Difference between revisions of "2021 AIME I Problems/Problem 9"
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~Math_Genius_164 | ~Math_Genius_164 | ||
− | ==Solution 2(LOC)== | + | ==Solution 2(LOC and Trig)== |
− | Call AD and BC <math>a</math>. Draw diagonal AC and call the foot of the perpendicular from B to AC <math>G</math>. Call the foot of the perpendicular from A to line BC F. Triangles CBG and CAF are similar, and we get that <math>\frac{10}{15}</math>=<math>frac{a}{AC}</math> Therefore, <math>AC=1.5a</math>. | + | Call AD and BC <math>a</math>. Draw diagonal AC and call the foot of the perpendicular from B to AC <math>G</math>. Call the foot of the perpendicular from A to line BC F, and call the foot of the perpindicular from A to DC H. Triangles CBG and CAF are similar, and we get that <math>\frac{10}{15}</math>=<math>\frac{a}{AC}</math> Therefore, <math>AC=1.5a</math>. It then follows that triangles ABF and ADH are similar. Using similar triangles, we can then find that <math>AB=1.2a</math>. Using the Law of Cosine on ABC, We can find that the cosine of angle ABC is <math>-\frac{1}{3}. Since angles ABF and ADH are equivalent and supplementary to angle ABC, we know that the cosine of angle ADH is 1/3. It then follows that </math>a=\frac{27\sqrt{2}}{2}<math>. Then it can be found that the area </math>K<math> is </math>\frac{567\sqrt{2}}{2}<math>. Multiplying this by </math>\sqrt{2}<math>, the answer is </math>\boxed{567}$. |
==See also== | ==See also== | ||
{{AIME box|year=2021|n=I|num-b=8|num-a=10}} | {{AIME box|year=2021|n=I|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 13:45, 12 March 2021
Problem
Let be an isosceles trapezoid with
and
Suppose that the distances from
to the lines
and
are
and
respectively. Let
be the area of
Find
Solution 1
Construct your isosceles trapezoid. Let, for simplicity, ,
, and
. Extend the sides
and
mark the intersection as
. Following what the question states, drop a perpendicular from
to
labeling the foot as
. Drop another perpendicular from
to
, calling the foot
. Lastly, drop a perpendicular from
to
, labeling it
. In addition, drop a perpendicular from
to
calling its foot
.
--DIAGRAM COMING SOON--
Start out by constructing a triangle congruent to
with its side of length
on line
. This works because all isosceles triangles are cyclic and as a result,
.
Notice that by AA similarity. We are given that
and by symmetry we can deduce that
. As a result,
. This gives us that
.
The question asks us along the lines of finding the area, , of the trapezoid
. We look at the area of
and notice that it can be represented as
. Substituting
, we solve for
, getting
.
Now let us focus on isosceles triangle , where
. Since,
is an altitude from
to
of an isosceles triangle,
must be equal to
. Since
and
, we can solve to get that
and
.
We must then set up equations using the Pythagorean Theorem, writing everything in terms of ,
, and
. Looking at right triangle
we get
Looking at right triangle
we get
Now rearranging and solving, we get two equation
Those are convenient equations as
which gives us
After some "smart" calculation, we get that
.
Notice that the question asks for , and
by applying the trapezoid area formula. Fortunately, this is just
, and plugging in the value of
, we get that
.
~Math_Genius_164
Solution 2(LOC and Trig)
Call AD and BC . Draw diagonal AC and call the foot of the perpendicular from B to AC
. Call the foot of the perpendicular from A to line BC F, and call the foot of the perpindicular from A to DC H. Triangles CBG and CAF are similar, and we get that
=
Therefore,
. It then follows that triangles ABF and ADH are similar. Using similar triangles, we can then find that
. Using the Law of Cosine on ABC, We can find that the cosine of angle ABC is
a=\frac{27\sqrt{2}}{2}
K
\frac{567\sqrt{2}}{2}
\sqrt{2}
\boxed{567}$.
See also
2021 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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.