Problems & Puzzles: Puzzles

Puzzle 224. Eccentric prime diagonals ^(*)

Let P be a point inside a rectangle axb so that the distances from P to the four vertices, in rotational order, are the primes p, q, r, and s.

a) Find the minimal solution if a=b, reals.
b) Find the minimal solution if a<>b, reals.
c) Find the minimal solution if a=b, rationals.
d) Find the minimal solution if a<>b, rationals.
e) Find the minimal solution if a=b, integers.
f) Find the minimal solution if a<>b, integers.

(In each case send a, b, p, q, r, and s)

_________
^(*) based in another puzzle published out there in a place that I will let you know later...

Solution:

J. C. Rosa and independently Jan van Delden sent more or less by the same time the algebraic solution for this figure. One more partial approach came from Ken Wilke.

J. C. Rosa wrote:

Let's start by drawing horizontal and vertical lines through P. Then the whole rectangle can be divided in four smaller rectangles having each as diagonals p, q, r, & s, respectively. Let's call x and y the horizontal and vertical sides of the right-angled triangle of the hypotenuse s. Then we may obtain the following four equations:

• s^2=x^2+y^2 ..................(1)

• p^2=x^2+(b-y)^2..............(2)

• q^2=(a-x)^2+(b-y)^2.........(3)

• r=(a-x)^2+y^2...................(4)

From (1),(3),(2)&(4) we have:

p^2+r^2 = s^2+q^2 ...........(5)

This equation shows that in fact there are only 3 independent eccentric diagonals. The importance of this will be evident later soon.

Case I. a=b:

From (1) & (2) we obtain:

y=(s^2+a^2-p^2)/(2.a) ...........(6)

Similarly, from (1) & (4) we obtain:

x=(s^2+a^2-r^2)/(2.a) ............(7)

Substituting (6) & (7) in (3) we obtain after a long a winding road:

a=b=sqrt[(p^2+r^2+/-sqrt(4.s^2.q^2-(p^2-r^2)^2))/2] ..........(8)

Case II. a<>b

Let's suppose that we have s, p & q as the independent variables. By observing (1), (2) & (3) we know that we have 3 equations and 4 unknowns (a, b, x & y). So in order we have a determined system we need introduce one more independent data. This independent data can not be r, because by (5) r is dependent of s, p & q. So we may introduce a as another independent data, and in this very moment, b, x & y are determined by the following system of equations in cascade:

Form (1), (4) & (5):

x = (p^2+a^2-q^2)/(2.a)  ............ (9)

From (1):

y = sqrt(s^2-x^2)..................(1')

From (2):

b = y+sqrt(p^2-x^2)...............(2')

The equation (9) shows that as x must be positive, then a>sqrt(q^2-p^2). On the other hand a must be less than the minimal of the following two quantities: (p+q) and (r+s). In short:

sqrt(q^2-p^2)<= a <=MIN{(p+q),(r+s)}...........(10)

Something similar to (10) can be stated regarding b:

sqrt(r^2-s^2)<= b <=MIN{(p+s),(q+r)}...........(11)

J. C. Rosa found the smallest solution for a rectangle with integer sides:

(p, q, r, s, a, b) = (37, 37, 13, 13, 24, 40)

(Note from the editor: The a smallest solution - if we are permitted that the point P is located not inside the rectangle but over one of its sides- is: (p, q, r, s, a, b) = (5, 5, 3, 3, 6, 4)

 

Jan van Delden wrote:

First of all I only considered the easy case, a and b real. Applying Pythagoras 4 times we get the following equations:

x1^2+y1^2=p^2 with
x1^2+y2^2=s^2 x1+x2=a
x2^2+y1^2=q^2 (**) y1+y2=b
x2^2+y2^2=r^2

Upon subtracting these in pairs we get:

y1^2-y2^2=p^2-s^2=q^2-r^2
x1^2-x2^2=p^2-q^2=s^2-r^2

Giving in either case the following equation p^2+r^2=q^2+s^2. (*)

Since an eccentric solution is required not all the p, q, r, s are equal.
Several cases need to be considered.

All 4 primes are distinct.
Suppose x1>x2 and y1>y2 gives the following implied orders:
(This does not limit the possibilities, it only rotates/flips the rectangle in question).

r<s<q<p or r<q<s<p.

I programmed a little routine in Maple to do a search, for say the
first 100 primes (2 not included, since then we would have a problem
mod 2 in (*)) complying to (*).

If we call the indices of the primes used (in order) i,,j,k,l
I searched as follows (n maximum index):

i in [3,n-3]
j in [i+1,n-2]
k in [j+1,n-1]
l in [k+1,n]
Test equality (*)

This can obviously be improved upon since a different order might
diminish the number of tests involved considerably. For instance if
we suppose that: r<s<q<p

It might be smarter to do our loops in the following order:
choose r, choose p, choose q where we could give sharper bounds
for at least q and test whether p^2+r^2-q^2 is a square of a prime.

I applied this procedure on a list which contained the first 100
primes, which gave 1572 different solutions in p, q, r, s.

The smallest being 7,11,17,19 for which I will show an answer to
your questions a) and b).
The biggest in this list 431,479,499,541.

It's straightforward to show that our set of equations (**) are
dependent, it's possible to choose 1 of our 4 variables.

Let's take y2 as remaining variable and express the rest into y2:
Giving
x2^2=r^2-y2^2
y1^2=q^2-r^2+y2^2
x1^2=s^2-y2^2 (use p^2+r^2=q^2+s^2)

Since r is my smallest value we can take 0<y2<r. It's a simple matter
to show this is a sufficient condition to make sure the other variables
are all real and positive.

Filling in the values 7,11,17,19 for r, s, q, p respectively gives:
(the other order r, q, s, p doesn't seem to give an answer to our
problem, I have the distinct impression this is always the case)

a = x1+x2 = sqrt(49-y2^2)+sqrt(121-y2^2) (***)
b = y1+y2 = sqrt(240+y2^2)+y2

'Answer' to b)

(One of them, there is an infinite number of them!)
Take y2=4 then sqrt(240+y2^2)=16 (at least one side integer)

a=sqrt(33)+sqrt(105)
b=20

So our average is equal to 13.5

'Answer' to a)

Same set of primes.

Solve a=b in (***) gives the answer:

y2 =2.11915406981173228081...
a=b=17.465462617094377467797...

I used Maple to solve this equation, hope you don't mind. The
answer for a, b can be expressed as a radical, I will however omit this
detail, because among other things it doesn't gain a lot of insight in
the answer obtained.

Two pairs of distinct primes (r=s<q=p, only case considering presumed order)
Since p^2+r^2=q^2+s^2 is always fulfilled, we get a sharper bound on the answer to question a and b.

Answer a, b)

r=s=2,p=q=3 with minimal average 2.5

Solving for a=b as before we get:
y2=sqrt(38-2*sqrt(119))/4 about equal to 1.005689308718
a=b=sqrt(26+2*sqrt(119))/2 about 3.457507202786

By the way these radicals are much easier to deal with. This is not so strange since we now have solved:

x2^2=r^2-y2^2
y1^2=p^2-r^2+y2^2
and x1=x2

So:

a=x1+x2=2*x2 gives:
a=2*sqrt(r^2-y2^2)
b=y2+sqrt(p^2-r^2+y2^2)

Giving a quadratic equation in y2^2.
This will only have an answer if 8*r^2*p^2>=(p^2+r^2)^2 (inner root)
Or (p+r)^2<=(2+sqrt(8))*r*p. Could be drawn to make a nice picture.

These equations will be the easiest to handle and obtain solutions in rationals/ integers.

(With different order r, q, s, p we get the same solution, but mirrored).

One pair of primes (r<s=q<p, only case considering presumed order).
It is possible to establish a 'relation' to limit the search. First of all r can not be 2. Using the fact that r,q,p are odd, one could get an expression of the sort:
(k^2+l^2)=s*(k-l) with r=s-2*k, l=s+2*l, so s and k-l should be a divisor of k^2+l^2.

I didn't pursue this any further.
In any case a quick calculation gives 7 13 13 17 as smallest result, and a total of 17 using the same list of primes.

One pair of primes (r=s<q<p or r<s<q=p).
Obviously no solution since then (*) not fulfilled.

Some remaining questions:
The order r<q<s<p doesn't lead to a solution, given x1>x2 and
y1>y2? (Different primes, geometric problem..).

How to see immediately that there's a solution of the type a=b, given
the solution p,q,r,s (or not). (Different primes or one identical pair).

And for your questions c through f, for which p,q,r,s does (***) in
general have rational/integer solutions.

***

At the end from both approaches we conclude that when we are given a quadruplet of valid eccentric prime diagonals (p, q, r & s), that is to say, a quadruplet that satisfies the equation (5), we are sure that these an be arranged into infinite rectangles and at the much in only one square, or none. Here are some examples:

Example 1:

The minimal solution with all the eccentric diagonals prime and distinct is this one:
(p, q, r, s) = (11, 19, 17, 7)

• The ranges for a and b are: 15.492<=a<=24, 8.485<=b<=18
• The square solution is a=b=17.553
• The rectangle solutions for some selected integer values of a, are:

a=16, b=17.971, x=0.500, y=6.982
a=17, b=17.755, x=1.441, y=6.850
...
a=23, b=12.116, x=6.283, y=3.087
a=24, b=  8.485, x=7.000, y=0.000

***

In short the  Ken Wilke's solution is:

x=p.s/k
y=s.r/k
a=(p.s+q.r)/k
b=(p.q+s.r)/k
k=sqrt(p^2+r^2)

It corresponds to a particular rectangle axb such that the inscribed quadrilateral that can be formed using the complementary eccentric diagonals  is a 'cyclic quadrilateral', that is to say, a quadrilateral that, in its turn, can be inscribed in a circle. A complementary diagonal is simply the other diagonal that is present in each smaller rectangle for each original eccentric diagonal (please see my emergency sketch-drawing below).

This solution is interesting because algebraically is evidently easy to calculate with its own evident arithmetical symmetry, and  geometrically has its definitive charming.

***