2015 AIME II Problems/Problem 4
Contents
[hide]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. Let 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
Solution 2 (gratuitous wishful thinking)
Set the base of the log as 2. Then call the trapezoid with
as the longer base. Then have the two feet of the altitudes be
and
, with
and
in position from left to right respectively. Then,
and
are
(from the log subtraction identity. Then
(isosceles trapezoid and
being 6. Then the 2 legs of the trapezoid is
.
And we have the answer:
-dragoon
Solution 3
Let be the trapezoid, where
and
and
. Draw altitudes from
and
to
with feet at
and
, respectively.
, so
. Now, we attempt to find
, or what's left of
after we take out
. We make use of the two logarithmic rules:
Thus, since ,
.
Now, why was finding important? Absolutely no reason! Just kidding, lol 🤣 Now, we essentially "glue" triangles
and
together to get
, where
is the point where
and
became one. Note we can do this because
and
are both right triangles with a common leg length (the altitude of trapezoid
).
Triangle has a base of $C'D'%, which is just equal to$ (Error compiling LaTeX. Unknown error_msg)DE + FC = 6\log 2
DE + FC
\triangle DAE
\triangle BFC
CD
EF
XC' = XD'
AD = BC
X
C'D'
\log 16 = 4\log 2
C'D'
\triangle XC'D'
M
MD' = 3\log 2
6\log 2
XD'
5\log 2
\log 2
XD' = XC' = BC = AD
BC = AD = 5\log 2 = \log 2^5$.
Now, we have$ (Error compiling LaTeX. Unknown error_msg)CD = \log (3 \cdot 2^6)AB = \log 3
BC = AD = \log 2^5$. Thus, their sum is
<cmath> \log (3 \cdot 2^6) + \log 3 + \log 2^5 + \log 2^5 = \log (2^16 \cdot 3^2)</cmath>
Thus,$ (Error compiling LaTeX. Unknown error_msg)p + q = 16 + 2 = \boxed{18}$. ~Extremelysupercooldude
Video Solution
https://www.youtube.com/watch?v=9re2qLzOKWk&t=226s
~MathProblemSolvingSkills.com
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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