Difference between revisions of "2014 AIME I Problems/Problem 15"
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+ | ==Solution 4== | ||
+ | See inside the <math>\triangle DEF</math>, we can find that <math>AG>AF</math> since if <math>AG<AF</math>, we can see that Ptolemy Theorem inside cyclic quadrilateral <math>EFGD</math> doesn't work. Now let's see when <math>AG>AF</math>, since <math>\frac{DG}{EG} = \frac{3}{4}</math>, we can assume that <math>EG=4x;GD=3x and ED=5x</math>, since we know <math>EF=FD</math> so <math>\triangle EFD </math> is isosceles right triangle. We can denote <math>DF=EF=\frac{5x\sqrt{2}}{2}</math>.Applying Ptolemy Theorem inside the cyclic quadrilateral <math>EFGD</math> we can get the length of <math>FG</math> can be represented as <math>\frac{x\sqrt{2}}{2}</math>. After observing, we can see <math>\angle AFE=\angle EDG</math>, whereas <math>\angle A=\angle EDG</math> so we can see <math>\triangel AEF</math> is isosceles triangle. It is a <math>3-4-5</math> triangle so we can directly know that the length of AF can be written in the form of <math>3x\sqrt{2}</math>. Denoting a point <math>J</math> on side <math>AC</math> with that <math>DJ</math> is perpendicular to side <math>AC</math>. Now with the same reason, we can see that <math>\triangle DJG</math> is a isosceles right triangle, so we can get <math>GJ=\frac{3x\sqrt{2}}{2}</math> while the segment <math>CJ</math> is <math>2x\sqrt{2}</math> since its 3-4-5 again. Now adding all those segments together we can find that <math>AC=5=7x\sqrt{2}</math> and <math>x=\frac{5\sqrt{2}}{14}</math> and the desired <math>ED=5x=\frac{25\sqrt{2}}{14}</math> | ||
+ | which our answer is <math>\boxed{041}</math> ~bluesoul | ||
== 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 22:11, 24 September 2021
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 1
Since ,
is the diameter of
. Then
. But
, so
is a 45-45-90 triangle. Letting
, we have that
,
, and
.
Note that by SAS similarity, so
and
. Since
is a cyclic quadrilateral,
and
, implying that
and
are isosceles. As a result,
, so
and
.
Finally, using the Pythagorean Theorem on ,
Solving for
, we get that
, so
. Thus, the answer is
.
Solution 2
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
:
Thus
.
Solution 3
Call and as a result
. Since
is cyclic we just need to get
and using LoS(for more detail see the
nd paragraph of Solution
) we get
and using a similar argument(use LoS again) and subtracting you get
so you can use Ptolemy to get
.
~First
Solution 4
See inside the , we can find that
since if
, we can see that Ptolemy Theorem inside cyclic quadrilateral
doesn't work. Now let's see when
, since
, we can assume that
, since we know
so
is isosceles right triangle. We can denote
.Applying Ptolemy Theorem inside the cyclic quadrilateral
we can get the length of
can be represented as
. After observing, we can see
, whereas
so we can see $\triangel AEF$ (Error compiling LaTeX. Unknown error_msg) is isosceles triangle. It is a
triangle so we can directly know that the length of AF can be written in the form of
. Denoting a point
on side
with that
is perpendicular to side
. Now with the same reason, we can see that
is a isosceles right triangle, so we can get
while the segment
is
since its 3-4-5 again. Now adding all those segments together we can find that
and
and the desired
which our answer is
~bluesoul
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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