1993 AHSME Problems/Problem 25
Problem
Let be the set of points on the rays forming the sides of a
angle, and let
be a fixed point inside the angle
on the angle bisector. Consider all distinct equilateral triangles
with
and
in
.
(Points
and
may be on the same ray, and switching the names of
and
does not create a distinct triangle.)
There are
Solution
Take the "obvious" equilateral triangle , where
is the vertex,
is on the upper ray, and
is our central point. Slide
down on the top ray to point
, and slide
down an equal distance on the bottom ray to point
.
Now observe and
. We have
and
, therefore
. By our construction of moving the points the same distance, we have
. Also,
by the original equilateral triangle. Therefore, by SAS congruence,
.
Now, look at . We have
from the above congruence. We also have the included angle
is
. To prove that, start with the
angle
, subtract the angle
, and add the congruent angle
.
Since is an isosceles triangle with vertex of
, it is equilateral.
See also
1993 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Problem 26 | |
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