42nd International Mathematical Olympiad

IMO 2001 Problems and Solutions

Washinton, DC, USA
July 1-14, 2001

Problem 1

Let ABC be an acute-angled triangle with circumcentre O. Let P on BC be the foot of the altitude from A.

Suppose that BCA >= ABC + 30°.

Prove that CAB + COP < 90°.

Solution 1

Let = CAB, = ABC, = BCA, and = COP. Let K and Q be the reflections of A and P, respectively, across the perpendicular bisector of BC. Let R denote the circumradius of ABC. Then OA = OB = OC = OK = R. Furthermore, we have QP = KA because KQPA is a rectangle. Now note that AOK = AOB - KOB = AOB - AOC = 2 - 2 >= 60°.

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 COP, PCO > . Now since = 1\2 BOC = 1\2 (180° - 2PCO) = 90° - PCO, it indeed follows that + < 90°.

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 and AC = 2Rsin. Therefore, we have BP - PC = ABcos - ACcos = 2R( sincos - sincos) = 2Rsin( - ). It follows from this and from

30° <= - < < 90°

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 and CP = ACcos = 2Rsincos, it suffices to show that 1\4 > sinsincos. We note that 1 > sin = sin( + ) = sincos + sincos and 1\2 <= sin( - ) = sincos - sincos since 30° <= - < 90°. It follows that 1\4 > sincos and that 1\4 > sinsincos.

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 OBC > OJC. Since OC/CJ = PC/CO and JCO = OCP, we have JCO OCP and OJC = POC = . It follows that < OBC = 90° - or + < 90°.

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 ABC is BP · PC = R² - OP². From these two equations we find that

OP² = R² - BP · PC > PC · CB - BP · PC = PC²,

from which OP > PC. Therefore, as in the first solution, we conclude that + < 90°.