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Difference between revisions of "2024 AMC 10B Problems/Problem 10"
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==Solution 6 (barycentrics)==
 
==Solution 6 (barycentrics)==
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Let <math>A=(1,0,0), B=(0,1,0), C=(0,0,1), D=(1,-1,1)</math>. Since <math>E</math> is the midpoint of <math>\overline{AD}</math>, <math>E=(1,-0.5,0.5)</math>. The equation of <math>\overline{EB}</math> is:
 +
<cmath>
 +
0 =
 +
\begin{vmatrix}
 +
x & y & z \\
 +
1 & -0.5 & 0.5 \\
 +
0 & 1 & 0
 +
\end{vmatrix}
 +
</cmath>
 +
The equation of <math>\overline{AC}</math> is:
 +
<cmath>
 +
0 =
 +
\begin{vmatrix}
 +
x & y & z \\
 +
1 & 0 & 0 \\
 +
0 & 0 & 1
 +
\end{vmatrix}
 +
</cmath>
 +
We also know that <math>x+y+z=1</math>. To find the intersection, we can solve the system of equations. Solving, we get <math>x=2/3,y=0,z=1/3</math>. Therefore, <math>F=\left(\frac{2}{3}, 0, \frac{1}{3}\right)</math>. Using barycentric area formula,
 +
<cmath>
 +
\frac{[CFB]}{[ABC]} =
 +
\begin{vmatrix}
 +
0 & 0 & 1 \\
 +
2/3 & 0 & 1/3 \\
 +
0 & 1 & 0
 +
\end{vmatrix}
 +
=\frac{2}{3}
 +
</cmath>
 +
<cmath>
 +
\frac{[CDEF]}{[ABC]} =
 +
\begin{vmatrix}
 +
0 & 0 & 1 \\
 +
1 & -0.5 & 0.5 \\
 +
2/3 & 0 & 1/3
 +
\end{vmatrix}
 +
+
 +
\begin{vmatrix}
 +
0 & 0 & 1 \\
 +
1 & -1 & 1 \\
 +
1 & -0.5 & 0.5
 +
\end{vmatrix}
 +
=\frac{5}{6}
 +
</cmath>
 +
<math>\frac{[CDEF]}{[CFB]}=\frac{\frac{5}{6}}{\frac{2}{3}}=\boxed{\textbf{(A) } 5:4}</math>
  
−
Let <math>A=(1,0,0), B=(0,1,0), C=(0,0,1), D=(1,-1,1)</math>
+
==🎥✨ Video Solution by Scholars Foundation ➡️ (Easy-to-Understand 💡✔️)==
 +
 
 +
https://youtu.be/T_QESWAKUUk?si=TG7ToQnDsYKsNSSJ&t=648
  
 
==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)==
 
==Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)==

Latest revision as of 17:35, 24 November 2024

Problem

Quadrilateral $ABCD$ is a parallelogram, and $E$ is the midpoint of the side $\overline{AD}$. Let $F$ be the intersection of lines $EB$ and $AC$. What is the ratio of the area of quadrilateral $CDEF$ to the area of $\triangle CFB$?

$\textbf{(A) } 5:4 \qquad\textbf{(B) } 4:3 \qquad\textbf{(C) } 3:2 \qquad\textbf{(D) } 5:3 \qquad\textbf{(E) } 2:1$

Solution 1

Let $AB = CD$ have length $b$ and let the altitude of the parallelogram perpendicular to $\overline{AD}$ have length $h$.

The area of the parallelogram is $bh$ and the area of $\triangle ABE$ equals $\frac{(b/2)(h)}{2} = \frac{bh}{4}$. Thus, the area of quadrilateral $BCDE$ is $bh - \frac{bh}{4} = \frac{3bh}{4}$.

We have from $AA$ that $\triangle CBF \sim \triangle AEF$. Also, $CB/AE = 2$, so the length of the altitude of $\triangle CBF$ from $F$ is twice that of $\triangle AEF$. This means that the altitude of $\triangle CBF$ is $2h/3$, so the area of $\triangle CBF$ is $\frac{(b)(2h/3)}{2} = \frac{bh}{3}$.

Then, the area of quadrilateral $CDEF$ equals the area of $BCDE$ minus that of $\triangle CBF$, which is $\frac{3bh}{4} - \frac{bh}{3} = \frac{5bh}{12}$. Finally, the ratio of the area of $CDEF$ to the area of triangle $CFB$ is $\frac{\frac{5bh}{12}}{\frac{bh}{3}} = \frac{\frac{5}{12}}{\frac{1}{3}} = \frac{5}{4}$, so the answer is $\boxed{\textbf{(A) } 5:4}$.

2024 AMC 10B 10.png

Solution 2

Let $[AFE]=1$. Since $\triangle AFE\sim\triangle CFB$ with a scale factor of $2$, $[CFB]=4$. The scale factor of $2$ also means that $\dfrac{AF}{FC}=\dfrac{1}{2}$, therefore since $\triangle BCF$ and $\triangle BFA$ have the same height, $[BFA]=2$. Since $ABCD$ is a parallelogram, \[[BCA]=[DAC]\implies4+2=1+[CDEF]\implies [CDEF]=5\implies\boxed{\text{(A) }5:4}\]

vladimir.shelomovskii@gmail.com, vvsss

Solution 3 (Techniques)

We assert that $ABCD$ is a square of side length $6$. Notice that $\triangle AFE\sim\triangle CFB$ with a scale factor of $2$. Since the area of $\triangle ABC$ is $18 \implies$ the area of $\triangle CFB$ is $12$, so the area of $\triangle AFE$ is $3$. Thus the area of $CDEF$ is $18-3=15$, and we conclude that the answer is $\frac{15}{12}\implies\boxed{\text{(A) }5:4}$ 

~Tacos_are_yummy_1

Solution 4

Let $ABCE$ be a square with side length $1$, to assist with calculations. We can put this on the coordinate plane with the points $D = (0,0)$, $C = (1, 0)$, $B = (1, 1)$, and $A = (0, 1)$. We have $E = (0, 0.5)$. Therefore, the line $EB$ has slope $0.5$ and y-intercept $0.5$. The equation of the line is then $y = 0.5x + 0.5$. The equation of line $AC$ is $y = -x + 1$. The intersection is when the lines are equal to each other, so we solve the equation. $0.5x + 0.5 = -x + 1$, so $x = \frac{1}{3}$. Therefore, plugging it into the equation, we get $y= \frac{2}{3}$. Using the shoelace theorem, we get the area of $CDEF$ to be $\frac{5}{12}$ and the area of $CFB$ to be $\frac{1}{3}$, so our ratio is $\frac{\frac{5}{12}}{\frac{1}{3}} = \boxed{(A) 5:4}$

Solution 5 (wlog)

Let $ABCE$ be a square with side length $2$. We see that $\triangle AFE \sim \triangle CFB$ by a Scale factor of $2$. Let the altitude of $\triangle AFE$ and altitude of $\triangle CFB$ be $h$ and $2h$, respectively. We know that $h+2h$ is equal to $2$, as the height of the square is $2$. Solving this equation, we get that $h = \frac{2}3.$ This means $[\triangle CFB] = \frac{4}3,$ we can also calculate the area of $\triangle ABE$. Adding the area we of $\triangle CFB$ and $\triangle ABE$ we get $\frac{7}3.$ We can then subtract this from the total area of the square: $4$, this gives us $\frac{5}3$ for the area of quadrilateral $CFED.$ Then we can compute the ratio which is equal to $\boxed{\textbf{(A) } 5:4}.$

~yuvag

(why does the $\LaTeX$ always look so bugged.)

Solution 6 (barycentrics)

Let $A=(1,0,0), B=(0,1,0), C=(0,0,1), D=(1,-1,1)$. Since $E$ is the midpoint of $\overline{AD}$, $E=(1,-0.5,0.5)$. The equation of $\overline{EB}$ is: \[0 = \begin{vmatrix} x & y & z \\ 1 & -0.5 & 0.5 \\ 0 & 1 & 0 \end{vmatrix}\] The equation of $\overline{AC}$ is: \[0 = \begin{vmatrix} x & y & z \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{vmatrix}\] We also know that $x+y+z=1$. To find the intersection, we can solve the system of equations. Solving, we get $x=2/3,y=0,z=1/3$. Therefore, $F=\left(\frac{2}{3}, 0, \frac{1}{3}\right)$. Using barycentric area formula, \[\frac{[CFB]}{[ABC]} = \begin{vmatrix} 0 & 0 & 1 \\ 2/3 & 0 & 1/3 \\ 0 & 1 & 0 \end{vmatrix} =\frac{2}{3}\] \[\frac{[CDEF]}{[ABC]} = \begin{vmatrix} 0 & 0 & 1 \\ 1 & -0.5 & 0.5 \\ 2/3 & 0 & 1/3 \end{vmatrix} + \begin{vmatrix} 0 & 0 & 1 \\ 1 & -1 & 1 \\ 1 & -0.5 & 0.5 \end{vmatrix} =\frac{5}{6}\] $\frac{[CDEF]}{[CFB]}=\frac{\frac{5}{6}}{\frac{2}{3}}=\boxed{\textbf{(A) } 5:4}$

🎥✨ Video Solution by Scholars Foundation ➡️ (Easy-to-Understand 💡✔️)

https://youtu.be/T_QESWAKUUk?si=TG7ToQnDsYKsNSSJ&t=648

Video Solution 1 by Pi Academy (Fast and Easy ⚡🚀)

https://youtu.be/QLziG_2e7CY?feature=shared

~ Pi Academy

Video Solution 2 by SpreadTheMathLove

https://www.youtube.com/watch?v=24EZaeAThuE

See also

2024 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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