Difference between revisions of "2015 AIME I Problems/Problem 11"
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Thus the perimeter of <math>\triangle ABC</math> must be <math>2(x+y) = \boxed{108}</math>. | Thus the perimeter of <math>\triangle ABC</math> must be <math>2(x+y) = \boxed{108}</math>. | ||
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+ | |||
+ | ==Solution 3 (More Trig)== -ccx09 (Someone please help clean this up. Thanks) | ||
+ | |||
+ | Let x be the measure of <ABI. Thus, <ABI=<CBI=<ACI=<BCI=x, and x<45 degrees. Note that the angle bisector of <BAC will pass through I and since AB=AC, will also be the perpendicular bisector of BC. Call the point of intersection of segments AI and BC point M. | ||
+ | Let BM=CM=a, and AB=AC=b. Thus, we want to minimize p=2(a+b), or minimize a+b. | ||
+ | |||
+ | We now consider triangle BAM. We note that cos(x)=BM/BI=a/8, and cos(2x)=a/b. | ||
+ | We can now simplify as follows. cos(2x)=2(cos x)(cos x)-1, and since cos(x)=a/8, we have | ||
+ | a/b = 2(a/8)^2-1 = a^2/32-1 = (a^2-32)/32. Therefore b=32a/(a^2-32). | ||
+ | And thus a + b = a + 32a / (a^2-32). Substituting 8cos(x) for a, we obtain | ||
+ | a + b = 8cos(x) + 32(8cos x) / ((8cos x)^2)-32). | ||
+ | |||
+ | Since a and b are integers, we must have cos(x) = 0, 1/8, 2/8, 3/8, 4/8, 5/8. 6/8, 7/8, or 1. | ||
+ | However, since x<45 degrees, we must have cos(x)>=6/8. To minimize a + b, we thus let cos(x)=6/8. Plugging cos(x)=6/8 into the previous expression of a + b yields a + b = 54. | ||
+ | This is the minimum possible value of a + b. | ||
+ | |||
+ | Thus, the minimum perimeter of triangle ABC is 54*2 = 108. | ||
==See Also== | ==See Also== |
Revision as of 21:29, 18 August 2017
Problem
Triangle has positive integer side lengths with
. Let
be the intersection of the bisectors of
and
. Suppose
. Find the smallest possible perimeter of
.
Solution 1 (No Trig)
Let and the foot of the altitude from
to
be point
and
. Since ABC is isosceles,
is on
. By Pythagorean Theorem,
. Let
and
. By Angle Bisector theorem,
. Also,
. Solving for
, we get
. Then, using Pythagorean Theorem on
we have
. Simplifying, we have
. Factoring out the
, we have
. Adding 1 to the fraction and simplifying, we have
. Crossing out the
, and solving for
yields
. Then, we continue as Solution 2 does.
Solution 2 (Trig)
Let be the midpoint of
. Then by SAS Congruence,
, so
.
Now let ,
, and
.
Then
and .
Cross-multiplying yields .
Since ,
must be positive, so
.
Additionally, since has hypotenuse
of length
,
.
Therefore, given that is an integer, the only possible values for
are
,
,
, and
.
However, only one of these values, , yields an integral value for
, so we conclude that
and
.
Thus the perimeter of must be
.
==Solution 3 (More Trig)== -ccx09 (Someone please help clean this up. Thanks)
Let x be the measure of <ABI. Thus, <ABI=<CBI=<ACI=<BCI=x, and x<45 degrees. Note that the angle bisector of <BAC will pass through I and since AB=AC, will also be the perpendicular bisector of BC. Call the point of intersection of segments AI and BC point M. Let BM=CM=a, and AB=AC=b. Thus, we want to minimize p=2(a+b), or minimize a+b.
We now consider triangle BAM. We note that cos(x)=BM/BI=a/8, and cos(2x)=a/b. We can now simplify as follows. cos(2x)=2(cos x)(cos x)-1, and since cos(x)=a/8, we have a/b = 2(a/8)^2-1 = a^2/32-1 = (a^2-32)/32. Therefore b=32a/(a^2-32). And thus a + b = a + 32a / (a^2-32). Substituting 8cos(x) for a, we obtain a + b = 8cos(x) + 32(8cos x) / ((8cos x)^2)-32).
Since a and b are integers, we must have cos(x) = 0, 1/8, 2/8, 3/8, 4/8, 5/8. 6/8, 7/8, or 1. However, since x<45 degrees, we must have cos(x)>=6/8. To minimize a + b, we thus let cos(x)=6/8. Plugging cos(x)=6/8 into the previous expression of a + b yields a + b = 54. This is the minimum possible value of a + b.
Thus, the minimum perimeter of triangle ABC is 54*2 = 108.
See Also
2015 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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