Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2005 AMC 10A Problems/Problem 8: Difference between revisions

← Older edit
Epicbeast (talk | contribs)
37 edits
Sevenoptimus (talk | contribs)
editor
698 edits
Improved formatting
 
(15 intermediate revisions by 11 users not shown)
Line 1: Line 1:
== Problem ==
== Problem ==
In the figure, the length of side <math>AB</math> of square <math>ABCD</math> is <math>\sqrt{50}</math> and <math>BE</math>=1. What is the area of the inner square <math>EFGH</math>?
In the figure, the length of side <math>AB</math> of square <math>ABCD</math> is <math>\sqrt{50}</math>, <math>E</math> is between <math>B</math> and <math>H</math>, and <math>BE = 1</math>. What is the area of the inner square <math>EFGH</math>?


[[File:AMC102005Aq.png]]
<asy>
unitsize(4cm);
defaultpen(linewidth(.8pt)+fontsize(10pt));


<math> \textbf{(A)}\ 25\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 40\qquad\textbf{(E)}\ 42 </math>
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>
\textbf{(A) } 25\qquad\textbf{(B) } 32\qquad\textbf{(C) } 36\qquad\textbf{(D) } 40\qquad\textbf{(E) } 42
</math>


==Solution==
==Solution==
'''(C)''' We see that side <math>BE</math>, which we know is 1, is also the shorter leg of one of the four right triangles (which are congruent, I'll not prove this). So, <math>AH = 1</math>. Then <math>HB = HE + BE = HE + 1</math>, and <math>HE</math> is one of the sides of the square whose area we want to find. So:
We see that side <math>BE</math>, which we know is <math>1</math>, is also the shorter leg of one of the four right triangles (which are obviously congruent, using the symmetry of the diagram). So <math>AH = 1</math>, and hence <math>HB = HE + BE = HE + 1</math>. Since <math>HE</math> is one of the sides of the square whose area we want to find, we can now simply apply Pythagoras' theorem:
 
<math>1^2 + (HE+1)^2=\sqrt{50}^2</math>


<math>1 + (HE+1)^2=50</math>
<cmath>\begin{align*}1^2 + (HE+1)^2 = \left(\sqrt{50}\right)^2 &\iff 1 + (HE+1)^2 = 50 \\ &\iff (HE+1)^2 = 49 \\&\iff HE+1 = 7 \qquad \text{(as } HE \text{ is a length, so must be positive)} \\ &\iff HE = 6.\end{align*}</cmath>  


<math>(HE+1)^2=49</math>
Thus the area of the square is <math>6^2 = \boxed{\textbf{(C) }36}</math>.


<math>HE+1=7</math>
==See also==
{{AMC10 box|year=2005|ab=A|num-b=7|num-a=9}}


<math>HE=6</math> So, the area of the square is <math>6^2=\boxed{36}</math>.
{{MAA Notice}}

Latest revision as of 16:09, 1 July 2025

Problem

In the figure, the length of side $AB$ of square $ABCD$ is $\sqrt{50}$, $E$ is between $B$ and $H$, and $BE = 1$. What is the area of the inner square $EFGH$?

[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]

$\textbf{(A) } 25\qquad\textbf{(B) } 32\qquad\textbf{(C) } 36\qquad\textbf{(D) } 40\qquad\textbf{(E) } 42$

Solution

We see that side $BE$, which we know is $1$, is also the shorter leg of one of the four right triangles (which are obviously congruent, using the symmetry of the diagram). So $AH = 1$, and hence $HB = HE + BE = HE + 1$. Since $HE$ is one of the sides of the square whose area we want to find, we can now simply apply Pythagoras' theorem:

\begin{align*}1^2 + (HE+1)^2 = \left(\sqrt{50}\right)^2 &\iff 1 + (HE+1)^2 = 50 \\ &\iff (HE+1)^2 = 49 \\&\iff HE+1 = 7 \qquad \text{(as } HE \text{ is a length, so must be positive)} \\ &\iff HE = 6.\end{align*}

Thus the area of the square is $6^2 = \boxed{\textbf{(C) }36}$.

See also

2005 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_10A_Problems/Problem_8&oldid=253057"