Difference between revisions of "2015 AIME II Problems/Problem 4"
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+ | ==Solution 3== | ||
+ | Let <math>ABCD</math> be the trapezoid, where <math>\overline{AB} || \overline{CD}</math> and <math>AB = \log 3</math> and <math>CD = \log 192</math>. Draw altitudes from <math>A</math> and <math>B</math> to <math>\overline{CD}</math> with feet at <math>E</math> and <math>F</math>, respectively. <math>AB = \log 3</math>, so <math>EF = \log 3</math>. Now, we attempt to find <math>DE + FC</math>, or what's left of <math>CD</math> after we take out <math>EF</math>. We make use of the two logarithmic rules: | ||
+ | |||
+ | <cmath>\log(xy) = \log x + \log y</cmath> | ||
+ | |||
+ | <cmath>\log(x^a) = a\log(x)</cmath> | ||
+ | |||
+ | <cmath>CD = \log 192 = \log (3 \cdot 2^6) = \log 3 + \log(2^6) = \log 3 + 6\log 2</cmath> | ||
+ | |||
+ | Thus, since <math>CD = DE + EF + FC = \log 3 + 6\log 2</math>, <math>CD - EF = \log 3 + 6\log 2 - \log 3 = 6\log 2 = DE + FC</math>. | ||
+ | |||
+ | Now, why was finding <math>DE + FC</math> important? Absolutely no reason! Just kidding, lol 🤣 Now, we essentially "glue" triangles <math>\triangle DAE</math> and <math>\triangle BFC</math> together to get <math>\triangle XC'D'</math>, where <math>X</math> is the point where <math>A</math> and <math>B</math> became one. Note we can do this because <math>\triangle DAE</math> and <math>\triangle BFC</math> are both right triangles with a common leg length (the altitude of trapezoid <math>ABCD</math>). | ||
+ | |||
+ | Triangle <math>XC'D'</math> has a base of <math>C'D'</math>, which is just equal to <math>DE + FC = 6\log 2</math>. It is equal to <math>DE + FC</math> because when we brought triangles <math>\triangle DAE</math> and <math>\triangle BFC</math> together, the length of <math>CD</math> was not changed except for taking out <math>EF</math>. | ||
+ | |||
+ | <math>XC' = XD'</math> since <math>AD = BC</math> because the problem tells us we have an isosceles trapezoid. Drop and altitude from <math>X</math> to <math>C'D'</math> The altitude has length <math>\log 16 = 4\log 2</math>. The altitude also bisects <math>C'D'</math> since <math>\triangle XC'D'</math> is isosceles. Let the foot of the altitude be <math>M</math>. Then <math>MD' = 3\log 2</math> (Remember that C'D' was <math>6\log 2</math>, and then it got bisected by the altitude). Thus, the hypotenuse, <math>XD'</math> must be <math>5\log 2</math> from the Pythagorean Theorem or by noticing that you have a 3-4-5 right triangle with a similarity ratio of <math>\log 2</math>. Since <math>XD' = XC' = BC = AD</math>, <math>BC = AD = 5\log 2 = \log 2^5</math>. | ||
+ | |||
+ | Now, we have <math>CD = \log (3 \cdot 2^6)</math>, <math>AB = \log 3</math>, and <math>BC = AD = \log 2^5</math>. 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, <math>p + q = 16 + 2 = \boxed{18}</math>. ~Extremelysupercooldude | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=9re2qLzOKWk&t=226s | ||
+ | |||
+ | ~MathProblemSolvingSkills.com | ||
+ | |||
+ | |||
==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}} |
Latest revision as of 14:17, 28 June 2023
Contents
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
, which is just equal to
. It is equal to
because when we brought triangles
and
together, the length of
was not changed except for taking out
.
since
because the problem tells us we have an isosceles trapezoid. Drop and altitude from
to
The altitude has length
. The altitude also bisects
since
is isosceles. Let the foot of the altitude be
. Then
(Remember that C'D' was
, and then it got bisected by the altitude). Thus, the hypotenuse,
must be
from the Pythagorean Theorem or by noticing that you have a 3-4-5 right triangle with a similarity ratio of
. Since
,
.
Now, we have ,
, and
. Thus, their sum is
Thus, . ~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.