Difference between revisions of "2019 AIME I Problems/Problem 13"
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==Problem 13== | ==Problem 13== | ||
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>. | ||
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Notice that <cmath>\angle DFE=\angle CFE-\angle CFD=\angle CBE-\angle CAD=180-B-A=C.</cmath>By the Law of Cosines, <cmath>\cos C=\frac{AC^2+BC^2-AB^2}{2\cdot AC\cdot BC}=\frac34.</cmath>Then, <cmath>DE^2=DF^2+EF^2-2\cdot DF\cdot EF\cos C=32\implies DE=4\sqrt2.</cmath>Let <math>X=\overline{AB}\cap\overline{CF}</math>, <math>a=XB</math>, and <math>b=XD</math>. Then, <cmath>XA\cdot XD=XC\cdot XF=XB\cdot XE\implies b(a+4)=a(b+4\sqrt2)\implies b=a\sqrt2.</cmath>However, since <math>\triangle XFD\sim\triangle XAC</math>, <math>XF=\tfrac{4+a}3</math>, but since <math>\triangle XFE\sim\triangle XBC</math>, <cmath>\frac75=\frac{4+a}{3a}\implies a=\frac54\implies BE=a+a\sqrt2+4\sqrt2=\frac{5+21\sqrt2}4,</cmath>and the requested sum is <math>5+21+2+4=\boxed{032}</math>. | Notice that <cmath>\angle DFE=\angle CFE-\angle CFD=\angle CBE-\angle CAD=180-B-A=C.</cmath>By the Law of Cosines, <cmath>\cos C=\frac{AC^2+BC^2-AB^2}{2\cdot AC\cdot BC}=\frac34.</cmath>Then, <cmath>DE^2=DF^2+EF^2-2\cdot DF\cdot EF\cos C=32\implies DE=4\sqrt2.</cmath>Let <math>X=\overline{AB}\cap\overline{CF}</math>, <math>a=XB</math>, and <math>b=XD</math>. Then, <cmath>XA\cdot XD=XC\cdot XF=XB\cdot XE\implies b(a+4)=a(b+4\sqrt2)\implies b=a\sqrt2.</cmath>However, since <math>\triangle XFD\sim\triangle XAC</math>, <math>XF=\tfrac{4+a}3</math>, but since <math>\triangle XFE\sim\triangle XBC</math>, <cmath>\frac75=\frac{4+a}{3a}\implies a=\frac54\implies BE=a+a\sqrt2+4\sqrt2=\frac{5+21\sqrt2}4,</cmath>and the requested sum is <math>5+21+2+4=\boxed{032}</math>. | ||
Revision as of 22:01, 16 October 2020
Contents
Problem 13
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)
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.