2020 AMC 10B Problems/Problem 21
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 
?
![[asy] real x=2sqrt(2); real y=2sqrt(16-8sqrt(2))-4+2sqrt(2); real z=2sqrt(8-4sqrt(2)); pair A, B, C, D, E, F, G, H, I, J; A = (0,0); B = (4,0); C = (4,4); D = (0,4); E = (x,0); F = (4,y); G = (y,4); H = (0,x); I = F + z * dir(225); J = G + z * dir(225); draw(A--B--C--D--A); draw(H--E); draw(J--G^^F--I); draw(rightanglemark(G, J, I), linewidth(.5)); draw(rightanglemark(F, I, E), linewidth(.5)); dot("$A$", A, S); dot("$B$", B, S); dot("$C$", C, dir(90)); dot("$D$", D, dir(90)); dot("$E$", E, S); dot("$F$", F, dir(0)); dot("$G$", G, N); dot("$H$", H, W); dot("$I$", I, SW); dot("$J$", J, SW); [/asy]](http://latex.artofproblemsolving.com/4/8/9/4892da26cb6334b359f58d4007f89460883779ab.png)
Solution
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 it's area to be 
. Now, we notice that 
is also a right isosceles triangle 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
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 | ||
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