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

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Sqrt <math>ABCD</math> is rotated <math>20^{\circ}</math> clockwise about its center to obtain square <math>EFGH</math>, as shown below(Please help me add diagram and then remove this). What is the degree measure of <math>\angle EAB</math>?
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==Problem==
 +
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>?
 +
 
 +
<asy>
 +
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);
 +
</asy>
  
 
<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>
 
<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==
 
== 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{35 \text{(B)}}</math>
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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>
 +
 
 +
~jonathanzhou18
 +
 
 +
== Solution 2 ==
 +
 
 +
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)
 +
 
 +
== 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>.
 +
 
 +
~hpotter2021
 +
 
 +
== 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)
 +
 
 +
== Solution 5 (Educated Guess) ==
 +
 
 +
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>.
 +
 
 +
~aleyang
 +
 
 +
==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)
  
−
== Solution 2 ==
+
==Video Solution by Interstigation==
 +
https://youtu.be/gDnmvcOzxjg?si=cYB6uChy7Ue0UT4L
  
−
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} 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 {APE} is an isosceles triangle, we get that </math>\angle{EAP} = (180 - 110)/2 = 35<math>. Therefore, </math>\angle{EAP} = 35$.
+
==See also==
 +
{{AMC10 box|year=2023|ab=B|num-b=6|num-a=8}}
 +
{{MAA Notice}}

Latest revision as of 11:22, 20 October 2024

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

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

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