Difference between revisions of "2018 AIME II Problems/Problem 12"
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So, let the area of <math>4</math> triangles <math>\triangle {ABP}=S_{1}</math>, <math>\triangle {BCP}=S_{2}</math>, <math>\triangle {CDP}=S_{3}</math>, <math>\triangle {DAP}=S_{4}</math>. Suppose <math>S_{1}>S_{3}</math> and <math>S_{2}>S_{4}</math>, then it is easy to show that <cmath>S_{1}\cdot S_{3}=S_{2}\cdot S_{4}</cmath>. Also, because <cmath>S_{1}+S_{3}=S_{2}+S_{4}</cmath>, we will have <cmath>(S_{1}+S_{3})^2=(S_{2}+S_{4})^2</cmath>. So <cmath>(S_{1}+S_{3})^2=S_{1}^2+S_{3}^2+2\cdot S_{1}\cdot S_{3}=(S_{2}+S_{4})^2=S_{2}^2+S_{4}^2+2\cdot S_{2}\cdot S_{4}</cmath>. So <cmath>S_{1}^2+S_{3}^2=S_{2}^2+S_{4}^2</cmath>. So <cmath>S_{1}^2+S_{3}^2-2\cdot S_{1}\cdot S_{3}=S_{2}^2+S_{4}^2-2\cdot S_{2}\cdot S_{4}</cmath>. So <cmath>(S_{1}-S_{3})^2=(S_{2}-S_{4})^2</cmath>. As a result, <cmath>S_{1}-S_{3}=S_{2}-S_{4}</cmath>. Then, we have <cmath>S_{1}+S_{4}=S_{2}+S_{3}</cmath>. Combine the condition <cmath>S_{1}+S_{3}=S_{2}+S_{4}</cmath>, we can find out that <cmath>S_{3}=S_{4}</cmath>. So <math>P</math> is the middle point of <math>\overline {AC}</math> | So, let the area of <math>4</math> triangles <math>\triangle {ABP}=S_{1}</math>, <math>\triangle {BCP}=S_{2}</math>, <math>\triangle {CDP}=S_{3}</math>, <math>\triangle {DAP}=S_{4}</math>. Suppose <math>S_{1}>S_{3}</math> and <math>S_{2}>S_{4}</math>, then it is easy to show that <cmath>S_{1}\cdot S_{3}=S_{2}\cdot S_{4}</cmath>. Also, because <cmath>S_{1}+S_{3}=S_{2}+S_{4}</cmath>, we will have <cmath>(S_{1}+S_{3})^2=(S_{2}+S_{4})^2</cmath>. So <cmath>(S_{1}+S_{3})^2=S_{1}^2+S_{3}^2+2\cdot S_{1}\cdot S_{3}=(S_{2}+S_{4})^2=S_{2}^2+S_{4}^2+2\cdot S_{2}\cdot S_{4}</cmath>. So <cmath>S_{1}^2+S_{3}^2=S_{2}^2+S_{4}^2</cmath>. So <cmath>S_{1}^2+S_{3}^2-2\cdot S_{1}\cdot S_{3}=S_{2}^2+S_{4}^2-2\cdot S_{2}\cdot S_{4}</cmath>. So <cmath>(S_{1}-S_{3})^2=(S_{2}-S_{4})^2</cmath>. As a result, <cmath>S_{1}-S_{3}=S_{2}-S_{4}</cmath>. Then, we have <cmath>S_{1}+S_{4}=S_{2}+S_{3}</cmath>. Combine the condition <cmath>S_{1}+S_{3}=S_{2}+S_{4}</cmath>, we can find out that <cmath>S_{3}=S_{4}</cmath>. So <math>P</math> is the middle point of <math>\overline {AC}</math> | ||
+ | |||
+ | ~Solution by <math>BladeRunnerAUG</math> (Frank FYC) | ||
{{AIME box|year=2018|n=II|num-b=11|num-a=13}} | {{AIME box|year=2018|n=II|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 21:01, 6 October 2018
Problem
Let be a convex quadrilateral with
,
, and
. Assume that the diagonals of
intersect at point
, and that the sum of the areas of triangles
and
equals the sum of the areas of triangles
and
. Find the area of quadrilateral
.
Solution 1
For reference, , so
is the longest of the four sides of
. Let
be the length of the altitude from
to
, and let
be the length of the altitude from
to
. Then, the triangle area equation becomes
.
What an important finding! Note that the opposite sides and
have equal length, and note that diagonal
bisects diagonal
. This is very similar to what happens if
were a parallelogram with
, so let's extend
to point
, such that
is a parallelogram. In other words,
and
. Now, let's examine
. Since
, the triangle is isosceles, and
. Note that in parallelogram
,
and
are congruent, so
and thus
. Define
, so
. We use the Law of Cosines on
and
:
Subtracting the second equation from the first yields
This means that dropping an altitude from to some foot
on
gives
and therefore
. Seeing that
, we conclude that
is a 3-4-5 right triangle, so
. Then, the area of
is
. Since
, points
and
are equidistant from
, so
and hence
. -kgator
Solution 2 (Another way to get the middle point)
So, let the area of triangles
,
,
,
. Suppose
and
, then it is easy to show that
. Also, because
, we will have
. So
. So
. So
. So
. As a result,
. Then, we have
. Combine the condition
, we can find out that
. So
is the middle point of
~Solution by (Frank FYC)
2018 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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