__Four Photograph
Kiosk__

** **

**Introduction.**

My father has been working in the high end tradeshow business for the past thirty years. His company Exhibit Systems Inc. has been designing and producing booths for numerous tradeshows and conventions throughout the world.

After years of experience, he has become very knowledgeable on the geometry and mathematics incorporated in his profession. One particular case in which geometry skills come in handy is determining the length of a photograph presented in a kiosk. There are two cases: one in which the dimensions of the photograph are known, and what needs to be determined are the lengths of the framework needed to construct the kiosk, and the other in which the measurements of the frame is known, and the dimension of the photograph need to be determined.

**The Framework.**

The framework of each kiosk is essentially the same. There are four radius bars connecting at a center post. There are four congruent arc frames that encase four photographs of the same size. Then, there are four posts situated equidistant from each other along the perimeter of the kiosk, and each is attached to the center post by a radius bars. Lastly, there are two different types of posts: round and square. Either a kiosk uses all round posts or all square posts.

__Scenario
for Kiosk A:__

- Each round post measures 1.791 inches in diameter
- Each square post measures 1.260 inches x 1.260 inches
- Each radius bar measures 0.748 inches in width
- Each arc frame measures 0.591 inches in width, and is a composition of two separate frames
- The photograph fits within 1/32 of an inch between the two arc frames

The dimensions of the kiosk vary from booth to booth depending on the taste and preferences of the company and product being shown.

**The Geometry.**

Given the dimensions of the photograph, the constructor of the kiosk is mainly focused on determining the length of the photograph. The length of the photograph helps to determine the lengths and measurements of the arcs and rods used in the circular framework of the kiosk. The following spreadsheet can help determine the lengths of the pieces of metal used in a kiosk with square posts.

o The photograph reaches each square post at the midpoint of the edge. Therefore, each photograph hits the midpoint of an edge of each square post along the perimeter of the kiosk. Since the square post measures 1.26 in. x 1.26 in., the square post must then extend inwards to the center post. Similarly, the center post extends outward toward each square post along the perimeter.

o Then the length of a radius rod is the inches less than the actual radius measurement of the circle.

o Each arc frame essentially measures 1.26 inches less than the actual measurement of ¼ of the perimeter.

o Since each arc frame consists of two concentric frames (or circles), the inner frame will have a shorter arc length than the outer frame.

o The photograph has 1/32 inch = 0.03125 inch of space on either side of it.

o The inner frame meets the square post at inches from the inner corner of the square post. Therefore, the radius of the inner concentric circle (the distance from the edge of the inner arc frame surrounding the photograph to the center of the kiosk) is 0.03125 inches less than the radius of the circle created by the photograph.

o The arc frame has a width of 0.591 inches, the photograph lies directly in between the two frames composing the arc. Therefore, inches lie between the photograph and the outer arc of the outer frame edge. So the radius of the outer concentric circle will be 0.2955 inches more than the radius of the circle created by the photograph.

**Problem.**

Given the length of a photograph, how can someone determine the lengths of the metal pieces needed to construct the kiosk so that the photographs will fit perfectly. In other words, we need to find the measurements at which to cut the rods and arc frames such that the least amount of material is used.

Each of the four photographs is 48 inches wide and 60 inches tall.

Each arc frame (inner and outer frame) is 0.591 inches thick.

Each photograph is 48in long. The perimeter of the circle created by the photographs is found below; keep in mind the inclusion of the four square posts.

The radius *R *of the
circle created by the photographs:

To find the length at which the inner radius rods should be cut at:

*Radius –
space created by center post and perimeter post*

When the arc frame is 0.591in thick, to find the measurement of the arc length of the inner concentric frame:

*Radius – *0.03125 = radius *r1* of inner frame

Then consider the
perimeter of a circle with this radius *r1*.

Subtract the perimeter distance of the four square posts:

And divide this perimeter measurement into four equal segments:

The arc length of the inner frame surrounding the photograph is approximately 47.95 inches long.

To find the arc length of the outer frame:

*Radius + *0.2955 = radius *r2 *of outer frame

Then consider the
perimeter of the circle with this radius *r2*:

Subtract the perimeter distance of the four square posts:

And divide this perimeter into four equal segments:

The arc length of the outer frame surrounding the photograph is approximately 48.46 inches long.

**Spreadsheet.**

The following spread sheet application will allow one to easily find the length of each radius rod, the arc length of the inner arc frame, and the arc length of the outer arc frame for creating a circular kiosk with square posts.

- Insert the given value for the length of the photograph in inches. Hit ÒEnterÓ to find calculations for unknown lengths.
- Link to Spreadsheet
Application for Square Post.

**The Framework.**

The constructor of the kiosk also has the choice of using round posts in the perimeter of the kiosk.

- The photograph bisects an edge of the round post. Since the diameter of the round post above is 1.791 inches. The round post extends inches inward and 0.8955 inches outward from the original radius measurement of the circle created by the photograph.
- Then the length of a radius rod is the 1.791 inches less than the actual radius measurement of the circle.
- Each arc frame essentially measures 1.791 inches less than the actual measurement of ¼ of the perimeter.
- Since each arc frame consists of two concentric frames (or circles), the inner frame will have a shorter arc length than the outer frame.
- The photograph has 1/32 inch = 0.03125 inch of space on either side of it.
- The inner frame meets the square post at inches from the center of the round post. Therefore, the radius of the inner concentric circle (the distance from the edge of the inner arc frame surrounding the photograph to the center of the kiosk) is 0.03125 inches less than the radius of the circle created by the photograph.
- The
arc frame has a width of 0.591 inches, the photograph lies directly in
between the two frames composing the arc. Therefore, inches lie between the photograph and the outer arc of
the outer frame edge. So the
radius of the outer concentric circle will be 0.2955 inches more than the
radius of the circle created by the photograph.

** **

**Problem.**

We
will now consider the same problem stated before, except this time the
constructor has decided to use *round*
posts instead of square posts.

Each photograph is 48in long. The perimeter of the circle created by the photographs is found below; keep in mind the inclusion of the four round posts.

The radius *R *of the
circle created by the photographs:

To find the length at which the inner radius rods should be cut at:

*Radius –
space created by center post and perimeter post*

When the arc frame is 0.591in thick, to find the measurement of the arc length of the inner concentric frame:

*Radius – *0.03125 = radius *r1* of inner frame

Then consider the
perimeter of a circle with this radius *r1*.

Subtract the perimeter distance of the four square posts:

And divide this perimeter measurement into four equal segments:

The arc length of the inner frame surrounding the photograph is approximately 47.95 inches long. (Note that this is the same length as using a square post).

To find the arc length of the outer frame:

*Radius + *0.2955 = radius *r2 *of outer frame

Then consider the
perimeter of the circle with this radius *r2*:

Subtract the perimeter distance of the four square posts:

And divide this perimeter into four equal segments:

The arc length of the outer frame surrounding the photograph is approximately 48.46 inches long. (Note that this is the same length as using a square post).

**Spreadsheet.**

The following spread sheet application will allow one to easily find the length of each radius rod, the arc length of the inner arc frame, and the arc length of the outer arc frame for creating a circular kiosk with round posts.

- Insert the given value for the length of the photograph in inches. Hit ÒEnterÓ to find calculations for unknown lengths.
- Link to Spreadsheet
Application for Round Post.

**Conclusion.**

The two spreadsheet applications will aid constructors in determining the lengths of the frames used in constructing a four photograph kiosk. The spreadsheet allows constructors to easily enter the dimensions and lengths of the posts and photograph in order to find the necessary lengths at which to cut radius rods and arc frames.