Difference between revisions of "2013 AIME I Problems/Problem 12"
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First, find that <math>\angle R = 45^\circ</math>. | First, find that <math>\angle R = 45^\circ</math>. | ||
Draw <math>ABCDEF</math>. Now draw <math>\bigtriangleup PQR</math> around <math>ABCDEF</math> such that <math>Q</math> is adjacent to <math>C</math> and <math>D</math>. The height of <math>ABCDEF</math> is <math>\sqrt{3}</math>, so the length of base <math>QR</math> is <math>2+\sqrt{3}</math>. Let the equation of <math>\overline{RP}</math> be <math>y = x</math>. Then, the equation of <math>\overline{PQ}</math> is <math>y = -\sqrt{3} (x - (2+\sqrt{3})) \to y = -x\sqrt{3} + 2\sqrt{3} + 3</math>. Solving the two equations gives <math>y = x = \frac{\sqrt{3} + 3}{2}</math>. The area of <math>\bigtriangleup PQR</math> is <math>\frac{1}{2} * (2 + \sqrt{3}) * \frac{\sqrt{3} + 3}{2} = \frac{5\sqrt{3} + 9}{4}</math>. <math>a + b + c + d = 9 + 5 + 3 + 4 = \boxed{021}</math> | Draw <math>ABCDEF</math>. Now draw <math>\bigtriangleup PQR</math> around <math>ABCDEF</math> such that <math>Q</math> is adjacent to <math>C</math> and <math>D</math>. The height of <math>ABCDEF</math> is <math>\sqrt{3}</math>, so the length of base <math>QR</math> is <math>2+\sqrt{3}</math>. Let the equation of <math>\overline{RP}</math> be <math>y = x</math>. Then, the equation of <math>\overline{PQ}</math> is <math>y = -\sqrt{3} (x - (2+\sqrt{3})) \to y = -x\sqrt{3} + 2\sqrt{3} + 3</math>. Solving the two equations gives <math>y = x = \frac{\sqrt{3} + 3}{2}</math>. The area of <math>\bigtriangleup PQR</math> is <math>\frac{1}{2} * (2 + \sqrt{3}) * \frac{\sqrt{3} + 3}{2} = \frac{5\sqrt{3} + 9}{4}</math>. <math>a + b + c + d = 9 + 5 + 3 + 4 = \boxed{021}</math> | ||
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+ | Note (different ending): When you have the length of the base <math>QR</math>, you don't need to find the equations of the lines <math>QP</math> and <math>PR</math>. Instead, make an altitude from <math>P</math> to <math>QR</math>, and call the foot <math>M</math>. <math>QPM</math> is a <math>45, 45, 90</math> triangle and <math>PMR</math> is a <math>30, 60, 90</math> triangle. And, they both share <math>PM</math>. So, we can set <math>RM</math> as <math>x</math>, so <math>PM</math> is <math>x\sqrt{3}</math>. Since <math>QPM</math> is a <math>45, 45, 90</math> triangle, <math>PM=MQ=x\sqrt{3}</math>. The base <math>QR</math> can be written as <math>QM+MR=x+x\sqrt{3}=2+\sqrt{3}</math>. Solve this equation and <math>x=\frac{\sqrt{3}+1}{2}</math> and <math>PM=x\sqrt{3}=\frac{\sqrt{3}+3}{2}</math>. Multiply this by base <math>QR</math> and divide by <math>2</math> to get the area of triangle <math>PQR</math> which is <math>\frac{9+5\sqrt{3}}{4}</math>. So, <math>a + b + c + d = 9 + 5 + 3 + 4 = \boxed{021}</math> | ||
+ | -hwan | ||
==Solution 2 (Cartesian Variation)== | ==Solution 2 (Cartesian Variation)== | ||
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[[File:Better_photo_for_2013_aime_i_problem_12.png]] | [[File:Better_photo_for_2013_aime_i_problem_12.png]] | ||
− | As we can see, the <math>75^\circ</math> angle of <math>\angle P</math> can be split into a <math>45^\circ</math> angle and a <math>30^\circ</math> angle. This allows us to drop an altitude from point <math>P</math> for <math>\triangle RPQ</math> which intersects <math>\overline{AF}</math> at point <math>a</math> and <math>\overline{RQ}</math> at point <math>b</math>. The main idea of our solution is to obtain enough sides of <math>\triangle RPQ</math> that we are able to directly figure out its area (specifically by figuring out side <math>\overline{RQ}</math> and <math>\overline{Pb}</math>. | + | As we can see, the <math>75^\circ</math> angle of <math>\angle P</math> can be split into a <math>45^\circ</math> angle and a <math>30^\circ</math> angle. This allows us to drop an altitude from point <math>P</math> for <math>\triangle RPQ</math> which intersects <math>\overline{AF}</math> at point <math>a</math> and <math>\overline{RQ}</math> at point <math>b</math>. The main idea of our solution is to obtain enough sides of <math>\triangle RPQ</math> that we are able to directly figure out its area (specifically by figuring out side <math>\overline{RQ}</math> and <math>\overline{Pb}</math>). |
We first begin by figuring out the length of <math>\overline{PQ}</math>. This can be easily done, since <math>\overline{AB}</math> is simply <math>1</math> (given in the problem) and <math>\overline{BQ}=1</math> because <math>\triangle BCQ</math> is an equilateral after some simple angle calculations. Now we need to find <math>\overline{PA}</math>. This is when we bring in some simple algebra. | We first begin by figuring out the length of <math>\overline{PQ}</math>. This can be easily done, since <math>\overline{AB}</math> is simply <math>1</math> (given in the problem) and <math>\overline{BQ}=1</math> because <math>\triangle BCQ</math> is an equilateral after some simple angle calculations. Now we need to find <math>\overline{PA}</math>. This is when we bring in some simple algebra. |
Latest revision as of 18:54, 19 July 2024
Contents
Problem
Let be a triangle with
and
. A regular hexagon
with side length 1 is drawn inside
so that side
lies on
, side
lies on
, and one of the remaining vertices lies on
. There are positive integers
and
such that the area of
can be expressed in the form
, where
and
are relatively prime, and c is not divisible by the square of any prime. Find
.
Solution 1
First, find that .
Draw
. Now draw
around
such that
is adjacent to
and
. The height of
is
, so the length of base
is
. Let the equation of
be
. Then, the equation of
is
. Solving the two equations gives
. The area of
is
.
Note (different ending): When you have the length of the base , you don't need to find the equations of the lines
and
. Instead, make an altitude from
to
, and call the foot
.
is a
triangle and
is a
triangle. And, they both share
. So, we can set
as
, so
is
. Since
is a
triangle,
. The base
can be written as
. Solve this equation and
and
. Multiply this by base
and divide by
to get the area of triangle
which is
. So,
-hwan
Solution 2 (Cartesian Variation)
Use coordinates. Call the origin and
be on the x-axis. It is easy to see that
is the vertex on
. After labeling coordinates (noting additionally that
is an equilateral triangle), we see that the area is
times
times the coordinate of
. Draw a perpendicular of
, call it
, and note that
after using the trig functions for
degrees.
Now, get the lines for and
:
and
, whereupon we get the ordinate of
to be
, and the area is
, so our answer is
.
Solution 3 (Trig)
Angle chasing yields that both triangles and
are
-
-
triangles. First look at triangle
. Using Law of Sines, we find:
Simplifying, we find .
Since
, WLOG assume triangle
is equilateral, so
. So
.
Apply Law of Sines again,
Simplifying, we find .
.
Evaluating and reducing, we get thus the answer is
Solution 4 (Special Triangles)
As we can see, the angle of
can be split into a
angle and a
angle. This allows us to drop an altitude from point
for
which intersects
at point
and
at point
. The main idea of our solution is to obtain enough sides of
that we are able to directly figure out its area (specifically by figuring out side
and
).
We first begin by figuring out the length of . This can be easily done, since
is simply
(given in the problem) and
because
is an equilateral after some simple angle calculations. Now we need to find
. This is when we bring in some simple algebra.
PREPARATION:
(45-45-90 Right Triangle)
(30-60-90 Right Triangle)
SOLVING:
so
Finally,
Now, we can finally get the length of by adding up
, which is simply
To get and
, we first work bit by bit.
(30-60-90 Right Triangle)
(same 30-60-90 Right Triangle)
Since because of 45-45-90 right triangles,
too.
Now, we can finally calculate , and it is
.
Finally, the area of can be calculated by
, which is equal to
. So the final answer is
.
-by What do Humanitarians Eat?
Solution 5 (Trig)
With some simple angle chasing we can show that and
are congruent. This means we have a large equilateral triangle with side length
and quadrilateral
. We know that
. Using Law of Sines and the fact that
we know that
and the height to that side is
so
. Using an extremely similar process we can show that
which means the height to
is
. So the area of
. This means the area of quadrilateral
. So the area of our larger triangle is
. Therefore
.
Solution 6 (Elementary Geo)
We can find that . This means that the perpendicular from
to
is perpendicular to
as well, so let that perpendicular intersect
at
, and the perpendicular intersect
at
. Set
. Note that
, so
and
. Also,
, so
. It's easy to calculate the area now, because the perpendicular from
to
splits
into a
(PHQ) and a
(PHR). From these triangles' ratios, it should follow that
, so the area is
.
.
By Mathscienceclass
Solution 7 (Combination of 1 & 2)
We can observe that (because
&
are both
). Thus we know that
is equivalent to the height of the hexagon, which is
. Now we look at triangle
and apply the Law of Sines to it.
. From here we can solve for
and get that
. Now we use the Sine formula for the area of a triangle with sides
,
, and
to get the answer. Setting
and
we get the expression
which is
. Thus our final answer is
.
By AwesomeLife_Math
See also
2013 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.