Keeping up with the Johnson's

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Four homes are neighbors. The Alberto’s and Chang’s are adjacent to the Brunson’s and Johnson’s
but only meet each other at a point. The Brunson’s and Johnson’s properties also only meet at a point. The combine yards of the properties are surrounded by three roads. There is a road that runs by Alberto and Johnson, another road going by Johnson and Chang, and a road that runs by the Brunson’s. The fence used to divide the Brunson’s property runs perfectly straight with his adjacent neighbors. The Albertson’s claim their property is 500
sq. ft., the Chang’s state that their property is only 300 sq. ft., and the Brunson’s property is 1000 sq. ft. However, the Johnson’s do not know how big their property is, all they know is that their property is the biggest property and a quadrilateral. Assuming his neighbors are correct, how big is their property?

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** to clarify: the Brunson's fence runs perfectly straight with their adjacent neighbor's fences.

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Are the roads supposed to be straight?

There is no possible layout that would allow 4 properties inside a triangle and satisfy all the conditions of the OP. I'm assuming that if a road is said to run by a property or two then these are the ONLY properties exposed to that road.

If the roads can curve or turn then there are any number of solutions to this puzzle.

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Posted · Report post

Yes. The roads are straight

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Yes. The roads are straight

Ok, so then the Brunson's fence cannot be straight. Did you mean that it consists of 2 straight sections?

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Posted · Report post

I think I get what you were trying to say. Brunson's fence forms straight lines with other fences, so essentially the puzzle is

post-9659-0-38481300-1361900500_thumb.gi
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Posted (edited) · Report post

I think I get what you were trying to say. Brunson's fence forms straight lines with other fences, so essentially the puzzle is

attachicon.gifJohnson's.gif

yeah that is the picture i am looking for. i didn't do the best with this OP. Thank you for helping me say it better. :)

Edited by BMAD
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... is not the largest of the four properties.It's not even the second largest. The area of Johnson's property is 494+2/17 sq. feet. Here is how I figured it out.

I drew a line to split Johnson's property into 2 triangles and called them J1 and J2. I will be using capital letters A, B, C, etc. to denote the areas of the corresponding triangles and lower case letters w, x, y and z to denote the lengths of the corresponding line segments.

post-9659-0-58160000-1361938643_thumb.gi

So, we know that A=500, B=1000 and C=300. Let's look at the triangles A and (A+B). They share a common side and a common angle. The ratio of their areas is the same as the ratio of the opposite side of a common angle - A/(A+B) = w/(w+z). Also can be expressed as A/B=w/z. So, w/z = 0.5

Similarly, C/B = x/y and x/y = 0.3

Now, let's use these ratios to calculate the areas of J1 and J2:

J1 = 0.5 (J2 + C)

J2 = 0.3 (J1 + A)

Solving this system we get J1 = 264+12/17 and J2 = 229+7/17. Add them up and the total is 494+2/17.

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Posted (edited) · Report post

Very nice! I guess the Johnson's really didn't know their property size!!

I used collinear bases but otherwise the same strategy to find the answer

Edited by BMAD
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