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:
$C_0$: $x^2 + y^2 = 25$
$C_1$: $(x + 4)^2 + (y – 4)^2 = 1$
$C_2$: $(x – 12)^2 + (y – 4)^2 = 65$

The circles $C_0$, $C_1$ and $C_2$ are drawn in the picture below.

$C_0$ and $C_1$ form a lenticular hole, as well as $C_0$ and $C_2$.

We call an ordered pair of positive real numbers $(r_1, r_2)$ a lenticular pair if there exist two circles with radii $r_1$ and $r_2$ that form a lenticular hole.
We can verify that $(1, 5)$ and $(5, \sqrt{65})$ are the lenticular pairs of the example above.

Let $L(N)$ be the number of distinct lenticular pairs $(r_1, r_2)$ for which $0 \lt r_1 \le r_2 \le N$.
We can verify that $L(10) = 30$ and $L(100) = 3442$.

Find $L(100\,000)$.

This problem is about enumeration of Pythagorean triplets with additional constraints. An answer to this can be computed using several different approaches, but one of them might be through use of computer programming and basic arithmetic knowledge. An efficient algorithm for the problem might be:

1. Calculate an upper bound for the radius of the larger circle. Each circle center is a lattice point, and they intersect at two lattice points, so all involved points are integral. By Pythagorean theorem, radii squared are also integral. If their difference is 1 (smallest possible), r1^2 + 2*r1 + 1 <= limit, we get the limit for r1 as (sqrt(4*limit + 1) - 1) / 2. 2. For each possible r1 in the range (from 1 to this limit calculated), calculate the possible square differences (as r2^2 - r1^2 = dist^2), where dist is the squared distance between the centres of two circle. The minimal distance is 1 (adjacent lattice points), and the maximum is (2*r1+1)^2 (diagonally). 3. These square differences should also be a square (another Pythagorean triple), so, for each possible r1, we start from dist = 1 and adjust the dist until reach the max dist, and count the number of possible r2 which are perfect square and <= r2 upper bound (N). 4. Repeat until we have gone through all possible r1. Remember when implementing this on a computer, efficient methods for checking perfect squares and for iterating through r1 and distances should be developed to have a reasonable running time. This algorithm should eventually give the solution for L(100,000) after repeating these steps for all possible r1. However, the actual result is fairly large, and a precise quick hand-calculation isn't feasible. This problem is designed to be solved with a math-oriented programming language like Python or Mathematica due to the large numbers of possibilities.

More Answers:
Digital Signature
Panaitopol Primes
Pseudo-Fortunate Numbers

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