（A)2(B)68(C)66(D)65(E)67

答案：Suppose that P has coordinates P(0;2a) for some real number a.SinceP has y-coordinate greater than 0 and less than 100, then 0<2a <100 or 0< a <50.We determine an expression for the radius of the circle in terms of a and then determine how many values of a give an integer radius.We determine the desired expression by rst nding the coordinates of the centre,C, of the circle in terms of a, and then calculating the distance from C to one of the points O, P, Q.If a circle passes through the three vertices O,Pand Qof a triangle, then its centre is the point of intersection of the perpendicular bisectors of the sides OP,OQ, and PQ of the triangle.We determine the centre of the circle by nding the point of intersection of the perpendicular bisectors of OP and OQ. (We could use PQ instead, but this would be more complicated algebraically.)Since O has coordinates (0;0) andP has coordinates (0;2a), then OP is vertical so its perpen-dicular bisector is horizontal.The midpoint ofOPis (12(0 + 0);12(0 + 2a)) = (0;a).Therefore, the perpendicular bisector of OPis the horizontal line through (0;a), and so has equation y=a.Since O has coordinates (0;0) and Q has coordinates (4;4), then OQ has slope 4040= 1.Therefore, a line perpendicular toOQhas slope1.The midpoint ofOQ is (12(0 + 4);12(0 + 4)) = (2;2).Therefore, the perpendicular bisector ofOQhas slope1 and passes through (2;2), so has equationy2 = (1)(x2) ory=x+ 4.The centre of the desired circle is thus the point of intersection of the lines with equationsy=a and y=x+ 4.They-coordinate of this point isaand thex-coordinate is obtained by solvinga=x+ 4 andobtainingx= 4a.Therefore, the coordinates ofCare (4a;a).The radius,r, of the circle is the distance fromCto any of the three pointsO,PandQ. It iseasiest to nd the distance fromOtoC, which isr=p(4a)2+a2=pa28a+ 16 +a2=p2a28a+ 16We rewrite this asr=p2(a24a+ 8) =p2(a24a+ 4 + 4) =p2((a2)2+ 4) =p2(a2)2+ 8Since (a2)20 and (a2)2= 0 only whena= 2, then the minimum value of 2(a2)2+ 8is 8 and this occurs whena= 2. Thus,rp8.The expressionp2(a2)2+ 8 is decreasing froma= 0 toa= 2 and then increasing froma= 2 toa= 50.Whena= 0,r=p2(a2)2+ 8 =p2(2)2+ 8 = 4.Whena= 2,r=p2(a2)2+ 8 =p2(0)2+ 8 =p82:83.Whena= 50,r=p2(a2)2+ 8 =p2(48)2+ 8 =p461667:94.Therefore, when 0< a2, we havep8r <4 and when 2a <50, we havep8r <p4616.The expressionr=p2(a2)2+ 8 will take every real number value in each of these ranges,becauseb= 2(a2)2+ 8 represents the equation of a parabola which is a \smooth" curve.Betweenp82:83 and 4, there is one integer value (namely, 3) which is achieved by theexpression. (We do not count 4 since it is an endpoint that is not included.)Betweenp82:83 andp461667:94, there are 65 integer values (namely, 3 to 67, inclusive)which are achieved by the expression.In total, there are 1 + 65 = 66 integer values achieved by the expression in the allowable rangefora, so there are 66 positions ofPfor which the radius is an integer.Answer:(C)

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4。16号，应该是亚洲区的