speed
by math_explorer, Jan 27, 2011, 3:57 AM
Let the triangle be 
. Suppose we reflect about the angle bisector of 
and get 
. Suppose 
and 
. The overlap is quadrilateral, 
, which is symmetric about the angle bisector and therefore has area twice that of 
, whose area is clearly 
times that of . By the angle bisector theorem 
so we've succeeded if 
, which is equivalent to 
.
Not always true of course, but angle A was arbitrary from among the three angles; let's pick it so that
is shortest. Then by the triangle inequality 
, and 
so .
Okay that was too easy.
USAMO 1997.4. A regular hexagon is given with area 1. We can pick two adjacent sides
and and cut off the triangle 
where 
and 
are the midpoints of and , and we can repeat this process indefinitely. Prove that the area is always at least 1/3.
. Suppose we reflect about the angle bisector of 
and get 
. Suppose 
and 
. The overlap is quadrilateral, 
, which is symmetric about the angle bisector and therefore has area twice that of 
, whose area is clearly 
times that of . By the angle bisector theorem 
so we've succeeded if 
, which is equivalent to 
.Not always true of course, but angle A was arbitrary from among the three angles; let's pick it so that
is shortest. Then by the triangle inequality 
, and 
so .Okay that was too easy.
USAMO 1997.4. A regular hexagon is given with area 1. We can pick two adjacent sides
and and cut off the triangle 
where 
and 
are the midpoints of and , and we can repeat this process indefinitely. Prove that the area is always at least 1/3.
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