2024 AMC 12A Problem 23What is the value oftan2π16tan23π16+tan2π16tan25π16+tan23π16tan27π16+tan25π16tan27π16?\tan^2\frac{\pi}{16}\tan^2\frac{3\pi}{16}+\tan^2\frac{\pi}{16}\tan^2\frac{5\pi}{16}+\tan^2\frac{3\pi}{16}\tan^2\frac{7\pi}{16}+\tan^2\frac{5\pi}{16}\tan^2\frac{7\pi}{16}?tan216πtan2163π+tan216πtan2165π+tan2163πtan2167π+tan2165πtan2167π?(A) 28\text{(A)}\;28(A)28(B) 68\text{(B)}\;68(B)68(C) 70\text{(C)}\;70(C)70(D) 72\text{(D)}\;72(D)72(E) 84\text{(E)}\;84(E)84Related TopicsCoreTrigonometric IdentitiesTrigonometric Values of Specific AnglesMinorTrigonometric TransformationsCheck AnswerYour answer:ABCDECheckHints (5)Hint 1tan2π16tan23π16+tan2π16tan25π16+tan23π16tan27π16+tan25π16tan27π16\tan^2\frac{\pi}{16}\tan^2\frac{3\pi}{16}+\tan^2\frac{\pi}{16}\tan^2\frac{5\pi}{16}+\tan^2\frac{3\pi}{16}\tan^2\frac{7\pi}{16}+\tan^2\frac{5\pi}{16}\tan^2\frac{7\pi}{16}tan216πtan2163π+tan216πtan2165π+tan2163πtan2167π+tan2165πtan2167π=tan2π16(tan23π16+tan25π16)+tan27π16(tan23π16+tan25π16)=\tan^2\frac{\pi}{16}\left(\tan^2\frac{3\pi}{16}+\tan^2\frac{5\pi}{16}\right)+\tan^2\frac{7\pi}{16}\left(\tan^2\frac{3\pi}{16}+\tan^2\frac{5\pi}{16}\right)=tan216π(tan2163π+tan2165π)+tan2167π(tan2163π+tan2165π)=(tan23π16+tan25π16)(tan2π16+tan27π16)=\left(\tan^2\frac{3\pi}{16}+\tan^2\frac{5\pi}{16}\right)\left(\tan^2\frac{\pi}{16}+\tan^2\frac{7\pi}{16}\right)=(tan2163π+tan2165π)(tan216π+tan2167π)Hint 23π16+5π16=8π16=π2⇒tan5π16=cot3π16\frac{3\pi}{16}+\frac{5\pi}{16}=\frac{8\pi}{16}=\frac{\pi}{2}\quad\Rightarrow\quad\tan\frac{5\pi}{16}=\cot\frac{3\pi}{16}163π+165π=168π=2π⇒tan165π=cot163πHint 3Let α=3π16\alpha=\frac{3\pi}{16}α=163π.tan23π16+tan25π16=tan2α+cot2α\tan^2\frac{3\pi}{16}+\tan^2\frac{5\pi}{16}=\tan^2\alpha+\cot^2\alphatan2163π+tan2165π=tan2α+cot2α=(tanα+cotα)2−2tanαcotα=(\tan\alpha+\cot\alpha)^2-2\tan\alpha\cot\alpha=(tanα+cotα)2−2tanαcotα=(2sin2α)2−2=4sin22α−2=\left(\frac{2}{\sin2\alpha}\right)^2-2=\frac{4}{\sin^2 2\alpha}-2=(sin2α2)2−2=sin22α4−2=42+24−2=162+2−2=16(2−2)2−2=\frac{4}{\frac{2+\sqrt2}{4}}-2=\frac{16}{2+\sqrt2}-2=\frac{16(2-\sqrt2)}{2}-2=42+24−2=2+216−2=216(2−2)−2=8(2−2)−2=14−82=8(2-\sqrt2)-2=14-8\sqrt2=8(2−2)−2=14−82Hint 4Let β=π16\beta=\frac{\pi}{16}β=16π.tan2π16+tan27π16=tan2β+cot2β\tan^2\frac{\pi}{16}+\tan^2\frac{7\pi}{16}=\tan^2\beta+\cot^2\betatan216π+tan2167π=tan2β+cot2β=(tanβ+cotβ)2−2tanβcotβ=(\tan\beta+\cot\beta)^2-2\tan\beta\cot\beta=(tanβ+cotβ)2−2tanβcotβ=(2sin2β)2−2=4sin22β−2=\left(\frac{2}{\sin2\beta}\right)^2-2=\frac{4}{\sin^2 2\beta}-2=(sin2β2)2−2=sin22β4−2=42−24−2=162−2−2=\frac{4}{\frac{2-\sqrt2}{4}}-2=\frac{16}{2-\sqrt2}-2=42−24−2=2−216−2=16(2+2)2−2=8(2+2)−2=\frac{16(2+\sqrt2)}{2}-2=8(2+\sqrt2)-2=216(2+2)−2=8(2+2)−2=14+82=14+8\sqrt2=14+82Hint 5By Hints 1, 3, and 4,tan2π16tan23π16+tan2π16tan25π16+tan23π16tan27π16+tan25π16tan27π16\tan^2\frac{\pi}{16}\tan^2\frac{3\pi}{16}+\tan^2\frac{\pi}{16}\tan^2\frac{5\pi}{16}+\tan^2\frac{3\pi}{16}\tan^2\frac{7\pi}{16}+\tan^2\frac{5\pi}{16}\tan^2\frac{7\pi}{16}tan216πtan2163π+tan216πtan2165π+tan2163πtan2167π+tan2165πtan2167π=(14−82)(14+82)=(14-8\sqrt2)(14+8\sqrt2)=(14−82)(14+82)=196−128=68=196-128=68=196−128=68Related Problems (7)AMC 10B/12B 2023 (Problem 25)AMC 12B 2025 (Problem 16)AIME II 2026 (Problem 8)2024 AMC 12A Problem 102024 AMC 12A Problem 18MathCounts 2026 Chapter Sprint Round (Problem 28)AIME I 2026 (Problem 14)Final Answer(B) 686868