Maths Olympiad Prep

Track / Stage 4 / 146 of 340 #406 of 1964

Problem 406

AMC 12 late, AIME early
Algebra Difficulty 4.7 Find the answer

tan⁡(3α−2β)=12,tan⁡(5a−4β)=14\tan (3 \alpha-2 \beta)=\frac{1}{2}, \tan (5 a-4 \beta)=\frac{1}{4}, then tan⁡α=\tan \alpha=

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

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Official solution

8. 1316\frac{13}{16}

So tana tan⁡[(6α−4.3)(5a+β)−−tan⁡(6a4.3)−tan⁡(5α−4.3)1⋅tan⁡16α4β)tan⁡(5α−4.3)\tan \left[(6 \alpha-4.3) \quad(5 a+\beta)^{-}-\begin{array}{ll}\tan (6 a & 4.3)-\tan (5 \alpha-4.3) \\ 1 \cdot \tan 16 \alpha & 4 \beta) \tan (5 \alpha-4.3)\end{array}\right.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.