A Point Inside a Square

Date: 09/18/2003 at 00:41:37
From: Scott
Subject: a point inside a square

Point P lies inside square ABCD such that P's distance from A is 1, 
P's distance from B is 4 and P's distance from C is 5. What is the 
area of the square?

Date: 09/18/2003 at 10:31:14
From: Doctor Floor
Subject: Re: a point inside a square

Hi, Scott,

Thanks for your question.

Let x be the length of the sides of ABCD, let angle ABP = t, and thus 
angle PBC = 90 - t.

The Law of Cosines in triangle ABP gives

 1^2 = 4^2 + x^2 - 8x cos t

and thus

 cos t = (x^2 + 15)/8x.

In triangle PBC we find

  5^2 = 4^2 + x^2 - 8x cos(90 - t)

and thus, since cos(90 - t)=sin t,

  sin t = (x^2 - 9)/8x.

From sin^2 t + cos^2 t = 1 and substitution of sin t and cos t as 
above we find, after some rewriting which I leave for you

  2x^4 - 52x^2 + 306 = 0

and with the quadratic formula

  x^2 = 9 or x^2 = 17.

The solution x^2 = 9 would mean AB = x = 3, and then P cannot be
inside the square since it could not be 5 units from C.  Thus P is not
inside ABCD, as was given. We can see this also from the above formula
for sin t - it gives that sin t = 0.

That leaves that the only other possible solution is x^2 = 17, so the 
area asked for is 17.

Interestingly, with this solution ABP is a right triangle.

If you have more questions, just write back.

Best regards,

- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/
 

Date: 09/18/2003 at 21:29:41
From: Scott
Subject: Thank you (a point inside a square)

Thanks!  Very cool solution.  I got what you did and it 
was really cool.  Thanks again!  Math rocks!

Date: 04/18/2004 at 18:43:55
From: George
Subject: A Point Inside a Square

I am wondering if there is a way to solve this problem using only
geometry, without the use of trigonometry?

Date: 04/20/2004 at 19:26:33
From: Doctor Link
Subject: Re: A Point Inside a Square

Hey George, 

Thanks for writing in, and great question!  Yes - there is a method of
solving this problem without the trigonometry.  Let's take a look.

Let the bottom left hand corner of the square be the point (0,0):

  (0,x)__________(x, x)
      |          |
      |   P(a,b) |
      |          |     
      |          | 
      |__________| 
  (0,0)          (x, 0)

Note that x is the length of each side of the square.

We know that the distance from P(a,b) to the point (0,x) is 1.  So,
making use of the distance formula, we get the equation:
 
         2          2
  (i)   a  + (b - x)  = 1

We know that the distance from P(a,b) to the point (x,x) is 4.  So,
making use of the distance formula, we get the equation:

               2          2
  (ii)  (a - x)  + (b - x)  = 16

Finally, we know that the distance from P(a,b) to the point (x,0) is
5.  So, making use of the distance formula, we get the equation:

               2    2
  (iii) (a - x)  + b  = 25

We now have three equations, each with three unknowns.  If you are
unclear as to where I got these equations, then refer to the following
website regarding the distance formula:

  The Distance Formula
    

I will let you try and use that system of equations to solve for the
variables.  As a hint, try to express each of 'a' and 'b' in terms of
x to get started, then see if you can solve for x.  When you have 
solved it (or if you can't), then write back and we can discuss this 
further.

Hope to hear from you soon.

- Doctor Link, The Math Forum
  

Date: 04/21/2004 at 17:16:55
From: George
Subject: Re: A Point Inside a Square

Thank you, Doctor!  I found that my equation for x contained x^4, but
I was able to factor it by starting each factor with x^2 instead of x,
and I got the correct answer.  Thanks for your help.