AMC12 2019 12B, Problem 12

Right triangle $ACD$ with right angle at $C$ is constructed outwards on the hypotenuse $\overline{AC}$ of isosceles right triangle $ABC$ with leg length $1$, as shown, so that the two triangles have equal perimeters. What is $\sin(2\angle BAD)$? [asy]  /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(8.016233639805293cm);  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 = -4.001920114613276, xmax = 4.014313525192017, ymin = -2.552570341575814, ymax = 5.6249093771911145;  /* image dimensions */  draw((-1.6742337260757447,-1.)--(-1.6742337260757445,-0.6742337260757447)--(-2.,-0.6742337260757447)--(-2.,-1.)--cycle, linewidth(2.));  draw((-1.7696484586262846,2.7696484586262846)--(-1.5392969172525692,3.)--(-1.7696484586262846,3.2303515413737154)--(-2.,3.)--cycle, linewidth(2.));   /* draw figures */ draw((-2.,3.)--(-2.,-1.), linewidth(2.));  draw((-2.,-1.)--(2.,-1.), linewidth(2.));  draw((2.,-1.)--(-2.,3.), linewidth(2.));  draw((-0.6404058554606791,4.3595941445393205)--(-2.,3.), linewidth(2.));  draw((-0.6404058554606791,4.3595941445393205)--(2.,-1.), linewidth(2.));  label("$D$",(-0.9382446143428628,4.887784444795223),SE*labelscalefactor,fontsize(14));  label("$A$",(1.9411496528285788,-1.0783204767840298),SE*labelscalefactor,fontsize(14));  label("$B$",(-2.5046350956841272,-0.9861798602345433),SE*labelscalefactor,fontsize(14));  label("$C$",(-2.5737405580962416,3.5747806589650395),SE*labelscalefactor,fontsize(14));  label("$1$",(-2.665881174645728,1.2712652452278765),SE*labelscalefactor,fontsize(14));  label("$1$",(-0.3393306067712029,-1.3547423264324894),SE*labelscalefactor,fontsize(14));   /* dots and labels */ dot((-2.,3.),linewidth(4.pt) + dotstyle);  dot((-2.,-1.),linewidth(4.pt) + dotstyle);  dot((2.,-1.),linewidth(4.pt) + dotstyle);  dot((-0.6404058554606791,4.3595941445393205),linewidth(4.pt) + dotstyle);  clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);   /* end of picture */ [/asy]

Solution: Firstly, note by the Pythagorean Theorem in $\triangle ABC$ that $AC = \sqrt{2}$. Now, the equal perimeter condition means that $BC + BA = 2 = CD + DA$, since side $AC$ is common to both triangles and thus can be discounted. This relationship, in combination with the Pythagorean Theorem in $\triangle ACD$, gives $AC^2+CD^2=\left(\sqrt{2}\right)^2+\left(2-DA\right)^2=DA^2$. Hence $2 + 4 - 4DA + DA^2 = DA^2$, so $DA = \frac{3}{2}$, and thus $CD = \frac{1}{2}$. Next, since $\angle BAC = 45^{\circ}$, $\sin{\left(\angle BAC\right)} = \cos{\left(\angle BAC\right)} = \frac{1}{\sqrt{2}}$. Using the lengths found above, $\sin{\left(\angle CAD\right)} = \frac{\left(\frac{1}{2}\right)}{\left(\frac{3}{2}\right)} = \frac{1}{3}$, and $\cos{\left(\angle CAD\right)} = \frac{\sqrt{2}}{\left(\frac{3}{2}\right)} = \frac{2 \sqrt{2}}{3}$. Thus, by the addition formulae for $\sin$ and $\cos$, we have \[\begin{split}\sin{\left(\angle BAD\right)}&=\sin{\left(\angle BAC + \angle CAD\right)}\\&=\sin{\left(\angle BAC\right)}\cos{\left(\angle CAD\right)}+\cos{\left(\angle BAC\right)}\sin{\left(\angle CAD\right)}\\&=\frac{1}{\sqrt{2}}\cdot\frac{2 \sqrt{2}}{3} + \frac{1}{\sqrt{2}}\cdot\frac{1}{3} \\ &= \frac{2 \sqrt{2} + 1}{3 \sqrt{2}}\end{split}\] and \[\begin{split}\cos{\left(\angle BAD\right)}&=\cos{\left(\angle BAC + \angle CAD\right)}\\&=\cos{\left(\angle BAC\right)}\cos{\left(\angle CAD\right)}-\sin{\left(\angle BAC\right)}\sin{\left(\angle CAD\right)}\\&=\frac{1}{\sqrt{2}}\cdot\frac{2 \sqrt{2}}{3} - \frac{1}{\sqrt{2}}\cdot\frac{1}{3} \\ &= \frac{2 \sqrt{2} - 1}{3 \sqrt{2}}\end{split}\] Hence, by the double angle formula for $\sin$, $\sin{\left(2\angle BAD\right)} = 2\sin{\left(\angle BAD\right)}\cos{\left(\angle BAD\right)} = \frac{2(8-1)}{18} = \fbox{\dfrac{7}{9}}$.

Answer: \dfrac{7}{9}