AMC 12B 2025 (Problem 16)An analog clock starts at midnight and runs for 202520252025 minutes before stopping. What is the tangent of the acute angle between the hour hand and the minute hand when the clock stops?(A) 0\text{(A)}\;0(A)0(B) 2−1\text{(B)}\;\sqrt2-1(B)2−1(C) 2−2\text{(C)}\;2-\sqrt2(C)2−2(D) 22\text{(D)}\;\frac{\sqrt2}{2}(D)22(E) 3−2\text{(E)}\;3-\sqrt2(E)3−2Related TopicsCoreTrigonometric IdentitiesMinorQuadratic FunctionsConsider the Unit Circle in TrigonometryCheck AnswerYour answer:ABCDECheckHints (3)Hint 12025 minutes=33 hours and 45 minutes2025\text{ minutes}=33\text{ hours and }45\text{ minutes}2025 minutes=33 hours and 45 minutesHint 233=12×2+933=12\times2+933=12×2+9So the time is 9:459:459:45.Hint 3α=4560⋅36012=34⋅30=452\alpha=\frac{45}{60}\cdot\frac{360}{12}=\frac34\cdot30=\frac{45}{2}α=6045⋅12360=43⋅30=2451=tan45∘=tan(2α)=tanα+tanα1−tan2α=2tanα1−tan2α1=\tan45^\circ=\tan(2\alpha)=\frac{\tan\alpha+\tan\alpha}{1-\tan^2\alpha}=\frac{2\tan\alpha}{1-\tan^2\alpha}1=tan45∘=tan(2α)=1−tan2αtanα+tanα=1−tan2α2tanα⇒1−tan2α=2tanα\Rightarrow1-\tan^2\alpha=2\tan\alpha⇒1−tan2α=2tanα⇒tan2α+2tanα−1=0\Rightarrow\tan^2\alpha+2\tan\alpha-1=0⇒tan2α+2tanα−1=0⇒tanα=−2±22−4(1)(−1)2=−2±82=−1±2\Rightarrow\tan\alpha=\frac{-2\pm\sqrt{2^2-4(1)(-1)}}{2}=\frac{-2\pm\sqrt8}{2}=-1\pm\sqrt2⇒tanα=2−2±22−4(1)(−1)=2−2±8=−1±20<α<90∘⇒tanα>0⇒tanα=−1+20<\alpha<90^\circ\Rightarrow\tan\alpha>0\Rightarrow\tan\alpha=-1+\sqrt20<α<90∘⇒tanα>0⇒tanα=−1+2Final Answer(B) 2−1\sqrt2-12−1