2015 AIME II Problems/Problem 4

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. Let 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 $p + q = \boxed{018}$

Solution 2 (gratuitous wishful thinking

Set the base of the log as 2, as it is the general log. Then call the trapezoid $ABCD$ with $CD$ as the longer base. Then have the two feet of the altitudes be $E$ and $F$ , with $E$ and $F$ in position from left to right respectively. Then, $CF$ and $ED$ are $log 192- log 3 = log 64$ (from the log subtraction identity. Then $CF=EF=3$ (isosceles trapezoid and $log 64$ being 6. Then the 2 legs of the trapezoid is $\sqrt{3^2+4^2}=5=log 32$ .

And we have the answer:

$log 192 + log 32 + log 32 + log 3 = log(192*32*32*3) = log(2^6*3*2^5*2^5*3) = log(2^16*3^2) = 16+2 = \boxed{18}$

2015 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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2015 AIME II Problems/Problem 4

Contents

Problem

Solution

Solution 2 (gratuitous wishful thinking

See also