# Find the smallest possible inradius

Inside triangle ABC there are three circles with radii $$r_1$$, $$r_2$$, and $$r_3$$. Each is tangent to two sides of the triangle and to its incircle. The incircle has radius $$r$$. All of $$r$$, $$r_1$$, $$r_2$$, and $$r_3$$ are distinct perfect square integers. Find the smallest possible value of inradius $$r$$.

I hope you enjoy this problem.

Note by Patrick Brown
5 years, 1 month ago

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r = 81, r1 = 9, r2 = 36, r3 = 49 is valid, and is probably the smallest possible value (pending some checks on the smaller cases) , by some trigonometric/algebraic arguments.

Letting r, r1, r2, r3 be x^2, a^2, b^2, c^2 respectively, you can force out x^2 = ab + bc + ca; that's all i'll let on

edit: these a,b,c are different from the diagram. (the a,b,c in there are irrelevant). To make it clearer:

r^2 = rt(r1r2) + rt(r2r3) + rt(r3*r1)

- 5 years, 1 month ago

Is it that r=81, r1=16, r2=36, r3=49??

- 5 years, 1 month ago

Hmm.. Gabriel. If r = 81, r1= 9, r2 = 36, and r3 = 49 and by your post: r = x^2 r1 = a^2 r2 = b^2 and r3 =c^2. Wouldn't that make each of the other variables (x, a, b, and c) be the smallest values? As in 81 = x^2 so x = 9, 9= a^2 so a =3, 36 = b^2 so be = 6, and 49 = c^2 so c= 7. With that would those be the new smallest values? Also by looking at the picture we can say that r1 and r3 are the same because the triangle is isosceles. In that case is r1 = 9, r3 = 9... And a and c = 3? So if they were not the same, Plugging the numbers in fit a b and c (3,6,7) we can find the angles of each side. So A = 60.038 B = 92.729 and C = 27.271. From there you can go deeper and find the arc length then find the true radius of each...? I may have done something wrong but, this is just my inquiry for this question.

- 5 years, 1 month ago

x^2 = ab + bc + ca

81 = 36 + 67 + 7*3

but this does not imply

x = rt(ab) + rt(bc) + rt(ca)

rt(18) + rt(42) + rt(21) does not equal 9

- 5 years, 1 month ago

1/r=1/r1+1/r2+1/r3

- 5 years, 1 month ago