Difference between revisions of "2021 AMC 10B Problems/Problem 21"
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− | Then, we can find we can find the length of <math>AE</math> by expressing the length of <math>EF</math> in two different ways, in terms of <math>AE</math>. If let <math>AE = a</math>, by the Pythagorean Theorem we have that <math>EC = \sqrt{a^2 + \left(\frac{2}{3}\right)^2} = \sqrt{a^2 + \frac{4}{9}}.</math> Therefore, since we know that <math>\angle EC'F</math> is right, by Pythagoras again we have that <math>EF = \sqrt{\left(\sqrt{a^2+\frac{4}{9}\right)^2}</math> | + | Then, we can find we can find the length of <math>\overline{AE}</math> by expressing the length of <math>\overline{EF}</math> in two different ways, in terms of <math>AE</math>. If let <math>AE = a</math>, by the Pythagorean Theorem we have that <math>EC = \sqrt{a^2 + \left(\frac{2}{3}\right)^2} = \sqrt{a^2 + \frac{4}{9}}.</math> Therefore, since we know that <math>\angle EC'F</math> is right, by Pythagoras again we have that <math>EF = \sqrt{\left(\sqrt{a^2+\frac{4}{9}}\right)^2 + \left(\frac{5}{9}\right)^2} = \sqrt{a^2 + \frac{61}{81}}.</math> |
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+ | Another way we can express <math>EF</math> is by using Pythagoras on <math>\triangle XEF</math>, where <math>X</math> is the foot of the perpendicular from <math>F</math> to <math>\overline{AE}.</math> We see that <math>ADFX</math> is a rectangle, so we know that <math>AD = 1 = FX</math>. Secondly, since <math>FD = \frac{4}{9}, EX = a - \frac{4}{9}</math>. Therefore, through the Pythagorean Theorem, we find that <math>EF = \sqrt{\left(a-\frac{4}{9}\right)^2 + 1^2} = \sqrt{a^2 - \frac{8}{9}a +\frac{97}{81}}.</math> | ||
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+ | Since we have found two expressions for the same length, we have the equation <math>\sqrt{a^2 + \frac{61}{81}} = \sqrt{a^2 - \frac{8}{9}a +\frac{97}{81}}.</math> Solving this, we find that <math>a=\frac{1}{2}</math>. | ||
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+ | 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> | ||
== Video Solution by OmegaLearn (Using Pythagoras Theorem and Similar Triangles) == | == Video Solution by OmegaLearn (Using Pythagoras Theorem and Similar Triangles) == |
Revision as of 10:32, 13 February 2021
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 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
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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