The three distances problem revisited

by luimichael, Mar 25, 2019, 3:34 PM

P is a point inside a square ABCD. The distance from P to the vertices A, B and C are a, b and c respectively.
The problem is to determine the length of a side of the square.
Here is a Geometric approach:
Let $PD =d$. Then $d^2+b^2=a^2+c^2$ $\implies$ $d^2=a^2+c^2-b^2$. (by British Flag theorem)
Now, we rotate triangle PCB about the point B through 90 degrees so that BC is moved to BA , while P is moved to P'.
Note that triangle PBP' is a right-angled isosceles triangle with two sides equal to b, and triangle PAP' is a triangle with sides a, b and c.
The area of quadrilateral PAP'B is therefore $\frac { b^2} 2 + Heron(a,\sqrt 2 ,b,c)$
We can apply rotations of figure to triangle PDC, triangle PAD and triangle PBA one by one, and get the followings:
$\frac {c^2} 2 + Heron(b, {\sqrt 2} c, d)$
$\frac {d^2} 2 + Heron(c,{\sqrt 2} d, a)$
$\frac {a^2} 2 + Heron(d,{\sqrt 2}a ,b)$

The sum of the above four expressions actually represent the double of the area of the given square ABCD, and hence the length of a side of ABCD can be found readily.
This post has been edited 1 time. Last edited by luimichael, Sep 14, 2020, 3:42 PM
Reason: Typo

Comment

0 Comments

a