Difference between revisions of "2019 AMC 8 Problems/Problem 24"
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− | Then <math>\triangle ADG \sim \triangle CDB</math> and <math>\triangle AEG \sim \triangle FEB</math>. Since <math> | + | Then <math>\triangle ADG \sim \triangle CDB</math> and <math>\triangle AEG \sim \triangle FEB</math>. Since <math>CD = 2AD</math>, triangle <math>CDB</math> has four times the area of triangle <math>ADG</math>. Since <math>[CDB] = 240</math>, we get <math>[ADG] = 60</math>. |
Since <math>[AED]</math> is also <math>60</math>, we have <math>ED = DG</math> because triangles <math>AED</math> and <math>ADG</math> have the same height and same areas and so their bases must be the congruent. Thus triangle <math>AEG</math> has twice the side lengths and therefore four times the area of triangle <math>BEF</math>, giving <math>[BEF] = (60+60)/4 = \boxed{\textbf{(B) }30}</math>. | Since <math>[AED]</math> is also <math>60</math>, we have <math>ED = DG</math> because triangles <math>AED</math> and <math>ADG</math> have the same height and same areas and so their bases must be the congruent. Thus triangle <math>AEG</math> has twice the side lengths and therefore four times the area of triangle <math>BEF</math>, giving <math>[BEF] = (60+60)/4 = \boxed{\textbf{(B) }30}</math>. | ||
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</asy> | </asy> | ||
(Credit to MP8148 for the idea) | (Credit to MP8148 for the idea) | ||
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==Solution 5 (Area Ratios)== | ==Solution 5 (Area Ratios)== | ||
<asy> | <asy> |
Revision as of 00:05, 21 November 2019
Contents
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 on such 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
Solution 3
is equal to . The area of triangle is equal to because it is equal to on 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 4 (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 5 (Area Ratios)
As before we figure out the areas labeled in the diagram. Then we note that Solving gives . (Credit to scrabbler94 for the idea)
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.