A. The answer as per this document is No, it cannot exist keeping all distances rational.

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1 Rational Distance Q. Given a unit square, can you find any point in the same plane, either inside or outside the square, that is a rational distance from all four corners? Or, put another way, given a square ABCD of any size, can you find a point P in the same plane such that the distances AB, PA, PB, PC, and PD are all integers? A. The answer as per this document is No, it cannot exist keeping all distances rational. This is the proof: Part 1: Ouside the square Notation AP = a, (distance from A to P also the radius of circle Ca) BP = b, (distance from B to P also the radius of circle Cb) CP = c, (distance from C to P also the radius of circle Cc) DP = d, (distance from D to P also the radius of circle Cd) P (X1, Y1) is the point of intersection of all the circles with X1, Y1 as co-ordinates. L = length of the side of square, Sqrt(2)L = Length of the diagonal of the square (irrational if L is rational) Ca is Circle with a as radius Cb is Circle with b as radius Cc is the Circle with c as radius Cd is the Circle with d as radius

2 Proof Map all the circles and squares back towards 0,0 And assuming A > D > B > C > 0 and X1 > L, Y1 > L & L >= 1 This keeps it in the top quadrant (meaning P will be above the side BC, nearest C. (slightly different assumptions can be made to push it into any of the other 3 quadrants but effectively the same proof will apply (with rotation...).) The formulas from Pythagoras Theorem/Circle formula are: Ca: (centre point (0,0), radius a, point on circle ( X1, Y1)): a 2 = X1 2 + Y1 2 Cb: (centre point (0, L), radius b, point on circle (X1, Y1)): b 2 = X1 2 + (Y1 L) 2 Cc: (centre point (L, L), radius c, point on circle (X1, Y1)): c 2 = (L - X1) 2 + (Y1 L) 2 Cd: (centre point (L, 0), radius d, point on circle (X1, Y1)): d 2 = (L - X1) 2 + Y1 2 Ca: a 2 = X1 2 + Y1 2 Cb: b 2 = X1 2 + (Y1 L) 2 = X1 2 + Y1 2-2LY1 + L 2 Cc: c 2 = (X1 L) 2 + (Y1 L) 2 = X1 2 + Y1 2-2LX1 + L 2-2LY1 + L 2 Cd: d 2 = (X1 L) 2 + Y1 2 = X1 2 + Y1 2-2LX1 + L 2 Putting Ca into Cb and Cd to find X1 and Y1 in terms of a, b, c, d Ca -> Cb: b 2 = a 2-2LY1 + L 2 -> (b 2 a 2 - L 2 ) /-2L = Y1 Ca -> Cd: d 2 = a 2-2LX1 + L 2 -> (d 2 - a 2 -L 2 )/ -2L = X1 Ca,Cb,Cd -> Cc: c 2 = a 2 + (b 2 a 2 ) + (d 2 a 2 ) c 2 = b 2 + d 2 a 2 c 2 + a 2 = b 2 + d 2 Putting the values of X1 and Y1 back into Ca gives: Ca: a 2 = ((d 2 - a 2 -L 2 )/ -2L) 2 + ((b 2 a 2 - L 2 ) /-2L) 2 Expanding gives: 4L 2 a 2 = (d 2 a 2 ) 2 + L 4-2L 2 (d 2 - a 2 ) + (b 2 a 2 ) 2 + L 4-2L 2 (b 2 - a 2 ) Grouping L 4 and L 2 and L 0 and bringing the left term to the right gives: 0 = 2L 4-2L 2 ((d 2 - a 2 ) + (b 2 - a 2 ) + 2a 2 ) + ((d 2 - a 2 ) 2 + (b 2 a 2 ) 2 ) 0 = 2L 4-2L 2 (d 2 + b 2 ) + ((d 2 - a 2 ) 2 + (b 2 a 2 ) 2 ) Divide by 2 gives: L 4 -L 2 (d 2 +b 2 ) + ((b 2 - a 2 ) 2 + (d 2 a 2 ) 2 )/2 Using the binomial theorem to solve for L 2 : (-b 1 +/- sqrt(b 1 2-4a 1 c 1 ))/2a 1 (using slightly different nomenclature as my variable names are also a, b and c) where in this case: a 1 = 1 b 1 = -(d 2 + b 2 ) c 1 = ((b 2 - a 2 ) 2 + (d 2 a 2 ) 2 )/2 L 2 = ((d 2 + b 2 ) +/- sqrt((-(d 2 + b 2 )) 2-4(1) ((b 2 - a 2 ) 2 + (d 2 a 2 ) 2 )/2 ))/ 2(1) = ((d 2 + b 2 ) +/- sqrt((-(d 2 + b 2 )) 2-2 ((b 2 - a 2 ) 2 + (d 2 a 2 ) 2 ))) /2 = ((d 2 + b 2 )/2 +/- sqrt( (-(d 2 + b 2 )) 2-2 ((b 2 - a 2 ) 2 + (d 2 a 2 ) 2 )))/4 Let Z be the part with the sqrt as follows: L 2 = ((d 2 + b 2 )/2 +/- Z) Z = sqrt( (-(d 2 + b 2 )) 2-2 ( (b 2 - a 2 ) 2 + (d 2 a 2 ) 2 ))/4)

3 Lets try and simply by initially expanding the right hand side... Z 2 = ( (-(d 2 + b 2 )) 2-2 ( (b 2 - a 2 ) 2 + (d 2 a 2 ) 2 ) /4 Z 2 = (d 4 + b b 2 d 2-2 (b 4 + a 4-2 a 2 b 2 + d 4 + a 4-2 a 2 d 2 ))/4 Z 2 = (d 4 + b b 2 d 2-2 b 4-2a 4 +4 a 2 b 2-2 d 4-2a 4 +4 a 2 d 2 )/4 Z 2 = ( + 2 b 2 d 2 - b 4-4a 4 +4 a 2 b 2 -d 4 +4 a 2 d 2 )/4 Z 2 = (-4a 4 - b 4 -d b 2 d 2 +4 a 2 b 2 +4 a 2 d 2 )/4 Grouping in a 4, a 2, a 0 Z 2 = -4a 4 +4 a 2 (b 2 + d 2 ) + - b 4 -d b 2 d 2 )/4 Z 2 = (+4 (-a 4 + a 2 (b 2 + d 2 ) + ( - b 4 - d b 2 d 2 )/4) /4) Z 2 (-a 4 + a 2 (b 2 + d 2 ) + ( - b 4 - d b 2 d 2 )/4) Z 2 = - a 4 + a 2 (b 2 + d 2 ) + ( - b 4 - d b 2 d 2 )/4 Z 2 = - a 4 + a 2 (b 2 + d 2 ) - ( b 2 - d 2 ) 2 /4 Factorizing again for a on the right hand side by letting Z 2 ->0... a1 = -1 b1 = b 2 + d 2 c1 = -(b 2 d 2 ) 2 /4 a 2 = (-(b 2 + d 2 ) +/- sqrt((b 2 + d 2 ) 2 4(-1)-(b 2 d 2 ) 2 /4))/-2 a 2 = (b 2 + d 2 )/2 +/- sqrt(b 4 + d 4 +2b 2 d 2 b 4 d 4 +2b 2 d 2 )/2 a 2 = (b 2 + d 2 )/2 +/- sqrt( +4b 2 d 2 )/2 a 2 = (b 2 + d 2 )/2 +/- sqrt( +b 2 d 2 ) a 2 = (b 2 + d 2 )/2 +/- bd a 2 = (b 2 + d 2 +/- 2bd)/2 a 2 = (b +/- d) 2 /2 a = +/-(b+/-d) /sqrt(2) so Z 2 = (a -(b+d)/sqrt(2)) (a +(b+d)/sqrt(2)) (a + (b-d)/sqrt(2)) (a - (b-d)/sqrt(2)) therefore Z 2 cannot be rational, as the sum or difference of b and d is being divided by the irrational sqrt(2) therefore Z cannot be rational as the sqrt of an irrational is irrational. And L 2 = ((d 2 + b 2 )/2 +/- Z) Z being irrational pushes L 2 to irrational which means that L is irrational. Therefore all lengths cannot be rational, and as such P cannot exist given the initial conditions.

4 Part 2: P inside the square Looking at point P inside the square: Notation AP = a, (distance from A to P also the radius of circle Ca) BP = b, (distance from B to P also the radius of circle Cb) CP = c, (distance from C to P also the radius of circle Cc) DP = d, (distance from D to P also the radius of circle Cd) P (X1, Y1) is the point of intersection of all the circles with X1, Y1 as co-ordinates. L = length of the side of square, Sqrt(2)L = Length of the diagonal of the square (irrational if L is rational) Ca is Circle with a as radius Cb is Circle with b as radius Cc is the Circle with c as radius Cd is the Circle with d as radius Proof Map all the circles and squares back towards 0,0 And assuming A > D > B > C > 0 and X1 < L and Y1 < L & L >= 0

5 This keeps it in the top quadrant of the square (P nearest point C), (slightly different assumptions can be made to push it into any of the other 3 quadrants(of the inside of the square) but effectively the same proof will apply.) The formulas from Pythagoras Theorem are: Ca: (centre point (0,0), radius a, point (X1,Y1)): a 2 = X1 2 + Y1 2 Cb: (centre point (0, L), radius b, point (X1,Y1)): b 2 = X1 2 + (L Y1) 2 Cc: (centre point (L, L), radius c, point (X1,Y1)): c 2 = (L - X1) 2 + (L - Y1) 2 Cd: (centre point (L, 0), radius d, point (X1,Y1)): d 2 = (L - X1) 2 + Y1 2 Ca: a 2 = X1 2 + Y1 2 Cb: b 2 = X1 2 + (L - Y1) 2 = X1 2 + Y1 2-2LY1 + L 2 Cc: c 2 = (L - X1) 2 + (L - Y1) 2 = X1 2 + Y1 2-2LX1 + L 2-2LY1 + L 2 Cd: d 2 = (X1 L) 2 + Y1 2 = X1 2 + Y1 2-2LX1 + L 2 So... they are the same formulas, the only difference is that they intersect inside the square based on the side of L verses X1, Y1... therefore the proof is the same as for inside the circle as per Part 1... concluding that a all rational a,b,c,d,l cannot exist, for both inside and outside the circle.