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Difference between revisions of "1996 IMO Problems/Problem 5"

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(→‎Solution)
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Let <math>d_{1}=\left| FB \right|,\;d_{2}=\left| BD \right|,\;d_{1}=\left| DF \right|</math>
 
Let <math>d_{1}=\left| FB \right|,\;d_{2}=\left| BD \right|,\;d_{1}=\left| DF \right|</math>
  
−
Let <math>\alpha_{1}=\angle FAB,\;\alpha_{2}=\angle ABC,\;\alpha_{3}=\angle BCD,\;\alpha_{4}=\angle CDE,\;\alpha_{5}=\angle DEF,\;\alpha_{6}=\angle EFA</math>
+
Let <math>\alpha_{1}=\angle FAB,\;\alpha_{2}=\angle ABC,\;\alpha_{3}=\angle BCD,\;\alpha_{4}=\angle CDE,\;\alpha_{5}=\angle DEF,\;\alpha_{6}=\angle EFA\;</math> [Equations 1]
  
 
From the parallel lines on the hexagon we get:
 
From the parallel lines on the hexagon we get:
Line 17: Line 17:
 
<math>\alpha_{1}=\alpha_{4},\;\alpha_{2}=\alpha_{5},\;\alpha_{3}=\alpha_{6}</math>
 
<math>\alpha_{1}=\alpha_{4},\;\alpha_{2}=\alpha_{5},\;\alpha_{3}=\alpha_{6}</math>
  
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{{alternate solutions}}
+
So now we look at <math>\Delta FAB</math>.  We construct a perpendicular from <math>A</math> to <math>FE</math> and a perpendicular from <math>A</math> to <math>BC</math>.
 +
 
 +
We find out the length of these two perpendiculars and add them to get the distance between parallel lines <math>FE</math> and <math>BC</math> and because of the triangle inequality the distance <math>\left| FB \right|</math> is greater or equal to tha the distance between parallel lines <math>FE</math> and <math>BC</math>:
 +
 
 +
This provides the following inequality:
 +
 
 +
<math>d_{1} \ge s_{6}sin(\alpha{6})+s_{1}sin(\alpha{2})</math>
 +
 
 +
Using the [Equations 1] we simplify to:
 +
 
 +
<math>d_{1} \ge s_{6}sin(\alpha{3})+s_{1}sin(\alpha{2})</math> [Equation 2]
 +
 
 +
We now construct a perpendicular from <math>D</math> to <math>FE</math> and a perpendicular from <math>D</math> to <math>BC</math>. Then we find out the length of these two perpendiculars and add them to get the distance between parallel lines <math>FE</math> and <math>BC</math> and get:
 +
 
 +
<math>d_{1} \ge s_{3}sin(\alpha{3})+s_{4}sin(\alpha{5})</math>
 +
 
 +
Using the [Equations 1] we simplify to:
 +
 
 +
<math>d_{1} \ge s_{3}sin(\alpha{3})+s_{4}sin(\alpha{2})</math> [Equation 3]
 +
 
 +
We now add [Equation 2] and [Equation 3] to get:

Revision as of 14:57, 13 November 2023

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:

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