Difference between revisions of "2022 AIME I Problems/Problem 11"
(→Solution 5) |
(→Solution 3) |
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Line 117: | Line 117: | ||
Hence, <math>\left( 6 + x \right)^2 = 4 r^2 + \left( 6 - x \right)^2</math>. | Hence, <math>\left( 6 + x \right)^2 = 4 r^2 + \left( 6 - x \right)^2</math>. | ||
This can be simplified as | This can be simplified as | ||
− | \[ | + | <cmath>\[ |
6 x = r^2 . \hspace{1cm} (1) | 6 x = r^2 . \hspace{1cm} (1) | ||
− | \] | + | \]</cmath> |
In <math>\triangle ACB</math>, by applying the law of cosines, we have | In <math>\triangle ACB</math>, by applying the law of cosines, we have | ||
− | \begin{align*} | + | <cmath>\begin{align*} |
AC^2 & = AB^2 + CB^2 - 2 AB \cdot CB \cos B \\ | 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 \cos A \\ | ||
Line 129: | Line 129: | ||
& = \left( 6 + x \right)^2 + \left( 20 + x \right)^2 + 2 \left( 20 + x \right) \left( 6 - x \right) \\ | & = \left( 6 + x \right)^2 + \left( 20 + x \right)^2 + 2 \left( 20 + x \right) \left( 6 - x \right) \\ | ||
& = 24 x + 676 . | & = 24 x + 676 . | ||
− | \end{align*} | + | \end{align*}</cmath> |
Because <math>AC = AP + PQ + QC = 28</math>, we get <math>x = \frac{9}{2}</math>. | Because <math>AC = AP + PQ + QC = 28</math>, we get <math>x = \frac{9}{2}</math>. | ||
Line 135: | Line 135: | ||
Therefore, | Therefore, | ||
− | \begin{align*} | + | <cmath>\begin{align*} |
{\rm Area} \ ABCD & = CB \cdot EF \\ | {\rm Area} \ ABCD & = CB \cdot EF \\ | ||
& = \left( 20 + x \right) \cdot 2r \\ | & = \left( 20 + x \right) \cdot 2r \\ | ||
& = 147 \sqrt{3} . | & = 147 \sqrt{3} . | ||
− | \end{align*} | + | \end{align*}</cmath> |
Therefore, the answer is <math>147 + 3 = \boxed{\textbf{(150) }}</math>. | Therefore, the answer is <math>147 + 3 = \boxed{\textbf{(150) }}</math>. |
Revision as of 12:01, 31 May 2022
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
.
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
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
~vvsss (www.deoma-cmd.ru)
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
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.