Difference between revisions of "1971 AHSME Problems/Problem 17"
Coolmath34 (talk | contribs) (Created page with "== Problem == A circular disk is divided by <math>2n</math> equally spaced radii(<math>n>0</math>) and one secant line. The maximum number of non-overlapping areas into whi...") |
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\textbf{(E) }3n+1 </math> | \textbf{(E) }3n+1 </math> | ||
| − | == Solution == | + | == Solution 1 == |
| + | Note that, before adding the secant line, we have <math>2n</math> regions, and, starting from the edge of the disk, every time the secant line intersects a radius or the other end of the disk, a new region is created. Thus, we desire to maximize the number of radii that the secant line passes through. | ||
| + | |||
| + | The angle between any two adjacent radii must be <math>\tfrac{360^{\circ}}{2n}=\tfrac{180^{\circ}}n</math>. A given secant cannot pass through more than <math>180^{\circ}</math> of the radii, and passing through exactly <math>180^{\circ}</math> would imply that the secant must go through the center of the circle, and so it would not produce any new regions. Thus, to maximize the number of regions, the secant line must pass through <math>n</math> radii (spanning <math>\tfrac{180}n(n-1)</math> degrees) and thereby create <math>n+1</math> new regions (one for each radius and one for exiting the disk). Thus, we have a maximum of <math>2n+n+1=\boxed{\textbf{(E) }3n+1}</math> regions. | ||
| + | |||
| + | == Solution 2 == | ||
We can draw the cases for small values of <math>n.</math> | We can draw the cases for small values of <math>n.</math> | ||
<cmath>n = 0 \rightarrow \text{areas} = 1</cmath> | <cmath>n = 0 \rightarrow \text{areas} = 1</cmath> | ||
| Line 18: | Line 23: | ||
It seems that for <math>2n</math> radii, there are <math>3n+1</math> distinct areas. The secant line must pass through <math>n</math> radii for this to occur. | It seems that for <math>2n</math> radii, there are <math>3n+1</math> distinct areas. The secant line must pass through <math>n</math> radii for this to occur. | ||
| − | The answer is <math>\textbf{(E)} | + | The answer is <math>\boxed{\textbf{(E) } 3n+1}.</math> |
-edited by coolmath34 | -edited by coolmath34 | ||
| + | |||
| + | == See Also == | ||
| + | {{AHSME 35p box|year=1971|num-b=16|num-a=18}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 09:17, 5 August 2024
Contents
Problem
A circular disk is divided by 
equally spaced radii(
) and one secant line. The maximum number of non-overlapping
areas into which the disk can be divided is
Solution 1
Note that, before adding the secant line, we have regions, and, starting from the edge of the disk, every time the secant line intersects a radius or the other end of the disk, a new region is created. Thus, we desire to maximize the number of radii that the secant line passes through.
The angle between any two adjacent radii must be 
. A given secant cannot pass through more than 
of the radii, and passing through exactly would imply that the secant must go through the center of the circle, and so it would not produce any new regions. Thus, to maximize the number of regions, the secant line must pass through 
radii (spanning 
degrees) and thereby create 
new regions (one for each radius and one for exiting the disk). Thus, we have a maximum of 
regions.
Solution 2
We can draw the cases for small values of 
![\[n = 0 \rightarrow \text{areas} = 1\]](http://latex.artofproblemsolving.com/1/b/4/1b400bb2c5a506970dea9b4f262de1533b485433.png)
![\[n = 1 \rightarrow \text{areas} = 4\]](http://latex.artofproblemsolving.com/0/c/4/0c4c477dd467136ec568600e06bc98e87353069f.png)
![\[n = 2 \rightarrow \text{areas} = 7\]](http://latex.artofproblemsolving.com/6/a/1/6a1082b2c4e1d929333f2dd11247d7d82142c9d2.png)
![\[n = 3 \rightarrow \text{areas} = 10\]](http://latex.artofproblemsolving.com/6/e/9/6e911b745e326ea1cf16f35a0aa4617c910c8143.png)
It seems that for radii, there are 
distinct areas. The secant line must pass through radii for this to occur.
The answer is 
-edited by coolmath34
See Also
| 1971 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 16 |
Followed by Problem 18 | |
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
