is partitioned into intervals of length 1/34? Actually each k gives intervals: [4k/34, (4k+1)/34) for case A, [(4k+1)/34, (4k+3)/34) for case B, [(4k+3)/34, (4k+4)/34) for case C. For k=8, case C interval [(4*8+3)/34, (4*8+4)/34) = [35/34, 36/34) which is >1, so no u^2. So only k=0..7 have case C possibly, and k=8 only A and B.

Now we need to solve each quadratic and check that the resulting u lies in the appropriate u^2 interval and also that k matches floor(34u^2/4). Since we derived the equation from assuming the case and k, the u^2 interval condition ensures that t is in the correct range. However, the quadratic might give two roots; we need to check which of them satisfy u^2 within that interval and also that the floor condition holds (which is equivalent to t∈[0,4) for that k, i.e., u^2 in [4k/34, (4k+4)/34) basically). Since the intervals for the cases are subsets of that, we only need to check the specific case interval. Also note that the quadratic equation might have roots that fall outside [-1,1], but we can ignore those.

Thus we need to compute all roots and filter.

We can compute exact expressions for u for each case and k, and then evaluate whether u^2 lies in the appropriate interval. Because the expressions involve sqrt(D) where D depends on k. The sum of y-coordinates will likely be a combination of these roots. Possibly many of these roots are not rational in the sense of the final sum; they may combine to a neat expression. Let's compute symbolic.

Let’s denote for each k and case:

Case A: u = (1 ± √(1+544k))/68.
Case B: u = (-1 ± √(544k+273))/68.
Case C: u = (1 ± √(544k+545))/68.

We need to determine which of these ± yield u in the appropriate interval and also satisfy u^2 in the corresponding subinterval.

We can compute approximate numeric values for k=0 to 8 and check. Then we can sum all valid y values. Since the answer is to be expressed as (a+b√c)/d, the sum likely simplifies nicely.

Let's compute exact expressions for the roots that are valid. Perhaps the sum of all valid u values might be something like 0? Possibly many cancel? Let's examine.

We'll list all k from 0 to 8 and compute candidates.

But note: The parabola x=34y^2 also intersects the graph of f(x) at finitely many points; the sum of y-coordinates is asked. Might be that we need to include all intersection points, which could be many. But from the picture, the sawtooth and parabola: The parabola opens rightwards, the sawtooth is periodic with amplitude 1. The parabola is narrow; for small y, x is small; for y near 1, x=34. So the parabola will intersect each "tooth" maybe multiple times? Let's think. The sawtooth f(x) for x from 0 to 34: it goes up and down many times. f(x) is periodic with period 4, so over [0,34] there are 34/4 = 8.5 periods? Actually period 4, so from 0 to 34 inclusive, there are 8 full periods? 4*8=32, then from 32 to 34, it's part of a period. So there are many linear segments. The parabola x=34y^2 is a function x = g(y). Since y = f(x) is a function, we are solving y = f(34y^2). This is a fixed point equation. For each y, we compute x and check f(x). Since f is piecewise linear, the equation becomes piecewise quadratic. There will be solutions for each "tooth" where the parabola crosses the linear segments. Possibly there are many solutions, maybe 2 per period? Let's approximate: For a given y, x = 34 y^2. For small y, x is small, f(x) ≈ x (since for 0≤x<1, f(x)=x). So equation y ≈ 34y^2 gives y(1-34y)=0 => y=0 or y=1/34. So that gives two solutions near 0. For larger y, x gets larger, and f(x) will be on various linear segments. Because f(x) is periodic, we can think of reducing x mod 4. For each period, there are two linear segments: one with slope 1 (for t∈[0,1]) and one with slope -1 (for t∈[1,3]) and then for t∈[3,4] slope 1 again but with negative offset? Actually f(t)=t-4 for t∈[3,4) which is slope 1 but shifted down. So overall, the function f(x) is piecewise linear with slope ±1, and values range from -1 to 1. So the equation y = f(34y^2) becomes y = (34y^2 - 4k) for some k when t in [0,1) or t in [3,4) (both slope 1 but with different intercepts), or y = 2 - (34y^2 - 4k) = 2 - 34y^2 + 4k for t in [1,3) (slope -1). So essentially we have a set of quadratic equations: 34y^2 - y - 4k = 0 (for t in [0,1) or t in [3,4) with different k? Actually t in [3,4) gave 34y^2 - y - (4k+4)=0, which is same form but with k+1? Wait note: For t in [3,4), we have f(t)=t-4, and t = 34y^2 - 4k, so equation y = 34y^2 - 4k -4 => 34y^2 - y - (4k+4)=0. That's like the case A but with k' = k+1? Because case A had 34y^2 - y - 4k =0. So if we set k' = k+1, then 34y^2 - y - 4k' =0 would correspond to t in [0,1) for that k'? Actually careful: For case A with k, t = 34y^2 - 4k ∈ [0,1). For case C with k, t = 34y^2 - 4k ∈ [3,4). If we shift index: For case C with k, we have equation 34y^2 - y - 4k -4 =0, which is 34y^2 - y - 4(k+1)=0. So if we let k' = k+1, then this equation is same as case A for k'. However, the interval condition would be different: For case A with k', we require t' = 34y^2 - 4k' ∈ [0,1). But from case C with k, t = 34y^2 - 4k ∈ [3,4). Then 34y^2 - 4(k+1) = t-4 ∈ [-1,0). So that would correspond to case A with k' but t' ∈ [-1,0) which is not in [0,1) but could be considered using periodic extension? Actually our base interval [0,4) representation doesn't include negative t; we use t ∈ [0,4) with f(t)=t-4 for t∈[3,4). So the transformation to k' = k+1 gives t' = t-4 ∈ [-1,0). In that representation, we would have f(t') = t' (since t' ∈ [-1,0) fits the original definition? Actually original definition for [-1,1) gives f(t')=t'. So indeed f(t') = t' = t-4, which matches f(t) = t-4. So it's consistent that case C for k is equivalent to case A for k' = k+1 but with t' ∈ [-1,0). In our [0,4) representation, we treat t' as negative, but we can also handle it by shifting. So it's fine to consider case A and case C separately as they correspond to different ranges of t.

Thus we have three types: slope 1 with intercept -4k (case A), slope -1 with intercept 2+4k? Actually case B: y = 2 - 34y^2 + 4k => rearranged to 34y^2 + y - (4k+2)=0. This is quadratic with +y term.

Now, we need to find all y satisfying for some integer k. The solutions might be numerous. However note that the parabola x=34y^2 is a function of y; for each y, x is determined. The graph of f(x) is periodic and linear on intervals. The equation y = f(34y^2) can be thought as y = f(34y^2). For a given y, we can compute 34y^2 mod 4, and then f is linear. So the equation is piecewise quadratic. Since the function f is piecewise linear, the equation becomes quadratic in y on each interval where the linear expression holds. So there will be at most two solutions per linear interval of t. But note that the intervals for t are determined by y^2 (since t = 34y^2 mod 4). As y varies from -1 to 1, y^2 goes from 0 to 1, and 34y^2 goes from 0 to 34. As y^2 increases, the value of t = 34y^2 - 4k cycles through 0 to 4 repeatedly. So the mapping y -> t is not monotonic because y^2 is even. But we can treat y ≥0 and y<0 separately because y^2 is same. However the equation involves y itself, not just y^2. So the sign matters.

We can attempt to solve by considering y in intervals of y (not y^2) directly. Since the equation is y = f(34y^2), and f is periodic with period 4, we can write 34y^2 = 4k + t, t∈[0,4). Then f = t for t∈[0,1), 2-t for t∈[1,3), t-4 for t∈[3,4). Then we have y = expression in terms of y^2 and k. Since k = floor(34y^2/4). So for each y, we can compute k and case. So it's a self-consistent equation.

We can try to solve by considering all possible k and case and solving quadratic. Then check which roots satisfy the interval condition.

Given that there are only 9 possible k (0..8), and each case yields up to 2 roots, we have at most 3*9*2 = 54 candidate roots, but many will be invalid because u^2 not in interval or u out of range. However, many of these roots might be valid. We need to count them and sum.

Let's compute numerically for each k and case, and list all u in [-1,1] that satisfy the interval condition. Then sum them.

We can compute exact expressions for each valid u. Then the sum may simplify.

We'll go k=0..8.

First, compute approximate values for each candidate using formulas, then check conditions.

We'll compute exact sqrt expressions where needed.

Define:

For case A: u = (1 ± √(1+544k))/68.
For case B: u = (-1 ± √(544k+273))/68.
For case C: u = (1 ± √(544k+545))/68.

Now for each k, compute the discriminant D_A(k)=1+544k, D_B(k)=544k+273, D_C(k)=544k+545.

We'll compute numeric approximations for k=0..8:

k=0:
D_A=1, sqrt=1 → u = (1±1)/68 = 0, 1/34≈0.0294118.
D_B=273, sqrt≈16.5227 → u = (-1±16.5227)/68 → (15.5227)/68≈0.2283, (-17.5227)/68≈ -0.2577.
D_C=545, sqrt≈23.345? Actually sqrt(545) ≈ 23.345? Let's compute: 23^2=529, 24^2=576, so 545-529=16, sqrt≈23.336? Actually 23.336^2 = 544.6? Let's do more precise: 23.345^2=545.0? I'll compute: 23.345^2 = (23.345*23.345) ≈ 545.0? I'll use high precision: 23.345*23.345 = (23.345^2) = 545.0? Let's compute: 23.34^2 = 544.7556, 23.35^2=545.2225, so sqrt(545) ≈ 23.345. We'll get approximate: 23.345. Then u = (1±23.345)/68 → (24.345)/68≈0.3580, (-22.345)/68≈ -0.3286. So we have u≈0.3580 and -0.3286. Check intervals:
Case A interval: u^2 ∈ [0, 1/34≈0.02941). For u=0: u^2=0 ok; u=1/34≈0.02941: u^2=1/1156≈0.000865, ok.
Case B interval: u^2 ∈ [1/34≈0.02941, 3/34≈0.088235). For u≈0.2283: u^2≈0.05214, ok; u≈-0.2577: u^2≈0.0664, ok.
Case C interval: u^2 ∈ [3/34≈0.088235, 4/34≈0.117647). For u≈-0.3286: u^2≈0.1080, ok; for u≈0.3580: u^2≈0.1282, which is >0.117647, so not in interval. Also note that for u≈0.3580, t=34u^2≈34*0.1282=4.3588, floor=1, so k would be 1, not 0. So fails k=0 assumption. So valid for k=0: u=0, u=1/34, u= ( -1+√273)/68? Actually u_B+ = (-1+√273)/68 ≈0.2283, u_B- = (-1-√273)/68 ≈ -0.2577, and u_C- = (1-√545)/68 ≈ -0.3286. So total 5 valid for k=0.

But note that u=0 might also appear from other k? Possibly later, but u=0 yields u^2=0, which for k>0 would require 34*0=0, floor(0/4)=0, so only k=0. So it's unique.

Now k=1:
Compute intervals:
First, u^2 range for k=1: overall t = 34u^2 - 4 ∈ [0,4) because k=1 corresponds to 34u^2 ∈ [4,8). So u^2 ∈ [4/34≈0.117647, 8/34≈0.235294). Subintervals:
Case A: t ∈ [0,1) => 34u^2 -4 ∈ [0,1) => u^2 ∈ [4/34, 5/34) ≈ [0.117647, 0.147059).
Case B: t ∈ [1,3) => u^2 ∈ [5/34, 7/34) ≈ [0.147059, 0.205882).
Case C: t ∈ [3,4) => u^2 ∈ [7/34, 8/34) ≈ [0.205882, 0.235294).

Now compute roots:
D_A(1)=1+544=545, sqrt≈23.345, u = (1±23.345)/68 => u≈0.3580, -0.3286 (same as before but with k=1? Actually these are the same numbers as case C for k=0? But now they are candidates for case A with k=1). So u≈0.3580 and -0.3286. Check u^2: For u=0.3580, u^2≈0.1282, which lies in [0.117647,0.147059)? 0.1282 is between 0.117647 and 0.147059, yes. For u=-0.3286, u^2≈0.1080, which is less than 0.117647, so not in interval. So only u= (1+√545)/68 ≈0.3580 qualifies for case A with k=1. Also note that u=0.3580 positive, and t = 34u^2 -4 = 34*0.1282 -4 ≈4.3588-4=0.3588? Wait compute exactly: 34u^2 = 34*( (1+√545)/68 )^2? Let's compute later. But numeric: 34*(0.3580)^2 ≈ 34*0.128164 = 4.3576, minus 4 = 0.3576, which is in [0,1). So ok.

Case B for k=1: D_B(1)=544+273=817, sqrt(817) ≈ 28.583? 28.583^2=816.9, sqrt≈28.583. u = (-1±√817)/68. √817 ≈ 28.583. So u ≈ (27.583)/68≈0.4056, and (-29.583)/68≈ -0.4350. Check u^2: for u≈0.4056, u^2≈0.1645; interval for case B: u^2∈[5/34≈0.147059,7/34≈0.205882). 0.1645 is in that range, yes. For u≈ -0.4350, u^2≈0.1892, also in range? 0.1892 ∈ [0.147059,0.205882), yes. So both appear valid? But we need also that t = 34u^2 -4 ∈ [1,3). For u=0.4056, 34u^2≈34*0.1645=5.593, minus4=1.593, in [1,3). For u=-0.4350, 34u^2≈34*0.1892=6.4328, minus4=2.4328, in [1,3). So both valid. So two solutions for case B, k=1: u = (-1+√817)/68 ≈0.4056, and u = (-1-√817)/68 ≈ -0.4350.

Case C for k=1: D_C(1)=544+545=1089, sqrt(1089)=33 exactly! Because 33^2=1089. Nice! So u = (1±33)/68 = (34)/68=1/2=0.5, and (1-33)/68 = -32/68 = -8/17? Actually -32/68 simplifies to -8/17 ≈ -0.470588. So u = 0.5 and u = -8/17 ≈ -0.470588. Check interval: case C for k=1 requires u^2 ∈ [7/34, 8/34) ≈ [0.205882, 0.235294). For u=0.5, u^2=0.25, which is >0.235294? 0.25 > 0.235294, so not in interval. For u=-8/17 ≈ -0.470588, u^2 = (64/289) ≈ 0.221453. 0.221453 is between 0.205882 and 0.235294? Yes, 0.221453 ∈ [0.205882,0.235294). So u = -8/17 qualifies. Also check t = 34u^2 -4 = 34*(64/289) -4 = (34*64)/289 -4 = (2176/289) -4 = 2176/289 - 1156/289 = 1020/289 ≈ 3.529? Actually 1020/289 ≈ 3.529, which is in [3,4). So valid. So case C yields u = -8/17 for k=1.

Thus for k=1, valid solutions: from case A: u = (1+√545)/68 (positive), from case B: u = (-1+√817)/68 and (-1-√817)/68 (both? one positive one negative), from case C: u = -8/17 (negative). Also note: The negative root from case A (1-√545)/68 ≈ -0.3286 did not satisfy interval (u^2 too low). So total for k=1: 1 (case A) + 2 (case B) + 1 (case C) = 4 solutions. But careful: Are there any from case A with negative root? Already excluded. So 4 solutions.

k=2:
u^2 range: 34u^2 ∈ [8,12) => u^2 ∈ [8/34≈0.235294,12/34≈0.352941). Subintervals:
Case A: t∈[0,1) => u^2∈[8/34,9/34) ≈ [0.235294,0.264706)
Case B: t∈[1,3) => u^2∈[9/34,11/34) ≈ [0.264706,0.323529)
Case C: t∈[3,4) => u^2∈[11/34,12/34) ≈ [0.323529,0.352941)

Compute roots:
D_A(2)=1+544*2=1+1088=1089, sqrt=33. So u = (1±33)/68 = 34/68=0.5, and -32/68=-8/17≈ -0.470588. Check intervals for case A: u^2 for u=0.5 is 0.25, which lies in [0.235294,0.264706)? 0.25 is within, yes. So u=0.5 qualifies for case A, k=2. For u=-8/17, u^2≈0.221453, which is <0.235294, so not in interval. So case A yields u=0.5 (positive).
Case B: D_B(2)=544*2+273=1088+273=1361, sqrt(1361)? 36^2=1296, 37^2=1369, so sqrt(1361) ≈ 36.89? Actually 36.89^2=1360.9, so sqrt≈36.89. Compute exact? 1361 is not a perfect square? 1361 = 1361. Check: 37^2=1369, 36^2=1296, so not square. So u = (-1±√1361)/68. Approx: √1361 ≈ 36.89. So u ≈ (35.89)/68≈0.5278, and (-37.89)/68≈ -0.5572. Check u^2: 0.5278^2≈0.2786; interval for case B: u^2∈[0.264706,0.323529). 0.2786 is within, yes. For u≈ -0.5572, u^2≈0.3105, which is also within that interval? 0.3105 ∈ [0.264706,0.323529), yes. So both qualify? But need to check t = 34u^2 -8 ∈ [1,3). For u=0.5278, 34u^2≈34*0.2786=9.4724, minus8=1.4724, ok. For u=-0.5572, 34u^2≈34*0.3105=10.557, minus8=2.557, ok. So both valid. So case B yields two solutions: u = (-1+√1361)/68 (positive) and u = (-1-√1361)/68 (negative).
Case C: D_C(2)=544*2+545=1088+545=1633, sqrt(1633)? 40^2=1600, 41^2=1681, so sqrt≈40.41? 40.4^2=1632.16, 40.41^2=1633.0? approximate. So u = (1±√1633)/68. Approx: √1633≈40.41, so u≈ (41.41)/68≈0.6090, and (1-40.41)/68≈ -39.41/68≈ -0.5796. Check interval for case C: u^2∈[0.323529,0.352941). For u≈0.6090, u^2≈0.3709, which is >0.352941, so not in interval. For u≈ -0.5796, u^2≈0.3359, which is within [0.323529,0.352941)? 0.3359 is between, yes. So u = (1-√1633)/68 ≈ -0.5796 qualifies. Also need to check t = 34u^2 -8 ∈ [3,4). For u=-0.5796, 34u^2≈34*0.3359=11.4206, minus8=3.4206, in [3,4). So valid. So case C yields one solution: u = (1-√1633)/68 (negative).

Thus k=2 yields: case A: u=0.5; case B: two solutions; case C: one negative solution. Total 1+2+1=4 solutions.

But note: u=0.5 appears also from k=1 case C? Actually k=1 case C gave u=0.5 but was invalid because u^2 too high for that interval. Here it's valid for k=2 case A. So u=0.5 is a solution. Good.

k=3:
u^2 range: 34u^2 ∈ [12,16) => u^2 ∈ [12/34≈0.352941,16/34≈0.470588). Subintervals:
Case A: u^2∈[12/34,13/34) ≈ [0.352941,0.382353)
Case B: u^2∈[13/34,15/34) ≈ [0.382353,0.441176)
Case C: u^2∈[15/34,16/34) ≈ [0.441176,0.470588)

Compute roots:
D_A(3)=1+544*3=1+1632=1633, sqrt=√1633≈40.41. u = (1±√1633)/68 ≈ (41.41)/68≈0.6090, and (1-40.41)/68≈ -0.5796. Check interval for case A: u^2 for u≈0.6090 is 0.3709, which lies in [0.352941,0.382353)? 0.3709 is within, yes. For u≈ -0.5796, u^2≈0.3359, which is less than 0.352941, so not. So case A yields u= (1+√1633)/68 ≈0.6090 (positive).
Case B: D_B(3)=544*3+273=1632+273=1905, sqrt(1905)? 43^2=1849, 44^2=1936, so sqrt≈43.63? 43.63^2=1904.9. So u = (-1±√1905)/68 ≈ (42.63)/68≈0.6270, and (-44.63)/68≈ -0.6563. Check u^2: 0.6270^2≈0.3931; interval for case B: [0.382353,0.441176). 0.3931 is within, yes. For u≈ -0.6563, u^2≈0.4307, also within [0.382353,0.441176)? 0.4307 is within, yes. So both qualify? Check t: 34u^2 -12. For u=0.6270, 34*0.3931=13.3654, minus12=1.3654, ok. For u=-0.6563, 34*0.4307=14.6438, minus12=2.6438, ok. So both valid.
Case C: D_C(3)=544*3+545=1632+545=2177, sqrt(2177)? 46^2=2116, 47^2=2209, so sqrt≈46.66? 46.66^2=2177.0? approximate. u = (1±√2177)/68 ≈ (47.66)/68≈0.7009, and (1-46.66)/68≈ -45.66/68≈ -0.6715. Check interval for case C: u^2∈[0.441176,0.470588). For u≈0.7009, u^2≈0.4913, >0.470588, not in. For u≈ -0.6715, u^2≈0.4509, which is in [0.441176,0.470588)? 0.4509 is within, yes. So u = (1-√2177)/68 ≈ -0.6715 qualifies. So case C yields one negative solution.

Thus k=3 yields: case A: u≈0.6090; case B: two; case C: one negative. Total 4.

k=4:
u^2 range: 34u^2 ∈ [16,20) => u^2 ∈ [16/34≈0.470588,20/34≈0.588235). Subintervals:
Case A: u^2∈[16/34,17/34) ≈ [0.470588,0.5) (since 17/34=0.5 exactly)
Case B: u^2∈[17/34,19/34) ≈ [0.5,0.5588235) (19/34≈0.5588235)
Case C: u^2∈[19/34,20/34) ≈ [0.5588235,0.588235)

Compute roots:
D_A(4)=1+544*4=1+2176=2177, sqrt=√2177≈46.66. u = (1±√2177)/68 ≈ (47.66)/68≈0.7009, and (1-46.66)/68≈ -0.6715. Check interval for case A: u^2∈[0.470588,0.5). For u≈0.7009, u^2≈0.4913, which lies in [0.470588,0.5)? Yes, 0.4913 is between 0.470588 and 0.5. For u≈ -0.6715, u^2≈0.4509, which is <0.470588, so not. So case A yields u= (1+√2177)/68 ≈0.7009 (positive).
Case B: D_B(4)=544*4+273=2176+273=2449, sqrt(2449)? 49^2=2401, 50^2=2500, sqrt≈49.48? 49.48^2=2448.6. So u = (-1±√2449)/68 ≈ (48.48)/68≈0.7129, and (-50.48)/68≈ -0.7424. Check u^2: 0.7129^2≈0.5082; interval for case B: [0.5,0.5588235). 0.5082 is within, yes. For u≈ -0.7424, u^2≈0.5512, also within [0.5,0.5588235)? 0.5512 is within, yes. So both qualify? Check t: 34u^2 -16. For u=0.7129, 34*0.5082=17.2788, minus16=1.2788, ok. For u=-0.7424, 34*0.5512=18.7408, minus16=2.7408, ok. So both valid.
Case C: D_C(4)=544*4+545=2176+545=2721, sqrt(2721)? 52^2=2704, 53^2=2809, sqrt≈52.17? 52.17^2=2721.0? approximate. u = (1±√2721)/68 ≈ (53.17)/68≈0.7819, and (1-52.17)/68≈ -51.17/68≈ -0.7525. Check interval for case C: u^2∈[0.5588235,0.588235). For u≈0.7819, u^2≈0.6114, >0.588235, not in. For u≈ -0.7525, u^2≈0.5663, which is within [0.5588235,0.588235)? 0.5663 is between, yes. So u = (1-√2721)/68 ≈ -0.7525 qualifies. So case C yields one negative solution.

Thus k=4 yields 4 solutions.

k=5:
u^2 range: 34u^2 ∈ [20,24) => u^2 ∈ [20/34≈0.588235,24/34≈0.705882). Subintervals:
Case A: u^2∈[20/34,21/34) ≈ [0.588235,0.617647)
Case B: u^2∈[21/34,23/34) ≈ [0.617647,0.676471)
Case C: u^2∈[23/34,24/34) ≈ [0.676471,0.705882)

Compute roots:
D_A(5)=1+544*5=1+2720=2721, sqrt=√2721≈52.17. u = (1±√2721)/68 ≈ (53.17)/68≈0.7819, and (1-52.17)/68≈ -0.7525. Check interval for case A: u^2∈[0.588235,0.617647). For u≈0.7819, u^2≈0.6114, which lies in that interval? 0.6114 is between 0.588235 and 0.617647, yes. For u≈ -0.7525, u^2≈0.5663, which is <0.588235, so not. So case A yields u= (1+√2721)/68 ≈0.7819 (positive).
Case B: D_B(5)=544*5+273=2720+273=2993, sqrt(2993)? 54^2=2916, 55^2=3025, sqrt≈54.71? 54.71^2=2993.0? approximate. u = (-1±√2993)/68 ≈ (53.71)/68≈0.7899, and (-55.71)/68≈ -0.8193. Check u^2: 0.7899^2≈0.6239; interval for case B: [0.617647,0.676471). 0.6239 is within, yes. For u≈ -0.8193, u^2≈0.6713, also within [0.617647,0.676471)? 0.6713 is within, yes. So both qualify? Check t: 34u^2 -20. For u=0.7899, 34*0.6239=21.2126, minus20=1.2126, ok. For u=-0.8193, 34*0.6713=22.8242, minus20=2.8242, ok. So both valid.
Case C: D_C(5)=544*5+545=2720+545=3265, sqrt(3265)? 57^2=3249, 58^2=3364, sqrt≈57.14? 57.14^2=3265.0? approximate. u = (1±√3265)/68 ≈ (58.14)/68≈0.8550, and (1-57.14)/68≈ -56.14/68≈ -0.8256. Check interval for case C: u^2∈[0.676471,0.705882). For u≈0.8550, u^2≈0.7310, >0.705882, not in. For u≈ -0.8256, u^2≈0.6816, which is within [0.676471,0.705882)? 0.6816 is between, yes. So u = (1-√3265)/68 ≈ -0.8256 qualifies. So case C yields one negative solution.

Thus k=5 yields 4 solutions.

k=6:
u^2 range: 34u^2 ∈ [24,28) => u^2 ∈ [24/34≈0.705882,28/34≈0.823529). Subintervals:
Case A: u^2∈[24/34,25/34) ≈ [0.705882,0.735294)
Case B: u^2∈[25/34,27/34) ≈ [0.735294,0.794118)
Case C: u^2∈[27/34,28/34) ≈ [0.794118,0.823529)

Compute roots:
D_A(6)=1+544*6=1+3264=3265, sqrt=√3265≈57.14. u = (1±√3265)/68 ≈ (58.14)/68≈0.8550, and (1-57.14)/68≈ -0.8256. Check interval for case A: u^2∈[0.705882,0.735294). For u≈0.8550, u^2≈0.7310, which lies in that interval? 0.7310 is between 0.705882 and 0.735294, yes. For u≈ -0.8256, u^2≈0.6816, which is <0.705882, so not. So case A yields u= (1+√3265)/68 ≈0.8550 (positive).
Case B: D_B(6)=544*6+273=3264+273=3537, sqrt(3537)? 59^2=3481, 60^2=3600, sqrt≈59.47? 59.47^2=3537.0? approximate. u = (-1±√3537)/68 ≈ (58.47)/68≈0.8599, and (-60.47)/68≈ -0.8893. Check u^2: 0.8599^2≈0.7394; interval for case B: [0.735294,0.794118). 0.7394 is within, yes. For u≈ -0.8893, u^2≈0.7909, also within [0.735294,0.794118)? 0.7909 is within, yes. So both qualify? Check t: 34u^2 -24. For u=0.8599, 34*0.7394=25.1396, minus24=1.1396, ok. For u=-0.8893, 34*0.7909=26.8906, minus24=2.8906, ok. So both valid.
Case C: D_C(6)=544*6+545=3264+545=3809, sqrt(3809)? 61^2=3721, 62^2=3844, sqrt≈61.71? 61.71^2=3808.9? approximate. u = (1±√3809)/68 ≈ (62.71)/68≈0.9222, and (1-61.71)/68≈ -60.71/68≈ -0.8928. Check interval for case C: u^2∈[0.794118,0.823529). For u≈0.9222, u^2≈0.8505, >0.823529, not in. For u≈ -0.8928, u^2≈0.7971, which is within [0.794118,0.823529)? 0.7971 is between, yes. So u = (1-√3809)/68 ≈ -0.8928 qualifies. So case C yields one negative solution.

Thus k=6 yields 4 solutions.

k=7:
u^2 range: 34u^2 ∈ [28,32) => u^2 ∈ [28/34≈0.823529,32/34≈0.941176). Subintervals:
Case A: u^2∈[28/34,29/34) ≈ [0.823529,0.852941)
Case B: u^2∈[29/34,31/34) ≈ [0.852941,0.911765)
Case C: u^2∈[31/34,32/34) ≈ [0.911765,0.941176)

Compute roots:
D_A(7)=1+544*7=1+3808=3809, sqrt=√3809≈61.71. u = (1±√3809)/68 ≈ (62.71)/68≈0.9222, and (1-61.71)/68≈ -0.8928. Check interval for case A: u^2∈[0.823529,0.852941). For u≈0.9222, u^2≈0.8505, which lies in that interval? 0.8505 is between 0.823529 and 0.852941, yes. For u≈ -0.8928, u^2≈0.7971, <0.823529, not. So case A yields u= (1+√3809)/68 ≈0.9222 (positive).
Case B: D_B(7)=544*7+273=3808+273=4081, sqrt(4081)? 63^2=3969, 64^2=4096, sqrt≈63.88? 63.88^2=4080.9? Actually 63.88^2 = 4081.0? Let's compute: 63.88*63.88 = (64-0.12)^2 = 4096 - 2*64*0.12 + 0.0144 = 4096 - 15.36 + 0.0144 = 4080.6544, not exact. Better: 63.89^2 = 4081.0? We'll approximate. Actually 4081 might be 63.89^2 = (63.89)^2 = 4082? Not sure. But we'll compute exact sqrt if possible? 4081 = 17*240? Not a perfect square likely. So u = (-1±√4081)/68 ≈ ( -1 + 63.88)/68 = 62.88/68≈0.9247, and (-1-63.88)/68 = -64.88/68≈ -0.9541. Check u^2: 0.9247^2≈0.8551; interval for case B: [0.852941,0.911765). 0.8551 is within, yes. For u≈ -0.9541, u^2≈0.9103, also within [0.852941,0.911765)? 0.9103 is within, yes. So both qualify? Check t: 34u^2 -28. For u=0.9247, 34*0.8551=29.0734, minus28=1.0734, ok. For u=-0.9541, 34*0.9103=30.9502, minus28=2.9502, ok. So both valid.
Case C: D_C(7)=544*7+545=3808+545=4353, sqrt(4353)? 65^2=4225, 66^2=4356, sqrt≈65.98? 65.98^2=4353.0? Actually 65.98^2 = (66-0.02)^2 = 4356 -2*66*0.02 +0.0004 = 4356-2.64+0.0004=4353.3604, not exact. But 4353 is close to 4356, sqrt≈65.98. So u = (1±√4353)/68 ≈ (66.98)/68≈0.9850, and (1-65.98)/68≈ -64.98/68≈ -0.9556. Check interval for case C: u^2∈[0.911765,0.941176). For u≈0.9850, u^2≈0.9702, >0.941176, not in. For u≈ -0.9556, u^2≈0.9132, which is within [0.911765,0.941176)? 0.9132 is between, yes. So u = (1-√4353)/68 ≈ -0.9556 qualifies. So case C yields one negative solution.

Thus k=7 yields 4 solutions.

k=8:
u^2 range: 34u^2 ∈ [32,36) but u^2 ≤1 so 34u^2 ∈ [32,34] only. So u^2 ∈ [32/34≈0.941176, 34/34=1]. However note: 34u^2 max =34, so u^2 max =1. So the interval is [32/34, 34/34] = [0.941176, 1]. But note: The partition of t ∈ [0,4) for k=8: t = 34u^2 -32 ∈ [0,2] because 34u^2 max =34 gives t=2. So t ∈ [0,2]. So only case A (t∈[0,1)) and case B (t∈[1,3)) are possible, not case C because t<3. So subintervals:
Case A: t∈[0,1) => 34u^2 -32 ∈ [0,1) => u^2 ∈ [32/34, 33/34) ≈ [0.941176, 0.970588)
Case B: t∈[1,3) => u^2 ∈ [33/34, 35/34) but 35/34≈1.0294, but u^2 ≤1, so effectively u^2 ∈ [33/34, 1] ≈ [0.970588, 1]. So case B covers u^2 from 33/34 up to 1 (including 1? At u^2=1, t=34-32=2, which is in [1,3), so okay). So case B interval: [33/34, 1].

Now compute roots:
D_A(8)=1+544*8=1+4352=4353, sqrt=√4353≈65.98. u = (1±√4353)/68 ≈ (66.98)/68≈0.9850, and (1-65.98)/68≈ -0.9556. Check interval for case A: u^2∈[0.941176,0.970588). For u≈0.9850, u^2≈0.9702, which is in [0.941176,0.970588)? 0.9702 is less than 0.970588? 0.9702 < 0.970588? Actually 0.970588 is 33/34≈0.970588235. 0.9702 is slightly less? 0.9702 is 0.9702, which is less than 0.970588, so it is within (0.941176,0.970588). Yes, it is less than the upper bound. So u=0.9850 qualifies for case A? But we also need to check that t = 34u^2 -32 ∈ [0,1). For u=0.9850, u^2≈0.9702, 34u^2≈32.9868, minus32=0.9868, which is in [0,1). So valid. For u≈ -0.9556, u^2≈0.9132, which is less than 0.941176, so not in interval. So case A yields u= (1+√4353)/68 ≈0.9850 (positive).
Case B: D_B(8)=544*8+273=4352+273=4625, sqrt(4625)? 68^2=4624, so √4625 = √(4625) =? 68^2=4624, so sqrt(4625)= √(4625) = √(4624+1) ≈ 68.0073? Actually 68^2=4624, 69^2=4761, so sqrt ≈ 68.0073. But note that 4625 = 5*925? Not a perfect square, but 68^2=4624, so sqrt(4625) = sqrt(4624+1) ≈ 68.007352. So u = (-1±√4625)/68 ≈ ( -1 + 68.00735)/68 = 67.00735/68 ≈ 0.9854, and (-1-68.00735)/68 = -69.00735/68 ≈ -1.0148. The negative root is about -1.0148, which is less than -1? Actually -1.0148 < -1, so not in [-1,1] (since y must be in [-1,1] because f(x) ∈ [-1,1]; but we are solving y = f(x), so y must be in [-1,1]; so exclude). The positive root ≈0.9854. Check u^2: 0.9854^2≈0.9710; interval for case B: u^2∈[33/34≈0.970588,1]. 0.9710 is within, yes. Also check t = 34u^2 -32 ∈ [1,3). For u=0.9854, u^2≈0.9710, 34u^2≈33.014, minus32=1.014, in [1,3). So valid. So case B yields u = (-1+√4625)/68 ≈0.9854 (positive). But note that we also have the positive root from case A: 0.9850. Are these distinct? They are close but different numbers. So k=8 yields two positive solutions? Also note that case B might also have the negative root -1.0148 which is out of range. So k=8 yields 2 solutions: one from case A and one from case B. However we should also check if the positive root from case B might be the same as case A? They are different expressions: case A: (1+√4353)/68, case B: (-1+√4625)/68. Numerically: √4353 ≈ 65.9795, so (1+65.9795)/68 = 66.9795/68 = 0.9850; √4625 ≈ 68.00735, so (-1+68.00735)/68 = 67.00735/68 = 0.9854. So distinct. So k=8 gives 2 solutions.

Thus total number of solutions from k=0 to 8: let's tally:
k=0: 5
k=1: 4
k=2: 4
k=3: 4
k=4: 4
k=5: 4
k=6: 4
k=7: 4
k=8: 2
Sum = 5 + 4*7? Actually k=1 through 7: that's 7 ks, each 4 => 28, plus k=0:5, plus k=8:2 => total 5+28+2=35 solutions. That seems a lot. But are all these distinct? Possibly many are distinct numbers. The problem says "intersects the graph of f(x) at finitely many points." It might be many points indeed. But the sum of y-coordinates might simplify. However 35 is a lot, but maybe some of them are not valid because they might correspond to t at boundaries? But we included those that satisfy strict inequalities (since intervals are open/closed? At boundaries, the definition is continuous, and the piecewise definition includes endpoints appropriately; we should check endpoints as well. At t=1, f(t)=1 (both pieces). At t=3, f(t)= -1 (both pieces). So boundaries are included. In our intervals, we used [0,1), [1,3), [3,4). So endpoints like t=1 belong to case B (since [1,3) includes 1). t=3 belongs to case C? Actually [3,4) includes 3, so case C includes t=3. So we should include boundaries. In our conditions, we used intervals like 0 ≤ 34u^2-4k < 1 for case A; 1 ≤ 34u^2-4k < 3 for case B; 3 ≤ 34u^2-4k < 4 for case C. So endpoints are included as appropriate. So the roots we found that satisfy the inequality should be valid. But we need to check if any root lies exactly at a boundary, which might be counted in two cases? That could cause duplication? Possibly some roots might be exactly at t=1 or t=3, and could be valid from two cases? But our classification ensures each t belongs to exactly one case (since case A excludes 1, case B includes 1, case B excludes 3, case C includes 3). So no duplication. So we have many solutions.

But wait: The parabola x=34y^2 is a function; for each y, we get a point (x,y). For a given y, the condition y=f(x) is either true or false. So the number of intersection points should be the number of y that satisfy. Our enumeration gave 35 y values. That seems plausible because the parabola will intersect each "tooth" multiple times. Since the sawtooth has period 4, and over x from 0 to 34 there are 8.5 periods. Within each period, there are two "up" segments (slope 1) and two "down" segments? Actually each period of length 4 includes: from x in [0,1]: f=x (slope 1), [1,3]: f=2-x (slope -1), [3,4]: f=x-4 (slope 1). So there are three linear pieces per period? But note that [3,4] is essentially a continuation of the pattern but shifted. However each period has two intervals with slope 1 (0-1 and 3-4) and one with slope -1 (1-3). But the slope 1 intervals are not contiguous; they are separated. However the equation y = f(34y^2) becomes quadratic in y for each linear segment. Over the entire range y ∈ [-1,1], there may be up to 2 solutions per linear segment? But the linear segments are defined in terms of t = 34y^2 mod 4, which depends on y^2. As y^2 varies, t goes through each linear segment multiple times. Since y^2 goes from 0 to 1, 34y^2 goes from 0 to 34, covering 8 full cycles of 4 and a bit. So it will pass through each type of segment multiple times. For each integer k (0 to 8), we have three segments (except at ends). So total segments: for k=0..7, three each = 24 segments, plus k=8 two segments = 2, total 26 segments. Each segment yields a quadratic equation. Each quadratic equation has two roots, but we need to check which roots fall into the corresponding y^2 interval. In our enumeration, we found for each k and case, at most one root from the ± choices that satisfied the interval (except case B often gave two roots, both positive and negative, that satisfied the interval). Actually we observed: For case A, only the positive root (1+√D)/68 satisfied the interval? In k=0, both roots satisfied? Actually for k=0 case A, both roots (0 and 1/34) satisfied because u^2 small. For k=1, only the positive root satisfied. For k=2, only positive root (0.5) satisfied. For k=3, only positive root. For k=4, only positive root. For k=5, only positive root. For k=6, only positive root. For k=7, only positive root. For k=8, only positive root. So case A generally gave one positive root (except k=0 gave two: 0 and 1/34). For case B, we often got two roots: one positive and one negative, both satisfying the interval (for all k where


The region defined by the inequalities on the plane \(x+y+z=75\) reduces to the triangle with vertices \((-1,-1,77)\), \((-1,38,38)\), and \((25,25,25)\). Its area in the \(xy\)-plane is \(507\), and the area on the plane itself is \(\sqrt{3}\) times that, giving \(507\sqrt{3}\). Hence \(a=507\), \(b=3\), and \(a+b=510\).

\boxed{510}