Difference between revisions of "2014 AIME I Problems/Problem 15"
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In <math>\triangle ABC</math>, <math>AB = 3</math>, <math>BC = 4</math>, and <math>CA = 5</math>. Circle <math>\omega</math> intersects <math>\overline{AB}</math> at <math>E</math> and <math>B</math>, <math>\overline{BC}</math> at <math>B</math> and <math>D</math>, and <math>\overline{AC}</math> at <math>F</math> and <math>G</math>. Given that <math>EF=DF</math> and <math>\frac{DG}{EG} = \frac{3}{4}</math>, length <math>DE=\frac{a\sqrt{b}}{c}</math>, where <math>a</math> and <math>c</math> are relatively prime positive integers, and <math>b</math> is a positive integer not divisible by the square of any prime. Find <math>a+b+c</math>. | In <math>\triangle ABC</math>, <math>AB = 3</math>, <math>BC = 4</math>, and <math>CA = 5</math>. Circle <math>\omega</math> intersects <math>\overline{AB}</math> at <math>E</math> and <math>B</math>, <math>\overline{BC}</math> at <math>B</math> and <math>D</math>, and <math>\overline{AC}</math> at <math>F</math> and <math>G</math>. Given that <math>EF=DF</math> and <math>\frac{DG}{EG} = \frac{3}{4}</math>, length <math>DE=\frac{a\sqrt{b}}{c}</math>, where <math>a</math> and <math>c</math> are relatively prime positive integers, and <math>b</math> is a positive integer not divisible by the square of any prime. Find <math>a+b+c</math>. | ||
− | == Solution == | + | == Solution 1 == |
− | First we note that triangle DEF is an isosceles right triangle with hypotenuse DE the same as the diameter of | + | Since <math>\angle DBE = 90^\circ</math>, <math>DE</math> is the diameter of <math>\omega</math>. Then <math>\angle DFE=\angle DGE=90^\circ</math>. But <math>DF=FE</math>, so <math>\triangle DEF</math> is a 45-45-90 triangle. Letting <math>DG=3x</math>, we have that <math>EG=4x</math>, <math>DE=5x</math>, and <math>DF=EF=\frac{5x}{\sqrt{2}}</math>. |
+ | |||
+ | Note that <math>\triangle DGE \sim \triangle ABC</math> by SAS similarity, so <math>\angle BAC = \angle GDE</math> and <math>\angle ACB = \angle DEG</math>. Since <math>DEFG</math> is a cyclic quadrilateral, <math>\angle BAC = \angle GDE=180^\circ-\angle EFG = \angle AFE</math> and <math>\angle ACB = \angle DEG = \angle GFD</math>, implying that <math>\triangle AFE</math> and <math>\triangle CDF</math> are isosceles. As a result, <math>AE=CD=\frac{5x}{\sqrt{2}}</math>, so <math>BE=3-\frac{5x}{\sqrt{2}}</math> and <math>BD =4-\frac{5x}{\sqrt{2}}</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> | ||
+ | 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 == | ||
+ | |||
+ | <asy> | ||
+ | pair A = (0,3); | ||
+ | pair B = (0,0); | ||
+ | pair C = (4,0); | ||
+ | draw(A--B--C--cycle); | ||
+ | dotfactor = 3; | ||
+ | dot("$A$",A,dir(135)); | ||
+ | dot("$B$",B,dir(215)); | ||
+ | dot("$C$",C,dir(305)); | ||
+ | pair D = (2.21, 0); | ||
+ | pair E = (0, 1.21); | ||
+ | pair F = (1.71, 1.71); | ||
+ | pair G = (2, 1.5); | ||
+ | dot("$D$",D,dir(270)); | ||
+ | dot("$E$",E,dir(180)); | ||
+ | dot("$F$",F,dir(90)); | ||
+ | dot("$G$",G,dir(0)); | ||
+ | draw(Circle((1.109, 0.609), 1.28)); | ||
+ | draw(D--E); | ||
+ | draw(E--F); | ||
+ | draw(D--F); | ||
+ | draw(E--G); | ||
+ | draw(D--G); | ||
+ | draw(B--F); | ||
+ | draw(B--G); | ||
+ | </asy> | ||
+ | |||
+ | First we note that <math>\triangle DEF</math> is an isosceles right triangle with hypotenuse <math>\overline{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, <math>\triangle BGC</math> is an isosceles triangle with <math>BG = GC</math>, so <math>G</math> is the midpoint of <math>\overline{AC}</math> and <math>AG = GC = 5/2</math>. Similarly, we can find from angle chasing that <math>\angle ABF = \angle EDF = \frac{\pi}4</math>. Therefore, <math>\overline{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}</math>, so <math>AF = 15/7</math> and <math>CF = 20/7</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>d</math>: | ||
+ | <cmath>DE\cdot FG+DG\cdot EF=DF\cdot EG</cmath> | ||
+ | <cmath>d\cdot FG+\frac{3d}{5}\cdot \frac{d}{\sqrt{2}}=\frac{4d}{5}\cdot \frac{d}{\sqrt{2}}</cmath> | ||
+ | <cmath>d\cdot FG+\frac{3d^2}{5\sqrt{2}}=\frac{4d^2}{5\sqrt{2}}\implies FG=\frac{d}{5\sqrt{2}}</cmath> | ||
+ | Thus <math>\frac{d}{5\sqrt{2}}=\frac{5}{14}\rightarrow d=5\sqrt{2}\cdot\frac{5}{14}=\frac{25\sqrt{2}}{14}</math>. <math>a+b+c=25+2+14= \boxed{041}</math> | ||
+ | |||
+ | ===Solution 3=== | ||
+ | Call <math>DE=x</math> and as a result <math>DF=EF=\frac{x\sqrt{2}}{2}, EG=\frac{4x}{5}, GD=\frac{3x}{5}</math>. Since <math>EFGD</math> is cyclic we just need to get <math>DG</math> and using LoS(for more detail see the <math>2</math>nd paragraph of Solution <math>2</math>) we get <math>AG=\frac{5}{2}</math> and using a similar argument(use LoS again) and subtracting you get <math>FG=\frac{5}{14}</math> so you can use Ptolemy to get <math>x=\frac{25\sqrt{2}}{14} \implies \boxed{041}</math>. | ||
+ | ~First | ||
+ | |||
+ | ==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;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>\triangle AEF</math> is isosceles triangle. Since <math>\triangle ABC</math> 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 | ||
+ | |||
+ | ==Solution 5== | ||
+ | [[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> | ||
+ | |||
+ | <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> | ||
+ | BF is bisector of <math>\angle ABC\implies BF = \frac {2AB \cdot BC}{AB+BC} \cos 45^\circ =\frac {12 \cdot \sqrt{2}}{7}.</math> | ||
+ | <cmath>\angle EGD = 90^\circ, \frac {EG}{GD}=\frac{4}{3} \implies</cmath> | ||
+ | <cmath>\angle GED = \angle GCD =\gamma \implies \overset{\Large\frown} {DG} = 2\gamma.</cmath> | ||
+ | <cmath>2\angle ACB = \overset{\Large\frown} {BEF} - \overset{\Large\frown} {DG} \implies \overset{\Large\frown} {BEF} = 4 \gamma \implies</cmath> | ||
+ | <cmath>\angle BOF = 4 \gamma \implies \angle OBF = \angle OFB = 90^\circ – 2 \gamma.</cmath> | ||
+ | Let <math>BO = EO = DO = r \implies BF = 2 r \cos(90^\circ – 2\gamma) =</math> | ||
+ | <cmath>=2 r \sin 2\gamma = 4r \sin \gamma \cdot \cos \gamma = 4 r\cdot \frac {3}{5} \cdot \frac {4}{5} = \frac {48}{25} = \frac {12 \cdot \sqrt{2}}{7}\implies</cmath> | ||
+ | <cmath>r = \frac {25 \cdot \sqrt{2}}{28}\implies ED = 2r = \frac {25 \cdot \sqrt{2}}{14}\implies \boxed{\textbf{041}}.</cmath> | ||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Solution 6 == | ||
+ | [[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> | ||
+ | |||
+ | <math>\triangle BCG </math> is isosceles triangle with <math>BG=CG.</math> | ||
+ | |||
+ | Similarly <math>BG = AG \implies AG = CG = BG = \frac {AC}{2} =\frac {5}{2}.</math> | ||
+ | |||
+ | <cmath>\overset{\Large\frown} {FG} = 90^\circ – \overset{\Large\frown} {GD} = 90^\circ – 2\gamma \implies</cmath> | ||
+ | <cmath>\overset{\Large\frown} {BFG} = 4\gamma + 90^\circ – 2\gamma = 90^\circ + 2\gamma \implies</cmath> | ||
+ | <cmath>\angle BOG = 90^\circ + 2\gamma \implies \angle BGO = \angle GBO = 45^\circ - \gamma.</cmath> | ||
+ | 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>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> | ||
+ | '''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 | ||
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