# 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,