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Category: | Math |

Submitted By: | javaguru |

Fun: | (2.38) |

Difficulty: | (3.22) |

Two mathematicians, Rex and Ralph, have an ongoing competition to stump each other. Ralph has just finished building a custom house and invites Rex to dinner.

He tells Rex, "The lot my house sits on is a regular polygon. My house is a matching regular polygon sitting on a circular foundation in the middle of the yard. The closest part of the foundation to the edge of my property is exactly the same as the diameter of the foundation. The house is two stories, built around a central circular atrium with a diameter that is exactly one tenth the longest measurement of my property."

"Sounds like quite the house!" remarked Rex.

"Yes, I've been building it for almost two years. When I was excavating the foundation I found a section of an old irrigation pipe. The pipe exactly bisected the area of the house, with each end of the pipe at an edge of the house. After the house was completed while I was preparing to put up a fence around the edge of the property, I discovered that the length of the pipe I found was exactly the length of one side of the property. I used the pipe as a gate across the side of the property with the driveway and built the fence around the other sides. I built a total of 400 feet of fence."

"So what's the square footage of my house?" Ralph asked.

"Are you counting the space occupied by the walls in the square footage?" asked Rex.

"Yes of course," Ralph replied. "It also includes the stairways, hallways, and basically all of the space inside of the exterior walls."

Rex smiles and says, "Nice try, Ralph! Assuming the information you gave me is correct, it'll take just a minute or two to calculate it."

How did Rex know the square footage, and what was his answer?

Rex realized that he would need to know the number of sides on Ralph's house in order to make use of the only measurement Ralph gave him, which was the perimeter minus the length of one side for the property. The length of a side of the property needed to be the length of the pipe that bisected the area of the house. The length of the pipe is the key to figuring out how many sides the house has.

The house is a regular polygon inscribed in the circle that is the foundation. Since the closest edge of the property is equal to the diameter of the foundation, then the property must be a similar polygon circumscribed around a circle with a diameter three times the size of the foundation.

The minimum and maximum length line segments that can bisect the area of a regular polygon are calculated differently for even- and odd-sided polygons.

For an even-sided polygon inscribed in a circle of radius r, the maximum line segment has its ends at two opposite vertices and the length is 2r. The minimum line segment goes from the middle of one edge to the middle of the opposite edge. The distance from the edge to the center of a polygon is the apothem, so the minimum line segment is two times the apothem a.

For an odd-sided polygon the maximum bisector goes from a vertex to the middle of the opposite side, so the maximum is r + a. As with an even-sided polygon, the minimum bisector for an odd-sided polygon extends through the center of the polygon and at an angle halfway between two adjacent maximum bisectors.

2Pi radians = 360 degrees. The central angle between any two adjacent vertices on a polygon with n sides is then 2Pi/n and the angle between a vertex radius and the apothem is half that or Pi/n. The apothem and the radius form two sides of a right triangle, with the third side formed by one half of the side bisected by the apothem. For a regular polygon with n sides of length s this gives the following relationships:

a = r*Cos(Pi/n)

r = a/Cos(Pi/n)

s = 2*a*Tan(Pi/n)

Rex assigns one as the radius of the foundation. This makes the apothem of the property three and the side of the property is:

2*3*Tan(Pi/n) = 6*Tan(Pi/n)

So where Min(n) and Max(n) are the minimum and maximum length of a bisector for an n-sided polygon inscribed in a unit circle, Rex needs to find an integer value for n that satisfies the inequality:

Max(n) >= 6*Tan(Pi/n) >= Min(n)

Since Rex is pretty sure there's only one solution, he figures first he'll just use the minimum and maximum functions for even-sided polygons. The minimum and maximum for odd-sided polygons will be within this range, so he might find an odd-sided polygon that doesn't satisfy the equation, but he won't miss any that do. This gives:

2 >= 6*Tan(Pi/n) >= 2*Cos(Pi/n)

Taking just the first part of the inequality gives:

2 >= 6*Tan(Pi/n)

1/3 >= Tan(Pi/n)

atan(1/3) >= Pi/n

atan(1/3) / Pi >= 1/n

n >= Pi/atan(1/3)

n >= 9.8

He tries 10 and finds that:

Max(10) = 2 >

6*Tan(Pi/10) = 1.95 >

Min(10) = Cos(Pi/10) = 1.90

Now Rex knows that Ralph's house has ten sides and is a regular decagon. That makes the length of the pipe and of a side of the larger decagon 400/9 = 44 4/9 = 44.444.. feet. From the equation above:

44.444 = 2*a*Tan(Pi/10)

a = 44.444 / Tan(Pi/10) ~ 68.393

So the radius of the larger circle is 68.393 feet and the radius of the foundation is 68.393/3 = 22.798, which is also the radius of the house decagon. The apothem and length of a side for the house are:

a = 22.798 * Cos(Pi/10) = 21.682

s = 2 * 21.682 * Tan(Pi/10) = 14.090

The side and apothem are the base and height of one of the ten triangles formed by two vertex radii and a side. The area of one of these triangles is then

14.090 * 21.682 / 2 = 152.75

and the area of the decagon is 10 * 152.75 = 1527.5. Now Rex just needs to know the size of the atrium to calculate the square footage of the house. The radius of the atrium is one-tenth the radius of the large decagon, which is

r = 68.393 / Cos(Pi/10) = 71.913

so the area of the atrium is

(71.913 / 10)^2 * Pi = 162.47

and the area of the two floors in Ralph's house is

(1527.5 - 162.47) * 2 = 2730.0 square feet.

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