1989 IMO Problems/Problem 4
Let be a convex quadrilateral such that the sides satisfy . There exists a point inside the quadrilateral at a distance from the line such that and . Show that:
Without loss of generality, assume .
Draw the circle with center and radius as well as the circle with center and radius . Since , the intersection of circle with the line is the same as the intersection of circle with the line . Thus, the two circles are externally tangent to each other.
Now draw the circle with center and radius . Since is the shortest distance from to , we have that the circle is tangent to . From the conditions and , by the same reasoning that we found circles are tangent, we find that circles and are externally tangent also.
Fix the two externally tangent circles with centers and and let the points and vary about their respective circles (without loss of generality, assume that one can read the letters in counter clockwise order). Notice that the value is fixed. Thus, in order to prove the claim presented in the problem, we must maximize the value of . It is clear that all possible circles are contained within the region created by circles as well as their common tangent line. Now it is clear that is maximized when circle is tangent to the common tangent of circles as well as to the circles . We shall prove that the value of at this point is .
Fix so that is the common tangent to circles . Since are right, the length is equal to the length of the segment from to the projection of onto which by the Pythagorean Theorem is . Let the projection of onto be units away from . Then, as we found the length of , we find
Subtracting the second equation from the first yields
Since , we have
Solving for yields
Since , we have
And since this is the maximimal value of , the minimal value of is as desired.