# Straightedge and Compass Geometry The figures above are not necessarily drawn in accurate proportion. In the figure on the left, CB intersects the circle, centered at A and containing C at D. EB contains A. The length of DB is equal to the length of AC. What is the ratio, r, of angle ABC to angle EAC?

The angle on the right, XYZ, is given separately with no specified relationship to the figure on the left. Your only tools allowed are a straightedge to construct the line containing two points and a compass to draw a circle centered at one point and containing another point. You are allowed to use these tools only in the traditionally accepted ways of Euclidean geometry. For example, you can pick an arbitrary point on the plane or on an already constructed item to use for further construction. Can you describe a finite number of allowed constructive steps to construct a new angle Q whose ratio to angle XYZ is r? It is not mandatory that the complete resulting construction resembles the figure on the left and the fewer steps the better.

## Just Multiplying Some Positive Integers

Once again, the mighty Bob Conger comes to our rescue with an industrial strength solution. Since A(n), B(n) and C(n) are defined as positive integers, k must be positive, since A(k) would not be an integer for integer values of k<1.

Solutions for Each of the Given Conditions

1. Does not exist.
2. k = 6.
3. Does not exist.

CONDITIONS 1 and 3

Condition 1 or 3 would require that B(k) < A(k) for some positive value of k, since GP! and C(k) are positive integers. However, we can demonstrate that B(k) > A(k) for all integers k > 1, as shown below. Thus, Conditions 1 and 3 will never be satisfied.

For k = 1, 2, 3, B(k) > A(k) is easily seen by inspection. B(3) = 720 > A(3) = 256 and, furthermore, B(3) > 2 × A(3).

We also can show that if B(k) > 2 × A(k), then B(k + 1) > 2 × A(k + 1) and, by induction, B(k) > 2 × A(k) for k > 2.

By definition, (2 × A(k))! = (2 × A(k)) × (2 × A(k) – 1) × … × 1.

There are easily A(k) terms in this product that are greater than A(k) and many other terms greater than 2. So, B(k + 1) = B(k)! > (2 × A(k))! > 2 × A(k)A(k) = 2 × A(k + 1).

CONDITION 2

We are able to find the smallest positive integer value of k such that A(k) ≥ GP! × C(k), or equivalently, since all are positive integers, we are to find the smallest positive integer k such that A(k) / C(k) ≥ GP!.

The numbers quickly spiral out of control as k grows. To keep a handle on them we can work with base-10 logarithms:

Log10(G) = 100
Log10(GP) = G = 10100
Log10(GP!) = Log10(GP) + Log10(GP-1) + … + 0.

This sum has GP = 10G terms, with an average value >1, since all terms but the final term are ≥ 1, and most of the terms are a lot bigger. The average value of a term is < Log10(GP) = 10100, since only the largest (first) term is 10100. So:

(10G) × (1) < Log10(GP!) < (10G) × (10100)
(10G) < Log(GP!) < (10G +100).

For the first few values of k, we can handle the problem easily with arithmetic:

A(1) = 2                C(1) = 1
A(1)/C(1) = 2 < GP!

A(2) = 22 = 4        C(2) = 21 = 2
A(2)/C(2) = 2 < GP!

A(3) = 44 = 256   C(3) = 22 = 4
A(3)/C(3) = 64 < GP!

A(3) = 44  = 256  C(3) = 22  = 4
A(3)/C(3) = 64 < GP!.

By k = 4 we have to deploy logarithms:

A(4) = 256256       C(4) = 24 = 16

Log10(A(4)) = 256 × Log10(256) ≈ 616.5
Log10(C(4)) ≈ 1.2

Log10(A(4)/C(4)) = Log10(A(4)) – Log10(C(4)) ≈ 615.3 < 10100 = Log10(GP)

(A(4)/(C(4) < GP
For k = 5:
A(5) = A(4)A(4)
Log10(A(5)) = A(4) × Log10(A(4) ≈ 10616.5 × 616.5 ≈ 10616.5 × 102.8 = 10619.3 > Log10(GP).

A(5) >> GP + 1   and A(5) >> GP
C(5) = 216 = 65,536
Log10(C(5)) ≈  4.8 ≈  100.7
Log10(A(5)/C(5)) ≈ 10619.3 – 100.7 = 10619.3 << 10G < Log10(GP!)

A(5)/C(5) << GP!.
For k = 6:
A(6) = A(5)A(5) and from above we know A(5) >> GP.
Each term of this product is A(5) >> GP; but GP ≥ each term in GP!.
The number of these terms in A(6) is A(5) >> GP + 1, which is the number of terms in GP! +1.
Therefore, A(6) >> GPGP+1 >> GP!.
C(6) = 2C(5) = 265,536
Log10(C(6)) = 65,536 × Log10(2) ≈ 19,728 ≈ 104.3 < 10619.3 ≈ Log10(A(5)).

Thus:
C(6) < A(5).

Thus:
A(6)/C(6) > A(5)A(5)/A(5) = A(5)A(5)-1 > GPGP > GP!.

Therefore, Condition 2 is met with k = 6.