Difference between revisions of "2024 AMC 10B Problems/Problem 10"
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[[File:2024 AMC 10B 10.png|300px|right]] | [[File:2024 AMC 10B 10.png|300px|right]] | ||
==Solution 2== | ==Solution 2== | ||
− | Let <math>[AFE]=1</math>. Since <math>\triangle AFE\sim\triangle CFB</math> with a scale factor of <math>2</math>, <math>[CFB]=4</math>. The scale factor of <math>2</math> also means that <math>\dfrac{AF}{FC}=\dfrac{1}{2}</math>, therefore since <math>\triangle BCF</math> and <math>\triangle BFA</math> have the same height, <math>[BFA]=2</math>. Since <math>ABCD</math> is a parallelogram, <cmath>[BCA]=[DAC]\implies4+2=1+[CDEF]\implies [CDEF]=5\implies\boxed{\text{(A) }5:4}</cmath> | + | Let <math>[AFE]=1</math>. Since <math>\triangle AFE\sim\triangle CFB</math> with a scale factor of <math>2</math>, <math>[CFB]=4</math>. The scale factor of <math>2</math> also means that <math>\dfrac{AF}{FC}=\dfrac{1}{2}</math>, therefore since <math>\triangle BCF</math> and <math>\triangle BFA</math> have the same height, <math>[BFA]=2</math>. Since <math>ABCD</math> is a parallelogram, <cmath>[BCA]=[DAC]\implies4+2=1+[CDEF]\implies [CDEF]=5\implies\boxed{\text{(A) }5:4}</cmath> |
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
==Solution 3 (Techniques)== | ==Solution 3 (Techniques)== | ||
We assert that <math>ABCD</math> is a square of side length <math>6</math>. Notice that <math>\triangle AFE\sim\triangle CFB</math> with a scale factor of <math>2</math>. Since the area of <math>\triangle ABC</math> is <math>18 \implies</math> the area of <math>\triangle CFB</math> is <math>12</math>, so the area of <math>\triangle AFE</math> is <math>3</math>. Thus the area of <math>CDEF</math> is <math>18-3=15</math>, and we conclude that the answer is <math>\frac{15}{12}\implies\boxed{\text{(A) }5:4}</math> | We assert that <math>ABCD</math> is a square of side length <math>6</math>. Notice that <math>\triangle AFE\sim\triangle CFB</math> with a scale factor of <math>2</math>. Since the area of <math>\triangle ABC</math> is <math>18 \implies</math> the area of <math>\triangle CFB</math> is <math>12</math>, so the area of <math>\triangle AFE</math> is <math>3</math>. Thus the area of <math>CDEF</math> is <math>18-3=15</math>, and we conclude that the answer is <math>\frac{15}{12}\implies\boxed{\text{(A) }5:4}</math> | ||
+ | |||
+ | ~Tacos_are_yummy_1 | ||
==Solution 4== | ==Solution 4== | ||
Let <math>ABCE</math> be a square with side length <math>1</math>, to assist with calculations. We can put this on the coordinate plane with the points <math>D = (0,0)</math>, <math>C = (1, 0)</math>, <math>B = (1, 1)</math>, and <math>A = (0, 1)</math>. We have <math>E = (0, 0.5)</math>. Therefore, the line <math>EB</math> has slope <math>0.5</math> and y-intercept <math>0.5</math>. The equation of the line is then <math>y = 0.5x + 0.5</math>. The equation of line <math>AC</math> is <math>y = -x + 1</math>. The intersection is when the lines are equal to each other, so we solve the equation. <math>0.5x + 0.5 = -x + 1</math>, so <math>x = \frac{1}{3}</math>. Therefore, plugging it into the equation, we get <math>y= \frac{2}{3}</math>. Using the shoelace theorem, we get the area of <math>CDEF</math> to be <math>\frac{5}{12}</math> and the area of <math>CFB</math> to be <math>\frac{1}{3}</math>, so our ratio is <math>\frac{\frac{5}{12}}{\frac{1}{3}} = \boxed{(A) 5:4}</math> | Let <math>ABCE</math> be a square with side length <math>1</math>, to assist with calculations. We can put this on the coordinate plane with the points <math>D = (0,0)</math>, <math>C = (1, 0)</math>, <math>B = (1, 1)</math>, and <math>A = (0, 1)</math>. We have <math>E = (0, 0.5)</math>. Therefore, the line <math>EB</math> has slope <math>0.5</math> and y-intercept <math>0.5</math>. The equation of the line is then <math>y = 0.5x + 0.5</math>. The equation of line <math>AC</math> is <math>y = -x + 1</math>. The intersection is when the lines are equal to each other, so we solve the equation. <math>0.5x + 0.5 = -x + 1</math>, so <math>x = \frac{1}{3}</math>. Therefore, plugging it into the equation, we get <math>y= \frac{2}{3}</math>. Using the shoelace theorem, we get the area of <math>CDEF</math> to be <math>\frac{5}{12}</math> and the area of <math>CFB</math> to be <math>\frac{1}{3}</math>, so our ratio is <math>\frac{\frac{5}{12}}{\frac{1}{3}} = \boxed{(A) 5:4}</math> | ||
+ | |||
+ | ==Solution 5 (wlog)== | ||
+ | Let <math>ABCE</math> be a square with side length <math>2</math>. We see that <math>\triangle AFE \sim \triangle CFB</math> by a Scale factor of <math>2</math>. Let the altitude of <math>\triangle AFE</math> and altitude of <math>\triangle CFB</math> be <math>h</math> and <math>2h</math>, respectively. We know that <math>h+2h</math> is equal to <math>2</math>, as the height of the square is <math>2</math>. Solving this equation, we get that <math>h = \frac{2}3.</math> This means <math>[\triangle CFB] = \frac{4}3,</math> we can also calculate the area of <math>\triangle ABE</math>. Adding the area we of <math>\triangle CFB</math> and <math>\triangle ABE</math> we get <math>\frac{7}3.</math> We can then subtract this from the total area of the square: <math>4</math>, this gives us <math>\frac{5}3</math> for the area of quadrilateral <math>CFED.</math> Then we can compute the ratio which is equal to <math>\boxed{\textbf{(A) } 5:4}.</math> | ||
+ | |||
+ | ~yuvag | ||
+ | |||
+ | (why does the <math>\LaTeX</math> always look so bugged.) | ||
+ | |||
+ | ==Solution 6 (barycentrics)== | ||
+ | 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> | ||
+ | |||
+ | ==🎥✨ 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
Contents
[hide]Problem
Quadrilateral is a parallelogram, and is the midpoint of the side . Let be the intersection of lines and . What is the ratio of the area of quadrilateral to the area of ?
Solution 1
Let have length and let the altitude of the parallelogram perpendicular to have length .
The area of the parallelogram is and the area of equals . Thus, the area of quadrilateral is .
We have from that . Also, , so the length of the altitude of from is twice that of . This means that the altitude of is , so the area of is .
Then, the area of quadrilateral equals the area of minus that of , which is . Finally, the ratio of the area of to the area of triangle is , so the answer is .
Solution 2
Let . Since with a scale factor of , . The scale factor of also means that , therefore since and have the same height, . Since is a parallelogram,
vladimir.shelomovskii@gmail.com, vvsss
Solution 3 (Techniques)
We assert that is a square of side length . Notice that with a scale factor of . Since the area of is the area of is , so the area of is . Thus the area of is , and we conclude that the answer is
~Tacos_are_yummy_1
Solution 4
Let be a square with side length , to assist with calculations. We can put this on the coordinate plane with the points , , , and . We have . Therefore, the line has slope and y-intercept . The equation of the line is then . The equation of line is . The intersection is when the lines are equal to each other, so we solve the equation. , so . Therefore, plugging it into the equation, we get . Using the shoelace theorem, we get the area of to be and the area of to be , so our ratio is
Solution 5 (wlog)
Let be a square with side length . We see that by a Scale factor of . Let the altitude of and altitude of be and , respectively. We know that is equal to , as the height of the square is . Solving this equation, we get that This means we can also calculate the area of . Adding the area we of and we get We can then subtract this from the total area of the square: , this gives us for the area of quadrilateral Then we can compute the ratio which is equal to
~yuvag
(why does the always look so bugged.)
Solution 6 (barycentrics)
Let . Since is the midpoint of , . The equation of is: The equation of is: We also know that . To find the intersection, we can solve the system of equations. Solving, we get . Therefore, . Using barycentric area formula,
🎥✨ 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.