Difference between revisions of "2009 AIME I Problems/Problem 12"
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(More elegant solution that shows the answer to be independent of the lengths of the sides of triangle) |
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The quadratic formula now yields <math>x=37/3</math>. Plugging this back in, the perimeter of <math>ABI</math> is <math>2s=2(37+x)=2(37+37/3) = 37(8/3)</math> so the ratio of the perimeter to <math>AB</math> is <math>8/3</math> and our answer is <math>8+3=\boxed{011}</math> | The quadratic formula now yields <math>x=37/3</math>. Plugging this back in, the perimeter of <math>ABI</math> is <math>2s=2(37+x)=2(37+37/3) = 37(8/3)</math> so the ratio of the perimeter to <math>AB</math> is <math>8/3</math> and our answer is <math>8+3=\boxed{011}</math> | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | As in Solution <math>2</math>, let <math>P</math> and <math>Q</math> be the intersections of <math>\omega</math> with <math>BI</math> and <math>AI</math> respectively. | ||
+ | |||
+ | Recall that the distance from a point outside a circle to that circle is the same along both tangent lines to the circle drawn from the point. | ||
+ | |||
+ | Recall also that the length of the altitude to the hypotenuse of a right-angle triangle is the geometric mean of the two segments into which it cuts the hypotenuse. | ||
+ | |||
+ | Let <math>x = \overline{AD} = \overline{AQ}</math>. Let <math>y = \overline{BD} = \overline{BQ}</math>. Let <math>z = \overline{PI} = \overline{QI}</math>. The semi-perimeter of <math>ABI</math> is <math>x + y + z</math>. | ||
+ | Since the lengths of the sides of <math>ABI</math> are <math>x + y</math>, <math>y + z</math> and <math>x + z</math>, the square of its area by Heron's formula is <math>(x+y+z)xyz</math>. | ||
+ | |||
+ | The radius <math>r</math> of <math>\omega</math> is <math>\overline{CD}/2</math>. Therefore <math>r^2 = xy/4</math>. As <math>\omega</math> is the in-circle of <math>ABI</math>, the area of <math>ABI</math> is also <math>r(x+y+z)</math>, and so the square area is <math>r^2(x+y+z)^2</math>. | ||
+ | |||
+ | Therefore <math>(x+y+z)xyz = r^2(x+y+z)^2 = xy(x+y+z)^2/4</math>. Dividing both sides by <math>xy(x+y+z)/4</math> we get: <math>4z = (x+y+z)</math>, and so <math>z = (x+y)/3</math>. The semi-perimeter of <math>ABI</math> is therefore <math>\frac{4}{3}(x+y)</math> and the whole perimeter is <math>\frac{8}{3}(x+y)</math>. Now <math>x + y = \overline{AB}</math>, so the ratio of the perimeter of <math>ABI</math> to the hypotenuse <math>\overline{AB}</math> is <math>8/3</math> and our answer is <math>8+3=\boxed{011}</math> | ||
== See also == | == See also == | ||
{{AIME box|year=2009|n=I|num-b=11|num-a=13}} | {{AIME box|year=2009|n=I|num-b=11|num-a=13}} |
Revision as of 17:17, 27 February 2010
Contents
[hide]Problem
In right with hypotenuse
,
,
, and
is the altitude to
. Let
be the circle having
as a diameter. Let
be a point outside
such that
and
are both tangent to circle
. The ratio of the perimeter of
to the length
can be expressed in the form
, where
and
are relatively prime positive integers. Find
.
Solution 1
Let be center of the circle and
,
be the two points of tangent such that
is on
and
is on
. We know that
.
Since the ratios between corresponding lengths of two similar diagrams are equal, we can let and
. Hence
and the radius
.
Since we have and
, we have
.
Hence . let
, then we have Area
=
=
. Then we get
.
Now the equation looks very complex but we can take a guess here. Assume that is a rational number
(If it's not then the answer to the problem would be irrational which can't be in the form of
)
that can be expressed as
such that
. Look at both sides; we can know that
has to be a multiple of
and not of
and it's reasonable to think that
is divisible by
so that we can cancel out the
on the right side of the equation.
Let's see if fits. Since
, and
. Amazingly it fits!
Since we know that , the other solution of this equation is negative which can be ignored. Hence
.
Hence the perimeter is , and
is
. Hence
,
.
Solution 2
As in Solution , let
and
be the intersections of
with
and
respectively.
First, by pythagorean theorem, . Now the area of
is
, so
and the inradius of
is
.
Now from we find that
so
and similarly,
.
Note ,
, and
. So we have
,
. Now we can compute the area of
in two ways: by heron's formula and by inradius times semiperimeter, which yields
The quadratic formula now yields . Plugging this back in, the perimeter of
is
so the ratio of the perimeter to
is
and our answer is
Solution 3
As in Solution , let
and
be the intersections of
with
and
respectively.
Recall that the distance from a point outside a circle to that circle is the same along both tangent lines to the circle drawn from the point.
Recall also that the length of the altitude to the hypotenuse of a right-angle triangle is the geometric mean of the two segments into which it cuts the hypotenuse.
Let . Let
. Let
. The semi-perimeter of
is
.
Since the lengths of the sides of
are
,
and
, the square of its area by Heron's formula is
.
The radius of
is
. Therefore
. As
is the in-circle of
, the area of
is also
, and so the square area is
.
Therefore . Dividing both sides by
we get:
, and so
. The semi-perimeter of
is therefore
and the whole perimeter is
. Now
, so the ratio of the perimeter of
to the hypotenuse
is
and our answer is
See also
2009 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |