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

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(additional solution)
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[[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 = \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=021</math>.
  
 
==Solution 4 (Elementary Geo)==
 
==Solution 4 (Elementary Geo)==
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By Mathscienceclass
 
By Mathscienceclass
  
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==Solution 5 (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
 
== 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 19:33, 15 March 2020

Problem 12

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 $\frac{1}{2} * (2 + \sqrt{3}) * \frac{\sqrt{3} + 3}{2} = \frac{5\sqrt{3} + 9}{4}$. $a + b + c + d = 9 + 5 + 3 + 4 = \boxed{021}$

Cartesian Variation Solution

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 ordinate 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 2 (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 3(Trig with Diagram)

2013 AIME I Problem 12.png

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 $9+5+3+4=021$.

Solution 4 (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 5 (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

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

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