Challenge (+) Problem 32, Chapter 1 (Pigeonhole Principle), A Walk Through Combinatorics, Miklos Bona Question:
Let T be a triangle with angles of 30, 60, and 90, and a hypotenuse of 1.
We choose ten points inside T at random. Prove that there will be four points among them that can be covered by a half-circle of radius 0.42.
 A: I'm not convinced by point #4. Suppose the ten points placed originally happened to be in triangle T, and so did the circle. Then how do I know if cutting the circle in half won't cut out some of the points?
A: My proof:
1) I took the rectangle composed of two such triangles (Area = 0.433).
2) By the pigeonhole principle, if I divide the rectangle into three equal areas (Area = 0.144), one among the three must hold four of ten points placed within the larger, undivided rectangle.
3) Because the radius of the half circle (r = 0.42) is larger than 1/2 of both sides of the smaller rectangle (0.42 > 0.25, 0.42 > 0.288675), and larger than 1/2 of its diagonal (0.42 > 0.288675), placing a half circle of r = 0.42 whose diameter is aligned with the side of length 1/2 and center is placed at its midpoint will encompass the whole of the smaller rectangle and cover a larger area. 
4) Because we can divide Triangle T into three (unequal) portions, each of which corresponds to one of the three smaller rectangles (for each of the smaller rectangles will contain one of the three portions of Triangle T), and because, by the PHP, one of these three components must have 4 points, a half-circle of r = 0.42 contains the portion with 4 points fully. Therefore, any four of ten points placed in triangle T can be covered by a semi-circle of r = 0.42.

Information for you that you may find useful:
A) Area of half-circle with radius 0.42 = 0.277088
B) Area of triangle = 0.2165
C) Area of rectangle = 0.433
D) Area of a third of rectangle = 0.144 (with sides 1/2 and radical(3)/6, that is, a third of radical(3)/2)
E) Lengths of sides of smaller rectangle are 1/2 and radical(3)/6 or 0.288675
F) Length of diagonal of smaller rectangle is 1/radical(3) or 0.57735
G) Length of half this diagonal = 1/[2*radical(3)] or 0.288675
