Difference between revisions of "1984 AIME Problems/Problem 15"
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== Problem == | == Problem == | ||
− | Determine <math> | + | Determine <math>x^2+y^2+z^2+w^2</math> if |
<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> | <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> | ||
== Solution 1 == | == Solution 1 == | ||
− | + | Rewrite the system of equations as <math> \frac{x^{2}}{t-1}+\frac{y^{2}}{t-3^{2}}+\frac{z^{2}}{t-5^{2}}+\frac{w^{2}}{t-7^{2}}=1. </math> This equation is satisfied when <math>t = 4,16,36,64</math>, as then the equation is equivalent to the given equations. | |
+ | After clearing fractions, for each of the values <math>t=4,16,36,64</math>, we have the [[equation]] <math>x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)</math> <math>+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25) = (t-1)(t-9)(t-25)(t-49)</math>. We can move the expression <math>(t-1)(t-9)(t-25)(t-49)</math> to the left hand side to obtain the difference of the polynomials: <math>x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)</math> <math>+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25)</math> and <math>(t-1)(t-9)(t-25)(t-49)</math> | ||
+ | |||
+ | Since the polynomials are equal at <math>t=4,16,36,64</math>, we can express the difference of the two polynomials with a quartic polynomial that has roots at <math>t=4,16,36,64</math>, so | ||
− | <div style="text-align:center;"><math>x^2 | + | <div style="text-align:center;"><math>x^2(t-9)(t-25)(t-49)+y^2(t-1)(t-25)(t-49)</math> <math>+z^2(t-1)(t-9)(t-49)+w^2(t-1)(t-9)(t-25) - (t-1)(t-9)(t-25)(t-49) = -(t-4)(t-16)(t-36)(t-64) </math> |
− | |||
</div> | </div> | ||
− | + | Note the leading coefficient of the RHS is <math>-1</math> because it must match the leading coefficient of the LHS, which is <math>-1</math>. | |
Now we can plug in <math>t=1</math> into the polynomial equation. Most terms drop, and we end up with | Now we can plug in <math>t=1</math> into the polynomial equation. Most terms drop, and we end up with | ||
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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}}\\ | 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}}\\ | ||
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}}\\ | 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}}\\ | ||
− | 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}}</cmath> | + | 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}}\end{align*}</cmath> |
Now adding them up, | Now adding them up, | ||
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<cmath>\frac{3^2(3\cdot 11\cdot 13+5\cdot 7\cdot 17)}{2^8}=3^2\cdot 4=\boxed{036}.</cmath> | <cmath>\frac{3^2(3\cdot 11\cdot 13+5\cdot 7\cdot 17)}{2^8}=3^2\cdot 4=\boxed{036}.</cmath> | ||
+ | |||
+ | /*Lengthy proof that any two cubic polynomials in <math>t</math> which are equal at 4 values of <math>t</math> are themselves equivalent: | ||
+ | Let the two polynomials be <math>A(t)</math> and <math>B(t)</math> and let them be equal at <math>t=a,b,c,d</math>. Thus we have <math>A(a) - B(a) = 0, A(b) - B(b) = 0, A(c) - B(c) = 0, A(d) - B(d) = 0</math>. Also the polynomial <math>A(t) - B(t)</math> is cubic, but it equals 0 at 4 values of <math>t</math>. Thus it must be equivalent to the polynomial 0, since if it were nonzero it would necessarily be able to be factored into <math>(t-a)(t-b)(t-c)(t-d)(</math>some nonzero polynomial<math>)</math> which would have a degree greater than or equal to 4, contradicting the statement that <math>A(t) - B(t)</math> is cubic. Because <math>A(t) - B(t) = 0, A(t)</math> and <math>B(t)</math> are equivalent and must be equal for all <math>t</math>. | ||
+ | |||
+ | '''Post script 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 == | == Solution 2 == | ||
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</div> | </div> | ||
Now the coefficient of <math>t^3</math> on both sides must be equal. Therefore we have <math>1+9+25+49+x^2+y^2+z^2+w^2=4+16+36+64\implies x^2+y^2+z^2+w^2=\boxed{036}</math>. | Now the coefficient of <math>t^3</math> on both sides must be equal. Therefore we have <math>1+9+25+49+x^2+y^2+z^2+w^2=4+16+36+64\implies x^2+y^2+z^2+w^2=\boxed{036}</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. | ||
+ | |||
+ | <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, maniuplate 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. | ||
+ | |||
+ | == Solution 4 (easy sol, not much algebra)== | ||
+ | |||
+ | Note that each equation is of the form <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> for <math>t=2^2, 4^2, 6^2, 8^2</math>. Removing fractions, we get | ||
+ | |||
+ | <cmath>x^2(t-3^2)(t-5^2)(t-7^2)+y^2(t-1^2)(t-5^2)(t-7^2) + z^2(t-1^2)(t-3^2)(t-7^2)+w^2(t-1^2)(t-3^2)(t-5^2) = (t-1^2)(t-3^2)(t-5^2)(t-7^2).</cmath> | ||
+ | |||
+ | Instead of expanding it fully, we will predict what the coefficients of the <math>t^4</math> and <math>t^3</math> term would be, since that's only what's important to us. So, 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>. We rearrange this to get <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 Formulas, 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>\text{-thinker123}</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}} | ||
− | [[Category:Intermediate Algebra Problems]] | + | [[Category: Intermediate Algebra Problems]] |
Latest revision as of 16:23, 18 August 2021
Contents
Problem
Determine if
Solution 1
Rewrite the system of equations as This equation is satisfied when , as then the equation is equivalent to the given equations. After clearing fractions, for each of the values , we have the equation . We can move the expression to the left hand side to obtain the difference of the polynomials: and
Since the polynomials are equal at , we can express the difference of the two polynomials with a quartic polynomial that has roots at , so
Note the leading coefficient of the RHS is because it must match the leading coefficient of the LHS, which is .
Now we can plug in into the polynomial equation. Most terms drop, and we end up with
so that
Similarly, we can plug in and get
Now adding them up,
with a sum of
/*Lengthy proof that any two cubic polynomials in which are equal at 4 values of are themselves equivalent: Let the two polynomials be and and let them be equal at . Thus we have . Also the polynomial is cubic, but it equals 0 at 4 values of . Thus it must be equivalent to the polynomial 0, since if it were nonzero it would necessarily be able to be factored into some nonzero polynomial which would have a degree greater than or equal to 4, contradicting the statement that is cubic. Because and are equivalent and must be equal for all .
Post script 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
Now the coefficient of on both sides must be equal. Therefore we have .
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, maniuplate 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 (easy sol, not much algebra)
Note that each equation is of the form for . Removing fractions, we get
Instead of expanding it fully, we will predict what the coefficients of the and term would be, since that's only what's important to us. So, we can write the long equation as . We rearrange this to get By Vieta's Formulas, 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,
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 |