Difference between revisions of "2012 AIME I Problems/Problem 12"
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== Solutions == | == Solutions == | ||
=== Solution 1 === | === Solution 1 === | ||
− | Let <math>CD = 2a</math>. Using angle bisector theorem on <math>\triangle CDB</math>, <math>\frac{2a}{8}=\frac{CB}{15}</math>, so <math>CB = \frac{15a}{4}</math>. Then, drop the altitude from <math>D</math> to <math>CB</math> and call the foot <math>F</math>. Thus, <math>CF = a</math>, <math> | + | Let <math>CD = 2a</math>. Using angle bisector theorem on <math>\triangle CDB</math>, <math>\frac{2a}{8}=\frac{CB}{15}</math>, so <math>CB = \frac{15a}{4}</math>. Then, drop the altitude from <math>D</math> to <math>CB</math> and call the foot <math>F</math>. Thus, <math>CF = a</math>, <math>FD = a\sqrt{3}</math>, and <math>FB = \frac{11a}{4}</math>. Finally, <math>\tan{B} = \frac{a\sqrt{3}}{\frac{11a}{4}} = \frac{4\sqrt{3}}{11}</math>. Our answer is <math>\boxed{018}</math>. |
=== Solution 2 === | === Solution 2 === |
Revision as of 15:58, 26 February 2016
Contents
Problem 12
Let be a right triangle with right angle at
Let
and
be points on
with
between
and
such that
and
trisect
If
then
can be written as
where
and
are relatively prime positive integers, and
is a positive integer not divisible by the square of any prime. Find
Solutions
Solution 1
Let . Using angle bisector theorem on
,
, so
. Then, drop the altitude from
to
and call the foot
. Thus,
,
, and
. Finally,
. Our answer is
.
Solution 2
Without loss of generality, set . Then, by the Angle Bisector Theorem on triangle
, we have
. We apply the Law of Cosines to triangle
to get
, which we can simplify to get
.
Now, we have by another application of the Law of Cosines to triangle
, so
. In addition,
, so
.
Our final answer is .
Solution 3
(This solution does not use the Angle Bisector Theorem or the Law of Cosines, but it uses the Law of Sines and more trig)
Find values for all angles in terms of .
,
,
,
, and
.
Use the law of sines on and
:
In ,
. This simplifies to
.
In ,
. This simplifies to
.
Solve for and equate them so that you get
.
From this, .
Use a trig identity on the denominator on the right to obtain:
This simplifies to
This gives
Dividing by
, we have
. Our final answer is
.
Solution 4
(This solution avoids advanced trigonometry)
Let be the foot of the perpendicular from
to
, and let
be the foot of the perpendicular from
to
.
Now let . Clearly, triangles
and
are similar with
, so
.
Since triangles and
are 30-60-90 right triangles, we can easily find other lengths in terms of
. For example, we see that
and
. Therefore
.
Again using the fact that triangles and
are similar, we see that
, so
.
Thus , and our answer is
.
Solution 5
(Another solution without trigonometry)
Extend to point
such that
. It is then clear that
is similar to
.
Let ,
. Then
.
With the Angle Bisector Theorem, we get that . From 30-60-90
, we get that
and
.
From , we have that
. Simplifying yields
, and
, so our answer is
.
See also
2012 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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.