Combinatorics and Probability 789= 2 E′(A(OP R))−
2
3
E′(χ A(OP R))=29
36 π.
Finally, note that the case in whichP′,Q′,R′lie on a semicircle in some order occurs
with probability^34. (The case in which they lie on a semicircle proceeding clockwise
fromP′to its antipode has probability^14 ; this case and its two analogues are exclusive
and exhaustive.) Hence
E(A(P QR))=E(A(OP Q)+A(OQR)+A(ORP ))−3
4
E′(A(OP Q)+A(OQR)+A(ORP )−A(P QR))
=
4
3 π−
29
48 π=
35
48 π.
We conclude that the original probability is
1 −4 E(A(P QR))
π= 1 −
35
12 π^2.
Second solution: As in the first solution, it suffices to check that forP , Q, Rchosen
uniformly at random in the disk,E(A(P QR))= 4835 π. Draw the linesPQ,QR,RP,
which with probability 1 divide the interior of the circle into seven regions. Seta =
A(P QR), letb 1 ,b 2 ,b 3 denote the areas of the other three regions sharing a side with
the triangle, and letc 1 ,c 2 ,c 3 denote the areas of the other three regions. SetA=E(a),
B=E(b 1 ),C=E(c 1 ), so thatA+ 3 B+ 3 C=π.
Note thatc 1 +c 2 +c 3 +ais the area of the region in which we can choose a fourth
pointSsuch that the quadrilateralPQRSfails to be convex. By comparing expectations
we find that 3C+A= 4 A,soA=Cand 4A+ 3 B=π.
We will computeB+ 2 A=B+ 2 C, which is the expected area of the part of the
circle cut off by a chord through two random pointsD, E, on the side of the chord not
containing a third random pointF. Lethbe the distance from the centerOof the circle
to the lineDE. We now determine the distribution ofh.
Setr=OD. As seen before, the distribution ofris 2ron[ 0 , 1 ]. Without loss of
generality, we may assume thatOis the origin andDlies on the positivex-axis. For
fixedr, the distribution ofhruns over[ 0 ,r], and can be computed as the area of the
infinitesimal region in whichEcan be chosen so the chord throughDEhas distance to
Obetweenhandh+dh, divided byπ. This region splits into two symmetric pieces,
one of which lies between chords making angles of arcsin(hr)and arcsin(h+rdh)with the
x-axis. The angle between these isdθ=r 2 dh−h 2. Draw the chord throughDat distanceh
toO, and letL 1 ,L 2 be the lengths of the parts on opposite sides ofD; then the area we
are looking for is^12 (L^21 +L^22 )dθ. Because
{L 1 ,L 2 }={√
1 −h^2 +√
r^2 −h^2 ,√
1 −h^2 −√
r^2 −h^2 },