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Difference between revisions of "2023 AMC 10B Problems/Problem 7"

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==Problem==
 
==Problem==
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HOG RIDER
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Square <math>ABCD</math> is rotated <math>20^{\circ}</math> clockwise about its center to obtain square <math>EFGH</math>, as shown below. What is the degree measure of <math>\angle EAB</math>?
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god emperor alexander
 
  
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==Answer==
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<asy>
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🐷🐖🐖🐖🔨🔨🔨
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size(170);
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defaultpen(linewidth(0.6));
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real r = 25;
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draw(dir(135)--dir(45)--dir(315)--dir(225)--cycle);
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draw(dir(135-r)--dir(45-r)--dir(315-r)--dir(225-r)--cycle);
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label("$A$",dir(135),NW);
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label("$B$",dir(45),NE);
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label("$C$",dir(315),SE);
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label("$D$",dir(225),SW);
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label("$E$",dir(135-r),N);
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label("$F$",dir(45-r),E);
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label("$G$",dir(315-r),S);
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label("$H$",dir(225-r),W);
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</asy>
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<math>\text{(A)}\ 24^{\circ} \qquad \text{(B)}\ 35^{\circ} \qquad \text{(C)}\ 30^{\circ} \qquad \text{(D)}\ 32^{\circ} \qquad \text{(E)}\ 20^{\circ}</math>
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 +
== Solution 1==
 +
 
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First, let's call the center of both squares <math>I</math>. Then, <math>\angle{AIE} = 20</math>, and since <math>\overline{EI} = \overline{AI}</math>, <math>\angle{AEI} = \angle{EAI} = 80</math>. Then, we know that <math>AI</math> bisects angle <math>\angle{DAB}</math>, so <math>\angle{BAI} = \angle{DAI} = 45</math>. Subtracting <math>45</math> from <math>80</math>, we get <math>\boxed{\text{(B)}  35}</math>
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 +
~jonathanzhou18
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== Solution 2 ==
 +
 
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First, label the point between <math>A</math> and <math>H</math> point <math>O</math> and the point between <math>A</math> and <math>H</math> point <math>P</math>. We know that <math>\angle{AOP} = 20</math> and that <math>\angle{A} = 90</math>. Subtracting <math>20</math> and <math>90</math> from <math>180</math>, we get that <math>\angle{APO}</math> is <math>70</math>. Subtracting <math>70</math> from <math>180</math>, we get that <math>\angle{OPB} = 110</math>. From this, we derive that <math>\angle{APE} = 110</math>. Since triangle <math>APE</math> is an isosceles triangle, we get that <math>\angle{EAP} = (180 - 110)/2 = 35</math>. Therefore, <math>\angle{EAB} = 35</math>. The answer is <math>\boxed{\text{(B)}  35}</math>.
 +
 
 +
~Stead (a.k.a. Aaron)
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 +
== Solution 3 ==
 +
 
 +
Call the center of both squares point <math>O</math>, and draw circle <math>O</math> such that it circumscribes the squares. <math>\angle{EOF} = 90</math> and <math>\angle{BOF} = 20</math>, so <math>\angle{EOB} = 70</math>. Since <math>\angle{EAB}</math> is inscribed in arc <math>\overset \frown {EB}</math>, <math>\angle{EAB} = 70/2 = \boxed{\textbf{(B) }35}</math>.
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 +
~hpotter2021
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== Solution 4 ==
 +
 
 +
Draw <math>EA</math>: we want to find <math>\angle EAB</math>. Call <math>P</math> the point at which <math>AB</math> and <math>EH</math> intersect. Reflecting <math>\triangle APE</math> over <math>EA</math>, we have a parallelogram. Since <math>\angle EPB = 70^{\circ}</math>, angle subtraction tells us that two of the angles of the parallelogram are <math>110^{\circ}</math>. The other two are equal to <math>2\angle EAB</math> (by properties of reflection).
 +
 
 +
Since angles on the transversal of a parallelogram sum to <math>180^{\circ}</math>, we have <math>2\angle EAB + 110 = 180</math>, yielding <math>\angle EAB = \boxed{\textbf{(B) }35}</math>
 +
 
 +
-Benedict T (countmath1)
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 +
== Solution 5 (Educated Guess) ==
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We call the point where <math>AB</math> and <math>EH</math> intersect I. We can make an educated guess that triangle AEI is isosceles so <math>AI=EI</math>, <math> \angle AIE = 110^{\circ} </math> , <math> \angle AIH = 20^{\circ} </math> , and <math>\angle EIB = 70^{\circ} </math> . So, we get <math> \angle EAI </math> is <math> (180^{\circ} - 110^{\circ})/2  = \boxed{\textbf{(B) }35}</math>.
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 +
~aleyang
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 +
==Solution 6 (Solution 1 Remastered)==
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Like in solution 1, we label the center of the squares \( R \).
 +
 
 +
Draw the lines \( AR \) and \( RE \). Because Vertex \( A \) is jusyt Vertex \( E \) shifted \( 20^\circ \) about the center \( R \), \( \angle ARE = 20^\circ \).
 +
 
 +
Now, notice how \( AM \) and \( ME \) are the diagonals (or half the diagonals, doesn't really matter) of the squares \( ABCD \) and \( EFGH \) respectively. This implies that \( AM = ME \), and, connecting a new segment \( AE \), we see that \( \triangle ARE \) is isoseles, and \( AR = AE \), implying that \( \angle EAM \cong \angle MEA \). We can easily solve this through the sum of interior angles of a triangle, and we get \( m \angle EAM = m \angle MEA = 80^\circ \).
 +
 
 +
Because \( AM \) is a diagonal of \( \square ABCD \), that means that \( \angle BAC \) is bisected, and \( \angle MAD = 45^\circ \).
 +
 +
Finally, to find \( \angle EAB \), we are just subtracting \( 80^\circ \) and \( 45^\circ \), giving us \( 80^\circ - 45^\circ = 35^\circ \) or <math>\boxed{B}</math>.
 +
 
 +
~Pinotation
 +
 
 +
==Video Solution by MegaMath==
 +
 
 +
https://www.youtube.com/watch?v=KsAxW53-P0A&t=4s
 +
 
 +
~megahertz13
 +
 
 +
==Video Solution 2 by OmegaLearn==
 +
https://youtu.be/LI1Xq2onHHg
 +
 
 +
==Video Solution 3 by SpreadTheMathLove==
 +
https://www.youtube.com/watch?v=cT-0V4a3FYY
 +
 
 +
==Video Solution by Math-X (First understand the problem!!!)==
 +
https://youtu.be/EuLkw8HFdk4?si=Te_9kmP_bmBoKrTn&t=1393
 +
 
 +
~Math-X
 +
 
 +
==Video Solution==
 +
 
 +
https://youtu.be/R9uCV2KsXc8
 +
 
 +
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 +
 
 +
==Video Solution by Interstigation==
 +
https://youtu.be/gDnmvcOzxjg?si=cYB6uChy7Ue0UT4L
 +
 
 +
==See also==
 +
{{AMC10 box|year=2023|ab=B|num-b=6|num-a=8}}
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{{MAA Notice}}

Latest revision as of 19:24, 22 August 2025

Problem

Square $ABCD$ is rotated $20^{\circ}$ clockwise about its center to obtain square $EFGH$, as shown below. What is the degree measure of $\angle EAB$?

[asy] size(170); defaultpen(linewidth(0.6)); real r = 25; draw(dir(135)--dir(45)--dir(315)--dir(225)--cycle); draw(dir(135-r)--dir(45-r)--dir(315-r)--dir(225-r)--cycle); label("$A$",dir(135),NW); label("$B$",dir(45),NE); label("$C$",dir(315),SE); label("$D$",dir(225),SW); label("$E$",dir(135-r),N); label("$F$",dir(45-r),E); label("$G$",dir(315-r),S); label("$H$",dir(225-r),W); [/asy]

$\text{(A)}\ 24^{\circ} \qquad \text{(B)}\ 35^{\circ} \qquad \text{(C)}\ 30^{\circ} \qquad \text{(D)}\ 32^{\circ} \qquad \text{(E)}\ 20^{\circ}$

Solution 1

First, let's call the center of both squares $I$. Then, $\angle{AIE} = 20$, and since $\overline{EI} = \overline{AI}$, $\angle{AEI} = \angle{EAI} = 80$. Then, we know that $AI$ bisects angle $\angle{DAB}$, so $\angle{BAI} = \angle{DAI} = 45$. Subtracting $45$ from $80$, we get $\boxed{\text{(B)} 35}$

~jonathanzhou18

Solution 2

First, label the point between $A$ and $H$ point $O$ and the point between $A$ and $H$ point $P$. We know that $\angle{AOP} = 20$ and that $\angle{A} = 90$. Subtracting $20$ and $90$ from $180$, we get that $\angle{APO}$ is $70$. Subtracting $70$ from $180$, we get that $\angle{OPB} = 110$. From this, we derive that $\angle{APE} = 110$. Since triangle $APE$ is an isosceles triangle, we get that $\angle{EAP} = (180 - 110)/2 = 35$. Therefore, $\angle{EAB} = 35$. The answer is $\boxed{\text{(B)} 35}$.

~Stead (a.k.a. Aaron)

Solution 3

Call the center of both squares point $O$, and draw circle $O$ such that it circumscribes the squares. $\angle{EOF} = 90$ and $\angle{BOF} = 20$, so $\angle{EOB} = 70$. Since $\angle{EAB}$ is inscribed in arc $\overset \frown {EB}$, $\angle{EAB} = 70/2 = \boxed{\textbf{(B) }35}$.

~hpotter2021

Solution 4

Draw $EA$: we want to find $\angle EAB$. Call $P$ the point at which $AB$ and $EH$ intersect. Reflecting $\triangle APE$ over $EA$, we have a parallelogram. Since $\angle EPB = 70^{\circ}$, angle subtraction tells us that two of the angles of the parallelogram are $110^{\circ}$. The other two are equal to $2\angle EAB$ (by properties of reflection).

Since angles on the transversal of a parallelogram sum to $180^{\circ}$, we have $2\angle EAB + 110 = 180$, yielding $\angle EAB = \boxed{\textbf{(B) }35}$

-Benedict T (countmath1)

Solution 5 (Educated Guess)

We call the point where $AB$ and $EH$ intersect I. We can make an educated guess that triangle AEI is isosceles so $AI=EI$, $\angle AIE = 110^{\circ}$ , $\angle AIH = 20^{\circ}$ , and $\angle EIB = 70^{\circ}$ . So, we get $\angle EAI$ is $(180^{\circ} - 110^{\circ})/2 = \boxed{\textbf{(B) }35}$.

~aleyang

Solution 6 (Solution 1 Remastered)

Like in solution 1, we label the center of the squares \( R \).

Draw the lines \( AR \) and \( RE \). Because Vertex \( A \) is jusyt Vertex \( E \) shifted \( 20^\circ \) about the center \( R \), \( \angle ARE = 20^\circ \).

Now, notice how \( AM \) and \( ME \) are the diagonals (or half the diagonals, doesn't really matter) of the squares \( ABCD \) and \( EFGH \) respectively. This implies that \( AM = ME \), and, connecting a new segment \( AE \), we see that \( \triangle ARE \) is isoseles, and \( AR = AE \), implying that \( \angle EAM \cong \angle MEA \). We can easily solve this through the sum of interior angles of a triangle, and we get \( m \angle EAM = m \angle MEA = 80^\circ \).

Because \( AM \) is a diagonal of \( \square ABCD \), that means that \( \angle BAC \) is bisected, and \( \angle MAD = 45^\circ \).

Finally, to find \( \angle EAB \), we are just subtracting \( 80^\circ \) and \( 45^\circ \), giving us \( 80^\circ - 45^\circ = 35^\circ \) or $\boxed{B}$.

~Pinotation

Video Solution by MegaMath

https://www.youtube.com/watch?v=KsAxW53-P0A&t=4s

~megahertz13

Video Solution 2 by OmegaLearn

https://youtu.be/LI1Xq2onHHg

Video Solution 3 by SpreadTheMathLove

https://www.youtube.com/watch?v=cT-0V4a3FYY

Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/EuLkw8HFdk4?si=Te_9kmP_bmBoKrTn&t=1393

~Math-X

Video Solution

https://youtu.be/R9uCV2KsXc8

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution by Interstigation

https://youtu.be/gDnmvcOzxjg?si=cYB6uChy7Ue0UT4L

See also

2023 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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