Difference between revisions of "2018 AIME II Problems/Problem 12"
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==Solution 8 (Simple Geometry)== | ==Solution 8 (Simple Geometry)== | ||
− | [[File: | + | [[File:AIME-II-2018-12.png|400px|right]] |
<math>BP = PD</math> as in another solutions. | <math>BP = PD</math> as in another solutions. | ||
− | Let <math>D'</math> be the | + | Let <math>D'</math> be the reflection of <math>D</math> across <math>C</math>. |
− | Let points <math>E, E',</math> and <math>H</math> be the foot of perpendiculars from <math>D,D' | + | Let points <math>E, E',</math> and <math>H</math> be the foot of perpendiculars on <math>AC</math> from <math>D,D'</math>, and <math>B</math> respectively. |
− | <cmath>AB = CD = CD', BH = DE = D'E' \ | + | <cmath>\begin{align*} |
− | + | &\qquad AB = CD = CD', \quad \textrm{and} \quad BH = DE = D'E' \\ | |
− | + | \Rightarrow &\qquad \triangle ABH = \triangle CDE = \triangle CD'E' \\ | |
+ | \Rightarrow &\qquad \angle BAC = \angle ACD' \\ | ||
+ | \Rightarrow &\qquad \triangle ABC = \triangle AD'C \\ | ||
+ | \Rightarrow &\qquad BC = AD'. | ||
+ | \end{align*}</cmath> | ||
The area of quadrilateral <math>ABCD</math> is equal to the area of triangle <math>ADD'</math> with sides <math>AD' = 14, AD = 2\sqrt{65}, DD' = 2 \cdot 10 = 20</math>. | The area of quadrilateral <math>ABCD</math> is equal to the area of triangle <math>ADD'</math> with sides <math>AD' = 14, AD = 2\sqrt{65}, DD' = 2 \cdot 10 = 20</math>. | ||
The semiperimeter is <math>s = 17 + \sqrt{65},</math> the area <cmath>[ADD'] = \sqrt {(17 + \sqrt{65}) (17 - \sqrt{65})(3 + \sqrt{65})(\sqrt{65}-3)} = \sqrt{(289 – 65)\cdot(65-9)} =\sqrt{56 \cdot 4 \cdot 56} = \boxed{112}.</cmath> | The semiperimeter is <math>s = 17 + \sqrt{65},</math> the area <cmath>[ADD'] = \sqrt {(17 + \sqrt{65}) (17 - \sqrt{65})(3 + \sqrt{65})(\sqrt{65}-3)} = \sqrt{(289 – 65)\cdot(65-9)} =\sqrt{56 \cdot 4 \cdot 56} = \boxed{112}.</cmath> |
Revision as of 23:01, 26 December 2022
Contents
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
.
Diagram
Let and let
. Let
and let
.
Solution 1
Let and let
. Let
and let
. We easily get
and
.
We are given that , which we can now write as
Either
or
. The former would imply that
is a parallelogram, which it isn't; therefore we conclude
and
is the midpoint of
. Let
and
. Then
. On one hand, since
, we have
whereas, on the other hand, using cosine formula to get the length of
, we get
Eliminating
in the above two equations and solving for
we get
which finally yields
.
Solution 2
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
Just to be complete -- and
can actually be equal. In this case,
, but
must be equal to
. We get the same result. -Mathdummy.
Solution 3 (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 midpoint of
~Solution by (Frank FYC)
Solution 4 (With yet another way to get the middle point)
Denote by
. Then
.
Using the formula for the area of a triangle, we get
so
Hence
(note that
makes no difference here).
Now, assume that
,
, and
. Using the cosine rule for
and
, it is clear that
or
Likewise, using the cosine rule for triangles
and
,
It follows that
Since
,
which simplifies to
Plugging this back to equations
,
, and
, it can be solved that
. Then, the area of the quadrilateral is
--Solution by MicGu
Solution 5
As in all other solutions, we can first find that either or
, but it's an AIME problem, we can take
, and assume the other choice will lead to the same result (which is true).
From , we have
, and
, therefore,
By Law of Cosines,
Square
and
, and add them, to get
Solve,
,
-Mathdummy
Solution 6
Either or
. Let
. Applying Stewart's Theorem on
and
, dividing by
and rearranging,
Applying Stewart on
and
,
Substituting equations 1 and 2 into 3 and rearranging,
. By Law of Cosines on
,
so
. Using
to find unknown areas,
.
-Solution by Gart
Solution 7
Now we prove P is the midpoint of . Denote the height from
to
as
, height from
to
as
.According to the problem,
implies
. Then according to basic congruent triangles we get
Firstly, denote that
. Applying Stewart theorem, getting that
, denote
Applying Stewart Theorem, getting
solve for a, getting
Now everything is clear, we can find
using LOC,
, the whole area is
~bluesoul
Solution 8 (Simple Geometry)
as in another solutions.
Let be the reflection of
across
.
Let points
and
be the foot of perpendiculars on
from
, and
respectively.
The area of quadrilateral
is equal to the area of triangle
with sides
.
The semiperimeter is
the area
vladimir.shelomovskii@gmail.com, vvsss
See Also
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 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.