Difference between revisions of "2021 AIME I Problems/Problem 9"
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Thus, <math>a=\frac{27}{\sqrt{2}}</math>, and <math>[ABCD]=\frac{AB+CD}{2} \cdot 18 = \frac{\frac{5a}{6} +\frac{9a}{6}}{2} \cdot 18 = 18 \cdot \frac{7}{6} \cdot \frac{27}{\sqrt{2}} = \frac{567}{\sqrt{2}}</math>, yielding <math>\boxed{567}</math>. | Thus, <math>a=\frac{27}{\sqrt{2}}</math>, and <math>[ABCD]=\frac{AB+CD}{2} \cdot 18 = \frac{\frac{5a}{6} +\frac{9a}{6}}{2} \cdot 18 = 18 \cdot \frac{7}{6} \cdot \frac{27}{\sqrt{2}} = \frac{567}{\sqrt{2}}</math>, yielding <math>\boxed{567}</math>. | ||
+ | |||
+ | ==Solution 6 (Similar Triangles, Two Variables, Two Equations) | ||
+ | <b><i>I am saving progress as I type this solution. Will finish up very soon. Please do not edit this solution. A million thanks.</i></b> | ||
+ | |||
+ | ~MRENTHUSIASM | ||
==See also== | ==See also== |
Revision as of 17:40, 16 March 2021
Contents
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
Diagram
~MRENTHUSIASM (by Geometry Expressions)
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
. 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
. Then it can be found that the area
is
. Multiplying this by
, the answer is
.
-happykeeper
Solution 3 (Similarity)
Let the foot of the altitude from A to BC be P, to CD be Q, and to BD be R.
Note that all isosceles trapezoids are cyclic quadrilaterals; thus, is on the circumcircle of
and we have that
is the Simson Line from
. As
, we have that
, with the last equality coming from cyclic quadrilateral
. Thus,
and we have that
or that
, which we can see gives us that
. Further ratios using the same similar triangles gives that
and
.
We also see that quadrilaterals and
are both cyclic, with diameters of the circumcircles being
and
respectively. The intersection of the circumcircles are the points
and
, and we know
and
are both line segments passing through an intersection of the two circles with one endpoint on each circle. By Fact 5, we know then that there exists a spiral similarity with center A taking
to
. Because we know a lot about
but very little about
and we would like to know more, we wish to find the ratio of similitude between the two triangles.
To do this, we use the one number we have for : we know that the altitude from
to
has length 10. As the two triangles are similar, if we can find the height from
to
, we can take the ratio of the two heights as the ratio of similitude. To do this, we once again note that
. Using this, we can drop the altitude from
to
and let it intersect
at
. Then, let
and thus
. We then have by the Pythagorean Theorem on
and
:
Then, . This gives us then from right triangle
that
and thus the ratio of
to
is
. From this, we see then that
and
The Pythagorean Theorem on
then gives that
Then, we have the height of trapezoid is
, the top base is
, and the bottom base is
. From the equation of a trapezoid,
, so the answer is
.
- lvmath
Solution 4 (Cool Solution by advanture)
First, draw the diagram. Then, notice that since is isosceles,
, and the length of the altitude from
to
is also
. Let the foot of this altitude be
, and let the foot of the altitude from
to
be denoted as
. Then,
. So,
. Now, notice that
, where
denotes the area of triangle
. Letting
, this equality becomes
. Also, from
, we have
. Now, by the Pythagorean theorem on triangles
and
, we have
and
. Notice that
, so
. Squaring both sides of the equation once, moving
and
to the right, dividing both sides by
, and squaring the equation once more, we are left with
. Dividing both sides by
(since we know
is positive), we are left with
. Solving for
gives us
.
Now, let the foot of the perpendicular from to
be
. Then let
. Let the foot of the perpendicular from
to
be
. Then,
is also equal to
. Notice that
is a rectangle, so
. Now, we have
. By the Pythagorean theorem applied to
, we have
. We know that
, so we can plug this into this equation. Solving for
, we get
.
Finally, to find , we use the formula for the area of a trapezoid:
. The problem asks us for
, which comes out to be
.
~advanture
Solution 5 (Compact similarity solution)
Let and
be the feet of the altitudes from
to
and
, respectively.
Claim: We have pairs of similar right triangles:
and
.
Proof: Note that is cyclic. We need one more angle, and we get this from this cyclic quad:
Let . We obtain from the similarities
and
.
By Ptolemy, , so
.
We obtain , so
.
Applying the Pythagorean theorem on , we get
.
Thus, , and
, yielding
.
==Solution 6 (Similar Triangles, Two Variables, Two Equations) I am saving progress as I type this solution. Will finish up very soon. Please do not edit this solution. A million thanks.
~MRENTHUSIASM
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.