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Problem 295


Problem 295


Lenticular holes
We call the convex area enclosed by two circles a lenticular hole if:

  • The centres of both circles are on lattice points.
  • The two circles intersect at two distinct lattice points.
  • The interior of the convex area enclosed by both circles does not contain any lattice points.

Consider the circles:
C0: x2+y2=25
C1: (x+4)2+(y-4)2=1
C2: (x-12)2+(y-4)2=65

The circles C0, C1 and C2 are drawn in the picture below.

C0 and C1 form a lenticular hole, as well as C0 and C2.

We call an ordered pair of positive real numbers (r1, r2) a lenticular pair if there exist two circles with radii r1 and r2 that form a lenticular hole. We can verify that (1, 5) and (5, √65) are the lenticular pairs of the example above.

Let L(N) be the number of distinct lenticular pairs (r1, r2) for which 0 < r1 ≤ r2 ≤ N.
We can verify that L(10) = 30 and L(100) = 3442.

Find L(100 000).


透镜孔洞
若两圆重叠而成的凸区域满足以下条件,我们称之为透镜孔洞

  • 两圆的中心都是格点。
  • 两圆在两个格点处相交。
  • 两圆重叠而成的凸区域内部不包含任何格点。

考虑如下的圆:
C0: x2+y2=25
C1: (x+4)2+(y-4)2=1
C2: (x-12)2+(y-4)2=65

圆C0、C1和C2如下图所示。

C0和C1以及C0和C2均生成了一个透镜孔洞。

如果存在两个半径分别为r1和r2的圆能够生成透镜孔洞,我们就称有序正实数对(r1, r2)为透镜数对。由上述例子,我们可以验证(1, 5)和(5, √65)均为透镜数对。

记L(N)是所有满足0 < r1 ≤ r2 ≤ N的不同透镜数对的数目。
我们可以验证L(10) = 30以及L(100) = 3442。

求L(100 000)。