Points $A,B,C,D$ have been marked on checkered paper (see fig.). Find the tangent of the angle $ABD$.
Problem
Source: 2021 Oral Moscow Geometry Olympiad grades 8-9 p1
Tags: geometry, trigonometry, grid
new_to_mew_too
24.05.2021 16:26
REDACTED
ZIXILI
24.05.2021 18:47
Use the vectors. $\overrightarrow {BA}=(-2,1), \ \overrightarrow{BD}=(-1,\frac{4}{3})$ $cos<\overrightarrow {BA}, \overrightarrow{BD}>=\frac{\overrightarrow{BA}\cdot\overrightarrow{BD}}{|\overrightarrow{BA}||\overrightarrow{BD}|}=\frac{2+\frac{4}{3}}{\sqrt{5} \times\frac{5}{3}}=\frac{2}{\sqrt{5}}$ ∴$tan\angle ABD=\frac{1}{2}$
Bexultan
14.04.2023 13:53
By Pythagorean theorem, it is easy to get that the triangle $ABC$ is right angled at $B$ and that $AB=BC$. By similar triangles you can get that $\dfrac{AD}{DC}=\dfrac{1}{2}$. Let $\angle ABD=\alpha$. We have: $$\text{tg } \alpha=\dfrac{\text{sin }\alpha}{\text{cos } \alpha}=\dfrac{\text{sin } \alpha}{\text{sin } (90-\alpha)}=\dfrac{\text{sin } \alpha}{\text{sin }CBD}=\dfrac{AD}{DC}\cdot\dfrac{AB}{BC}=\dfrac{1}{2}$$
