2020 AMC 10A Problems/Problem 20
- The following problem is from both the 2020 AMC 12A #18 and 2020 AMC 10A #20, so both problems redirect to this page.
Contents
Problem
Quadrilateral satisfies and Diagonals and intersect at point and What is the area of quadrilateral
Solution 1 (Just Drop An Altitude)
It's crucial to draw a good diagram for this one. Since and , we get . Now we need to find to get the area of the whole quadrilateral. Drop an altitude from to and call the point of intersection . Let . Since , then . By dropping this altitude, we can also see two similar triangles, and . Since is , and , we get that . Now, if we redraw another diagram just of , we get that because of the altitude geometric mean theorem which states that the altitude squared is equal to the product of the two lengths that it divides the base into. Now expanding, simplifying, and dividing by the GCF, we get . This factors to . Since lengths cannot be negative, . Since , . So
(I'm very sorry if you're a visual learner but now you have a diagram by ciceronii)
~ Solution by Ultraman
~ Diagram by ciceronii
Solution 2 (coordinates)
Let the points be , , , ,and , respectively. Since lies on line , we know that . Furthermore, since , lies on the circle with diameter , so . Solving for and with these equations, we get the solutions and . We immediately discard the solution as should be negative. Thus, we conclude that .
Solution 3 (Trigonometry)
Let and Using Law of Sines on we get and LoS on yields Divide the two to get Now, and solve the quadratic, taking the positive solution (C is acute) to get So if then and By Pythagorean Theorem, and the answer is
(This solution is incomplete, can someone complete it please-Lingjun) ok Latex edited by kc5170
We could use the famous m-n rule in trigonometry in with Point [Unable to write it here.Could anybody write the expression] . We will find that is an angle bisector of (because we will get ). Therefore by converse of angle bisector theorem . By using Pythagorean theorem, we have values of and . Computing . Adding the areas of and , hence the answer is .
By: Math-Amaze Latex: Catoptrics.
Solution 4 (Answer Choices)
We know that the big triangle has area 300. Use the answer choices which would mean that the area of the little triangle is a multiple of 10. Thus the product of the legs is a multiple of 20. Guess that the legs are equal to and , and because the hypotenuse is 20 we get . Testing small numbers, we get that when and , is indeed a square. The area of the triangle is thus 60, so the answer is .
~tigershark22
Video Solution
On The Spot STEM https://www.youtube.com/watch?v=hIdNde2Vln4
Video Solution
https://www.youtube.com/watch?v=sHrjx968ZaM&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=2 ~ MathEx
See Also
2020 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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 |
2020 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 |
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