Difference between revisions of "1962 AHSME Problems/Problem 5"
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<math> \textbf{(A)}\ \pi+2\qquad\textbf{(B)}\ \frac{2\pi+1}{2}\qquad\textbf{(C)}\ \pi\qquad\textbf{(D)}\ \frac{2\pi-1}{2}\qquad\textbf{(E)}\ \pi-2 </math> | <math> \textbf{(A)}\ \pi+2\qquad\textbf{(B)}\ \frac{2\pi+1}{2}\qquad\textbf{(C)}\ \pi\qquad\textbf{(D)}\ \frac{2\pi-1}{2}\qquad\textbf{(E)}\ \pi-2 </math> | ||
| − | ==Solution== | + | ==Solution (Intuitive)== |
The ratio of a circumference to a diameter always is the same so the answer is obviously C. | The ratio of a circumference to a diameter always is the same so the answer is obviously C. | ||
| + | |||
| + | ==Solution 2 (Full Proof)== | ||
| + | |||
| + | Let us say that the radius of a circle is <math>r</math>. When the radius is increased by <math>1</math>, the new radius is <math>r+1</math> so the diameter is <math>2r+2</math>. We know that the circumference of a circle is <math>2\pi r</math> so <math>2 \cdot \pi \cdot (r+1) = \pi \cdot (2r+2)</math>. Finally, the problem asked for the ratio of the new circumference to the new diameter is <math>\frac{\pi \cdot (2r+2)}{2r+2}=\boxed{\pi}</math>. | ||
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| + | ~Mathfun1000 | ||
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| + | ==See Also== | ||
| + | {{AHSME 40p box|year=1962|num-b=4|num-a=6}} | ||
| + | |||
| + | [[Category:Introductory Geometry Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 10:24, 17 August 2021
Problem
If the radius of a circle is increased by 
unit, the ratio of the new circumference to the new diameter is:
Solution (Intuitive)
The ratio of a circumference to a diameter always is the same so the answer is obviously C.
Solution 2 (Full Proof)
Let us say that the radius of a circle is 
. When the radius is increased by , the new radius is 
so the diameter is 
. We know that the circumference of a circle is 
so 
. Finally, the problem asked for the ratio of the new circumference to the new diameter is 
.
~Mathfun1000
See Also
| 1962 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 4 |
Followed by Problem 6 | |
| 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 • 36 • 37 • 38 • 39 • 40 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
