Rational Distance Problem for the Unit Square

Transcription

1 Rational Distance Problem for the Unit Square Ameet Sharma November 18, 015 Abstract We present a proof of the non-existence of points at a rational distance from all 4 corners of a square with rational lengths. A Preliminaries This problem is described in [3]. We're given a square ABCD with rational length g. We assume the existence of a point S at a rational distance from all of its corners. The coordinates of S are (x, y). We assume the square has corners (0,0) and (g,g). We call such a point S a rational-distance point. We note that for two dierent squares length g 1, and g, if we have a rationaldistance point for g 1, say (x 1, y 1 ), then we can immediately get a rational distance point for the g square (x, y ) by scaling the coordinates. x = g g 1 x 1 and y = g g 1 y 1. So the non-existence of a rational-distance point for any one rational square proves the result for all rational squares. We show that the existence of such a rational distance point S(x, y) leads to a contradiction. B Outline of the argument 1. Prove that x and y are rational. 1

2 . Prove that the existence of S on a line containing a side of the square gives a contradiction. 3. In preparation for the next parts we shift coordinates and perform a change of variables. We set our point of interest S as the origin (0,0). And we set the lower left corner of the square as having coordinates (x,y). 4. Get equations relating the slopes of the lines from the origin to the 4 corners. Slopes are h a,h b,h c,h d 5. Scale the unit square situation up by the minimum value so that x,y and g are all integers (so the length of the square and all the distances between points are scaled up. But the slopes remain the same). x,y and g have no common factor other than 1. Now we have 4 pythagorean triples. We prove one of x and y is odd and the other is even. We prove g is even. 6. Divide the plane into 9 regions. Prove that S cannot exist in any of the outer regions. 7. Prove S cannot exist inside the square. C Details C.1 Proof of rationality of coordinates. Given our square with corners (0,0) and (g,g), if S lies on a line containing a side of the square, its coordinates are rational. For example, suppose S lies on the line joining (g,0) and (g,g). Its x coordinate is g which is rational by assumption. And its y coordinate is plus or minus the distance from the point (g,0). By assumption this distance is rational so y must be rational. Same argument applies to any line containing any side of the square. The gures on (F.1) illustrate the other possible scenarios. Remaining scenarios are covered by rotational symmetry. In general, (E.1) Area( DSA) = xg

3 (F.1) Figure 1: Figure : Figure 3: Figure 4: 3

4 Area( ASB) = yg Area( DSC) =± (y g)g Area( CSB) = ± (x g)g We set g = 1, and go to the unit square situation. We call coordinates of S, (x 0, y 0 ). (E.) Area( DSA) = x 0 Area( ASB) = y 0 Area( DSC) = ± y 0 1 Area( CSB) = ± x 0 1 If we establish that the areas of the above triangles are rational, then we've established that x 0 and y 0 are rational. [] and [5] discuss square roots and linearly independence. We summarize the result we need from these articles: (E.3) Given rational numbers r 1, r, r 3...r n, which need not be distinct, and given that r 1, r, r 3... r n are all irrational, if ± r 1 ± r ± r3... ± r n = q, where q is a rational number, then q = 0. To put it simply, the only way a sum of irrational square roots or their negatives add to a rational number, is if that rational number is 0. As shown by R. Barbara in [1], given a point S separate from a polygon, the area of the polygon can be formed by adding or subtracting the area of the triangles formed by S and each of the sides of the polygon. Also shown in [1], the area of each triangle has the form r i where r i is rational. So applying this to our situation with 4 triangles and a unit square: Let r 1 = Area( DSA), r = Area( ASB), r 3 = Area( DSC), r 4 4

5 = Area( CSB) (E.4) ± r 1 ± r ± r 3 ± r 4 = 1 This applies to Figure 1,, 3 or 4. We investigate whether any of these ± r i can be irrational. From (E.), we can see that areas have an x 0 and areas have a y 0. If one of the areas with an x 0 is irrational, then x 0 is irrational and so is the other area with an x 0. Similarly with y 0. So we have 3 possibilities: 0 irrational terms, irrational terms (corresponding to the x 0 -pair areas or the y 0 -pair areas alone), or 4 irrational areas (both x 0 and y 0 pairs). Can we have all 4 terms in (E.4) be irrational? No, because then, by (E.3) the left side of (E.4) would be 0, whereas the right side of (E.4) is 1. This would be a contradiction. Suppose we have just irrational terms. By (E.3) they must add to zero. Either it's the two x 0 -terms or the two y 0 -terms. Suppose it's the two x 0 - terms. x 0 +( x 0 1 ) = 0, gives no solution for x 0. That only leaves the possibility that x 0 + x 0 1 = 0. Solving this equation gives a rational x 0 value, which contradicts our assumption of irrational x 0 terms. By exactly symmetrical arguments the two y 0 terms alone being irrational also leads to a contradiction. We are left with the conclusion that we have 0 irrational terms in (E.3), hence x 0 and y 0 are rational. So by scaling, S has rational coordinates for all squares. C. Prove the existence of S on a line containing a side of the square gives a contradiction. We stay in the unit square situation. Can S be a corner? No, because then its distance from the opposite corner is which is irrational. We check other points on a line going through a side of the square, as per gure 5 (F.). Assume S is on line l1 going through DC. S can be on DC or outside of it. Without loss of generality we assume the x-coordinate is positive. We 5

6 assume length SA and SB are rational. This will lead to a contradiction. (F.) Figure 5: let: x = a b, y = 1, SA = c d Where a,b,c,d are integers. The fractions are in lowest terms so a and b are coprime. So are c and d. SA = ( a b ) + 1 ( c d ) = ( a b ) + 1 b c = d (a + b ) We know that b and a + b are coprime. Hence b d. We know that d and c are coprime, hence d b. This means that b = d so b = d. And c = a + b. let SB = e f, where e and f are coprime. SB = 1 + (1 a b ) ( e f ) = 1 + (1 a b ) ( e f ) = ( a b ) + a b b e = (b ab + a )f. Using the same argument as before we get that f = b. We can divide both sides of the equation by b. To summarize our results so far: 6

7 (E.5) x = a b, y = 1 SA = c b SB = e b c = a + b e = b ab + a. (a, b, c) forms a primitive pythagorean triple. We can use the pythagorean triple parametrization given here: [6]. We need to check possibilities. First when a is the even leg, then when a is the odd leg. a = mn b = m n c = m + n where m and n are coprime positive integers, m > n and m n is odd. Substitute these into the e equation in (E.5) e = (m n ) (mn)(m n ) + (mn). e = (m 4 m 3 n + mn 3 + n 4 ) We know that m n is odd, hence one of m and n is odd, and the other is even. Therefore the term in the parentheses must be odd. Therefore only occurs an odd number of times in the prime factorization of e hence there's no rational solution for e. We also need to check when a is the odd leg: b = mn a = m n c = m + n Substitute these into the e equation in (E.5) e = (mn) (m n )(mn) + (m n ). e = m 4 4m 3 n + 6m n + 4mn 3 + n 4 e = (m + n) 4 8m 3 n 7

8 math.stackexchange.com user Steven Gregory pointed out the following algebraic manipulation which turns e into a sum of a squares. [4]: e = (m mn n ) + (mn) We have another pythagorean triple. We assume m mn n > 0. We examine whether or not it is primitive. m mn n is odd, so does not divide it. any factor of m does not divide it. any factor of n does not divide it. So m mn n and mn are coprime. So we have another primitive pythagorean triple. We can give similar parameters to: (E.6) mn = m n m mn n = m n m mn n = (m n )(m + n ) So m n m mn n and m + n m mn n (m + n ) + (m n ) m mn n (m + n ) (m n ) m mn n m m mn n n m mn n m m mn n 8

9 n m mn n Since m and n are coprime, we can multiply the above two to get: m n (m mn n ) And then using (E.6) mn (m mn n ) mn (m n ) Any prime factor p of m divides the left side. So p divides the right side. So p divides n and p divides n. So m and n have a common prime factor. But this is impossible since m and n are coprime by our pythagorean triple parametrization. We've arrived at a contradiction. In the case of m mn n < 0 we perform the same argument, except using n +mn m = m n. In the exact same manner we arrive at a contradiction again. We don't need to worry about m mn n = 0 since m mn n is always odd and 0 is even. This completes our proof that S cannot exist on a line containing a side of the square. C.3 Shift coordinates, change of variables. For convenience, we now set S as the origin, and the coordinates (x,y) as the lower left corner of the square. As an example, Figure 6 (F.3) shows the situation when when the origin is on the lower left side of the bottom left corner of the square.. 9

10 (F.3) Figure 6: C.4 Get equations for the slopes of the lines from the origin to the corners. Let h a, h b, h c, h d be the slopes of the lines from the origin to A,B,C,D respectively. (E.7) h a = y x h b = y x+g h a = y+g x+g h a = y+g x We know that x 0 x + g 0 y 0 y + g 0 The reason is because if any of these values were 0, then that would mean the x or y axis contains a side of the square. But the x and y axes contain 10

11 the origin which is our point of interest S. This gives a contradiction since we've proven that a line containing a side of the square cannot contain our point of interest. So all our slopes in (E.7) are well dened and not equal to zero. Suppose h c = h a. This would mean that the line through the origin through C also goes through A. If S lies outside the square then either SA = SC+g or SC = SA + g. This is impossible since is irrational but g, SA and SC are rational. If the origin is inside the square then SA + SC = g. This is also impossible for the same reason. So we get a contradiction and h c h a. Similarly h d h b. We also know that h d h c and h b h c and h a h d and h a h b because these possibilities lead to zero slopes or undened slopes which we've proven don't occur in our situation. So all four of these slopes are dierent. Rewrite (E.7): (E.8) h a x y = 0 h b x y + h b g = 0 h d x y g = 0 h c x y + (h c 1)g = 0 Suppose we are given arbitrary nonzero values of h a, h b, h c and h d that t our situation (i.e.: nonzero slopes, all slopes dierent etc.). Now we try to nd values of x, y, g which t the above equations. We have a set of 4 linear equations in terms of x,y and g. These equations are consistent because they have a solution x = 0, y = 0, g = 0 but this solution does not t our situation since we can't have a square with length 0. We need other solutions. So this system of equations has to be dependent (only a dependent system has multiple solutions). We can set the equations in augmented matrix form, and reduce them to row echelon form. We have 3 variables, so for the system to be dependent, the number of nonzero rows has to be or less. 11

12 h a h b 1 h b 0 h d h c 1 h c 1 0 In row echelon form we have (h c 1) h c h a h b h b h a 0 h a h b h b h c 0 h d h a + h b h b h a 0 h b h b h a 0 We require two zero rows for a dependent system, so 1 h d h a = h b h b h a (h c 1) h c h a = h b h b h a h a h b = h b (h d h a ) (h c 1)(h b h a ) = h b (h c h a ) Solve for h d and h c in terms of h a and h b (E.9) h d = h a h b + h a 1 h c = h b h a + h b + 1 1

13 C.5 Scale up the situation to the minimum value where x,y and g are all integers. So suppose we're in the unit square situation. g=1. We multiply the coordinates of all the corners of the squares by the minimum positive integer required to make x and y integers if they both aren't integers already. So x,y and g are all integers now. But they don't share a common factor other than 1. We have the following equations: (E.10) SA = x + y SB = (x + g) + y SC = (x + g) + (y + g) SD = x + (y + g) Looking above, we see that SA, SD, SC and SB are algebraic integers (i.e.: they are zero's of polynomials with integer coecients). Since we already know they are rational, then they must be regular integers since regular integers are the only rational algebraic integers. This means that (E.10) presents a set of 4 pythagorean triples. Not necessarily primitive. We know that at least one leg of a triple is even. So looking at the rst triple either x or y are even, or both. Suppose both x and y are even. Then looking at the third triple, g must be even. But this would make x,y and g all even, so they'd share a common factor. This is impossible by our starting assumption. So either x is even or y is even but not both. Suppose g is odd. Then either the nd or 4th triples consists of two odd integers as legs. This is impossible, so g must be even. There are cases 1. x is odd, g is even, x+g is odd, y is even, y+g is even.. x is even, g is even, x+g is even, y is odd, y+g is odd. We only need to consider the rst case. The second case is automatically taken care of by symmetry across the line y=x. 13

14 C.6 Divide the plane into 9 regions. Prove that S cannot exist on any of the outer regions. We divide the plane into 9 regions as per gure 7, (F.4). Our square encloses region 5 in the center. (F.4) Figure 7: First we assume h a and h b are positive. So the origin is located in either regions 1,6 or 9. x is odd, y is even and g is even. We give a pythagorean triple parametrization to our points A and B, as per [6]. Although we may have non-primitive pythagorean triples, we avoid bringing in the scale factor k, by simply dealing with ratios and slopes. (E.11) h a = y x = m an a h b = m a n y a x+g = m bn b m b n b Then we substitute these expressions into our equation for h d in (E.9), and simplify. h d = y+g x = m an a (m b +m bn b n b ) m bn b (m a n a) m b n b (m a n a) 14

15 Note that m a n a is even, m b n b is even, m b m bn b n b is odd and m a n a is odd. Let GCD(m a n a, m b n b ) = r s where where s is odd. Let m b n b = r s where s is odd. r is the highest power of that divides the numerator. power of that divides the denominator. r is the highest r r So we know that h d in lowest terms is either an odd integer divided by an odd integer, or an odd integer divided by an even integer. We know y + g = h d x. Since x is odd, the right hand side is either an odd integer times an odd integer divided by an even integer. This result can never be an integer. Or it's an odd integer times an odd integer divided by an odd integer. This result is either an odd integer or not an integer at all. But by our starting assumption, y + g is an even integer. So we have a contradiction. This eliminates regions 1,6 and 9. By rotational symmetry we automatically eliminate regions,3,4,7 and 8. We are only left with region 5, inside the square. Nothing of signicance changes. Only the sign of h b. C.7 Prove S exist cannot inside the square. Here h a is positive and h b is negative (E.1) h a = y x = m an a h b = m a n a y x+g = m bn b m b n b Again using (E.9) we get: 15

16 h d = y+g x = m an a (m b m bn b n b ) m bn b (m a n a) m b n b (m a n a) Again, the exact same argument in C.6 eliminates this case. This completes our proof for the non-existence of a point at a rational distance from the corners of a rational square. References [1] Roy Barbara. Points at rational distance from the vertices of a unit polygon. Bulletin of the Iranian Mathematical Society, Vol. 35(No. ):pp. 0915, 009. URL [] The Math Drexel. Linear independence of square roots of primes, URL html. [3] Richard K. Guy. Unsolved Problems in Number Theory. Springer, 005. [4] Ameet Sharma ( sharma). Solutions of diophantine eq: x 4 x 3 y + xy 3 + y 4 = s. Mathematics Stack Exchange. URL q/ URL: (version: ). [5] Martin Klazar. A question on linear indepdence of square roots, 009. URL [6] Wikipedia. Pythagorean triple, 014. URL wiki/pythagorean_triple. 16