Maths Olympiad Prep

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, 2017

Geometry Difficulty 4.6 AIME Prove it United States

Problem:

Let ABCABC be a triangle with circumradius R=17R=17 and inradius r=7r=7. Find the maximum possible value of sinA2\sin \frac{A}{2}.

Solution

Solution:

Letting II and OO denote the incenter and circumcenter of triangle ABCABC we have by the triangle inequality that
AOAI+OIRrsinA2+R(R2r) AO \leq AI + OI \Longrightarrow R \leq \frac{r}{\sin \frac{A}{2}} + \sqrt{R(R-2r)}
and by plugging in our values for rr and RR we get
sinA217+5134 \sin \frac{A}{2} \leq \frac{17+\sqrt{51}}{34}
as desired. Equality holds when ABCABC is isosceles and II lies between AA and OO.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.