Difference between revisions of "2018 AIME I Problems/Problem 4"
m (→Solution 2 (Law of Cosines)) |
m (→Solution 3) |
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<math>\angle BAC = 180-2\cdot\angle ABC</math>. And hence, | <math>\angle BAC = 180-2\cdot\angle ABC</math>. And hence, | ||
+ | |||
<math>\begin{align*} \cos \angle BAC = \cos \angle (180-2\cdot\angle ABC) | <math>\begin{align*} \cos \angle BAC = \cos \angle (180-2\cdot\angle ABC) | ||
&= -\cos (2\cdot\angle ABC) | &= -\cos (2\cdot\angle ABC) | ||
&= \sin^2 \angle ABC - \cos^2 \angle ABC | &= \sin^2 \angle ABC - \cos^2 \angle ABC | ||
− | &= \frac{16}{25}-\frac{9}{25}=\frac{7}{25}</math> | + | &= \frac{16}{25}-\frac{9}{25}=\frac{7}{25} |
+ | \end{align*}</math> | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2018|n=I|num-b=3|num-a=5}} | {{AIME box|year=2018|n=I|num-b=3|num-a=5}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 04:08, 8 March 2018
Problem 4
In and
. Point
lies strictly between
and
on
and point
lies strictly between
and
on
) so that
. Then
can be expressed in the form
, where
and
are relatively prime positive integers. Find
.
Solution 1
We draw the altitude from to
to get point
. We notice that the triangle's height from
to
is 8 because it is a
Right Triangle. To find the length of
, we let
be the height and set up an equation by finding two ways to express the area. The equation is
, which leaves us with
. We then solve for the length
, which is done through pythagorean theorm and get
=
. We can now see that
is a
Right Triangle. Thus, we set
as
, and yield that
. Now, we can see
, so we have
. Solving this equation, we yield
, or
. Thus, our final answer is
.
~bluebacon008
Solution 2 (Law of Cosines)
As shown in the diagram, let denote
. Let us denote the foot of the altitude of
to
as
. Note that
can be expressed as
and
is a
triangle . Therefore,
and
. Before we can proceed with the Law of Cosines, we must determine
. Using LOC, we can write the following statement:
Thus, the desired answer is
~ blitzkrieg21
Solution 3
In isosceles triangle, draw the altitude from onto
. Let the point of intersection be
. Clearly,
, and hence
.
Now, we recognise that the perpendicular from onto
gives us two
-
-
triangles. So, we calculate
and
. And hence,
$
See Also
2018 AIME I (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.