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 TopicsCoreToolkit 38 — Trigonometric IdentitiesMinorToolkit 21 — Quadratic FormulaToolkit 62 — Consider the Unit Circle in TrigonometryHints (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
