Difference between revisions of "2021 AMC 10B Problems/Problem 21"
(→Problem) |
(→Solution 2) |
||
(2 intermediate revisions by one other user not shown) | |||
Line 8: | Line 8: | ||
pair CC=(0.666666666666,1); | pair CC=(0.666666666666,1); | ||
pair D=(1,1); | pair D=(1,1); | ||
− | pair F=(1,0. | + | pair F=(1,0.440062); |
pair C=(1,0); | pair C=(1,0); | ||
pair B=(0,0); | pair B=(0,0); | ||
− | pair G=(0,0. | + | pair G=(0,0.22005); |
pair H=(-0.13,0.41); | pair H=(-0.13,0.41); | ||
pair E=(0,0.5); | pair E=(0,0.5); | ||
Line 39: | Line 39: | ||
==Solution 2== | ==Solution 2== | ||
− | Let line we're reflecting over be <math>\ell</math>, and let the points where it hits <math>AB</math> and <math>CD</math>, be <math>M</math> and <math>N</math>, 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 <math>\ell</math>. The segment <math>CC'</math> has slope <math>\frac{0 - 1}{1 - 2/3} = -3</math>, implying line <math>\ell</math> has a slope of <math>\frac{1}{3}</math>. Also, the midpoint of segment <math>CC'</math> is <math>\left( \frac{5}{6}, \frac{1}{2} \right)</math>, so line <math>\ell</math> passes through this point. Then, we get the equation of line <math>\ell</math> is simply <math>y = \frac{1}{3} x + \frac{2}{9}</math>. Then, if the point where <math>B</math> is reflected over line <math>\ell</math> is <math>B'</math>, then we get <math>BB'</math> is the line <math>y = -3x</math>. The intersection of <math>\ell</math> and segment <math>BB'</math> is <math>\left( - \frac{1}{15}, \frac{1}{5} \right)</math>. So, we get <math>B' = \left(- \frac{2}{15}, \frac{2}{5} \right)</math>. Then, line segment <math>B'C'</math> has equation <math>y = \frac{3}{4} x + \frac{1}{2}</math>, so the point <math>E</math> is the <math>y</math>-intercept, or <math>\left(0, \frac{1}{2} \right)</math>. This implies that <math>AE = \frac{1}{2}, AC' = \frac{2}{3}</math>, and by the Pythagorean Theorem, <math>EC' = \frac{5}{6}</math> (or you could notice <math>\triangle AEC'</math> is a <math>3-4-5</math> right triangle). Then, the perimeter is <math>\frac{1}{2} + \frac{2}{3} + \frac{5}{6} = 2</math>, so our answer is <math>\boxed{\textbf{(A)} ~2}</math>. ~rocketsri | + | Let the line we're reflecting over be <math>\ell</math>, and let the points where it hits <math>AB</math> and <math>CD</math>, be <math>M</math> and <math>N</math>, 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 <math>\ell</math>. The segment <math>CC'</math> has slope <math>\frac{0 - 1}{1 - 2/3} = -3</math>, implying line <math>\ell</math> has a slope of <math>\frac{1}{3}</math>. Also, the midpoint of segment <math>CC'</math> is <math>\left( \frac{5}{6}, \frac{1}{2} \right)</math>, so line <math>\ell</math> passes through this point. Then, we get the equation of line <math>\ell</math> is simply <math>y = \frac{1}{3} x + \frac{2}{9}</math>. Then, if the point where <math>B</math> is reflected over line <math>\ell</math> is <math>B'</math>, then we get <math>BB'</math> is the line <math>y = -3x</math>. The intersection of <math>\ell</math> and segment <math>BB'</math> is <math>\left( - \frac{1}{15}, \frac{1}{5} \right)</math>. So, we get <math>B' = \left(- \frac{2}{15}, \frac{2}{5} \right)</math>. Then, line segment <math>B'C'</math> has equation <math>y = \frac{3}{4} x + \frac{1}{2}</math>, so the point <math>E</math> is the <math>y</math>-intercept, or <math>\left(0, \frac{1}{2} \right)</math>. This implies that <math>AE = \frac{1}{2}, AC' = \frac{2}{3}</math>, and by the Pythagorean Theorem, <math>EC' = \frac{5}{6}</math> (or you could notice <math>\triangle AEC'</math> is a <math>3-4-5</math> right triangle). Then, the perimeter is <math>\frac{1}{2} + \frac{2}{3} + \frac{5}{6} = 2</math>, so our answer is <math>\boxed{\textbf{(A)} ~2}</math>. ~rocketsri |
==Solution 3 (Fakesolve):== | ==Solution 3 (Fakesolve):== | ||
Line 59: | Line 59: | ||
==Solution 5 (Trig)== | ==Solution 5 (Trig)== | ||
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 <math>2\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 | 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 <math>2\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 | ||
+ | ==Solution 6== | ||
+ | It is easy to prove that the ratio of the sum of the larger leg and hypotenuse to the smaller leg depends monotonically on the angle of a right triangle, which means: | ||
+ | <cmath>C' F + DF = CF + DF = CD = AD = 3 C'D \implies C' D : DF : C' F = 3 : 4 : 5.</cmath> | ||
+ | |||
+ | For a similar triangle, the ratio of the perimeter to the larger leg is <math>\frac {3 + 4 +5}{4} = 3.</math> | ||
+ | |||
+ | <math>\triangle AEC' \sim \triangle DC'F \implies</math> the perimeter of triangle <math>\bigtriangleup AEC'</math> is <math>3 AC' = \boxed{\textbf{(A)} ~ 2}.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
== Video Solution by OmegaLearn (Using Pythagoras Theorem and Similar Triangles) == | == Video Solution by OmegaLearn (Using Pythagoras Theorem and Similar Triangles) == |
Latest revision as of 23:49, 3 March 2024
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 the 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
, 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
Solution 6
It is easy to prove that the ratio of the sum of the larger leg and hypotenuse to the smaller leg depends monotonically on the angle of a right triangle, which means:
For a similar triangle, the ratio of the perimeter to the larger leg is
the perimeter of triangle
is
vladimir.shelomovskii@gmail.com, vvsss
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.