Source: Created by reader Sudipta Das.
TS1 TD DS2 DS1 TH 9 8 9 7 14 11 12 11 7 14 13 16 17 11 18 17 16 13 9 22 18 16 18 14 28 19 12 19 17 34
Give the points coordinates as follows: T (0, 0), D (d, 0), S1 (u, v), H (2u - d, 2v), S2 (u - d/2, v). Then (DS2)^2 = (u - 3d/2)^2 + v^2 = m^2 (say) (DS1)^2 = (u - d)^2 + v^2 = n^2 (say) (TS1)^2 = u^2 + v^2 = 64 = k^2 (say) Eliminating u and v from these three equations: d^2 = (4 * m^2 - 6 * n^2 +2 * k^2)/3..... (1) To ensure that no straight line can go through T, D, and H the sum of any two sides of the triangle TDH must be greater than the third side. With that condition and with k = 8 the only values for which the RHS is a square are m = 7, n = 6 and hence d = 6. This is the result for part 1. We have (TH)^2 = (2u - d)^2 + (2v)^2 and with the earlier equations we can eliminate u and v giving: (TH)^2 = 2k^2 - d^2 + 2n^2 ..... (2) For n = 6, k = 8 and d = 6 (the results of part 1), (TH)^2 = 164 and TH = 12.8... and is not an integer. This is the result for part 2. For part 3 we require integral values of k, m and n such that d, given by (1) and the not-in-line conditions, and TH, given by (2), are integers. This gives rise to the tabular result for part 3. Note that all the numbers in any row can be multiplied by an integer to give a result for part 3. For example row 5 is twice row 1. Some results give a symmetrical diagram with DH = 2(DS1) = TH and TS1 = DS2.
Suppose the positions of Tom , Dick and Harry are T , D and H respectively. Let TD = x yards DS1 = HS1 = y yards TS2 = HS2 = z yards DS2 = t yards Now, TS1 and DS2 are the medians of the triangle TDH. So, x ^ 2 + ( 2*y ) ^2 = 2 * ( t^2 + z^2 ) => z ^2 = x^2 /2 + 2* y^2 - t^2 And , x^2 + ( 2*z ) ^2 = 2* ( 8^2 + y^2 ) => x ^2 + 2*x^2 + 8*y^2 - 4*t^2 = 128 + 2 * y^2 => ( x^2 + 2*y^2 ) /4 = ( t^2 + 32 ) / 3 ...(1) Since x, y and t are all integers, ( t^2 + 32 ) must be a multiple of 3, and ( x^2 + 2*y ^ 2 ) must be a multiple of 4. This is possible, if t is not of the form 3*k ( i.e. if t = 1,2,4,5,7,8,... ), and, if x and y are both even. Let TS1 and DS2 meet at the point O, which is then the centroid of the triangle TDH . So, TO = (2/3) * TS1 DO = (2/3) * DS2 OS1 = (1/3) * TS1 Now, since sum of two sides of a triangle is greater than the third side, x + y > 8 TO + DO > TD => x < (2/3) * ( t + 8 ) ...(2) DO + OS1 > DS1 => y < (2/3) * ( t + 4 ) ...(3) Using relations (1), (2) and (3), we find that the only solution for which x, y and t are all integers is t = 7 x = 6 y = 6 and z = sqrt (41) So, Dick is 6 yards from Tom , 12 yards from Harry and 7 yards from S2. Since this is the only solution, Tom cannot be an integral number of yards from Harry.