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> | ||
| + | ==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 09:13, 27 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 
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
