Difference between revisions of "2021 AMC 10B Problems/Problem 21"
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==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} | + | 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 |
==Solution 3 (Fakesolve):== | ==Solution 3 (Fakesolve):== | ||
− | Assume that E is the midpoint of < | + | Assume that E is the midpoint of <math>\overline{AB}</math>. Then, <math>\overline{AE}=\frac{1}{2}</math> and since <math>C'D=\frac{1}{3}</math>, <math>\overline{AC'}=\frac{2}{3}</math>. By the Pythagorean Theorem, <math>\overline{EC'}=\frac{5}{6}</math>. It easily follows that our desired perimeter is <math>2 \rightarrow \boxed{A}</math> ~samrocksnature |
== Video Solution by OmegaLearn (Using Pythagoras Theorem and Similar Triangles) == | == Video Solution by OmegaLearn (Using Pythagoras Theorem and Similar Triangles) == |
Revision as of 17:19, 12 February 2021
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
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
. Thats 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
Video Solution by OmegaLearn (Using Pythagoras Theorem and Similar Triangles)
~ pi_is_3.14
2021 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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All AMC 10 Problems and Solutions |