Difference between revisions of "2021 Fall AMC 10B Problems/Problem 25"
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label("$a$",(7/2,81/8),N); | label("$a$",(7/2,81/8),N); | ||
</asy> | </asy> | ||
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| + | We see that this creates two congruent triangles. Let the smaller side of the triangle have length <math>a</math> and let the larger side of the triangle have length <math>b</math>. Now we see by AAS congruency that if we draw perpendiculars that surround the smaller square, each outside triangle will be congruent to these two triangles. | ||
Solution in progress | Solution in progress | ||
Revision as of 13:11, 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 
![[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;
![[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((1,6)--(0,6)); draw((1,6)--(4,6)); draw((4,6)--(4,10)); 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); label("$b$",(-1/8,9/2),W); label("$b$",(-1/8,7),W); label("$a$",(1/3,6),N); label("$a$",(-1/8,28/3),W); label("$b$",(1,81/8),N); label("$b$",(3,6),N); label("$a$",(401/100,13/2),W); label("$b$",(401/100,17/2),E); label("$a$",(7/2,81/8),N); [/asy]](http://latex.artofproblemsolving.com/2/2/6/226e46eb891426ea1456c080917bcbff67949695.png)
We see that this creates two congruent triangles. Let the smaller side of the triangle have length 
and let the larger side of the triangle have length 
. Now we see by AAS congruency that if we draw perpendiculars that surround the smaller square, each outside triangle will be congruent to these two triangles.
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. 
