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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>
  
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==Solution==
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==Solution (Intuitive)==
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{{solution}}
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The ratio of a circumference to a diameter always is the same so the answer is obviously C.
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==Solution 2 (Full Proof)==
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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==
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{{AHSME 40p box|year=1962|num-b=4|num-a=6}}
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[[Category:Introductory Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 10:24, 17 August 2021

Problem

If the radius of a circle is increased by $1$ unit, the ratio of the new circumference to the new diameter is:

$\textbf{(A)}\ \pi+2\qquad\textbf{(B)}\ \frac{2\pi+1}{2}\qquad\textbf{(C)}\ \pi\qquad\textbf{(D)}\ \frac{2\pi-1}{2}\qquad\textbf{(E)}\ \pi-2$

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 $r$. When the radius is increased by $1$, the new radius is $r+1$ so the diameter is $2r+2$. We know that the circumference of a circle is $2\pi r$ so $2 \cdot \pi \cdot (r+1) = \pi \cdot (2r+2)$. Finally, the problem asked for the ratio of the new circumference to the new diameter is $\frac{\pi \cdot (2r+2)}{2r+2}=\boxed{\pi}$.

~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

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