Find a point R, such that angle increases 3 times

Question

Let $X=(4,0), Y=(4,3), O=(0,0) $ are points. I have to find point R with integer coordinates, such that $3|\angle$$XOY|$=$\angle$$|XOR|$. I think it's $R=(-3,8)$, but I am not sure. How can I prove it? Thanks for your help.

Answer 1

If $v, w$ are vectors in $\mathbb{R}^2$, the cosine of the angle between them is equal to $$\frac{v \cdot w}{||v|| \cdot ||w||}$$

Also, the triple angle identity formula can be derived from the formula for $\cos x+ y$: $$\cos 3x = 4 \cos^3 x - 3 \cos x$$

Let $\phi$ be the angle between $X$ and $Y$, and $\theta$ be the angle between $X$ and $R$. Try using these formulas to see whether $\cos 3 \phi = \cos \theta$.

Answer 2

You can either evaluate $$\tan(3\arctan(\frac 43))$$ which is $-\frac{44}{117}$ and thus obtain the line on which $R$ must lie. A solution is therefore $$(-44,117)$$

Or you can use a rotation matrix for rotating by angle $\arctan (\frac 43)$ applied twice to the coordinates of $Y$ and obtain the point $$(-\frac{44}{25},\frac{117}{25})$$ and thus draw the same conclusion about $R$