2023 AMC 10B Problems/Problem 7
Contents
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
.
~yourmomisalosinggame (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)
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
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.