Can you help this draftsman find the point on the circle?

First, find the point. Take a point (a, b) on the circle with a, b > 0. To reflect, decompose the direction from (6, 0) to (a, b), namely (a-6, b), into components parallel and perpendicular to the tangent line, which have directions (a, b) and (b, -a), respectively. Flip the sign of the perpendicular component.

Compute the projections:

proj of (a-6, b) onto (a, b)

= (a-6, b) dot (a, b)/|(a, b)|^2 (a, b)

= [a^2-6a+b^2]/[a^2+b^2] (a, b)

= [4-6a]/4 (a, b)

= (1-3a/2) (a, b)

proj of (a-6, b) onto (b, -a)

= (a-6, b) dot (b, -a)/|(b, -a)|^2 (b, -a)

= (ab-6b-ab)/4 (b, -a)

= -3b/2 (b, -a)

(I double checked; they add up to (a-6, b).) The direction after reflection is then

(1-3a/2) (a, b) + 3b/2 (b, -a)

= (a - 3a^2/2, b - 3ab/2) + (3b^2/2, -3ab/2)

= (a + 3(b^2 - a^2)/2, b - 3ab)

= (a + 3(2-a^2), b - 3ab)

Next compute where the line traveling in this direction from (a, b) hits the y-axis. This line is

L(t) = (a, b) + t(a + 3(2-a^2), b - 3ab)

The x-component is zero when

a + t(a + 3(2-a^2)) = 0

=> t = a/(3a^2 - a - 6)

(If the denominator is zero, the line is vertical, unless a=0 it can't intersect the y-axis, so we can ignore this case.)

In all, the intersection with the y-axis is at

y = b + a(b-3ab)/(3a^2 - a - 6)

Solving the system y = 10/3 and a^2 + b^2 = 4 subject to a, b >= 0 using Mathematica gives a single solution, (a, b) = (6/5, 8/5). So, indeed, the point, being rational, can be found using a straightedge and compass. It's not clear to me if the problem can be "solved", in some sense, using only a straightedge and compass, rather than algebra. Technically I should check to make sure this point is in the line of sight of both (6, 0) and (0, 10/3), though I won't take the time since I don't find that part interesting.

Your question is great. Indeed we need a frame of reference for these. But to the question: shortest distance is the geometrical straight line distance(I guess). But if the space-time curvature is more curved, there is a possibilty that the geometrical distance there would be short than how we calculate the normal distance. For practical purposes, the normal geometrical distance might not help.

Tangent to the circle at (x' y') on the circle has the equation

xx' yy' = 4 ------------------------------ (1)

Equation to the normal at (x', y') is given by

y = (y'/x')*x or x'y - y'x = 0 ----------------------------- given by (2)

Calculate the slopes of the lines joining (0, 10/3) with (x', y') and (6, 0) with (x', y') and equate the slopes of the angle that these lines make with the normal. If there exists solutions to this then the draftsman can do it.

slope of normal at (x', y') on circle = y'/x' = m1 ----------------- (1) say

Slope, m2 of line joining (x', y') and (1, 10/3) is given by

m2 = [{y'-(10/3)}/(x'-0)] = [(3y' - 10)/(3x')] = (y'/x') - {10/(3x')} --------------------------(2)

tangent of angle formed between these two lines say θ is given by

tan θ = [(m1-m2)/{1 (m1*m2)}] ------------------------------------ (3)

From (1) and (2), m1 - m2 = {10/(3x')} ----------------------- (4) and

{1 (m1*m2)} = 1 (y'/x')² -{(10y')/(3x'²)} = {(3x'² 3y'² -10y'}/(3x'²)} but we know

x'² y'² = 4, so

{1 (m1*m2)} = {(12-10y')/(3x'²)} -------------------------------(5)

Substituting from (4) and (5) in (3) we get

tan θ = [{10/(3x')}/{(12-10y')/(3x'²)}] = {(5x')/(6-5y')} --------------(5a)

Similarly

slope of normal at (x', y') on circle = y'/x' = m1 ----------------- (6) say

Slope, M2 of line joining (x', y') and (6, 0) is given by

M2 = [y' /(x'-6)] --------------------------(7)

tangent of angle formed between these two lines say φ is given by

tan φ = [(m1-M2)/{1 (m1*M2)}] ------------------------------------ (8)

From (6) and (7), (m1 - M2) = (y'/x') - {y' /(x'-6)} or

(m1 - M2) = [{y'(x'-6)-(y'x')}/{x'(x'-6)}] = [(-6y')/{x'(x'-6)}]----------------------- (9) and

{1 (m1*M2)} = [1 {(y'²)/(x'² -6x')] or

{1 (m1*M2)} = [(x'² y'²-6x') /(x'² - 6x')}] = [{(4-(6x')}/(x'² - 6x')}] --------------(10)

Substituting from (9) and (10) in (8) we get

tan φ = [(-6y')/{x'(x'-6)}]/[{(4-(6x')}/(x'² - 6x')}] = [(-3y')/{2-(3x')}] ------------------(11)

as θ need to be equal to φ, from (5a) and (11) we get

{(5x')/(6-5y')} = [(-3y')/{2-(3x')}] or

10x'-15x'² = -18y' 15y'² or

10x' 18y' = 15(x'² y'²) = 60 or

5x' 9y' = 30 or 5x' = 30 -9y'--------------- (12)

(5x')² (5y')² = 100; substituting from 12 we get

(30-9y')² 25y'² = 100

106y'² -540y' 800 = 0 or

53y'² -270y' 400 = 0

as determinant of this quadratic equation is < 0

There is no real solution. Hence it is not possible for the draftsman to get the task done any way.

I would request the questioner to please find flaw in my solution.

I am working on a geometric method.