Difference between revisions of "2021 AMC 10B Problems/Problem 21"
(→Solution 2) |
m (→Problem) |
||
Line 2: | Line 2: | ||
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{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 */ | /* Made by samrocksnature */ |
Revision as of 17:58, 12 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
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
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 |