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royprosen · Aug 19, 2008, 02:30 AM

How can it be proved that among all the triangles of equal base and area, the perimeter of the isosceles triangle is the least? Please give the detailed proof.

ebaines · Aug 19, 2008, 02:56 PM

If you drop a vertical line from the vertex of the triangle to the base (B), the length of that line is the height of the triangle (H), and the point where it intersects the base can be defined as the distance x from one end of the base. For a given area A and base length, the height h must be constant, since the area of a triangle is equal to $A = \frac 1 2 BH$ . The distance from that point to the other end of the base is B-x. Note that for an isoceles triangle the value of x is B/2. So the job here is to show that the length of the perimeter is a minimum when x = B/2.

The perimeter of the triangle is simply the sum of the lengths of the three sides:

$<br /> P(x) = B + \sqrt { x^2 + H^2} + \sqrt { (B-x)^2 + H^2 }<br />$

Take the derivative of P with respect to x, and set it to zero - that will define the value for x where the length of the perimeter is either at a max or a min.

$<br /> \frac {dP} {dx} = \frac x { \sqrt { x^2 + H^2}} - \frac {B-x} {\sqrt{B^2 -2Bx + x^2 + H^2}}<br />$

Now if you substitute in x = B/2, you find that dP/dx is zero. Hence the perimeter is either at a max or a min for an isoceles triangle. To prove it's a min, you could take the 2nd derivative of P(x) and show that it's negative at x = B/2. But that's a very messy bit of work. Another way is to show that there is only one value where dP/dx is zero, hence that point is either an absolute max or an absolute min. Then show that for $x = \pm \infty$ the value of P(x) goes to $\infty$ . Hence the value of P at x = B/2 must be a min, and the isoceles traingle has the minimum length perimeter for a given base and area.

Unknown008 · Aug 20, 2008, 05:04 AM

That reminds me of shearing of triangles... Having a constant and invariable base, the triangle can be sheared parallel to the y-axis (base of triangle parallel to x-axis) while conserving its area. I can picture that the isosceles has the smallest perimeter.

royprosen · Aug 20, 2008, 10:16 PM

When I asked for the proof of the theorem that among all the triangles of equal base and area,the perimeter of the isosceles triangle is the least,you proved it using calculus.But for a student of tenth standard,it is beyond reach.Please give the proof using only the theorems of pure geometry.