2018 AMC 10A Problems/Problem 24
- The following problem is from both the 2018 AMC 10A #24 and 2018 AMC 12A #18, so both problems redirect to this page.
Contents
Problem
Triangle with and has area . Let be the midpoint of , and let be the midpoint of . The angle bisector of intersects and at and , respectively. What is the area of quadrilateral ?
Solution 1
Let , , , and the length of the perpendicular from through be . By angle bisector theorem, we have that where . Therefore substituting we have that . By similar triangles, we have that , and the height of this trapezoid is . Then, we have that . We wish to compute , and we have that it is by substituting.
Solution 2
For this problem, we have because of SAS and . Therefore, is a quarter of the area of , which is . Subsequently, we can compute the area of quadrilateral to be . Using the angle bisector theorem in the same fashion as the previous problem, we get that is times the length of . We want the larger piece, as described by the problem. Because the heights are identical, one area is times the other, and .
Solution 3
The ratio of the to is by the Angle Bisector Theorem, so area of to the area of is also (They have the same height). Therefore, the area of is . Since is the midsegment of , so is the midsegment of . Thus, the ratio of the area of to the area of is , so the area of is . Therefore, the area of quadrilateral is
Solution 4
The area of quadrilateral is the area of minus the area of . Notice, , so , and since , the area of . Given that the area of is , using on side yields . Using the Angle Bisector Theorem, , so the height of . Therefore our answer is
Solution 5 (Trigonometry)
We try to find the area of quadrilateral by subtracting the area outside the quadrilateral but inside triangle . Note that the area of is equal to and the area of triangle is equal to . The ratio is thus equal to and the area of triangle is . Let side be equal to , then by the angle bisector theorem. Similarly, we find the area of triangle to be and the area of triangle to be . A ratio between these two triangles yields , so . Now we just need to find the area of triangle and subtract it from the combined areas of and , since we count it twice. Note that the angle bisector theorem also applies for and , so thus and we find , and the area outside must be , and we finally find , and we are done.
Solution 6 (Areas)
Give triangle area X. Then, by similarity, since , has area 4X. Thus, has area 3X. Doing the same for triangle , we get that triangle has area Y and quadrilateral has area 3Y. Since has the same height as , the ratios of the areas is equal to the ratios of the bases. Because of the Angle Bisector Theorem, . So, . Since has area X, we can write the equation 5X = Y and substitute 5X for Y. Now we can solve for X by adding up all the sums. X + 3X + 5X + 15X = 120, so X = 5. Since we want to find , we substitute 5 for 15X to get . krishkhushi09
Solution 7 (Barycentrics)
Let our reference triangle be . Consequently, we have , , Since is the midpoint of , we have that . Similarly, we have Hence, the line through and is given by the equation
Additionally, since all points on are characterized by , we may plug in for to get . Thus, we have Now, we homogenize the coordinates for to get , , ,
Splitting into we may now evaluate the two determinants:
After simplification, we get and , respectively. Summing, we get Hence, Math0323
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2018amc10a/469
~ dolphin7
Video Solution by OmegaLearn
https://youtu.be/4_x1sgcQCp4?t=4898
~ pi_is_3.14
See Also
2018 AMC 10A (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 AMC 10 Problems and Solutions |
2018 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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 AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.