Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2024 AMC 10B Problems/Problem 10"

(→‎Solution 6 (barycentrics))
(→‎Solution 6 (barycentrics))
 
(8 intermediate revisions by the same user not shown)
Line 37: Line 37:
 
==Solution 6 (barycentrics)==
 
==Solution 6 (barycentrics)==
  
−
<div><font color="red", font size="101px"> NOTE: this solution is complete overkill, do this to waste time or you are a mopper that forgot how to do intro to geo math</font></div>
+
<div style="color: red; font-size: 24px;">
 +
<strong>NOTE: This solution is complete overkill. Do this to waste time, or if you are a mopper who forgot how to do intro to geometry math. If you do this you are either orz or trying to act orz when you really aren't and wasting time on problem 10</strong>
 +
</div>
 
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:
 
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>
 
<cmath>

Latest revision as of 19:40, 19 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)

NOTE: This solution is complete overkill. Do this to waste time, or if you are a mopper who forgot how to do intro to geometry math. If you do this you are either orz or trying to act orz when you really aren't and wasting time on problem 10

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AMC_10B_Problems/Problem_10&oldid=235092"