Difference between revisions of "2009 AIME I Problems/Problem 12"
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Since the ratios between corresponding lengths of two similar diagrams are equal, we can let <math>AD = 144, CD = 420</math> and <math>BD = 1225</math>. Hence <math>BP = 144, AQ = 1225, AB = 1369</math> and the radius <math>r = OD = 210</math>. | Since the ratios between corresponding lengths of two similar diagrams are equal, we can let <math>AD = 144, CD = 420</math> and <math>BD = 1225</math>. Hence <math>BP = 144, AQ = 1225, AB = 1369</math> and the radius <math>r = OD = 210</math>. | ||
− | Since we have <math>\tan OAB = \frac {35}{24}</math> and <math>\tan OBA = \frac{6}{35}</math> , we have <math>\sin {(OAB + OBA)} = \frac {1369}{\sqrt {(1801*1261)}}, \cos {(OAB + OBA)} = \frac {630}{\sqrt {(1801*1261)}}</math>. Hence <math>\sin I = \sin {(2OAB + 2OBA)} = \frac {2*1369*630}{1801*1261}</math>. let <math>IP = IQ = x</math> , then we have Area<math>(IBC)</math> = <math>(2x + 1225*2 + 144*2)*\frac {210}{2}</math> = <math>(x + 144)(x + 1225)* \sin {\frac {I}{2}}</math>. Then we get <math>x + 1369 = \frac {3*1369*(x + 144)(x + 1225)}{1801*1261}</math>. | + | Since we have <math>\tan OAB = \frac {35}{24}</math> and <math>\tan OBA = \frac{6}{35}</math> , we have <math>\sin {(OAB + OBA)} = \frac {1369}{\sqrt {(1801*1261)}},</math><math>\cos {(OAB + OBA)} = \frac {630}{\sqrt {(1801*1261)}}</math>. |
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+ | Hence <math>\sin I = \sin {(2OAB + 2OBA)} = \frac {2*1369*630}{1801*1261}</math>. let <math>IP = IQ = x</math> , then we have Area<math>(IBC)</math> = <math>(2x + 1225*2 + 144*2)*\frac {210}{2}</math> = <math>(x + 144)(x + 1225)* \sin {\frac {I}{2}}</math>. Then we get <math>x + 1369 = \frac {3*1369*(x + 144)(x + 1225)}{1801*1261}</math>. | ||
Now the equation looks very complex but we can take a guess here. Assume that <math>x</math> is a rational number | Now the equation looks very complex but we can take a guess here. Assume that <math>x</math> 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 m/ | + | (If it's not then the answer to the problem would be irrational which can't be in the form of <math>\frac {m}{n}</math>) |
− | that can be expressed as a/ | + | that can be expressed as <math>\frac {a}{b}</math> such that <math>(a,b) = 1</math>. Look at both sides; we can know that <math>a</math> has to be a multiple of <math>1369</math> and not of <math>3</math> and it's reasonable to think that <math>b</math> is divisible by <math>3</math> so that we can cancel out the <math>3</math> on the right side of the equation. |
− | Let's | + | Let's see if <math>x = \frac {1369}{3}</math> fits. Since <math>\frac {1369}{3} + 1369 = \frac {4*1369}{3}</math>, and <math>\frac {3*1369*(x + 144)(x + 1225)}{1801*1261} = \frac {3*1369* \frac {1801}{3} * \frac {1261*4}{3}} {1801*1261} = \frac {4*1369}{3}</math>. Amazingly it fits! |
− | Since we know that 3*1369*144*1225 - 1369*1801*1261 < 0, the other solution of this equation is negative which can be ignored. Hence x = 1369/3. | + | Since we know that <math>3*1369*144*1225 - 1369*1801*1261 < 0</math>, the other solution of this equation is negative which can be ignored. Hence <math>x = 1369/3</math>. |
− | Hence the perimeter is 1225*2 + 144*2 + | + | Hence the perimeter is <math>1225*2 + 144*2 + \frac {1369}{3} *2 = 1369* \frac {8}{3}</math>, and <math>BC</math> is <math>1369</math>. Hence <math>\frac {m}{n} = \frac {8}{3}</math>, <math>m + n = 11</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 19:16, 22 March 2009
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
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
,
.
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 |