Difference between revisions of "2019 AIME I Problems/Problem 13"
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− | ==Problem | + | ==Problem== |
Triangle <math>ABC</math> has side lengths <math>AB=4</math>, <math>BC=5</math>, and <math>CA=6</math>. Points <math>D</math> and <math>E</math> are on ray <math>AB</math> with <math>AB<AD<AE</math>. The point <math>F \neq C</math> is a point of intersection of the circumcircles of <math>\triangle ACD</math> and <math>\triangle EBC</math> satisfying <math>DF=2</math> and <math>EF=7</math>. Then <math>BE</math> can be expressed as <math>\tfrac{a+b\sqrt{c}}{d}</math>, where <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> are positive integers such that <math>a</math> and <math>d</math> are relatively prime, and <math>c</math> is not divisible by the square of any prime. Find <math>a+b+c+d</math>. | Triangle <math>ABC</math> has side lengths <math>AB=4</math>, <math>BC=5</math>, and <math>CA=6</math>. Points <math>D</math> and <math>E</math> are on ray <math>AB</math> with <math>AB<AD<AE</math>. The point <math>F \neq C</math> is a point of intersection of the circumcircles of <math>\triangle ACD</math> and <math>\triangle EBC</math> satisfying <math>DF=2</math> and <math>EF=7</math>. Then <math>BE</math> can be expressed as <math>\tfrac{a+b\sqrt{c}}{d}</math>, where <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> are positive integers such that <math>a</math> and <math>d</math> are relatively prime, and <math>c</math> is not divisible by the square of any prime. Find <math>a+b+c+d</math>. | ||
Line 70: | Line 70: | ||
==Solution 3== | ==Solution 3== | ||
− | Construct <math>FC</math> and let <math>FC\cap AE=K</math>. Let <math>FK=x</math>. Using <math>\triangle FKE\sim \triangle BKC</math>, <cmath>BK=\frac{5}{7}x</cmath> Using <math>\triangle FDK\sim ACK</math>, it can be found that <cmath>3x=AK=4+\frac{5}{7}x\to x=\frac{7}{4}</cmath> This also means that <math>BK=\frac{21}{4}-4=\frac{5}{4}</math>. It suffices to find <math>KE</math>. It is easy to see the following: <cmath>180-\angle ABC=\angle KBC=\angle KFE</cmath> Using reverse Law of Cosines on <math>\triangle ABC</math>, <math>\cos{\angle ABC}=\frac{1}{8}\to \cos{180-\angle ABC}=\frac{-1}{8}</math>. Using Law of Cosines on <math>\triangle EFK</math> gives <math>KE=\frac{21\sqrt 2}{4}</math>, so <math>BE=\frac{5+21\sqrt 2}{4}\to \textbf{032}</math>. | + | Construct <math>FC</math> and let <math>FC\cap AE=K</math>. Let <math>FK=x</math>. Using <math>\triangle FKE\sim \triangle BKC</math>, <cmath>BK=\frac{5}{7}x</cmath> Using <math>\triangle FDK\sim ACK</math>, it can be found that <cmath>3x=AK=4+\frac{5}{7}x\to x=\frac{7}{4}</cmath> This also means that <math>BK=\frac{21}{4}-4=\frac{5}{4}</math>. It suffices to find <math>KE</math>. It is easy to see the following: <cmath>180-\angle ABC=\angle KBC=\angle KFE</cmath> Using reverse Law of Cosines on <math>\triangle ABC</math>, <math>\cos{\angle ABC}=\frac{1}{8}\to \cos{180-\angle ABC}=\frac{-1}{8}</math>. Using Law of Cosines on <math>\triangle EFK</math> gives <math>KE=\frac{21\sqrt 2}{4}</math>, so <math>BE=\frac{5+21\sqrt 2}{4}\to \boxed{\textbf{032}}</math>. |
-franchester | -franchester | ||
+ | |||
==Solution 4 (No <C = <DFE, no LoC)== | ==Solution 4 (No <C = <DFE, no LoC)== | ||
Let <math>P=AE\cap CF</math>. Let <math>CP=5x</math> and <math>BP=5y</math>; from <math>\triangle{CBP}\sim\triangle{EFP}</math> we have <math>EP=7x</math> and <math>FP=7y</math>. From <math>\triangle{CAP}\sim\triangle{DFP}</math> we have <math>\frac{6}{4+5y}=\frac{2}{7y}</math> giving <math>y=\frac{1}{4}</math>. So <math>BP=\frac{5}{4}</math> and <math>FP=\frac{7}{4}</math>. These similar triangles also gives us <math>DP=\frac{5}{3}x</math> so <math>DE=\frac{16}{3}x</math>. Now, Stewart's Theorem on <math>\triangle{FEP}</math> and cevian <math>FD</math> tells us that <cmath>\frac{560}{9}x^3+28x=\frac{49}{3}x+\frac{245}{3}x,</cmath>so <math>x=\frac{3\sqrt{2}}{4}</math>. Then <math>BE=\frac{5}{4}+7x=\frac{5+21\sqrt{2}}{4}</math> so the answer is <math>\boxed{032}</math> as desired. (Solution by Trumpeter, but not added to the Wiki by Trumpeter) | Let <math>P=AE\cap CF</math>. Let <math>CP=5x</math> and <math>BP=5y</math>; from <math>\triangle{CBP}\sim\triangle{EFP}</math> we have <math>EP=7x</math> and <math>FP=7y</math>. From <math>\triangle{CAP}\sim\triangle{DFP}</math> we have <math>\frac{6}{4+5y}=\frac{2}{7y}</math> giving <math>y=\frac{1}{4}</math>. So <math>BP=\frac{5}{4}</math> and <math>FP=\frac{7}{4}</math>. These similar triangles also gives us <math>DP=\frac{5}{3}x</math> so <math>DE=\frac{16}{3}x</math>. Now, Stewart's Theorem on <math>\triangle{FEP}</math> and cevian <math>FD</math> tells us that <cmath>\frac{560}{9}x^3+28x=\frac{49}{3}x+\frac{245}{3}x,</cmath>so <math>x=\frac{3\sqrt{2}}{4}</math>. Then <math>BE=\frac{5}{4}+7x=\frac{5+21\sqrt{2}}{4}</math> so the answer is <math>\boxed{032}</math> as desired. (Solution by Trumpeter, but not added to the Wiki by Trumpeter) | ||
+ | |||
+ | ==Solution 5== | ||
+ | Connect <math>CF</math> meeting <math>AE</math> at <math>J</math>. We can observe that <math>\triangle{ACJ}\sim \triangle{FJD}</math> Getting that <math>\frac{AJ}{FJ}=\frac{AC}{FD}=3</math>. We can also observe that <math>\triangle{CBJ}\sim \triangle{EFJ}</math>, getting that <math>\frac{CB}{EF}=\frac{BJ}{FJ}=\frac{CJ}{EJ}=\frac{5}{7}</math> | ||
+ | |||
+ | Assume that <math>BJ=5x;FJ=7x</math>, since <math>\frac{AJ}{FJ}=3</math>, we can get that <math>\frac{AJ}{FJ}=\frac{AB+BJ}{FJ}=\frac{4+5x}{7x}=3</math>, getting that <math>x=\frac{1}{4}; BJ=\frac{5}{4}; FJ=\frac{7}{4}</math> | ||
+ | |||
+ | Using Power of Point, we can get that <math>BJ * EJ=CJ*FJ; DJ * AJ=CJ * FJ</math> Assume that <math>DJ=5k,CJ=15k</math>, getting that <math>JE=21k, DE=16k</math> | ||
+ | |||
+ | Now applying Law of Cosine on two triangles, <math>\triangle{ACJ};\triangle{FJE}</math> separately, we can get two equations | ||
+ | |||
+ | <math>(1): (15k)^2+(\frac{21}{4})^2-2*15k *\frac{21}{4} * cos\angle{CJA}=36</math> | ||
+ | |||
+ | <math>(2):(21k)^2+(\frac{7}{4})^2-2*\frac{7}{4} * 21k*cos\angle{FJE}=49</math> | ||
+ | |||
+ | Since <math>\angle{CJA}=\angle{FJE}</math>, we can use <math>15(2)-7(1)</math> to eliminate the <math>cos</math> term | ||
+ | |||
+ | Then we can get that <math>5040k^2=630</math>, getting <math>k=\frac{\sqrt{2}}{4}</math> | ||
+ | |||
+ | <math>BE=21k=\frac{21\sqrt{2}}{4}; BJ=\frac{5}{4}</math>, so the desired answer is <math>\frac{21\sqrt{2}+5}{4}</math>, which leads to the answer <math>\boxed{032}</math> | ||
+ | |||
+ | ~bluesoul | ||
+ | |||
+ | ==Solution 6== | ||
+ | |||
+ | Nice problem! | ||
+ | |||
+ | First, let <math>AE</math> and <math>CF</math> intersect at <math>X</math>. Our motivation here is to introduce cyclic quadrilaterals and find useful relationships in terms of angles. Observe that | ||
+ | <cmath>\angle DFE = \angle XFE - \angle XFD = \angle CBE - \angle CAB = 180 - \angle ABC - \angle CAB = \angle BAC</cmath> | ||
+ | By the so-called "Reverse Law of Cosines" on <math>\triangle ABC</math> we have | ||
+ | <cmath>\cos(\angle BAC) = \frac{4^2 - 5^2 - 6^2}{-2 \cdot 5 \cdot 6} = \frac{3}{4}</cmath> | ||
+ | Applying on <math>\triangle DFE</math> gives | ||
+ | <cmath>DE^2 = 2^2 + 7^2 - 2 \cdot 2 \cdot 7 \cos(\angle DFE)</cmath> | ||
+ | <cmath>= 2^2 + 7^2 - 2 \cdot 2 \cdot 7 \cdot \frac{3}{4}</cmath> | ||
+ | <cmath>=32</cmath> | ||
+ | So <math>DE = 4 \sqrt{2}</math>, now by our cyclic quadrilaterals again, we are motivated by the multiple appearances of similar triangles throughout the figure. We want some that are related to <math>BX</math> and <math>XD</math>, which are crucial lengths in the problem. Suppose <math>BX = r, XD = s</math> for simplicity. We have: | ||
+ | |||
+ | <math>\bullet~~~~\triangle AXC \sim \triangle FXD</math> | ||
+ | <math>\bullet~~~~\triangle BXC \sim \triangle FXE</math> | ||
+ | |||
+ | So | ||
+ | <cmath>\frac{AX}{FX} = \frac{XC}{XD} = \frac{AC}{FD} \implies \frac{4 + r}{FX} = \frac{XC}{s} = 3</cmath> | ||
+ | <cmath>\frac{BX}{FX} = \frac{XC}{XE} = \frac{BC}{FE} \implies \frac{r}{FX} = \frac{XC}{s + 4 \sqrt{2}} = \frac{5}{7}</cmath> | ||
+ | <cmath>\implies \frac{4 + r}{r} = \frac{s + 4 \sqrt{2}}{s} = \frac{21}{5}</cmath> | ||
+ | <cmath>\implies r = \frac{5}{4}, s = \frac{5 \sqrt{2}}{4}</cmath> | ||
+ | So <math>BE = r + s + 4 \sqrt{2} = \frac{5 + 21 \sqrt{2}}{4}</math>. The requested sum is <math>5 + 21 + 2 + 4 = \boxed{032}</math>. | ||
+ | |||
+ | ~CoolJupiter | ||
+ | |||
+ | ==Video Solution by MOP 2024== | ||
+ | https://youtube.com/watch?v=B7rFw05AYQ0 | ||
+ | |||
+ | ~r00tsOfUnity | ||
==See Also== | ==See Also== |
Latest revision as of 20:44, 20 December 2023
Contents
Problem
Triangle has side lengths
,
, and
. Points
and
are on ray
with
. The point
is a point of intersection of the circumcircles of
and
satisfying
and
. Then
can be expressed as
, where
,
,
, and
are positive integers such that
and
are relatively prime, and
is not divisible by the square of any prime. Find
.
Solution 1
Notice that By the Law of Cosines,
Then,
Let
,
, and
. Then,
However, since
,
, but since
,
and the requested sum is
.
(Solution by TheUltimate123)
Solution 2
Define to be the circumcircle of
and
to be the circumcircle of
.
Because of exterior angles,
But because
is cyclic. In addition,
because
is cyclic. Therefore,
. But
, so
. Using Law of Cosines on
, we can figure out that
. Since
,
. We are given that
and
, so we can use Law of Cosines on
to find that
.
Let be the intersection of segment
and
. Using Power of a Point with respect to
within
, we find that
. We can also apply Power of a Point with respect to
within
to find that
. Therefore,
.
Note that is similar to
.
. Also note that
is similar to
, which gives us
. Solving this system of linear equations, we get
. Now, we can solve for
, which is equal to
. This simplifies to
, which means our answer is
.
Solution 3
Construct and let
. Let
. Using
,
Using
, it can be found that
This also means that
. It suffices to find
. It is easy to see the following:
Using reverse Law of Cosines on
,
. Using Law of Cosines on
gives
, so
.
-franchester
Solution 4 (No <C = <DFE, no LoC)
Let . Let
and
; from
we have
and
. From
we have
giving
. So
and
. These similar triangles also gives us
so
. Now, Stewart's Theorem on
and cevian
tells us that
so
. Then
so the answer is
as desired. (Solution by Trumpeter, but not added to the Wiki by Trumpeter)
Solution 5
Connect meeting
at
. We can observe that
Getting that
. We can also observe that
, getting that
Assume that , since
, we can get that
, getting that
Using Power of Point, we can get that Assume that
, getting that
Now applying Law of Cosine on two triangles, separately, we can get two equations
Since , we can use
to eliminate the
term
Then we can get that , getting
, so the desired answer is
, which leads to the answer
~bluesoul
Solution 6
Nice problem!
First, let and
intersect at
. Our motivation here is to introduce cyclic quadrilaterals and find useful relationships in terms of angles. Observe that
By the so-called "Reverse Law of Cosines" on
we have
Applying on
gives
So
, now by our cyclic quadrilaterals again, we are motivated by the multiple appearances of similar triangles throughout the figure. We want some that are related to
and
, which are crucial lengths in the problem. Suppose
for simplicity. We have:
So
So
. The requested sum is
.
~CoolJupiter
Video Solution by MOP 2024
https://youtube.com/watch?v=B7rFw05AYQ0
~r00tsOfUnity
See Also
2019 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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.