2006 iTest Problems/Problem U5
The following problem is from the Ultimate Question of the 2006 iTest, where solving this problem required the answer of a previous problem. When the problem is rewritten, the T-value is substituted.
Problem
In triangle , points
,
, and
are the feet of the angle bisectors of
,
,
respectively. Let point
be the intersection of segments
and
, and let
denote the perimeter of
. If
,
, and
, then the value of
can be expressed uniquely as
where
and
are positive integers such that
is not divisible by the square of any prime. Find
.
Solution
According to the Steiner-Lehmus Theorem, . Thus,
by SAS Congruency, so
and
.
Let and
. Since
and
share an altitude and
and
share an altitude, we know that
and
. Also, since
,
. Similarly,
and
.
Since and
share an altitude, we have
Now let
and
. By the Angle Bisector Theorem,
, so
. By using the Pythagorean Theorem on
, we have
, so
. By using the Pythagorean Theorem on
,
Therefore,
the
. The perimeter of the triangle equals
, and
. Thus,
, and
.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
2006 iTest (Problems, Answer Key) | ||
Preceded by: Problem U4 |
Followed by: Problem U6 | |
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • U1 • U2 • U3 • U4 • U5 • U6 • U7 • U8 • U9 • U10 |