# Question related to tan in a ratio in a triangle

If in a triangle $\tan A:\tan B:\tan C = 1:2:3$ then, what are the ratio of the sides $a,b,c$?

As shown in this answer, if $A+B+C=\pi$, then $$\tan(A)+\tan(B)+\tan(C)=\tan(A)\tan(B)\tan(C)\tag{1}$$ Suppose $\tan(A)=x$, then $(1)$ becomes $$6x=(x+2x+3x)=(x\cdot2x\cdot3x)=6x^3\tag{2}$$ whose solutions are $x\in\{-1,0,1\}$. Therefore, $\tan(A)=1$, $\tan(B)=2$, and $\tan(C)=3$. Therefore, $$\sin(A)=\frac1{\sqrt2},\sin(B)=\frac2{\sqrt5},\sin(C)=\frac3{\sqrt{10}}\tag{3}$$ and by the Law of Sines, those are the ratios of the sides opposite those angles.

Let $AD, BE, CF$ be altitudes of triangle $ABC$. Denote $BC=a, CA=b, AB=c$.

Observe that $$\frac 23=\frac{\tan B}{\tan C} = \frac{\frac{AD}{BD}}{\frac{AD}{DC}} = \frac{DC}{BD}$$ so $BD=\frac 35 a$ and $DC=\frac 25 a$.

On the other hand Pythagorean theorem says that $$c^2-b^2=AB^2-AC^2=(BD^2+AD^2)-(CD^2+AD^2)=BD^2-CD^2=\left(\frac 35 a\right)^2 - \left(\frac 25 a\right)^2 = \frac 15 a^2$$

Analogously we find $CE=\frac 14 b$, $AE=\frac 34 b$ and $$c^2-a^2=\frac 12 b^2$$

The above results imply that $b^2=\frac 85 a^2$ and $c^2=\frac 95 a^2$. Thus $a^2:b^2:c^2=5:8:9$ which leads to the answer: $$a:b:c=\sqrt 5 : 2\sqrt 2 : 3.$$

$$sinA = \frac{a}{2R}$$ where R is the radius of the circumcircle. It'll be cancelled out later. $$cosA = \frac{b^2+c^2-a^2}{2bc}$$ So, $$tanA = \frac{abc}{R(b^2+c^2-a^2)}$$ And similarly, $$tanB = \frac{abc}{R(a^2+c^2-b^2)},tanC = \frac{abc}{R(a^2+b^2-c^2}$$ Canceling out $\frac{abc}{R}$ from the ratio, we get, $$\frac{1}{b^2+c^2-a^2}:\frac{1}{a^2+c^2-b^2}:\frac{1}{a^2+b^2-c^2} = 1:2:3$$ Equating each of these terms to k, 2k and 3k respectively (so that the ratio 1:2:3 is maintained), we get the following equations $$b^2+c^2-a^2 = \frac{1}{k}$$ $$a^2+c^2-b^2 = \frac{1}{2k}$$ $$a^2+b^2-c^2 = \frac{1}{3k}$$ On solving these, we get $a=\frac{\surd{5}}{2\surd{3}k}$, $b=\frac{\surd{2}}{\surd{3}k}$ and $c=\frac{\surd{3}}{2k}$

Therefore $a:b:c = \surd{5}:2\surd{2}:3$