1992 OIM Problems/Problem 5
Problem
The circumference 
and the positive numbers 
and 
are given so that there is a trapezoid 
inscribed in , of height and in which the sum of the bases 
and 
is . Build the trapezoid .
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
Solution
First we realize that the trapezoid is isosceles because it is inscribed in a circle. If we draw a rhombus using the midpoints of the trapezoid, the diagonals of the rhombus will be 
and We can construct such rhombus. Then, with similar triangles and the midpoints we know that the diagonal of the trapezoid is twice the length of the rhombus. Now we have all we need to start constructing:
Given the height in blue and length in red, we begin by bisecting with straight edge and compass the line of length and then bisecting it again as shown in the drawing above. We also bisect the blue line. Using the center of the blue line we measure 
with the compass and draw a circle at the center of the point of the second bisection of the red line as shown. Then we can construct the rhombus by joining the intersection of the second perpendicular bisector with the circles and the edge of red line with the middle as shown above.
- Note. I actually competed at this event in Venezuela when I was in High School representing Puerto Rico. I'm proud to say that I got full points on this one and I solved it very quickly. I had a straight rule and compass kit which I used to solve it as we're supposed to build the trapezoid with it. Now, 3 decades later, I attempted this and spent a full 3 hours on it and couldn't solve it nor I remember what I did. I will attempt again some other time.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
