Difference between revisions of "2024 AMC 10B Problems/Problem 10"
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− | <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.</strong> | + | <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> |
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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: | 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: |
Revision as of 19:40, 19 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)
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 . 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.