Question 1085409
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Obviously, it is an Olimpiad level problem.


Usually, typical Olympiad level problem requires one non-trivial idea.


This one requires <U>TWO</U> non-trivial ideas.


These ideas are:


<pre>
1.  The given triangle is <U>SIMILAR</U> to the second triangle formed by the uniform border.

    It is obvious: the sides are parallel, so the angles are congruent.

        * * * This is <U>IDEA #1</U> * * * 


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 <U>IDEA #2</U> * * * 

    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 = {{{sqrt(12.5*5.5*4.5*2.5)}}} = 27.81 {{{m^2}}}.

    Then the radius of the inscribed circle is  r = {{{A/(semi-perimeter)}}} = {{{27.81/12.5}}} = 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 <U>whether I should prove</U> that <U>the incentres of these triangles coincide</U>: is is <U>SO OBVIOUS</U> . . . 


5.  Now the similarity coefficient is {{{(r+1)/r}}} = {{{3.225/2.225}}} = 1.45 (larger to smaller) approximately with two correct decimal digits after the decimal dot.


<U>Answer</U>.  The perimeter under the question is 25*1.45 = 36.24 m.
</pre>

Solved.



It is a nice problem.  &nbsp;&nbsp;Thanks for submitting it.  &nbsp;&nbsp;It was a pleasure to solve it.



It is my reward after the daily fighting with idiotic, semi-idiotic, regular and routine problems . . . 



Thanks again.