Difference between revisions of "1968 AHSME Problems/Problem 1"
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== Solution == | == Solution == | ||
| − | Let <math>d</math> be the diameter of the original circle. If <math>d</math> is increased by <math>\pi</math>, then the new circumference is <math>\pi d + \pi^2 | + | Let <math>d</math> be the diameter of the original circle. If <math>d</math> is increased by <math>\pi</math>, then the new circumference is <math>\pi d + \pi^2</math>. The difference in circumference is therefore <math>\pi d + \pi^2 - \pi d = \pi^2</math> |
| − | <math>\fbox{D}</math> | + | |
| + | Therefore, the answer is <math>\fbox{D}</math> | ||
| + | |||
| + | Solution by VivekA | ||
== See also == | == See also == | ||
| − | {{AHSME box|year=1968|num-b=1|num-a=2}} | + | {{AHSME 35p box|year=1968|num-b=1|num-a=2}} |
[[Category: Introductory Geometry Problems]] | [[Category: Introductory Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 00:52, 16 August 2023
Problem
Let 
units be the increase in circumference of a circle resulting from an increase in 
units in the diameter. Then equals:
Solution
Let 
be the diameter of the original circle. If is increased by , then the new circumference is 
. The difference in circumference is therefore 
Therefore, the answer is 
Solution by VivekA
See also
| 1968 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 2 | |
| 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. 
