Distances to Rectangle Vertices

Let vertices of rectangle ABCD be (0,0), (0,a), (b,0),(a,b). Let P be the point (x,y) Then the square of the distances AP, BP, CP, DP are (x^2+y^2), (x^2+(y-a)^2), ((x-b)^2+y^2) and ((x-b)^2 + (y-a)^2) in some order. ((x-b)^2 + (y-a)^2) + (x^2+y^2) = (x^2+(y-a)^2)+((x-b)^2+y^2) and three of these values are 7^2 = 49, 15^2 = 225, 24^2 = 576. The only combination giving the fourth value as an integer is: 576 + 49 = 225 + N^2 N = 20

Let [7], [15] and [24] denote the vertices at distance 7, 15 and 24 respectively. [7][15] cannot be a diagonal, since a diagonal in a rectangle is the longest distance contained within it and 7+15 = 22 < 24. Thus either [7][24] or [15][24] must be a diagonal...

Assume [7][24] to be diagonal. Then triangle [7][15][24] has a right angle at [15]. Let s denote the length of the side [7][15] and t [15][24]. If x denotes the angle at [15] in triangle [15]P[24] the law of cosines leads to (refer to the dynamic diagram

24² = 15²+t²-2∙15t∙cos(x) and 7² = 15²+s²-2∙15s∙sin(x)

In turn this leads to

N² = (t-15cos(x))²+(s-15sin(x))² = [30²cos²(x)-4∙(15²-24²)]/4 +[30²sin²(x)-4∙(15²-7²)]/4 = [30²-4∙(15²-24²+15²-7²)]/4 = 400

So here N = 20. That is one solution. Doing the similar calculations assuming [15][24] to be a diagonal leads to

N² = [14²-4∙(7²-24²+7²-15²)]/4 = 752

which is not a perfect square - so then N is not an integer. Problem solved ;).

It surprised me to see that N does not depend on the angle x so any rectangle it is possible to form with [7][24] as a diagonal has N = 20. And any rectangle with [15][24] as a diagonal has N = 4√47.

See if there is a point P in a rectangle ABCD then we have a property that PA^2 + PC^2 = PB^2 + PD^2 (easy to prove by dropping perpendiculars from P to the sides and using Pythagorean Theorem). So we want to find an integer x such that the squares of 2 pairs of 7, 15, 24 and X add up to the same value. Now we can either do it by trial and error (only 3 possible combos to check anyway!) or if alert we can see that 7, 24 already are 2 parts of a Pyth Triplet (7, 24, 25) and if we take X as 20, then (15, 20, 25) also forms a Pyth Triplet, and hence 20 should be the required value..

Draw perpendicular lines from P to AB, BC, CD and DA and label them W, X, Y and Z respectively.

If

PA = 7

PB = 24

PD = 15

PC = N

Then

PC^2 = N^2

       = X^2 + Y^2

       = (PB^2 - W^2) + (PD^2 - Z^2)

       = 24^2 - W^2 + 15^2 - Z^2

       = 24^2 + 15^2 - (W^2 + Z^2)

       = 24^2 + 15^2 - PA^2

       = 24^2 + 15^2 - 7^2

       = 576 + 225 - 49

So we have

N^2 = 576 + 225 - 49 = 752

Similarly, we have

N^2 = 576 + 49 - 225 = 400

N^2 = 225 + 49 - 576= -302

Clearly, only

N^2 = 400

has a positive integer solution

Hence N = 20

Draw two segments parallel to the sides of the rectangle. The intersection of these two segments must be Point P.

Each of the 2 segments are now divided into 2 parts. Let the two parts of the segment parallel to CD and AB be a and b and we also let the two parts of the segment parallel to AD and BC be c and d.

AP, BP, CP, and DP became the diagonals of the four rectangles formed. Assume that DP < AP < CP < BP.

We will now have four cases.

Case 1: DP = N

                                    a^2+d^2 = N^2

(a^2+c^2)-(b^2+c^2)+(b^2+d^2) = N^2 49-576+225 = N^2 -302= N^2 But there is no imaginary length, therefore N > 7.

Case 2: AP = N

                                    a^2+c^2 = N^2

(a^2+d^2)-(b^2+d^2)+(b^2+c^2) = N^2 49-225+576 = N^2 400 = N^2 N = 20 For each of the remaining cases, we will get the same result. Therefore, N = 20.

Because P is in rectangle ABCD, we can write down consecutively: PA = 7, PB = 15, PC = 24, and PD = N.

From British Flag theorem, we get PA^2 + PC^2 = PB^2 + PD^2 \leftrightarrow 7^2 + 24^2 = 15^2 + N^2 N^2 = 7^2 + 24^2 - 15^2 = 400 \leftrightarrow N = 20

Let AB = DC = a and BC = AD = b, now if we draw a line parallel to BC containing the point P, let E and H be the intersection points of this line with the sides AB and DC respectively. Of course EP is perpendicular to AB, becauuse EP is parallel to BC, so we can use the pythagorean theorem to find out the distances from P to A,B,C and D. PA² = PE² + AE² PB² = PE² + (a-AE)² PC² = (b - PE)² + (a-AE)² PD² = (b-PE)² + AE² So we can say: PA² + PC² = PB² + PD² which is(PE² + AE² + (a - PE)² + (b-PE)²) If PA = 7 , PC = 24, PB = 15 and PD = N 7² + 24² = 15² + N² N² = 7² + 24² - 15² = 400, therefore N=20

Let the projection of $P$ onto $AB$ and $BC$ be $E$ and $F$ , respectively. By the Pythagorean theorem, $PA^2 + PC^2 = PE^2 + EA^2 + PF^2 +FC^2 = PE^2+PF^2 +AE^2 +FC^2 = PB^2 + PD^2$ . Since we are not given which lengths correspond to which edges, we have the following cases:

Case 1: $24^2 + N^2 = 7^2 + 15^2$ . This gives $N^2 = 7^2+15^2 - 24^2 < 0$ which has no solution.

Case 2: $24^2 + 7^2 = 15^2 + N^2$ . This gives $N^2 = 400$ so $N=20$ .

Case 3: $24^2 + 15^2 = 7^2 + N^2$ . This gives $N^2 = 752$ , which is not a perfect square (since no perfect square ends with 2 as the units digit).

Hence, the only solution is $N=20$ .

Note: The equality $PA^2 + PC^2 = PB^2 + PD^2$ is sometimes known as the British Flag Theorem.

When one's only tool is a hammer everything looks like a nail. Anyway, this looked like an interesting exercise for which to use sympy's geometry routines. (Or that's my excuse this time.) The first permutation of possibilities yielded an answer, consequently I never finished coding. Nevertheless my line of reasoning is probably obvious.

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from sympy import *
from itertools import permutations

P = Point ( 0, 0 )
UL_corner = Point ( -4*sqrt(3), 1 )

top_line = Line ( Point ( -4*sqrt(3), 1 ), Point ( 10 + -4*sqrt(3), 1 ) )
left_line = Line ( Point ( -4*sqrt(3), 1 ), Point ( -4*sqrt(3), -10 ) )

perm = permutations ( [ 15, 24, 'N' ] )
for p in perm :
    print '--->', p
    if p . index ( 'N' ) == 2 :
        c = Circle ( P, p [ 0 ] )
        corner = top_line . intersection ( c ) [ 1 ]
        trial_right_line = Line ( corner, corner - Point ( 0, 10 ) )
        c = Circle ( P, p [ 1 ] )
        trial_corner = trial_right_line . intersection ( c )
        if trial_corner :
            trial_lower_left_corner = left_line . intersection ( Line ( trial_corner [ 1 ], trial_corner [ 1 ] - Point ( 1, 0 ) ) )
            trial_N = trial_lower_left_corner [ 0 ] . distance ( P )
            print trial_N 
        else :
            print 'not feasible'
    elif p . index ( 'N' ) == 1 :
        pass
    else :
        pass

The following claim is proved first.

If P is any point in rectangle ABCD, PA^2 + PC^2= PB^2 + PD^2 Proof

Let the co-ordinates of A,B,C, and D be respectively (-l, b), ( -l, -b), (l, -b), and (l, b).

Let the co-ordinates of P be (m, n).

      PB^2= (m+l)^2 + (n+b)^2

Then, PA^2= (m+l)^2 + (n-b)^2 PC^2= (m-l)^2 + (n+b)^2 PD^2= (m-l)^2 + (n-b)^2

So, PA^2 + PC^2= (m+l)^2 + (m-l)^2 + (n+b)^2 + (n-b)^2 PB^2 + PD^2= (m+l)^2 + (m-l)^2 + (n+b)^2 + (n-b)^2

Therefore, PA^2 + PC^2= PB^2 + PD^2 [proven].

Now let N=PA.

So, N^2= PA^2 + PC^2= PB^2 + PD^2 - PC^2

The possible values of N are listed below.

a)7^2 + 15^2 - 24^2= 49+225-576= -302 [not possible because N^2 cannot be negative] b) 24^2 + 15^2 - 7^2= 576 + 225 - 49= 752 [not possible because N is an integer and 752 is not as perfect square] c) 24^2 + 7^2 - 15^2 = 576 + 49 - 225 = 400= 20^2 [the only possibility]

So the value of N is 20.