Wierd Area Problem (difficult)

Oct 2015
2
0
UK
Hi all,
I am a maths teacher from the UK. A week or so ago I found a problem that looked challenging so I set it to one of my genius pupils as a bit of a "try this in your own time" extension. He came to me a few days later and said "I have tried to solve this problem for 2 days now and I can't do it. Can you show me the solution?" I said yes of course! Drew a diagram on the board, put in a few extra lines and got stuck. I sent him away saying "there are many different ways you could approach it, give me until tomorrow." And here we are a week later...

I have asked the rest of my dept. and nobody has a clue. We have managed to construct it accurately on Geogebra and measure the side. We have managed to make an excel document that solves it by force.

But as Mathematicians, we are looking for a nice, elegant solution that just works.

Here is the problem.

Inside a square there is a point. This point is 30m, 40m and 50m away from 3 consecutive corners. How big is the square?

We have interpreted "big" as either length or area as you can derive one from the other.

We have managed to find that the distance away from the 4th corner would be 10root18

We have, by construction, confirmed that the diagonal of the square is NOT 80m (50+30). Therefore, the point does not lie on the diagonal of the square

We know the length of one side is going to be 56.54-ish, therefore, the area is about 3196.8 (this solution has come from both the construction and the solve by force.)

Any help would be much appreciated. As I said, we are convinced that there is an elegant solution to the problem, we just can't find it.

Diagram below.
mathsproblem.png
 

skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
I've seen a technique for a similar problem before that involves rotating the square 90 degrees to create an adjacent image of the original. I'll neatly sketch out the details once I have access to my PC (I'm using my iphone now)

fyi, the square's exact side length is $10\sqrt{4\sqrt{14}+17}$, so your estimate was very close.
 
Feb 2015
2,255
510
Ottawa Ontario
Same diagram (I like the "30" line at top left!):
Code:
B        u=?         C


   b=30         c=40

       P


   a=?      d=50


A                    D
Square ABCD, u = side length
Point P inside, such that: BP = b =30, CP = c, DP = d = 50
Let AP = a

Step 1: calculate a
a = SQRT(b^2 + d^2 - c^2) = ~42.426

Step 2: calculate u:
u = SQRT[(a^2 + c^2 +- k) / 2] where k = SQRT[(2bd)^2 - (a^2 - c^2)^2]
u = ~56.539 or u = ~14.259
Note: ~14.259 is solution if P was outside the square.

So area = ~56.539^2 = ~3196.66