Difference between revisions of "2007 Cyprus MO/Lyceum/Problem 15"
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==Solution== | ==Solution== | ||
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+ | The problem statement apparently misses one crucial piece of information: the fact that '''all the sides of the octagon are equal'''. Without this fact the octagon area is not uniquely determined. For example, we could move point <math>B</math> along a suitable arc (the locus of all points <math>X</math> such that <math>AXC</math> is <math>120^\circ</math>), and as this would change the height from <math>B</math> to <math>AC</math>, it would change the area of the triangle <math>ABC</math>, and hence the area of the octagon. | ||
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+ | With this additional assumption we can compute the area of <math>ABCDEFGH</math> as the area of <math>ACEG</math> (which is obviously <math>2</math>), minus four times the area of <math>ABC</math>. | ||
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+ | In the triangle <math>ABC</math>, we have <math>AC=\sqrt 2</math>. Let <math>B'</math> be the foot of the height from <math>B</math> onto <math>AC</math>. As <math>AB=BC</math>, <math>B'</math> bisects <math>AC</math>. As the angle <math>ABC</math> is <math>120^\circ</math>, the angle <math>ABB'</math> is <math>60^\circ</math>. | ||
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+ | (to be finished in a few hours) | ||
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{{solution}} | {{solution}} | ||
==See also== | ==See also== | ||
{{CYMO box|year=2007|l=Lyceum|num-b=14|num-a=16}} | {{CYMO box|year=2007|l=Lyceum|num-b=14|num-a=16}} |
Revision as of 07:59, 29 January 2009
Problem
The reflex angles of the concave octagon measure
each. Diagonals
and
are perpendicular, bisect each other, and are both equal to
.
The area of the octagon is
Solution
The problem statement apparently misses one crucial piece of information: the fact that all the sides of the octagon are equal. Without this fact the octagon area is not uniquely determined. For example, we could move point along a suitable arc (the locus of all points
such that
is
), and as this would change the height from
to
, it would change the area of the triangle
, and hence the area of the octagon.
With this additional assumption we can compute the area of as the area of
(which is obviously
), minus four times the area of
.
In the triangle , we have
. Let
be the foot of the height from
onto
. As
,
bisects
. As the angle
is
, the angle
is
.
(to be finished in a few hours)
This problem needs a solution. If you have a solution for it, please help us out by adding it.
See also
2007 Cyprus MO, Lyceum (Problems) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 |