Difference between revisions of "2021 AMC 10B Problems/Problem 21"
m (→Solution 5) |
m (→Solution 5 (Trig)) |
||
Line 57: | Line 57: | ||
==Solution 5 (Trig)== | ==Solution 5 (Trig)== | ||
− | Draw a perpendicular line from <math>AB</math> at <math>E</math>, and let it intersect <math>DC</math> at <math>E'</math>. The angle between <math>AB</math> and <math>EE'</math> is 2<math>\theta</math>, where <math>\theta</math> is the angle between the fold and a line perpendicular to <math>AD</math>. The slope of the fold is <math>\frac{1}{3}</math> because it is perpendicular to <math>CC'</math> (<math>CC'</math> has a slope of <math>-3</math> using points <math>C'</math> and <math>C</math>, and perpendicular lines have slopes negative inverses of each other). Using tangent double angle formula, the slope of <math>EC'</math> is 3/ | + | Draw a perpendicular line from <math>\overline{AB}</math> at <math>E</math>, and let it intersect <math>\overline{DC}</math> at <math>E'</math>. The angle between <math>\overline{AB}</math> and <math>\overline{EE'}</math> is 2<math>\theta</math>, where <math>\theta</math> is the angle between the fold and a line perpendicular to <math>\overline{AD}</math>. The slope of the fold is <math>\frac{1}{3}</math> because it is perpendicular to <math>\overline{CC'}</math> (<math>\overline{CC'}</math> has a slope of <math>-3</math> using points <math>C'</math> and <math>C</math>, and perpendicular lines have slopes negative inverses of each other). Using tangent double angle formula, the slope of <math>\overline{EC'}</math> is <math>\frac{3}{4}</math>, which implies <math>\overline{AE}</math> = <math>\frac{1}{2}</math>. By the Pythagorean Theorem, <math>\overline{EC'}=\frac{5}{6}</math>. It easily follows that our desired perimeter is <math>\boxed{\textbf{(A)} ~2}</math> ~forrestc |
== Video Solution by OmegaLearn (Using Pythagoras Theorem and Similar Triangles) == | == Video Solution by OmegaLearn (Using Pythagoras Theorem and Similar Triangles) == |
Revision as of 02:20, 8 November 2022
Contents
Problem
A square piece of paper has side length and vertices
and
in that order. As shown in the figure, the paper is folded so that vertex
meets edge
at point
, and edge
intersects edge
at point
. Suppose that
. What is the perimeter of triangle
Solution 1
We can set the point on where the fold occurs as point
. Then, we can set
as
, and
as
because of symmetry due to the fold. It can be recognized that this is a right triangle, and solving for
, we get,
We know this is a 3-4-5 triangle because the side lengths are . We also know that
is similar to
because angle
is a right angle. Now, we can use similarity to find out that the perimeter is just the perimeter of
. That's just
. Therefore, the final answer is
~Tony_Li2007
Solution 2
Let line we're reflecting over be , and let the points where it hits
and
, be
and
, respectively. Notice, to reflect over a line we find the perpendicular passing through the midpoint of the two points (which are the reflected and the original). So, we first find the equation of the line
. The segment
has slope
, implying line
has a slope of
. Also, the midpoint of segment
is
, so line
passes through this point. Then, we get the equation of line
is simply
. Then, if the point where
is reflected over line
is
, then we get
is the line
. The intersection of
and segment
is
. So, we get
. Then, line segment
has equation
, so the point
is the
-intercept, or
. This implies that
, and by the Pythagorean Theorem,
(or you could notice
is a
right triangle). Then, the perimeter is
, so our answer is
. ~rocketsri
Solution 3 (Fakesolve):
Assume that E is the midpoint of . Then,
and since
,
. By the Pythagorean Theorem,
. It easily follows that our desired perimeter is
~samrocksnature
Solution 4
As described in Solution 1, we can find that , and
Then, we can find we can find the length of by expressing the length of
in two different ways, in terms of
. If let
, by the Pythagorean Theorem we have that
Therefore, since we know that
is right, by Pythagoras again we have that
Another way we can express is by using Pythagoras on
, where
is the foot of the perpendicular from
to
We see that
is a rectangle, so we know that
. Secondly, since
. Therefore, through the Pythagorean Theorem, we find that
Since we have found two expressions for the same length, we have the equation Solving this, we find that
.
Finally, we see that the perimeter of is
which we can simplify to be
. Thus, the answer is
~laffytaffy
Solution 5 (Trig)
Draw a perpendicular line from at
, and let it intersect
at
. The angle between
and
is 2
, where
is the angle between the fold and a line perpendicular to
. The slope of the fold is
because it is perpendicular to
(
has a slope of
using points
and
, and perpendicular lines have slopes negative inverses of each other). Using tangent double angle formula, the slope of
is
, which implies
=
. By the Pythagorean Theorem,
. It easily follows that our desired perimeter is
~forrestc
Video Solution by OmegaLearn (Using Pythagoras Theorem and Similar Triangles)
~ pi_is_3.14
Video Solution by Interstigation
~Interstigation
Video Solution by The Power of Logic
https://www.youtube.com/watch?v=5kbQHcx1FfE
~The Power of Logic
See Also
2021 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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.