Complicated maths problem

[nikki1234]

I've been looking at this question for a few days now and I cant seem to figure it out. Please help.

Standing at a location which is exactly 3m, 4m and 5m (can be in any order) from 3 corners of a rectangle. What are the possible distances from the fourth corner?

Looking at the problem I know there are 3 different ways this can be drawn. One of the answer is o as it doesnt move and this can be proved using Pythagoras triples. I know there are 3 possible solutions. With one of the diagrams you can get a set of four equations with four unknowns using good principle of using perpendicular distances from point to corners

 

[Deleted member 4993]

I've been looking at this question for a few days now and I cant seem to figure it out. Please help.

Standing at a location which is exactly 3m, 4m and 5m (can be in any order) from 3 corners of a rectangle. What are the possible distances from the fourth corner?

Looking at the problem I know there are 3 different ways this can be drawn. One of the answer is o as it doesnt move and this can be proved using Pythagoras triples. I know there are 3 possible solutions. With one of the diagrams you can get a set of four equations with four unknowns using good principle of using perpendicular distances from point to corners

DUPLICATE POST:

http://www.freemathhelp.com/forum/viewtopic.php?f=9&t=43832&p=170363#p170363

Assume one of the corners is origin and the length of the room is 'L" and the width is "W".

Assign location of Person (point P) is (x,y)

Now write the length equations.

(L-x)^2 + y^2 = 25 .................................(1)

(W-y)^2 + x^2 = 16................................(2)

x^2 + y^2 = 9 .....................................(3)

You need to find

(L-x)^2 + (W-y)^2 = ???

If you subtract (3) from (1) you get,

(L-x)^2 - x^2 = 16...............................(4)

adding

(2) and (4) we get

(L-x)^2 + (W-y)^2 = 32

That is one of your solutions.

The rest can be found by changing the numerical parts of equations (1), (2) and (3)

 

[Denis]

Same thing; a bit shorter:

C             a              D


            
b          P(x,y)


B                            A

a = AB = CD, b = BC = AD, AP = 5, CP = 4, BP = 3, DP = d = ?

Triangle BPC:
p - y^2 = 16 - (b - y)^2 : leads to b^2 - 2by = 7 [1]

Triangle APD:
25 - y^2 = d^2 - (b - y)^2 : leads to b^2 - 2by = d^2 - 25 [2]

[1][2]: d^2 - 25 = 7 : d = SQRT(32) : as per The Honorable S. Khan :wink:

 

[Denis]

Here's one you can "experiment" with, Nikki; ALL integers:

Position rectangle ABCD with A(0,56), B(63,56), C(63,0) and D(0,0).
Place the inside point P(15,20).
Results: AP = 39, BP = 60, CP = 52, DP = 25.

Btw, that's the smallest "all-integer" case I could find.