Difference between revisions of "2013 AIME I Problems/Problem 12"
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==Solution 4 (Elementary Geo)== | ==Solution 4 (Elementary Geo)== | ||
− | We can find that <math>AF || CD || QR</math>. This means that the perpendicular from <math>P</math> to <math>QR</math> is perpendicular to <math>AF</math> as well, so let that perpendicular intersect <math>AF</math> at <math>G</math>, and the perpendicular intersect <math>QR</math> at <math>H</math>. Set <math>AP=x</math>. Note that <math>\angle {PAG} = 60^\circ</math>, so <math>AG=\frac{x}{2}</math> and <math>PG = GF = \frac{x\sqrt3}{2}</math>. Also, <math>1=AF=AG+GF=\frac{x}{2} + \frac{x\sqrt{3}}{2}</math>, so<math>x=\sqrt{3} - 1</math>. It's easy to calculate the area now, because the perpendicular from <math>P</math> to <math>QR</math> splits <math>\triangle{PQR}</math> into a <math>30-60-90</math> (PHQ) and a <math>45-45-90</math> (PHR). From these triangles' ratios, it should follow that <math>QH=\frac{\sqrt{3} + 1}{2}, PH=HR=\frac{sqrt{3}+3}{2}</math>, so the area is <math>\frac{1}{2} * PH * QR = \frac{1}{2} * PH * (QH + HR) = \frac{1}{2} * \frac{\sqrt{3} + 3}{2} * \frac{2\sqrt{3}+4}{2} = \boxed{\frac{9+5\sqrt{3}}{4}}</math>. | + | We can find that <math>AF || CD || QR</math>. This means that the perpendicular from <math>P</math> to <math>QR</math> is perpendicular to <math>AF</math> as well, so let that perpendicular intersect <math>AF</math> at <math>G</math>, and the perpendicular intersect <math>QR</math> at <math>H</math>. Set <math>AP=x</math>. Note that <math>\angle {PAG} = 60^\circ</math>, so <math>AG=\frac{x}{2}</math> and <math>PG = GF = \frac{x\sqrt3}{2}</math>. Also, <math>1=AF=AG+GF=\frac{x}{2} + \frac{x\sqrt{3}}{2}</math>, so<math>x=\sqrt{3} - 1</math>. It's easy to calculate the area now, because the perpendicular from <math>P</math> to <math>QR</math> splits <math>\triangle{PQR}</math> into a <math>30-60-90</math> (PHQ) and a <math>45-45-90</math> (PHR). From these triangles' ratios, it should follow that <math>QH=\frac{\sqrt{3} + 1}{2}, PH=HR=\frac{\sqrt{3}+3}{2}</math>, so the area is <math>\frac{1}{2} * PH * QR = \frac{1}{2} * PH * (QH + HR) = \frac{1}{2} * \frac{\sqrt{3} + 3}{2} * \frac{2\sqrt{3}+4}{2} = \boxed{\frac{9+5\sqrt{3}}{4}}</math>. <math>9+5+3+4=021</math>. |
By Mathscienceclass | By Mathscienceclass | ||
Revision as of 15:26, 15 June 2019
Contents
Problem 12
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
.
Cartesian Variation Solution
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 ordinate 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 2 (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 3(Trig with Diagram)
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 4 (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
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.