Difference between revisions of "2001 Pan African MO Problems/Problem 3"
Rockmanex3 (talk | contribs) (Solution to Problem 3 -- coordinate geometry is OP!) |
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Latest revision as of 10:09, 27 March 2020
Problem
Let be an equilateral triangle and let
be a point outside this triangle, such that
is an isosceles triangle with a right angle at
. A grasshopper starts from
and turns around the triangle as follows. From
the grasshopper jumps to
, which is the symmetric point of
with respect to
. From
, the grasshopper jumps to
, which is the symmetric point of
with respect to
. Then the grasshopper jumps to
which is the symmetric point of
with respect to
, and so on. Compare the distance
and
.
.
Solution
We can use coordinate geometry to solve the problem. Let ,
, and
, making
. To calculate the coordinates of
, note that
since
is a kite. Thus,
bissects
, so
. Additionally,
because
bissects
. Thus, the coordinates of
are
.
By repeatedly applying the Midpoint Formula, we can determine the coordinates of
,
,
, and so on. We can also use the Distance Formula to calculate the distance of
,
, and so on. The values are shown in the below table.
![]() |
Coordinates of ![]() |
![]() |
1 | ![]() |
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2 | ![]() |
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3 | ![]() |
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4 | ![]() |
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5 | ![]() |
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6 | ![]() |
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7 | ![]() |
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8 | ![]() |
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Note that the coordinates of as well as the distance
cycle after
. Thus,
if
,
if
,
if
, and
if
.
See Also
2001 Pan African MO (Problems) | ||
Preceded by Problem 2 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 4 |
All Pan African MO Problems and Solutions |