2021 AIME I Problems/Problem 9
Contents
- 1 Problem
- 2 Diagram
- 3 Solution 1 (Similar Triangles and Pythagorean Theorem)
- 4 Solution 2 (Similar Triangles and Pythagorean Theorem)
- 5 Solution 3 (Similar Triangles and Pythagorean Theorem)
- 6 Solution 4 (Cyclic Quadrilaterals, Similar Triangles, Pythagorean Theorem)
- 7 Solution 5 (Cyclic Quadrilaterals, Similar Triangles, Pythagorean Theorem)
- 8 Solution 6 (Similar Triangles and Trigonometry)
- 9 Solution 7 (Similar Triangles and Trigonometry)
- 10 Solution 8 (Heron's Formula)
- 11 Video Solution
- 12 See Also
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
Solution 1 (Similar Triangles and Pythagorean Theorem)
Let and
be the perpendiculars from
to
and
respectively. Next, let
be the intersection of
and
We set and
as shown below.
From here, we obtain
by segment subtraction, and
and
by the Pythagorean Theorem.
Since and
are both complementary to
we have
from which
by AA. It follows that
so
or
Since
by vertical angles, we have
by AA, with the ratio of similitude
It follows that
Since by angle chasing, we have
by AA, with the ratio of similitude
It follows that
By the Pythagorean Theorem on right we have
or
Solving this system of equations (
and
), we get
and
so
and
Finally, the area of
is
from which
~MRENTHUSIASM
Solution 2 (Similar Triangles and Pythagorean Theorem)
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 3 (Similar Triangles and Pythagorean Theorem)
Make perpendicular to
;
perpendicular to
;
perpendicular
.
It's obvious that . Let
. Then make
perpendicular to
, it's easy to get
.
Since parallel to
,
, so
. After drawing the altitude, it's obvious that
, so
. According to the property of similar triangles,
. So,
, or
.
Now, we see the , pretty easy to find that
, then we get
, then express
into
form that
we put the length of
back to
:
. So,
After calculating, we can have a final equation of
. It's easy to find
then
. So,
.
~bluesoul
Solution 4 (Cyclic Quadrilaterals, Similar Triangles, Pythagorean Theorem)
Let the foot of the altitude from to
be
, to
be
, and to
be
.
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
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
. 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 5 (Cyclic Quadrilaterals, Similar Triangles, Pythagorean Theorem)
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 quadrilateral:
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 and Trigonometry)
Let . Draw diagonal
and let
be the foot of the perpendicular from
to
,
be the foot of the perpendicular from
to line
, and
be the foot of the perpendicular from
to
.
Note that , and we get that
. Therefore,
. It then follows that
. Using similar triangles, we can then find that
. Using the Law of Cosines on
, We can find that the
. Since
, and each is supplementary to
, we know that the
. It then follows that
. Then it can be found that the area
is
. Multiplying this by
, the answer is
.
~happykeeper
Solution 7 (Similar Triangles and Trigonometry)
Draw the distances in terms of , as shown in the diagram. By similar triangles,
. As a result, let
, then
and
. The triangle
is
which
. By angle subtraction,
. Therefore,
and
. By trapezoid area formula, the area of
is equal to
which
.
~math2718281828459
Solution 8 (Heron's Formula)
Let the points formed by dropping altitudes from to the lines
,
, and
be
,
, and
, respectively.
We have
and
For convenience, let . By Heron's formula on
, we have sides
and semiperimeter
, so
so
.
Then,
and
Finally, recalling that is isosceles,
so
.
Video Solution
https://www.youtube.com/watch?v=6rLnl8z7lnM
See Also
2021 AIME I (Problems • Answer Key • Resources) | ||
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Followed by Problem 10 | |
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