Acute Problem

PROPOSITION:  There do not exist four distinct points A,B,C,D on the
plane such that the interior angles of the triangles ABC, ABD, ACD, 
and BCD are all acute.

PROOF 1: If one of the four points is inside the triangle formed by
the other three, then the angles of the three triangles meeting at
that point sum to 2pi, so they cannot each be acute (i.e., less than
pi/2).  If none of the four points is inside the triangle formed by
the other three, then we have a convex quadrilateral ABCD, and the
four interior angles ABC, BCD, CDA, and DAB sum to 2pi, so they 
cannot each be acute.

      

PROOF 2: Given any two points A and B on the plane, construct the 
lines L_ab and L_ba perpendicular to the segment AB through the points
A and B respectively.  Also, construct the circle on the diameter AB.
The triangle formed by A, B, and a third point C will have all acute 
interior angle if and only if C falls between L_ab and L_ba and 
outside the circle on the diameter AB.

Assuming such a point C has been selected, we can now construct
the lines L_ca, L_cb, and L_bc.  Since ACB forms an acute interior
angle, it follows that the lines L_ac and L_bc must intersect
inside the circle with diameter AB (because their interior angle
must be obtuse).  Similarly, the intersection of L_ca and L_ba is
inside the circle with diameter BC, and the intersection of L_ab 
and L_cb is inside the circle with diameter AC.  These six lines
form a convex hexagonal region, and any fourth point would have to
fall inside this hexagon to make all the angles acute.

However, each of the vertices of this hexagon is on or inside the
three overlapping circles with diameters AB, AC, and BC, and the
fourth point would have to fall outside all of these circles.
Therefore, if the circle completely cover the hexagon, there is 
no place that a fourth point could be located such that all the 
conditions of the problem are satisfied.

    

To prove that the three circles with diameters AB, AC, and BC do
in fact cover the entire region inside the hexagon, note that the
isosceles right triangles with the hypotenuses AB, BC, and CA
necessarily overlap because the angles ABC, BCA, and CAB are all
acute and therefore less than the sum of two 45 degree angles (and
the three circles certainly cover those right triangles).  Also,
the hexagon cannot extend outside the circles, because (for example)
the angle between the line perpendicular to BC (at C) and the line
perpendiculat to BA (at A) equals the angle ABC, which is acute, so
the complementary angles are obtuse, from which it follows that the
vertex of the exterior triangle on AC is inside the right triangle, 
and therefore inside the circle on AC.