Difference between revisions of "2009 AIME I Problems/Problem 12"
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== Solution 4 == | == Solution 4 == | ||
+ | <asy> | ||
+ | size(300); | ||
+ | defaultpen(linewidth(0.4)+fontsize(10)); | ||
+ | pen s = linewidth(0.8)+fontsize(8); | ||
+ | |||
+ | pair A,B,C,D,O,X; | ||
+ | C=origin; | ||
+ | A=(0,12); | ||
+ | B=(18,0); | ||
+ | D=foot(C,A,B); | ||
+ | O = (C+D)/2; | ||
+ | real r = length(D-C)/2; | ||
+ | path c = CR(O, r); | ||
+ | pair OA = (O+A)/2; | ||
+ | real rA = length(A-O)/2; | ||
+ | pair Ap = OP(CR(OA,rA), c); | ||
+ | pair OB = (O+B)/2; | ||
+ | real rB = length(B-O)/2; | ||
+ | pair Bp = OP(CR(OB,rB), c); | ||
+ | X=extension(A,Ap,B,Bp); | ||
+ | draw(A--B--C--A, s); | ||
+ | draw(C--D^^B--O--A^^Ap--O--X, gray+0.25); | ||
+ | draw(c^^A--X--B); | ||
+ | |||
+ | dot("$A$", A, N); | ||
+ | dot("$B$", B, SE); | ||
+ | dot("$C$", C, SW); | ||
+ | dot("$D$", D, 0.2*(D-C)); | ||
+ | dot("$I$", X, 0.5*(X-C)); | ||
+ | dot("$P$", Ap, 0.3*(Ap-O)); | ||
+ | dot("$Q$", Bp, 0.3*(Bp-O)); | ||
+ | dot("$O$", O, W); | ||
+ | label("$\beta$",B,10*dir(157)); | ||
+ | label("$\alpha$",A,5*dir(-55)); | ||
+ | label("$\theta$",X,5*dir(55)); | ||
+ | </asy> | ||
We shall yet again let <math>P</math> and <math>Q</math> be the intersections of <math>AI</math> and <math>BI</math> to <math>\omega</math>, respectively. We want to find the perimeter of <math>ABI</math>, which is <math>AD+BD+BQ+QI+IP+PA</math>. We can easily find <math>AD</math> and <math>BD</math> using the fact that <math>ABC</math>, <math>ACD</math>, and <math>BCD</math> are all similar triangles. We get <math>AD=\frac{144}{37}</math> and <math>\frac{1225}{37}</math>. Since <math>AP</math> and <math>AD</math> are tangents to <math>\omega</math>, <math>AP=AD=\frac{144}{37}</math>, and similarly <math>BQ=BD=\frac{1225}{37}</math>. We now wish to find <math>IP</math> and <math>IQ</math>. | We shall yet again let <math>P</math> and <math>Q</math> be the intersections of <math>AI</math> and <math>BI</math> to <math>\omega</math>, respectively. We want to find the perimeter of <math>ABI</math>, which is <math>AD+BD+BQ+QI+IP+PA</math>. We can easily find <math>AD</math> and <math>BD</math> using the fact that <math>ABC</math>, <math>ACD</math>, and <math>BCD</math> are all similar triangles. We get <math>AD=\frac{144}{37}</math> and <math>\frac{1225}{37}</math>. Since <math>AP</math> and <math>AD</math> are tangents to <math>\omega</math>, <math>AP=AD=\frac{144}{37}</math>, and similarly <math>BQ=BD=\frac{1225}{37}</math>. We now wish to find <math>IP</math> and <math>IQ</math>. | ||
Revision as of 12:27, 26 July 2021
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
Solution 4
We shall yet again let
and
be the intersections of
and
to
, respectively. We want to find the perimeter of
, which is
. We can easily find
and
using the fact that
,
, and
are all similar triangles. We get
and
. Since
and
are tangents to
,
, and similarly
. We now wish to find
and
.
Let the center of the given circle be . We know that
,
, and
. Since all six angles sum to
,
. If we knew the radius of circle
now, then we could find
and
, and then we can use the sum (or difference) of tangents formula to find
, which reveals
. This means we should find the radius of
. We can easily see that the height of triangle
from
has length
, so the radius of
is
. Now we shall proceed with the above plan.
.
.
.
Therefore , and the perimeter of
is
. Since
, the desired ratio is
, and
.
Solution 5
This solution is not a real solution and is solving the problem with a ruler and compass.
Draw . Then, drawing the tangents and intersecting them, we get that
is around
and
is around
. We then find the ratio to be around
. Using long division, we find that this ratio is approximately 2.666, which you should recognize as
. Since this seems reasonable, we find that the answer is
~ilp
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.