Difference between revisions of "2018 AMC 10A Problems/Problem 24"
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Revision as of 20:43, 24 January 2021
- The following problem is from both the 2018 AMC 12A #18 and 2018 AMC 10A #24, so both problems redirect to this page.
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 ?
Let , , , and the length of the perpendicular to 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.
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 .
The area of to the area of is by Law of Sines. So the area of is . Since is the midsegment of , so is the midsegment of . So the area of to the area of is , so the area of is , by similar triangles. Therefore the area of quad is
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: Trig
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
Video Solution by Richard Rusczyk
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