1984 IMO Problems/Problem 4
Contents
[hide]Problem
Let be a convex quadrilateral with the line
being tangent to the circle on diameter
. Prove that the line
is tangent to the circle on diameter
if and only if the lines
and
are parallel.
Solution
First, we prove that if and
are parallel then the claim is true: Let
and
intersect at
(assume
is closer to
, the other case being analogous). Let
be the midpoints of
respectively. Let the length of the perpendicular from
to
be
. It is known that the length of the perpendicular from
to
is
. Let the foot of the perpendicular from
to
be
, and similarly define
for side
. Then, since triangles
and
are similar, we have
. This gives an expression for
:
Noticing that simplifies the expression to
By the Law of Sines, . Since triangles
are similar, we have
and thus we have
and we are done.
Now to prove the converse. Suppose we have the quadrilateral with parallel to
, and with all conditions satisfied. We shall prove that there exists no point
on
such that
is a midpoint of a side
of a quadrilateral
which also satisfies the condition. Suppose there was such a
. Like before, define the points
for quadrilateral
. Let
be the length of the perpendicular from
to
. Then, using similar triangles,
. This gives
But, we must have . Thus, we have
Since , we have
as desired.
Solution 2
Let and
be midpoints of
and
, respectively. Let
denote the area of figure
.
If and the circle centered at
with diameter
touches
at
, then
. It follows that
, so
, where
is the distance from
to
. But we have
, so
. It follows that the circle centered at
with diameter
touches
.
If on the other hand we have the circle with diameter touching
at
(as well as the circle with diameter
touching
at
), we must have
because of the equivalence of radii in circles. Hence
, so
and
are equidistant from
(as
is the common base). Hence
. Similarly
, and so
, as desired.