Difference between revisions of "2009 AIME II Problems/Problem 10"
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== Solution == | == Solution == | ||
− | Let <math>O</math> be the intersection of <math>BC</math> and <math>AD</math>. By the [[Angle Bisector Theorem]], <math>5 | + | Let <math>O</math> be the intersection of <math>BC</math> and <math>AD</math>. By the [[Angle Bisector Theorem]], <math>\frac {5}{BO}</math> = <math>\frac {13}{CO}, so </math>BO<math> = </math>5x<math> and </math>CO<math> = </math>13x<math>, and </math>BO<math> + </math>OC<math> = </math>BC<math> = </math>12<math>, so </math>x<math> = </math>\frac {2}{3}<math>, and </math>OC<math> = </math>\frac {26}{3}<math>. Let </math>P<math> be the altitude from </math>D<math> to </math>OC<math>. It can be seen that triangle </math>DOP<math> is similar to triangle </math>AOB<math>, and triangle </math>DPC<math> is similar to triangle </math>ABC<math>. If </math>DP<math> = </math>15y<math>, then </math>CP<math> = </math>36y<math>, </math>OP<math> = </math>10y<math>, and </math>OD<math> = </math>5y\sqrt {13}<math>. Since </math>OP<math> + </math>CP<math> = </math>46y<math> = </math>\frac {26}{3}<math>, </math>y<math> = </math>\frac {13}{69}<math>, and </math>AD<math> = </math>\frac {60\sqrt{13}}{23}<math>. The answer is </math>60<math> + </math>13<math> + </math>23<math> = </math>\boxed{096}$. |
== See Also == | == See Also == | ||
{{AIME box|year=2009|n=II|num-b=9|num-a=11}} | {{AIME box|year=2009|n=II|num-b=9|num-a=11}} |
Revision as of 20:52, 17 April 2009
Four lighthouses are located at points ,
,
, and
. The lighthouse at
is
kilometers from the lighthouse at
, the lighthouse at
is
kilometers from the lighthouse at
, and the lighthouse at
is
kilometers from the lighthouse at
. To an observer at
, the angle determined by the lights at
and
and the angle determined by the lights at
and
are equal. To an observer at
, the angle determined by the lights at
and
and the angle determined by the lights at
and
are equal. The number of kilometers from
to
is given by
, where
,
, and
are relatively prime positive integers, and
is not divisible by the square of any prime. Find
+
+
.
Solution
Let be the intersection of
and
. By the Angle Bisector Theorem,
=
BO
5x
CO
13x
BO
OC
BC
12
x
\frac {2}{3}
OC
\frac {26}{3}
P
D
OC
DOP
AOB
DPC
ABC
DP
15y
CP
36y
OP
10y
OD
5y\sqrt {13}
OP
CP
46y
\frac {26}{3}
y
\frac {13}{69}
AD
\frac {60\sqrt{13}}{23}
60
13
23
\boxed{096}$.
See Also
2009 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |