2020 AMC 10B Problems/Problem 21
- The following problem is from both the 2020 AMC 10B #21 and 2020 AMC 12B #18, so both problems redirect to this page.
Contents
Problem
In square , points and lie on and , respectively, so that Points and lie on and , respectively, and points and lie on so that and . See the figure below. Triangle , quadrilateral , quadrilateral , and pentagon each has area What is ?
Solution 1
Since the total area is , the side length of square is . We see that since triangle is a right isosceles triangle with area 1, we can determine sides and both to be . Now, consider extending and until they intersect. Let the point of intersection be . We note that is also a right isosceles triangle with side and find its area to be . Now, we notice that is also a right isosceles triangle (because ) and find it's area to be . This is also equal to or . Since we are looking for , we want two times this. That gives .~TLiu
Solution 2
Draw the auxiliary line . Denote by the point it intersects with , and by the point it intersects with . Last, denote by the segment , and by the segment . We will find two equations for and , and then solve for .
Since the overall area of is , and . In addition, the area of .
The two equations for and are then:
Length of :
Area of CMIF: .
Substituting the first into the second, yields
Solving for gives ~DrB
Solution 3
Plot a point such that and are parallel and extend line to point such that forms a square. Extend line to meet line and point is the intersection of the two. The area of this square is equivalent to . We see that the area of square is , meaning each side is of length 2. The area of the pentagon is . Length , thus . Triangle is isosceles, and the area of this triangle is . Adding these two areas, we get . --OGBooger
Solution 4 (HARD Calculation)
We can easily observe that the area of square is 4 and its side length is 2 since all four regions that build up the square has area 1. Extend and let the intersection with be . Connect , and let the intersection of and be . Notice that since the area of triangle is 1 and , , therefore . Let , then . Also notice that , thus . Now use the condition that the area of quadrilateral is 1, we can set up the following equation: We solve the equation and yield . Now notice that . Hence . -HarryW
Solution 5
Easily, we can find that: quadrilateral and are congruent with each other, so we can move to the shaded area ( and , and overlapping) to form a square ( = , = , = = so ). Then we can solve = = , = , = .
= of of of =
--Ryan Zhang @BRS
Solution 6
, , , , ,
Because is a square and , is the line of symmetry of pentagon . Because , is the reflection of about line
Let ,
Video Solution 1
https://www.youtube.com/watch?v=AKJXB07Sat0&list=PLLCzevlMcsWNcTZEaxHe8VaccrhubDOlQ&index=7 ~ MathEx
Video Solution 2 by the Beauty of Math
https://youtu.be/VZYe3Hu88OA?t=189
See Also
2020 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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 12B (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.