Difference between revisions of "2021 AIME I Problems/Problem 2"
MRENTHUSIASM (talk | contribs) m (Video solutions should be at the end. Also, the order of solutions should be first-come-first-serve.) |
|||
Line 97: | Line 97: | ||
Multiplying, (35/4)*3 = 105/4. m+n = 105+4 = 109. ~ishanvannadil2008 | Multiplying, (35/4)*3 = 105/4. m+n = 105+4 = 109. ~ishanvannadil2008 | ||
+ | |||
+ | Any integer can only be <math>0mod3</math>, <math>1mod3</math>, or <math>2mod3</math>. There are <math>C(5, 3) = 10</math> sums of 3 of the integers out of 5. | ||
+ | |||
+ | |||
+ | ==Solution 6 (Trigbash)== | ||
+ | |||
+ | Let the intersection of <math>AE</math> and <math>BC</math> be <math>G</math>. It is useful to find <math>tan(\angle DAE)</math>, because <math>tan(\angle DAE)=\frac{3}{BG}</math> and <math>\frac{3}{tan(\angle DAE)}=BG</math>. From there, subtracting the areas of the two triangles from the larger rectangle, we get Area = <math>33-3BG=33-\frac{9}{tan(\angle DAE)}</math>. | ||
+ | |||
+ | let <math>\angle CAD</math> = α. Let <math>\angle CAE</math> = β. Note, α+β=<math>\angle DAE</math>. | ||
+ | |||
+ | α = arctan(<math>\frac{3}{11}</math>) | ||
+ | |||
+ | β = arctan(<math>\frac{7}{9}</math>) | ||
+ | |||
+ | tan(<math>\angle DAE</math>)= tan(arctan(<math>\frac{3}{11}</math>)+arctan(<math>\frac{7}{9}</math>)) = | ||
+ | |||
+ | <math>\frac{\frac{3}{11}+\frac{7}{9}}{1-\frac{3}{11}\cdot\frac{7}{9}}</math> = <math>\frac{\frac{104}{99}}{\frac{78}{99}}</math> = <math>\frac{4}{3}</math> | ||
+ | |||
+ | Area= <math>33-\frac{9}{\frac{4}{3}}</math> = <math>33-\frac{27}{4 }= \frac{105}{4}</math>. The answer is <math>105+4=109</math>. | ||
+ | ~ twotothetenthis1024 | ||
==Video Solution by Punxsutawney Phil== | ==Video Solution by Punxsutawney Phil== |
Revision as of 21:23, 12 March 2021
Contents
Problem
In the diagram below, is a rectangle with side lengths
and
, and
is a rectangle with side lengths
and
as shown. The area of the shaded region common to the interiors of both rectangles is
, where
and
are relatively prime positive integers. Find
.
Solution 1 (Similar Triangles)
Let be the intersection of
and
.
From vertical angles, we know that
. Also, given that
and
are rectangles, we know that
.
Therefore, by AA similarity, we know that triangles
and
are similar.
Let . Then, we have
. By similar triangles, we know that
and
. We have
.
Solving for , we have
.
The area of the shaded region is just
.
Thus, the answer is
.
~yuanyuanC
Solution 2 (Coordinate Geometry Bash)
Suppose It follows that
Since is a rectangle, we have
and
The equation of the circle with center
and radius
is
and the equation of the circle with center
and radius
is
We now have a system of two equations with two variables. Expanding and rearranging respectively give
Subtracting
from
we get
Simplifying and rearranging produce
Substituting
into
gives
which is a quadratic of
We clear fractions by multiplying both sides by
then solve by factoring:
Since
is in Quadrant IV, we have
It follows that the equation of
is
Let be the intersection of
and
and
be the intersection of
and
Since
is the
-intercept of
we obtain
By symmetry, quadrilateral is a parallelogram. Its area is
and the requested sum is
~MRENTHUSIASM
Solution 3 (Pythagorean Theorem)
Let the intersection of and
be
, and let
, so
.
By the Pythagorean theorem, , so
, and thus
.
By the Pythagorean theorem again, :
Solving, we get , so the area of the parallelogram is
, and
.
~JulianaL25
Solution 4 (Similar triangles and area)
Again, let the intersection of and
be
. By AA similarity,
with a
ratio. Define
as
. Because of similar triangles,
. Using
, the area of the parallelogram is
. Using
, the area of the parallelogram is
. These equations are equal, so we can solve for
and obtain
. Thus,
, so the area of the parallelogram is
.
~mathboy100
Solution 5
The intersection of AD and FC = P. The intersection of AE and BC = K. Let's set AP to x. CK also has to be x because of the properties of a parallelogram. Then PD and BK must be 11-x because the sum of the segments has to be 11. We can easily solve for PC by the Pythagorean Theorem. DC^2 + PD^2 = PC^2. 9 + (11-x)^2 = PC^2. After 10 seconds of simplifying, we get that PC = sqrt(x^2-22x+30).
FC = 9, and FP + PC = 9. PC = sqrt(x^2-22x+30), so FP = 9 - sqrt(x^2-22x+30).
Now we can apply the Pythagorean Theorem to triangle AFP. AF^2 + FP^2 = AP^2. 49 + (9 - (sqrt(x^2-22x+30)))^2 = x^2. After simplifying (took me 2 minutes on the test), we get that x = 35/4. If you don't believe me, then plug it into WolframAlpha.
Now we have to solve for the area of APCK. We know that the height is 3 because the height of the parallelogram is the same as the height of the smaller rectangle.
The area of a parallelogram is base * height. The base is x (or 35/4), and the height is 3.
Multiplying, (35/4)*3 = 105/4. m+n = 105+4 = 109. ~ishanvannadil2008
Any integer can only be ,
, or
. There are
sums of 3 of the integers out of 5.
Solution 6 (Trigbash)
Let the intersection of and
be
. It is useful to find
, because
and
. From there, subtracting the areas of the two triangles from the larger rectangle, we get Area =
.
let = α. Let
= β. Note, α+β=
.
α = arctan()
β = arctan()
tan()= tan(arctan(
)+arctan(
)) =
=
=
Area= =
. The answer is
.
~ twotothetenthis1024
Video Solution by Punxsutawney Phil
https://youtube.com/watch?v=H17E9n2nIyY&t=289s
Video Solution
https://youtu.be/M3DsERqhiDk?t=275
See also
2021 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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.