Olympiad Maths Prep

Track / Stage 5 / 248 of 400 #848 of 2000

Problem 848

AIME late
Geometry Difficulty 5.6 Prove it

8. In ABC\triangle A B C, prove:
(1) s=bcos2C2+ccos2B2s=b \cos ^{2} \frac{C}{2}+c \cos ^{2} \frac{B}{2};
(2) s=a+bsin2C2+csin2B2s=a+b \sin ^{2} \frac{C}{2}+c \sin ^{2} \frac{B}{2};
(3) s=4RcosA2cosB2cosC2s=4 R \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2};
(4) s=r(cotA2+cotB2+cotC2)s=r\left(\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2}\right).

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

8. (1) Reduce the order and use the projection theorem.
(2) Use sin2x+cos2x=1\sin ^{2} x+\cos ^{2} x=1 to transform into (1).
(3) s=Δrs=\frac{\Delta}{r}, proved by the conclusion of the 11th sub-question of the 1st question.
(4) s=Δrs=\frac{\Delta}{r}, from the conclusion of the 2nd sub-question of the 1st question, we know cotA2cotB2cotC2=cotA2+cotB2+cotC2\cot \frac{A}{2} \cot \frac{B}{2} \cot \frac{C}{2}=\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2}.

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