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{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> | |
− | 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> | ||
− | <math>\textbf{(A)} ~2 \qquad\textbf{(B)} ~1+\frac{2}{3}\sqrt{3} \qquad\textbf{(C)} ~\sqrt{ | + | <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> |
<asy> | <asy> | ||
+ | /* Made by samrocksnature */ | ||
pair A=(0,1); | pair A=(0,1); | ||
pair CC=(0.666666666666,1); | pair CC=(0.666666666666,1); | ||
Line 24: | Line 24: | ||
label("D",D,NE); | label("D",D,NE); | ||
label("E",E,NW); | label("E",E,NW); | ||
− | label("C",CC,N); | + | label("C'",CC,N); |
</asy> | </asy> | ||
− | ==Solution ( | + | |
+ | ==Solution 1== | ||
+ | |||
+ | 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 + \left(\frac{1}{3}\right)^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> | ||
+ | |||
+ | ~Tony_Li2007 | ||
+ | |||
+ | ==Solution 2== | ||
+ | Let line we're reflecting over be <math>\ell</math>, and let the points where it hits <math>AB</math> and <math>CD</math>, be <math>M</math> and <math>N</math>, 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 <math>\ell</math>. The segment <math>CC'</math> has slope <math>\frac{0 - 1}{1 - 2/3} = -3</math>, implying line <math>\ell</math> has a slope of <math>\frac{1}{3}</math>. Also, the midpoint of segment <math>CC'</math> is <math>\left( \frac{5}{6}, \frac{1}{2} \right)</math>, so line <math>\ell</math> passes through this point. Then, we get the equation of line <math>\ell</math> is simply <math>y = \frac{1}{3} x + \frac{2}{9}</math>. Then, if the point where <math>B</math> is reflected over line <math>\ell</math> is <math>B'</math>, then we get <math>BB'</math> is the line <math>y = -3x</math>. The intersection of <math>\ell</math> and segment <math>BB'</math> is <math>\left( - \frac{1}{15}, \frac{1}{5} \right)</math>. So, we get <math>B' = \left(- \frac{2}{15}, \frac{2}{5} \right)</math>. Then, line segment <math>B'C'</math> has equation <math>y = \frac{3}{4} x + \frac{1}{2}</math>, so the point <math>E</math> is the <math>y</math>-intercept, or <math>\left(0, \frac{1}{2} \right)</math>. This implies that <math>AE = \frac{1}{2}, AC' = \frac{2}{3}</math>, and by the Pythagorean Theorem, <math>EC' = \frac{5}{6}</math> (or you could notice <math>\triangle AEC'</math> is a <math>3-4-5</math> right triangle). Then, the perimeter is <math>\frac{1}{2} + \frac{2}{3} + \frac{5}{6} = 2</math>, so our answer is <math>\boxed{\textbf{(A)} ~2}</math>. ~rocketsri | ||
+ | |||
+ | ==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== | ||
+ | As described in Solution 1, we can find that <math>DF=\frac{4}{9}</math>, and <math>C'F = \frac{5}{9}.</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> | ||
+ | |||
+ | |||
+ | 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> | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | 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 | ||
+ | |||
+ | == Video Solution by OmegaLearn (Using Pythagoras Theorem and Similar Triangles) == | ||
+ | https://youtu.be/cagzLmdbqYQ | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | ==Video Solution by Interstigation== | ||
+ | https://youtu.be/0sEQOjLG-V4 | ||
+ | |||
+ | ~Interstigation | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC10 box|year=2021|ab=B|num-b=20|num-a=22}} | ||
+ | {{MAA Notice}} |
Latest revision as of 01:01, 20 April 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
Video Solution by Interstigation
~Interstigation
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 |
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