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 "2018 UNCO Math Contest II Problems/Problem 3"
Line 3: Line 3:
 
then so does the point <math>(3x + 4y, 2x + By)</math>.
 
then so does the point <math>(3x + 4y, 2x + By)</math>.
  
−
== Solution ==
+
== Solution 1 ==
 
We can write a system of equations -  
 
We can write a system of equations -  
 
<cmath>2y^2-x^2 = 1</cmath>
 
<cmath>2y^2-x^2 = 1</cmath>
Line 18: Line 18:
 
<cmath>2(2x + By)^2 - (3x+4y)^2 = 1</cmath>
 
<cmath>2(2x + By)^2 - (3x+4y)^2 = 1</cmath>
  
−
Through expanding the second equation, we get <math>-x^2+8Bxy-24xy+2B^2y^2-16y^2 = 1</math>. Since <math>2y^2-x^2 = 1</math>, we have <cmath>-x^2+8Bxy-24xy+2B^2y^2-16y^2 = 2y^2-x^2</cmath>.
+
Through expanding the second equation, we get <math>-x^2+8Bxy-24xy+2B^2y^2-16y^2 = 1</math>. Since <math>2y^2-x^2 = 1</math>, we have <cmath>-x^2+8Bxy-24xy+2B^2y^2-16y^2 = 2y^2-x^2</cmath>  
 
The <math>x^2</math> terms on each side cancel out, so the equation becomes <math>(8B-24)xy + (2B^2-16)y^2 = 2y^2.</math> The coefficient of <math>xy</math> on the RHS is 0 and the coefficient of <math>y^2</math> is 2. From these two observations, we now create two new equations.
 
The <math>x^2</math> terms on each side cancel out, so the equation becomes <math>(8B-24)xy + (2B^2-16)y^2 = 2y^2.</math> The coefficient of <math>xy</math> on the RHS is 0 and the coefficient of <math>y^2</math> is 2. From these two observations, we now create two new equations.
 
<cmath>8B-24 = 0</cmath>
 
<cmath>8B-24 = 0</cmath>

Revision as of 02:16, 11 June 2022

Problem

Find all values of $B$ that have the property that if $(x, y)$ lies on the hyperbola $2y^2-x^2 = 1$, then so does the point $(3x + 4y, 2x + By)$.

Solution 1

We can write a system of equations - \[2y^2-x^2 = 1\] \[2(2x + By)^2 - (3x+4y)^2 = 1\]

Expanding the second equation, we get $-x^2+8Bxy-24xy+2B^2y^2-16y^2=1$.

Since we want this to look like $2y^2-x^2=1$, we plug in B's that would put it into that form. If we plug in $B=3$, things cancel, and we get $-x^2+24xy-24xy+18y^2-16y^2=1 \rightarrow 2y^2-x^2=1$. So $\boxed{B=3}$ ~Ultraman

Solution 2 (Grinding)

As with Solution 1, we create a system of equations. \[2y^2-x^2 = 1\] \[2(2x + By)^2 - (3x+4y)^2 = 1\]

Through expanding the second equation, we get $-x^2+8Bxy-24xy+2B^2y^2-16y^2 = 1$. Since $2y^2-x^2 = 1$, we have \[-x^2+8Bxy-24xy+2B^2y^2-16y^2 = 2y^2-x^2\] The $x^2$ terms on each side cancel out, so the equation becomes $(8B-24)xy + (2B^2-16)y^2 = 2y^2.$ The coefficient of $xy$ on the RHS is 0 and the coefficient of $y^2$ is 2. From these two observations, we now create two new equations. \[8B-24 = 0\] \[2B^2-16 = 2\] Solving either equation and then checking with the other will reveal that $\boxed{B=3}$. ~kingme271

See also

2018 UNCO Math Contest II (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10
All UNCO Math Contest Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2018_UNCO_Math_Contest_II_Problems/Problem_3&oldid=174660"