{"status": "success", "data": {"description_md": "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 $\\sin(2\\angle BAD)$?<br><center><img class=\"problem-image\" alt='[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]' class=\"latexcenter\" height=\"402\" src=\"https://latex.artofproblemsolving.com/c/d/5/cd54aedb94e876573c078d8b61aba6bbbeb68b71.png\" width=\"305\"/></center>\n\n$\\textbf{(A) } \\dfrac{1}{3}  \\qquad\\textbf{(B) } \\dfrac{\\sqrt{2}}{2} \\qquad\\textbf{(C) } \\dfrac{3}{4} \\qquad\\textbf{(D) } \\dfrac{7}{9} \\qquad\\textbf{(E) }  \\dfrac{\\sqrt{3}}{2}$\n___\nFull credit goes to [MAA](https://maa.org/) for authoring these problems. These problems were taken on the [AOPS](https://artofproblemsolving.com/) website.", "description_html": "<p>Right triangle  <span class=\"katex--inline\">ACD</span>  with right angle at  <span class=\"katex--inline\">C</span>  is constructed outwards on the hypotenuse  <span class=\"katex--inline\">\\overline{AC}</span>  of isosceles right triangle  <span class=\"katex--inline\">ABC</span>  with leg length  <span class=\"katex--inline\">1</span> , as shown, so that the two triangles have equal perimeters. What is  <span class=\"katex--inline\">\\sin(2\\angle BAD)</span> ?<br/><center><img class=\"latexcenter\" alt=\"[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(&#34;$D$&#34;,(-0.9382446143428628,4.887784444795223),SE*labelscalefactor,fontsize(14));  label(&#34;$A$&#34;,(1.9411496528285788,-1.0783204767840298),SE*labelscalefactor,fontsize(14));  label(&#34;$B$&#34;,(-2.5046350956841272,-0.9861798602345433),SE*labelscalefactor,fontsize(14));  label(&#34;$C$&#34;,(-2.5737405580962416,3.5747806589650395),SE*labelscalefactor,fontsize(14));  label(&#34;$1$&#34;,(-2.665881174645728,1.2712652452278765),SE*labelscalefactor,fontsize(14));  label(&#34;$1$&#34;,(-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]\" height=\"402\" src=\"https://latex.artofproblemsolving.com/c/d/5/cd54aedb94e876573c078d8b61aba6bbbeb68b71.png\" width=\"305\"/></center></p>&#10;<p> <span class=\"katex--inline\">\\textbf{(A) } \\dfrac{1}{3}  \\qquad\\textbf{(B) } \\dfrac{\\sqrt{2}}{2} \\qquad\\textbf{(C) } \\dfrac{3}{4} \\qquad\\textbf{(D) } \\dfrac{7}{9} \\qquad\\textbf{(E) }  \\dfrac{\\sqrt{3}}{2}</span> </p>&#10;<hr><p>Full credit goes to <a href=\"https://maa.org/\">MAA</a> for authoring these problems. These problems were taken on the <a href=\"https://artofproblemsolving.com/\">AOPS</a> website.</p>", "hints_md": "", "hints_html": "", "editorial_md": "", "editorial_html": "", "flag_hint": "", "point_value": 3, "problem_name": "2019 AMC 12B Problem 12", "can_next": true, "can_prev": true, "nxt": "/problem/19_amc12B_p13", "prev": "/problem/19_amc12B_p11"}}