Difference between revisions of "2024 AMC 10B Problems/Problem 10"
Jonathan.li (talk | contribs) (→Solution 6 (barycentrics)) |
Jonathan.li (talk | contribs) (→Solution 6 (barycentrics)) |
||
Line 38: | Line 38: | ||
<div style="color: red; font-size: 24px;"> | <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.</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. (oren bad)</strong> |
</div> | </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: |
Revision as of 19:32, 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. (oren bad)
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.