Difference between revisions of "2022 AIME I Problems/Problem 11"
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~Steven Chen (www.professorchenedu.com) | ~Steven Chen (www.professorchenedu.com) | ||
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+ | ==Solution 3== | ||
+ | Let <math>\omega</math> be the circle, let <math>r</math> be the radius of <math>\omega</math>, and let the points at which <math>\omega</math> is tangent to <math>AB</math>, <math>BC</math>, and <math>AD</math> be <math>X</math>, <math>Y</math>, and <math>Z</math>, respectively. Note that PoP on <math>A</math> and <math>C</math> with respect to <math>\omega</math> yields <math>AX=6</math> and <math>CY=20</math>. We can compute the area of <math>ABC</math> in two ways: | ||
+ | |||
+ | 1. By the half-base-height formula, <math>[ABC]=r(20+BX)</math>. | ||
+ | |||
+ | 2. We can drop altitudes from the center <math>O</math> of <math>\omega</math> to <math>AB</math>, <math>BC</math>, and <math>AC</math>, which have lengths <math>r</math>, <math>r</math>, and <math>\sqrt{r^2-81/4}</math>. Thus, <math>[ABC]=[OAB]+[OBC]+[OAC]=r(BX+13)+14\sqrt{r^2-81/4}</math>. | ||
+ | |||
+ | Equating the two expressions for <math>[ABC]</math> and solving for <math>r</math> yields <math>r=3\sqrt{3}</math>. Let <math>BX=BY=a</math>. By the Pythagorean Theorem, <math>(6-a)^2+(2r)^2=(a+6)^2</math>. Solving for <math>a</math> yields <math>a=9/2</math>. Thus, <math>[ABCD]=2[ABC]=2r(20+a)=147\sqrt{3}</math>, for a final answer of <math>\boxed{150}</math>. | ||
==Video Solution== | ==Video Solution== |
Revision as of 14:16, 18 February 2022
Problem
Let be a parallelogram with
. A circle tangent to sides
,
, and
intersects diagonal
at points
and
with
, as shown. Suppose that
,
, and
. Then the area of
can be expressed in the form
, where
and
are positive integers, and
is not divisible by the square of any prime. Find
.
Solution 1
Let the circle tangent to at
separately, denote that
Using POP, it is very clear that , let
, using LOC in
,
, similarly, use LOC in
, getting that
. We use the second equation to minus the first equation, getting that
, we can get
.
Now applying LOC in , getting
, solving this equation to get
, then
,
, the area is
leads to
~bluesoul
Solution 2
Denote by the center of the circle. Denote by
the radius of the circle.
Denote by
,
,
the points that the circle meets
,
,
at, respectively.
Because the circle is tangent to ,
,
,
,
,
,
.
Because ,
,
,
are collinear.
Following from the power of a point, . Hence,
.
Following from the power of a point, . Hence,
.
Denote . Because
and
are tangents to the circle,
.
Because is a right trapezoid,
.
Hence,
.
This can be simplified as
\[
6 x = r^2 . \hspace{1cm} (1)
\]
In , by applying the law of cosines, we have
\begin{align*}
AC^2 & = AB^2 + CB^2 - 2 AB \cdot CB \cos B \\
& = AB^2 + CB^2 + 2 AB \cdot CB \cos A \\
& = AB^2 + CB^2 + 2 AB \cdot CB \cdot \frac{AE - BF}{AB} \\
& = AB^2 + CB^2 + 2 CB \left( AE - BF \right) \\
& = \left( 6 + x \right)^2 + \left( 20 + x \right)^2 + 2 \left( 20 + x \right) \left( 6 - x \right) \\
& = 24 x + 676 .
\end{align*}
Because , we get
.
Plugging this into Equation (1), we get
.
Therefore, \begin{align*} {\rm Area} \ ABCD & = CB \cdot EF \\ & = \left( 20 + x \right) \cdot 2r \\ & = 147 \sqrt{3} . \end{align*}
Therefore, the answer is .
~Steven Chen (www.professorchenedu.com)
Solution 3
Let be the circle, let
be the radius of
, and let the points at which
is tangent to
,
, and
be
,
, and
, respectively. Note that PoP on
and
with respect to
yields
and
. We can compute the area of
in two ways:
1. By the half-base-height formula, .
2. We can drop altitudes from the center of
to
,
, and
, which have lengths
,
, and
. Thus,
.
Equating the two expressions for and solving for
yields
. Let
. By the Pythagorean Theorem,
. Solving for
yields
. Thus,
, for a final answer of
.
Video Solution
https://www.youtube.com/watch?v=FeM_xXiJj0c&t=1s
~Steven Chen (www.professorchenedu.com)
See Also
2022 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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.