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Geometry Level 2

Find the minimum value of the expression R r \frac{R}{r} ,where R R and r r are the circumradius and inradius of any triangle.

Image Credit: Wikimedia Régis Lachaume .


The answer is 2.

Siddharth Singh by Siddharth Singh , 20, India

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4 solutions

From Euler's theorem in geometry, let d d the distance between the incentre and the circumcentre, then d 2 = R ( R − 2 r ) d^2=R(R-2r) .

From that, it follows that R ( R − 2 r ) ≥ 0 R(R-2r) \geq 0 or R − 2 r ≥ 0 R-2r \geq 0 , or R ≥ 2 r R \geq 2r .

Dividing by r r we get R r ≥ 2 \dfrac{R}{r} \geq \boxed{2} .

5 Helpful 0 Interesting 0 Brilliant 0 Confused
Alan Yan
Sep 2, 2015

By Euler you know that R ≥ 2 r ⟹ R r ≥ 2 R \geq 2r \implies \frac{R}{r} \geq 2

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Ramiel To-ong
Sep 15, 2015

the least value according to euler is 2

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Yash Dharme
Sep 4, 2015

The triangle is equilateral triangle Draw perpandicular to tangent BC then, According to 30-60-90 theorem, Side opp. to 30=hypo/2 And in this case hypo=R and side opp. to 30 is r. Then, r=R/2 Hence, R/r=2

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