Usually, typical Olympiad level problem requires one non-trivial idea.
This one requires TWO non-trivial ideas.
These ideas are:
1. The given triangle is SIMILAR to the second triangle formed by the uniform border.
It is obvious: the sides are parallel, so the angles are congruent.
* * * This is IDEA #1 * * *
2. So, the only thing to discover is to find the proportionality (similarity) coefficient.
Then we simply multiply the perimeter of the given triangle, 7 + 8 + 10 = 25 m, by the similarity coefficient.
3. How to find the similarity coefficient ? ?? ??? It is the question: TO BE OR NOT TO BE ???
Use the radius of the inscribed circle. * * * It is the IDEA #2 * * *
You can calculate the area of the given triangle using the Heron's formula.
We all know this formula, so I will not bore with calculations and simply will give the answer:
its area is A = = 27.81 .
Then the radius of the inscribed circle is r = = = 2.225 (approximately; with 3 correct decimal digits after the decimal dot).
4. Now only one step remains to the finish: The radius of the inscribed circle to the larger triangle is r + 1 = 2.225 + 1 = 3.225 m.
I don't know whether I should prove that the incentres of these triangles coincide: is is SO OBVIOUS . . .
5. Now the similarity coefficient is = = 1.45 (larger to smaller) approximately with two correct decimal digits after the decimal dot.
Answer. The perimeter under the question is 25*1.45 = 36.24 m.
Solved.
It is a nice problem. Thanks for submitting it. It was a pleasure to solve it.
It is my reward after the daily fighting with idiotic, semi-idiotic, regular and routine problems . . .
