Difference between revisions of "1990 AIME Problems/Problem 3"
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Let <math>P_1^{}</math> be a [[Regular polygon|regular]] <math>r~\mbox{gon}</math> and <math>P_2^{}</math> be a regular <math>s~\mbox{gon}</math> <math>(r\geq s\geq 3)</math> such that each [[interior angle]] of <math>P_1^{}</math> is <math>\frac{59}{58}</math> as large as each interior angle of <math>P_2^{}</math>. What's the largest possible value of <math>s_{}^{}</math>? | Let <math>P_1^{}</math> be a [[Regular polygon|regular]] <math>r~\mbox{gon}</math> and <math>P_2^{}</math> be a regular <math>s~\mbox{gon}</math> <math>(r\geq s\geq 3)</math> such that each [[interior angle]] of <math>P_1^{}</math> is <math>\frac{59}{58}</math> as large as each interior angle of <math>P_2^{}</math>. What's the largest possible value of <math>s_{}^{}</math>? | ||
− | == Solution == | + | == Solution 1== |
The formula for the interior angle of a regular sided [[polygon]] is <math>\frac{(n-2)180}{n}</math>. | The formula for the interior angle of a regular sided [[polygon]] is <math>\frac{(n-2)180}{n}</math>. | ||
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This is achievable because the denominator is <math>1</math>, making <math>r</math> a positive number <math>116 \cdot 117</math> and <math>s = \boxed{117}</math>. | This is achievable because the denominator is <math>1</math>, making <math>r</math> a positive number <math>116 \cdot 117</math> and <math>s = \boxed{117}</math>. | ||
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+ | == Solution 2== | ||
+ | Like above, use the formula for the interior angles of a regular sided [[polygon]]. | ||
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+ | <math>\frac{(r-2)180}{r} = \frac{59}{58} * \frac{(s-2)180}{s}</math> | ||
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+ | <math>59 * 180 * (s-2) * r = 58 * 180 * (r-2) * s</math> | ||
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+ | <math>59 * (rs - 2r) = 58 * (rs - 2s)</math> | ||
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+ | <math>rs - 118r = -116s</math> | ||
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+ | <math>rs = 118r-116s</math> | ||
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+ | This equation tells us <math>s</math> divides <math>118r</math>. If <math>s</math> specifically divides 118 then the highest it can be is 118. However, this gives an equation with no solution. The second largest possibility in this case is <math>s=59</math>, which does give a solution: <math>s=59, r=116</math>. Although, the problem asks for <math>s</math>, not <math>r</math>. The only conceivable reasoning behind this is that <math>r</math> is greater than 1000. This prompts us to look into the second case, where <math>s</math> divides <math>r</math>. Make <math>r = s * k</math>. Rewrite the equation using this new information. | ||
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+ | <math>s * s * k = 118 * s * k - 116 * s</math> | ||
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+ | <math>s * k = 118 * k - 116</math> | ||
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+ | Now we now k divides 116. The larger k is, the larger s will be, so we set k to be the maximum, 116. | ||
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+ | <math>s * 116 = 118 * 116 - 116</math> | ||
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+ | <math>s = 118 - 1</math> | ||
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+ | <math>s = \boxed{117}</math> | ||
== See also == | == See also == |
Revision as of 14:20, 29 August 2020
Contents
Problem
Let be a regular and be a regular such that each interior angle of is as large as each interior angle of . What's the largest possible value of ?
Solution 1
The formula for the interior angle of a regular sided polygon is .
Thus, . Cross multiplying and simplifying, we get . Cross multiply and combine like terms again to yield . Solving for , we get .
and , making the numerator of the fraction positive. To make the denominator positive, ; the largest possible value of is .
This is achievable because the denominator is , making a positive number and .
Solution 2
Like above, use the formula for the interior angles of a regular sided polygon.
This equation tells us divides . If specifically divides 118 then the highest it can be is 118. However, this gives an equation with no solution. The second largest possibility in this case is , which does give a solution: . Although, the problem asks for , not . The only conceivable reasoning behind this is that is greater than 1000. This prompts us to look into the second case, where divides . Make . Rewrite the equation using this new information.
Now we now k divides 116. The larger k is, the larger s will be, so we set k to be the maximum, 116.
See also
1990 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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