Difference between revisions of "2005 AMC 12A Problems/Problem 7"
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== Problem == | == Problem == | ||
[[Square]] <math>EFGH</math> is inside the square <math>ABCD</math> so that each side of <math>EFGH</math> can be extended to pass through a vertex of <math>ABCD</math>. Square <math>ABCD</math> has side length <math>\sqrt {50}</math> and <math>BE = 1</math>. What is the area of the inner square <math>EFGH</math>? | [[Square]] <math>EFGH</math> is inside the square <math>ABCD</math> so that each side of <math>EFGH</math> can be extended to pass through a vertex of <math>ABCD</math>. Square <math>ABCD</math> has side length <math>\sqrt {50}</math> and <math>BE = 1</math>. What is the area of the inner square <math>EFGH</math>? | ||
+ | |||
+ | <asy> | ||
+ | unitsize(4cm); | ||
+ | defaultpen(linewidth(.8pt)+fontsize(10pt)); | ||
+ | pair D=(0,0), C=(1,0), B=(1,1), A=(0,1); | ||
+ | pair F=intersectionpoints(Circle(D,2/sqrt(5)),Circle(A,1))[0]; | ||
+ | pair G=foot(A,D,F), H=foot(B,A,G), E=foot(C,B,H); | ||
+ | draw(A--B--C--D--cycle); | ||
+ | draw(D--F); | ||
+ | draw(C--E); | ||
+ | draw(B--H); | ||
+ | draw(A--G); | ||
+ | label("$A$",A,NW); | ||
+ | label("$B$",B,NE); | ||
+ | label("$C$",C,SE); | ||
+ | label("$D$",D,SW); | ||
+ | label("$E$",E,NNW); | ||
+ | label("$F$",F,ENE); | ||
+ | label("$G$",G,SSE); | ||
+ | label("$H$",H,WSW); | ||
+ | </asy> | ||
<math> | <math> | ||
(\mathrm {A}) \ 25 \qquad (\mathrm {B}) \ 32 \qquad (\mathrm {C})\ 36 \qquad (\mathrm {D}) \ 40 \qquad (\mathrm {E})\ 42 | (\mathrm {A}) \ 25 \qquad (\mathrm {B}) \ 32 \qquad (\mathrm {C})\ 36 \qquad (\mathrm {D}) \ 40 \qquad (\mathrm {E})\ 42 | ||
</math> | </math> | ||
+ | |||
== Solution == | == Solution == | ||
[[Image:2005_12A_AMC-7b.png]] | [[Image:2005_12A_AMC-7b.png]] | ||
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Notice the [[right triangle]] (in blue) with legs <math>1, x+1</math> and [[hypotenuse]] <math>\sqrt{50}</math>. By the [[Pythagorean Theorem]], we have <math>1^2 + (x+1)^2 = (\sqrt{50})^2 \Longrightarrow (x+1)^2 = 49 \Longrightarrow x = 6</math>. Thus, <math>[EFGH] = x^2 = 36\ \mathrm{(C)}</math> | Notice the [[right triangle]] (in blue) with legs <math>1, x+1</math> and [[hypotenuse]] <math>\sqrt{50}</math>. By the [[Pythagorean Theorem]], we have <math>1^2 + (x+1)^2 = (\sqrt{50})^2 \Longrightarrow (x+1)^2 = 49 \Longrightarrow x = 6</math>. Thus, <math>[EFGH] = x^2 = 36\ \mathrm{(C)}</math> | ||
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== See also == | == See also == | ||
{{AMC12 box|year=2005|num-b=6|num-a=8|ab=A}} | {{AMC12 box|year=2005|num-b=6|num-a=8|ab=A}} |
Latest revision as of 11:15, 19 July 2024
Problem
Square is inside the square
so that each side of
can be extended to pass through a vertex of
. Square
has side length
and
. What is the area of the inner square
?
Solution
Arguable the hardest part of this question is to visualize the diagram. Since each side of can be extended to pass through a vertex of
, we realize that
must be tilted in such a fashion. Let a side of
be
.
Notice the right triangle (in blue) with legs and hypotenuse
. By the Pythagorean Theorem, we have
. Thus,
See also
2005 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 6 |
Followed by Problem 8 |
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.