Difference between revisions of "2022 AIME I Problems/Problem 11"
m |
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Line 5: | Line 5: | ||
defaultpen(linewidth(0.6)+fontsize(11)); | defaultpen(linewidth(0.6)+fontsize(11)); | ||
size(8cm); | size(8cm); | ||
− | pair A,B,C,D,P,Q; | + | |
+ | pair A,B,C,D,P,Q,O; | ||
A=(0,0); | A=(0,0); | ||
label("$A$", A, SW); | label("$A$", A, SW); | ||
− | B=(6 | + | B=(1.5,6*sqrt(3)); |
label("$B$", B, NW); | label("$B$", B, NW); | ||
− | C=( | + | C=(26,6*sqrt(3)); |
label("$C$", C, NE); | label("$C$", C, NE); | ||
− | D=(24,0); | + | D=(24.5,0); |
label("$D$", D, SE); | label("$D$", D, SE); | ||
− | P=( | + | P=(3.1,2.9); |
− | label("$P$", (5 | + | label("$P$", P, S); |
− | + | Q=(9.5,6); | |
− | label("$ | + | label("$Q$", Q, SE); |
+ | O=(6,3*sqrt(3)); | ||
+ | |||
+ | label("$3$", (1.3,2.2), S); | ||
+ | label("$9$", (6.5,4.3), S); | ||
+ | label("$16$", (17.5,8.9), S); | ||
+ | |||
draw(A--B--C--D--cycle); | draw(A--B--C--D--cycle); | ||
draw(C--A); | draw(C--A); | ||
− | draw( | + | draw(circle(O, 3*sqrt(3))); |
− | dot(A^^B^^C^^D | + | |
+ | dot(A^^B^^C^^D); | ||
+ | dot(intersectionpoints(circle(O, 3*sqrt(3)), A--C)); | ||
</asy> | </asy> | ||
Revision as of 01:52, 29 December 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
.
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
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
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.