Difference between revisions of "2023 AMC 10B Problems/Problem 7"
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==Problem== | ==Problem== | ||
− | + | Square <math>ABCD</math> is rotated <math>20^{\circ}</math> clockwise about its center to obtain square <math>EFGH</math>, as shown below. | |
+ | [[Image:IMG 1031.jpeg]] | ||
− | + | What is the degree measure of <math>\angle EAB</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> | ||
− | this is a | + | == Solution 1== |
+ | |||
+ | 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) | ||
+ | |||
+ | ==Video Solution by Interstigation== | ||
+ | https://youtu.be/gDnmvcOzxjg?si=cYB6uChy7Ue0UT4L | ||
+ | |||
+ | ==See also== | ||
+ | {{AMC10 box|year=2023|ab=B|num-b=6|num-a=8}} | ||
+ | {{MAA Notice}} |
Latest revision as of 07:18, 7 July 2024
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3
- 5 Solution 4
- 6 Solution 5 (Educated Guess)
- 7 Video Solution by MegaMath
- 8 Video Solution 2 by OmegaLearn
- 9 Video Solution 3 by SpreadTheMathLove
- 10 Video Solution by Math-X (First understand the problem!!!)
- 11 Video Solution
- 12 Video Solution by Interstigation
- 13 See also
Problem
Square is rotated
clockwise about its center to obtain square
, as shown below.
What is the degree measure of ?
Solution 1
First, let's call the center of both squares . Then,
, and since
,
. Then, we know that
bisects angle
, so
. Subtracting
from
, we get
~jonathanzhou18
Solution 2
First, label the point between and
point
and the point between
and
point
. We know that
and that
. Subtracting
and
from
, we get that
is
. Subtracting
from
, we get that
. From this, we derive that
. Since triangle
is an isosceles triangle, we get that
. Therefore,
. The answer is
.
~Stead (a.k.a. Aaron)
Solution 3
Call the center of both squares point , and draw circle
such that it circumscribes the squares.
and
, so
. Since
is inscribed in arc
,
.
~hpotter2021
Solution 4
Draw : we want to find
. Call
the point at which
and
intersect. Reflecting
over
, we have a parallelogram. Since
, angle subtraction tells us that two of the angles of the parallelogram are
. The other two are equal to
(by properties of reflection).
Since angles on the transversal of a parallelogram sum to , we have
, yielding
-Benedict T (countmath1)
Solution 5 (Educated Guess)
We call the point where and
intersect I. We can make an educated guess that triangle AEI is isosceles so
,
,
, and
. So, we get
is
.
~aleyang
Video Solution by MegaMath
https://www.youtube.com/watch?v=KsAxW53-P0A&t=4s
~megahertz13
Video Solution 2 by OmegaLearn
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
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.