Difference between revisions of "2018 AIME II Problems/Problem 12"
(→Solution 1) |
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Let <math>AP=x</math> and let <math>PC=\rho x</math>. Let <math>[ABP]=\Delta</math> and let <math>[ADP]=\Lambda</math>. | Let <math>AP=x</math> and let <math>PC=\rho x</math>. Let <math>[ABP]=\Delta</math> and let <math>[ADP]=\Lambda</math>. | ||
<asy> | <asy> | ||
− | size( | + | size(300); |
defaultpen(linewidth(0.4)+fontsize(10)); | defaultpen(linewidth(0.4)+fontsize(10)); | ||
pen s = linewidth(0.8)+fontsize(8); | pen s = linewidth(0.8)+fontsize(8); | ||
Line 13: | Line 13: | ||
pair A, B, C, D, P; | pair A, B, C, D, P; | ||
+ | real theta = 10; | ||
A = origin; | A = origin; | ||
− | D = (2*sqrt(65), 0); | + | D = dir(theta)*(2*sqrt(65), 0); |
− | B = 10*dir( | + | B = 10*dir(theta + 147.5); |
− | C = D | + | C = IP(CR(D,10), CR(B,14)); |
P = extension(A,C,B,D); | P = extension(A,C,B,D); | ||
− | draw(A--B--C--D--cycle, | + | draw(A--B--C--D--cycle, s); |
draw(A--C^^B--D); | draw(A--C^^B--D); | ||
Line 26: | Line 27: | ||
dot("$C$", C, NE); | dot("$C$", C, NE); | ||
dot("$D$", D, SE); | dot("$D$", D, SE); | ||
− | dot("$P$", P, | + | dot("$P$", P, 2*dir(115)); |
− | label("$10$", A--B, | + | label("$10$", A--B, SW); |
− | label("$10$", C--D, | + | label("$10$", C--D, 2*N); |
label("$14$", B--C, N); | label("$14$", B--C, N); | ||
label("$2\sqrt{65}$", A--D, S); | label("$2\sqrt{65}$", A--D, S); | ||
label("$x$", A--P, SE); | label("$x$", A--P, SE); | ||
− | |||
label("$\rho x$", P--C, SE); | label("$\rho x$", P--C, SE); | ||
− | label("$\Delta$", B--P/2, | + | label("$\Delta$", B--P/2, 3*dir(-25)); |
− | label("$\Lambda$", D--P/2, | + | label("$\Lambda$", D--P/2, 7*dir(180)); |
</asy> | </asy> | ||
Revision as of 17:53, 29 November 2021
Contents
[hide]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
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.