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1995 AIME Problems

Revision as of 21:00, 21 January 2007 by Minsoens (talk | contribs)

Problem 1

Square $\displaystyle S_{1}$ is $1\times 1.$ For $i\ge 1,$ the lengths of the sides of square $\displaystyle S_{i+1}$ are half the lengths of the sides of square $\displaystyle S_{i},$ two adjacent sides of square $\displaystyle S_{i}$ are perpendicular bisectors of two adjacent sides of square $\displaystyle S_{i+1},$ and the other two sides of square $\displaystyle S_{i+1},$ are the perpendicular bisectors of two adjacent sides of square $\displaystyle S_{i+2}.$ The total area enclosed by at least one of $\displaystyle S_{1}, S_{2}, S_{3}, S_{4}, S_{5}$ can be written in the form $\displaystyle m/n,$ where $\displaystyle m$ and $\displaystyle n$ are relatively prime positive integers. Find $\displaystyle m-n.$

AIME 1995 Problem 1.png

Solution

Problem 2

Find the last three digits of the product of the positive roots of $\sqrt{1995}x^{\log_{1995}x}=x^2.$

Solution

Problem 3

Starting at $\displaystyle (0,0),$ an object moves in the coordinate plane via a sequence of steps, each of length one. Each step is left, right, up, or down, all four equally likely. Let $\displaystyle p$ be the probability that the object reaches $\displaystyle (2,2)$ in six or fewer steps. Given that $\displaystyle p$ can be written in the form $\displaystyle m/n,$ where $\displaystyle m$ and $\displaystyle n$ are relatively prime positive integers, find $\displaystyle m+n.$

Solution

Problem 4

Circles of radius $\displaystyle 3$ and $\displaystyle 6$ are externally tangent to each other and are internally tangent to a circle of radius $\displaystyle 9$. The circle of radius $\displaystyle 9$ has a chord that is a common external tangent of the other two circles. Find the square of the length of this chord.

Solution

Problem 5

Solution

Problem 6

Solution

Problem 7

Solution

Problem 8

Solution

Problem 9

Solution

Problem 10

Solution

Problem 11

Solution

Problem 12

Solution

Problem 13

Solution

Problem 14

Solution

Problem 15

Solution

See also

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