Difference between revisions of "2014 AMC 10A Problems/Problem 18"
Isabelchen (talk | contribs) |
Isabelchen (talk | contribs) |
||
Line 38: | Line 38: | ||
<math>\sqrt{(- \frac{4}{x_1})^2 +4^2} = \sqrt{(x_1)^2+1^2}</math> | <math>\sqrt{(- \frac{4}{x_1})^2 +4^2} = \sqrt{(x_1)^2+1^2}</math> | ||
<math>\frac{16}{x_1^2}+16=x_1^2+1</math> | <math>\frac{16}{x_1^2}+16=x_1^2+1</math> | ||
− | <math>16 | + | <math>\frac{16}{x_1^2}+15=x_1^2</math> |
+ | <math>16+15x_1^2=x_1^4</math> | ||
<math>Let</math> <math>y=x_1^2</math> | <math>Let</math> <math>y=x_1^2</math> | ||
− | <math>16+ | + | <math>16+15y=y^2</math> |
<math>y^2-15y-16=0</math> | <math>y^2-15y-16=0</math> | ||
<math>(y-16)(y+1)=0</math> | <math>(y-16)(y+1)=0</math> | ||
Line 48: | Line 49: | ||
Note that the square with <math>x_1=-4</math> is just the reflection of square with <math>x_1=4</math> over the origin. I will use <math>x_1=4</math>. | Note that the square with <math>x_1=-4</math> is just the reflection of square with <math>x_1=4</math> over the origin. I will use <math>x_1=4</math>. | ||
<math>B=(4,1), AB=\sqrt{17}, [ABCD]=\boxed{\textbf{(B) }17}</math> | <math>B=(4,1), AB=\sqrt{17}, [ABCD]=\boxed{\textbf{(B) }17}</math> | ||
+ | |||
+ | ~isabelchen | ||
+ | |||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | In this solution, we will use the fact that the diagonals of a square bisect each other, they are perpendicular to each other, and they are equal in length. | ||
+ | |||
+ | Using the fact that the diagonals bisect each other, we get the equation: | ||
+ | <math>\frac{x_2}{2}=\frac{x_1+x_3}{2}</math> | ||
+ | <math>x_2=x_1+x_3</math> | ||
+ | |||
+ | Now we use the fact that the diagonals are perpendicular to each other: | ||
+ | <math>Slope</math> <math>of</math> <math>AC=\frac{5-0}{x_2}=\frac{5}{x_2}</math> | ||
+ | <math>Slope</math> <math>of</math> <math>BD=\frac{1-4}{x_1-x_3}=- \frac{3}{x_1-x_3}</math> | ||
+ | <math>\frac{5}{x_2} \cdot (- \frac{3}{x_1-x_3}) = -1</math> | ||
+ | <math>x_2(x_1-x_3)=15</math> | ||
+ | |||
+ | Using the fact that the diagonals are equal in length, we get the equation: | ||
+ | <math>BD=AC</math> | ||
+ | <math>(4-1)^2+(x_3-x_1)^2=5^2+x_2^2</math> | ||
+ | <math>9+(x_3-x_1)^2=25+x_2^2</math> | ||
+ | <math>(x_3-x_1)^2-x_2^2=16</math> | ||
+ | |||
+ | Now we have 3 equations with 3 variables: | ||
+ | |||
+ | <math> | ||
+ | |||
+ | We substitute <math>x_2</math> into the 2 other equations: | ||
+ | <math>(x_1+x_3)(x_1-x_3)=15</math> | ||
+ | <math>x_1^2-x_3^2=15</math> | ||
+ | |||
+ | <math>(x_3-x_1)^2-(x_3+x_1)^2=16</math> | ||
+ | <math>x_3^2-2x_1x_3+x_3^2-x_1^2-2x_1x_3-x_3^2=16</math> | ||
+ | <math>-4x_1x_3=16</math> | ||
+ | <math>x_1x_3=-4</math> | ||
+ | |||
+ | Now we have 2 equations of <math>x_1</math> and <math>x_3</math>: | ||
+ | |||
+ | <math> | ||
+ | |||
+ | <math>x_3=- \frac{4}{x_1}</math> | ||
+ | <math>x_1^2-(- \frac{4}{x_1})^2=15</math> | ||
+ | <math>x_1^2- \frac{16}{x_1^2}=15</math> | ||
+ | |||
+ | This is the same equation as solution <math>2</math>. So <math>x_1= \pm 4, AB=\sqrt{17}, [ABCD]=\boxed{\textbf{(B) }17}</math> | ||
~isabelchen | ~isabelchen |
Revision as of 08:10, 1 October 2021
Problem
A square in the coordinate plane has vertices whose -coordinates are , , , and . What is the area of the square?
Solution 1
Let the points be , , , and
Note that the difference in value of and is . By rotational symmetry of the square, the difference in value of and is also . Note that the difference in value of and is . We now know that , the side length of the square, is equal to , so the area is .
Solution 2
By translation, we can move the square with point at the origin. Then, . We will use the relationship among the 4 sides of being perpendicular and equal.
The slope of is .
Because is perpendicular to , the slope of . From the information above we could have the equation:
Because is perpendicular to , the slope of . From the information above we could have the equation:
Because
Note that the square with is just the reflection of square with over the origin. I will use .
~isabelchen
Solution 3
In this solution, we will use the fact that the diagonals of a square bisect each other, they are perpendicular to each other, and they are equal in length.
Using the fact that the diagonals bisect each other, we get the equation:
Now we use the fact that the diagonals are perpendicular to each other:
Using the fact that the diagonals are equal in length, we get the equation:
Now we have 3 equations with 3 variables:
We substitute into the 2 other equations:
Now we have 2 equations of and :
This is the same equation as solution . So
~isabelchen
Video Solution
https://www.youtube.com/watch?v=iPPQUrNE4RE
~ naren_pr
See Also
2014 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.