Difference between revisions of "2018 AIME I Problems/Problem 4"
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==Solution 6 (Easiest way- Coordinates without bash)== | ==Solution 6 (Easiest way- Coordinates without bash)== | ||
− | Let <math>B=(0, 0)</math>, and <math>C=(12, 0)</math>. From there, we know that <math>A=(6, 8)</math>, so line <math>AB</math> is <math>y=\frac{4}{3}x</math>. Hence, <math>D=(a, \frac{4}{3}a)</math> for some <math>a</math>, and <math>BD=\frac{5}{3}a</math> so <math>AD=10-\frac{5}{3}a</math>. Now, notice that by symmetry, <math>E=(6+a, 8-\frac{4}{3}a)</math>, so <math>ED^2=6^2+(8-\frac{8}{3}a)^2</math>. Because <math>AD=ED</math>, we now have <math>(10-\frac{5}{3})^2=6^2+(8-\frac{8}{3}a)^2</math>, which simplifies to <math>\frac{64}{9}a^2-\frac{128}{3}a+100=\frac{25}{9}a^2-\frac{100}{3}a+100</math>, so <math>\frac{39}{9}a=\frac{13} | + | Let <math>B=(0, 0)</math>, and <math>C=(12, 0)</math>. From there, we know that <math>A=(6, 8)</math>, so line <math>AB</math> is <math>y=\frac{4}{3}x</math>. Hence, <math>D=(a, \frac{4}{3}a)</math> for some <math>a</math>, and <math>BD=\frac{5}{3}a</math> so <math>AD=10-\frac{5}{3}a</math>. Now, notice that by symmetry, <math>E=(6+a, 8-\frac{4}{3}a)</math>, so <math>ED^2=6^2+(8-\frac{8}{3}a)^2</math>. Because <math>AD=ED</math>, we now have <math>(10-\frac{5}{3})^2=6^2+(8-\frac{8}{3}a)^2</math>, which simplifies to <math>\frac{64}{9}a^2-\frac{128}{3}a+100=\frac{25}{9}a^2-\frac{100}{3}a+100</math>, so <math>\frac{39}{9}a=\frac{13}{3}a=\frac{28}{3}</math>, and <math>a=\frac{28}{13}</math>. |
It follows that <math>AD=10-\frac{5}{3}\times\frac{28}{13}=10-\frac{140}{39}=\frac{390-140}{39}=\frac{250}{39}</math>, and our answer is <math>250+39=\boxed{289}</math>. | It follows that <math>AD=10-\frac{5}{3}\times\frac{28}{13}=10-\frac{140}{39}=\frac{390-140}{39}=\frac{250}{39}</math>, and our answer is <math>250+39=\boxed{289}</math>. | ||
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-Stormersyle | -Stormersyle | ||
==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 17:04, 23 November 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
=
. 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
Solution 5 (Fastest via Law of Cosines)
We can have 2 Law of Cosines applied on (one from
and one from
),
and
Solving for in both equations, we get
and
, so the answer is
-RootThreeOverTwo
Solution 6 (Easiest way- Coordinates without bash)
Let , and
. From there, we know that
, so line
is
. Hence,
for some
, and
so
. Now, notice that by symmetry,
, so
. Because
, we now have
, which simplifies to
, so
, and
.
It follows that
, and our answer is
.
-Stormersyle
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.