Four Incircles in an Equilateral Triangle, Third Solution

Elsewhere we described and solved the following sangaku:

In an equilateral triangle \(ABC\) the lines \(BEF\), \(CFD\), \(ADE\) are drawn making equal angles with \(AB\), \(BC\), \(CA\), respectively, forming the triangle \(DEF\), and so that the radius of the incircle of triangle \(DEF\) is equal to the radii of the incircles of triangles $ABE,$ $BCF,$ and $ACD.$

four incircles sangaku

Find \(DE\) in terms of \(AB\).

The solution below is by J. John Samuel (7 MAy, 2014). Like George Zettler, John first establishes the basic facts, viz.,

  1. The area $A$ of a triangle with the semiperimeter $p$ and the inradius $r$ equals $A=pr.$
  2. In an equilateral triangle the side length $a$ and the inradius $r$ are related through $a = 2\sqrt{3}r.$
  3. With the reference to the above diagram, in $\Delta ABE,$ the altitude $h$ from vertex $B$ equals $\displaystyle h=\frac{\sqrt{3}}{2}BE.$

Next, with a change of notations reflected in the diagram below,

four incircles sangaku - J. John Samuel's proof

John derives the expression

$\displaystyle \frac{S_{o}^{2}}{S_{i}^{2}}=3k(k+1)+1,$

where $S_{o}$ and $S_{i}$ are the side lengths of the $o$uter and the $i$nner equilateral triangles. The easiest way to see that is to observe that, say, in $\Delta ABE,$ the Law of Cosines gives

$S_{o}^{2}=((kS_{i})^{2}+(kS_{i})((k+1)S_{i})+((k+1)S_{i})^{2}=(S_{i})^{2}(k^{2}+k(k+1)+(k+1)^{2}).$

The final step is illustrated by a diagram and a moving observation:

four incircles sangaku - J. John Samuel's proof, the final step

We may find the relationship between $S_{o}$ and $S_{i}$ in yet another way, through the area formula $A=pr:$

$\displaystyle\begin{align} Area(\Delta ABE) &= \frac{1}{2}(S_{o}+kS_{i}+(k+1)S_{i})R\\ &=\frac{1}{2}\frac{S_{i}}{2\sqrt{3}}(S_{o}+(2k+1)S_{i})\\ &=\frac{S_{i}}{4\sqrt{3}}(S_{o}+(2k+1)S_{i}). \end{align}$

On the other hand,

$\displaystyle\begin{align} Area(\Delta ABE) &= \frac{1}{2}h\cdot AE\\ &=\frac{1}{2}\frac{\sqrt{3}}{2}(kS_{i})((k+1)S_{i})\\ &=\frac{\sqrt{3}}{4}S_{i}^{2}k(k+1). \end{align}$

Comparing the two expressions, we obtain $S_{o}=S_{i}(3k^{2}+k-1),$ which leads to an equation for k:

$\displaystyle\frac{S_{o}^{2}}{S_{i}^{2}}=(3k^{2}+k-1)^{2}=3k(k+1)+1.$

This is a fourth degree equation. Undeterred, we proceed to simplify:

$\displaystyle\begin{align} 3k(k+1)+1 &= (3k^{2}+k-1)^{2}\\ &=9k^{4}+6k^{3}-5k^{2}-2k+1, \end{align}$

or, $9k^{4}+6k^{3}-8k^{2}-5k=0.$ $k=0,$ not fitting the context of the problem can be cancelled out: $9k^{3}+6k^{2}-8k-5=0.$ In addition, we may notice (and John does) that $-9+6+8-5=0,$ implying that $k=-1$ is another irrelevant root of the equation. Dividing by $k+1$ leaves $9k^{2}-3k-5=0,$ with the roots $\displaystyle\frac{3\pm\sqrt{189}}{18}=\frac{1\pm\sqrt{21}}{6}.$ Dropping the negative root, we are left with the only suitable solution

$\displaystyle k=\frac{1+\sqrt{21}}{6}. $

Now we are in a position to solve the problem:

$\displaystyle \frac{S_{o}}{S_{i}}=3k^{2}+k-1=\frac{3+\sqrt{21}}{3}. $

From this

$\displaystyle \frac{S_{i}}{S_{o}}=\frac{3}{3+\sqrt{21}}=\frac{3(\sqrt{21}-3)}{21-9}=\frac{\sqrt{21}-3}{4}. $

This agrees with the solutions of George Zettler and Angela Drei.

    [an error occurred while processing this directive]

|Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny
[an error occurred while processing this directive]
[an error occurred while processing this directive]