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

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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 $\boxed{010}$ .

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{aligned}
   2x+3 &= 7 & 2x+3-3 &= 7-3 \\
   2x &= 4 & \frac{2x}2 &= \frac42\\
   x &= 2
\end{aligned}
\end{align}

@@ Line 86: / Line 86: @@
-\begin{align*}
+ \begin{align}
-\begin{cases}
+ \begin{aligned}
-&=|EC|^2 \\
+x+3 &= 7 & 2x+3-3 &= 7-3 \\
-&=|EB|^2 \\
+x &= 4 & \frac{2x}2 &= \frac42\\
-\end{cases}
+    x &= 2
-\end{align*}
+ \end{aligned}
+ \end{align}
 ==See Also==
 {{AIME box|year=2013|n=II|num-b=12|num-a=14}}
 {{MAA Notice}}

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