Difference between revisions of "2019 AIME II Problems/Problem 1"
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Thus, our answer is <math>059</math>. ~a.y.711 | Thus, our answer is <math>059</math>. ~a.y.711 | ||
+ | == Solution 6 == | ||
+ | |||
+ | Let <math>A=(0,0), B=(9,0)</math>. Now consider <math>C</math>, and if we find the coordinates of <math>C</math>, by symmetry about <math>x=4.5</math>, we can find the coordinates of D. | ||
+ | |||
+ | So let <math>C=(a,b)</math>. So the following equations hold: | ||
+ | |||
+ | <math>\sqrt{(a-9)^2+(b)^2}=17</math>. | ||
+ | |||
+ | <math>\sqrt{a^2+b^2}=10</math>. | ||
+ | |||
+ | Solving by squaring both equations and then subtracting one from the other to eliminate <math>b^2</math>, we get <math>C=(-6,8)</math> because <math>C</math> is in the second quadrant. | ||
+ | |||
+ | Now by symmetry, <math>D=(16, 8)</math>. | ||
+ | |||
+ | So now you can proceed by finding the intersection and then calculating the area directly. We get <math>\boxed{059}</math>. | ||
+ | |||
+ | ~hastapasta | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2019|n=II|before=First Problem|num-a=2}} | {{AIME box|year=2019|n=II|before=First Problem|num-a=2}} | ||
[[Category: Intermediate Geometry Problems]] | [[Category: Intermediate Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 14:05, 12 January 2023
Contents
Problem
Two different points, and
, lie on the same side of line
so that
and
are congruent with
,
, and
. The intersection of these two triangular regions has area
, where
and
are relatively prime positive integers. Find
.
Solution
- Diagram by Brendanb4321
Extend to form a right triangle with legs
and
such that
is the hypotenuse and connect the points
so
that you have a rectangle. (We know that
is a
, since
is an
.) The base
of the rectangle will be
. Now, let
be the intersection of
and
. This means that
and
are with ratio
. Set up a proportion, knowing that the two heights add up to 8. We will let
be the height from
to
, and
be the height of
.
This means that the area is . This gets us
-Solution by the Math Wizard, Number Magician of the Second Order, Head of the Council of the Geometers
Solution 2
Using the diagram in Solution 1, let be the intersection of
and
. We can see that angle
is in both
and
. Since
and
are congruent by AAS, we can then state
and
. It follows that
and
. We can now state that the area of
is the area of
the area of
. Using Heron's formula, we compute the area of
. Using the Law of Cosines on angle
, we obtain
(For convenience, we're not going to simplify.)
Applying the Law of Cosines on yields
This means
. Next, apply Heron's formula to get the area of
, which equals
after simplifying. Subtracting the area of
from the area of
yields the area of
, which is
, giving us our answer, which is
-Solution by flobszemathguy
Solution 3 (Very quick)
- Diagram by Brendanb4321 extended by Duoquinquagintillion
Begin with the first step of solution 1, seeing is the hypotenuse of a
triangle and calling the intersection of
and
point
. Next, notice
is the hypotenuse of an
triangle. Drop an altitude from
with length
, so the other leg of the new triangle formed has length
. Notice we have formed similar triangles, and we can solve for
.
So has area
And
- Solution by Duoquinquagintillion
Solution 4
Let . By Law of Cosines,
And
- by Mathdummy
Solution 5
Because and
, quadrilateral
is cyclic. So, Ptolemy's theorem tells us that
From here, there are many ways to finish which have been listed above. If we let , then
Using Heron's formula on , we see that
Thus, our answer is . ~a.y.711
Solution 6
Let . Now consider
, and if we find the coordinates of
, by symmetry about
, we can find the coordinates of D.
So let . So the following equations hold:
.
.
Solving by squaring both equations and then subtracting one from the other to eliminate , we get
because
is in the second quadrant.
Now by symmetry, .
So now you can proceed by finding the intersection and then calculating the area directly. We get .
~hastapasta
See Also
2019 AIME II (Problems • Answer Key • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
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.