Difference between revisions of "2013 AMC 8 Problems/Problem 24"
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Therefore, <math>\dfrac{[AJICB]}{[ABCIJFED]}= \boxed{\textbf{(C)}\hspace{.05in}\frac{1}{3}}</math>. | Therefore, <math>\dfrac{[AJICB]}{[ABCIJFED]}= \boxed{\textbf{(C)}\hspace{.05in}\frac{1}{3}}</math>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | <asy> | ||
+ | pair A,B,C,D,E,F,G,H,I,J,K; | ||
+ | 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); | ||
+ | K= (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(K,red); | ||
+ | label("$A$", A, NW); | ||
+ | label("$B$", B, NE); | ||
+ | label("$C$", C, NE); | ||
+ | label("$D$", D, NW); | ||
+ | label("$E$", E, NW); | ||
+ | label("$F$", F, SW); | ||
+ | label("$G$", G, S); | ||
+ | label("$H$", H, N); | ||
+ | label("$I$", I, NE); | ||
+ | label("$K$", K,SW,red); | ||
+ | label("$J$", J, SE);</asy> | ||
+ | |||
+ | Let the intersection of <math>AJ</math> and <math>EI</math> be <math>K</math>. | ||
+ | |||
+ | Now we have <math>\triangle ADK</math> and <math>\triangle KIJ</math>. | ||
+ | |||
+ | Because both triangles has a side on congruent squares therefore <math>AD \cong IJ</math>. | ||
+ | |||
+ | Because <math>\angle AKD</math> and <math>\angle JKI</math> are vertical angles <math>\angle AKD \cong \angle JKI</math>. | ||
+ | |||
+ | Also both <math>\angle ADK</math> and <math>\angle JIK</math> are right angles so <math>\angle ADK \cong \angle JIK</math>. | ||
+ | |||
+ | Therefore by AAS(Angle, Angle, Side) <math>\triangle ADK \cong \triangle KIJ</math>. | ||
+ | |||
+ | Then translating/rotating the shaded <math>\triangle JIK</math> into the position of <math>\triangle ADK</math> | ||
+ | |||
+ | So the shaded area now completely covers the square ABCD | ||
+ | |||
+ | Set the area of a square as <math>x</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 09:27, 28 November 2013
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?
Solution 1
First let (where
is the side length of the squares) for simplicity. We can extend
until it hits the extension of
. Call this point
. The area of triangle
then is
The area of rectangle
is
. Thus, our desired area is
. Now, the ratio of the shaded area to the combined area of the three squares is
.
Solution 2
Let the side length of each square be .
Let the intersection of and
be
.
Since ,
. Since
and
are vertical angles, they are congruent. We also have
by definition.
So we have by
congruence. Therefore,
.
Since and
are midpoints of sides,
. This combined with
yields
.
The area of trapezoid is
.
The area of triangle is
.
So the area of the pentagon is
.
The area of the squares is
.
Therefore, .
Solution 3
Let the intersection of and
be
.
Now we have and
.
Because both triangles has a side on congruent squares therefore .
Because and
are vertical angles
.
Also both and
are right angles so
.
Therefore by AAS(Angle, Angle, Side) .
Then translating/rotating the shaded into the position of
So the shaded area now completely covers the square ABCD
Set the area of a square as
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.