Difference between revisions of "1984 AIME Problems/Problem 15"
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== Solution 1 == | == Solution 1 == | ||
− | Rewrite the system of equations as < | + | 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> |
− | After clearing fractions, for each of the values <math>t=4,16,36,64</math>, we have the | + | 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>. | |
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− | < | + | 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>. | ||
− | + | 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>. | ||
− | + | '''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. | |
− | <cmath>x^ | + | == 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>. | |
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− | < | + | 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> |
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== Solution 3 (Highly Unrecommended) == | == Solution 3 (Highly Unrecommended) == | ||
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\frac{x^2}{63}+\frac{y^2}{55}+\frac{z^2}{39}+\frac{w^2}{15}=1\\ | \frac{x^2}{63}+\frac{y^2}{55}+\frac{z^2}{39}+\frac{w^2}{15}=1\\ | ||
\end{align*}</cmath> | \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, | + | 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>. |
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. | 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 ( | + | ==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-3)(a-7)</math> |
− | + | <math>+ w^2(a-1)(a-9)(a-25)</math>. | |
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+ | Now, we have arrived at this huge expression, but what do we do with it? | ||
− | <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>. |
== See also == | == See also == |
Revision as of 04:52, 8 May 2024
Contents
Problem
Determine if
![$\frac{x^2}{2^2-1}+\frac{y^2}{2^2-3^2}+\frac{z^2}{2^2-5^2}+\frac{w^2}{2^2-7^2}=1$](http://latex.artofproblemsolving.com/3/f/a/3faf3bd2051d28e84510522f1060d255643f46fe.png)
![$\frac{x^2}{4^2-1}+\frac{y^2}{4^2-3^2}+\frac{z^2}{4^2-5^2}+\frac{w^2}{4^2-7^2}=1$](http://latex.artofproblemsolving.com/9/d/2/9d236639c64e0cd9eeb0bc55dcadef28a7fbb48a.png)
![$\frac{x^2}{6^2-1}+\frac{y^2}{6^2-3^2}+\frac{z^2}{6^2-5^2}+\frac{w^2}{6^2-7^2}=1$](http://latex.artofproblemsolving.com/d/9/0/d907da2778c251c183225c514cdab8b8f6a62e97.png)
![$\frac{x^2}{8^2-1}+\frac{y^2}{8^2-3^2}+\frac{z^2}{8^2-5^2}+\frac{w^2}{8^2-7^2}=1$](http://latex.artofproblemsolving.com/c/3/f/c3fed85bdaf16d0b43eb458a87be68b5d851b31b.png)
Solution 1
Rewrite the system of equations as
This equation is satisfied when
. After clearing fractions, for each of the values
, we have the equation
where
and
, for
.
Since the polynomials on each side are equal at , we can express the difference of the two polynomials by a quartic polynomial that has roots at
, so
The leading coefficient of the RHS is
because the leading coefficient of the LHS is
.
Plug in in succession, into
. In each case, most terms drop, and we end up with
Adding them up we get the sum as
.
Postscript for the puzzled: This solution which is seemingly unnecessarily redundant in that it computes and
separately before adding them to obtain the final answer is appealing because it gives the individual values of
and
which can be plugged into the given equations to check.
Solution 2
As in Solution 1, we have where
and
, for
.
Now the coefficient of on both sides must be equal. So instead of expanding it fully, we will find what the coefficients of the
and
terms are, so we can eventually apply Vieta's. We can write the long equation as
Rearranging gives us
By Vieta's, we know that the sum of the roots of this equation is
(recall that the roots of the original and this manipulated form of it had roots
and
). Thus,
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.
can be rewritten as
You might be able to see where this is going. First off, find
and
. Then, multiply by the respective lcm to clear all of the denominators. Once you do that, manipulate the equations to solve for
.
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 . We can let this be a polynomial, with respect to
. We can say that
,
,
, and
are coefficients, and not variables. So, we can now expand the fractions to get
.
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 - . So, we want the sum of
,
,
, and
. Looking back to our expression, we can note how on the right hand side, the
terms add to
. Also, on the left hand side, the
coefficient is
(which is achievable by Vieta's formulas rather than expanding if you want to save a few seconds). So, moving all the
terms to the left hand side, then we have that by Vieta's formulas, the sum of the roots is
. Then, we can solve to find that
.
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 |