Difference between revisions of "2019 AMC 8 Problems/Problem 24"
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==Solution 1== | ==Solution 1== | ||
Draw <math>X</math> so that <math>XD</math> is parallel to <math>BC</math>. That makes triangles <math>BEF</math> and <math>EXD</math> congruent since <math>BE</math>=<math>ED</math>. <math>FC</math>=3<math>XD</math> so <math>BC</math>=4<math>BC</math>. Since <math>AF</math>=3<math>EF</math>( <math>XE</math>=<math>EF</math> and <math>AX</math>=1/3<math>AF</math>, so <math>XE</math>=<math>EF</math>=1/3<math>AF</math>), the altitude of triangle <math>BEF</math> is equal to 1/3 of the altitude of <math>ABC</math>. The area of <math>ABC</math> is 360, so the area of <math>BEF</math>=1/3*1/4*360=<math>\boxed{(B) 30}</math>~heeeeeeeheeeee | Draw <math>X</math> so that <math>XD</math> is parallel to <math>BC</math>. That makes triangles <math>BEF</math> and <math>EXD</math> congruent since <math>BE</math>=<math>ED</math>. <math>FC</math>=3<math>XD</math> so <math>BC</math>=4<math>BC</math>. Since <math>AF</math>=3<math>EF</math>( <math>XE</math>=<math>EF</math> and <math>AX</math>=1/3<math>AF</math>, so <math>XE</math>=<math>EF</math>=1/3<math>AF</math>), the altitude of triangle <math>BEF</math> is equal to 1/3 of the altitude of <math>ABC</math>. The area of <math>ABC</math> is 360, so the area of <math>BEF</math>=1/3*1/4*360=<math>\boxed{(B) 30}</math>~heeeeeeeheeeee | ||
+ | |||
+ | ==Solution 2 (Mass Points)== | ||
+ | <asy> | ||
+ | /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ | ||
+ | import graph; size(7cm); | ||
+ | real labelscalefactor = 0.5; /* changes label-to-point distance */ | ||
+ | pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ | ||
+ | pen dotstyle = black; /* point style */ | ||
+ | real xmin = -6.28, xmax = 6.28, ymin = -5.49, ymax = 5.73; /* image dimensions */ | ||
+ | pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); | ||
+ | /* draw figures */ | ||
+ | draw((0.28,2.39)--(-2.8,-1.17), linewidth(2) + wrwrwr); | ||
+ | draw((-2.8,-1.17)--(3.78,-1.05), linewidth(2) + wrwrwr); | ||
+ | draw((3.78,-1.05)--(0.28,2.39), linewidth(2) + wrwrwr); | ||
+ | draw((-2.8,-1.17)--(1.2887445398528459,1.3985482236874887), linewidth(2) + wrwrwr); | ||
+ | draw((0.28,2.39)--(-0.7199623188673492,-1.1320661821070033), linewidth(2) + wrwrwr); | ||
+ | draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); | ||
+ | draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); | ||
+ | draw(circle((4.48,-1.28), 0.46818799642878506), linewidth(2) + wrwrwr); | ||
+ | draw(circle((1.98,1.56), 0.46818799642878495), linewidth(2) + wrwrwr); | ||
+ | draw(circle((-3.36,-1.62), 0.46818799642878517), linewidth(2) + wrwrwr); | ||
+ | draw(circle((0.16,0.14), 0.46818799642878495), linewidth(2) + wrwrwr); | ||
+ | draw(circle((-0.74,-1.81), 0.46818799642878495), linewidth(2) + wrwrwr); | ||
+ | /* dots and labels */ | ||
+ | dot((0.28,2.39),dotstyle); | ||
+ | label("$A$", (0.36,2.59), NE * labelscalefactor); | ||
+ | dot((-2.8,-1.17),dotstyle); | ||
+ | label("$B$", (-2.72,-0.97), NE * labelscalefactor); | ||
+ | dot((3.78,-1.05),dotstyle); | ||
+ | label("$C$", (3.86,-0.85), NE * labelscalefactor); | ||
+ | dot((1.2887445398528459,1.3985482236874887),dotstyle); | ||
+ | label("$D$", (1.36,1.59), NE * labelscalefactor); | ||
+ | dot((-0.7199623188673492,-1.1320661821070033),dotstyle); | ||
+ | label("$F$", (-0.64,-0.93), NE * labelscalefactor); | ||
+ | dot((-0.2815567696989588,0.41208536204620183),linewidth(4pt) + dotstyle); | ||
+ | label("$E$", (-0.2,0.57), NE * labelscalefactor); | ||
+ | label("2", (-0.18,2.81), NE * labelscalefactor,wrwrwr); | ||
+ | label("1", (4.4,-1.39), NE * labelscalefactor,wrwrwr); | ||
+ | label("3", (1.9,1.45), NE * labelscalefactor,wrwrwr); | ||
+ | label("3", (-3.44,-1.73), NE * labelscalefactor,wrwrwr); | ||
+ | label("6", (0.08,0.03), NE * labelscalefactor,wrwrwr); | ||
+ | label("4", (-0.82,-1.93), NE * labelscalefactor,wrwrwr); | ||
+ | clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); | ||
+ | /* end of picture */ | ||
+ | </asy> | ||
+ | |||
+ | 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. | ||
+ | |||
+ | First, we assign a mass of <math>2</math> to point <math>A</math>. We figure out that <math>C</math> has a mass of <math>1</math> since <math>2\times1 = 1\times2</math>. Then, by adding <math>1+2 = 3</math>, we get that point <math>D</math> has a mass of 3. By equality, point <math>B</math> has a mass of 3 also. | ||
+ | |||
+ | Now, we add <math>3+3 = 6</math> for point <math>E</math> and <math>3+1 = 4</math> for point <math>F</math>. | ||
+ | |||
+ | Now, <math>BF</math> is a common base for triangles <math>ABF</math> and <math>EBF</math>, so we figure out that the ratios of the areas is the ratios of the heights which is <math>\frac{AE}{EF} = 2:1</math>. So, <math>EBF</math>'s area is one third the area of <math>ABF</math>, and we know the area of <math>ABF</math> is <math>\frac{1}{4}</math> the area of <math>ABC</math> since they have the same heights but different bases. | ||
+ | |||
+ | So we get the area of <math>EBF</math> as <math>\frac{1}{3}\times\frac{1}{4}\times360 = \boxed{B}30)</math> | ||
+ | -Brudder | ||
+ | Note: We can also find the ratios of the areas using the reciprocal of the product of the mass points of <math>EBF</math> over the product of the mass points of <math>ABC</math> which is <math>\frac{2\times3\times1}{3\times6\times4}\times360</math> which also yields <math>\boxed{B}</math> | ||
+ | -Brudder | ||
==See Also== | ==See Also== |
Revision as of 20:37, 20 November 2019
Problem 24
In triangle , point
divides side
s that
. Let
be the midpoint of
and left
be the point of intersection of line
and line
. Given that the area of
is
, what is the area of
?
Solution 1
Draw so that
is parallel to
. That makes triangles
and
congruent since
=
.
=3
so
=4
. Since
=3
(
=
and
=1/3
, so
=
=1/3
), the altitude of triangle
is equal to 1/3 of the altitude of
. The area of
is 360, so the area of
=1/3*1/4*360=
~heeeeeeeheeeee
Solution 2 (Mass Points)
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.
First, we assign a mass of to point
. We figure out that
has a mass of
since
. Then, by adding
, we get that point
has a mass of 3. By equality, point
has a mass of 3 also.
Now, we add for point
and
for point
.
Now, is a common base for triangles
and
, so we figure out that the ratios of the areas is the ratios of the heights which is
. So,
's area is one third the area of
, and we know the area of
is
the area of
since they have the same heights but different bases.
So we get the area of as
-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
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.