Difference between revisions of "2018 AIME I Problems/Problem 4"
Bluebacon008 (talk | contribs) (→Solution 1) |
Mathwiz0803 (talk | contribs) |
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~bluebacon008 | ~bluebacon008 | ||
− | ==Solution 2 (Law of Cosines)== | + | ==Solution 2 (Coordinates)== |
+ | Let <math>B = (0, 0)</math>, <math>C = (12, 0)</math>, and <math>A = (6, 8)</math>. Then, let <math>x</math> be in the interval <math>0<x<2</math> and parametrically define <math>D</math> and <math>E</math> as <math>(6-3x, 8-4x)</math> and <math>(12-3x, 4x)</math> respectively. Note that <math>AD = 5x</math>, so <math>DE = 5x</math>. This means that | ||
+ | <cmath>\begin{align*} | ||
+ | \sqrt{36+(8x-8)^2} &= 5x\ | ||
+ | 36+(8x-8)^2 &= 25x^2\ | ||
+ | 64x^2-128x+100 &= 25x^2\ | ||
+ | 39x^2-128x+100 &= 0\ | ||
+ | x &= \dfrac{128\pm\sqrt{16384-15600}}{78}\ | ||
+ | x &= \dfrac{100}{78}, 2\ | ||
+ | \end{align*}</cmath> | ||
+ | However, since <math>2</math> is extraneous by definition, <math>x=\dfrac{50}{39}\implies AD = \dfrac{250}{39}\implies\boxed{289}</math> ~ mathwiz0803 | ||
+ | |||
+ | ==Solution 3 (Law of Cosines)== | ||
As shown in the diagram, let <math>x</math> denote <math>\overline{AD}</math>. Let us denote the foot of the altitude of <math>A</math> to <math>\overline{BC}</math> as <math>F</math>. Note that <math>\overline{AE}</math> can be expressed as <math>10-x</math> and <math>\triangle{ABF}</math> is a <math>6-8-10</math> triangle . Therefore, <math>\sin(\angle{BAF})=\frac{3}{5}</math> and <math>\cos(\angle{BAF})=\frac{4}{5}</math>. Before we can proceed with the Law of Cosines, we must determine <math>\cos(\angle{BAC})=\cos(2\cdot \angle{BAF})=\cos^2(\angle{BAF})-\sin^2(\angle{BAF})=\frac{7}{25}</math>. Using LOC, we can write the following statement: | As shown in the diagram, let <math>x</math> denote <math>\overline{AD}</math>. Let us denote the foot of the altitude of <math>A</math> to <math>\overline{BC}</math> as <math>F</math>. Note that <math>\overline{AE}</math> can be expressed as <math>10-x</math> and <math>\triangle{ABF}</math> is a <math>6-8-10</math> triangle . Therefore, <math>\sin(\angle{BAF})=\frac{3}{5}</math> and <math>\cos(\angle{BAF})=\frac{4}{5}</math>. Before we can proceed with the Law of Cosines, we must determine <math>\cos(\angle{BAC})=\cos(2\cdot \angle{BAF})=\cos^2(\angle{BAF})-\sin^2(\angle{BAF})=\frac{7}{25}</math>. Using LOC, we can write the following statement: | ||
<cmath>(\overline{DE})^2=(\overline{AD})^2+\overline{AE}^2-2(\overline{AD})(\overline{AE})\cos(\angle{BAC})\implies</cmath> | <cmath>(\overline{DE})^2=(\overline{AD})^2+\overline{AE}^2-2(\overline{AD})(\overline{AE})\cos(\angle{BAC})\implies</cmath> | ||
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Thus, the desired answer is <math>\boxed{289}</math> ~ blitzkrieg21 | Thus, the desired answer is <math>\boxed{289}</math> ~ blitzkrieg21 | ||
− | ==Solution | + | ==Solution 4== |
In isosceles triangle, draw the altitude from <math>D</math> onto <math>\overline{AD}</math>. Let the point of intersection be <math>X</math>. Clearly, <math>AE=10-AD</math>, and hence <math>AX=\frac{10-AD}{2}</math>. | In isosceles triangle, draw the altitude from <math>D</math> onto <math>\overline{AD}</math>. Let the point of intersection be <math>X</math>. Clearly, <math>AE=10-AD</math>, and hence <math>AX=\frac{10-AD}{2}</math>. | ||
Revision as of 14:10, 9 March 2018
Contents
[hide]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 (No Trig)
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
represent
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 (Coordinates)
Let ,
, and
. Then, let
be in the interval
and parametrically define
and
as
and
respectively. Note that
, so
. This means that
However, since
is extraneous by definition,
~ mathwiz0803
Solution 3 (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 4
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,
Inspecting gives us
Solving the equation
gives
~novus677
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.