Difference between revisions of "2015 AIME II Problems/Problem 4"
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==Solution== | ==Solution== | ||
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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. 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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| + | 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>. | ||
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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{18}</math> | ||
{{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 15:54, 26 March 2015
Problem
In an isosceles trapezoid, the parallel bases have lengths 
and 
, and the altitude to these bases has length 
. The perimeter of the trapezoid can be written in the form 
, where 
and 
are positive integers. Find 
.
Solution
Call the trapezoid 
with 
as the smaller base and 
as the longer. The point where an altitude intersects the larger base be 
where is closer to 
.
Subtract the two bases and divide to find that 
is 
. The altitude can be expressed as 
. Therefore, the two legs are 
, or 
.
The perimeter is thus 
which is 
. So 
| 2015 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
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