Difference between revisions of "2019 AIME I Problems/Problem 13"
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+ | |||
+ | ==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 | ||
==See Also== | ==See Also== |
Revision as of 12:27, 2 January 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
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.