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Difference between revisions of "1997 AJHSME Problems/Problem 15"

(Created page with "==Problem== Each side of the large square in the figure is trisected (divided into three equal parts). The corners of an inscribed square are at these trisection points, as sho...")
 
(→‎Solution 2)
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==Solution 2==
 
==Solution 2==
  
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Let the side of the big square be <math>3x</math>.
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Let the side of the big square be <math>3x</math>. The area of this square is <math>3x\cdot 3x = 9x^2</math>
  
 
You can use the Pythagorean Theorem on the any of the right triangles to find the length of the side of the inscribed square.  One leg is <math>x</math>, while the other leg is <math>2x</math>.  Thus, the length of the hypotenuse <math>c</math>, which is also the side of the inscrubed square, is:
 
You can use the Pythagorean Theorem on the any of the right triangles to find the length of the side of the inscribed square.  One leg is <math>x</math>, while the other leg is <math>2x</math>.  Thus, the length of the hypotenuse <math>c</math>, which is also the side of the inscrubed square, is:
Line 36: Line 36:
 
<math>5x^2 = c^2</math>
 
<math>5x^2 = c^2</math>
  
−
Once you get <math>c^2</math>, you can stop, because <math>c^2</math> is the area of the inscribed square.  The area of the large square is <math>3x\cdot 3x = 9x^2</math>. The ratio of these areas is <math>\frac{5x^2}{9x^2} = \frac{5}{9}</math>, giving option <math>\boxed{B}</math>.
+
Once you get <math>c^2</math>, you can stop, because <math>c^2</math> is the area of the inscribed square.  The ratio of the areas of the small square to the big square is <math>\frac{5x^2}{9x^2} = \frac{5}{9}</math>, giving option <math>\boxed{B}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 14:39, 31 July 2011

Problem

Each side of the large square in the figure is trisected (divided into three equal parts). The corners of an inscribed square are at these trisection points, as shown. The ratio of the area of the inscribed square to the area of the large square is

[asy] draw((0,0)--(3,0)--(3,3)--(0,3)--cycle); draw((1,0)--(1,0.2)); draw((2,0)--(2,0.2)); draw((3,1)--(2.8,1)); draw((3,2)--(2.8,2)); draw((1,3)--(1,2.8)); draw((2,3)--(2,2.8)); draw((0,1)--(0.2,1)); draw((0,2)--(0.2,2)); draw((2,0)--(3,2)--(1,3)--(0,1)--cycle); [/asy]

$\text{(A)}\ \dfrac{\sqrt{3}}{3} \qquad \text{(B)}\ \dfrac{5}{9} \qquad \text{(C)}\ \dfrac{2}{3} \qquad \text{(D)}\ \dfrac{\sqrt{5}}{3} \qquad \text{(E)}\ \dfrac{7}{9}$

Solution 1

Since we are dealing with ratios, let the big square have sides of $3$ and thus an area of $3^2 = 9$. Chosing a multiple of $3$ will avoid fractions in the rest of the answer.

To find the area of the inscribed square, subtract off the areas of the four triangles. Each triangle has an area of $\frac{1}{2}\cdot 2\cdot 1 = 1$.

Thus, the area of the inscribed square is $9 - 4\cdot 1 = 5$, and the ratio of areas is $\frac{5}{9}$, giving option $\boxed{B}$.

Solution 2

Let the side of the big square be $3x$. The area of this square is $3x\cdot 3x = 9x^2$

You can use the Pythagorean Theorem on the any of the right triangles to find the length of the side of the inscribed square. One leg is $x$, while the other leg is $2x$. Thus, the length of the hypotenuse $c$, which is also the side of the inscrubed square, is:

$a^2 + b^2 = c^2$

$x^2 + (2x)^2 = c^2$

$x^2 + 4x^2 = c^2$

$5x^2 = c^2$

Once you get $c^2$, you can stop, because $c^2$ is the area of the inscribed square. The ratio of the areas of the small square to the big square is $\frac{5x^2}{9x^2} = \frac{5}{9}$, giving option $\boxed{B}$.

See also

1997 AJHSME (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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
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