How can I improve my picture proof of Reverse Triangle Inequality?

Question

Diagram beneath reappears on standardized tests IN BLACK AND WHITE with different lengths, letters, and orientation that require students to label in terms of $\vec{b}, \vec{r}$ ( = circle's radius ) the:

1. long line from $(-4, 0)$ to B.

2. short line segment from a point on the circumference to B.

So I added color.

But many parents and students STILL fail to distinguish $\color{deepskyblue}{|\vec{b}| - |\vec{r}|}$ from $\color{red}{\vec{b} - \vec{r}}$, solely by eye! How can I improve my diagram? How can I better contrast the differences between $\color{cornflowerblue}{|\vec{b}| - |\vec{r}|}$ vs. $\color{red}{\vec{b} - \vec{r}}$ ?

Answer 1

Some people may never be convinced. But I would have thought this might help others:

• $\vec{r}$ is the vector $OR$ and $R$ is $|\vec{r}|$ from $O$
• $\vec{b}$ is the vector $OB$ and $B$ is $|\vec{b}|$ from $O$
• So $RB$ is $\vec{b}-\vec{r}$ and $B$ is $|\vec{b}-\vec{r}|$ from $R$
• The black circle about $O$ containing $R$ is of radius $|\vec{r}|$
• The point $D$ is where the black circle crosses $OB$ and is $|\vec{r}|$ from $O$ so $|\vec{b}|-|\vec{r}|$ from $B$.
• The red circle about $B$ containing $R$ is of radius $|\vec{b}-\vec{r}|$ and the red radius $RB$ shown is $\vec{b}-\vec{r}$
• The blue circle about $B$ containing $D$ is of radius $|\vec{b}|-|\vec{r}|$ and the blue radius $DB$ shown is of length $|\vec{b}-\vec{r}|$
• $D$ is the closest point on the black circle to $B$
• So $|\vec{b}|-|\vec{r}| \le |\vec{b}-\vec{r}|$
• with equality when $D$ is $R$, i.e when $R$ lies on $OB$

This is not a complete proof, and in particular does not deal with the case where $|\vec{b}|<|\vec{r}|$.

Another diagram may help for that, and shows that we need more absolute signs in the argument above. The key point is that $D$ remains the closest point on the black circle to $B$ but better to say $D$ is $\Big||\vec{b}|-|\vec{r}|\Big|$ from $B$ and so $\Big||\vec{b}|-|\vec{r}|\Big| \le |\vec{b}-\vec{r}|$, with equality $D$ is $R$ i.e. when $R$ lies on the ray $\vec{OB}$.