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:
Solution
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.