Difference between revisions of "2004 JBMO Problems/Problem 2"
(Created page with "==Problem== Let <math>ABC</math> be an isosceles triangle with <math>AC=BC</math>, let <math>M</math> be the midpoint of its side <math>AC</math>, and let <math>Z</math> be t...") |
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<math>m . (2 a \cos C/2) = (3/2 x) . a = (3/2).(2Ra) \cos C/2 = 3Ra \cos C/2</math> or, | <math>m . (2 a \cos C/2) = (3/2 x) . a = (3/2).(2Ra) \cos C/2 = 3Ra \cos C/2</math> or, | ||
− | <math> R = 2/ | + | <math> R = (2/3)m</math> |
<math>Kris17</math> | <math>Kris17</math> |
Revision as of 00:22, 18 December 2018
Problem
Let be an isosceles triangle with
, let
be the midpoint of its side
, and let
be the line through
perpendicular to
. The circle through the points
,
, and
intersects the line
at the points
and
. Find the radius of the circumcircle of the triangle
in terms of
.
Solution
Let length of side =
and length of
. We shall first prove that
.
Let be the circumcenter of
which must lie on line
as
is a perpendicular bisector of isosceles
.
So, we have .
Now is a cyclic quadrilateral by definition, so we have:
and,
, thus
, so
.
Therefore in isosceles we have that
.
Let be the circumradius of
.
So we have
or
Now applying Ptolemy's theorem in cyclic quadrilateral , we get:
or,
or,