As shown in the following figure, $D$, $E$, $F$ are on the sides of $\triangle{ABC}$, $AC$, $AB$, and $BC$ respectively. $AE=BE$, $AD=6$, $CD=7$, $BF=2$, $CF=9$. $DEFG$ is a square. The length of $AB$ can be expressed as $\ \ \ a\sqrt{\dfrac{b}{c}}\ \ \ $ in its simplest form. Find the value of $a+b+c$. Click here for the solution.
Solution: Draw line $AF$, $BD$ and $DF$. Let $S$ denote the area of a shape. We have:
$S_{\triangle{ADE}}=\dfrac{1}{2}S_{\triangle{ABD}}=\dfrac{1}{2}\cdot\dfrac{AD}{AD+CD}S_{\triangle{ABC}}$$ $$=\dfrac{1}{2}\cdot\dfrac{6}{6+7}S_{\triangle{ABC}}=\dfrac{3}{13}S_{\triangle{ABC}}$
$S_{\triangle{BEF}}=\dfrac{1}{2}S_{\triangle{BAF}}=\dfrac{1}{2}\cdot\dfrac{BF}{BF+CF}S_{\triangle{ABC}}=$$ $$\dfrac{1}{2}\cdot\dfrac{1}{2+9}S_{\triangle{ABC}}=\dfrac{1}{11}S_{\triangle{ABC}}$
$S_{\triangle{ADF}}=\dfrac{BD}{AD+BD}S_{\triangle{AFC}}=\dfrac{7}{6+7}\cdot\dfrac{CF}{BF+CF}S_{\triangle{ABC}}$$ $$=\dfrac{7}{13}\cdot\dfrac{9}{2+9}S_{\triangle{ABC}}=\dfrac{63}{143}S_{\triangle{ABC}}$
$S_{\triangle{DEF}}=S_{\triangle{ABC}}-S_{\triangle{ADE}}-S_{\triangle{BEF}}-S_{\triangle{ADF}}$$ $$=(1-\dfrac{3}{13}-\dfrac{1}{11}-\dfrac{63}{143})S_{\triangle{ABC}}=\dfrac{34}{143}S_{\triangle{ABC}}$
$S_{CDGF}=S_{\triangle{ADF}}-S_{\triangle{GDF}}=S_{\triangle{ADF}}-S_{\triangle{DEF}}$$ $$=(\dfrac{63}{143}-\dfrac{34}{143})S_{\triangle{ABC}}=\dfrac{29}{143}S_{\triangle{ABC}}$
Rotate $A$ $90^\circ$ counter-clockwise around $D$ to point $H$, and draw line $CH$, $DH$, $FH$, $GH$:
Because $\angle{ADE}+\angle{HDE}=90^\circ$, $\angle{HDG}+\angle{HDE}=90^\circ$, $\angle{ADE}=\angle{HDG}$. Given $AD=HD$, $DE=DG$, $\triangle{ADE}\cong\triangle{HDG}$.
Because $\angle{FGH}+\angle{DGH}=90^\circ$, $\angle{FEB}+\angle{DEA}=90^\circ$, and $\angle{DGH}=\angle{DEA}$, we have $\angle{FGH}=\angle{FEB}$.
Additionally, because $GF=EF$, $HG=AE=BE$, we have $\triangle{BEF}\cong\triangle{HGF}$. Therefore $H$ is also the point when $B$ is rotated $90^\circ$ clockwise around $F$. Both $\triangle{CDH}$ and $\triangle{CFH}$ are right triangles.
$S_{CDHF}=S_{CDGF}+S_{\triangle{HDG}}+S_{\triangle{HGF}}=S_{CDGF}+S_{\triangle{ADE}}+S_{\triangle{BEF}}$$ $$=(\dfrac{29}{143}+\dfrac{3}{13}+\dfrac{1}{11})S_{\triangle{ABC}}=\dfrac{75}{143}S_{\triangle{ABC}}$
Because
$S_{CDHF}=S_{\triangle{CDH}}+S_{\triangle{CFH}}=\dfrac{1}{2}CD\cdot HD+\dfrac{1}{2}CF\cdot HF=\dfrac{1}{2}(7\cdot 6+9\cdot 2)=30$
Therefore
$\dfrac{75}{143}S_{\triangle{ABC}}=30$
$S_{\triangle{ABC}}=\dfrac{286}{5}$
$S_{DEFG}=2\cdot S_{\triangle{DEF}}=2\cdot\dfrac{34}{143}\cdot S_{\triangle{ABC}}=2\cdot\dfrac{34}{143}\cdot \dfrac{286}{5}=\dfrac{136}{5}$
Let $x$ be the length of $AE$, $s$ the length of semi-perimeter of $\triangle{ABC}$, we have:
$x=\dfrac{AB+BC+CA}{2}=\dfrac{AE+CE+BF+CF+CD+AD}{2}$$ $$=\dfrac{x+x+2+9+7+6}{2}=x+12$
Per Heron’s theorem, $S_{\triangle{ABC}}=\sqrt{(x-12)(x+1)(x-1)(12-x)}=\dfrac{286}{5}$
$$(x^2-1)(x^2-144)+\dfrac{286^2}{25}=0$$
$$(5x^2)^2-5\cdot 145\cdot(5x^2)+25\cdot 144+286^2=0$$
$$(5x^2-148)(5x^2-577)=0$$
$$x=\pm 2\sqrt{\dfrac{37}{5}}\ \ \ \ \ \ \ \ x=\pm\sqrt{\dfrac{577}{5}}$$
Because $x=\sqrt{\dfrac{577}{5}}$ would result in a triangle that does not meet the description, and $x$ must be positive, therefore $x=2\sqrt{\dfrac{37}{5}}$ and $AE=2x=4\sqrt{\dfrac{37}{5}}$. We have $$a+b+c=4+37+5=\boxed{046}$$
Note: Assume $AD=a$, $CD=b$, $BF=c$, $CF=d$:
Let $$t=\dfrac{a}{a+b}+\dfrac{c}{c+d}+\dfrac{2bd}{(a+b)(c+d)}$$
We have $$S_{DEFG}=\dfrac{2-t}{2(t-1)}(a\cdot b+c\cdot d)$$
$$S_{\triangle{ABC}}=\dfrac{a\cdot b+c\cdot d}{2(t-1)}$$