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Difference between revisions of "Mock AIME II 2012 Problems/Problem 9"

(Created page with "==Problem== In <math>\triangle ABC</math>, <math>AB=12</math>, <math>AC=20</math>, and <math>\angle ABC=120^\circ</math>. <math>D, E,</math> and <math>F</math> lie on <math>\ove...")
 
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Start out by finding <math>BC</math>.  Remark that by the law of sines, <math>\frac{\sin(120^\circ)}{20}=\frac{\sin(\angle ACB)}{12}</math>.  Therefore <math>\sin(\angle ACB)=\frac{3}{5}*\frac{\sqrt{3}}{2}=\frac{3\sqrt3}{10}</math>.  We know <math>90^\circ>\angle ACB>0^\circ</math> because <math>\angle ABC>90^\circ</math> in <math>\triangle ABC</math>, therefore <math>\angle ACB</math> is in the first quadrant.  Use the Pythagorean Identity to give us <math>\cos^2(\angle ACB)+\sin^2(\angle ACB)=1\implies \cos(\angle ACB)=\frac{\sqrt{73}}{10}</math>.   
 
Start out by finding <math>BC</math>.  Remark that by the law of sines, <math>\frac{\sin(120^\circ)}{20}=\frac{\sin(\angle ACB)}{12}</math>.  Therefore <math>\sin(\angle ACB)=\frac{3}{5}*\frac{\sqrt{3}}{2}=\frac{3\sqrt3}{10}</math>.  We know <math>90^\circ>\angle ACB>0^\circ</math> because <math>\angle ABC>90^\circ</math> in <math>\triangle ABC</math>, therefore <math>\angle ACB</math> is in the first quadrant.  Use the Pythagorean Identity to give us <math>\cos^2(\angle ACB)+\sin^2(\angle ACB)=1\implies \cos(\angle ACB)=\frac{\sqrt{73}}{10}</math>.   
  
−
Now, note that <math>\angle BAC</math> is the same thing as <math>60^\circ-\angle ACB</math>, from <math>\triangle ABC</math>, therefore we have <math>\sin(\angle BAC)=\sin(60^\circ-\angle ACB)=\sin(60^\circ)\cos(\angle ACB)-\sin(\angle ACB)\cos(60^\circ)= \frac{\sqrt3*\sqrt{73}}{20}-\frac{3\sqrt3}{20}=\frac{\sqrt3*\sqrt{73}-3\sqrt3}{20}</math>.
+
Now, note that <math>\angle BAC</math> is the same thing as <math>60^\circ-\angle ACB</math>, from <math>\triangle ABC</math>, therefore we have <math>\sin(\angle BAC)=\sin(60^\circ-\angle ACB)=\sin(60^\circ)\cos(\angle ACB)-\sin(\angle ACB)\cos(60^\circ)</math>
 +
<math>\equal{} \frac{\sqrt3*\sqrt{73}}{20}-\frac{3\sqrt3}{20}=\frac{\sqrt3*\sqrt{73}-3\sqrt3}{20}</math>.
  
 
Next, use the law of sines to give us <math>\frac{\sin(120^\circ)}{20}=\frac{\sin(\angle BAC)}{BC}</math>.  This gives us <math>\frac{\sqrt3}{40}=\frac{\sqrt3*\sqrt{73}-3\sqrt3}{20BC}</math>.  This gives us <math>2\sqrt{73}-6=BC</math>.   
 
Next, use the law of sines to give us <math>\frac{\sin(120^\circ)}{20}=\frac{\sin(\angle BAC)}{BC}</math>.  This gives us <math>\frac{\sqrt3}{40}=\frac{\sqrt3*\sqrt{73}-3\sqrt3}{20BC}</math>.  This gives us <math>2\sqrt{73}-6=BC</math>.   

Revision as of 02:13, 5 April 2012

Problem

In $\triangle ABC$, $AB=12$, $AC=20$, and $\angle ABC=120^\circ$. $D, E,$ and $F$ lie on $\overline{AC}, \overline{AB}$, and $\overline{BC}$, respectively. If $AE=\frac{1}{4}AB, BF=\frac{1}{4}BC$, and $AD=\frac{1}{4}AC$, the area of $\triangle DEF$ can be expressed in the form $\frac{a\sqrt{b}-c\sqrt{d}}{e}$ where $a, b, c, d, e$ are all positive integers, and $b$ and $d$ do not have any perfect squares greater than $1$ as divisors. Find $a+b+c+d+e$.

Solution

Here is a diagram (note that D should be on AC and F should be on BC): [asy] /* File unicodetex not found. */ /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -24, xmax = 24, ymin = -15, ymax = 22; /* image dimensions */ pen qqwuqq = rgb(0,0,0); draw(arc((12,0),2,60,180)--(12,0)--cycle, qqwuqq); /* draw figures */ draw((0,0)--(12,0)); draw((0,0)--(18,10)); draw((18,10)--(12,0)); label("12",(6,-1),SE*labelscalefactor); label("20",(8,7),SE*labelscalefactor); draw((5,3)--(14,3)); draw((5,3)--(3,0)); draw((3,0)--(14,3)); /* dots and labels */ dot((0,0),dotstyle); label("$A$", (0,0), NE * labelscalefactor); dot((12,0),dotstyle); label("$B$", (12,0), NE * labelscalefactor); dot((18,10),dotstyle); label("$C$", (18,10), NE * labelscalefactor); label("$120^\circ$", (11,1), NE * labelscalefactor,qqwuqq); dot((3,0),dotstyle); label("$E$", (3,0), NE * labelscalefactor); dot((14,3),dotstyle); label("$F$", (14,3), NE * labelscalefactor); dot((5,3),dotstyle); label("$D$", (5,3), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy] . Start out by finding $BC$. Remark that by the law of sines, $\frac{\sin(120^\circ)}{20}=\frac{\sin(\angle ACB)}{12}$. Therefore $\sin(\angle ACB)=\frac{3}{5}*\frac{\sqrt{3}}{2}=\frac{3\sqrt3}{10}$. We know $90^\circ>\angle ACB>0^\circ$ because $\angle ABC>90^\circ$ in $\triangle ABC$, therefore $\angle ACB$ is in the first quadrant. Use the Pythagorean Identity to give us $\cos^2(\angle ACB)+\sin^2(\angle ACB)=1\implies \cos(\angle ACB)=\frac{\sqrt{73}}{10}$.

Now, note that $\angle BAC$ is the same thing as $60^\circ-\angle ACB$, from $\triangle ABC$, therefore we have $\sin(\angle BAC)=\sin(60^\circ-\angle ACB)=\sin(60^\circ)\cos(\angle ACB)-\sin(\angle ACB)\cos(60^\circ)$ $\equal{} \frac{\sqrt3*\sqrt{73}}{20}-\frac{3\sqrt3}{20}=\frac{\sqrt3*\sqrt{73}-3\sqrt3}{20}$ (Error compiling LaTeX. Unknown error_msg).

Next, use the law of sines to give us $\frac{\sin(120^\circ)}{20}=\frac{\sin(\angle BAC)}{BC}$. This gives us $\frac{\sqrt3}{40}=\frac{\sqrt3*\sqrt{73}-3\sqrt3}{20BC}$. This gives us $2\sqrt{73}-6=BC$.

Now, we use coordinates to find that the coordinates of $C$, $E$, $F$, and $D$. $C$ is going to be the point $(9+\sqrt{73}, \sqrt{219}-3\sqrt3)$ from adding point $H$ to create $30^\circ, 60^\circ, 90^\circ$ triangle $BCH$ as shown in this diagram (where $H$ is the right angle, $B$ is the $30^\circ$ angle, and $C$ is the $60^\circ$ angle):

[asy] import graph; usepackage("amsmath"); size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -24, xmax = 24, ymin = -15, ymax = 22; /* image dimensions */ pen qqwuqq = rgb(0,0,0); draw(arc((12,0),2,60,180)--(12,0)--cycle, qqwuqq); /* draw figures */ draw((0,0)--(12,0)); draw((0,0)--(18,10)); draw((18,10)--(12,0)); label("12",(6,-1),SE*labelscalefactor); label("$ 2\sqrt{76}-6 $",(14,5),SE*labelscalefactor); label("20",(8,7),SE*labelscalefactor); draw((12,0)--(12,10)); draw((12,10)--(18,10)); /* dots and labels */ dot((0,0),dotstyle); label("$A$", (0,0), NE * labelscalefactor); dot((12,0),dotstyle); label("$B$", (12,0), NE * labelscalefactor); dot((18,10),dotstyle); label("$C$", (18,10), NE * labelscalefactor); label("$120^\circ$", (11,1), NE * labelscalefactor,qqwuqq); dot((12,10),dotstyle); label("$H$", (12,10), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]

Now, note that we have to find the coordinates of $E,F$ and $D$. Assume WLOG that $A$ is $(0,0)$ and $B$ be $(12,0)$. $E$ is obviously $(3,0)$, $F$ is going to be $(12+\frac{9+\sqrt{73}-12}{4}, \frac{\sqrt{219}-3\sqrt3}{4})$ and the coordinates of $D$ is going to be $(\frac{9+\sqrt{73}}{4}, \frac{\sqrt{219}-3\sqrt3}{4})$. Since $D$ and $F$ have the same $y$ value, we can find $[EDF]$ by using $\frac12*b*h$. The base is going to be equal to the distance from $D$ to $F$, which is the same thing as $12+\frac{9+\sqrt{73}-12}{4}-\frac{9+\sqrt{73}}{4}=9=b$, and the height is the change in the $y$ coordinate from $E$ to $D$. This is the same thing as $\frac{\sqrt{219}-3\sqrt3}{4}-0=\frac{\sqrt{219}-3\sqrt3}{4}=h$. Hence, plugging these into $[EDF]=\frac12*b*h$ give us $\frac12*9*\frac{\sqrt{219}-3\sqrt3}{4}=\frac{9\sqrt{219}-27\sqrt3}{8}$. The answer is thus $9+219+27+3+8=\boxed{266}$.

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