Difference between revisions of "2019 AMC 8 Problems/Problem 24"
(→Solution 11 (Graph Paper)) |
(Added another solution.) |
||
Line 424: | Line 424: | ||
<br> | <br> | ||
<b>Additional note:</b> There are many subtle variations of this triangle; this method is one of the more compact ones. ~[[User:i_equal_tan_90|i_equal_tan_90]] | <b>Additional note:</b> There are many subtle variations of this triangle; this method is one of the more compact ones. ~[[User:i_equal_tan_90|i_equal_tan_90]] | ||
+ | |||
+ | ==Solution 12== | ||
+ | <asy> | ||
+ | unitsize(2cm); | ||
+ | pair A,B,C,DD,EE,FF,G; | ||
+ | B = (0,0); C = (3,0); | ||
+ | A = (1.2,1.7); | ||
+ | DD = (2/3)*A+(1/3)*C; | ||
+ | EE = (B+DD)/2; | ||
+ | FF = intersectionpoint(B--C,A--A+2*(EE-A)); | ||
+ | G = (1.5,0); | ||
+ | draw(A--B--C--cycle); | ||
+ | draw(A--FF); | ||
+ | draw(B--DD); | ||
+ | draw(G--DD); | ||
+ | label("$A$",A,N); | ||
+ | label("$B$", | ||
+ | B,SW); | ||
+ | label("$C$",C,SE); | ||
+ | label("$D$",DD,NE); | ||
+ | label("$E$",EE,NW); | ||
+ | label("$F$",FF,S); | ||
+ | label("$G$",G,S); | ||
+ | </asy> | ||
+ | We know that <math>AD = \dfrac{1}{3} AC</math>, so <math>[ABD] = \dfrac{1}{3} [ABC] = 120</math>. Using the same method, since <math>BE = \dfrac{1}{2} BD</math>, <math>[ABE] = \dfrac{1}{2} [ABD] = 60</math>. Next, we draw <math>G</math> on <math>\overline{BC}</math> such that <math>\overline{DG}</math> is parallel to <math>\overline{AF}</math> and create segment <math>DG</math>. We then observe that <math>\triangle AFC \sim \triangle DGC</math>, and since <math>AD:DC = 1:2</math>, <math>FG:GC</math> is also equal to <math>1:2</math>. Similarly (no pun intended), <math>\triangle DBG \sim \triangle EBF</math>, and since <math>BE:ED = 1:1</math>, <math>BF:FG</math> is also equal to <math>1:1</math>. Combining the information in these two ratios, we find that <math>BF:FG:GC = 1:1:2</math>, or equivalently, <math>BF = \dfrac{1}{4} BC</math>. Thus, <math>[BFA] = \dfrac{1}{4} [BCA] = 90</math>. We already know that <math>[ABE] = 60</math>, so the area of <math>\triangle EBF</math> is <math>[BFA] - [ABE] = \boxed{\textbf{(B) }30}</math>. ~[[User:i_equal_tan_90|i_equal_tan_90]] | ||
==See Also== | ==See Also== |
Revision as of 10:13, 10 January 2020
Contents
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
Draw on such that is parallel to . That makes triangles and congruent since . so . Since ( and , so ), the altitude of triangle is equal to of the altitude of . The area of is , so the area of ~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 . By equality, point has a mass of 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)
Solution 6 (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 7
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 8
The diagram is very inaccurate.
Note: All numbers above and are ratios, not actual side lengths. If somebody could edit this to make the diagram more accurate it would be greatly appreicated.
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 9
Labeling the areas in the diagram, we have:
so .
So our answer is . ~~RWhite
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. ~i_equal_tan_90
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 . ~i_equal_tan_90
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.