1967 IMO Problems/Problem 1
Let be a parallelogram with side lengths , and with . If is acute, prove that the four circles of radius with centers , , , cover the parallelogram if and only if
To start our proof we draw a parallelogram with the requested sides. We notice that by drawing the circles with centers A, B, C, D that the length of must not exceed 2 (the radius for each circle) or the circles will not meet and thus not cover the parallelogram.
To prove our conjecture we draw a parallelogram with and draw a segment so that
This is the parallelogram which we claim has the maximum length on and the highest value on any one angle.
We now have two triangles inside a parallelogram with lengths and , being segment . Using the Pythagorean theorem we conclude:
Using trigonometric functions we can compute:
Notice that by applying the and functions, we can conclude that our angle
To conclude our proof we make sure that our values match the required values for maximum length of
Notice that as decreases, the value of (1) increases beyond 2. We can prove this using the law of sines. Similarly as increases, the value of (1) decreases below 2, confirming that (1) is only implied when is acute.
--Bjarnidk 02:16, 17 May 2013 (EDT)
Solution can also be found here 
|1967 IMO (Problems) • Resources|
|1 • 2 • 3 • 4 • 5 • 6||Followed by|
|All IMO Problems and Solutions|