Problem:
In triangle ABC,A′,B′, and C′ are on sides BC,AC, and AB, respectively. Given that AA′,BB′, and CC′ are concurrent at the point O, and that
OA′AO​+OB′BO​+OC′CO​=92
find the value of
OA′AO​⋅OB′BO​⋅OC′CO​
Solution:
Since △AOB and △A′OB share an altitude, as do △AOC and △A′OC, we have
OA′AO​=[A′OB][AOB]​​=[COA′][COA]​=[A′OB]+[COA′][AOB]+[COA]​=[BOC][AOB]+[COA]​=xz+y​​
where x=[BOC],y=[COA], and z=[AOB]. Similarly,
OB′BO​=yx+z​ and OC′CO​=zy+x​
We then have
OA′AO​OB′BO​OC′CO​​=xyz(z+y)(x+z)(y+x)​=xyzyz2+y2z+x2z+xz2+xy2+x2y+2xyz​=xyzyz(z+y)+xz(x+z)+xy(y+x)​+2=xz+y​+yx+z​+zy+x​+2​
Hence,
OA′AO​OB′BO​OC′CO​=(OA′AO​+OB′BO​+OC′CO​)+2=92+2=94​
The problems on this page are the property of the MAA's American Mathematics Competitions
