Difference between revisions of "2014 AIME I Problems/Problem 15"
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Finally, using the Pythagorean Theorem on <math>\triangle BDE</math>, | Finally, using the Pythagorean Theorem on <math>\triangle BDE</math>, | ||
<cmath> \left(3-\frac{5x}{\sqrt{2}}\right)^2 + \left(4-\frac{5x}{\sqrt{2}}\right)^2 = (5x)^2</cmath> | <cmath> \left(3-\frac{5x}{\sqrt{2}}\right)^2 + \left(4-\frac{5x}{\sqrt{2}}\right)^2 = (5x)^2</cmath> | ||
− | Solving for <math>x</math>, we get that <math>x=\frac{5\sqrt{2}}{14}</math>, so <math>DE=5x=\frac{25\sqrt{2}}{14}</math>. Thus, the answer is <math>25+2+14=\boxed{041}</math>. | + | Solving for <math>x</math>, we get that <math>x=\frac{5\sqrt{2}}{14}</math>, so <math>DE= 5x =\frac{25 \sqrt{2}}{14}</math>. Thus, the answer is <math>25+2+14=\boxed{041}</math>. |
== Solution 2 == | == Solution 2 == | ||
Line 65: | Line 65: | ||
==Solution 5== | ==Solution 5== | ||
[[File:2014 AIME II 15.png|450px|right]] | [[File:2014 AIME II 15.png|450px|right]] | ||
+ | The main element of the solution is the proof that <math>BF</math> is bisector of <math>\angle B.</math> | ||
+ | |||
Let <math>O</math> be the midpoint of <math>DE.</math> <math>\angle EBF = 90^\circ \implies</math> | Let <math>O</math> be the midpoint of <math>DE.</math> <math>\angle EBF = 90^\circ \implies</math> | ||
+ | |||
<math>O</math> is the center of the circle <math>BDGFE.</math> | <math>O</math> is the center of the circle <math>BDGFE.</math> | ||
<math>\angle EOF = 90^\circ \implies \overset{\Large\frown} {EF} = 90^\circ \implies \angle EBF = 45^\circ \implies</math> | <math>\angle EOF = 90^\circ \implies \overset{\Large\frown} {EF} = 90^\circ \implies \angle EBF = 45^\circ \implies</math> | ||
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==Solution 6 == | ==Solution 6 == | ||
[[File:2014 AIME II 15a.png|450px|right]] | [[File:2014 AIME II 15a.png|450px|right]] | ||
+ | The main element of the solution is the proof that <math>G</math> is midpoint of <math>AC.</math> | ||
+ | |||
As in Solution 5 we get <math>\angle GED = \angle DBG =\gamma \implies</math> | As in Solution 5 we get <math>\angle GED = \angle DBG =\gamma \implies</math> | ||
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Let <math>\hspace{10mm} BO = EO = DO = r \implies</math> | Let <math>\hspace{10mm} BO = EO = DO = r \implies</math> | ||
<cmath>BG = 2 r \cos(45^\circ – \gamma) = 2 r (\sin \gamma + \cos \gamma)\frac {\sqrt {2}}{2} =</cmath> | <cmath>BG = 2 r \cos(45^\circ – \gamma) = 2 r (\sin \gamma + \cos \gamma)\frac {\sqrt {2}}{2} =</cmath> | ||
− | <cmath>r (\frac {3}{5} + \frac {4}{5}) \sqrt {2} = r \frac {7 \sqrt{2}}{5} = \frac {5}{2}\implies</cmath> | + | <cmath>r \biggl(\frac {3}{5} + \frac {4}{5}\biggr) \sqrt {2} = r \frac {7 \sqrt{2}}{5} = \frac {5}{2}\implies</cmath> |
<cmath>r = \frac {25 \cdot \sqrt{2}}{28}\implies ED = \frac {25 \cdot \sqrt{2}}{14}\implies \boxed{\textbf{041}}.</cmath> | <cmath>r = \frac {25 \cdot \sqrt{2}}{28}\implies ED = \frac {25 \cdot \sqrt{2}}{14}\implies \boxed{\textbf{041}}.</cmath> | ||
'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Solution 7== | ||
+ | Let <math>(BEFGD) = \omega</math>. | ||
+ | By Incenter-Excenter(Fact <math>5</math>), <math>F</math> is the angle bisector of <math>\angle B</math>. | ||
+ | Then by Ratio Lemma we have | ||
+ | <cmath>\frac{AG}{CG} = \frac{\sin(ABG)}{\sin(CBG)} \cdot \frac{AB}{BC} = \frac{\sin(GDE)}{\sin(DEG)} \cdot \frac{3}{4} = 1</cmath> | ||
+ | Thus, <math>G</math> is the midpoint of <math>AC</math>. | ||
+ | |||
+ | We can calculate <math>AF</math> and <math>CF</math> to be <math>\frac{15}{7}</math> and <math>\frac{20}{7}</math> respectively. | ||
+ | And then by Power of a Point, we have | ||
+ | <math>\newline</math> | ||
+ | <cmath>\operatorname{Pow}_{\omega}(A) = AE \cdot AB = AF \cdot AG \implies AE = \frac{25}{14}</cmath> | ||
+ | And then similarly, we have <math>CD = AE = \frac{25}{14}</math>. | ||
+ | <math>\newline</math> | ||
+ | |||
+ | Then <math>EB = \frac{17}{14}</math> and <math>DB = \frac{31}{14}</math> and by Pythagorean we have <math>DE = \frac{25\sqrt{2}}{14}</math>, so our answer is <math>\boxed{\textbf{041}}.</math> | ||
+ | |||
+ | ~dolphinday | ||
+ | |||
+ | ==Video Solution by mop 2024== | ||
+ | https://youtu.be/GxxZYZrQl2A | ||
+ | |||
+ | ~r00tsOfUnity | ||
== 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}} |
Latest revision as of 22:30, 28 January 2024
Contents
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
is isosceles triangle. Since
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
Solution 5
The main element of the solution is the proof that is bisector of
Let be the midpoint of
is the center of the circle
BF is bisector of
Let
vladimir.shelomovskii@gmail.com, vvsss
Solution 6
The main element of the solution is the proof that is midpoint of
As in Solution 5 we get
is isosceles triangle with
Similarly
Let
vladimir.shelomovskii@gmail.com, vvsss
Solution 7
Let .
By Incenter-Excenter(Fact
),
is the angle bisector of
.
Then by Ratio Lemma we have
Thus,
is the midpoint of
.
We can calculate and
to be
and
respectively.
And then by Power of a Point, we have
And then similarly, we have
.
Then and
and by Pythagorean we have
, so our answer is
~dolphinday
Video Solution by mop 2024
~r00tsOfUnity
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.