Difference between revisions of "2013 AMC 8 Problems/Problem 24"
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| + | ==Solution 2 (extremely simple)== | ||
| + | Extend line AD such that it intersects line FG. Call the intersection K. If we assign an arbitrary value to the side lengths of the squares, such as 2, the base of the now formed triangle AJK would have a length of 3, while the height would be 4. The area of triangle AJK would then be 6, and the rectangle EDKF would have an area of 2. 6 + 2 = 8, and since all of the squares' areas add up to a total of 12 (if each has side length 2), we obtain (12 - 8)/12 for the shaded part of the diagram which yields 4/12 = <math>\boxed{\textbf{(C)}\ \frac {1}{3}}</math> | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2013|num-b=23|num-a=25}} | {{AMC8 box|year=2013|num-b=23|num-a=25}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 21:01, 6 September 2024
Problem
Squares 
, 
, and 
are equal in area. Points 
and 
are the midpoints of sides 
and 
, respectively. What is the ratio of the area of the shaded pentagon 
to the sum of the areas of the three squares?
![[asy] pair A,B,C,D,E,F,G,H,I,J; A = (0.5,2); B = (1.5,2); C = (1.5,1); D = (0.5,1); E = (0,1); F = (0,0); G = (1,0); H = (1,1); I = (2,1); J = (2,0); draw(A--B); draw(C--B); draw(D--A); draw(F--E); draw(I--J); draw(J--F); draw(G--H); draw(A--J); filldraw(A--B--C--I--J--cycle,grey); draw(E--I); dot("$A$", A, NW); dot("$B$", B, NE); dot("$C$", C, NE); dot("$D$", D, NW); dot("$E$", E, NW); dot("$F$", F, SW); dot("$G$", G, S); dot("$H$", H, N); dot("$I$", I, NE); dot("$J$", J, SE); [/asy]](http://latex.artofproblemsolving.com/1/6/9/169139bda69e81f8342f246f95cc3f636c1f9abe.png)
Solution 1 (shortcut)
It can be proven that 
(where 
is the point where 
intersects 
) which also means quadrilaterals 
(due to the squares being equal in area which means the squares are congruent, and since the triangles earlier mentioned are congruent). The area of the shaded region is equal to the area of one square since the quadrilaterals and triangles are congruent. The total area of the shape is the area of three squares. Putting these two pieces of information together, the answer is 
.
![[asy] pair A,B,C,D,E,F,G,H,I,J,X; A = (0.5,2); B = (1.5,2); C = (1.5,1); D = (0.5,1); E = (0,1); F = (0,0); G = (1,0); H = (1,1); I = (2,1); J = (2,0); X = (1.25,1); draw(A--B); draw(C--B); draw(D--A); draw(F--E); draw(I--J); draw(J--F); draw(G--H); draw(A--J); filldraw(A--B--C--I--J--cycle,grey); draw(E--I); dot("$A$", A, NW); dot("$B$", B, NE); dot("$C$", C, NE); dot("$D$", D, NW); dot("$E$", E, NW); dot("$F$", F, SW); dot("$G$", G, S); dot("$H$", H, N); dot("$I$", I, NE); dot("$J$", J, SE); dot("$X$", X, NE); [/asy]](http://latex.artofproblemsolving.com/3/d/f/3dfb6045a3e24d57f2c87bf58a726a025043d955.png)
~ julia333
Solution 2 (extremely simple)
Extend line AD such that it intersects line FG. Call the intersection K. If we assign an arbitrary value to the side lengths of the squares, such as 2, the base of the now formed triangle AJK would have a length of 3, while the height would be 4. The area of triangle AJK would then be 6, and the rectangle EDKF would have an area of 2. 6 + 2 = 8, and since all of the squares' areas add up to a total of 12 (if each has side length 2), we obtain (12 - 8)/12 for the shaded part of the diagram which yields 4/12 =
See Also
| 2013 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 23 |
Followed by Problem 25 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
