lastochka wrote:

Emmanuel wrote:

perimeter of equilateral triangle = 3R*3^1/2 (R - external circle) and perimeter = 6r*3^1/2 (r - internal circle)=> p(small) = 3R*3^1/2, p(large) = 6R*3^1/2 => p(large) = 2*p(small).

Can you explain that in plainer terms?

Lastochka, if you have an equilateral triangle and consider an inscribed circle, you get: p = 6*r*[3^1/2], where r is the radius of the circle, p is perimeter of the triangle. This is true because (how to explain it in simplier words?

) the radius of inscribed circle r = (dist^2 - (a/2)^2)^1/2, where dist - distance between any vertex of the triangle and its center, a = length of its side. dist = a/3^1/2, since there is a triangle with sides: (dist, dist, a) and greatest angle of 120. => r = a/(2*3^1/2). Since p = 3a => p = 6r*3^1/2.

if you have an equilateral triangle which is CIRCUMscribed in a circle, then if R - its radius, a is the length of the side of this trianle, => R = a/3^1/2. The logis is similar: R = dist(which was defined previously), R = a/(3^1/2). => p = 3*a {sum of all sides} = 3*R*3^1/2.

The solution is straightforward now, since we have 1 circle, which is INscrideb in big triangle and at the same time a triangle which is CIRCUMscribed in that same circle. => p(big) = 6*r*3^1/2, p(small) = 3*r*3^1/2.

I know, this solution lacks illustratin, but maybe this can help you:

Attachments

3.GIF [ 7.64 KiB | Viewed 1366 times ]