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
- 1 Problem
- 2 Solution 1
- 3 Solution 2 (Coordinates)
- 4 Solution 3 (Trigonometry)
- 5 Solution 4 (Law of Cosines)
- 6 Solution 5 (Vectors / Coordinates)
- 7 Solution 6 (Power of a Point)
- 8 Solution 7 (Solving Equations)
- 9 Solution 8
- 10 Video Solution by Education, The Study of Everything
- 11 Video Solution by On The Spot STEM
- 12 Video Solution by MathEx
- 13 Video Solution by TheBeautyOfMath
- 14 Video Solution by Triviality
- 15 Video Solution by OmegaLearn
- 16 See Also
Problem
Quadrilateral satisfies and Diagonals and intersect at point and What is the area of quadrilateral
Solution 1
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, . 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 in any right triangle, the altitude squared is equal to the product of the two lengths that it divides the base into.
Expanding, simplifying, and dividing by the GCF, we get . This factors to , which has roots of . Since lengths cannot be negative, . Since , that means the altitude , or . Thus
~ 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 (Law of Cosines)
Denote as . By the Law of Cosines:
Adding these up yields: By the quadratic formula, .
Observe: .
Thus the desired area is
~qwertysri987
Solution 5 (Vectors / Coordinates)
Let and . Then and lies on the line So the coordinates of are
We can make this a vector problem. We notice that point forms a right angle, meaning vectors and are orthogonal, and their dot-product is .
We determine and to be and , respectively. (To get this, we use the fact that and similarly, )
Equating the cross-product to gets us the quadratic The solutions are Since clearly has a more negative x-coordinate than , we take . So
From here, there are multiple ways to get the area of to be , and since the area of is , we get our final answer to be
-PureSwag
Solution 6 (Power of a Point)
Let be the midpoint of , and draw where is on . We have .
. Therefore is a cyclic quadrilateral.
Notice that via Power of a Point.
The altitude from to is then equal to .
Finally, the total area of is equal to
~asops
Solution 7 (Solving Equations)
Let ,
Looking at the diagram we have , ,
Because ,
, substituting , we get
Because and share the same base,
By , . So,
Let , , , ,
Because , can only equal 40. , ,
Solution 8
Drop perpendiculars and to Notice that since (since they are vertical angles) and triangles and are similar. Therefore, we have
where Therefore,
Additionally, angle chasing shows that triangles and are also similar. This gives so Thus, applying the Pythagorean Theorem to triangle gives
so Our pairs of similar triangles then allow us to fill in the following lengths (in this order):
Now, let Angle chasing shows that triangle and are similar, so Plugging in known lengths gives
This gives Now we know all the lengths that make up which allows us to find
Therefore,
--vaporwave
Video Solution by Education, The Study of Everything
Video Solution by On The Spot STEM
https://www.youtube.com/watch?v=hIdNde2Vln4
Video Solution by MathEx
https://www.youtube.com/watch?v=sHrjx968ZaM
Video Solution by TheBeautyOfMath
https://www.youtube.com/watch?v=RKlG6oZq9so
Video Solution by Triviality
https://youtu.be/R220vbM_my8?t=658 (amritvignesh0719062.0)
Video Solution by OmegaLearn
https://youtu.be/hDsoyvFWYxc?t=1224 ~ pi_is_3.14
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.