## Two Scarecrows

1. Tom, Dick and Harry are sitting in a cornfield, in which there are two scarecrows. One of the scarecrows (S1) lies mid-way between Dick and Harry, while the other one (S2) lies mid-way between Tom and Harry. Tom is 8 yards from S1, while Dick is an integral number of yards from each of Tom, Harry, S1 and S2. Assume that no straight line can go through all the 3 guys. What are the possible distances of Tom, Harry and S2 from Dick?
2. If Tom is also an integral number of yards from Harry, can you find a solution?
3. If no answer exists for part 2, can you find a distance between Tom and S1, such that all distances are integral?

Source: Created by reader Sudipta Das.

Denis Borris suggested an extension to this puzzle:
Tom, Dick and Harry are sitting in a cornfield, in which there are three scarecrows. One of the scarecrows (S1) lies mid-way between Dick and Harry, another (S2) lies mid-way between Tom and Harry, and the 3rd one (S3) lies mid-way between Tom and Dick. The distance between any of the 3 guys is even. The distance between any scarecrow and any guy is also even.
DistanceTom:Dick > distanceTom:Harry > distanceDick:Harry.
Distance Dick:Harry is at minimum. Find the above 3 distances.
Solutions were received from Denis Borris, John Hewson, and Sudipta Das. The answers are here and the more thorough solutions are below.
1. Tom(T), Harry(H), and S2 are 6, 12, and 7 yards respectively from Dick(D).
2. There is no solution with Tom 8 yards from S1, and an integral number of yards from Harry.
3. There are numerous solutions with all distances integral. The table shows the first few in increasing order of TS1 (yards between Tom and S1).
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

From John Hewson:
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.

From Sudipta Das:
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.

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