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
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3
- 5 Solution 4 (HARD Calculation)
- 6 Solution 5 (Basically Same as Solution 3)
- 7 Solution 6
- 8 Solution 7 (Easy to See)
- 9 Video Solution (HOW TO THINK CREATIVELY!!!)
- 10 Video Solution by MathEx
- 11 Video Solution by TheBeautyOfMath
- 12 Really Good Vid Explanation
- 13 See Also
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 (Basically Same as Solution 3)
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 ,
Solution 7 (Easy to See)
Note that the side length of is 2 and thus the diagonal is of length . However, the height to side in triangle is 1, implying that where is the midpoint of . From here suppose that is the midpoint of and let , which means . The area of the pentagon is then Solving this quadratic for yields (technically the smaller value is the correct one but it doesn’t matter for our purposes). We can then calculate .
~Dhillonr25
Video Solution (HOW TO THINK CREATIVELY!!!)
~Education, the Study of Everything
Video Solution by MathEx
https://www.youtube.com/watch?v=AKJXB07Sat0
Video Solution by TheBeautyOfMath
https://youtu.be/VZYe3Hu88OA?t=189
Really Good Vid Explanation
https://www.youtube.com/watch?v=AUndgrOH8U8&ab_channel=ReachTheStars
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.