1951 AHSME Problems/Problem 32
Problem
If is inscribed in a semicircle whose diameter is
, then
must be
Solution
Because is the diameter of the semi-circle, it follows that
. Now we can try to eliminate all the solutions except for one by giving counterexamples.
Set point
anywhere on the perimeter of the semicircle except on
. By triangle inequality,
, so
is wrong.
Set point
on the perimeter of the semicircle infinitesimally close to
, and so
almost equals
, therefore
is wrong.
Because we proved that
can be very close to
in case
, it follows that
is wrong.
Because we proved that
can be very close to
in case
, it follows that
is wrong.
Therefore, the only possible case is