Olympiad Maths Prep

Track / Stage 5 / 17 of 400 #617 of 2000

Problem 617

AIME late
Algebra Difficulty 5.0 Find the answer

5. Let vectors aa and bb satisfy a=1,b=2|a|=1,|b|=2, and the angle between aa and bb is 6060^{\circ}. If the angle between the vectors 7a+2tb7 a+2 t b and ta+bt a+b is obtuse, then the range of the real number tt is \qquad

Official solution

5. 7<t<12-7<t<-\frac{1}{2}, and t142t \neq-\frac{\sqrt{14}}{2}.

From (7a+2tb)(ta+b)<0(7 a+2 t b) \cdot(t a+b)<0, we get 7<t<12-7<t<-\frac{1}{2}.
Moreover, the angle between 7a+2tb7 a+2 t b and ta+bt a+b must not be a straight angle, hence, t142t \neq-\frac{\sqrt{14}}{2}.

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