Difference between revisions of "2013 AIME I Problems/Problem 12"

Latest revision as of 17:32, 22 November 2024

Problem

Let $\bigtriangleup PQR$ be a triangle with $\angle P = 75^\circ$ and $\angle Q = 60^\circ$ . A regular hexagon $ABCDEF$ with side length 1 is drawn inside $\triangle PQR$ so that side $\overline{AB}$ lies on $\overline{PQ}$ , side $\overline{CD}$ lies on $\overline{QR}$ , and one of the remaining vertices lies on $\overline{RP}$ . There are positive integers $a, b, c,$ and $d$ such that the area of $\triangle PQR$ can be expressed in the form $\frac{a+b\sqrt{c}}{d}$ , where $a$ and $d$ are relatively prime, and c is not divisible by the square of any prime. Find $a+b+c+d$ .

Solution 1

First, find that $\angle R = 45^\circ$ . Draw $ABCDEF$ . Now draw $\bigtriangleup PQR$ around $ABCDEF$ such that $Q$ is adjacent to $C$ and $D$ . The height of $ABCDEF$ is $\sqrt{3}$ , so the length of base $QR$ is $2+\sqrt{3}$ . Let the equation of $\overline{RP}$ be $y = x$ . Then, the equation of $\overline{PQ}$ is $y = -\sqrt{3} (x - (2+\sqrt{3})) \to y = -x\sqrt{3} + 2\sqrt{3} + 3$ . Solving the two equations gives $y = x = \frac{\sqrt{3} + 3}{2}$ . The area of $\bigtriangleup PQR$ is $\frac{1}{2} * (2 + \sqrt{3}) * \frac{\sqrt{3} + 3}{2} = \frac{5\sqrt{3} + 9}{4}$ . $a + b + c + d = 9 + 5 + 3 + 4 = \boxed{021}$

Note (different ending): When you have the length of the base $QR$ , you don't need to find the equations of the lines $QP$ and $PR$ . Instead, make an altitude from $P$ to $QR$ , and call the foot $M$ . $QPM$ is a $45, 45, 90$ triangle and $PMR$ is a $30, 60, 90$ triangle. And, they both share $PM$ . So, we can set $RM$ as $x$ , so $PM$ is $x\sqrt{3}$ . Since $QPM$ is a $45, 45, 90$ triangle, $PM=MQ=x\sqrt{3}$ . The base $QR$ can be written as $QM+MR=x+x\sqrt{3}=2+\sqrt{3}$ . Solve this equation and $x=\frac{\sqrt{3}+1}{2}$ and $PM=x\sqrt{3}=\frac{\sqrt{3}+3}{2}$ . Multiply this by base $QR$ and divide by $2$ to get the area of triangle $PQR$ which is $\frac{9+5\sqrt{3}}{4}$ . So, $a + b + c + d = 9 + 5 + 3 + 4 = \boxed{021}$ -hwan

Solution 2 (Cartesian Variation)

Use coordinates. Call $Q$ the origin and $QP$ be on the x-axis. It is easy to see that $F$ is the vertex on $RP$ . After labeling coordinates (noting additionally that $QBC$ is an equilateral triangle), we see that the area is $QP$ times $0.5$ times the coordinate of $R$ . Draw a perpendicular of $F$ , call it $H$ , and note that $QP = 1 + \sqrt{3}$ after using the trig functions for $75$ degrees.

Now, get the lines for $QR$ and $RP$ : $y=\sqrt{3}x$ and $y=-(2+\sqrt{3})x + (5+\sqrt{3})$ , whereupon we get the ordinate of $R$ to be $\frac{3+2\sqrt{3}}{2}$ , and the area is $\frac{5\sqrt{3} + 9}{4}$ , so our answer is $\boxed{021}$ .

Solution 3 (Trig)

Angle chasing yields that both triangles $PAF$ and $PQR$ are $75$ - $60$ - $45$ triangles. First look at triangle $PAF$ . Using Law of Sines, we find:

$\frac{\frac{\sqrt{6} + \sqrt{2}}{4}}{1} = \frac{\frac{\sqrt{2}}{2}}{PA}$

Simplifying, we find $PA = \sqrt{3} - 1$ . Since $\angle{Q} = 60^\circ$ , WLOG assume triangle $BQC$ is equilateral, so $BQ = 1$ . So $PQ = \sqrt{3} + 1$ .

Apply Law of Sines again,

$\frac{\frac{\sqrt{2}}{2}}{\sqrt{3} + 1} = \frac{\frac{\sqrt{3}}{2}}{PR}$

Simplifying, we find $PR = \frac{\sqrt{6}}{2} \cdot (1 + \sqrt{3})$ .

$[PQR] = \frac{1}{2} \cdot PQ \cdot PR \cdot \sin 75^\circ$ .

Evaluating and reducing, we get $\frac{9 + 5\sqrt{3}}{4},$ thus the answer is $\boxed{021}$

Solution 4 (Special Triangles)

As we can see, the $75^\circ$ angle of $\angle P$ can be split into a $45^\circ$ angle and a $30^\circ$ angle. This allows us to drop an altitude from point $P$ for $\triangle RPQ$ which intersects $\overline{AF}$ at point $a$ and $\overline{RQ}$ at point $b$ . The main idea of our solution is to obtain enough sides of $\triangle RPQ$ that we are able to directly figure out its area (specifically by figuring out side $\overline{RQ}$ and $\overline{Pb}$ ).

We first begin by figuring out the length of $\overline{PQ}$ . This can be easily done, since $\overline{AB}$ is simply $1$ (given in the problem) and $\overline{BQ}=1$ because $\triangle BCQ$ is an equilateral after some simple angle calculations. Now we need to find $\overline{PA}$ . This is when we bring in some simple algebra.

PREPARATION: $\overline{aF}=\overline{Pa}$ (45-45-90 Right Triangle)

$\overline{Pa}=\sqrt{3}\overline{Aa}$ (30-60-90 Right Triangle)

$\overline{PA}=2\overline{Aa}$

$\overline{Aa}+\overline{aF}=1$

SOLVING: $\overline{Aa}+\sqrt{3}\overline{aF}=1$

so $\overline{Aa}(\sqrt{3}+1)=1$

$\overline{Aa}=\frac{1}{\sqrt{3}+1}=\frac{\sqrt{3}-1}{2}$

Finally, $\overline{PA}=2\cdot\frac{\sqrt{3}-1}{2}=\sqrt{3}-1$

Now, we can finally get the length of $\overline{PQ}$ by adding up $\overline{PA}+\overline{AB}+\overline{BQ}$ , which is simply $(\sqrt{3}-1)+(1)+(1)=\sqrt{3}+1$

To get $\overline{RQ}$ and $\overline{Pb}$ , we first work bit by bit.

$\overline{Qb}=\frac{\overline{PQ}}{2}=\frac{\sqrt{3}+1}{2}$ (30-60-90 Right Triangle)

$\overline{Pb}=\sqrt{3}\overline{Qb}=\frac{\sqrt{3}+3}{2}$ (same 30-60-90 Right Triangle)

Since $\overline{Pb}=\overline{Rb}$ because of 45-45-90 right triangles,

$\overline{Rb}=\frac{\sqrt{3}+3}{2}$ too.

Now, we can finally calculate $\overline{RQ}$ , and it is $\overline{Rb}+\overline{Qb}=\frac{\sqrt{3}+3}{2}+\frac{\sqrt{3}+1}{2}=\sqrt{3}+2$ .

Finally, the area of $\triangle PRQ$ can be calculated by $\frac{1}{2}\cdot\overline{RQ}\cdot\overline{Pb}$ , which is equal to $[\triangle PRQ]=\frac{1}{2} \cdot (\sqrt{3}+2) \cdot \frac{\sqrt{3}+3}{2} =\frac{9+5\sqrt{3}}{4}$ . So the final answer is $9+5+3+4=\fbox{021}$ .

-by What do Humanitarians Eat?

Solution 5 (Trig)

With some simple angle chasing we can show that $\triangle OJL$ and $\triangle MPL$ are congruent. This means we have a large equilateral triangle with side length $3$ and quadrilateral $OJQN$ . We know that $[OJQN] = [\triangle NQL] - [\triangle OJL]$ . Using Law of Sines and the fact that $\angle N = 45^{\circ}$ we know that $\overline{NL} = \sqrt{6}$ and the height to that side is $\frac{\sqrt{3} -1}{\sqrt{2}}$ so $[\triangle NQL] = \frac{3-\sqrt{3}}{2}$ . Using an extremely similar process we can show that $\overline{OJ} = 2-\sqrt{3}$ which means the height to $\overline{LJ}$ is $\frac{2\sqrt{3}-3}{2}$ . So the area of $\triangle OJL = \frac{2\sqrt{3}-3}{4}$ . This means the area of quadrilateral $OJQN = \frac{3-\sqrt{3}}{2} - \frac{2\sqrt{3}-3}{4} = \frac{9-4\sqrt{3}}{4}$ . So the area of our larger triangle is $\frac{9-4\sqrt{3}}{4} + \frac{9\sqrt{3}}{4} = \frac{9+5\sqrt{3}}{4}$ . Therefore $9+5+3+4=\fbox{021}$ .

Solution 6 (Elementary Geo)

We can find that $AF || CD || QR$ . This means that the perpendicular from $P$ to $QR$ is perpendicular to $AF$ as well, so let that perpendicular intersect $AF$ at $G$ , and the perpendicular intersect $QR$ at $H$ . Set $AP=x$ . Note that $\angle {PAG} = 60^\circ$ , so $AG=\frac{x}{2}$ and $PG = GF = \frac{x\sqrt3}{2}$ . Also, $1=AF=AG+GF=\frac{x}{2} + \frac{x\sqrt{3}}{2}$ , so $x=\sqrt{3} - 1$ . It's easy to calculate the area now, because the perpendicular from $P$ to $QR$ splits $\triangle{PQR}$ into a $30-60-90$ (PHQ) and a $45-45-90$ (PHR). From these triangles' ratios, it should follow that $QH=\frac{\sqrt{3} + 1}{2}, PH=HR=\frac{\sqrt{3}+3}{2}$ , so the area is $\frac{1}{2} * PH * QR = \frac{1}{2} * PH * (QH + HR) = \frac{1}{2} * \frac{\sqrt{3} + 3}{2} * \frac{2\sqrt{3}+4}{2} = \boxed{\frac{9+5\sqrt{3}}{4}}$ . $9+5+3+4=021$ . By Mathscienceclass

Solution 7 (Combination of 1 & 2)

We can observe that $RD=DF$ (because $\angle R$ & $\angle RFD$ are both $45^\circ$ ). Thus we know that $RD$ is equivalent to the height of the hexagon, which is $\sqrt3$ . Now we look at triangle $\triangle AFP$ and apply the Law of Sines to it. $\frac{1}{\sin{75}}=\frac{AP}{\sin{45}}$ . From here we can solve for $AP$ and get that $AP=\sqrt{3}-1$ . Now we use the Sine formula for the area of a triangle with sides $RQ$ , $PQ$ , and $\angle {RQP}$ to get the answer. Setting $PQ=\sqrt{3}+1$ and $QR=\sqrt{3}+2$ we get the expression $\frac{(\sqrt{3}+1)(\sqrt{3}+2)(\frac{\sqrt{3}}{2})}{2}$ which is $\frac{9 + 5\sqrt{3}}{4}$ . Thus our final answer is $9+5+3+4=\fbox{021}$ . By AwesomeLife_Math

Solution 8 (Area Ratios + Trig)

Note that $\triangle PAF \sim \triangle PQR$ . Let $X$ be the foot of an altitude dropped from $F$ to $PQ$ . Using trigonometry, we find that $PA = \sqrt3 - 1$ and $FX = \frac{\sqrt3}{2}$ , thus the area of $\triangle PAF$ is $\frac{3 - \sqrt3}{4}$ . Since $QA$ is clearly $2$ , the scale factor between $\triangle PAF$ and $\triangle PQR$ is $\frac{\sqrt3 + 1}{\sqrt3 - 1}$ , and thus $[PQR] = \frac{3 - \sqrt3}{4} \cdot \left( \frac{\sqrt3 + 1}{\sqrt3 - 1} \right)^2 = \frac{9 + 5\sqrt3}{4}$ , and we extract $\boxed{021}$ . ~ostriches88

@@ Line 1: / Line 1: @@
-== Problem 12 ==
+== 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>.
@@ Line 6: / Line 6: @@
 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==
-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.
+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)==
+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 coordinate 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>.
-== Solution 2 (Trig) ==
+== 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:
@@ Line 26: / Line 31: @@
 Simplifying, we find <math>PR = \frac{\sqrt{6}}{2} \cdot (1 + \sqrt{3})</math>.
-<math>[PQR] = \frac{1}{2} \cdot PQ \cdot PR \cdot sin 75^\circ</math>.
+<math>[PQR] = \frac{1}{2} \cdot PQ \cdot PR \cdot \sin 75^\circ</math>.
 Evaluating and reducing, we get <math>\frac{9 + 5\sqrt{3}}{4}, </math>thus the answer is <math> \boxed{021}</math>
-==Solution 3(Trig with Diagram)==
+==Solution 4 (Special Triangles)==
+[[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>).
+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.
+PREPARATION:
+<math>\overline{aF}=\overline{Pa}</math> (45-45-90 Right Triangle)
+<math>\overline{Pa}=\sqrt{3}\overline{Aa}</math> (30-60-90 Right Triangle)
+<math>\overline{PA}=2\overline{Aa}</math>
+<math>\overline{Aa}+\overline{aF}=1</math>
+SOLVING:
+<math>\overline{Aa}+\sqrt{3}\overline{aF}=1</math>
+so <math>\overline{Aa}(\sqrt{3}+1)=1</math>
+<math>\overline{Aa}=\frac{1}{\sqrt{3}+1}=\frac{\sqrt{3}-1}{2}</math>
+Finally, <math>\overline{PA}=2\cdot\frac{\sqrt{3}-1}{2}=\sqrt{3}-1</math>
+Now, we can finally get the length of <math>\overline{PQ}</math> by adding up <math>\overline{PA}+\overline{AB}+\overline{BQ}</math>, which is simply <math>(\sqrt{3}-1)+(1)+(1)=\sqrt{3}+1</math>
+To get <math>\overline{RQ}</math> and <math>\overline{Pb}</math>, we first work bit by bit.
+<math>\overline{Qb}=\frac{\overline{PQ}}{2}=\frac{\sqrt{3}+1}{2}</math> (30-60-90 Right Triangle)
+<math>\overline{Pb}=\sqrt{3}\overline{Qb}=\frac{\sqrt{3}+3}{2}</math> (same 30-60-90 Right Triangle)
+Since <math>\overline{Pb}=\overline{Rb}</math> because of 45-45-90 right triangles,
+<math>\overline{Rb}=\frac{\sqrt{3}+3}{2}</math> too.
+Now, we can finally calculate <math>\overline{RQ}</math>, and it is <math>\overline{Rb}+\overline{Qb}=\frac{\sqrt{3}+3}{2}+\frac{\sqrt{3}+1}{2}=\sqrt{3}+2</math>.
+Finally, the area of <math>\triangle PRQ</math> can be calculated by <math>\frac{1}{2}\cdot\overline{RQ}\cdot\overline{Pb}</math>, which is equal to <math>[\triangle PRQ]=\frac{1}{2} \cdot (\sqrt{3}+2) \cdot \frac{\sqrt{3}+3}{2} =\frac{9+5\sqrt{3}}{4}</math>. So the final answer is <math>9+5+3+4=\fbox{021}</math>.
+-by What do Humanitarians Eat?
+==Solution 5 (Trig)==
 [[File:2013_AIME_I_Problem_12.png]]
-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 = \boxed{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=\fbox{021}</math>.
+==Solution 6 (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>.
+By Mathscienceclass
+==Solution 7 (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>.
+By AwesomeLife_Math
+==Solution 8 (Area Ratios + Trig)==
+Note that <math>\triangle PAF \sim \triangle PQR</math>. Let <math>X</math> be the foot of an altitude dropped from <math>F</math> to <math>PQ</math>. Using trigonometry, we find that <math>PA = \sqrt3 - 1</math> and <math>FX = \frac{\sqrt3}{2}</math>, thus the area of <math>\triangle PAF</math> is <math>\frac{3 - \sqrt3}{4}</math>. Since <math>QA</math> is clearly <math>2</math>, the scale factor between <math>\triangle PAF</math> and <math>\triangle PQR</math> is <math>\frac{\sqrt3 + 1}{\sqrt3 - 1}</math>, and thus <math>[PQR] = \frac{3 - \sqrt3}{4} \cdot \left( \frac{\sqrt3 + 1}{\sqrt3 - 1} \right)^2 = \frac{9 + 5\sqrt3}{4}</math>, and we extract <math>\boxed{021}</math>.
+~ostriches88
 == See also ==
 {{AIME box|year=2013|n=I|num-b=11|num-a=13}}
 {{MAA Notice}}

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