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### 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)

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

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