Difference between revisions of "2021 AMC 10B Problems/Problem 21"
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==Problem== | ==Problem== | ||
− | A square piece of paper has side length <math>1</math> and vertices <math>A,B,C,</math> and <math>D</math> in that order. As shown in the figure, the paper is folded so that vertex <math>C</math> meets edge <math>\overline{AD}</math> at point <math>C'</math>, and edge <math>\overline{AB}</math> at point <math>E</math>. Suppose that <math>C'D = \frac{1}{3}</math>. What is the perimeter of triangle <math>\bigtriangleup AEC' ?</math> | + | A square piece of paper has side length <math>1</math> and vertices <math>A,B,C,</math> and <math>D</math> in that order. As shown in the figure, the paper is folded so that vertex <math>C</math> meets edge <math>\overline{AD}</math> at point <math>C'</math>, and edge <math>\overline{BC}</math> intersects edge <math>\overline{AB}</math> at point <math>E</math>. Suppose that <math>C'D = \frac{1}{3}</math>. What is the perimeter of triangle <math>\bigtriangleup AEC' ?</math> |
<math>\textbf{(A)} ~2 \qquad\textbf{(B)} ~1+\frac{2}{3}\sqrt{3} \qquad\textbf{(C)} ~\sqrt{136} \qquad\textbf{(D)} ~1 + \frac{3}{4}\sqrt{3} \qquad\textbf{(E)} ~\frac{7}{3}</math> | <math>\textbf{(A)} ~2 \qquad\textbf{(B)} ~1+\frac{2}{3}\sqrt{3} \qquad\textbf{(C)} ~\sqrt{136} \qquad\textbf{(D)} ~1 + \frac{3}{4}\sqrt{3} \qquad\textbf{(E)} ~\frac{7}{3}</math> |
Revision as of 23:16, 24 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 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 . 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
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
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 |