Difference between revisions of "2023 AMC 10A Problems/Problem 19"

(Video Solution by Math-X (First fully understand the problem!!!))
(Solution 4)
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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==Solution 5 (Trigonometry)==
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 +
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First, we need to find the coordinates of point P(r,s) such that rotating the line segment from A(1,2) to B(3,3) about P maps it to the line segment from A(3,1) to B(4,3). After finding r and s, we will compute |rs|.
 +
 +
When a point (x,y) is rotated about another point (r,s) by an angle θ, the new coordinates (x,y) are given by:
 +
 +
\[
 +
\begin{cases}
 +
x' = r + (x - r) \cos \theta - (y - s) \sin \theta \
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y' = s + (x - r) \sin \theta + (y - s) \cos \theta
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\end{cases}
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\]
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 +
We apply the rotation to points A and B:
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 +
For A(1,2) mapping to A(3,1):
 +
 +
\[
 +
\begin{cases}
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3 = r + (1 - r) \cos \theta - (2 - s) \sin \theta \quad \text{(1)} \
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1 = s + (1 - r) \sin \theta + (2 - s) \cos \theta \quad \text{(2)}
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\end{cases}
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\]
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For B(3,3) mapping to B(4,3):
 +
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\[
 +
\begin{cases}
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4 = r + (3 - r) \cos \theta - (3 - s) \sin \theta \quad \text{(3)} \
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3 = s + (3 - r) \sin \theta + (3 - s) \cos \theta \quad \text{(4)}
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\end{cases}
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\]
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Now, we subtract equations to eliminate variables:
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 +
Subtract equation (1) from equation (3):
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\[
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(4 - 3) = [r + (3 - r) \cos \theta - (3 - s) \sin \theta] - [r + (1 - r) \cos \theta - (2 - s) \sin \theta]
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\]
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Simplifying yields:
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\[
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1 = 2 \cos \theta - \sin \theta \quad \text{(A)}
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\]
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Similarly, subtract equation (2) from equation (4):
 +
 +
\[
 +
(3 - 1) = [s + (3 - r) \sin \theta + (3 - s) \cos \theta] - [s + (1 - r) \sin \theta + (2 - s) \cos \theta]
 +
\]
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 +
Simplifying further:
 +
 +
\[
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2 = 2 \sin \theta + \cos \theta \quad \text{(B)}
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\]
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Now, using equations (A) and (B):
 +
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\[
 +
\begin{cases}
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2 \cos \theta - \sin \theta = 1 \
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\cos \theta + 2 \sin \theta = 2
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\end{cases}
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\]
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We now express cosθ from the second equation:
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 +
\[
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\cos \theta = 2 - 2 \sin \theta
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\]
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Substituting this into the first equation, we get:
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\[
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2(2 - 2 \sin \theta) - \sin \theta = 1 \
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4 - 4 \sin \theta - \sin \theta = 1 \
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-5 \sin \theta = -3 \
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\sin \theta = \frac{3}{5}
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\]
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Now, we find cosθ as follows:
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\[
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\cos \theta = 2 - 2 \left( \frac{3}{5} \right) = \frac{4}{5}
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\]
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Substituting cosθ and sinθ back into equations (1) and (2):
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Equation (1):
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\[
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3 - r = (1 - r) \left( \frac{4}{5} \right) - (2 - s) \left( \frac{3}{5} \right)
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\]
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Multiply both sides by 5:
 +
 +
\[
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5(3 - r) = 4(1 - r) - 3(2 - s)
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\]
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Simplify:
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\[
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15 - 5r = 4 - 4r - 6 + 3s \
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15 - 5r = -2 - 4r + 3s \
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-5r + 4r - 3s = -2 - 15 \
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-r - 3s = -17 \quad \text{(C)}
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\]
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Equation (2):
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\[
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1 - s = (1 - r) \left( \frac{3}{5} \right) + (2 - s) \left( \frac{4}{5} \right)
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\]
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Multiply both sides by 5:
 +
 +
\[
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5(1 - s) = 3(1 - r) + 4(2 - s)
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\]
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 +
Simplifying more:
 +
 +
\[
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5 - 5s = 3 - 3r + 8 - 4s \
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5 - 5s = 11 - 3r - 4s \
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-5s + 4s + 3r = 11 - 5 \
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3r - s = 6 \quad \text{(D)}
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\]
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From equations (C) and (D):
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\[
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\begin{cases}
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-r - 3s = -17 \
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3r - s = 6
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\end{cases}
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\]
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We then multiply the first equation by 3:
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\[
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-3r - 9s = -51
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\]
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Add this to the second equation multiplied by 9:
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\[
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27r - 9s = 54
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\]
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Subtract:
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\[
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(27r - 9s) - (-3r - 9s) = 54 - (-51) \
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30r = 105 \
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r = \frac{105}{30} = \frac{7}{2} = 3.5
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\]
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Now find s:
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\[
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3(3.5) - s = 6 \
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10.5 - s = 6 \
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s = 4.5
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\]
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Therefore, we have the following:
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\[
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|r - s| = |3.5 - 4.5| = |-1| = \boxed{\text{(E)}1}
 +
\]
  
 
==Video Solution by Little Fermat==
 
==Video Solution by Little Fermat==

Revision as of 13:27, 9 November 2024

Problem

The line segment formed by $A(1, 2)$ and $B(3, 3)$ is rotated to the line segment formed by $A'(3, 1)$ and $B'(4, 3)$ about the point $P(r, s)$. What is $|r-s|$?

$\textbf{(A) } \frac{1}{4} \qquad \textbf{(B) } \frac{1}{2} \qquad \textbf{(C) } \frac{3}{4}   \qquad \textbf{(D) } \frac{2}{3} \qquad   \textbf{(E) } 1$

Solution 1

Due to rotations preserving an equal distance, we can bash the answer with the distance formula. $D(A, P) = D(A', P)$, and $D(B, P) = D(B',P)$. Thus we will square our equations to yield: $(1-r)^2+(2-s)^2=(3-r)^2+(1-s)^2$, and $(3-r)^2+(3-s)^2=(4-r)^2+(3-s)^2$. Canceling $(3-s)^2$ from the second equation makes it clear that $r$ equals $3.5$.

Substituting will yield

\begin{align*}(2.5)^2+(2-s)^2 &= (-0.5)^2+(1-s)^2 \\ 6.25+4-4s+s^2 &= 0.25+1-2s+s^2 \\ 2s &= 9 \\ s &=4.5 \\ \end{align*}.

Now $|r-s| = |3.5-4.5| = \boxed{\textbf{(E) } 1}$.

-Antifreeze5420

Solution 2

Due to rotations preserving distance, we have that $BP = B^\prime P$, as well as $AP = A^\prime P$. From here, we can see that P must be on the perpendicular bisector of $\overline{BB^\prime}$ due to the property of perpendicular bisectors keeping the distance to two points constant.

From here, we proceed to find the perpendicular bisector of $\overline{BB^\prime}$. We can see that this is just a horizontal line segment with midpoint at $(3.5, 3)$. This means that the equation of the perpendicular bisector is $x = 3.5$.

Similarly, we find the perpendicular bisector of $\overline{AA^\prime}$. We find the slope to be $\frac{1-2}{3-1} = -\frac12$, so our new slope will be $2$. The midpoint of $A$ and $A^\prime$ is $(2, \frac32)$, which we can use with our slope to get another equation of $y = 2x - \frac52$.

Now, point P has to lie on both of these perpendicular bisectors, meaning that it has to satisfy both equations. Plugging in the value of $x$ we found earlier, we find that $y=4.5$. This means that $|r - s| = |3.5 - 4.5| = \boxed{\textbf{(E) } 1}$.

-DEVSAXENA

Solution 3 (Coordinate Geometry)

To find the center of rotation, we find the intersection point of the perpendicular bisectors of $\overline{AA^\prime}$ and $\overline{BB^\prime}$.

We can find that the equation of the line $\overline{AA^\prime}$ is $y = -\frac{1}{2}x + \frac{5}{2}$, and that the equation of the line $\overline{BB^\prime}$ is $y = 3$.

When we solve for the perpendicular bisector of $y = -\frac{1}{2}x + \frac{5}{2}$, we determine that it has a slope of 2, and it runs through $(2, 1.5)$. Plugging in $1.5 = 2(2)-n$, we get than $n = \frac{5}{2}$. Therefore our perpendicular bisector is $y=2x-\frac{5}{2}$. Next, we solve for the perpendicular of $y = 3$. We know that it has an undefined slope, and it runs through $(3.5, 3)$. We can determine that our second perpendicular bisector is $x = 3.5$.

Setting the equations equal to each other, we get $y = \frac{9}{2}$. Therefore, $|r - s| = |3.5 - 4.5| = \boxed{\textbf{(E) } 1}$.

[asy] pair A=(1,2); pair B=(3,3); pair A1=(3,1); pair B1=(4,3); dot("A",A,NW); dot("B",B,S); dot("A'",A1,S); dot("B'",B1,E); draw(A--A1); draw(B--B1); draw((3.5,0)--(3.5,6),BeginArrow(5),EndArrow(5)); draw((1,-0.5)--(4,5.5),BeginArrow(5),EndArrow(5)); pair P=(3.5,4.5); dot("P",P,NW); [/asy]

~aydenlee & wuwang2002

Solution 4

We use the complex numbers approach to solve this problem. Denote by $\theta$ the angle that $AB$ rotates about $P$ in the counterclockwise direction.

Thus, $A' - P = e^{i \theta} \left( A - P \right)$ and $B' - P = e^{i \theta} \left( B - P \right)$.

Taking ratio of these two equations, we get \[ \frac{A' - P}{A - P} = \frac{B' - P}{B - P} . \]

By solving this equation, we get $P = \frac{7}{2} + i \frac{9}{2}$. Therefore, $|s-t| = \left| \frac{7}{2} - \frac{9}{2} \right| = \boxed{\textbf{(E) 1}}$.

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Solution 5 (Trigonometry)

First, we need to find the coordinates of point P(r,s) such that rotating the line segment from A(1,2) to B(3,3) about P maps it to the line segment from A(3,1) to B(4,3). After finding r and s, we will compute |rs|.

When a point (x,y) is rotated about another point (r,s) by an angle θ, the new coordinates (x,y) are given by:

\[ {x=r+(xr)cosθ(ys)sinθy=s+(xr)sinθ+(ys)cosθ \]

We apply the rotation to points A and B:

For A(1,2) mapping to A(3,1):

\[ {3=r+(1r)cosθ(2s)sinθ(1)1=s+(1r)sinθ+(2s)cosθ(2) \]

For B(3,3) mapping to B(4,3):

\[ {4=r+(3r)cosθ(3s)sinθ(3)3=s+(3r)sinθ+(3s)cosθ(4) \]

Now, we subtract equations to eliminate variables:

Subtract equation (1) from equation (3):

\[ (4 - 3) = [r + (3 - r) \cos \theta - (3 - s) \sin \theta] - [r + (1 - r) \cos \theta - (2 - s) \sin \theta] \]

Simplifying yields:

\[ 1 = 2 \cos \theta - \sin \theta \quad \text{(A)} \]

Similarly, subtract equation (2) from equation (4):

\[ (3 - 1) = [s + (3 - r) \sin \theta + (3 - s) \cos \theta] - [s + (1 - r) \sin \theta + (2 - s) \cos \theta] \]

Simplifying further:

\[ 2 = 2 \sin \theta + \cos \theta \quad \text{(B)} \]

Now, using equations (A) and (B):

\[ {2cosθsinθ=1cosθ+2sinθ=2 \]

We now express cosθ from the second equation:

\[ \cos \theta = 2 - 2 \sin \theta \]

Substituting this into the first equation, we get:

\[ 2(2 - 2 \sin \theta) - \sin \theta = 1 \ 4 - 4 \sin \theta - \sin \theta = 1 \ -5 \sin \theta = -3 \ \sin \theta = \frac{3}{5} \]

Now, we find cosθ as follows:

\[ \cos \theta = 2 - 2 \left( \frac{3}{5} \right) = \frac{4}{5} \]

Substituting cosθ and sinθ back into equations (1) and (2):

Equation (1):

\[ 3 - r = (1 - r) \left( \frac{4}{5} \right) - (2 - s) \left( \frac{3}{5} \right) \]

Multiply both sides by 5:

\[ 5(3 - r) = 4(1 - r) - 3(2 - s) \]

Simplify:

\[ 15 - 5r = 4 - 4r - 6 + 3s \ 15 - 5r = -2 - 4r + 3s \ -5r + 4r - 3s = -2 - 15 \ -r - 3s = -17 \quad \text{(C)} \]

Equation (2):

\[ 1 - s = (1 - r) \left( \frac{3}{5} \right) + (2 - s) \left( \frac{4}{5} \right) \]

Multiply both sides by 5:

\[ 5(1 - s) = 3(1 - r) + 4(2 - s) \]

Simplifying more:

\[ 5 - 5s = 3 - 3r + 8 - 4s \ 5 - 5s = 11 - 3r - 4s \ -5s + 4s + 3r = 11 - 5 \ 3r - s = 6 \quad \text{(D)} \]

From equations (C) and (D):

\[ {r3s=173rs=6 \]

We then multiply the first equation by 3:

\[ -3r - 9s = -51 \]

Add this to the second equation multiplied by 9:

\[ 27r - 9s = 54 \]

Subtract:

\[ (27r - 9s) - (-3r - 9s) = 54 - (-51) \ 30r = 105 \ r = \frac{105}{30} = \frac{7}{2} = 3.5 \]

Now find s:

\[ 3(3.5) - s = 6 \ 10.5 - s = 6 \ s = 4.5 \]

Therefore, we have the following:

\[ |r - s| = |3.5 - 4.5| = |-1| = \boxed{\text{(E)}1} \]

Video Solution by Little Fermat

https://youtu.be/h2Pf2hvF1wE?si=8YX-h2OqRiF8j2y4&t=4238 ~little-fermat

Video Solution by Math-X (First fully understand the problem!!!)

https://youtu.be/GP-DYudh5qU?si=UCyIECgDXCoSakc0&t=5831

~Math-X

Video Solution

https://youtu.be/aYHwBpdWkAk

Video Solution by

https://www.youtube.com/watch?v=fIzCR4x4x-M

Video Solution 1 by OmegaLearn

https://youtu.be/88F18qth0xI

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=oHMJwiEwOS0

Video Solution

https://youtu.be/va3ZCFeKfzg

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution by TheBeautyofMath

https://youtu.be/Q_4uMxMbQlI

~IceMatrix

See Also

2023 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 18
Followed by
Problem 20
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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