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 "1984 AIME Problems/Problem 15"

m (→‎See also)
m (→‎Solution 4 (Fast, Efficient))
 
(43 intermediate revisions by 19 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
−
Determine <math>\displaystyle w^2+x^2+y^2+z^2</math> if
+
Determine <math>x^2+y^2+z^2+w^2</math> if
  
−
<center><math> \frac{x^2}{2^2-1}+\frac{y^2}{2^2-3^2}+\frac{z^2}{2^2-5^2}+\frac{w^2}{2^2-7^2}=1 </math></center>
+
<div style="text-align:center;"><math> \frac{x^2}{2^2-1}+\frac{y^2}{2^2-3^2}+\frac{z^2}{2^2-5^2}+\frac{w^2}{2^2-7^2}=1 </math><br /><math> \frac{x^2}{4^2-1}+\frac{y^2}{4^2-3^2}+\frac{z^2}{4^2-5^2}+\frac{w^2}{4^2-7^2}=1 </math><br /><math> \frac{x^2}{6^2-1}+\frac{y^2}{6^2-3^2}+\frac{z^2}{6^2-5^2}+\frac{w^2}{6^2-7^2}=1 </math><br /><math> \frac{x^2}{8^2-1}+\frac{y^2}{8^2-3^2}+\frac{z^2}{8^2-5^2}+\frac{w^2}{8^2-7^2}=1 </math></div>
−
<center><math> \frac{x^2}{4^2-1}+\frac{y^2}{4^2-3^2}+\frac{z^2}{4^2-5^2}+\frac{w^2}{4^2-7^2}=1 </math></center>
 
−
<center><math> \frac{x^2}{6^2-1}+\frac{y^2}{6^2-3^2}+\frac{z^2}{6^2-5^2}+\frac{w^2}{6^2-7^2}=1 </math></center>
 
−
<center><math> \frac{x^2}{8^2-1}+\frac{y^2}{8^2-3^2}+\frac{z^2}{8^2-5^2}+\frac{w^2}{8^2-7^2}=1 </math></center>
 
  
−
== Solution ==
+
== Solution 1 ==
−
For each of the values <math>t=4,16,36,64</math>, we have the equation
+
Rewrite the system of equations as <cmath>\frac{x^{2}}{t-1}+\frac{y^{2}}{t-3^{2}}+\frac{z^{2}}{t-5^{2}}+\frac{w^{2}}{t-7^{2}}=1.</cmath>
 +
This equation is satisfied when <math>t \in \{4, 16, 36, 64\}</math>. After clearing fractions, for each of the values <math>t=4,16,36,64</math>, we have the equation  
 +
<cmath>x^2P_1(t)+y^2P_3(t)+z^2P_5(t)+w^2P_7(t)=F(t),</cmath>where <math>F(t)=(t-1^2)(t-3^2)(t-5^2)(t-7^2)</math> and <math>P_k(t)=F(t)/(t-k^2)</math>, for <math>k=1,3,5,7</math>.
  
−
<math>x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25)</math>
+
Since the polynomials on each side are equal at <math>t=4,16,36,64</math>, we can express the difference of the two polynomials by a quartic polynomial that has roots at <math>t=4,16,36,64</math>, so
 +
<cmath>\begin{align} \tag{\dag}x^2P_1(t)+y^2P_3(t)+z^2P_5(t)+w^2P_7(t)-F(t) = -(t-4)(t-16)(t-36)(t-64)
 +
\end{align}</cmath>
 +
The leading coefficient of the RHS is <math>-1</math> because the leading coefficient of the LHS is <math>-1</math>.
  
−
<math>=(t-1)(t-9)(t-25)(t-49)-(t-4)(t-16)(t-36)(t-64)</math>
+
Plug in <math>t=1^2, 3^2, 5^2, 7^2</math> in succession, into <math>(\dag)</math>. In each case, most terms drop, and we end up with
 +
<cmath>\begin{align*}
 +
x^2=\frac{3^2\cdot 5^2\cdot 7^2}{2^{10}}, \quad y^2=\frac{3^3\cdot 5\cdot 7\cdot 11}{2^{10}},\quad z^2=\frac{3^2\cdot 7\cdot 11\cdot 13}{2^{10}},\quad w^2=\frac{3^2\cdot 5\cdot 11\cdot 13}{2^{10}}
 +
\end{align*}</cmath>
 +
Adding them up we get the sum as <math>3^2\cdot 4=\boxed{036}</math>.
  
−
However, each side of the equation is a polynomial in <math>t</math> of degree at most 3, and they have 4 common roots. Therefore, the polynomials must be equal.
+
'''Postscript for the puzzled''': This solution which is seemingly unnecessarily redundant in that it computes <math>x^2,y^2,z^2,</math> and <math>w^2</math> separately before adding them to obtain the final answer is appealing because it gives the individual values of <math>x^2,y^2,z^2,</math> and <math>w^2</math> which can be plugged into the given equations to check.
  
 +
== Solution 2 ==
 +
As in Solution 1, we have <cmath>x^2P_1(t)+y^2P_3(t)+z^2P_5(t)+w^2P_7(t)=F(t),</cmath>where <math>F(t)=(t-1^2)(t-3^2)(t-5^2)(t-7^2)</math> and <math>P_k(t)=F(t)/(t-k^2)</math>, for <math>k=1,3,5,7</math>.
  
−
Now we can plug in <math>t=1</math> into the polynomial equation. Most terms drop, and we end up with
+
Now the coefficient of <math>t^3</math> on both sides must be equal. So instead of expanding it fully, we will find what the coefficients of the <math>t^4</math> and <math>t^3</math> terms are, so we can eventually apply Vieta's. We can write the long equation as <cmath>(x^2 + y^2 + z^2 + w^2)t^3 + \dots = t^4 - (1^2 + 3^2 + 5^2 + 7^2)t^3 + \dots</cmath> Rearranging gives us <cmath>t^4 - (1^2 + 3^2 + 5^2 + 7^2 + x^2 + y^2 + z^2 + w^2)t^3 \dots = 0.</cmath> By Vieta's, we know that the sum of the roots of this equation is <cmath>1^2 + 3^2 + 5^2 + 7^2 + x^2 + y^2 + z^2 + w^2 = 2^2 + 4^2 + 6^2 + 8^2.</cmath> (recall that the roots of the original and this manipulated form of it had roots <math>2^2, 4^2, 6^2,</math> and <math>8^2</math>). Thus, <cmath>x^2 + y^2 + z^2 + w^2 = 2^2 + 4^2 + 6^2 + 8^2 - 1^2 - 3^2 - 5^2 - 7^2 = \boxed{36}.</cmath>
  
−
<math>x^2(-8)(-24)(-48)=-(-3)(-15)(-35)(-63)</math>
+
== Solution 3 (Highly Unrecommended) ==
 +
Before starting this solution, I highly recommend never following this unless you have no idea what to do with an hour of your time. Even so, learning the above solutions will be more beneficial.
  
−
so that
+
<cmath>\begin{align*}
 +
\frac{x^2}{2^2-1}+\frac{y^2}{2^2-3^2}+\frac{z^2}{2^2-5^2}+\frac{w^2}{2^2-7^2}=1\\
 +
\frac{x^2}{4^2-1}+\frac{y^2}{4^2-3^2}+\frac{z^2}{4^2-5^2}+\frac{w^2}{4^2-7^2}=1\\
 +
\frac{x^2}{6^2-1}+\frac{y^2}{6^2-3^2}+\frac{z^2}{6^2-5^2}+\frac{w^2}{6^2-7^2}=1\\
 +
\frac{x^2}{8^2-1}+\frac{y^2}{8^2-3^2}+\frac{z^2}{8^2-5^2}+\frac{w^2}{8^2-7^2}=1\\
 +
\end{align*}</cmath>
 +
can be rewritten as
 +
<cmath>\begin{align*}
 +
\frac{x^2}{3}-\frac{y^2}{5}-\frac{z^2}{21}-\frac{w^2}{45}=1\\
 +
\frac{x^2}{15}+\frac{y^2}{7}-\frac{z^2}{9}-\frac{w^2}{33}=1\\
 +
\frac{x^2}{35}+\frac{y^2}{27}+\frac{z^2}{11}-\frac{w^2}{13}=1\\
 +
\frac{x^2}{63}+\frac{y^2}{55}+\frac{z^2}{39}+\frac{w^2}{15}=1\\
 +
\end{align*}</cmath>
 +
You might be able to see where this is going. First off, find <math>\text{lcm}(3,5,21,45),\text{lcm}(15,7,9,33), \text{lcm}(35,27,11,13),</math> and <math>\text{lcm}(63,55,39,15)</math>. Then, multiply by the respective lcm to clear all of the denominators. Once you do that, manipulate the equations to solve for <math>w^2+x^2+y^2+z^2</math>.
  
−
<math>x^2=\frac{3\cdot 15\cdot 35\cdot 63}{8\cdot 24\cdot 48}=\frac{3^2\cdot 5^2\cdot 7^2}{2^{10}}</math>
+
Now, most of this is just a brainless bash, and reemphasizing, please try to learn the above solutions. This is only a last resort and only to be used if you have too much time left. The exact amount of time this bash takes depends on the person and how quickly they can manipulate the equations.
  
−
Similarly, we can plug in <math>t=9,25,49</math> and get
+
==Solution 4 (Fast, Efficient)==
 +
Notice how on each line, we have equations of the form <math>\frac{x^2}{a-1^2}+\frac{y^2}{a-3^2}+\frac{z^2}{a-5^2}+\frac{w^2}{a-7^2}=1</math>. We can let this be a polynomial, with respect to <math>a</math>. We can say that <math>w^2</math>, <math>x^2</math>, <math>y^2</math>, and <math>z^2</math> are coefficients, and not variables. So, we can now expand the fractions to get
 +
<math>(a-1)(a-9)(a-25)(a-49)=x^2(a-9)(a-25)(a-49)</math>
 +
<math>+ y^2(a-1)(a-25)(a-49)</math>
 +
<math>+ z^2(a-1)(a-9)(a-49)</math>
 +
<math>+ w^2(a-1)(a-9)(a-25)</math>.
  
−
<math>y^2=\frac{5\cdot 7\cdot 27\cdot 55}{8\cdot 16\cdot 40}=\frac{3^3\cdot 5\cdot 7\cdot 11}{2^{10}}</math>
+
Now, we have arrived at this huge expression, but what do we do with it?
  
−
<math>z^2=\frac{21\cdot 9\cdot 11\cdot 39}{24\cdot 16\cdot 24}=\frac{3^2\cdot 7\cdot 11\cdot 13}{2^{10}}</math>
+
Well, we can look at what we want to find - <math>x^2+y^2+z^2+w^2</math>. So, we want the sum of <math>x^2</math>, <math>y^2</math>, <math>z^2</math>, and <math>w^2</math>. Looking back to our expression, we can note how on the right hand side, the <math>a^3</math> terms add to <math>x^2+y^2+z^2+w^2</math>. Also, on the left hand side, the <math>a^3</math> coefficient is <math>-84</math> (which is achievable by Vieta's formulas rather than expanding if you want to save a few seconds). So, moving all the <math>a^3</math> terms to the left hand side, then we have that by Vieta's formulas, the sum of the roots is <math>-84-x^2-y^2-z^2-w^2=-(2^2+4^2+6^2+8^2)</math>. Then, we can solve to find that <math>x^2+y^2+z^2+w^2=120-84=\boxed{036}</math>.
  
−
<math>w^2=\frac{45\cdot 33\cdot 13\cdot 15}{48\cdot 40\cdot 24}=\frac{3^2\cdot 5\cdot 11\cdot 13}{2^{10}}</math>
 
−
 
−
 
−
Now add them up...
 
−
 
−
<math>z^2+w^2=\frac{3^2\cdot 11\cdot 13(7+5)}{2^{10}}=\frac{3^3\cdot 11\cdot 13}{2^8}</math>
 
−
 
−
<math>x^2+y^2=\frac{3^2\cdot 5\cdot 7(5\cdot 7+3\cdot 11)}{2^{10}}=\frac{3^2\cdot 5\cdot 7\cdot 17}{2^8}</math>
 
−
 
−
with a sum of
 
−
 
−
<math>\frac{3^2(3\cdot 11\cdot 13+5\cdot 7\cdot 17)}{2^8}=3^2\cdot 4=36</math>
 
 
== See also ==
 
== See also ==
 
{{AIME box|year=1984|num-b=14|after=Last Question}}
 
{{AIME box|year=1984|num-b=14|after=Last Question}}
−
* [[AIME Problems and Solutions]]
+
 
−
* [[American Invitational Mathematics Examination]]
+
[[Category: Intermediate Algebra Problems]]
−
* [[Mathematics competition resources]]
 

Latest revision as of 18:35, 1 August 2024

Problem

Determine $x^2+y^2+z^2+w^2$ if

$\frac{x^2}{2^2-1}+\frac{y^2}{2^2-3^2}+\frac{z^2}{2^2-5^2}+\frac{w^2}{2^2-7^2}=1$
$\frac{x^2}{4^2-1}+\frac{y^2}{4^2-3^2}+\frac{z^2}{4^2-5^2}+\frac{w^2}{4^2-7^2}=1$
$\frac{x^2}{6^2-1}+\frac{y^2}{6^2-3^2}+\frac{z^2}{6^2-5^2}+\frac{w^2}{6^2-7^2}=1$
$\frac{x^2}{8^2-1}+\frac{y^2}{8^2-3^2}+\frac{z^2}{8^2-5^2}+\frac{w^2}{8^2-7^2}=1$

Solution 1

Rewrite the system of equations as \[\frac{x^{2}}{t-1}+\frac{y^{2}}{t-3^{2}}+\frac{z^{2}}{t-5^{2}}+\frac{w^{2}}{t-7^{2}}=1.\] This equation is satisfied when $t \in \{4, 16, 36, 64\}$. After clearing fractions, for each of the values $t=4,16,36,64$, we have the equation \[x^2P_1(t)+y^2P_3(t)+z^2P_5(t)+w^2P_7(t)=F(t),\]where $F(t)=(t-1^2)(t-3^2)(t-5^2)(t-7^2)$ and $P_k(t)=F(t)/(t-k^2)$, for $k=1,3,5,7$.

Since the polynomials on each side are equal at $t=4,16,36,64$, we can express the difference of the two polynomials by a quartic polynomial that has roots at $t=4,16,36,64$, so \begin{align} \tag{\dag}x^2P_1(t)+y^2P_3(t)+z^2P_5(t)+w^2P_7(t)-F(t) = -(t-4)(t-16)(t-36)(t-64) \end{align} The leading coefficient of the RHS is $-1$ because the leading coefficient of the LHS is $-1$.

Plug in $t=1^2, 3^2, 5^2, 7^2$ in succession, into $(\dag)$. In each case, most terms drop, and we end up with \begin{align*} x^2=\frac{3^2\cdot 5^2\cdot 7^2}{2^{10}}, \quad y^2=\frac{3^3\cdot 5\cdot 7\cdot 11}{2^{10}},\quad z^2=\frac{3^2\cdot 7\cdot 11\cdot 13}{2^{10}},\quad w^2=\frac{3^2\cdot 5\cdot 11\cdot 13}{2^{10}} \end{align*} Adding them up we get the sum as $3^2\cdot 4=\boxed{036}$.

Postscript for the puzzled: This solution which is seemingly unnecessarily redundant in that it computes $x^2,y^2,z^2,$ and $w^2$ separately before adding them to obtain the final answer is appealing because it gives the individual values of $x^2,y^2,z^2,$ and $w^2$ which can be plugged into the given equations to check.

Solution 2

As in Solution 1, we have \[x^2P_1(t)+y^2P_3(t)+z^2P_5(t)+w^2P_7(t)=F(t),\]where $F(t)=(t-1^2)(t-3^2)(t-5^2)(t-7^2)$ and $P_k(t)=F(t)/(t-k^2)$, for $k=1,3,5,7$.

Now the coefficient of $t^3$ on both sides must be equal. So instead of expanding it fully, we will find what the coefficients of the $t^4$ and $t^3$ terms are, so we can eventually apply Vieta's. We can write the long equation as \[(x^2 + y^2 + z^2 + w^2)t^3 + \dots = t^4 - (1^2 + 3^2 + 5^2 + 7^2)t^3 + \dots\] Rearranging gives us \[t^4 - (1^2 + 3^2 + 5^2 + 7^2 + x^2 + y^2 + z^2 + w^2)t^3 \dots = 0.\] By Vieta's, we know that the sum of the roots of this equation is \[1^2 + 3^2 + 5^2 + 7^2 + x^2 + y^2 + z^2 + w^2 = 2^2 + 4^2 + 6^2 + 8^2.\] (recall that the roots of the original and this manipulated form of it had roots $2^2, 4^2, 6^2,$ and $8^2$). Thus, \[x^2 + y^2 + z^2 + w^2 = 2^2 + 4^2 + 6^2 + 8^2 - 1^2 - 3^2 - 5^2 - 7^2 = \boxed{36}.\]

Solution 3 (Highly Unrecommended)

Before starting this solution, I highly recommend never following this unless you have no idea what to do with an hour of your time. Even so, learning the above solutions will be more beneficial.

\begin{align*} \frac{x^2}{2^2-1}+\frac{y^2}{2^2-3^2}+\frac{z^2}{2^2-5^2}+\frac{w^2}{2^2-7^2}=1\\ \frac{x^2}{4^2-1}+\frac{y^2}{4^2-3^2}+\frac{z^2}{4^2-5^2}+\frac{w^2}{4^2-7^2}=1\\ \frac{x^2}{6^2-1}+\frac{y^2}{6^2-3^2}+\frac{z^2}{6^2-5^2}+\frac{w^2}{6^2-7^2}=1\\ \frac{x^2}{8^2-1}+\frac{y^2}{8^2-3^2}+\frac{z^2}{8^2-5^2}+\frac{w^2}{8^2-7^2}=1\\ \end{align*} can be rewritten as \begin{align*} \frac{x^2}{3}-\frac{y^2}{5}-\frac{z^2}{21}-\frac{w^2}{45}=1\\ \frac{x^2}{15}+\frac{y^2}{7}-\frac{z^2}{9}-\frac{w^2}{33}=1\\ \frac{x^2}{35}+\frac{y^2}{27}+\frac{z^2}{11}-\frac{w^2}{13}=1\\ \frac{x^2}{63}+\frac{y^2}{55}+\frac{z^2}{39}+\frac{w^2}{15}=1\\ \end{align*} You might be able to see where this is going. First off, find $\text{lcm}(3,5,21,45),\text{lcm}(15,7,9,33), \text{lcm}(35,27,11,13),$ and $\text{lcm}(63,55,39,15)$. Then, multiply by the respective lcm to clear all of the denominators. Once you do that, manipulate the equations to solve for $w^2+x^2+y^2+z^2$.

Now, most of this is just a brainless bash, and reemphasizing, please try to learn the above solutions. This is only a last resort and only to be used if you have too much time left. The exact amount of time this bash takes depends on the person and how quickly they can manipulate the equations.

Solution 4 (Fast, Efficient)

Notice how on each line, we have equations of the form $\frac{x^2}{a-1^2}+\frac{y^2}{a-3^2}+\frac{z^2}{a-5^2}+\frac{w^2}{a-7^2}=1$. We can let this be a polynomial, with respect to $a$. We can say that $w^2$, $x^2$, $y^2$, and $z^2$ are coefficients, and not variables. So, we can now expand the fractions to get $(a-1)(a-9)(a-25)(a-49)=x^2(a-9)(a-25)(a-49)$ $+ y^2(a-1)(a-25)(a-49)$ $+ z^2(a-1)(a-9)(a-49)$ $+ w^2(a-1)(a-9)(a-25)$.

Now, we have arrived at this huge expression, but what do we do with it?

Well, we can look at what we want to find - $x^2+y^2+z^2+w^2$. So, we want the sum of $x^2$, $y^2$, $z^2$, and $w^2$. Looking back to our expression, we can note how on the right hand side, the $a^3$ terms add to $x^2+y^2+z^2+w^2$. Also, on the left hand side, the $a^3$ coefficient is $-84$ (which is achievable by Vieta's formulas rather than expanding if you want to save a few seconds). So, moving all the $a^3$ terms to the left hand side, then we have that by Vieta's formulas, the sum of the roots is $-84-x^2-y^2-z^2-w^2=-(2^2+4^2+6^2+8^2)$. Then, we can solve to find that $x^2+y^2+z^2+w^2=120-84=\boxed{036}$.

See also

1984 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Last Question
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1984_AIME_Problems/Problem_15&oldid=224177"