Difference between revisions of "2014 AMC 10A Problems/Problem 18"
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[[Category: Introductory Geometry Problems]] | [[Category: Introductory Geometry Problems]] | ||
− | ==Solution== | + | ==Solution 1== |
Let the points be <math>A=(x_1,0)</math>, <math>B=(x_2,1)</math>, <math>C=(x_3,5)</math>, and <math>D=(x_4,4)</math> | Let the points be <math>A=(x_1,0)</math>, <math>B=(x_2,1)</math>, <math>C=(x_3,5)</math>, and <math>D=(x_4,4)</math> | ||
Note that the difference in <math>y</math> value of <math>B</math> and <math>C</math> is <math>4</math>. By rotational symmetry of the square, the difference in <math>x</math> value of <math>A</math> and <math>B</math> is also <math>4</math>. Note that the difference in <math>y</math> value of <math>A</math> and <math>B</math> is <math>1</math>. We now know that <math>AB</math>, the side length of the square, is equal to <math>\sqrt{1^2+4^2}=\sqrt{17}</math>, so the area is <math>\boxed{\textbf{(B) }17}</math>. | Note that the difference in <math>y</math> value of <math>B</math> and <math>C</math> is <math>4</math>. By rotational symmetry of the square, the difference in <math>x</math> value of <math>A</math> and <math>B</math> is also <math>4</math>. Note that the difference in <math>y</math> value of <math>A</math> and <math>B</math> is <math>1</math>. We now know that <math>AB</math>, the side length of the square, is equal to <math>\sqrt{1^2+4^2}=\sqrt{17}</math>, so the area is <math>\boxed{\textbf{(B) }17}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | By translation, we can move the square with point <math>A</math> at the origin. Then, <math>A=(0,0), B=(x_1,1), C=(x_2,5), D=(x_3,4)</math>. We will use the relationship among the 4 sides of being perpendicular and equal. | ||
+ | |||
+ | The slope of <math>AB</math> is <math>\frac{1-0}{x_1-0}=\frac{1}{x_1}</math>. | ||
+ | |||
+ | Because <math>BC</math> is perpendicular to <math>AB</math>, the slope of <math>BC=-x_1</math>. | ||
+ | From the information above we could have the equation: | ||
+ | |||
+ | <math>\frac{5-1}{x_2-x_1}=-x_1</math> | ||
+ | <math>-x_1 \cdot x_2+x_1^2=4</math> | ||
+ | <math>x_1 \cdot x_2=x_1^2-4</math> | ||
+ | <math>x_2=\frac{x_1^2-4}{x_1}</math> | ||
+ | |||
+ | Because <math>CD</math> is perpendicular to <math>BC</math>, the slope of <math>CD=\frac{1}{x_1}</math>. | ||
+ | From the information above we could have the equation: | ||
+ | |||
+ | <math>\frac{5-4}{x_2-x_3}=\frac{1}{x_1}</math> | ||
+ | <math>x_2-x_3=x_1</math> | ||
+ | <math>\frac{x_1^2-4}{x_1}-x_3=x_1</math> | ||
+ | <math>x_1^2-4-x_1 \cdot x_3 = x_1^2</math> | ||
+ | <math>x_1 \cdot x_3=-4</math> | ||
+ | <math>x_3=- \frac{4}{x_1}</math> | ||
+ | |||
+ | Because <math>AD=AB,</math> | ||
+ | |||
+ | <math>\sqrt{x_3^2 +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}+15=x_1^2</math> | ||
+ | <math>16+15x_1^2=x_1^4</math> | ||
+ | <math>Let</math> <math>y=x_1^2</math> | ||
+ | <math>16+15y=y^2</math> | ||
+ | <math>y^2-15y-16=0</math> | ||
+ | <math>(y-16)(y+1)=0</math> | ||
+ | <math>y=16</math> | ||
+ | <math>x_1=\pm4</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> | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen 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{4-1}{x_3-x_1}=\frac{3}{x_3-x_1}</math> | ||
+ | <math>\frac{5}{x_2} \cdot \frac{3}{x_3-x_1} = -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>\begin{cases} x_2=x_1+x_3 \\ x_2(x_1-x_3)=15 \\ (x_3-x_1)^2-x_2^2=16 \end{cases}</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>\begin{cases} x_1^2-x_3^2=15 \\ x_1x_3=-4 \end{cases}</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> | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
+ | |||
+ | ==Solution 4 (not rigorous)== | ||
+ | Draw it out, by inspection the coordinates are <math>(-1, 4)</math>, <math>(0, 0)</math>, <math>(4, 1)</math>, and <math>(3, 5)</math>. The side length is <math>\sqrt{17}, [ABCD]=\boxed{\textbf{(B) }17}</math> | ||
+ | ~JH. L | ||
+ | |||
+ | == Video Solution== | ||
+ | |||
+ | https://www.youtube.com/watch?v=iPPQUrNE4RE | ||
+ | |||
+ | ~ naren_pr | ||
==See Also== | ==See Also== |
Latest revision as of 03:18, 20 June 2022
Contents
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
.
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
Solution 4 (not rigorous)
Draw it out, by inspection the coordinates are ,
,
, and
. The side length is
~JH. L
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.