Difference between revisions of "2021 AMC 10B Problems/Problem 21"
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label("E",E,NW); | label("E",E,NW); | ||
label("C'",CC,N); | label("C'",CC,N); | ||
+ | dot(C); | ||
+ | dot(E); | ||
</asy> | </asy> | ||
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<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> | <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>EC'F</math> is a right angle. Now, we can use similarity to find out that the perimeter is just the perimeter of <math>C'DF | + | 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>EC'F</math> is a right angle. Now, we can use similarity to find out that the perimeter is just the perimeter of <math>C'DF \times \frac{AC'}{DF}</math>. That's just <math>\frac{4}{3} \times \frac{\frac{2}{3}}{\frac{4}{9}} = \frac{4}{3} \times \frac{3}{2} = 2</math>. Therefore, the final answer is <math>\boxed{\textbf{(A)} ~2}</math> |
~Tony_Li2007 | ~Tony_Li2007 | ||
==Solution 2== | ==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 | + | Let the 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):== | ==Solution 3 (Fakesolve):== | ||
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− | 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> | + | 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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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 (Trig)== | ||
+ | Draw a perpendicular line from <math>\overline{AB}</math> at <math>E</math>, and let it intersect <math>\overline{DC}</math> at <math>E'</math>. The angle between <math>\overline{AB}</math> and <math>\overline{EE'}</math> is <math>2\theta</math>, where <math>\theta</math> is the angle between the fold and a line perpendicular to <math>\overline{AD}</math>. The slope of the fold is <math>\frac{1}{3}</math> because it is perpendicular to <math>\overline{CC'}</math> (<math>\overline{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>\overline{EC'}</math> is <math>\frac{3}{4}</math>, which implies <math>\overline{AE}</math> = <math>\frac{1}{2}</math>. 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 | ||
+ | ==Solution 6== | ||
+ | It is easy to prove that the ratio of the sum of the larger leg and hypotenuse to the smaller leg depends monotonically on the angle of a right triangle, which means: | ||
+ | <cmath>C' F + DF = CF + DF = CD = AD = 3 C'D \implies C' D : DF : C' F = 3 : 4 : 5.</cmath> | ||
+ | |||
+ | For a similar triangle, the ratio of the perimeter to the larger leg is <math>\frac {3 + 4 +5}{4} = 3.</math> | ||
+ | |||
+ | <math>\triangle AEC' \sim \triangle DC'F \implies</math> the perimeter of triangle <math>\bigtriangleup AEC'</math> is <math>3 AC' = \boxed{\textbf{(A)} ~ 2}.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
== Video Solution by OmegaLearn (Using Pythagoras Theorem and Similar Triangles) == | == Video Solution by OmegaLearn (Using Pythagoras Theorem and Similar Triangles) == |
Latest revision as of 19:47, 9 November 2024
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 the 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 , 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 , which implies = . By the Pythagorean Theorem, . It easily follows that our desired perimeter is ~forrestc
Solution 6
It is easy to prove that the ratio of the sum of the larger leg and hypotenuse to the smaller leg depends monotonically on the angle of a right triangle, which means:
For a similar triangle, the ratio of the perimeter to the larger leg is
the perimeter of triangle is
vladimir.shelomovskii@gmail.com, vvsss
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 |
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