Difference between revisions of "2021 AMC 10B Problems/Problem 21"
(→Solution 5) |
m (→Solution 5) |
||
Line 56: | Line 56: | ||
Finally, we see that the perimeter of <math>\triangle AEC'</math> is <math>\frac{1}{2} + \frac{2}{3} + \sqrt{\left(\frac{1}{2}\right)^2 + \frac{4}{9}},</math> which we can simplify to be <math>2</math>. Thus, the answer is <math>\boxed{\textbf{(A)} ~2}.</math> ~laffytaffy | Finally, we see that the perimeter of <math>\triangle AEC'</math> is <math>\frac{1}{2} + \frac{2}{3} + \sqrt{\left(\frac{1}{2}\right)^2 + \frac{4}{9}},</math> which we can simplify to be <math>2</math>. Thus, the answer is <math>\boxed{\textbf{(A)} ~2}.</math> ~laffytaffy | ||
− | ==Solution 5== | + | ==Solution 5 (Trig)== |
− | Draw a perpendicular line from <math>AB</math> at <math>E</math>, and let it intersect <math>DC</math> at <math>E'</math>. The angle between <math>AB</math> and <math>EE'</math> is 2<math>\theta</math>, where <math>\theta</math> is the angle between the fold and a line perpendicular to <math>AD</math>. The slope of the fold is <math>\frac{1}{3}</math> because it is perpendicular to <math>CC'</math> (<math>CC'</math> has a slope of -3 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>EC'</math> is 3/4, which implies <math>AE</math> = 1/2. 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>AB</math> at <math>E</math>, and let it intersect <math>DC</math> at <math>E'</math>. The angle between <math>AB</math> and <math>EE'</math> is 2<math>\theta</math>, where <math>\theta</math> is the angle between the fold and a line perpendicular to <math>AD</math>. The slope of the fold is <math>\frac{1}{3}</math> because it is perpendicular to <math>CC'</math> (<math>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>EC'</math> is 3/4, which implies <math>AE</math> = 1/2. 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 |
== Video Solution by OmegaLearn (Using Pythagoras Theorem and Similar Triangles) == | == Video Solution by OmegaLearn (Using Pythagoras Theorem and Similar Triangles) == |
Revision as of 01:13, 8 November 2022
Contents
[hide]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 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 2, 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 3/4, which implies = 1/2. By the Pythagorean Theorem, . It easily follows that our desired perimeter is ~forrestc
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.