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Difference between revisions of "2021 AMC 12A Problems/Problem 9"
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{{duplicate|[[2021 AMC 10A Problems#Problem 10|2021 AMC 10A #10]] and [[2021 AMC 12A Problems#Problem 9|2021 AMC 12A #9]]}}
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==Problem==
 
==Problem==
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Triangle <math>ABC</math> lies in a plane with <math>AB=13</math>, <math>AC=14</math>, and <math>BC=15</math>. For any point <math>X</math> in the plane of <math>\triangle ABC</math>, let <math>f(X)</math> denote the sum of the three distances from <math>X</math> to the three vertices of <math>\triangle ABC</math>. Let <math>P</math> be the unique point in the plane of <math>\triangle ABC</math> where <math>f(X)</math> is minimized. Then <math>AP^2+BP^2+CP^2=\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. What is the value of <math>m+n</math>?
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Which of the following is equivalent to
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<cmath>(2+3)(2^2+3^2)(2^4+3^4)(2^8+3^8)(2^{16}+3^{16})(2^{32}+3^{32})(2^{64}+3^{64})?</cmath>
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<math>\textbf{(A)} ~3^{127} + 2^{127} \qquad\textbf{(B)} ~3^{127} + 2^{127} + 2 \cdot 3^{63} + 3 \cdot 2^{63} \qquad\textbf{(C)} ~3^{128}-2^{128} \qquad\textbf{(D)} ~3^{128} + 2^{128} \qquad\textbf{(E)} ~5^{127}</math>
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==Solution 1==
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All you need to do is multiply the entire equation by <math>(3-2)</math>. Then all the terms will easily simplify by difference of squares and you will get <math>3^{128}-2^{128}</math> or <math>\boxed{C}</math> as your final answer. Notice you don't need to worry about <math>3-2</math> because that's equal to <math>1</math>.
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-Lemonie
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==Solution 2==
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 +
If you weren't able to come up with the <math>(3 - 2)</math> insight, then you could just notice that the answer is divisible by
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<math>(2 + 3) = 5</math>, and <math>(2^2 + 3^2) = 13</math>. We can then use Fermat's Little Theorem for <math>p = 5, 13</math> on the answer choices to determine which of the answer choices are divisible by both <math>5</math> and <math>13</math>. This is <math>\boxed{C}</math>.
 +
 
 +
-mewto
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==Solution 3==
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 +
After expanding the first few terms, the result after each term appears to be <math>2^{2^n-1} + 2^{2^n-2}\cdot{3^1} + 2^{2^n-3}\cdot{3^2} + ... + 2^1\cdot{3^{2^n-2}} + 3^{2^n-1}</math> where n is the number of terms expanded. We can prove this using mathematical induction. The base step is trivial. When expanding another term, all of the previous terms multiplied by <math>2^{2^{n-1}}</math> would give <math>2^{2^n-1} + 2^{2^n-2}\cdot{3^1} + 2^{2^n-3}\cdot{3^2} + ... + 2^{2^{n-1}+1}\cdot{3^{2^{n-1}-1}} + 2^{2^{n-1}}\cdot{3^{2^{n-1}}}</math>, and all the previous terms multiplied by <math>3^{2^{n-1}}</math> would give <math>3^{2^n-1} + 3^{2^n-2}\cdot{2^1} + 3^{2^n-3}\cdot{2^2} + ... + 3^{2^{n-1}+1}\cdot{2^{2^{n-1}-1}} + 3^{2^{n-1}}\cdot{2^{2^{n-1}}}</math>. Their sum is equal to <math>2^{2^n-1} + 2^{2^n-2}\cdot{3^1} + 2^{2^n-3}\cdot{3^2} + ... + 2^1\cdot{3^{2^n-2}} + 3^{2^n-1}</math>, so the proof is complete. Since <math>\frac{3^{2^n}-2^{2^n}}{3-2}</math> is equal to <math>2^{2^n-1} + 2^{2^n-2}\cdot{3^1} + 2^{2^n-3}\cdot{3^2} + ... + 2^1\cdot{3^{2^n-2}} + 3^{2^n-1}</math>, the answer is <math>\frac{3^{2^7}-2^{2^7}}{3-2}=\boxed{C}</math>.
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 +
-SmileKat32
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 +
== Solution 4 (Engineer's Induction) ==
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We can compute some of the first few partial products, and notice that <math>\prod_{k = 0}^{2^n} (2^{2^n}+3^{2^n}) = 3^{2^{n+1}} - 2^{2^{n+1}}</math>. As we don't have to prove this, we get the product is <math>3^{2^7} - 2^{2^7} = 3^{128} - 2^{128}</math>, and smugly click <math>\boxed{\textbf{(C)} ~3^{128} - 2^{128}}</math>. ~rocketsri
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 +
 
 +
 
 +
==Video Solution by Aaron He==
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https://www.youtube.com/watch?v=xTGDKBthWsw&t=9m30s
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 +
==Video Solution(Conjugation, Difference of Squares)==
 +
https://www.youtube.com/watch?v=gXaIyeMF7Qo&list=PLexHyfQ8DMuKqltG3cHT7Di4jhVl6L4YJ&index=9
 +
 
 +
==Video Solution by Hawk Math==
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https://www.youtube.com/watch?v=P5al76DxyHY
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 +
== Video Solution by OmegaLearn(Factorizations/Telescoping& Meta-solving) ==
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https://youtu.be/H34IFMlq7Lk
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 +
~ pi_is_3.14
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==Video Solution (Quick and Simple)==
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https://youtu.be/Pm3euI3jyDk
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 +
~ Education, the Study of Everything
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==Video Solution 6==
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https://youtu.be/-MJXKZowfO0
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~savannahsolver
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==Video Solution by TheBeautyofMath==
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https://youtu.be/s6E4E06XhPU?t=771 (for AMC 10A)
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 +
https://youtu.be/cckGBU2x1zg?t=548 (for AMC 12A)
  
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==Solution==
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~IceMatrix
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Point <math>P</math> is the [[Brocard point|Brocard point]] of <math>\triangle ABC</math>, where <math>\angle APB = \angle BPC = \angle APC = 120^\circ</math> and <math>\triangle APB, \triangle BPC, \triangle APC</math> are all <math>30-30-120</math> triangles, and the squares of the side lengths are in the ratio <math>\frac{\frac{1}{1}}{3}</math> which can easily be seen by dividing this triangle into two smaller <math>30-60-90</math> triangles. It follows <math>AP^2+BP^2=\frac{2}{3}AB^2</math>, <math>AP^2+CP^2=\frac{2}{3}AC^2</math>, and <math>BP^2+CP^2=\frac{2}{3}BC^2</math>. Now <math>2AP^2+2BP^2+2CP^2=\frac{2}{3}(AB^2+AC^2+BC^2)</math> because each got counted twice, so <math>AP^2+BP^2+CP^2=\frac{1}{3}(13^2+14^2+15^2)=\frac{590}{3}</math>, and <math>m+n=\boxed{593}</math>.
 
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~icematrix2
 
  
 
==See also==
 
==See also==
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{{AMC10 box|year=2021|ab=A|num-b=9|num-a=11}}
 
{{AMC12 box|year=2021|ab=A|num-b=8|num-a=10}}
 
{{AMC12 box|year=2021|ab=A|num-b=8|num-a=10}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 17:44, 25 February 2021

The following problem is from both the 2021 AMC 10A #10 and 2021 AMC 12A #9, so both problems redirect to this page.

Problem

Which of the following is equivalent to \[(2+3)(2^2+3^2)(2^4+3^4)(2^8+3^8)(2^{16}+3^{16})(2^{32}+3^{32})(2^{64}+3^{64})?\] $\textbf{(A)} ~3^{127} + 2^{127} \qquad\textbf{(B)} ~3^{127} + 2^{127} + 2 \cdot 3^{63} + 3 \cdot 2^{63} \qquad\textbf{(C)} ~3^{128}-2^{128} \qquad\textbf{(D)} ~3^{128} + 2^{128} \qquad\textbf{(E)} ~5^{127}$

Solution 1

All you need to do is multiply the entire equation by $(3-2)$. Then all the terms will easily simplify by difference of squares and you will get $3^{128}-2^{128}$ or $\boxed{C}$ as your final answer. Notice you don't need to worry about $3-2$ because that's equal to $1$.

-Lemonie

Solution 2

If you weren't able to come up with the $(3 - 2)$ insight, then you could just notice that the answer is divisible by $(2 + 3) = 5$, and $(2^2 + 3^2) = 13$. We can then use Fermat's Little Theorem for $p = 5, 13$ on the answer choices to determine which of the answer choices are divisible by both $5$ and $13$. This is $\boxed{C}$.

-mewto

Solution 3

After expanding the first few terms, the result after each term appears to be $2^{2^n-1} + 2^{2^n-2}\cdot{3^1} + 2^{2^n-3}\cdot{3^2} + ... + 2^1\cdot{3^{2^n-2}} + 3^{2^n-1}$ where n is the number of terms expanded. We can prove this using mathematical induction. The base step is trivial. When expanding another term, all of the previous terms multiplied by $2^{2^{n-1}}$ would give $2^{2^n-1} + 2^{2^n-2}\cdot{3^1} + 2^{2^n-3}\cdot{3^2} + ... + 2^{2^{n-1}+1}\cdot{3^{2^{n-1}-1}} + 2^{2^{n-1}}\cdot{3^{2^{n-1}}}$, and all the previous terms multiplied by $3^{2^{n-1}}$ would give $3^{2^n-1} + 3^{2^n-2}\cdot{2^1} + 3^{2^n-3}\cdot{2^2} + ... + 3^{2^{n-1}+1}\cdot{2^{2^{n-1}-1}} + 3^{2^{n-1}}\cdot{2^{2^{n-1}}}$. Their sum is equal to $2^{2^n-1} + 2^{2^n-2}\cdot{3^1} + 2^{2^n-3}\cdot{3^2} + ... + 2^1\cdot{3^{2^n-2}} + 3^{2^n-1}$, so the proof is complete. Since $\frac{3^{2^n}-2^{2^n}}{3-2}$ is equal to $2^{2^n-1} + 2^{2^n-2}\cdot{3^1} + 2^{2^n-3}\cdot{3^2} + ... + 2^1\cdot{3^{2^n-2}} + 3^{2^n-1}$, the answer is $\frac{3^{2^7}-2^{2^7}}{3-2}=\boxed{C}$.

-SmileKat32

Solution 4 (Engineer's Induction)

We can compute some of the first few partial products, and notice that $\prod_{k = 0}^{2^n} (2^{2^n}+3^{2^n}) = 3^{2^{n+1}} - 2^{2^{n+1}}$. As we don't have to prove this, we get the product is $3^{2^7} - 2^{2^7} = 3^{128} - 2^{128}$, and smugly click $\boxed{\textbf{(C)} ~3^{128} - 2^{128}}$. ~rocketsri


Video Solution by Aaron He

https://www.youtube.com/watch?v=xTGDKBthWsw&t=9m30s

Video Solution(Conjugation, Difference of Squares)

https://www.youtube.com/watch?v=gXaIyeMF7Qo&list=PLexHyfQ8DMuKqltG3cHT7Di4jhVl6L4YJ&index=9

Video Solution by Hawk Math

https://www.youtube.com/watch?v=P5al76DxyHY

Video Solution by OmegaLearn(Factorizations/Telescoping& Meta-solving)

https://youtu.be/H34IFMlq7Lk

~ pi_is_3.14

Video Solution (Quick and Simple)

https://youtu.be/Pm3euI3jyDk

~ Education, the Study of Everything

Video Solution 6

https://youtu.be/-MJXKZowfO0

~savannahsolver

Video Solution by TheBeautyofMath

https://youtu.be/s6E4E06XhPU?t=771 (for AMC 10A)

https://youtu.be/cckGBU2x1zg?t=548 (for AMC 12A)

~IceMatrix

See also

2021 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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 10 Problems and Solutions
2021 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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