Point P in a Square

Draw $EG$ and $FH$ through $P$ such that $EG\parallel BC$ and $FH\parallel AB$ , with E, F, G, H on AB, BC, CD, AD, respectively. Then $AE^2+AH^2=1 \, (1)$ , $EB^2+AH^2=29^2 \, (2)$ , and $EB^2+HD^2=41^2 \, (3)$ . Using this system of equations, we can find that $AE^2+HD^2=29^2$ by adding (1) and (3) and subtracting (2). Noting that this is a square, we see that it has to be symmetrical about the diagonal $AC$ . Thus, the main diagonal has a length of $1+41=42$ , so the area is $\dfrac{42^2}{2}=\boxed{882}$ .

Let the rotation centered at $B$ that carries $A$ to $C$ also carry $P$ to $P'$ (Alternatively, Let $P'$ be a point outside the square such that $\angle CBP'=\angle ABP, \angle BCP'=\angle BAP$ ).

Then $\Delta ABP \cong \Delta CBP'$ , thus $BP'=BP=29, CP'=CP=1, \angle PBP'=\angle ABC=90^\circ$ .

So $PP'^2=BP^2+BP'^2=1682$ . Also since $CP^2+CP'^2=1682$ , we know that $\angle PCP'=90^\circ$ .

So $B, P, C, P'$ are concyclic. By Ptolemy's Theorem $CB=\frac{BP\cdot CP' + BP' \cdot CP }{PP'}=\frac{29+29\cdot 41}{\sqrt{1682}}$ , so $[ABCD]=CB^2=\frac{(29\cdot 42)^2}{1682}=882$ .

Construct $P'$ such that $\triangle BCP' \cong  \triangle BPA$

So, since $\angle PBP' = 90^{\circ}$ because $\triangle BCP' \cong \triangle BPA$

Thus, $\triangle BPP'$ is a right isosceles triangle since $BP = BP' = 29$

So, $PP' = 29\sqrt{2}$

Looking in $\triangle PP'C$ , by the converse of Pythagoras Theorem, $\triangle PP'C$ is a right triangle, since $41^{2} + 1^{2} = (29\sqrt{2})^{2}$

Thus, quadrilateral BPCP' is cyclic(due to opposite right angles.) and by Brahmagupta's Formula we get

$|PBP'C| = \sqrt{(50 - 41)(50 - 29)(50 - 29)(50 - 1)} = (3)(7)(21)$

$= 21^{2} = 441$

On the other hand,by Heron's Formula applied to $\triangle PBC$ and $\triangle BCP'$ and equating this to 441, we get $CD^2 = 882$

$|ABCD| = CD^{2} = 882$ .             Q.E.D

Using the British Flag Theorem, we have that $PA^2+PC^2=PB^2+PD^2$ . This gives us that PD=29. Thus, PDB is an isosceles triangle, so P is on the diagonal AC. Since P is on AC, we have AC=PA+PC=42. Thus, the area is $42\cdot42/2=21\cdot42=882$ .

By the British Flag theorem, PD = 29. So because PB = PD, P, A, and C are collinear, so the diagonal has length 1+41 = 42, so the area of the square is 882.

I used what my friends and I call gorilla math. I'm sure there is an easier way to solve this but here is what I did: After drawing the problem, I drew lines from point P that were perpendicular to AB (intersection at point D) and BC (intersection at point E). Then using the given lengths of PA, PB, and PC, I created 3 different equations using the length of our square (I will use AB or X), AD or Y, and BE or Z.

(X - Y)^2 + Z^2 = 29^2

(X - Y)^2 + (X - Z)^2 = 41^2

Y^2 + Z^2 = 1

After expanding getting rid of our exponents we get:

(Equation 1) X^2 - 2XY + Y^2 + Z^2 = 841

(Equation 2) X^2 - 2XY + Y^2 + X^2 - 2XZ + Z^2 = 1681

(Equation 3) Y^2 + Z^2 = 1

We can see Equation 1 is in Equation 2 so we can substitute to obtain:

X^2 - 2XZ = 840

and substituting Equation 3 into Equation 1 we obtain:

X^2 - 2XY = 840

From this, we see Y must equal Z. so from Equation 3 we can obtain

Y = Z = sqrt(1/2)

Now substituting back in and solving for X, we obtain sqrt(882) so therefore the area is 882

(i tried formatting it but when i previewed it didn't look right so i undid it and left it like this, sorry)

Let Q be the point so that $\triangle APB \equiv \triangle BQB$ Note $\angle PBQ = \angle PBC + \angle CBQ = \angle PBC + \angle ABP = \angle ABC = 90^/circ$ . By Pythagorean theorem, we note that $PQ = \sqrt{29^2+29^2} = \sqrt{1682} = \sqrt{41^2+1^2}$

By the Pythagorean theorem, we see that $\triangle PQC$ is a right triangle, with a right angle at C.

By cosine rule, we get:

$BC^2=BQ^2+QC^2-2 \times BQ \times QC \times \cos(\angle BQC)$

$=29^2+1^2-2 \times 29 \times 1 \times \cos(\angle CQP + 45^\circ)$

$=842-58 \times \cos(\arccos(\frac{1}{\sqrt{1682}}) + 45^\circ)$

$=842-58 \times (\cos(\arccos(\frac{1}{\sqrt{1682}}) \times \cos(45^\circ)$ $-\sin(\arccos(\frac{1}{\sqrt{1682}}) \times \sin(45^\circ))$

$=842-58 \times (\frac{1}{\sqrt{1682}} \times \frac{1}{\sqrt{2}} - \frac{41}{\sqrt{1682}} \times \frac{1}{\sqrt{2}}$

$=842-58 \times (\frac{1}{\sqrt{3364}} - \frac{41}{\sqrt{3364}}$

$=842-58 \times (\frac{1}{58} - \frac{41}{58})$

$=842+40$

$=882$

We call $l$ the side of the square. On the Cartesian plane we place $B$ in the origin. $A$ will be $(l, 0)$ and $C$ will be $(0, l)$ . We draw a circle of radius $1$ and centre $A$ , one of radius $29$ and centre $B$ and one of radius $41$ and centre $C$ . Therefore intersecting the three circles we get: $\begin{cases} (x-l)^2+y^2=1^2 \\ x^2+y^2=29^2\\ x^2+(y-l)^2=41^2 \end{cases}$ The only solution to this system where $x$ , $y$ and $l$ are positive is: $(\frac{41}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 21\sqrt{2})$ . We notice that in this situation the point is actually within the square, therefore the area of the square will be: $(21\sqrt{2})^2=\fbox{882}$

Let $A$ , $B$ , $C$ , $D$ , and $P$ be located at $(0,s)$ , $(0,0)$ , $(s,0)$ , $(s,s)$ , and $(x,y)$ respectively. Since $P$ is inside $ABCD$ , $0\leq x,y\leq s$ .

From the given conditions, we know that $x^2+(y-s)^2=1^2$ , $x^2+y^2=29^2$ , and $(x-s)^2+y^2=41^2$ . Solving this system of equations and considering the constraints on $x$ and $y$ , the area of the square, or $s^2$ , is 882.

Side AB will always be less than 30 , because P is within the square and therefore A - P - B is not possible .

Therefore , area of the square is always less than 30^2 ie 900 and also diagonal will always be less than 30*sqrt2 ie approx 42.42

Further , assume that the perpendicular from P on side BC divides the side BC into the lengths b and c , so that b + c = the side of the square ,

then by Pythagoras theorom , 29^2-b^2=41^2-c^2 therefore : c^2-b^2=41^2-29^2=840 ie ( b + c ) ( b - c ) = 840

Since b + c = side of square , say s

s ( b - c ) = 840

Similarly , for the side AB , if perpendicular from P divides AB into 2 lengths e and f , then

we get s ( f - e ) = 840

Since e cannot be greater than or equal to 1 and f will be only marginally less than 29 , ( f - e ) will be a little less than 29 .

Therefore , s should be little more than 29
Now , we get limits for the side to be little more than 29 and therefore the area to be little more than 841 .

Summarising , we have 29 < s < 30 and 841 < area < 900 and 41 < diagonal < 42.42 Assume that P , A and C all lie on straight line . Then , since A - P - C , AC = AP + PC = 1 + 41 = 42 .

If diagonal of a square is 42 , then its area can be easily calculated to be 882 and side approx. = 29.694 , which satisfy the limits shown above