Drexel dragonThe Math ForumDonate to the Math Forum

Ask Dr. Math - Questions and Answers from our Archives
_____________________________________________
Associated Topics || Dr. Math Home || Search Dr. Math
_____________________________________________

Using the Incenter

Date: 05/06/2003 at 08:48:54
From: Darryl
Subject: Triangles

I need to construct a triangle to fit inside a triangle. There needs 
to be 1/4" clearance on all sides. Normally I would undercut 1/2" from 
the width and height, but I have a friend with a CAD program who says 
that won't work. Is this true? What would be the formula to calculate 
the width and height?


Date: 05/06/2003 at 12:02:44
From: Doctor Peterson
Subject: Re: Triangles

Hi, Darryl.

The problem is that you want to move in 1/4" _perpendicular to each 
side_; for a square or circle, that decreases the whole height (or 
diameter) by 1/2", but for a triangle the altitude or width will be 
decreased by more than that, since the sides are moving in at an 
angle.

There is a nice way to do this calculation, using the concept of the 
incenter. This is a point in a triangle that is the same distance 
from all three sides, so that a circle tangent to the three sides has 
its center there. The radius of this inscribed circle is called the 
inradius, and the point is called the incenter.

Since the circle is tangent to the sides of the triangle, the radii to 
those points are perpendicular to the sides, and when you move each 
side in the same distance of 1/4", you are making a new triangle 
similar to the original, with the same incenter, and its inradius is 
1/4" less than the original.

So if you calculate the inradius, you will be proportionally 
decreasing that by 1/4" divided by the inradius, and the new sides 
will be multiplied by that same amount. This allows you to find the 
lengths of the new sides (or the altitude) by multiplying.

The formulas for the inradius are found in the Dr. Math Geometric 
Formulas FAQ (scroll down to "Scalene Triangle"):

   http://mathforum.org/dr.math/faq/formulas/faq.triangle.html 

Assuming that what you know about the triangle is the lengths of the 
three sides, you would use this formula:

   r = sqrt[(s-a)(s-b)(s-c)/s]

where a, b, and c are the lengths of the sides, and s=(a+b+c)/2.

Once you have done this, calculate

   x = 1 - d/r

where d is the distance you are moving each side in (1/4" in your 
example). Then you will multiply each side by x to find its new value.

Let's go through an example. Suppose we have an isosceles triangle 
with base 200" and height 53". We can use the Pythagorean theorem to 
find the lengths of the sides, namely

   sqrt(100^2 + 53^2) = sqrt(12809) = 113.18

Then

  s = (113.18+113.18+200)/2 = 213.18

  r = sqrt[(213.18-113.18)(213.18-113.18)(213.18-200)/213.18]
    = sqrt[100*100*13.18/213.18] = 24.86

  x = 1 - 0.25/24.86 = 0.9899

So the new base is

   200 * 0.9899 = 197.99

and the new height is

   53 * 0.9899 = 52.47

We did lose about 2" on the width, and over 1/2" in height.

In this case, where you know the base and height, you could instead 
use the fact that

   r = 2K/P

where K is the area and P is the perimeter:

   r = bh/(a+b+c) = bh/(2a+b)

since a=c are the sides of the triangle; with numbers, this gives

   r = 200*53/(226.36+200) = 24.86

as above, and the rest of the work would be the same.

Without this trick, the problem could have been solved with more 
complicated geometry and trigonometry.

This is a good example where basic but significant geometry turns out 
to be very useful for practical calculations. All I needed was the 
concept of the inradius and enough insight to see its relevance, 
together with formulas I looked up. It's too bad students so seldom 
see that school math is really worthwhile.

If you have any further questions, feel free to write back.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/ 


Date: 05/06/2003 at 13:41:47
From: Darryl
Subject: Thank you (Triangles)

Thank you so much for your time and effort. Your answer is simpler 
than I had expected. When I was in school I would say, "What do I need 
math for?" and I did the minimum to pass. Now I have a job where I use 
it every day, and I have to teach myself mostly. It is truly great to 
have a Web site like this, with people like you. I use this site often 
and I tell everybody who needs an answer about it. Thank you again.  
               

Darryl Isaacs
Associated Topics:
College Triangles and Other Polygons
High School Triangles and Other Polygons

Search the Dr. Math Library:


Find items containing (put spaces between keywords):
 
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

_____________________________________
Math Forum Home || Math Library || Quick Reference || Math Forum Search
_____________________________________

Ask Dr. MathTM
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/