Given a = PA = 19, b = PB = 29, c = PC = 23

(1) a² = x² + (t-y)²

(2) b² = x² + y²

(3) c² = (t-x)² + y²

(4) d² = (t-x)² + (t-y)²

adding a² and c² together on the one hand, adding b² and d² on the other hand, we get :

a² + c² = x² + (t-y)² + (t-x)² + y²

b² + d² = x² + y² + (t-x)² + (t-y)² hence :

a² + c² = b² + d² |

Note. This relation is true in any rectangle ABCD with sides u and v :

a² + c² = b² + d² = x² + (v-y)² + (u-x)² + y²

Value of PD ?

d² = a² + c² - b² = 19² + 23² - 29² = 49,
d = 7

This being possible only if a² + c² > b²

x = (t² - c² + b²)/2t

Similarily, substracting (1) - (2) : y = (t² - a² + b²)/2t

Substituting x and y into (2) :

(t² - c² + b²)² + (t² - a² + b²)² = 4b²t², and equation :

2t^{4} - 2(a²+c²)t^{2} + (b² - c²)² + (b² - a²)² = 0 |

This equation can take other forms as a² + c² = b² + d² :

exchanging b and d, or (a,c) with (b,d), or a and c.

With a=19, b=29, c=23 (and d=7), we get :
2t^{4} - 2×890 t^{2} + 327744 = 0

whose solutions are :
t = √( 445 ± √34153 )

that is t = 25,095922... and t = 16,130551...

Only the largest value results in a point P inside the square.

Condition for existence is (a²+c²)² - 2(b² - c²)² - 2(b² - a²)² > 0, and could also be written (a²+c²-2b²)² < 4a²c², hence :

|a - c| < b√2 < a+c |

For P to be inside the square, 0<x<t and 0<y<s

Using values of x and y results into :

0 < (t² - c² + b²)/2t < t,
that is 0 < t² - c² + b² < 2t², or |b² - c²| < t²

And similarily for y.

|b² - c²| < t²
|b² - a²| < t² |

Construct a right triangle with sides a and c, then its circumscribed circle (with diameter PQ). Construct M with PM = b.

By swapping the vertices, we can set for the sequel : a > (b,c,d) and b > d,

Because of the relation between a,b,c,d that means : a > b > d > c

look at construction from given a,b,c.

The vertices ABC build an isosceles right triangle, with B being the right angle.
Vertex A is deduced from C by rotating -π/2 with center B. Hence the construction :

With P as center, draw circles (A), (B),(C) with radii a, b, c.

We can then fix B anywhere on circle (B).

The locus of A is (A).
But the locus of A, which is image of C, is the image (C') of (C) by rotating -π/2 with center B.
Let's then draw this circle (C').
A is then the intersection point of (A) and (C').

C is deduced from A by rotating +π/2,
then complete the ABCD square.

There are two solutions A_{1} and A_{2}.
Only the largest BA_{1} may result into point P inside the square.

a-c<b√2<a+c we find the same as the analytic condition.

Ensuring P is inside the square is equivalent (after rotating by -π/2)
to compare position of M and angle BAx.

Angle BAM should be less than π/2, that is A outside the circle with diameter BM.
A should also be abovee BM.

First condition is a> PF, second one is a<PH

That last condition is more restrictive than the mere existence condition a<b√2+c because it could be written

a²<b²+(b+c)² = 2b² + c² + 2bc as compared with 2b² + c² + 2bc√2

Calculating PF is harder.

In right triangle BMF :
BF² = b² - c² and sin(α) = c/b

In triangle PBF :

PF² = b² + BF² -2b.BF.cos(π/2 -α)= 2b² - c² -2c√(b² - c²)

Hence :

{ 2b²+c²-2bc√2, 2b²-c²-2bc√(1-c²/b²) } < a² < 2b²+c²+2bc |

If (a,b,c,d) is a solution, (ka,kb,kc,kd) is also solution. We care then only of solutions with GCD(a,b,c,d) = 1.

N = a² + c² = b² + d² should then be written in at least two ways as sum of two squares. It is known (Fermat) that N is sum of two squares if and only if exponents of prime factors of the form 4k+3 in N are even.

Studying decompositions of N as sum of two squares is easier by using Gauss integers Z[i], in the form u+iv.

Then the 4k+3 primes are also primes in Z[i], hence a 4k+3 prime factor in N should divide a,b,c and d. Similarily 2 = -i(1+i)² is specific, and if exponent of 2 is even, 2 divides a,b,c,d.

Search of N being sum of two squares in at least two ways
N = a² + c² = b² + d²

with GCD(a,b,c,d)=1 is restricted to the set of numbers in the form
N = 2^{n}.∏p_{i}^{mi}

with all prime factors p_{i} being of the form 4k+1 :
5, 13, 17, 29...

4k+3 prime factors (3, 7, 11, 19...) being excluded, and exponent of 2 being n=0 or 1.

Number of decompositions of N as sum of two squares is d(N)=[(1+∏(mi+1))/2],
with [x] = integer part of x (floor function).

There are at least two decompositions if d(N) ≥ 2,
that is ∏(mi+1) ≥ 3,
hence ∑mi ≥ 2

In these are counted the uninterresting decomposition p² = p² + 0²

Squares of 4k+1 primes should then be excluded.

Doubles of these squares, giving the decomposition 2p² = p² + p², also uninterresting, will also be excluded.

First candidate is then N = 5*13 = 65 = 8² + 1² = 7² + 4²

then 5*17, 2*5*13, 5*29 ...

For every found pair of decomposition of N, remains to check the geometric conditions.

5*13 = 65 = 8² + 1² = 7² + 4² | rejected, no solution (a+c<b√2) |

2*5*13 = 130 = 11² + 3² = 9² + 7² | t = 9,688227 |

5*17 = 85 = 9² + 2² = 7² + 6² | t = 7,699702 |

2*5*17 = 170 = 13² + 1² = 11² + 7² | rejected, no solution |

5*29 = 145 = 12² + 1² = 9² + 8² | t = 8,998363 |

5*37 = 185 = 13² + 4² = 11² + 8² | t = 11,661657 |

5*41 = 205 = 14² + 3² = 13² + 6² | rejected, no solution |

13*17 = 221 = 14² + 5² = 11² + 10² | t = 13,405522 |

... |

All other values of N result into t > 7,699702

The smallest solution is then :

9² + 2² = 7² + 6²
t = √( (85+√1127)/2 ) = 7,6997019638... |

For side t to be also an integer, this problem is conjecture to be impossible.