Difference between revisions of "2019 AMC 8 Problems/Problem 24"
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<math>\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40</math> | <math>\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40</math> | ||
− | == Solution 1 | + | == Solution 1 == |
We use the line-segment ratios to infer area ratios and height ratios. | We use the line-segment ratios to infer area ratios and height ratios. | ||
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The area of <math>ABC</math> is <math>360</math>, so the area of<math> \triangle EBF=\frac{1}{3} \cdot \frac{1}{4} \cdot 360=\boxed{\textbf{(B) }30}</math>. | The area of <math>ABC</math> is <math>360</math>, so the area of<math> \triangle EBF=\frac{1}{3} \cdot \frac{1}{4} \cdot 360=\boxed{\textbf{(B) }30}</math>. | ||
− | ==Solution 3 == | + | ==Solution 3(Mass point method, AMC10 tricks) == |
<asy> | <asy> | ||
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ | /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ | ||
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Finally, <math>F</math> balances <math>B</math> and <math>C</math> so <math>F=B+C=3+1=4</math>. We can confirm we have done everything right by noting that <math>E</math> balances <math>A</math> and <math>F</math>, so <math>E</math> should equal <math>A+F</math>, which it does. | Finally, <math>F</math> balances <math>B</math> and <math>C</math> so <math>F=B+C=3+1=4</math>. We can confirm we have done everything right by noting that <math>E</math> balances <math>A</math> and <math>F</math>, so <math>E</math> should equal <math>A+F</math>, which it does. | ||
− | Now that our points have weights, we can solve the problem. <math>BF:FC=1:3</math> so <math>BF:BC=1:4</math> so <math>[ABF]=\frac{1}{4}[ABC]=90</math>. Also, <math>EF:EA=2:4=1:2</math> so <math>EF:AF=1:3</math> so <math>[EBF]=\frac{1}{3}[ABF]= | + | Now that our points have weights, we can solve the problem. <math>BF:FC=1:3</math> so <math>BF:BC=1:4</math> so <math>[ABF]=\frac{1}{4}[ABC]=90</math>. Also, <math>EF:EA=2:4=1:2</math> so <math>EF:AF=1:3</math> so <math>[EBF]=\frac{1}{3}[ABF]=\boxed{\textbf{(B) }30}</math>. |
− | |||
− | |||
-Firebolt360 and Brudder | -Firebolt360 and Brudder | ||
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==Solution 4== | ==Solution 4== | ||
− | <math>\frac{BF}{FC}</math> is equal to <math>\frac{\textrm{The area of triangle ABE}}{\textrm{The area of triangle ACE}}</math>. The area of triangle <math>ABE</math> is equal to <math>60</math> because it is equal to | + | <math>\frac{BF}{FC}</math> is equal to <math>\frac{\textrm{The area of triangle ABE}}{\textrm{The area of triangle ACE}}</math>. The area of triangle <math>ABE</math> is equal to <math>60</math> because it is equal to one-half of the area of triangle <math>ABD</math>, which is equal to one-third of the area of triangle <math>ABC</math>, which is <math>360</math>. The area of triangle <math>ACE</math> is the sum of the areas of triangles <math>AED</math> and <math>CED</math>, which is respectively <math>60</math> and <math>120</math>. So, <math>\frac{BF}{FC}</math> is equal to <math>\frac{60}{180}</math>=<math>\frac{1}{3}</math>, so the area of triangle <math>ABF</math> is <math>90</math>. That minus the area of triangle <math>ABE</math> is <math>\boxed{\textbf{(B) }30}</math>. |
~~SmileKat32 | ~~SmileKat32 | ||
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==Solution 13 (Fastest Solution if you have no time)== | ==Solution 13 (Fastest Solution if you have no time)== | ||
− | The picture is misleading. Assume that the triangle ABC is right. | + | The picture is misleading. Assume that the triangle ABC is a right triangle. |
Then, find two factors of <math>720</math> that are the closest together so that the picture becomes easier in your mind. Quickly searching for squares near <math>720</math> to use difference of squares, we find <math>24</math> and <math>30</math> as our numbers. Then, the coordinates of D are <math>(10,16)</math>(note, A=0,0). E is then <math>(5,8)</math>. Then the equation of the line AE is <math>-16x/5+24=y</math>. Plugging in <math>y=0</math>, we have <math>x=\dfrac{15}{2}</math>. Now notice that we have both the height and the base of EBF. | Then, find two factors of <math>720</math> that are the closest together so that the picture becomes easier in your mind. Quickly searching for squares near <math>720</math> to use difference of squares, we find <math>24</math> and <math>30</math> as our numbers. Then, the coordinates of D are <math>(10,16)</math>(note, A=0,0). E is then <math>(5,8)</math>. Then the equation of the line AE is <math>-16x/5+24=y</math>. Plugging in <math>y=0</math>, we have <math>x=\dfrac{15}{2}</math>. Now notice that we have both the height and the base of EBF. | ||
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==Video Solutions== | ==Video Solutions== | ||
+ | |||
+ | ==Video Solution by Math-X (Let's do this step by step!!!)== | ||
+ | https://youtu.be/IgpayYB48C4?si=ypjLOH-vfgQr8Neu&t=7358 | ||
+ | |||
+ | ~Math-X | ||
https://www.youtube.com/watch?v=AY4mByrL8v0 | https://www.youtube.com/watch?v=AY4mByrL8v0 | ||
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~Hayabusa1 | ~Hayabusa1 | ||
+ | |||
+ | ==Video Solution by SpreadTheMathLove== | ||
+ | https://www.youtube.com/watch?v=AY4mByrL8v0&feature=youtu.be | ||
==See Also== | ==See Also== |
Latest revision as of 15:26, 23 August 2024
Contents
- 1 Problem 24
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3(Mass point method, AMC10 tricks)
- 5 Solution 4
- 6 Solution 5 (Similar Triangles)
- 7 Solution 6 (Area Ratios)
- 8 Solution 7 (Coordinate Bashing)
- 9 Solution 8
- 10 Solution 9
- 11 Solution 10 (Menelaus's Theorem)
- 12 Solution 11 (Graph Paper)
- 13 Solution 12
- 14 Solution 13 (Fastest Solution if you have no time)
- 15 Solution 14
- 16 Solution 15 - Geometry & Algebra
- 17 Solution 16 (Straightfoward & Simple Solution)
- 18 Solution 17
- 19 Note
- 20 Video Solutions
- 21 Video Solution by Math-X (Let's do this step by step!!!)
- 22 Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)
- 23 Video Solution by SpreadTheMathLove
- 24 See Also
Problem 24
In triangle , point
divides side
so that
. Let
be the midpoint of
and let
be the point of intersection of line
and line
. Given that the area of
is
, what is the area of
?
Solution 1
We use the line-segment ratios to infer area ratios and height ratios.
Areas:
.
.
Heights:
Let = height (of altitude) from
to
.
from
to
is
.
from
to
is
.
Conclusion:
, and also
.
So, , and thus,
.
Solution 2
Draw on
such that
is parallel to
.
Triangles and
are similar, and since
, they are also congruent, and so
and
.
implies
, so
,
.
Since ,
, and since
, all of these are equal to
, and so the altitude of triangle
is equal to
of the altitude of
.
The area of is
, so the area of
.
Solution 3(Mass point method, AMC10 tricks)
First, when we see the problem, we see ratios, and we see that this triangle basically has no special properties (right, has medians, etc.) and this screams mass points at us.
The triangle we will consider is (obviously), and we will let
be the center of mass, so that
balances
and
(this is true since
balances
and
, but
also balances
and
and
so
balances
and
), and
balances
and
.
We know that and
balances
and
so we assign
to
and
to
. Then, since
balances
and
, we get
(by mass points addition).
Next, since balances
and
in a ratio of
, we know that
. Similarly, by mass points addition,
.
Finally, balances
and
so
. We can confirm we have done everything right by noting that
balances
and
, so
should equal
, which it does.
Now that our points have weights, we can solve the problem. so
so
. Also,
so
so
.
-Firebolt360 and Brudder
Note: We can also find the ratios of the areas using the reciprocal of the product of the mass points of over the product of the mass points of
which is
which also yields
.
-Brudder
Solution 4
is equal to
. The area of triangle
is equal to
because it is equal to one-half of the area of triangle
, which is equal to one-third of the area of triangle
, which is
. The area of triangle
is the sum of the areas of triangles
and
, which is respectively
and
. So,
is equal to
=
, so the area of triangle
is
. That minus the area of triangle
is
.
~~SmileKat32
Solution 5 (Similar Triangles)
Extend to
such that
as shown:
Then,
and
. Since
, triangle
has four times the area of triangle
. Since
, we get
.
Since is also
, we have
because triangles
and
have the same height and same areas and so their bases must be the congruent. Thus, triangle
has twice the side lengths and therefore four times the area of triangle
, giving
.
(Credit to MP8148 for the idea)
Solution 6 (Area Ratios)
As before, we figure out the areas labeled in the diagram. Then, we note that
Even simpler:
Solving gives
.
(Credit to scrabbler94 for the idea)
Solution 7 (Coordinate Bashing)
Let be a right triangle, and
Let
The line can be described with the equation
The line can be described with
Solving, we get and
Now we can find
-Trex4days
Solution 8
Let =
(the median divides the area of the triangle into two equal parts)
Construction: Draw a circumcircle around with
as is diameter. Extend
to
such that it meets the circle at
. Draw line
.
(Since
is cyclic)
But is common in both with an area of 60. So,
.
Therefore (SAS Congruency Theorem).
In , let
be the median of
,
which means .
Rotate to meet
at
and
at
.
will fit exactly in
(both are radii of the circle). From the above solutions,
.
is a radius and
is half of it implies
=
,
which means .
Thus, .
~phoenixfire & flamewavelight
Solution 9
Using the ratio of
and
, we find the area of
is
and the area of
is
. Also using the fact that
is the midpoint of
, we know
.
Let
be a point such
is parellel to
. We immediatley know that
by
. Using that we can conclude
has ratio
. Using
, we get
. Therefore using the fact that
is in
, the area has ratio
and we know
has area
so
is
.
- fath2012
Solution 10 (Menelaus's Theorem)
By Menelaus's Theorem on triangle
, we have
Therefore,
Solution 11 (Graph Paper)
Note: If graph paper is unavailable, this solution can still be used by constructing a small grid on a sheet of blank paper.
As triangle is loosely defined, we can arrange its points such that the diagram fits nicely on a coordinate plane. By doing so, we can construct it on graph paper and be able to visually determine the relative sizes of the triangles.
As point splits line segment
in a
ratio, we draw
as a vertical line segment
units long. Point
is thus
unit below point
and
units above point
. By definition, Point
splits line segment
in a
ratio, so we draw
units long directly left of
and draw
directly between
and
,
unit away from both.
We then draw line segments and
. We can easily tell that triangle
occupies
square units of space. Constructing line
and drawing
at the intersection of
and
, we can easily see that triangle
forms a right triangle occupying
of a square unit of space.
The ratio of the areas of triangle and triangle
is thus
, and since the area of triangle
is
, this means that the area of triangle
is
.
~emerald_block
Additional note: There are many subtle variations of this triangle; this method is one of the more compact ones.
Solution 12
We know that
, so
. Using the same method, since
,
. Next, we draw
on
such that
is parallel to
and create segment
. We then observe that
, and since
,
is also equal to
. Similarly (no pun intended),
, and since
,
is also equal to
. Combining the information in these two ratios, we find that
, or equivalently,
. Thus,
. We already know that
, so the area of
is
.
Solution 13 (Fastest Solution if you have no time)
The picture is misleading. Assume that the triangle ABC is a right triangle.
Then, find two factors of that are the closest together so that the picture becomes easier in your mind. Quickly searching for squares near
to use difference of squares, we find
and
as our numbers. Then, the coordinates of D are
(note, A=0,0). E is then
. Then the equation of the line AE is
. Plugging in
, we have
. Now notice that we have both the height and the base of EBF.
Solving for the area, we have .
Solution 14
, so
has area
and
has area
.
so the area of
is equal to the area of
.
Draw
parallel to
.
Set area of BEF = . BEF is similar to BDG in ratio of 1:2
so area of BDG = , area of EFDG=
, and area of CDG
.
CDG is similar to CAF in ratio of 2:3 so area CDG = area CAF, and area AFDG=
area CDG.
Thus, and
.
~EFrame
Solution 15 - Geometry & Algebra
We draw line so that we can define a variable
for the area of
. Knowing that
and
share both their height and base, we get that
.
Since we have a rule where 2 triangles, ( which has base
and vertex
), and (
which has Base
and vertex
)who share the same vertex (which is vertex
in this case), and share a common height, their relationship is : Area of
(the length of the two bases), we can list the equation where
. Substituting
into the equation we get:
and we now have that
.
~
Solution 16 (Straightfoward & Simple Solution)
Since
thus
Similarly,
Now, since is a midpoint of
,
We can use the fact that is a midpoint of
even further. Connect lines
and
so that
and
share 2 sides.
We know that since
is a midpoint of
Let's label
. We know that
is
since
Note that with this information now, we can deduct more things that are needed to finish the solution.
Note that because of triangles
and
We want to find
This is a simple equation, and solving we get
~mathboy282, an expanded solution of Solution 5, credit to scrabbler94 for the idea.
Solution 17
Because and
is the midpoint of
, we know that the areas of
and
are
and the areas of
and
are
.
Note
This question is extremely similar to 1971 AHSME Problems/Problem 26.
Video Solutions
Video Solution by Math-X (Let's do this step by step!!!)
https://youtu.be/IgpayYB48C4?si=ypjLOH-vfgQr8Neu&t=7358
~Math-X
https://www.youtube.com/watch?v=AY4mByrL8v0
Associated video
https://www.youtube.com/watch?v=DMNbExrK2oo
https://www.youtube.com/watch?v=nm-Vj_fsXt4
- Happytwin (Another video solution)
https://www.youtube.com/watch?v=nyevg9w-CCI&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=6
~ MathEx
https://www.youtube.com/watch?v=m04K0Q2SNXY&t=1s
~savannahsolver
Video Solution by The Power of Logic(Problem 1 to 25 Full Solution)
~Hayabusa1
Video Solution by SpreadTheMathLove
https://www.youtube.com/watch?v=AY4mByrL8v0&feature=youtu.be
See Also
2019 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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.