Difference between revisions of "2014 AIME I Problems/Problem 15"
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First we note that <math>\triangle DEF</math> is an isosceles right triangle with hypotenuse <math>DE</math> the same as the diameter of <math>\omega</math>. We also note that <math>\triangle DGE \sim \triangle ABC</math> since <math>\angle EGD</math> is a right angle and the ratios of the sides are <math>3:4:5</math>. | First we note that <math>\triangle DEF</math> is an isosceles right triangle with hypotenuse <math>DE</math> the same as the diameter of <math>\omega</math>. We also note that <math>\triangle DGE \sim \triangle ABC</math> since <math>\angle EGD</math> is a right angle and the ratios of the sides are <math>3:4:5</math>. | ||
− | From congruent arc intersections, we know that <math>\angle GED \cong \angle GBC</math>, and that from similar triangles <math>\angle GED</math> is also congruent to <math>\angle GCB</math>. Thus, triangle BGC is an isosceles triangle with BG = GC, so G is the midpoint of AC and AG = GC = 5/2. Similarly, we can find from angle chasing that BF is the angle bisector of B | + | From congruent arc intersections, we know that <math>\angle GED \cong \angle GBC</math>, and that from similar triangles <math>\angle GED</math> is also congruent to <math>\angle GCB</math>. Thus, <math>\triangle BGC</math> is an isosceles triangle with <math>BG = GC</math>, so G is the midpoint of <math>AC</math> and <math>AG = GC = 5/2</math>. Similarly, we can find from angle chasing that <math>BF</math> is the angle bisector of <math>\angle B</math>. From the angle bisector theorem, we have <math>\frac{AF}{AB} = \frac{CF}{CB}, so </math>AF = 15/7<math> and </math>CF = 20/7<math>. |
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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}$ | ||
== 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 02:02, 18 March 2014
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 G is the midpoint of
and
. Similarly, we can find from angle chasing that
is the angle bisector of
. From the angle bisector theorem, we have
AF = 15/7
CF = 20/7$.
Lastly, we apply power of a point from points$ (Error compiling LaTeX. Unknown error_msg)AC
\omega
AE \times AB=AF \times AG
CD \times CB=CG \times CF
EB = \frac{17}{14}
DB = \frac{31}{14}
DE = \frac{25 \sqrt{2}}{14}
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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