Difference between revisions of "2019 AMC 12A Problems/Problem 21"
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− | ==Problem== | + | == Problem == |
+ | Let <cmath>z=\frac{1+i}{\sqrt{2}}.</cmath>What is <cmath>\left(z^{1^2}+z^{2^2}+z^{3^2}+\dots+z^{{12}^2}\right) \cdot \left(\frac{1}{z^{1^2}}+\frac{1}{z^{2^2}}+\frac{1}{z^{3^2}}+\dots+\frac{1}{z^{{12}^2}}\right)?</cmath> | ||
− | |||
<math>\textbf{(A) } 18 \qquad \textbf{(B) } 72-36\sqrt2 \qquad \textbf{(C) } 36 \qquad \textbf{(D) } 72 \qquad \textbf{(E) } 72+36\sqrt2</math> | <math>\textbf{(A) } 18 \qquad \textbf{(B) } 72-36\sqrt2 \qquad \textbf{(C) } 36 \qquad \textbf{(D) } 72 \qquad \textbf{(E) } 72+36\sqrt2</math> | ||
− | == | + | == Solutions 1(Using Modular Functions) == |
− | |||
Note that <math>z = \mathrm{cis }(45^{\circ})</math>. | Note that <math>z = \mathrm{cis }(45^{\circ})</math>. | ||
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The term thus <math>\left(z^{1^2}+z^{2^2}+z^{3^2}+\dots+z^{{12}^2}\right)</math> simplifies to <math>6\mathrm{cis }(45^{\circ})</math>, while the term <math>\left(\frac{1}{z^{1^2}}+\frac{1}{z^{2^2}}+\frac{1}{z^{3^2}}+\dots+\frac{1}{z^{{12}^2}}\right)</math> simplifies to <math>\frac{6}{\mathrm{cis }(45^{\circ})}</math>. Upon multiplication, the <math>\mathrm{cis }(45^{\circ})</math> cancels out and leaves us with <math>\boxed{\textbf{(C) }36}</math>. | The term thus <math>\left(z^{1^2}+z^{2^2}+z^{3^2}+\dots+z^{{12}^2}\right)</math> simplifies to <math>6\mathrm{cis }(45^{\circ})</math>, while the term <math>\left(\frac{1}{z^{1^2}}+\frac{1}{z^{2^2}}+\frac{1}{z^{3^2}}+\dots+\frac{1}{z^{{12}^2}}\right)</math> simplifies to <math>\frac{6}{\mathrm{cis }(45^{\circ})}</math>. Upon multiplication, the <math>\mathrm{cis }(45^{\circ})</math> cancels out and leaves us with <math>\boxed{\textbf{(C) }36}</math>. | ||
− | ==Solution 2== | + | == Solution 2(Using Magnitudes and Conjugates to our Advantage) == |
+ | It is well known that if <math>|z|=1</math> then <math>\bar{z}=\frac{1}{z}</math>. Therefore, we have that the desired expression is equal to <cmath>\left(z^1+z^4+z^9+...+z^{144}\right)\left(\bar{z}^1+\bar{z}^4+\bar{z}^9+...+\bar{z}^{144}\right)</cmath> We know that <math>z=e^{\frac{i\pi}{4}}</math> so <math>\bar{z}=e^{\frac{i7\pi}{4}}</math>. Then, by De Moivre's Theorem, we have <cmath>\left(e^{\frac{i\pi}{4}}+e^{i\pi}+...+e^{2i\pi}\right)\left(e^{\frac{i7\pi}{4}}+e^{i7\pi}+...+e^{2i\pi}\right)</cmath> which can easily be computed as <math>\boxed{36}</math>. | ||
+ | |||
+ | == Solution 3 (Bashing) == | ||
+ | We first calculate that <math>z^4 = -1</math>. After a bit of calculation for the other even powers of <math>z</math>, we realize that they cancel out add up to zero. Now we can simplify the expression to <math>\left(z^{1^2} + z^{3^2} + ... + z^{11^2}\right)\left(\frac{1}{z^{1^2}} + \frac{1}{z^{3^2}} + ... + \frac{1}{z^{11^2}}\right)</math>. Then, we calculate the first few odd powers of <math>z</math>. We notice that <math>z^1 = z^9</math>, so the values cycle after every 8th power. Since all of the odd squares are a multiple of <math>8</math> away from each other, <math>z^1 = z^9 = z^{25} = ... = z^{121}</math>, so <math>z^{1^2} + z^{3^2} + ... + z^{11^2} = 6z^{1^2}</math>, and <math>\frac{1}{z^{1^2}} + \frac{1}{z^{3^2}} + ... + \frac{1}{z^{11^2}} = \frac{6}{z^{1^2}}</math>. When multiplied together, we get <math>6 \cdot 6 = \boxed{\textbf{(C) } 36}</math> as our answer. | ||
− | + | == Solution 4 (this is what people would write down on their scratch paper) == | |
+ | <math>z=\mathrm{cis }(\pi/4)</math> | ||
+ | |||
+ | Perfect squares mod 8: <math>1,4,1,0,1,4,1,0,1,4,1,0</math> | ||
+ | |||
+ | <math>1/z=\overline{z}=\mathrm{cis }(7\pi/4)</math> | ||
+ | |||
+ | <math>6\mathrm{cis }(\pi/4)\cdot 6\mathrm{cis }(7\pi/4)=\boxed{36}</math> | ||
+ | |||
+ | ~ MathIsFun286 | ||
+ | |||
+ | == Video Solution1 == | ||
+ | https://youtu.be/58pxV5Lkiks | ||
+ | ~ Education, the Study of Everything | ||
− | ==Solution | + | == Solution 5 == |
− | We | + | We notice that <math>z = e^{\pi i/4}</math> and <math>\frac1z = \overline{z} = e^{-\pi i/4}</math>. We then see that: <cmath>z^4 = e^{\pi i} = -1.</cmath> This means that: <cmath>z^{(2n)^2} = z^{4n^2} = (-1)^{n^2}.</cmath> In the first summation, there are <math>6</math> even exponents, and the <math>-1</math>'s will cancel among those. This means that: <cmath>\sum_{k=1}^{12} z^{k^2} = \sum_{m=0}^5 z^{(2m+1)^2}.</cmath> We can simplify <math>z^{(2m+1)^2}</math> to get: <cmath>z^{(2m+1)^2} = z^{4m^2} \cdot z^{4m} \cdot z = (-1)^{m^2} \cdot (-1)^{m} \cdot z.</cmath> We know that <math>m^2</math> and <math>m</math> will have the same parity so the <math>-1</math>'s multiply into a <math>1</math>, so what we get left is: <cmath>\sum_{m=0}^5 z^{(2m+1)^2} = \sum_{m=0}^5 z = 6z.</cmath> Now, for the conjugates, we notice that: <cmath>\overline{z}^4 = \overline{z^4} = -1.</cmath> This means that: <cmath>\overline{z}^{(2n)^2} = (-1)^{n^2}.</cmath> Therefore: <cmath>\sum_{k=1}^{12}\overline{z}^{k^2} = \sum_{m=0}^5 \overline{z}^{(2m+1)^2}.</cmath> Now, we see that: <cmath>\overline{z}^{(2m+1)^2} = \overline{z}^{4m^2} \cdot \overline{z}^{4m} \cdot z = (-1)^{m^2} \cdot (-1)^{m} \cdot \overline{z}.</cmath> Again, <math>m^2</math> and <math>m</math> have the same parity, so the <math>-1</math>'s multiply into a <math>1</math>, leaving us with <math>\overline{z}</math>. Therefore: <cmath>\sum_{m=0}^5 \overline{z}^{(2m+1)^2} = \sum_{m=0}^5 \overline{z} = 6\overline{z}.</cmath> Now, what we have is: <cmath>6z \cdot 6\overline{z} = 6e^{\pi i/4} \cdot 6e^{-\pi i/4} = 6 \cdot 6 = \boxed{\textbf{(C) }36}</cmath> |
− | ~ | + | ~ ap246 |
− | == | + | === Video Solution by Richard Rusczyk === |
+ | https://artofproblemsolving.com/videos/amc/2019amc12a/493 | ||
+ | == See Also == | ||
{{AMC12 box|year=2019|ab=A|num-b=20|num-a=22}} | {{AMC12 box|year=2019|ab=A|num-b=20|num-a=22}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 11:36, 3 December 2023
Contents
[hide]Problem
Let What is
Solutions 1(Using Modular Functions)
Note that .
Also note that for all positive integers because of De Moivre's Theorem. Therefore, we want to look at the exponents of each term modulo .
and are all
and are all
and are all
and are all
Therefore,
The term thus simplifies to , while the term simplifies to . Upon multiplication, the cancels out and leaves us with .
Solution 2(Using Magnitudes and Conjugates to our Advantage)
It is well known that if then . Therefore, we have that the desired expression is equal to We know that so . Then, by De Moivre's Theorem, we have which can easily be computed as .
Solution 3 (Bashing)
We first calculate that . After a bit of calculation for the other even powers of , we realize that they cancel out add up to zero. Now we can simplify the expression to . Then, we calculate the first few odd powers of . We notice that , so the values cycle after every 8th power. Since all of the odd squares are a multiple of away from each other, , so , and . When multiplied together, we get as our answer.
Solution 4 (this is what people would write down on their scratch paper)
Perfect squares mod 8:
~ MathIsFun286
Video Solution1
~ Education, the Study of Everything
Solution 5
We notice that and . We then see that: This means that: In the first summation, there are even exponents, and the 's will cancel among those. This means that: We can simplify to get: We know that and will have the same parity so the 's multiply into a , so what we get left is: Now, for the conjugates, we notice that: This means that: Therefore: Now, we see that: Again, and have the same parity, so the 's multiply into a , leaving us with . Therefore: Now, what we have is:
~ ap246
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2019amc12a/493
See Also
2019 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 20 |
Followed by Problem 22 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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