Difference between revisions of "2021 Fall AMC 10B Problems/Problem 25"
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Line 44: | Line 44: | ||
draw((1,6)--(0,9)); | draw((1,6)--(0,9)); | ||
draw((1,6)--(0,6)); | draw((1,6)--(0,6)); | ||
+ | draw((1,6)--(4,6)); | ||
+ | draw((4,6)--(4,10)); | ||
draw((0,9)--(3,10)); | draw((0,9)--(3,10)); | ||
draw((3,10)--(4,7)); | draw((3,10)--(4,7)); | ||
Line 69: | Line 71: | ||
label("$a$",(-1/8,28/3),W); | label("$a$",(-1/8,28/3),W); | ||
label("$b$",(1,81/8),N); | label("$b$",(1,81/8),N); | ||
+ | label("$b$",(3,6),N); | ||
</asy> | </asy> | ||
Revision as of 12:53, 23 November 2021
Contents
Problem
A rectangle with side lengths and a square with side length and a rectangle are inscribed inside a larger square as shown. The sum of all possible values for the area of can be written in the form , where and are relatively prime positive integers. What is
Solution
We see that the polygon bounded by the small square, large square, and rectangle of known lengths is an isosceles triangle. Let’s draw a perpendicular from the vertex of this triangle to its opposing side;
Solution in progress ~KingRavi
Video Solution
https://www.youtube.com/watch?v=5mPvkipCvhE
See Also
2021 Fall AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Problem 1 | |
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.