Difference between revisions of "2013 AMC 8 Problems/Problem 24"
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==Solution 1 (shortcut)== | ==Solution 1 (shortcut)== | ||
It can be proven that <math>\Delta ADX \cong \Delta JIX</math> (where <math>X</math> is the point where <math>\overline {AJ}</math> intersects <math>\overline {HC}</math>) which also means quadrilaterals <math>ABCX \cong JGHX</math> (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, <math>\boxed{\textbf{(C)}\ \frac {1}{3}}</math> | It can be proven that <math>\Delta ADX \cong \Delta JIX</math> (where <math>X</math> is the point where <math>\overline {AJ}</math> intersects <math>\overline {HC}</math>) which also means quadrilaterals <math>ABCX \cong JGHX</math> (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, <math>\boxed{\textbf{(C)}\ \frac {1}{3}}</math> | ||
| + | |||
| + | <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> | ||
| + | |||
~ julia333 | ~ julia333 | ||
Revision as of 21:24, 29 August 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, 
![[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
