Difference between revisions of "2022 AIME I Problems/Problem 11"
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− | ==Solution== | + | ==Solution 1== |
Let the circle tangent to <math>BC,AD,AB</math> at <math>P,Q,M</math> separately, denote that <math>\angle{ABC}=\angle{D}=\alpha</math> | Let the circle tangent to <math>BC,AD,AB</math> at <math>P,Q,M</math> separately, denote that <math>\angle{ABC}=\angle{D}=\alpha</math> | ||
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~bluesoul | ~bluesoul | ||
+ | |||
+ | ==Solution 2== | ||
+ | Denote by <math>O</math> the center of the circle. Denote by <math>r</math> the radius of the circle. | ||
+ | Denote by <math>E</math>, <math>F</math>, <math>G</math> the points that the circle meets <math>AB</math>, <math>CD</math>, <math>AD</math> at, respectively. | ||
+ | |||
+ | Because the circle is tangent to <math>AD</math>, <math>CB</math>, <math>AB</math>, <math>OE = OF = OG = r</math>, <math>OE \perp AD</math>, <math>OF \perp CB</math>, <math>OG \perp AB</math>. | ||
+ | |||
+ | Because <math>AD \parallel CB</math>, <math>E</math>, <math>O</math>, <math>F</math> are collinear. | ||
+ | |||
+ | Following from the power of a point, <math>AG^2 = AE^2 = AP \cdot AQ</math>. Hence, <math>AG = AE = 6</math>. | ||
+ | |||
+ | Following from the power of a point, <math>CF^2 = CQ \cdot CP</math>. Hence, <math>CF = 20</math>. | ||
+ | |||
+ | Denote <math>BG = x</math>. Because <math>DG</math> and <math>DF</math> are tangents to the circle, <math>BF = x</math>. | ||
+ | |||
+ | Because <math>AEFB</math> is a right trapezoid, <math>AB^2 = EF^2 + \left( AE - BF \right)^2</math>. | ||
+ | Hence, <math>\left( 6 + x \right)^2 = 4 r^2 + \left( 6 - x \right)^2</math>. | ||
+ | This can be simplified as | ||
+ | \[ | ||
+ | 6 x = r^2 . \hspace{1cm} (1) | ||
+ | \] | ||
+ | |||
+ | In <math>\triangle ACB</math>, 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 <math>AC = AP + PQ + QC = 28</math>, we get <math>x = \frac{9}{2}</math>. | ||
+ | Plugging this into Equation (1), we get <math>r = 3 \sqrt{3}</math>. | ||
+ | |||
+ | Therefore, | ||
+ | \begin{align*} | ||
+ | {\rm Area} \ ABCD & = CB \cdot EF \\ | ||
+ | & = \left( 20 + x \right) \cdot 2r \\ | ||
+ | & = 147 \sqrt{3} . | ||
+ | \end{align*} | ||
+ | |||
+ | Therefore, the answer is <math>147 + 3 = \boxed{\textbf{(150) }}</math>. | ||
+ | |||
+ | ~Steven Chen (www.professorchenedu.com) | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://www.youtube.com/watch?v=FeM_xXiJj0c&t=1s | ||
+ | |||
+ | ~Steven Chen (www.professorchenedu.com) | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2022|n=I|num-b=10|num-a=12}} | {{AIME box|year=2022|n=I|num-b=10|num-a=12}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 23:57, 17 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)
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.