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} = \frac{-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}</math>, 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 <math>\overline{AB}< | + | 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}$ ~samrocksnature |
== Video Solution by OmegaLearn (Using Pythagoras Theorem and Similar Triangles) == | == Video Solution by OmegaLearn (Using Pythagoras Theorem and Similar Triangles) == |
Revision as of 17:18, 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 $\frac{0 - 1}{1 - 2/3} = \frac{-3}$ (Error compiling LaTeX. ! Missing } inserted.), 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 , AC' = \frac{2}{3}EC' = \frac{5}{6}\triangle AEC'3-4-5\frac{1}{2} + \frac{2}{3} + \frac{5}{6} = 2\boxed{\textbf{(A)} ~2}$. ~rocketsri
==Solution 3 (Fakesolve):== Assume that E is the midpoint of$ (Error compiling LaTeX. ! Missing $ inserted.)\overline{AB}\overline{AE}=\frac{1}{2}C'D=\frac{1}{3}\overline{AC'}=\frac{2}{3}\overline{EC'}=\frac{5}{6}2 \rightarrow \boxed{A}$ ~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 |