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1996 IMO Problems/Problem 5

Revision as of 13:59, 13 November 2023 by Tomasdiaz (talk | contribs) (→‎Solution)

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

From the parallel lines on the hexagon we get:

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

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 $R_{A}$:

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