A circle with [radius] r is inside the isosceles triangle. Now the perimeter of the isosceles triangle=2*a+b. Now what to do?
Yes, what to do. We have three symbols: a,b,r
We need to express the perimeter in terms of a single variable, so that we can minimize that function using a derivative. One approach, started below, uses an angle measure for this variable. Also, they say, "
A circle", so the circle is fixed, whatever its radius. Therefore, our answer will be stated in terms of r.
For example, I can show the smallest possible perimeter is 6∙√3∙r
r ends up being a scaling factor, to adjust for whatever circle we're talking about.
Using symbols a and b won't work because there are many different isosceles triangles that circumscribe a fixed circle, so many different a,b pairs to consider. We need a single variable: x, and I chose to use trigonometry and make x an angle. (There are other approaches.)
I found a function for the triangle perimeter P(x) by:
(1) drawing a picture and dividing the triangle in half (vertically), resulting in two right triangles
(2) using a radius tangent to the isosceles triangle side (i.e., hypotenuse of right triangle) -- this forms a smaller, similar right triangle inside the larger right triangle
(3) defining a shared angle (x) within the two similar right triangles
(4) using symmetry and substitution
Here's a picture.
Symmetry: AE=AF and BE=BD=CD=CF
Similar right triangles: ADB (
∠ADB = 90°) and AEO (
∠AEO = 90°)
Note that r=OE=OD=OF
P = 2∙AE + 4∙BD
Now you can start substituting. Use relationships like tan(x)=OE/AE and sin(x)=OE/AO
Your goal is to get:
P(x) = [an expression which contains only symbol r and trigonometric functions of x]
If you get stuck, let us know what's up. Otherwise, please share your result. :cool:
PS: Does your class allow use of technology, for determining derivatives and solving equations?