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![]() | 42nd International Mathematical OlympiadIMO 2001 Problems and SolutionsWashinton, DC, USAJuly 1-14, 2001 Problem 1Let ABC be an acute-angled triangle with circumcentre O. Let P on BC be the foot of the altitude from A.
Suppose that
Prove that Solution 1
Let
It follows from this and from OA = OK = R that KA >= R and QP >= R. Therefore, using the Triangle Inequality, we have OP + R = OQ + OC > QC =
QP + PC >= R + PC. It follows that OP > PC, and hence in Solution 2
As in the previous solution, it is enough to show that OP > PC. To this end, recall that by the (Extended) Law of Sines, AB = 2Rsin
30° <= that BP - PC >= R. Therefore, we obtain that R + OP = BO + OP > BP >= R + PC, from which OP > OC, as desired. Solution 3
We first show that R2 > CP · CB. To this end, since CB = 2Rsin
Now we choose a point J on BC so that CJ · CP = R2. It follows from this and from R2 > CP · CB that CJ > CB, so that Solution 4
On the one hand, as in the third solution, we have R2 > CP · CB. On the other hand, the power of P with respect to the circumcircle of OP2 = R2 - BP · PC > PC · CB - BP · PC = PC2,
from which OP > PC. Therefore, as in the first solution, we conclude that |