Keep constructing triangles in the incircle of the previous triangle. What happens?
M is any point on the line AB. Squares of side length AM and MB are
constructed and their circumcircles intersect at P (and M). Prove
that the lines AD and BE produced pass through P.
What can you say about the lengths of the sides of a quadrilateral whose vertices are on a unit circle?
Tony Cardell, has sent in this solution,
which gives the correct answer. We're still not quite convinced:
how does he know
that the strips are parallel to the longer
side? If anyone can explain this, we'll add their explanation
We must find the maximum distance between a pair of opposite
sides (it doesn't matter which since this is a kite). If we extend
the two sides (coloured blue), the intersection is on the same side
of the kite as the equilateral triangle, as shown. So the maximum
distance between the two sides is the red line.
Now look at the isosceles triangle part of the kite. One formula
for the area of a triangle with sides a,b,c says that if s is the
semi-perimeter, (a+b+c)/2, then the area is the square root of
s(s-a)(s-b)(s-c). The isosceles triangle has sides 169, 169, 130,
so s=234 and the area is 10140 square feet. But we also know that
the area is 1/2 x base x height, so the height we want is 10140 x 2
/ 169=120. So 120 foot-wide strips will be needed.