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
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.