Difference between revisions of "2020 AMC 12B Problems/Problem 18"
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<math>\textbf{(A) } \frac{7}{3} \qquad \textbf{(B) } 8-4\sqrt2 \qquad \textbf{(C) } 1+\sqrt2 \qquad \textbf{(D) } \frac{7}{4}\sqrt2 \qquad \textbf{(E) } 2\sqrt2</math> | <math>\textbf{(A) } \frac{7}{3} \qquad \textbf{(B) } 8-4\sqrt2 \qquad \textbf{(C) } 1+\sqrt2 \qquad \textbf{(D) } \frac{7}{4}\sqrt2 \qquad \textbf{(E) } 2\sqrt2</math> | ||
− | + | == Solution 1 == | |
− | |||
Since the total area is <math>4</math>, the side length of square <math>ABCD</math> is <math>2</math>. We see that since triangle <math>HAE</math> is a right isosceles triangle with area 1, we can determine sides <math>HA</math> and <math>AE</math> both to be <math>\sqrt{2}</math>. Now, consider extending <math>FB</math> and <math>IE</math> until they intersect. Let the point of intersection be <math>K</math>. We note that <math>EBK</math> is also a right isosceles triangle with side <math>2-\sqrt{2}</math> and find it's area to be <math>3-2\sqrt{2}</math>. Now, we notice that <math>FIK</math> is also a right isosceles triangle and find it's area to be <math>\frac{1}{2}</math><math>FI^2</math>. This is also equal to <math>1+3-2\sqrt{2}</math> or <math>4-2\sqrt{2}</math>. Since we are looking for <math>FI^2</math>, we want two times this. That gives <math>\boxed{\textbf{(B)}\ 8-4\sqrt{2}}</math>.~TLiu | Since the total area is <math>4</math>, the side length of square <math>ABCD</math> is <math>2</math>. We see that since triangle <math>HAE</math> is a right isosceles triangle with area 1, we can determine sides <math>HA</math> and <math>AE</math> both to be <math>\sqrt{2}</math>. Now, consider extending <math>FB</math> and <math>IE</math> until they intersect. Let the point of intersection be <math>K</math>. We note that <math>EBK</math> is also a right isosceles triangle with side <math>2-\sqrt{2}</math> and find it's area to be <math>3-2\sqrt{2}</math>. Now, we notice that <math>FIK</math> is also a right isosceles triangle and find it's area to be <math>\frac{1}{2}</math><math>FI^2</math>. This is also equal to <math>1+3-2\sqrt{2}</math> or <math>4-2\sqrt{2}</math>. Since we are looking for <math>FI^2</math>, we want two times this. That gives <math>\boxed{\textbf{(B)}\ 8-4\sqrt{2}}</math>.~TLiu | ||
− | + | == Solution 2 == | |
Since this is a geometry problem involving sides, and we know that <math>HE</math> is <math>2</math>, we can use our ruler and find the ratio between <math>FI</math> and <math>HE</math>. Measuring(on the booklet), we get that <math>HE</math> is about <math>1.8</math> inches and <math>FI</math> is about <math>1.4</math> inches. Thus, we can then multiply the length of <math>HE</math> by the ratio of <math>\frac{1.4}{1.8},</math> of which we then get <math>FI= \frac{14}{9}.</math> We take the square of that and get <math>\frac{196}{81},</math> and the closest answer to that is <math>\boxed{\textbf{(B)}\ 8-4\sqrt{2}}</math>. ~Celloboy (Note that this is just a strategy I happened to use that worked. Do not press your luck with this strategy, for it was a lucky guess) | Since this is a geometry problem involving sides, and we know that <math>HE</math> is <math>2</math>, we can use our ruler and find the ratio between <math>FI</math> and <math>HE</math>. Measuring(on the booklet), we get that <math>HE</math> is about <math>1.8</math> inches and <math>FI</math> is about <math>1.4</math> inches. Thus, we can then multiply the length of <math>HE</math> by the ratio of <math>\frac{1.4}{1.8},</math> of which we then get <math>FI= \frac{14}{9}.</math> We take the square of that and get <math>\frac{196}{81},</math> and the closest answer to that is <math>\boxed{\textbf{(B)}\ 8-4\sqrt{2}}</math>. ~Celloboy (Note that this is just a strategy I happened to use that worked. Do not press your luck with this strategy, for it was a lucky guess) | ||
− | + | == Solution 3 == | |
Draw the auxiliary line <math>AC</math>. Denote by <math>M</math> the point it intersects with <math>HE</math>, and by <math>N</math> the point it intersects with <math>GF</math>. Last, denote by <math>x</math> the segment <math>FN</math>, and by <math>y</math> the segment <math>FI</math>. We will find two equations for <math>x</math> and <math>y</math>, and then solve for <math>y^2</math>. | Draw the auxiliary line <math>AC</math>. Denote by <math>M</math> the point it intersects with <math>HE</math>, and by <math>N</math> the point it intersects with <math>GF</math>. Last, denote by <math>x</math> the segment <math>FN</math>, and by <math>y</math> the segment <math>FI</math>. We will find two equations for <math>x</math> and <math>y</math>, and then solve for <math>y^2</math>. | ||
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Solving for <math>y^2</math> gives <math>\boxed{\textbf{(B)}\ 8-4\sqrt{2}}</math> ~DrB | Solving for <math>y^2</math> gives <math>\boxed{\textbf{(B)}\ 8-4\sqrt{2}}</math> ~DrB | ||
− | + | == Solution 4 == | |
Plot a point <math>F'</math> such that <math>F'I</math> and <math>AB</math> are parallel and extend line <math>FB</math> to point <math>B'</math> such that <math>FIB'F'</math> forms a square. Extend line <math>AE</math> to meet line <math>F'B'</math> and point <math>E'</math> is the intersection of the two. The area of this square is equivalent to <math>FI^2</math>. We see that the area of square <math>ABCD</math> is <math>4</math>, meaning each side is of length 2. The area of the pentagon <math>EIFF'E'</math> is <math>2</math>. Length <math>AE=\sqrt{2}</math>, thus <math>EB=2-\sqrt{2}</math>. Triangle <math>EB'E'</math> is isosceles, and the area of this triangle is <math>\frac{1}{2}(4-2\sqrt{2})(2-\sqrt{2})=6-4\sqrt{2}</math>. Adding these two areas, we get <cmath>2+6-4\sqrt{2}=8-4\sqrt{2}\rightarrow \boxed{\mathrm{(B)}}</cmath>. --OGBooger | Plot a point <math>F'</math> such that <math>F'I</math> and <math>AB</math> are parallel and extend line <math>FB</math> to point <math>B'</math> such that <math>FIB'F'</math> forms a square. Extend line <math>AE</math> to meet line <math>F'B'</math> and point <math>E'</math> is the intersection of the two. The area of this square is equivalent to <math>FI^2</math>. We see that the area of square <math>ABCD</math> is <math>4</math>, meaning each side is of length 2. The area of the pentagon <math>EIFF'E'</math> is <math>2</math>. Length <math>AE=\sqrt{2}</math>, thus <math>EB=2-\sqrt{2}</math>. Triangle <math>EB'E'</math> is isosceles, and the area of this triangle is <math>\frac{1}{2}(4-2\sqrt{2})(2-\sqrt{2})=6-4\sqrt{2}</math>. Adding these two areas, we get <cmath>2+6-4\sqrt{2}=8-4\sqrt{2}\rightarrow \boxed{\mathrm{(B)}}</cmath>. --OGBooger | ||
− | + | == Solution 5 (HARD Calculation) == | |
We can easily observe that the area of square <math>ABCD</math> is 4 and its side length is 2 since all four regions that build up the square has area 1. | We can easily observe that the area of square <math>ABCD</math> is 4 and its side length is 2 since all four regions that build up the square has area 1. | ||
Extend <math>FI</math> and let the intersection with <math>AB</math> be <math>K</math>. Connect <math>AC</math>, and let the intersection of <math>AC</math> and <math>HE</math> be <math>L</math>. | Extend <math>FI</math> and let the intersection with <math>AB</math> be <math>K</math>. Connect <math>AC</math>, and let the intersection of <math>AC</math> and <math>HE</math> be <math>L</math>. | ||
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-edit: annabelle0913 | -edit: annabelle0913 | ||
− | ==Video Solution== | + | ==Video Solution 1== |
https://www.youtube.com/watch?v=AKJXB07Sat0&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=7 ~ MathEx | https://www.youtube.com/watch?v=AKJXB07Sat0&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=7 ~ MathEx | ||
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==See Also== | ==See Also== | ||
+ | {{AMC10 box|year=2020|ab=B|num-b=20|num-a=22}} | ||
{{AMC12 box|year=2020|ab=B|num-b=17|num-a=19}} | {{AMC12 box|year=2020|ab=B|num-b=17|num-a=19}} | ||
− | |||
[[Category:Intermediate Geometry Problems]] | [[Category:Intermediate Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 11:07, 24 December 2020
Contents
[hide]Problem
In square , points
and
lie on
and
, respectively, so that
Points
and
lie on
and
, respectively, and points
and
lie on
so that
and
. See the figure below. Triangle
, quadrilateral
, quadrilateral
, and pentagon
each has area
What is
?
Solution 1
Since the total area is , the side length of square
is
. We see that since triangle
is a right isosceles triangle with area 1, we can determine sides
and
both to be
. Now, consider extending
and
until they intersect. Let the point of intersection be
. We note that
is also a right isosceles triangle with side
and find it's area to be
. Now, we notice that
is also a right isosceles triangle and find it's area to be
. This is also equal to
or
. Since we are looking for
, we want two times this. That gives
.~TLiu
Solution 2
Since this is a geometry problem involving sides, and we know that is
, we can use our ruler and find the ratio between
and
. Measuring(on the booklet), we get that
is about
inches and
is about
inches. Thus, we can then multiply the length of
by the ratio of
of which we then get
We take the square of that and get
and the closest answer to that is
. ~Celloboy (Note that this is just a strategy I happened to use that worked. Do not press your luck with this strategy, for it was a lucky guess)
Solution 3
Draw the auxiliary line . Denote by
the point it intersects with
, and by
the point it intersects with
. Last, denote by
the segment
, and by
the segment
. We will find two equations for
and
, and then solve for
.
Since the overall area of is
, and
. In addition, the area of
.
The two equations for and
are then:
Length of
:
Area of CMIF:
.
Substituting the first into the second, yields
Solving for gives
~DrB
Solution 4
Plot a point such that
and
are parallel and extend line
to point
such that
forms a square. Extend line
to meet line
and point
is the intersection of the two. The area of this square is equivalent to
. We see that the area of square
is
, meaning each side is of length 2. The area of the pentagon
is
. Length
, thus
. Triangle
is isosceles, and the area of this triangle is
. Adding these two areas, we get
. --OGBooger
Solution 5 (HARD Calculation)
We can easily observe that the area of square is 4 and its side length is 2 since all four regions that build up the square has area 1.
Extend
and let the intersection with
be
. Connect
, and let the intersection of
and
be
.
Notice that since the area of triangle
is 1 and
,
, therefore
.
Let
, then
.
Also notice that
, thus
.
Now use the condition that the area of quadrilateral
is 1, we can set up the following equation:
We solve the equation and yield
.
Now notice that
.
Hence
. -HarryW
-edit: annabelle0913
Video Solution 1
https://www.youtube.com/watch?v=AKJXB07Sat0&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=7 ~ MathEx
Video Solution 2 by the Beauty of Math
Solution starts at 3:09: https://youtu.be/VZYe3Hu88OA
See Also
2020 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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 |
2020 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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 12 Problems and Solutions |
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