Difference between revisions of "2021 Fall AMC 10B Problems/Problem 25"
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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; | 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; | ||
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Solution in progress | Solution in progress | ||
Revision as of 03:06, 23 November 2021
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 
![[asy] size(8cm); draw((0,0)--(10,0)); draw((0,0)--(0,10)); draw((10,0)--(10,10)); draw((0,10)--(10,10)); draw((1,6)--(0,9)); draw((0,9)--(3,10)); draw((3,10)--(4,7)); draw((4,7)--(1,6)); draw((0,3)--(1,6)); draw((1,6)--(10,3)); draw((10,3)--(9,0)); draw((9,0)--(0,3)); draw((6,13/3)--(10,22/3)); draw((10,22/3)--(8,10)); draw((8,10)--(4,7)); draw((4,7)--(6,13/3)); label("$3$",(9/2,3/2),N); label("$3$",(11/2,9/2),S); label("$1$",(1/2,9/2),E); label("$1$",(19/2,3/2),W); label("$1$",(1/2,15/2),E); label("$1$",(3/2,19/2),S); label("$1$",(5/2,13/2),N); label("$1$",(7/2,17/2),W); label("$R$",(7,43/6),W); [/asy]](http://latex.artofproblemsolving.com/7/3/d/73d3dad024e36cb7991fb09eac46f40b460c89a8.png)
=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. 
