Difference between revisions of "2014 AIME I Problems/Problem 15"
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Lastly, we apply power of a point from points <math>A</math> and <math>C</math> with respect to <math>\omega</math> and have <math>AE \times AB=AF \times AG</math> and <math>CD \times CB=CG \times CF</math>, so we can compute that <math>EB = \frac{17}{14}</math> and <math>DB = \frac{31}{14}</math>. From the Pythagorean Theorem, we result in <math>DE = \frac{25 \sqrt{2}}{14}</math>, so <math>a+b+c=25+2+14= \boxed{041}</math> | Lastly, we apply power of a point from points <math>A</math> and <math>C</math> with respect to <math>\omega</math> and have <math>AE \times AB=AF \times AG</math> and <math>CD \times CB=CG \times CF</math>, so we can compute that <math>EB = \frac{17}{14}</math> and <math>DB = \frac{31}{14}</math>. From the Pythagorean Theorem, we result in <math>DE = \frac{25 \sqrt{2}}{14}</math>, so <math>a+b+c=25+2+14= \boxed{041}</math> | ||
+ | |||
+ | Also: <math>FG=\frac{20}{7}-\frac{5}{2}=\frac{5}{2}-\frac{15}{7}=\frac{5}{14}</math>. We can also use Ptolemy's Theorem on quadrilateral <math>DEFG</math> to figure what <math>FG</math> is in terms of <math>\omega</math>: | ||
+ | <cmath>DG\cot EF+EF\cdot DG=DF\cdot EG</cmath> | ||
+ | <cmath>\frac{3\omega}{5}\cdot \frac{\omega\sqrt{2}{2}+\omega\cdot FG=\frac{4\omega}{5}\cdot \frac{\omega\sqrt{2}{2}</cmath> | ||
+ | <cmath>\frac{3\omega\sqrt{2}}{10}+EF\frac{4\omega\sqrt{2}}{10}\implies FG=\frac{\omega\sqrt{2}}{10}</cmath> | ||
+ | Thus <math>\frac{\omega\sqrt{2}}{10}=\frac{5}{14}\rightarrow \omega=5\sqrt{2}\cdot\frac{5}{14}=\frac{25\sqrt{2}}{14}. </math>a+b+c=25+2+14= \boxed{041}$ | ||
== See also == | == See also == | ||
{{AIME box|year=2014|n=I|num-b=14|after=Last Question}} | {{AIME box|year=2014|n=I|num-b=14|after=Last Question}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 14:58, 14 February 2016
Problem 15
In ,
,
, and
. Circle
intersects
at
and
,
at
and
, and
at
and
. Given that
and
, length
, where
and
are relatively prime positive integers, and
is a positive integer not divisible by the square of any prime. Find
.
Solution
First we note that is an isosceles right triangle with hypotenuse
the same as the diameter of
. We also note that
since
is a right angle and the ratios of the sides are
.
From congruent arc intersections, we know that , and that from similar triangles
is also congruent to
. Thus,
is an isosceles triangle with
, so
is the midpoint of
and
. Similarly, we can find from angle chasing that
. Therefore,
is the angle bisector of
. From the angle bisector theorem, we have
, so
and
.
Lastly, we apply power of a point from points and
with respect to
and have
and
, so we can compute that
and
. From the Pythagorean Theorem, we result in
, so
Also: . We can also use Ptolemy's Theorem on quadrilateral
to figure what
is in terms of
:
\[\frac{3\omega}{5}\cdot \frac{\omega\sqrt{2}{2}+\omega\cdot FG=\frac{4\omega}{5}\cdot \frac{\omega\sqrt{2}{2}\] (Error compiling LaTeX. Unknown error_msg)
Thus
a+b+c=25+2+14= \boxed{041}$
See also
2014 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Question | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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