How Do You Find AC/AB Given Tan Ratios in a Triangle?

[anemone]

Here is this week's POTW:

-----

In a triangle $ABC$, $\tan A:\tan B: \tan C=1:2:3$. Find $\dfrac{AC}{AB}$.

-----

Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to the following members for their correct solution::)

1. greg1313
2. kaliprasad

Solution from greg1313:

Spoiler

Using the triple tangent identity

$$\tan A+\tan B+\tan C=\tan A\tan B\tan C$$

and the given ratios we have

$$6\tan A=6\tan^3A\Rightarrow\tan A=1$$

Now we construct a triangle with base $\overline{AB}$ and an altitude from $C$ to $\overline{AB}$ at $P$.
WLOG, let $\overline{AP}=\overline{CP}=1$. Then, to form a tangent of $2$ at $B$, let $\overline{BP}=\frac12$.
We will now verify that $\tan C=3$.

Using the Pythagorean theorem with $\overline{AB}=\frac32$, $\overline{AC}=\sqrt2$ and $\overline{BC}=\frac{\sqrt5}{2}$,

$$\frac54-x^2=\frac94-(\sqrt2-x)^2\Rightarrow x=\frac{\sqrt2}{4}$$

where $x$ is the distance between the foot of an altitude from $B$ to $\overline{AC}$ and $C$.
Again with Pythagoras, the altitude from $B$ is $\frac{3\sqrt2}{4}$ and the tangent at $C$ is indeed $3$.

As $\overline{AC}=\sqrt2$ and $\overline{AB}=\frac32$, the desired ratio is $\frac{2\sqrt2}{3}$.