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Difference between revisions of "1952 AHSME Problems/Problem 43"

(→‎Problem)
(The previous solution had the correct answer, but it didn't explain in any way the actual calculation to be done, in fact it had an error, and just bla bla bla with no justification, look for yourself to check it.)
 
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<math>\textbf{(A) } \qquad</math> equal to the semi-circumference of the original circle
 
<math>\textbf{(A) } \qquad</math> equal to the semi-circumference of the original circle
 
<math>\textbf{(B) } \qquad</math> equal to the diameter of the original circle
 
<math>\textbf{(B) } \qquad</math> equal to the diameter of the original circle
−
<math>\textbf{(C) } (11 - k) \qquad</math> greater than the diameter, but less than the semi-circumference of the original circle
+
<math>\textbf{(C) } \qquad</math> greater than the diameter, but less than the semi-circumference of the original circle
−
<math>\textbf{(D) } (k - 1) \qquad</math> that is infinite
+
<math>\textbf{(D) } \qquad</math> that is infinite
−
<math>\textbf{(E) } (k+1) </math> greater than the semi-circumference
+
<math>\textbf{(E) } </math> greater than the semi-circumference
  
 
== Solution ==
 
== Solution ==
−
Let our two digit number be <math>AB</math>. Its value is <math>10A + B</math>. The number formed by interchanging its digits is BA and has value <math>10B + A</math>.
+
Note that the half the circumference of a circle with diameter <math>d</math> is <math>\frac{\pi*d}{2}</math>.  
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Setting AB equal to <math>k</math> times the sum of the digits yields
+
 
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<cmath>10A + B = k(A + B)</cmath>
+
Let's call the diameter of the circle D. Dividing the circle's diameter into n parts means that each semicircle has diameter <math>\frac{D}{n}</math>, and thus each semicircle measures <math>\frac{D*pi}{n*2}</math>. The total sum of those is <math>n*\frac{D*pi}{n*2}=\frac{D*pi}{2}</math>, and since that is the exact expression for the semi-circumference of the original circle, the answer is <math>\boxed{A}</math>.
−
We now must relate AB to BA. Note that
 
−
<cmath>11(A + B) - (10A + B) = 10B + A</cmath>
 
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Using this in the first equation yields
 
−
<cmath>10A + B = k(A + B)</cmath>
 
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<cmath>11(A + B) - (10A + B) = 11(A + B) - k(A + B)</cmath>
 
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<cmath>10B + A = (11 - k)(A + B)</cmath>
 
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Therefore, <math>BA</math> is <math>(11 - k)</math> times the sum of its digits and our answer is <math>\fbox{C}</math>.
 
  
 
== See also ==
 
== See also ==

Latest revision as of 17:53, 30 June 2021

Problem

The diameter of a circle is divided into $n$ equal parts. On each part a semicircle is constructed. As $n$ becomes very large, the sum of the lengths of the arcs of the semicircles approaches a length: $\textbf{(A) } \qquad$ equal to the semi-circumference of the original circle $\textbf{(B) } \qquad$ equal to the diameter of the original circle $\textbf{(C) } \qquad$ greater than the diameter, but less than the semi-circumference of the original circle $\textbf{(D) } \qquad$ that is infinite $\textbf{(E) }$ greater than the semi-circumference

Solution

Note that the half the circumference of a circle with diameter $d$ is $\frac{\pi*d}{2}$.

Let's call the diameter of the circle D. Dividing the circle's diameter into n parts means that each semicircle has diameter $\frac{D}{n}$, and thus each semicircle measures $\frac{D*pi}{n*2}$. The total sum of those is $n*\frac{D*pi}{n*2}=\frac{D*pi}{2}$, and since that is the exact expression for the semi-circumference of the original circle, the answer is $\boxed{A}$.

See also

1952 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 42 		Followed by
Problem 44
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All AHSME Problems and Solutions

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