SOLUTION: Find the perimeter of a triangle with sides 7 m, 8, and 10 m that has a uniform 1m border

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Question 1085409: Find the perimeter of a triangle with sides 7 m, 8, and 10 m that has a uniform 1m border
Answer by ikleyn(52798)   (Show Source): You can put this solution on YOUR website!
.
Obviously, it is an Olimpiad level problem.

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 . . .


Thanks again.



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