Maths Olympiad Prep

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Geometry Difficulty 6.0 National olympiad Prove it

7. (HUN 5) Prove that the product of the radii of three circles exscribed to a given triangle does not exceed 338\frac{3 \sqrt{3}}{8} times the product of the side lengths of the triangle. When does equality hold?

Solution

7. Let ra,rb,rc r_{a}, r_{b}, r_{c} denote the radii of the exscribed circles corresponding to the sides of lengths a,b,c a, b, c respectively, and R,p R, p and S S denote the circumradius, semiperimeter, and area of the given triangle. It is well-known that ra(pa)=rb(pb)=rc(pc)=S=p(pa)(pb)(pc)=abc4R r_{a}(p-a)=r_{b}(p-b)=r_{c}(p-c)=S=\sqrt{p(p-a)(p-b)(p-c)}=\frac{abc}{4R} . Hence, the desired inequality rarbrc338abc r_{a} r_{b} r_{c} \leq \frac{3 \sqrt{3}}{8} abc reduces to p332R p \leq \frac{3 \sqrt{3}}{2} R , which is by the law of sines equivalent to
sinα+sinβ+sinγ332 \sin \alpha + \sin \beta + \sin \gamma \leq \frac{3 \sqrt{3}}{2}
This inequality immediately follows from Jensen's inequality, since the sine is concave on [0,π][0, \pi]. Equality holds if and only if the triangle is equilateral.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.