Difference between revisions of "1996 IMO Problems/Problem 5"
(→Solution) |
|||
(2 intermediate revisions by the same user not shown) | |||
Line 11: | Line 11: | ||
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> |
From the parallel lines on the hexagon we get: | From the parallel lines on the hexagon we get: | ||
− | <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> [Equations 1] |
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>. | 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>. | ||
Line 80: | Line 80: | ||
{{alternate solutions}} | {{alternate solutions}} | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{IMO box|year=1996|num-b=4|num-a=6}} | ||
+ | [[Category:Olympiad Geometry Problems]] | ||
+ | [[Category:3D Geometry Problems]] |
Latest revision as of 16:32, 20 November 2023
Problem
Let be a convex hexagon such that
is parallel to
,
is parallel to
, and
is parallel to
. Let
,
,
denote the circumradii of triangles
,
,
, respectively, and let
denote the perimeter of the hexagon. Prove that
Solution
Let
Let
Let
From the parallel lines on the hexagon we get:
[Equations 1]
So now we look at . We construct a perpendicular from
to
and a perpendicular from
to
.
We find out the length of these two perpendiculars and add them to get the distance between parallel lines and
and because of the triangle inequality the distance
is greater or equal to tha the distance between parallel lines
and
:
This provides the following inequality:
Using the [Equations 1] we simplify to:
[Equation 2]
We now construct a perpendicular from to
and a perpendicular from
to
. Then we find out the length of these two perpendiculars and add them to get the distance between parallel lines
and
and get:
Using the [Equations 1] we simplify to:
[Equation 3]
We now add [Equation 2] and [Equation 3] to get:
[Equation 4]
We now use the Extended law of sines on with
to get:
[Equation 5]
Substitute [Equation 5] into [Equation 4]:
[Equation 6]
To find the equivalent inequality for and
we just need shift the indexes by two. That is to add
to each of the indexes of
and
and adjust the indexes so that for
the indexes are 1 through 6, and for
the indexes are 1 through 3.
[Equation 7]
[Equation 8]
Adding [Equation 6], [Equation 7], and [Equation 8] we get:
From AM-GM inequality we get:
Therefore,
for any index
and
Therefore,
Since , then
~ Tomas Diaz. orders@tomasdiaz.com
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
1996 IMO (Problems) • Resources | ||
Preceded by Problem 4 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 6 |
All IMO Problems and Solutions |