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Difference between revisions of "2013 AIME II Problems/Problem 13"
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In <math>\triangle ABC</math>, <math>AC = BC</math>, and point <math>D</math> is on <math>\overline{BC}</math> so that <math>CD = 3\cdot BD</math>. Let <math>E</math> be the midpoint of <math>\overline{AD}</math>. Given that <math>CE = \sqrt{7}</math> and <math>BE = 3</math>, the area of <math>\triangle ABC</math> can be expressed in the form <math>m\sqrt{n}</math>, where <math>m</math> and <math>n</math> are positive integers and <math>n</math> is not divisible by the square of any prime. Find <math>m+n</math>.
 
In <math>\triangle ABC</math>, <math>AC = BC</math>, and point <math>D</math> is on <math>\overline{BC}</math> so that <math>CD = 3\cdot BD</math>. Let <math>E</math> be the midpoint of <math>\overline{AD}</math>. Given that <math>CE = \sqrt{7}</math> and <math>BE = 3</math>, the area of <math>\triangle ABC</math> can be expressed in the form <math>m\sqrt{n}</math>, where <math>m</math> and <math>n</math> are positive integers and <math>n</math> is not divisible by the square of any prime. Find <math>m+n</math>.
 
==Solution==
 
==Solution==
−
After drawing the figure, we suppose <math>BD=a</math>, so that<math>CD=3a</math>,<math>AC=4a</math>, and <math>AE=ED=b</math>.
+
 
 +
=== Solution 1 ===
 +
After drawing the figure, we suppose <math>BD=a</math>, so that <math>CD=3a</math>, <math>AC=4a</math>, and <math>AE=ED=b</math>.
  
 
Using cosine law for <math>\triangle AEC</math> and <math>\triangle CED</math>,we get
 
Using cosine law for <math>\triangle AEC</math> and <math>\triangle CED</math>,we get
  
−
<math>b^2+7-2\sqrt{7}\cdot cos(\angle CED)=9a^2</math> ... <math>(1)</math>
+
<cmath>b^2+7-2\sqrt{7}\cdot \cos(\angle CED)=9a^2\qquad (1)</cmath>
 +
<cmath>b^2+7+2\sqrt{7}\cdot \cos(\angle CED)=16a^2\qquad (2)</cmath>
 +
So, <math>(1)+(2)</math>, we get<cmath>2b^2+14=25a^2. \qquad (3)</cmath>
  
−
<math>b^2+7+2\sqrt{7}\cdot cos(\angle CED)=16a^2</math> ...<math>(2)</math>
+
Using cosine law in <math>\triangle ACD</math>, we get
  
−
So, <math>(1)+(2)</math>, we get<math>2b^2+14=25a^2</math>...<math>(3)</math>
+
<cmath>b^2+9a^2-2\cdot 2b\cdot 3a\cdot \cos(\angle ADC)=16a^2</cmath>
  
−
Using cosine law in <math>\triangle ACD</math>,we get
+
So, <cmath>\cos(\angle ADC)=\frac{7a^2-4b^2}{12ab}.\qquad (4)</cmath>
  
−
<math>4b^2+9a^2-2\cdot 2b\cdot 3a\cdot cos(\angle ADC)=16a^2</math>
+
Using cosine law in <math>\triangle EDC</math> and <math>\triangle EDB</math>, we get
  
−
So, <math>cos(\angle ADC)=\frac{7a^2-4b^2}{12ab}</math>...<math>(4)</math>
+
<cmath>b^2+9a^2-2\cdot 3a\cdot b\cdot \cos(\angle ADC)=7\qquad (5)</cmath>
  
−
Using cosine law in <math>\triangle EDC</math> and <math>\triangle EDB</math>, we get
+
<cmath>b^2+a^2+2\cdot a\cdot b\cdot \cos(\angle ADC)=9.\qquad (6)</cmath>
  
−
<math>b^2+9a^2-2\cdot 3a\cdot b\cdot cos(\angle ADC)=7</math>...<math>(5)</math>
+
<math>(5)+(6)</math>, and according to <math>(4)</math>, we can get
 +
<cmath>37a^2+2b^2=48. \qquad (7)</cmath>
  
−
<math>b^2+a^2+2\cdot a\cdot b\cdot cos(\angle ADC)=9</math>...<math>(6)</math>
+
Using <math>(3)</math> and <math>(7)</math>, we can solve <math>a=1</math> and <math>b=\frac{\sqrt{22}}{2}</math>.
  
−
<math>(5)+(6)</math>, and according to <math>(4)</math>, we can get <math>37a^2+2b^2=48</math>...<math>(7)</math>
+
Finally, we use cosine law for <math>\triangle ADB</math>,  
  
−
Using <math>(3)</math> and <math>(7)</math>, we can solve <math>a=1</math> and <math>b=\frac{\sqrt{22}}{2}</math>
+
<cmath>4(\frac{\sqrt{22}}{2})^2+1+2\cdot\2(\frac{\sqrt{22}}{2})\cdot \cos(ADC)=AB^2</cmath>
  
−
Finally, we use cosine law for  <math>\triangle ADB</math>,
+
then <math>AB=2\sqrt{7}</math>, so the height of this <math>\triangle ABC</math> is <math>\sqrt{4^2-(\sqrt{7})^2}=3</math>.
  
−
<math>4(\frac{\sqrt{22}}{2})^2+1+2\cdot\2(\frac{\sqrt{22}}{2})\cdot cos(ADC)=AB^2</math>
+
Then the area of <math>\triangle ABC</math> is <math>3\sqrt{7}</math>, so the answer is <math>\boxed{010}</math>.
  
−
then <math>AB=2\sqrt{7}</math>
+
=== Solution 2 ===
 +
Let <math>X</math> be the foot of the altitude from <math>C</math> with other points labelled as shown below.
 +
<asy>
 +
size(200);
 +
pair A=(0,0),B=(2*sqrt(7),0),C=(sqrt(7),3),D=(3*B+C)/4,L=C/5,M=3*B/7;
 +
draw(A--B--C--cycle);draw(A--D^^B--L^^C--M);
 +
label("$A$",A,SW);label("$B$",B,SE);label("$C$",C,N);label("$D$",D,NE);label("$L$",L,NW);label("$M$",M,S);
 +
pair X=foot(C,A,B), Y=foot(L,A,B);
 +
pair EE=D/2;
 +
label("$X$",X,S);label("$E$",EE,NW);label("$Y$",Y,S);
 +
draw(C--X^^L--Y,dotted);
 +
draw(rightanglemark(B,X,C)^^rightanglemark(B,Y,L));
 +
</asy>
 +
Now we proceed using [[mass points]]. To balance along the segment <math>BC</math>, we assign <math>B</math> a mass of <math>3</math> and <math>C</math> a mass of <math>1</math>. Therefore, <math>D</math> has a mass of <math>4</math>.  As <math>E</math> is the midpoint of <math>AD</math>, we must assign <math>A</math> a mass of <math>4</math> as well. This gives <math>L</math> a mass of <math>5</math> and <math>M</math> a mass of <math>7</math>.
  
−
so the height of this <math>\triangle ABC</math> is <math>\sqrt{4^2-(\sqrt{7})^2}=3</math>
+
Now let <math>AB=b</math> be the base of the triangle, and let <math>CX=h</math> be the height. Then as <math>AM:MB=3:4</math>, and as <math>AX=\frac{b}{2}</math>, we know that <cmath>MX=\frac{b}{2}-\frac{3b}{7}=\frac{b}{14}.</cmath> Also, as <math>CE:EM=7:1</math>, we know that <math>EM=\frac{1}{\sqrt{7}}</math>. Therefore, by the Pythagorean Theorem on <math>\triangle {XCM}</math>, we know that <cmath>\frac{b^2}{196}+h^2=\left(\sqrt{7}+\frac{1}{\sqrt{7}}\right)^2=\frac{64}{7}.</cmath>
  
−
Then the area of <math>\triangle ABC</math> is <math>3\sqrt{7}</math>, so the answer is <math>\boxed{010}</math>.
+
Also, as <math>LE:BE=5:3</math>, we know that <math>BL=\frac{8}{5}\cdot 3=\frac{24}{5}</math>. Furthermore, as <math>\triangle YLA\sim \triangle XCA</math>, and as <math>AL:LC=1:4</math>, we know that <math>LY=\frac{h}{5}</math> and <math>AY=\frac{b}{10}</math>, so <math>YB=\frac{9b}{10}</math>. Therefore, by the Pythagorean Theorem on <math>\triangle BLY</math>, we get <cmath>\frac{81b^2}{100}+\frac{h^2}{25}=\frac{576}{25}.</cmath>
 +
Solving this system of equations yields <math>b=2\sqrt{7}</math> and <math>h=3</math>. Therefore, the area of the triangle is <math>3\sqrt{7}</math>, giving us an answer of <math>\boxed{010}</math>.
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2013|n=II|num-b=12|num-a=14}}
 
{{AIME box|year=2013|n=II|num-b=12|num-a=14}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 04:16, 22 July 2013

Problem 13

In $\triangle ABC$, $AC = BC$, and point $D$ is on $\overline{BC}$ so that $CD = 3\cdot BD$. Let $E$ be the midpoint of $\overline{AD}$. Given that $CE = \sqrt{7}$ and $BE = 3$, the area of $\triangle ABC$ can be expressed in the form $m\sqrt{n}$, where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $m+n$.

Solution

Solution 1

After drawing the figure, we suppose $BD=a$, so that $CD=3a$, $AC=4a$, and $AE=ED=b$.

Using cosine law for $\triangle AEC$ and $\triangle CED$,we get

\[b^2+7-2\sqrt{7}\cdot \cos(\angle CED)=9a^2\qquad (1)\] \[b^2+7+2\sqrt{7}\cdot \cos(\angle CED)=16a^2\qquad (2)\] So, $(1)+(2)$, we get\[2b^2+14=25a^2. \qquad (3)\]

Using cosine law in $\triangle ACD$, we get

\[b^2+9a^2-2\cdot 2b\cdot 3a\cdot \cos(\angle ADC)=16a^2\]

So, \[\cos(\angle ADC)=\frac{7a^2-4b^2}{12ab}.\qquad (4)\]

Using cosine law in $\triangle EDC$ and $\triangle EDB$, we get

\[b^2+9a^2-2\cdot 3a\cdot b\cdot \cos(\angle ADC)=7\qquad (5)\]

\[b^2+a^2+2\cdot a\cdot b\cdot \cos(\angle ADC)=9.\qquad (6)\]

$(5)+(6)$, and according to $(4)$, we can get \[37a^2+2b^2=48. \qquad (7)\]

Using $(3)$ and $(7)$, we can solve $a=1$ and $b=\frac{\sqrt{22}}{2}$.

Finally, we use cosine law for $\triangle ADB$,

\[4(\frac{\sqrt{22}}{2})^2+1+2\cdot\2(\frac{\sqrt{22}}{2})\cdot \cos(ADC)=AB^2\]

then $AB=2\sqrt{7}$, so the height of this $\triangle ABC$ is $\sqrt{4^2-(\sqrt{7})^2}=3$.

Then the area of $\triangle ABC$ is $3\sqrt{7}$, so the answer is $\boxed{010}$.

Solution 2

Let $X$ be the foot of the altitude from $C$ with other points labelled as shown below. [asy] size(200); pair A=(0,0),B=(2*sqrt(7),0),C=(sqrt(7),3),D=(3*B+C)/4,L=C/5,M=3*B/7; draw(A--B--C--cycle);draw(A--D^^B--L^^C--M); label("$A$",A,SW);label("$B$",B,SE);label("$C$",C,N);label("$D$",D,NE);label("$L$",L,NW);label("$M$",M,S); pair X=foot(C,A,B), Y=foot(L,A,B); pair EE=D/2; label("$X$",X,S);label("$E$",EE,NW);label("$Y$",Y,S); draw(C--X^^L--Y,dotted); draw(rightanglemark(B,X,C)^^rightanglemark(B,Y,L)); [/asy] Now we proceed using mass points. To balance along the segment $BC$, we assign $B$ a mass of $3$ and $C$ a mass of $1$. Therefore, $D$ has a mass of $4$. As $E$ is the midpoint of $AD$, we must assign $A$ a mass of $4$ as well. This gives $L$ a mass of $5$ and $M$ a mass of $7$.

Now let $AB=b$ be the base of the triangle, and let $CX=h$ be the height. Then as $AM:MB=3:4$, and as $AX=\frac{b}{2}$, we know that \[MX=\frac{b}{2}-\frac{3b}{7}=\frac{b}{14}.\] Also, as $CE:EM=7:1$, we know that $EM=\frac{1}{\sqrt{7}}$. Therefore, by the Pythagorean Theorem on $\triangle {XCM}$, we know that \[\frac{b^2}{196}+h^2=\left(\sqrt{7}+\frac{1}{\sqrt{7}}\right)^2=\frac{64}{7}.\]

Also, as $LE:BE=5:3$, we know that $BL=\frac{8}{5}\cdot 3=\frac{24}{5}$. Furthermore, as $\triangle YLA\sim \triangle XCA$, and as $AL:LC=1:4$, we know that $LY=\frac{h}{5}$ and $AY=\frac{b}{10}$, so $YB=\frac{9b}{10}$. Therefore, by the Pythagorean Theorem on $\triangle BLY$, we get \[\frac{81b^2}{100}+\frac{h^2}{25}=\frac{576}{25}.\] Solving this system of equations yields $b=2\sqrt{7}$ and $h=3$. Therefore, the area of the triangle is $3\sqrt{7}$, giving us an answer of $\boxed{010}$.

See Also

2013 AIME II (Problems • Answer Key • Resources)
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Problem 12 		Followed by
Problem 14
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