Difference between revisions of "2018 AMC 8 Problems/Problem 20"
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− | ==Problem | + | ==Problem== |
In <math>\triangle ABC,</math> a point <math>E</math> is on <math>\overline{AB}</math> with <math>AE=1</math> and <math>EB=2.</math> Point <math>D</math> is on <math>\overline{AC}</math> so that <math>\overline{DE} \parallel \overline{BC}</math> and point <math>F</math> is on <math>\overline{BC}</math> so that <math>\overline{EF} \parallel \overline{AC}.</math> What is the ratio of the area of <math>CDEF</math> to the area of <math>\triangle ABC?</math> | In <math>\triangle ABC,</math> a point <math>E</math> is on <math>\overline{AB}</math> with <math>AE=1</math> and <math>EB=2.</math> Point <math>D</math> is on <math>\overline{AC}</math> so that <math>\overline{DE} \parallel \overline{BC}</math> and point <math>F</math> is on <math>\overline{BC}</math> so that <math>\overline{EF} \parallel \overline{AC}.</math> What is the ratio of the area of <math>CDEF</math> to the area of <math>\triangle ABC?</math> | ||
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==Solution 1== | ==Solution 1== | ||
− | By similar triangles, we have <math>[ADE] = \frac{1}{9}[ABC]</math>. Similarly, we see that | + | By similar triangles, we have <math>[ADE] = \frac{1}{9}[ABC]</math>. Similarly, we see that [mathjax][BEF] = \dfrac{4}{9}[ABC][/mathjax]. Using this information, we get <cmath>[ACFE] = \frac{5}{9}[ABC].</cmath> Then, since <math>[ADE] = \frac{1}{9}[ABC]</math>, it follows that the [mathjax][CDEF] = \dfrac{4}{9}[ABC][/mathjax]. Thus, the answer would be <math>\boxed{\textbf{(A) } \frac{4}{9}}</math>. |
− | Sidenote: | + | Sidenote: [mathjax][ABC][/mathjax] denotes the area of triangle [mathjax]ABC[/mathjax]. Similarly, [mathjax][ABCD][/mathjax] denotes the area of figure [mathjax]ABCD[/mathjax]. |
==Solution 2== | ==Solution 2== | ||
− | We can extend | + | Let <math>a = DE</math> and <math>b =</math> the height of <math>\triangle ABC</math>. We can extend <math>\triangle ABC</math> to form a parallelogram, which would equal <math>3a \cdot 3b</math>. The smaller parallelogram is <math>a</math> times <math>2b</math>. The smaller parallelogram is <math>\frac{2}{9}</math> of the larger parallelogram, so the answer would be <math>\frac{2}{9} \cdot 2</math>, since the triangle is <math>\frac{1}{2}</math> of the parallelogram, so the answer is <math>\boxed{\textbf{(A) } \frac{4}{9}}</math>. |
By babyzombievillager with credits to many others who helped with the solution :D | By babyzombievillager with credits to many others who helped with the solution :D | ||
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==Solution 3== | ==Solution 3== | ||
− | <math>\triangle{ADE} \sim \triangle{ABC} \sim \triangle{EFB}</math>. We can substitute <math>\overline{DA}</math> as <math>\frac{1}{3}x</math> and <math>\overline{CD}</math> as <math>\frac{2}{3}x</math>, where <math>x</math> is <math>\overline{AC}</math>. Side <math>\overline{CB}</math> having, distance <math>y</math>, has <math>2</math> parts also. And <math>\overline{CF}</math> and <math>\overline{FB}</math> are <math>\frac{1}{3}y</math> and <math>\frac{2}{3}y</math> respectfully. You can consider the height of <math>\triangle{ADE}</math> and <math>\triangle{EFB}</math> as <math>z</math> and <math>2z</math> respectfully. The area of <math>\triangle{ADE}</math> is <math>\frac{1\cdot z}{2}=0.5z</math> because the area formula for a triangle is <math>\frac{1}{2}bh</math> or <math>\frac{bh}{2}</math>. The area of <math>\triangle{EFB}</math> will be <math>\frac{2\cdot 2z}{2}=2z</math>. So the area of <math>\triangle{ABC}</math> will be <math>\frac{3\cdot (2z+z)}{2}=\frac{3\cdot 3z}{2}=\frac{9z}{2}=4.5z</math>. The area of parallelogram <math>CDEF</math> will be <math>4.5z-(0.5z+2z)=4.5z-2.5z=2z</math>. Parallelogram <math>CDEF</math> to <math>\triangle{ABC}= \frac{2z}{4.5z}=\frac{2}{4.5}=\frac{4}{9}</math>. The answer is <math>\boxed{ | + | <math>\triangle{ADE} \sim \triangle{ABC} \sim \triangle{EFB}</math>. We can substitute <math>\overline{DA}</math> as <math>\frac{1}{3}x</math> and <math>\overline{CD}</math> as <math>\frac{2}{3}x</math>, where <math>x</math> is <math>\overline{AC}</math>. Side <math>\overline{CB}</math> having, distance <math>y</math>, has <math>2</math> parts also. And, <math>\overline{CF}</math> and <math>\overline{FB}</math> are <math>\frac{1}{3}y</math> and <math>\frac{2}{3}y</math> respectfully. You can consider the height of <math>\triangle{ADE}</math> and <math>\triangle{EFB}</math> as <math>z</math> and <math>2z</math> respectfully. The area of <math>\triangle{ADE}</math> is <math>\frac{1\cdot z}{2}=0.5z</math> because the area formula for a triangle is <math>\frac{1}{2}bh</math> or <math>\frac{bh}{2}</math>. The area of <math>\triangle{EFB}</math> will be <math>\frac{2\cdot 2z}{2}=2z</math>. So, the area of <math>\triangle{ABC}</math> will be <math>\frac{3\cdot (2z+z)}{2}=\frac{3\cdot 3z}{2}=\frac{9z}{2}=4.5z</math>. The area of parallelogram <math>CDEF</math> will be <math>4.5z-(0.5z+2z)=4.5z-2.5z=2z</math>. Parallelogram <math>CDEF</math> to <math>\triangle{ABC}= \frac{2z}{4.5z}=\frac{2}{4.5}=\frac{4}{9}</math>. The answer is <math>\boxed{\textbf{(A) } \frac{4}{9}}</math>. |
+ | ==Video Solution (CREATIVE ANALYSIS!!!)== | ||
+ | https://youtu.be/ayUmpmgFi3E | ||
− | ==Solution | + | ~Education, the Study of Everything |
− | + | ||
+ | == Video Solution (Meta-Solving Technique) == | ||
+ | https://youtu.be/GmUWIXXf_uk?t=1541 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | ==Video Solution 2== | ||
+ | https://youtu.be/V_-yIhs_Bps | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | ==Video Solution by SpreadTheMathLove== | ||
+ | https://www.youtube.com/watch?v=TpsuRedYOiM&t=250s | ||
==See Also== | ==See Also== |
Latest revision as of 01:41, 4 April 2024
Contents
Problem
In a point
is on
with
and
Point
is on
so that
and point
is on
so that
What is the ratio of the area of
to the area of
Solution 1
By similar triangles, we have . Similarly, we see that [mathjax][BEF] = \dfrac{4}{9}[ABC][/mathjax]. Using this information, we get
Then, since
, it follows that the [mathjax][CDEF] = \dfrac{4}{9}[ABC][/mathjax]. Thus, the answer would be
.
Sidenote: [mathjax][ABC][/mathjax] denotes the area of triangle [mathjax]ABC[/mathjax]. Similarly, [mathjax][ABCD][/mathjax] denotes the area of figure [mathjax]ABCD[/mathjax].
Solution 2
Let and
the height of
. We can extend
to form a parallelogram, which would equal
. The smaller parallelogram is
times
. The smaller parallelogram is
of the larger parallelogram, so the answer would be
, since the triangle is
of the parallelogram, so the answer is
.
By babyzombievillager with credits to many others who helped with the solution :D
Solution 3
. We can substitute
as
and
as
, where
is
. Side
having, distance
, has
parts also. And,
and
are
and
respectfully. You can consider the height of
and
as
and
respectfully. The area of
is
because the area formula for a triangle is
or
. The area of
will be
. So, the area of
will be
. The area of parallelogram
will be
. Parallelogram
to
. The answer is
.
Video Solution (CREATIVE ANALYSIS!!!)
~Education, the Study of Everything
Video Solution (Meta-Solving Technique)
https://youtu.be/GmUWIXXf_uk?t=1541
~ pi_is_3.14
Video Solution 2
~savannahsolver
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=TpsuRedYOiM&t=250s
See Also
2018 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.