Difference between revisions of "2020 AMC 12B Problems/Problem 18"
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Plot a point <math>F'</math> such that <math>F'</math> and <math>I</math> are collinear 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 F'B' 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 quadrilateral <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 $2+6-4*\sqrt{2}=8-4*\sqrt{2}} | Plot a point <math>F'</math> such that <math>F'</math> and <math>I</math> are collinear 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 F'B' 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 quadrilateral <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 $2+6-4*\sqrt{2}=8-4*\sqrt{2}} | ||
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+ | {{AMC12 box|year=2020|ab=B|num-b=12|num-a=14}} | ||
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Revision as of 22:31, 7 February 2020
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
Plot a point such that
and
are collinear and extend line
to point
such that
forms a square. Extend line
to meet line F'B' 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 quadrilateral
is
. Length
, thus
. Triangle
is isosceles, and the area of this triangle is
. Adding these two areas, we get $2+6-4*\sqrt{2}=8-4*\sqrt{2}}
2020 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 12 |
Followed by Problem 14 |
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