Difference between revisions of "2015 AIME II Problems/Problem 4"

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==Solution==
 
==Solution==
  
Call the trapezoid <math>ABCD</math> with <math>AB</math> as the smaller base and <math>CD</math> as the longer. The point where an altitude intersects the larger base be <math>E</math> where <math>E</math> is closer to <math>D</math>.  
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Call the trapezoid <math>ABCD</math> with <math>AB</math> as the smaller base and <math>CD</math> as the longer. Let the point where an altitude intersects the larger base be <math>E</math>, where <math>E</math> is closer to <math>D</math>.  
  
 
Subtract the two bases and divide to find that <math>ED</math> is <math>\log 8</math>. The altitude can be expressed as <math>\frac{4}{3} \log 8</math>. Therefore, the two legs are <math>\frac{5}{3} \log 8</math>, or <math>\log 32</math>.
 
Subtract the two bases and divide to find that <math>ED</math> is <math>\log 8</math>. The altitude can be expressed as <math>\frac{4}{3} \log 8</math>. Therefore, the two legs are <math>\frac{5}{3} \log 8</math>, or <math>\log 32</math>.

Latest revision as of 15:16, 28 March 2020

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 $p + q = \boxed{018}$

See also

2015 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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