Unfortunately it is a set of simultaneous nonlinear equations. Which tells me that you are extremely unlikely to find an analytical solution - you must go to numerical methods.

You can get an idea of the difficulties if you look at a few special cases.

As everyone has noted, we know that we can find the peak if it is in the middle but we can't find the height or slope. (I am assuming that the slope is positive, if you allow any slope then the same results could be achieved by having the slope be 0 and the position unknowable.)

A more interesting special case is where points 1 and 2 are at the same height, but 3 is not. From the fact that 1 and 2 are at the same height, we know that the answer must lie along the line bisecting them. Because it is on that line we know that points 3 and 4 are also at the same height. By assuming that the slope is positive we can figure out which half of that line the peak is on. But we can't find where it is! The problem is that we need to find 4 variables from 4 data points, but 2 pairs of data points are giving us redundant information, so we wind up with 3 pieces of useful information which is not enough to nail down 4 variables.

Now let's go for a more promising case. For ease of reference, let's say that points 1, 2, 3 and 4 are at (1,0), (0,1), (-1,0) and (0,-1) respectively. Let's suppose that the height at points 2 and 4 are the same, and 1 is higher than 2. From the symmetry of points 2 and 4, the peak must be on the x-axis. Since the slope is positive, it must be closer to point 1 and point 3. Can we find it?

Well let's label the slope S, the height H, and the location of the peak (x,y). Then in general we have:

h(u,v) = H - S*sqrt((x-u)**2 + (y-v)**2)
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h1 = h(1,0) = H - S*(1 - x)
h2 = h(0,1) = H - S*sqrt(1 + x*x)
h3 = h(-1,0) = H - S*(1 + x)
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-h3 + h1 = S*(1 + x) - S*(1 - x) = 2*S*x
-h3 + h2 = S*(1 + x) - S*sqrt(1 + x*x)
-h3 + h3 = 0
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2 * (-h3 + h1) / (h1 - h3) = 2
2 * (-h3 + h2) / (h1 - h3) = (1 + x - sqrt(1 + x*x)) / x
                           = 1/x + 1 - sqrt(1/x**2 + 1)
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-1/x**2 + 1/(x**3 * sqrt(1 + 1/x**2))
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Therefore it can be uniquely solved for x. And once you have x, then H and S are easy to find. But good luck finding an analytical solution for x!

So we have 2 degenerate cases that we can't solve. And 1 that we can't solve analytically but can numerically. What about the general case? Well as I've been saying, we have 4 data points and 4 variables to solve, so generally it should be solvable. But we probably can't find an analytic solution. So I'd expect from general principles that the general case, excepting the known degenerate ones, can be solved numerically. And the odds against an analytical solution are so high that I won't waste time looking.

Incidentally if you move just one point, for example make the first point be (2,0), then there are no degenerate cases and I'd be willing to bet that you can always find the answer numerically. But again, a simple analytical solution is unlikely to exist.

If you want to enjoy looking, go ahead. I think that there will be unlimited fun. (Unlimited because you'll never find an answer....) But if you actually need an answer, go with a numerical approach.

Geez, I was bein silly when I said that this special case cannot readily be solved. In fact in the special case we have:

2 * (-h3 + h2) / (h1 - h3) = 1/x + 1 - sqrt(1/x**2 + 1)
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Let's declare that value to be a constant K. Then:

K - 1 - 1/x = - sqrt(1/x**2 + 1)

K**2 - 2K - 2K/x + 1 + 2/x + 1/x**2 = 1/x**2 + 1
K**2 - 2K = (2K - 2)/x
x = 2(K-1)/(K**2 - 2K)
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and so we can calculate K, and then calculate X, and then calculate everything else.

But what about the general case? Well I half-way retract my analytic comment. There is an analytic approach - but it will be useless for practical purposes.

As above, you can get rid of H and S by subtracting one from the rest, then dividing the rest by one remaining. That gives you 2 complex equations in various square roots of x and y. With a couple of squarings, both those equations can be turned into 4'th degree polynomials in x and y. There is an analytic solution to 4th degree polynomials (but it is very complex), so you can find y in terms of x and the other junk, and get high-order equation in terms of x alone. With some more manipulation that can be turned into a polynomial, and there is a general solution known to n'th degree polynomials using integrals.

So in principle you can write down an exact solution. But the solution will be absurdly complex, and probably won't be as easy to calculate as a naive numerical approach.