Difference between revisions of "2021 AMC 10B Problems/Problem 21"
Tony li2007 (talk | contribs) m (→Solution 1) |
Tony li2007 (talk | contribs) (→Solution 1) |
||
Line 31: | Line 31: | ||
We can set the point on <math>CD</math> where the fold occurs as point <math>F</math>. Then, we can set <math>FD</math> as <math>x</math>, and <math>CF</math> as <math>1-x</math> because of symmetry due to the fold. It can be recognized that this is a right triangle, and solving for <math>x</math>, we get, | We can set the point on <math>CD</math> where the fold occurs as point <math>F</math>. Then, we can set <math>FD</math> as <math>x</math>, and <math>CF</math> as <math>1-x</math> because of symmetry due to the fold. It can be recognized that this is a right triangle, and solving for <math>x</math>, we get, | ||
− | < | + | <cmath>x^2 + (\frac{1}{3})^2 = (1-x)^2 \rightarrow x^2 + \frac{1}{9} = x^2 - 2x + 1 \rightarrow x=\frac{4}{9}</cmath> |
We know this is a 3-4-5 triangle because the side lengths are <math>\frac{3}{9}, \frac{4}{9}, \frac{5}{9}</math>. We also know that <math>EAC'</math> is similar to <math>C'DF</math> because angle <math>C'</math> is a right angle. Now, we can use similarity to find out that the perimeter is just the perimeter of <math>C'DF * \frac{AC'}{DF}</math>. Thats just <math>\frac{4}{3} * \frac{\frac{2}{3}}{\frac{4}{9}} = \frac{4}{3} * \frac{3}{2} = 2</math>. Therefore, the final answer is <math>\boxed{A}</math> | We know this is a 3-4-5 triangle because the side lengths are <math>\frac{3}{9}, \frac{4}{9}, \frac{5}{9}</math>. We also know that <math>EAC'</math> is similar to <math>C'DF</math> because angle <math>C'</math> is a right angle. Now, we can use similarity to find out that the perimeter is just the perimeter of <math>C'DF * \frac{AC'}{DF}</math>. Thats just <math>\frac{4}{3} * \frac{\frac{2}{3}}{\frac{4}{9}} = \frac{4}{3} * \frac{3}{2} = 2</math>. Therefore, the final answer is <math>\boxed{A}</math> |
Revision as of 00:58, 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 (Outlined)
double angle tangent to find angle ACE then trig
Solution (Quicksolve)
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 | |
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 |