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

Difference between revisions of "2005 AIME I Problems/Problem 14"

 
m (→‎Solution 1)
 
(12 intermediate revisions by 7 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
−
Consider the points <math> A(0,12), B(10,9), C(8,0), and D(-4,7). </math> There is a unique square <math> S </math> such that each of the four points is on a different side of <math> S. </math> Let <math> K </math> be the area of <math> S. </math> Find the remainder when <math> 10K </math> is divided by 1000.
+
Consider the [[point]]s <math> A(0,12), B(10,9), C(8,0),</math> and <math> D(-4,7). </math> There is a unique [[square]] <math> S </math> such that each of the four points is on a different side of <math> S. </math> Let <math> K </math> be the area of <math> S. </math> Find the remainder when <math> 10K </math> is divided by <math>1000</math>.
  
 +
__TOC__
 
== Solution ==
 
== Solution ==
 +
[[Image:2005_I_AIME-14.png|center]]
 +
 +
=== Solution 1 ===
 +
Consider a point <math>E</math> such that <math>AE</math> is [[perpendicular]] to <math>BD</math>, <math>AE</math> intersects <math>BD</math>, and <math>AE = BD</math>. E will be on the same side of the square as point <math>C</math>.
 +
 +
Let the coordinates of <math>E</math> be <math>(x_E,y_E)</math>. Since <math>AE</math> is perpendicular to <math>BD</math>, and <math>AE = BD</math>, we have <math>9 - 7 = x_E - 0</math> and <math>10 - ( - 4) = 12 - y_E</math>
 +
The coordinates of <math>E</math> are thus <math>(2, - 2)</math>.
 +
 +
Now, since <math>E</math> and <math>C</math> are on the same side, we find the slope of the sides going through <math>E</math> and <math>C</math> to be <math>\frac { - 2 - 0}{2 - 8} = \frac {1}{3}</math>. Because the other two sides are perpendicular, the slope of the sides going through <math>B</math> and <math>D</math> are now <math>- 3</math>.
 +
 +
Let <math>A_1,B_1,C_1,D_1</math> be the vertices of the square so that <math>A_1B_1</math> contains point <math>A</math>, <math>B_1C_1</math> contains point <math>B</math>, and etc. Since we know the slopes and a point on the line for each side of the square, we use the point slope formula to find the linear equations. Next, we use the equations to find <math>2</math> vertices of the square, then apply the distance formula.
 +
 +
We find the coordinates of <math>C_1</math> to be <math>(12.5,1.5)</math> and the coordinates of <math>D_1</math> to be <math>( - 0.7, - 2.9)</math>. Applying the distance formula, the side length of our square is <math>\sqrt {\left( \frac {44}{10} \right)^2 + \left( \frac {132}{10} \right)^2} = \frac {44}{\sqrt {10}}</math>.
 +
 +
Hence, the area of the square is <math>K = \frac {44^2}{10}</math>. The remainder when <math>10K</math> is divided by <math>1000</math> is <math>936</math>.
 +
 +
=== Solution 2 ===
 +
Let <math> (a,b)</math> denote a [[normal vector]] of the side containing <math> A</math>. Note that <math>\overline{AC}, \overline{BD}</math> intersect and hence must be opposite [[vertex|vertices]] of the square. The lines containing the sides of the square have the form <math> ax+by=12b</math>, <math> ax+by=8a</math>, <math> bx-ay=10b-9a</math>, and <math> bx-ay=-4b-7a</math>. The lines form a square, so the distance between <math>C</math> and the line through <math>A</math> equals the distance between <math>D</math> and the line through <math>B</math>, hence <math> 8a+0b-12b=-4b-7a-10b+9a</math>, or <math>-3a=b</math>. We can take <math>a=-1</math> and <math>b=3</math>. So the side of the square is <math>\frac{44}{\sqrt{10}}</math>, the area is <math>K=\frac{1936}{10}</math>, and the answer to the problem is <math>\boxed{936}</math>.
  
 
== See also ==
 
== See also ==
−
* [[2005 AIME I Problems]]
+
{{AIME box|year=2005|n=I|num-b=13|num-a=15}}
 +
 
 +
[[Category:Intermediate Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 13:11, 18 February 2018

Problem

Consider the points $A(0,12), B(10,9), C(8,0),$ and $D(-4,7).$ There is a unique square $S$ such that each of the four points is on a different side of $S.$ Let $K$ be the area of $S.$ Find the remainder when $10K$ is divided by $1000$.

Solution

2005 I AIME-14.png

Solution 1

Consider a point $E$ such that $AE$ is perpendicular to $BD$, $AE$ intersects $BD$, and $AE = BD$. E will be on the same side of the square as point $C$.

Let the coordinates of $E$ be $(x_E,y_E)$. Since $AE$ is perpendicular to $BD$, and $AE = BD$, we have $9 - 7 = x_E - 0$ and $10 - ( - 4) = 12 - y_E$ The coordinates of $E$ are thus $(2, - 2)$.

Now, since $E$ and $C$ are on the same side, we find the slope of the sides going through $E$ and $C$ to be $\frac { - 2 - 0}{2 - 8} = \frac {1}{3}$. Because the other two sides are perpendicular, the slope of the sides going through $B$ and $D$ are now $- 3$.

Let $A_1,B_1,C_1,D_1$ be the vertices of the square so that $A_1B_1$ contains point $A$, $B_1C_1$ contains point $B$, and etc. Since we know the slopes and a point on the line for each side of the square, we use the point slope formula to find the linear equations. Next, we use the equations to find $2$ vertices of the square, then apply the distance formula.

We find the coordinates of $C_1$ to be $(12.5,1.5)$ and the coordinates of $D_1$ to be $( - 0.7, - 2.9)$. Applying the distance formula, the side length of our square is $\sqrt {\left( \frac {44}{10} \right)^2 + \left( \frac {132}{10} \right)^2} = \frac {44}{\sqrt {10}}$.

Hence, the area of the square is $K = \frac {44^2}{10}$. The remainder when $10K$ is divided by $1000$ is $936$.

Solution 2

Let $(a,b)$ denote a normal vector of the side containing $A$. Note that $\overline{AC}, \overline{BD}$ intersect and hence must be opposite vertices of the square. The lines containing the sides of the square have the form $ax+by=12b$, $ax+by=8a$, $bx-ay=10b-9a$, and $bx-ay=-4b-7a$. The lines form a square, so the distance between $C$ and the line through $A$ equals the distance between $D$ and the line through $B$, hence $8a+0b-12b=-4b-7a-10b+9a$, or $-3a=b$. We can take $a=-1$ and $b=3$. So the side of the square is $\frac{44}{\sqrt{10}}$, the area is $K=\frac{1936}{10}$, and the answer to the problem is $\boxed{936}$.

See also

2005 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2005_AIME_I_Problems/Problem_14&oldid=91930"