Difference between revisions of "1952 AHSME Problems/Problem 43"
(Created page with "== Problem == If an integer of two digits is <math>k</math> times the sum of its digits, the number formed by interchanging the digits is the sum of the digits multiplied by: ...") |
Santropedro (talk | contribs) (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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== Problem == | == Problem == | ||
| − | + | The diameter of a circle is divided into <math>n</math> equal parts. On each part a semicircle is constructed. As <math>n</math> becomes very large, the sum of the lengths of the arcs of the semicircles approaches a length: | |
| + | <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{(C) } \qquad</math> greater than the diameter, but less than the semi-circumference of the original circle | ||
| + | <math>\textbf{(D) } \qquad</math> that is infinite | ||
| + | <math>\textbf{(E) } </math> greater than the semi-circumference | ||
| − | <math>\ | + | == Solution == |
| − | \ | + | 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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| − | = | + | 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>. |
| − | <math>\ | ||
== See also == | == See also == | ||
Latest revision as of 17:53, 30 June 2021
Problem
The diameter of a circle is divided into 
equal parts. On each part a semicircle is constructed. As becomes very large, the sum of the lengths of the arcs of the semicircles approaches a length:
equal to the semi-circumference of the original circle
equal to the diameter of the original circle
greater than the diameter, but less than the semi-circumference of the original circle
that is infinite
greater than the semi-circumference
Solution
Note that the half the circumference of a circle with diameter 
is 
.
Let's call the diameter of the circle D. Dividing the circle's diameter into n parts means that each semicircle has diameter 
, and thus each semicircle measures 
. The total sum of those is 
, and since that is the exact expression for the semi-circumference of the original circle, the answer is 
.
See also
| 1952 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 42 |
Followed by Problem 44 | |
| 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 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
