Difference between revisions of "Mock Geometry AIME 2011 Problems/Problem 13"
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− | Solving the problem involves the ambiguous case of law of sines. First part of solution is angle chasing. Let <math>L</math> be intersection of angle bisector <math>\ell</math> with <math>BC</math> Let <math>\angle BAL</math> be <math>\theta</math>, and <math>\angle LAC is \theta</math> as well, since angle bisector. Since line through <math>M</math> is parallel to <math>\ell</math>, <math>\angle MEC</math> is also <math>\theta</math>. Let <math>\angle BLA</math> then be <math>\alpha</math>, and by parallel lines, <math>\angle BME</math> is also <math>\alpha</math>. Doing further angle chasing, we find that <math>AFE</math> is isoceles with base <math>EF</math>. Using <math>30-60-90</math> triangle ratio, we find <math>\theta = 30^\circ</math> | + | Solving the problem involves the ambiguous case of law of sines. First part of solution is angle chasing. Let <math>L</math> be intersection of angle bisector <math>\ell</math> with <math>BC</math> Let <math>\angle BAL</math> be <math>\theta</math>, and <math>\angle LAC</math> is <math>\theta</math> as well, since angle bisector. Since line through <math>M</math> is parallel to <math>\ell</math>, <math>\angle MEC</math> is also <math>\theta</math>. Let <math>\angle BLA</math> then be <math>\alpha</math>, and by parallel lines, <math>\angle BME</math> is also <math>\alpha</math>. Doing further angle chasing, we find that <math>AFE</math> is isoceles with base <math>EF</math>. Using <math>30-60-90</math> triangle ratio, we find <math>\theta = 30^\circ</math> |
There are two possible configurations of the triangle, one such that <math>L</math> is to the left of <math>M</math>, and vice versa. In the first <math>A</math> falls between <math>B</math> and <math>F</math>, with <math>F</math> outside the triangle, and in the second <math>F</math> between <math>B</math> and <math>A</math>, with <math>E</math> outside the triangle. Using Law of Sines then: | There are two possible configurations of the triangle, one such that <math>L</math> is to the left of <math>M</math>, and vice versa. In the first <math>A</math> falls between <math>B</math> and <math>F</math>, with <math>F</math> outside the triangle, and in the second <math>F</math> between <math>B</math> and <math>A</math>, with <math>E</math> outside the triangle. Using Law of Sines then: |
Revision as of 16:30, 21 July 2018
Problem 13
In acute triangle
is the bisector of
.
is the midpoint of
. a line through
parallel to
meets
at
respectively. Given that
the sum of all possible values of
can be expressed as
where
are positive integers. What is
?
Solution
Solving the problem involves the ambiguous case of law of sines. First part of solution is angle chasing. Let be intersection of angle bisector
with
Let
be
, and
is
as well, since angle bisector. Since line through
is parallel to
,
is also
. Let
then be
, and by parallel lines,
is also
. Doing further angle chasing, we find that
is isoceles with base
. Using
triangle ratio, we find
There are two possible configurations of the triangle, one such that is to the left of
, and vice versa. In the first
falls between
and
, with
outside the triangle, and in the second
between
and
, with
outside the triangle. Using Law of Sines then:
Plugging in values, we find for acute and obtuse triangles denoted as and
, respectively,
, and
Using Law of Sines again and substituting the expression for the
and
for
,
, and
Solving for the ratio of on both triangles, and then applying Angle Bisector theorem yields a
with included angle
for
and
with included angle
for
. Solving using Law of Cosine yields answer of
and
, or
.