x = 15

Denote the origin of the circle O and the points A, B, C clockwise starting from the left. From the isosceles triangle OAB we get 2 r sin(alpha/2) = 24, where alpha is the angle between OA and OB.

Construct the line orthogonal to OB that goes through C. The length of the line, h, between C and the intersection is h = 7 sin(beta) = x sin(90 - alpha). Denote the lengths of the parts of OB a and b, where a is connected to B. We have a + b = r

Use Thales circle theorem to find that the triangle ABA’ completes the red shape, with A’ on the circle opposite to A. That means that the angle between A’A and A’B is alpha/2, but A’OB is also an isosceles triangle. So the angle on the other side, beta, has to be the same. Thus, beta = alpha/2.

Now, put everything together: a = 7 cos (alpha/2), b = h cot(90 - alpha) = 7 sin(alpha/2) tan(alpha), r = 12 / sin(alpha/2).

a + b = r <=> cos(alpha/2) sin(alpha/2) + sin^2(alpha/2) tan(alpha) = 12 / 7

1/2 sin(alpha) + 1/2(1 - cos(alpha)) tan(alpha) = 12/7 <=> tan(alpha) = 24/7

From the identity for h we know that x = 7 sin(alpha/2) / cos(alpha). Insert alpha = arctan(24/7)

Beltalowda best at rock tossing! Not Olympic discipline, sasa que. Damn inyalowda.

I’ve been a four-star programmer a few times. Imagine a blocked symmetric matrix where the rows and columns are indexed by triples (u,v,w). The entries are zero whenever u != u’ or v != v’, and because of symmetry you only store entries with w <= w’. But the range of v depends on the value of u and the range of w on the value of v. So you do

double ****mat = calloc (UMAX, sizeof(*mat));
for (int u = 0; u < UMAX; ++u) {
  mat[u] = calloc (u + 1, sizeof(**mat));
  for (int v = 0; v <= u; ++v) {
    mat[u][v] = calloc (v + 1, sizeof(***mat));
    for (int w = 0; w <= v; ++w) {
      mat[u][v][w] = calloc (w + 1, sizeof(****mat));
      for (int ww = 0; ww <= w; ++ww)
        mat[u][v][w][ww] = some_function (u, v, w, ww);
    }
  }
}

and weep a little. In reality, this gets a bit optimized by allocating a single chunk of memory and carving that up into the pointer and data arrays, so everything is reasonably close together in memory.

Yes, for the (anti-)electrons. Antiprotons and neutrons should be able to annihilate as well, and vice versa. They are composite particles made of (anti) up and down quarks, so processes like antiproton (anti u anti u anti d) + neutron (udd) -> photons + pi- (anti u d) ( + pi0 maybe) could happen. The pions are short-lived particles called mesons, made of a quark and an antiquark.

I doubt their soul would be willing to be revived just to be tortured and killed again.