Difference between revisions of "2021 AMC 10B Problems/Problem 21"
m (→Problem) |
Laffytaffy (talk | contribs) |
||
Line 42: | Line 42: | ||
==Solution 3 (Fakesolve):== | ==Solution 3 (Fakesolve):== | ||
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 | 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 | ||
+ | |||
+ | ==Solution 4== | ||
== Video Solution by OmegaLearn (Using Pythagoras Theorem and Similar Triangles) == | == Video Solution by OmegaLearn (Using Pythagoras Theorem and Similar Triangles) == |
Revision as of 10:57, 13 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
Solution 4
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 |