You have an ink stamp that is so amazingly precise that, when inked and pressed down on the plane, it makes every circle whose radius is an irrational number (centered at the center of the stamp) black.
Is it possible to use the stamp three times and make every point in the plane black?
If it is possible, where would you center the three stamps?
Given three points in the plane that define a triangle with distinct sides of lenght S1,S2,and S3 and obey the following restrictions on the lengths;
1. S1 is irrational of the form R1(2^1/2) where R1 is rational.
2. S2 is irrational of the form R2(2^1/2) where R2 is rational.
3. S3 is transcendental.
Then,the stamp when applied to the three points will ink the plane.
Proof: Let P be an uninked point. Then the straight line distances from P to the three points are all rational.There are three cases;
1. P is in the exterior of the triangle and the angles formed by the connections to the three points of the triangle can be labeled alpha,beta, and (alpha+beta).
2. P is on the perimeter of the triangle which leads to an immediate contradiction since every point on the perimeter is an irrational distance from at least one triangle point by the above restrictions.
3. P is in the interior of the triangle and the angles formed can be labeled alpha,beta,and (360-alpha+beta). This case can be mapped into case one for the purpose of this proof without any loss of generality.
In case one there are three equations from the law of cosines. The first two are of the form;
side^2=rational^2+rational^2-2(rational)(rational)cos(alpha or beta). The third equation is of the form;
side^2=rational^2+rational^2-2(rational)(rational)cos(alpha+beta).
The third equation can be put in terms of cos(alpha) and cos(beta) by the following steps;
1.cos(alpha+beta)=cos(alpha)cos(beta)-sin(alpha)sin(beta)
2. Collect sin(alpha)sin(beta) alone on one side of equation three and square both sides.
3. Replace sin^2(alpha) with 1-cos^2(alpha) and sin^2(beta) with 1-cos^2(beta).
Finally solve equations one and two for cos(alpha) and cos(beta) and substitutee in equation three.
The resulting equation has only rational terms and powers of S3. Note that S1^2 and S2^2 are rational. Therefore our third equation is algebraic in S3 with rational coefficients which is a contradiction.
QED
General solution
