Difference between revisions of "2006 IMO Problems/Problem 1"
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==Solution== | ==Solution== | ||
| − | . | + | We have angle(IBP)= angle(IBC) - angle(PBC)=1/2 angle(ABC) - angle(PBC)=1/2 [ angle(PBA) - angle(PBC)] (1) |
| + | and similarly angle(ICP)=angle(PCB) - angle(ICB) = angle(PCB) - 1/2 angle(ACB)= 1/2[ angle(PCB) - angle(PCA)] (2) | ||
| + | since angle(PBA)+angle(PCA)=angle(PBC)+angle(PCB) , then angle(PBA) - angle(PBC) = angle(PCB) - angle(PCA) (3) | ||
| + | (1) , (2) and (3) therefore : angle(IBP)=angle(ICP) i.e BIPC is con cyclic . | ||
| + | Let J be a point of intersection between (AI) and the circumcise of triangle ABC , then IJB is isosceles traingle because of angle(BIJ)=angle(IBJ) | ||
| + | so JB=JC=JI , then JI=JP (because BIPC con cyclic) . | ||
| + | |||
| + | In triangle APJ : AP+JP >= AJ=AI+IJ but JI=JP so AP >= AI . | ||
Revision as of 23:54, 13 May 2015
Problem
Let 
be triangle with incenter 
. A point 
in the interior of the triangle satisfies 
. Show that 
, and that equality holds if and only if 
Solution
We have angle(IBP)= angle(IBC) - angle(PBC)=1/2 angle(ABC) - angle(PBC)=1/2 [ angle(PBA) - angle(PBC)] (1) and similarly angle(ICP)=angle(PCB) - angle(ICB) = angle(PCB) - 1/2 angle(ACB)= 1/2[ angle(PCB) - angle(PCA)] (2) since angle(PBA)+angle(PCA)=angle(PBC)+angle(PCB) , then angle(PBA) - angle(PBC) = angle(PCB) - angle(PCA) (3) (1) , (2) and (3) therefore : angle(IBP)=angle(ICP) i.e BIPC is con cyclic . Let J be a point of intersection between (AI) and the circumcise of triangle ABC , then IJB is isosceles traingle because of angle(BIJ)=angle(IBJ) so JB=JC=JI , then JI=JP (because BIPC con cyclic) .
In triangle APJ : AP+JP >= AJ=AI+IJ but JI=JP so AP >= AI .
