1995 AIME Problems/Problem 9
Problem
Triangle is isosceles, with
and altitude
Suppose that there is a point
on
with
and
Then the perimeter of
may be written in the form
where
and
are integers. Find
Solution 1
Let , so
. Then,
. Expanding
using the angle sum identity gives
Thus,
. Solving, we get
. Hence,
and
by the Pythagorean Theorem. The total perimeter is
. The answer is thus
.
Solution 2
In a similar fashion, we encode the angles as complex numbers, so if , then
and
. So we need only find
such that
. This will happen when
, which simplifies to
. Therefore,
. By the Pythagorean Theorem,
, so the perimeter is
, giving us our answer,
.
Solution 3
Let , so
,
, and thus
We can then draw the angle bisector of
, and let it intersect
at
Since
,
Let
. Then we see by the Pythagorean Theorem,
,
,
, and
By the angle bisector theorem,
Substituting in what we know for the lengths of those segments, we see that
multiplying by both denominators and squaring both sides yields
which simplifies to
Substituting this in for x in the equations for
and
yields
and
Thus the perimeter is
, and the answer is
.
Solution 4
The triangle is symmetrical so we can split it in half ( and
).
Let and
. By the Law of Sines on triangle
,
. Using
we can get
. We can use this information to relate
to
by using the Law of Sines on triangle
.
(as
is a right angle), so
. Using the identity
, we can turn the equation into::
Now that we've found , we can look at the side lengths of
and
(since they are symmetrical, the perimeter of
is
.
We note that and
.
(Note it is positive since ).
The answer is .
See also
1995 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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