A Wee Incircle

Geometry Level pending

Find the smallest inradius of a triangle A B C ABC with vertex coordinates:

A = ( 0 , 0 ) , B = ( 2 + a , 2 − a ) , C = ( 2 a , 3 − a ) A = (0,0), B=(2+a, 2-a), C=(2a, 3-a) , where 0 < a < 3 0 < a < 3 .

If this inradius is R R , submit ⌊ 1 0 6 R ⌋ \lfloor 10^6 R\rfloor .


The answer is 437039.

Fletcher Mattox by Fletcher Mattox , 72, USA

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1 solution

Saya Suka
Apr 25, 2021

This is a long working since I don't know many formulas, but here goes :
1) find the length of each side in a variable.
==> √(2a² + 8)
==> √(5a² – 6a + 9)
==> √(a² – 4a + 5)
2) find the area of triangle in a variable. I used the lengthy √[s(s-x)(s-y)(s-z)] but in the end it turned out to be triangle's area A = (a² – 3a + 6)/2.
3) find the local minimum of { 2A / [ sum of side lengths] } for 0 < a < 3.
==> minimum of 0.437039559
==> at a = 2.14234


Answer
= floor (10^6 × 0.437039559)
= 437039

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