Difference between revisions of "1968 AHSME Problems/Problem 24"
(→Solution) |
m |
||
| (One intermediate revision by one other user not shown) | |||
| Line 18: | Line 18: | ||
Setting the area of the frame equal to the area of the picture, | Setting the area of the frame equal to the area of the picture, | ||
| − | <cmath> | + | <cmath>(2x+18)(4x+24)-18\cdot24 = 18\cdot24</cmath> |
| − | <cmath>x^2 = 18\ | + | Solving, |
| + | <cmath>8x^2+120x+432 = 864</cmath> | ||
| + | <cmath>x^2+15x-54=0</cmath> | ||
| + | <cmath>(x+18)(x-3)=0\Rightarrow x=3</cmath> | ||
| + | |||
| + | Therefore, the ratio of the smaller side to the larger side is <math>\frac{18+2\cdot3}{24+4\cdot3}=\frac{24}{36}=\boxed{2:3 (C)}</math>. | ||
| + | |||
| + | ~AOPS81619 | ||
== See also == | == See also == | ||
| − | {{AHSME box|year=1968|num-b=23|num-a=25}} | + | {{AHSME 35p box|year=1968|num-b=23|num-a=25}} |
[[Category: Intermediate Algebra Problems]] | [[Category: Intermediate Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 00:53, 16 August 2023
Problem
A painting 
" X 
" is to be placed into a wooden frame with the longer dimension vertical. The wood at the top and bottom is twice as wide as the wood on the sides. If the frame area equals that of the painting itself, the ratio of the smaller to the larger dimension of the framed painting is:
Solution
Let the width of the frame on the sides to be 
.
Then, the width of the frame on the top and bottom is 
.
The area of the frame is then 
Setting the area of the frame equal to the area of the picture,
![\[(2x+18)(4x+24)-18\cdot24 = 18\cdot24\]](http://latex.artofproblemsolving.com/0/5/c/05c72b7e113df5e72e4fe3b9a9d58019f72c1a08.png)
Solving,
![\[8x^2+120x+432 = 864\]](http://latex.artofproblemsolving.com/e/c/2/ec2c12f7deaef6bfff85dac64e663c1c985dbea6.png)
![\[x^2+15x-54=0\]](http://latex.artofproblemsolving.com/b/e/2/be2d51b6b99feada4d5ba2492015d3178b2a4f6a.png)
![\[(x+18)(x-3)=0\Rightarrow x=3\]](http://latex.artofproblemsolving.com/2/a/f/2af6bce53c411a7dc33267c587ea3c669a27625d.png)
Therefore, the ratio of the smaller side to the larger side is 
.
~AOPS81619
See also
| 1968 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 23 |
Followed by Problem 25 | |
| 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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
