Difference between revisions of "2022 AIME I Problems/Problem 11"
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− | + | Denote the center of the circle point <math>O</math>. Let points <math>M</math>, <math>N</math>, and <math>L</math> be the tangent points of lines <math>BC</math>, <math>AD</math>, and <math>AB</math> respectively to the circle. By Power of a Point, <math>16\cdot{25}=({MC})^2 \Longrightarrow MC=20</math>. Similarly, <math>({AL})^2=3\cdot{12} \Longrightarrow AL=6</math>. That means <math>AN</math> is also <math>6</math> since quadrilateral <math>LONA</math> is symmetrical. Let <math>AC</math> intersect <math>MN</math> at point <math>K</math>. Then, <math>\bigtriangleup{KMC}</math> is similar to <math>\bigtriangleup{AKN}</math>. Therefore, <math>\frac{CK}{MC}=\frac{AK}{AN}</math>. Let the length of <math>PK=l</math>, then <math>\frac{25-l}{20}=\frac{3+l}{6}</math>. Solving we get <math>l=\frac{45}{13}</math>. Doing the Pythagorean theorem on triangles <math>KMC</math> and <math>AKN</math> for sides <math>MK</math> and <math>KN</math> respectively, we obtain the equation <math>\sqrt{(\frac{280}{13})^2-400} +\sqrt{(\frac{84}{13})^2-36}=MN=2r_1</math> where <math>r_1</math> denotes the radius of the circle. Solving, we get <math>MN=6\sqrt{3}</math>. Additionally, quadrilateral <math>OLBM</math> is symmetrical so <math>OL=OM</math>. Let <math>OL=OM=x</math> and extend a perpendicular foot from <math>B</math> to <math>AD</math> and call it <math>R</math>. Then, <math>\bigtriangleup{ABR}</math> is right with <math>AR=6-x</math>, <math>AB=6+x</math>, and <math>RB=2r_1=MN=6\sqrt{3}</math>. Taking the difference of squares, we get <math>108=24x \Longrightarrow x=\frac{9}{2}</math>. The area of <math>ABCD</math> is <math>MN\cdot{BC}=(20+x)\cdot{MN} \Longrightarrow \frac{49}{2}\cdot{6\sqrt{3}}=147\sqrt{3}</math>. Therefore, the answer is <math>147+3=\boxed{150}</math> | |
− | Denote the center of the circle point <math>O</math>. Let points <math>M</math>, <math>N</math>, and <math>L</math> be the tangent points of lines <math>BC</math>, <math>AD</math>, and <math>AB</math> respectively to the circle. By Power of a Point, <math>16\cdot{25}=({MC})^2 | ||
~[https://artofproblemsolving.com/wiki/index.php/User:Magnetoninja Magnetoninja] | ~[https://artofproblemsolving.com/wiki/index.php/User:Magnetoninja Magnetoninja] |
Revision as of 00:11, 19 June 2023
Contents
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
Video Solution by Punxsutawney Phil
https://www.youtube.com/watch?v=1m3pqCgwLFE
Solution 1 (No trig)
Let's redraw the diagram, but extend some helpful lines.
We obviously see that we must use power of a point since they've given us lengths in a circle and there are intersection points. Let be our tangents from the circle to the parallelogram. By the secant power of a point, the power of
. Then
. Similarly, the power of
and
. We let
and label the diagram accordingly.
Notice that because . Let
be the center of the circle. Since
and
intersect
and
, respectively, at right angles, we have
is a right-angled trapezoid and more importantly, the diameter of the circle is the height of the triangle. Therefore, we can drop an altitude from
to
and
to
, and both are equal to
. Since
,
. Since
and
. We can now use Pythagorean theorem on
; we have
and
.
We know that because
is a parallelogram. Using Pythagorean theorem on
,
. Therefore, base
. Thus the area of the parallelogram is the base times the height, which is
and the answer is
~KingRavi
Solution 2
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,HarveyZhang
Solution 3
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
In , by applying the law of cosines, we have
Because , we get
.
Plugging this into Equation (1), we get
.
Therefore,
Therefore, the answer is .
~Steven Chen (www.professorchenedu.com)
Solution 4
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 Parallelogram Law,
. Solving for
yields
. Thus,
, for a final answer of
.
~ Leo.Euler
Solution 5
Let be the circle, let
be the radius of
, and let the points at which
is tangent to
,
, and
be
,
, and
, respectively. PoP on
and
with respect to
yields
Let
In
Area is
vladimir.shelomovskii@gmail.com, vvsss
Solution 6 (Short and Sweet)
Denote the center of the circle point . Let points
,
, and
be the tangent points of lines
,
, and
respectively to the circle. By Power of a Point,
. Similarly,
. That means
is also
since quadrilateral
is symmetrical. Let
intersect
at point
. Then,
is similar to
. Therefore,
. Let the length of
, then
. Solving we get
. Doing the Pythagorean theorem on triangles
and
for sides
and
respectively, we obtain the equation
where
denotes the radius of the circle. Solving, we get
. Additionally, quadrilateral
is symmetrical so
. Let
and extend a perpendicular foot from
to
and call it
. Then,
is right with
,
, and
. Taking the difference of squares, we get
. The area of
is
. Therefore, the answer is
Video Solution
https://www.youtube.com/watch?v=FeM_xXiJj0c&t=1s
~Steven Chen (www.professorchenedu.com)
Video Solution 2 (Mathematical Dexterity)
https://www.youtube.com/watch?v=1nDKQkr9NaU
Video Solution 3 by OmegaLearn
https://youtu.be/LpOegT0fKy8?t=740
~ pi_is_3.14
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.