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Difference between revisions of "2002 JBMO Problems/Problem 1"

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==Solution==
 
==Solution==
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Since <math>APBC</math> is a cyclic quadrilateral, <math>\angle CPA = \angle CBA</math>, and <math>\angle ABP = \angle ACP</math> --- (1)
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Since <math>APBC</math> is a cyclic quadrilateral, <math>\angle CPA = \angle CPB</math>, and <math>\angle ABP = \angle ACP</math> --- (1)
  
 
Now let <math>E</math> be the foot of the perpendicular from <math>C</math> to <math>AP</math>. Then we have, <math>CEPD</math> is a cyclic quadrilateral with <math>CP</math> as diameter of the circumcircle.
 
Now let <math>E</math> be the foot of the perpendicular from <math>C</math> to <math>AP</math>. Then we have, <math>CEPD</math> is a cyclic quadrilateral with <math>CP</math> as diameter of the circumcircle.

Latest revision as of 01:52, 6 June 2024

Problem

The triangle $ABC$ has $CA = CB$. $P$ is a point on the circumcircle between $A$ and $B$ (and on the opposite side of the line $AB$ to $C$). $D$ is the foot of the perpendicular from $C$ to $PB$. Show that $PA + PB = 2 \cdot PD$.


Solution

Since $APBC$ is a cyclic quadrilateral, $\angle CPA = \angle CPB$, and $\angle ABP = \angle ACP$ --- (1)

Now let $E$ be the foot of the perpendicular from $C$ to $AP$. Then we have, $CEPD$ is a cyclic quadrilateral with $CP$ as diameter of the circumcircle.

It follows that $\triangle CPE$ and $\triangle CDP$ are congruent (since $\angle CPA = \angle CPB$). So, we have $CE = CD$ and $PE = PD$ --- (2)

Also, in $\triangle CAE$ we have $\angle EAC = \angle CPA + \angle ACP = \angle CBA + \angle ABP$ ( from (1) above) Thus $\angle EAC = \angle DBC$ So from $SAS$, $\triangle CAE$ is congruent to $\triangle CBD$. Hence we have $AE = DB$.

Now, $PA + PB = PA + PD + DB = PA + PD + AE = PE + PD = 2 . PD$ (from (2) above)


$Kris17$

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