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

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The perimeter is thus <math>\log 32 + \log 32 + \log 192 + \log 3</math> which is <math>\log 2^{16} 3^2</math>. So <math>p + q = \boxed{018}</math>
 
The perimeter is thus <math>\log 32 + \log 32 + \log 192 + \log 3</math> which is <math>\log 2^{16} 3^2</math>. So <math>p + q = \boxed{018}</math>
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==Solution 2 (gratuitous wishful thinking==
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Set the base of the log as 2, as it is the general log. Then call the trapezoid <math>ABCD</math> with <math>CD</math> as the longer base. Then have the two feet of the altitudes be <math>E</math> and <math>F</math>, with <math>E</math> and <math>F</math> in position from left to right respectively. Then, <math>CF</math> and <math>ED</math> are <math>log 192- log 3 = log 64</math> (from the log subtraction identity. Then <math>CF=EF=3</math> (isosceles trapezoid and <math>log 64</math> being 6. Then the 2 legs of the trapezoid is <math>\sqrt{3^2+4^2}=5=log 32</math>.
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And we have the answer:
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<math>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}</math>
  
 
==See also==
 
==See also==
 
{{AIME box|year=2015|n=II|num-b=3|num-a=5}}
 
{{AIME box|year=2015|n=II|num-b=3|num-a=5}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 20:20, 18 October 2022

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:

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

See also

2015 AIME II (Problems • Answer Key • Resources)
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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