Difference between revisions of "2021 Fall AMC 10B Problems/Problem 25"
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~Steven Chen (www.professorchenedu.com) | ~Steven Chen (www.professorchenedu.com) | ||
+ | |||
+ | ==Solution 3== | ||
+ | <asy> | ||
+ | size(8cm); | ||
+ | label("D",(0,0),SW); | ||
+ | label("A",(0,10),NW); | ||
+ | label("B",(10,10),NE); | ||
+ | label("C",(10,0),SE); | ||
+ | label("H",(1,5.7),SE); | ||
+ | label("O",(0,6),W); | ||
+ | label("I",(0,3),W); | ||
+ | draw((0,0)--(10,0)); | ||
+ | draw((0,0)--(0,10)); | ||
+ | draw((10,0)--(10,10)); | ||
+ | draw((0,10)--(10,10)); | ||
+ | draw((1,6)--(0,9)); | ||
+ | draw((0,9)--(3,10)); | ||
+ | draw((3,10)--(4,7)); | ||
+ | draw((4,7)--(1,6)); | ||
+ | draw((0,3)--(1,6)); | ||
+ | draw((1,6)--(10,3)); | ||
+ | draw((10,3)--(9,0)); | ||
+ | draw((9,0)--(0,3)); | ||
+ | draw((6,13/3)--(10,22/3)); | ||
+ | draw((10,22/3)--(8,10)); | ||
+ | draw((8,10)--(4,7)); | ||
+ | draw((4,7)--(6,13/3)); | ||
+ | draw((1,6)--(0,6)); | ||
+ | label("$3$",(9/2,3/2),N); | ||
+ | label("$3$",(11/2,9/2),S); | ||
+ | label("$1$",(1/2,9/2),E); | ||
+ | label("$1$",(19/2,3/2),W); | ||
+ | label("$1$",(1/2,15/2),E); | ||
+ | draw((0.3,6)--(0.3,5.7)--(0,5.7)); | ||
+ | label("$1$",(3/2,19/2),S); | ||
+ | label("$1$",(5/2,13/2),N); | ||
+ | label("$1$",(7/2,17/2),W); | ||
+ | label("$R$",(7,43/6),W); | ||
+ | label("E",(0,9),W); | ||
+ | label("F",(3,10),N); | ||
+ | label("P",(4,10),N); | ||
+ | label("L",(8,10),N); | ||
+ | draw((4,7)--(4,10)); | ||
+ | draw((4,9.7)--(4.3,9.7)--(4.3,10)); | ||
+ | label("G",(3.9,6.8),S); | ||
+ | label("N",(6.2,4.1),S); | ||
+ | label("M",(10,22/3),E); | ||
+ | label("Q",(10,13/3),E); | ||
+ | draw((10,13/3)--(6,13/3)); | ||
+ | draw((9.7,13/3)--(9.7,4.033333333333333333333333)--(10,4.03333333333333333333333333)); | ||
+ | label("K",(10,3),E); | ||
+ | label("J",(9,0),S); | ||
+ | </asy> | ||
+ | We will scale every number up by a factor of <math>\sqrt{10}.</math> This implies our final area will be <math>\frac{1}{10}</math> of the answer we receive. | ||
+ | |||
+ | We have | ||
+ | <cmath>FAE\sim EOH \sim IOH\sim JDI\sim KCJ \sim NQK\sim GPF.</cmath> | ||
+ | Let <math>AE=a</math> and <math>FA=b.</math> We have | ||
+ | <cmath>FP=AE=OH=JC=\frac13ID=a</cmath> | ||
+ | and | ||
+ | <cmath>PG=AF=EO=OI=KC=\frac13 DJ=b</cmath> | ||
+ | As <math>ABCD</math> is a square, we have <math>AD=DC</math> or | ||
+ | <cmath>a+2b+3a=3b+a \Rightarrow 3a=b.</cmath> | ||
+ | Since <math>a^2+b^2=10,</math> we have | ||
+ | <cmath>a=1,b=3.</cmath> | ||
+ | We have <math>\triangle GPL\cong \triangle MQN</math> which implies | ||
+ | <cmath>MQ=GP=3.</cmath> | ||
+ | Denote <math>QK=x.</math> As <math>\triangle NQK\sim \triangle JDI,</math> we have <math>NQ=3x.</math> | ||
+ | |||
+ | We have | ||
+ | <cmath>\begin{align*}BM&=BC-(CK+QK+MQ)\\ &=4-x.\end{align*}</cmath> | ||
+ | |||
+ | In addition, | ||
+ | <cmath>\begin{align*}LB&=AB-(AF+FP+PL)\\&=6-3x.\end{align*}</cmath> | ||
+ | |||
+ | Since <math>\triangle LBM\sim \triangle MQN,</math> we have | ||
+ | <cmath>\frac{LB}{BM}=\frac{MQ}{QN}\Rightarrow \frac{6-3x}{4-x}=\frac{3}{3x}=\frac{1}{x}.</cmath> | ||
+ | Simplifying we have | ||
+ | <cmath>3x^2-7x+4=0 \Rightarrow x=\frac43, 1.</cmath> | ||
+ | We have | ||
+ | <cmath>\begin{align*}[GLMN]&=MN\cdot LM\\ &= 3\sqrt{x^2+1}\cdot \sqrt{10x^2-44x+52}.\end{align*}</cmath> | ||
+ | |||
+ | Plugging in <math>x=1,</math> we have <math>[GLMN]=18.</math> | ||
+ | |||
+ | Plugging in <math>x=\frac43,</math> we have <math>[GLMN]=\frac{50}{3}.</math> | ||
+ | |||
+ | The sum of the two possible <math>R</math>s is | ||
+ | <cmath>\frac{1}{10}\cdot\frac{104}{3}=\frac{52}{15}.</cmath> | ||
+ | Hence, <math>52+15=\boxed{\textbf{(E) }67}.</math> | ||
+ | |||
+ | ~ASAB | ||
== Video Solution == | == Video Solution == |
Revision as of 10:12, 1 December 2021
Problem
A rectangle with side lengths and
a square with side length
and a rectangle
are inscribed inside a larger square as shown. The sum of all possible values for the area of
can be written in the form
, where
and
are relatively prime positive integers. What is
Solution 1
We see that the polygon bounded by the small square, large square, and rectangle of known lengths is an isosceles triangle. Let’s draw a perpendicular from the vertex of this triangle to its opposing side;
We see that this creates two congruent triangles. Let the smaller side of the triangle have length and let the larger side of the triangle have length
. Now we see by AAS congruency that if we draw perpendiculars that surround the smaller square, each outer triangle will be congruent to these two triangles.
Now notice that these small triangles are also similar to the large triangle bounded by the bigger square and the rectangle by AA, and the ratio of the sides are 1:3, so we can fill in the lengths of that triangle. Similarly, the small triangle on the right bounded by the rectangle and the square is also congruent to the other small triangles by AAS, so we can fill in those sides;
Since the larger square by definition has all equal sides, we can set the sum of the lengths of the sides equal to each other.
. Now let's draw some more perpendiculars and rename the side lengths.
By AA similarity, when we draw a perpendicular from the intersection of the two rectangles to the large square, we create a triangle below that is similar to the small congruent triangles with length . Since we don't know it's scale, we'll label it's sides
.
The triangle that is created above the perpendicular is congruent to the triangle on the opposite of the rectangle with unknown dimensions because they share the same hypotenuse and have two angles in common. Thus we can label these two triangles accordingly.
The side length of the big square is , so we can find the remaining dimensions of the triangle bounded by the rectangle with unknown dimensions and the large square in terms of
and
:
This triangle with side lengths and
is similar to the triangle directly below it with side lengths
and
by AA similarity, so we can set up a ratio equation:
.
There are two solutions to this equation;
and
. For the first solution, the triangle in the corner has sides
and
. Using Pythagorean theorem on that triangle, the hypotenuse has length
.
The triangle directly below has side lengths
and
in this case, so special right triangle yields the hypotenuse to be
.
The area of the rectangle is thus
.
For the second solution, the side lengths of the corner triangle are
and
, so the hypotenuse of the triangle is
. The triangle below that also has side lengths
and
, so it's hypotenuse is the same. Then the area of the rectangle is
.
The sum of the possible areas of the rectangle is therefore .
Using Pythagorean theorem on the original small congruent triangles, or
. Therefore the sum of the possible areas of the rectangle is
. Therefore
,
, and
~KingRavi
Solution 2
We use Image:2021_AMC_10B_(Nov)_Problem_25,_sol.png to facilitate our analysis.
Denote . Thus,
.
Hence,
and
Because ,
.
Hence,
.
Hence,
and
.
Hence, .
Now, we put the graph to a coordinate plane by setting point as the origin, putting
in the
-axis and
on the
-axis.
Hence, ,
,
,
,
,
,
,
,
.
Denote ,
.
Because is a rectangle,
. Hence,
.
We have
and
.
Hence,
Because is a rectangle,
and
.
Hence,
.
The equation of line is
Because point
is on line
, plugging the coordinates of
into the equation of line
, we get
By solving Equations (1) and (2), we get or
.
Case 1: .
We have and
.
Thus,
and
.
Therefore, .
Case 2: .
We have and
.
Thus,
and
.
Therefore, .
Putting these two cases together, the sum of all possible values of the area of is
.
Therefore, the answer is .
~Steven Chen (www.professorchenedu.com)
Solution 3
We will scale every number up by a factor of
This implies our final area will be
of the answer we receive.
We have
Let
and
We have
and
As
is a square, we have
or
Since
we have
We have
which implies
Denote
As
we have
We have
In addition,
Since we have
Simplifying we have
We have
Plugging in we have
Plugging in we have
The sum of the two possible s is
Hence,
~ASAB
Video Solution
https://www.youtube.com/watch?v=5mPvkipCvhE
See Also
2021 Fall AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Problem 1 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.