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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 = 2/3m</math>
+
<math> R = (2/3)m</math>
  
  
 
<math>Kris17</math>
 
<math>Kris17</math>

Revision as of 23:22, 17 December 2018

Problem

Let $ABC$ be an isosceles triangle with $AC=BC$, let $M$ be the midpoint of its side $AC$, and let $Z$ be the line through $C$ perpendicular to $AB$. The circle through the points $B$, $C$, and $M$ intersects the line $Z$ at the points $C$ and $Q$. Find the radius of the circumcircle of the triangle $ABC$ in terms of $m = CQ$.


Solution

Let length of side $CB$ = $x$ and length of $QM = a$. We shall first prove that $QM = QB$.

Let $O$ be the circumcenter of $\triangle ACB$ which must lie on line $Z$ as $Z$ is a perpendicular bisector of isosceles $\triangle ACB$.

So, we have $\angle ACO = \angle BCO = \angle C/2$.

Now $MQBC$ is a cyclic quadrilateral by definition, so we have: $\angle QMB = \angle QCB = \angle C/2$ and, $\angle QBM = \angle QCM = \angle C/2$, thus $\angle QMB = \angle QBM$, so $QM = QB = a$.

Therefore in isosceles $\triangle QMB$ we have that $MB = 2 QB \cos C/2 = 2 a \cos C/2$.

Let $R$ be the circumradius of $\triangle ACB$. So we have $CM = x/2 = R \cos C/2$ or $x = 2R \cos C/2$

Now applying Ptolemy's theorem in cyclic quadrilateral $MQBC$, we get:

$m . MB = x . QM + (x/2) . QB$ or,

$m . (2 a \cos C/2) = (3/2 x) . a = (3/2).(2Ra) \cos C/2 = 3Ra \cos C/2$ or,

$R = (2/3)m$


$Kris17$

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