Difference between revisions of "2013 AIME II Problems/Problem 13"

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Then plugging gives <math>b=4</math> and <math>c=2\sqrt{7}</math>. Then the height from <math>C</math> is <math>3</math>, and the area is <math>3\sqrt{7}</math> and our answer is <math>\boxed{010}</math>.
 
Then plugging gives <math>b=4</math> and <math>c=2\sqrt{7}</math>. Then the height from <math>C</math> is <math>3</math>, and the area is <math>3\sqrt{7}</math> and our answer is <math>\boxed{010}</math>.
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===Solution 6 (Desperate for points)===
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Note that <math>\triangle BEC</math> uniquely determines <math>\triangle ABC</math>. Assume WTMLOG (without too much loss of generality) that <math>\angle BEC=90^\circ</math>. Then, the answer is <math>2\cdot \frac{1}{2}\cdot 3\cdot \sqrt{7}=3\sqrt{7}\rightarrow\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 02:53, 23 February 2018

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 Law of Cosines 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 Law of Cosines 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 Law of Cosines 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 Law of Cosines 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 $\boxed{010}$.

Solution 3

Let the coordinates of A, B and C be (-a, 0), (a, 0) and (0, h) respectively. Then $D = (\frac{3a}{4}, \frac{h}{4})$ and $E = (-\frac{a}{8},\frac{h}{8}).$ $EC^2 = 7$ implies $a^2 + 49h^2 = 448$; $EB^2 = 9$ implies $81a^2 + h^2 = 576.$ Solve this system of equations simultaneously, $a=\sqrt{7}$ and $h=3$. Area of the triangle is ah = $3\sqrt{7}$, giving us an answer of $\boxed{010}$.

Solution 4

[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 EE=D/2; label("$\sqrt{7}$", C--EE, W); label("$x$", D--B, E); label("$3x$", C--D, E); label("$l$", EE--D, N); label("$3$", EE--B, N); label("$E$",EE,NW); [/asy] (Thanks to writer of Solution 2)

Let $BD = x$. Then $CD = 3x$ and $AC = 4x$. Also, let $AE = ED = l$. Using Stewart's Theorem on $\bigtriangleup CEB$ gives us the equation $(x)(3x)(4x) + (4x)(l^2) = 27x + 7x$ or, after simplifying, $4l^2 = 34 - 12x^2$. We use Stewart's again on $\bigtriangleup CAD$: $(l)(l)(2l) + 7(2l) = (16x^2)(l) + (9x^2)(l)$, which becomes $2l^2 = 25x^2 - 14$. Substituting $2l^2 = 17 - 6x^2$, we see that $31x^2 = 31$, or $x = 1$. Then $l^2 = \frac{11}{2}$.

We now use Law of Cosines on $\bigtriangleup CAD$. $(2l)^2 = (4x)^2 + (3x)^2 - 2(4x)(3x)\cos C$. Plugging in for $x$ and $l$, $22 = 16 + 9 - 2(4)(3)\cos C$, so $\cos C = \frac{1}{8}$. Using the Pythagorean trig identity $\sin^2 + \cos^2 = 1$, $\sin^2 C = 1 - \frac{1}{64}$, so $\sin C = \frac{3\sqrt{7}}{8}$.

$[ABC] = \frac{1}{2} AC \cdot BC \sin C = (\frac{1}{2})(4)(4)(\frac{3\sqrt{7}}{8}) = 3\sqrt{7}$, and our answer is $3 + 7 = \boxed{010}$.

Solution 5 (Barycentric Coordinates)

Let ABC be the reference triangle, with $A=(1,0,0)$, $B=(0,1,0)$, and $C=(0,0,1)$. We can easily calculate $D=(0,\frac{3}{4},\frac{1}{4})$ and subsequently $E=(\frac{1}{2},\frac{3}{8},\frac{1}{8})$. Using distance formula on $\overline{EC}=(\frac{1}{2},\frac{3}{8},-\frac{7}{8})$ and $\overline{EB}=(\frac{1}{2},-\frac{5}{8},\frac{1}{8})$ gives

\begin{align*} \begin{cases} 7&=|EC|^2=-a^2 \cdot \frac{3}{8} \cdot (-\frac{7}{8})-b^2 \cdot \frac{1}{2} \cdot (-\frac{7}{8})-c^2 \cdot \frac{1}{2} \cdot \frac{3}{8} \\ 9&=|EB|^2=-a^2 \cdot (-\frac{5}{8}) \cdot \frac{1}{8}-b^2 \cdot \frac{1}{2} \cdot \frac{1}{8}-c^2 \cdot \frac{1}{2} \cdot (-\frac{5}{8}) \\ \end{cases} \end{align*}

But we know that $a=b$, so we can substitute and now we have two equations and two variables. So we can clear the denominators and prepare to cancel a variable:

\begin{align*} \begin{cases} 7\cdot 64&=3\cdot 7\cdot a^2+b^2\cdot 4\cdot 7-c^2\cdot 4\cdot 3\\ 9\cdot 64&=5a^2-4b^2+4\cdot 5\cdot c^2 \\ \end{cases} \end{align*}

\begin{align*} \begin{cases} 7\cdot 64&=49a^2-12c^2 \\ 9\cdot 64&=a^2+20c^2 \\ \end{cases} \end{align*}

\begin{align*} \begin{cases} 5\cdot 7\cdot 64&=245a^2-60c^2 \\ 3\cdot 9\cdot 64&=3a^2+60c^2 \\ \end{cases} \end{align*}

Then we add the equations to get

\begin{align*} 62\cdot 64&=248a^2 \\ a^2 &=16 \\ a &=4 \\ \end{align*}

Then plugging gives $b=4$ and $c=2\sqrt{7}$. Then the height from $C$ is $3$, and the area is $3\sqrt{7}$ and our answer is $\boxed{010}$.

Solution 6 (Desperate for points)

Note that $\triangle BEC$ uniquely determines $\triangle ABC$. Assume WTMLOG (without too much loss of generality) that $\angle BEC=90^\circ$. Then, the answer is $2\cdot \frac{1}{2}\cdot 3\cdot \sqrt{7}=3\sqrt{7}\rightarrow\boxed{010}$.

See Also

2013 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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