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AIME 1990 · 第 3 题

AIME 1990 — Problem 3

专题
Contest Math
难度
L4
来源
AIME

题目详情

Problem

Let P1P_1^{} be a regular r gonr~\text{gon} and P2P_2^{} be a regular s gons~\text{gon} (r≥s≥3)(r\geq s\geq 3) such that each interior angle of P1P_1^{} is 5958\frac{59}{58} as large as each interior angle of P2P_2^{}. What's the largest possible value of ss_{}^{}?

解析

Solution 1

The formula for the interior angle of a regular sided polygon is (n−2)180n\frac{(n-2)180}{n}.

Thus, (r−2)180r(s−2)180s=5958\frac{\frac{(r-2)180}{r}}{\frac{(s-2)180}{s}} = \frac{59}{58}. Cross multiplying and simplifying, we get 58(r−2)r=59(s−2)s\frac{58(r-2)}{r} = \frac{59(s-2)}{s}. Cross multiply and combine like terms again to yield 58rs−58⋅2s=59rs−59⋅2r⟹118r−116s=rs58rs - 58 \cdot 2s = 59rs - 59 \cdot 2r \Longrightarrow 118r - 116s = rs. Solving for rr, we get r=116s118−sr = \frac{116s}{118 - s}.

r≥0r \ge 0 and s≥0s \ge 0, making the numerator of the fraction positive. To make the denominator positive, s<118s < 118; the largest possible value of ss is 117117.

This is achievable because the denominator is 11, making rr a positive number 116⋅117116 \cdot 117 and s=117s = \boxed{117}.

Solution 2

Like above, use the formula for the interior angles of a regular sided polygon.

(r−2)180r=5958∗(s−2)180s\frac{(r-2)180}{r} = \frac{59}{58} * \frac{(s-2)180}{s} 59∗180∗(s−2)∗r=58∗180∗(r−2)∗s59 * 180 * (s-2) * r = 58 * 180 * (r-2) * s 59∗(rs−2r)=58∗(rs−2s)59 * (rs - 2r) = 58 * (rs - 2s) rs−118r=−116srs - 118r = -116s rs=118r−116srs = 118r-116s

This equation tells us ss divides 118r118r. If ss 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 s=59s=59, which does give a solution: s=59,r=116s=59, r=116. Although, the problem asks for ss, not rr. The only conceivable reasoning behind this is that rr is greater than 1000. This prompts us to look into the second case, where ss divides rr. Make r=s∗kr = s * k. Rewrite the equation using this new information.

s∗s∗k=118∗s∗k−116∗ss * s * k = 118 * s * k - 116 * s s∗k=118∗k−116s * k = 118 * k - 116

Now we now k divides 116. The larger k is, the larger s will be, so we set k to be the maximum: 116.

s∗116=118∗116−116s * 116 = 118 * 116 - 116 s=118−1s = 118 - 1 s=117s = \boxed{117}

-jackshi2006

Solution 3

As in above, we have rs=118r−116s.rs = 118r - 116s. This means that rs+116s−118r=0.rs + 116s - 118r = 0. Using SFFT we obtain s(r+116)−118(r+116)=−118⋅116  ⟹  (s−118)(r+116)=−118⋅116.s(r+116) - 118(r+116) = -118 \cdot 116 \implies (s-118)(r+116) = -118 \cdot 116. Since r+116r+116 is always positive, we know that s−118s-118 must be negative. Therefore the maximum value of ss must be 117\boxed{117} which indeed yields an integral value of r.r.

Video Solution

https://www.youtube.com/watch?v=9YwQlFAJqvc