2019 AMC 8 Problems/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 ?
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 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
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
The line can be described with the equation
The line can be described with
Solving, we get and
Now we can find
(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 .
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 = .
~phoenixfire & flamewavelight
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
Labeling the areas in the diagram, we have:
So our answer is . ~~RWhite
Solution 10 (Menelaus's Theorem)
By Menelaus's Theorem on triangle , we have Therefore,
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