2015 AIME II Problems/Problem 4

Revision as of 15:55, 26 March 2015 by XXQw3rtyXx (talk | contribs) (Solution)

Problem

In an isosceles trapezoid, the parallel bases have lengths $\log 3$ and $\log 192$, and the altitude to these bases has length $\log 16$. The perimeter of the trapezoid can be written in the form $\log 2^p 3^q$, where $p$ and $q$ are positive integers. Find $p + q$.

Solution

Call the trapezoid $ABCD$ with $AB$ as the smaller base and $CD$ as the longer. The point where an altitude intersects the larger base be $E$ where $E$ is closer to $D$.

Subtract the two bases and divide to find that $ED$ is $\log 8$. The altitude can be expressed as $\frac{4}{3} log 8$. Therefore, the two legs are $\frac{5}{3} \log 8$, or $\log 32$.

The perimeter is thus $\log 32 + \log 32 + \log 192 + \log 3$ which is $\log 2^{16} 3^2$. So $p + q = \boxed{18}$

2015 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
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