1996 IMO Problems/Problem 5

Problem

Let $ABCDEF$ be a convex hexagon such that $AB$ is parallel to $DE$, $BD$ is parallel to $EF$, and $CD$ is parallel to $FA$. Let $R_{A}$, $R_{C}$, $R_{E}$ denote the circumradii of triangles $FAB$, $BCD$, $DEF$, respectively, and let $P$ denote the perimeter of the hexagon. Prove that

$R_{A}+R_{C}+R_{E} \ge \frac{P}{2}$

Solution

Let $s_{1}=\left| AB \right|,\;s_{2}=\left| BC \right|,\;s_{3}=\left| CD \right|,\;s_{4}=\left| DE \right|,\;s_{5}=\left| EF \right|,\;s_{6}=\left| FA \right|$

Let $d_{1}=\left| FB \right|,\;d_{2}=\left| BD \right|,\;d_{1}=\left| DF \right|$

Let $\alpha_{1}=\angle FAB,\;\alpha_{2}=\angle ABC,\;\alpha_{3}=\angle BCD,\;\alpha_{4}=\angle CDE,\;\alpha_{5}=\angle DEF,\;\alpha_{6}=\angle EFA\;$

From the parallel lines on the hexagon we get:

$\alpha_{1}=\alpha_{4},\;\alpha_{2}=\alpha_{5},\;\alpha_{3}=\alpha_{6}$ [Equations 1]

So now we look at $\Delta FAB$. We construct a perpendicular from $A$ to $FE$ and a perpendicular from $A$ to $BC$.

We find out the length of these two perpendiculars and add them to get the distance between parallel lines $FE$ and $BC$ and because of the triangle inequality the distance $\left| FB \right|$ is greater or equal to tha the distance between parallel lines $FE$ and $BC$:

This provides the following inequality:

$d_{1} \ge s_{6}sin(\alpha{6})+s_{1}sin(\alpha{2})$

Using the [Equations 1] we simplify to:

$d_{1} \ge s_{6}sin(\alpha{3})+s_{1}sin(\alpha{2})$ [Equation 2]

We now construct a perpendicular from $D$ to $FE$ and a perpendicular from $D$ to $BC$. Then we find out the length of these two perpendiculars and add them to get the distance between parallel lines $FE$ and $BC$ and get:

$d_{1} \ge s_{3}sin(\alpha{3})+s_{4}sin(\alpha{5})$

Using the [Equations 1] we simplify to:

$d_{1} \ge s_{3}sin(\alpha{3})+s_{4}sin(\alpha{2})$ [Equation 3]

We now add [Equation 2] and [Equation 3] to get:

$2d_{1} \ge (s_{3}+s_{6})sin(\alpha_{3})+(s_{1}+s_{4})sin(\alpha_{2})$ [Equation 4]

We now use the Extended law of sines on $\Delta FAB$ with $R_{A}$ to get:

$\frac{d_{1}}{sin(\alpha_{1})}=2R_{A}$

$d_{1}=2R_{A}sin(\alpha_{1})$ [Equation 5]

Substitute [Equation 5] into [Equation 4]:

$4R_{A}sin(\alpha_{1}) \ge (s_{3}+s_{6})sin(\alpha_{3})+(s_{1}+s_{4})sin(\alpha_{2})$

$4R_{A} \ge (s_{3}+s_{6})\frac{sin(\alpha_{3})}{sin(\alpha_{1})}+(s_{1}+s_{4})\frac{sin(\alpha_{2})}{sin(\alpha_{1})}$ [Equation 6]

To find the equivalent inequality for $4R_{C}$ and $4R_{E}$ we just need shift the indexes by two. That is to add $2$ to each of the indexes of $s_{i}$ and $\alpha_{i}$ and adjust the indexes so that for $s$ the indexes are 1 through 6, and for $\alpha$ the indexes are 1 through 3.

$4R_{C} \ge (s_{2}+s_{5})\frac{sin(\alpha_{2})}{sin(\alpha_{3})}+(s_{3}+s_{6})\frac{sin(\alpha_{1})}{sin(\alpha_{3})}$ [Equation 7]

$4R_{E} \ge (s_{1}+s_{4})\frac{sin(\alpha_{1})}{sin(\alpha_{2})}+(s_{2}+s_{5})\frac{sin(\alpha_{3})}{sin(\alpha_{2})}$ [Equation 8]

Adding [Equation 6], [Equation 7], and [Equation 8] we get:

$4\left( R_{A}+R_{C}+R_{E} \right) \ge (s_{1}+s_{4})\left( \frac{sin(\alpha_{2})}{sin(\alpha_{1})}+ \frac{sin(\alpha_{1})}{sin(\alpha_{2})}\right)+(s_{2}+s_{5})\left( \frac{sin(\alpha_{2})}{sin(\alpha_{3})}+ \frac{sin(\alpha_{3})}{sin(\alpha_{2})}\right)+(s_{3}+s_{6})\left( \frac{sin(\alpha_{3})}{sin(\alpha_{1})}+ \frac{sin(\alpha_{1})}{sin(\alpha_{3})}\right)$

From AM-GM inequality we get:

$x+\frac{1}{x} \ge 2\;$ Therefore, $\left( \frac{sin(\alpha_{a})}{sin(\alpha_{b})}+ \frac{sin(\alpha_{b})}{sin(\alpha_{a})}\right) \ge 2\;$ for any index $a$ and $b$

Therefore,

$4\left( R_{A}+R_{C}+R_{E} \right) \ge 2(s_{1}+s_{4})+2(s_{2}+s_{5})+2(s_{3}+s_{6})$

$4\left( R_{A}+R_{C}+R_{E} \right) \ge 2\sum_{i=1}^{6}s_{i}$

$R_{A}+R_{C}+R_{E} \ge \frac{1}{2}\sum_{i=1}^{6}s_{i}$

Since $P=\sum_{i=1}^{6}s_{i}$, then $R_{A}+R_{C}+R_{E} \ge \frac{P}{2}$

~ Tomas Diaz. orders@tomasdiaz.com