Difference between revisions of "2013 AIME I Problems/Problem 12"
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− | == Problem | + | == Problem == |
Let <math>\bigtriangleup PQR</math> be a triangle with <math>\angle P = 75^\circ</math> and <math>\angle Q = 60^\circ</math>. A regular hexagon <math>ABCDEF</math> with side length 1 is drawn inside <math>\triangle PQR</math> so that side <math>\overline{AB}</math> lies on <math>\overline{PQ}</math>, side <math>\overline{CD}</math> lies on <math>\overline{QR}</math>, and one of the remaining vertices lies on <math>\overline{RP}</math>. There are positive integers <math>a, b, c, </math> and <math>d</math> such that the area of <math>\triangle PQR</math> can be expressed in the form <math>\frac{a+b\sqrt{c}}{d}</math>, where <math>a</math> and <math>d</math> are relatively prime, and c is not divisible by the square of any prime. Find <math>a+b+c+d</math>. | Let <math>\bigtriangleup PQR</math> be a triangle with <math>\angle P = 75^\circ</math> and <math>\angle Q = 60^\circ</math>. A regular hexagon <math>ABCDEF</math> with side length 1 is drawn inside <math>\triangle PQR</math> so that side <math>\overline{AB}</math> lies on <math>\overline{PQ}</math>, side <math>\overline{CD}</math> lies on <math>\overline{QR}</math>, and one of the remaining vertices lies on <math>\overline{RP}</math>. There are positive integers <math>a, b, c, </math> and <math>d</math> such that the area of <math>\triangle PQR</math> can be expressed in the form <math>\frac{a+b\sqrt{c}}{d}</math>, where <math>a</math> and <math>d</math> are relatively prime, and c is not divisible by the square of any prime. Find <math>a+b+c+d</math>. | ||
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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> | ||
− | ==Cartesian Variation | + | ==Solution 2 (Cartesian Variation)== |
Use coordinates. Call <math>Q</math> the origin and <math>QP</math> be on the x-axis. It is easy to see that <math>F</math> is the vertex on <math>RP</math>. After labeling coordinates (noting additionally that <math>QBC</math> is an equilateral triangle), we see that the area is <math>QP</math> times <math>0.5</math> times the ordinate of <math>R</math>. Draw a perpendicular of <math>F</math>, call it <math>H</math>, and note that <math>QP = 1 + \sqrt{3}</math> after using the trig functions for <math>75</math> degrees. | Use coordinates. Call <math>Q</math> the origin and <math>QP</math> be on the x-axis. It is easy to see that <math>F</math> is the vertex on <math>RP</math>. After labeling coordinates (noting additionally that <math>QBC</math> is an equilateral triangle), we see that the area is <math>QP</math> times <math>0.5</math> times the ordinate of <math>R</math>. Draw a perpendicular of <math>F</math>, call it <math>H</math>, and note that <math>QP = 1 + \sqrt{3}</math> after using the trig functions for <math>75</math> degrees. | ||
Now, get the lines for <math>QR</math> and <math>RP</math>: <math>y=\sqrt{3}x</math> and <math>y=-(2+\sqrt{3})x + (5+\sqrt{3})</math>, whereupon we get the ordinate of <math>R</math> to be <math>\frac{3+2\sqrt{3}}{2}</math>, and the area is <math>\frac{5\sqrt{3} + 9}{4}</math>, so our answer is <math>\boxed{021}</math>. | Now, get the lines for <math>QR</math> and <math>RP</math>: <math>y=\sqrt{3}x</math> and <math>y=-(2+\sqrt{3})x + (5+\sqrt{3})</math>, whereupon we get the ordinate of <math>R</math> to be <math>\frac{3+2\sqrt{3}}{2}</math>, and the area is <math>\frac{5\sqrt{3} + 9}{4}</math>, so our answer is <math>\boxed{021}</math>. | ||
− | == Solution | + | == Solution 3 (Trig) == |
Angle chasing yields that both triangles <math>PAF</math> and <math>PQR</math> are <math>75</math>-<math>60</math>-<math>45</math> triangles. First look at triangle <math>PAF</math>. Using Law of Sines, we find: | Angle chasing yields that both triangles <math>PAF</math> and <math>PQR</math> are <math>75</math>-<math>60</math>-<math>45</math> triangles. First look at triangle <math>PAF</math>. Using Law of Sines, we find: | ||
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Evaluating and reducing, we get <math>\frac{9 + 5\sqrt{3}}{4}, </math>thus the answer is <math> \boxed{021}</math> | Evaluating and reducing, we get <math>\frac{9 + 5\sqrt{3}}{4}, </math>thus the answer is <math> \boxed{021}</math> | ||
− | ==Solution | + | ==Solution 4 (Trig with Diagram)== |
[[File:2013_AIME_I_Problem_12.png]] | [[File:2013_AIME_I_Problem_12.png]] | ||
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With some simple angle chasing we can show that <math>\triangle OJL</math> and <math>\triangle MPL</math> are congruent. This means we have a large equilateral triangle with side length <math>3</math> and quadrilateral <math>OJQN</math>. We know that <math>[OJQN] = [\triangle NQL] - [\triangle OJL]</math>. Using Law of Sines and the fact that <math>\angle N = 45^{\circ}</math> we know that <math>\overline{NL} = \sqrt{6}</math> and the height to that side is <math>\frac{\sqrt{3} -1}{\sqrt{2}}</math> so <math>[\triangle NQL] = \frac{3-\sqrt{3}}{2}</math>. Using an extremely similar process we can show that <math>\overline{OJ} = 2-\sqrt{3}</math> which means the height to <math>\overline{LJ}</math> is <math>\frac{2\sqrt{3}-3}{2}</math>. So the area of <math>\triangle OJL = \frac{2\sqrt{3}-3}{4}</math>. This means the area of quadrilateral <math>OJQN = \frac{3-\sqrt{3}}{2} - \frac{2\sqrt{3}-3}{4} = \frac{9-4\sqrt{3}}{4}</math>. So the area of our larger triangle is <math>\frac{9-4\sqrt{3}}{4} + \frac{9\sqrt{3}}{4} = \frac{9+5\sqrt{3}}{4}</math>. Therefore <math>9+5+3+4=021</math>. | With some simple angle chasing we can show that <math>\triangle OJL</math> and <math>\triangle MPL</math> are congruent. This means we have a large equilateral triangle with side length <math>3</math> and quadrilateral <math>OJQN</math>. We know that <math>[OJQN] = [\triangle NQL] - [\triangle OJL]</math>. Using Law of Sines and the fact that <math>\angle N = 45^{\circ}</math> we know that <math>\overline{NL} = \sqrt{6}</math> and the height to that side is <math>\frac{\sqrt{3} -1}{\sqrt{2}}</math> so <math>[\triangle NQL] = \frac{3-\sqrt{3}}{2}</math>. Using an extremely similar process we can show that <math>\overline{OJ} = 2-\sqrt{3}</math> which means the height to <math>\overline{LJ}</math> is <math>\frac{2\sqrt{3}-3}{2}</math>. So the area of <math>\triangle OJL = \frac{2\sqrt{3}-3}{4}</math>. This means the area of quadrilateral <math>OJQN = \frac{3-\sqrt{3}}{2} - \frac{2\sqrt{3}-3}{4} = \frac{9-4\sqrt{3}}{4}</math>. So the area of our larger triangle is <math>\frac{9-4\sqrt{3}}{4} + \frac{9\sqrt{3}}{4} = \frac{9+5\sqrt{3}}{4}</math>. Therefore <math>9+5+3+4=021</math>. | ||
− | ==Solution | + | ==Solution 5 (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>. <math>9+5+3+4=021</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 | ||
− | ==Solution | + | ==Solution 6 (Combination of 1 & 2)== |
We can observe that <math>RD=DF</math> (because <math>\angle R</math> & <math>\angle RFD</math> are both <math>45^\circ</math>). Thus we know that <math>RD</math> is equivalent to the height of the hexagon, which is <math>\sqrt3</math>. Now we look at triangle <math>\triangle AFP</math> and apply the Law of Sines to it. <math>\frac{1}{\sin{75}}=\frac{AP}{\sin{45}}</math>. From here we can solve for <math>AP</math> and get that <math>AP=\sqrt{3}-1</math>. Now we use the Sine formula for the area of a triangle with sides <math>RQ</math>, <math>PQ</math>, and <math>\angle {RQP}</math> to get the answer. Setting <math>PQ=\sqrt{3}+1</math> and <math>QR=\sqrt{3}+2</math> we get the expression <math>\frac{(\sqrt{3}+1)(\sqrt{3}+2)(\frac{\sqrt{3}}{2})}{2}</math> which is <math>\frac{9 + 5\sqrt{3}}{4}</math>. Thus our final answer is <math>9+5+3+4=\fbox{021}</math>. | We can observe that <math>RD=DF</math> (because <math>\angle R</math> & <math>\angle RFD</math> are both <math>45^\circ</math>). Thus we know that <math>RD</math> is equivalent to the height of the hexagon, which is <math>\sqrt3</math>. Now we look at triangle <math>\triangle AFP</math> and apply the Law of Sines to it. <math>\frac{1}{\sin{75}}=\frac{AP}{\sin{45}}</math>. From here we can solve for <math>AP</math> and get that <math>AP=\sqrt{3}-1</math>. Now we use the Sine formula for the area of a triangle with sides <math>RQ</math>, <math>PQ</math>, and <math>\angle {RQP}</math> to get the answer. Setting <math>PQ=\sqrt{3}+1</math> and <math>QR=\sqrt{3}+2</math> we get the expression <math>\frac{(\sqrt{3}+1)(\sqrt{3}+2)(\frac{\sqrt{3}}{2})}{2}</math> which is <math>\frac{9 + 5\sqrt{3}}{4}</math>. Thus our final answer is <math>9+5+3+4=\fbox{021}</math>. | ||
By AwesomeLife_Math | By AwesomeLife_Math | ||
+ | |||
== See also == | == See also == | ||
{{AIME box|year=2013|n=I|num-b=11|num-a=13}} | {{AIME box|year=2013|n=I|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 20:11, 24 January 2021
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
.
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 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 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 (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 5 (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 6 (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.