2019 AMC 12B Problems/Problem 12

Revision as of 23:35, 14 February 2019 by Sahith1234567890 (talk | contribs) (Solution 1)


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]

$\textbf{(A) } \dfrac{1}{3}  \qquad\textbf{(B) } \dfrac{\sqrt{2}}{2} \qquad\textbf{(C) } \dfrac{3}{4} \qquad\textbf{(D) } \dfrac{7}{9} \qquad\textbf{(E) }  \dfrac{\sqrt{3}}{2}$

Solution 1

Observe that the "equal perimeter" part implies that $BC + BA = 2 = CD + DA$. A quick Pythagorean chase gives $CD = \frac{1}{2}, DA = \frac{3}{2}$. (Note: You could set up variables such as X, 2-X to denote the side lengths). Use the sine addition formula on angles $BAC$ and $CAD$ (which requires finding their cosines as well), and this gives the sine of $BAD$. Now, use $\sin{2x} = 2\sin{x}\cos{x}$ on angle $BAD$ to get $\boxed{\textbf{(D)} = \frac{7}{9}}$.

Feel free to elaborate if necessary.

Solution 1.5 (Little bit of coordinate bash)

After using Pythagorean to find $AD$ and $CD$, we can instead notice that the angle between the y-coordinate and $CD$ is $45$ degrees, and implies that the slope of that line is 1. If we draw a perpendicular from point $D$, we can then proceed to find the height and base of this new triangle (defined by $ADE$ where $E$ is the intersection of the altitude and $AB$) by coordinate-bashing, which turns out to be $1+\frac{\sqrt{2}}{4}$ and $1-\frac{\sqrt{2}}{4}$ respectively.

By double angle formula and difference of squares, it's easy to see that our answer is $\boxed{\textbf{(D) }\frac{7}{9}}$

~Solution by MagentaCobra

Solution 2

Let $x = \angle BAC$ and $y = \angle CAD$, so $\angle BAD = x+y$.

By the double-angle formula, $\sin(2\angle BAD)= 2\sin(\angle BAD)\cos(\angle BAD)$. To write this in terms of $x$ and $y$, we can say that we are looking for $2\sin(x+y)\cos(x+y)$.

Using trigonometric addition and subtraction formulas, we know that

$\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)$ $= (\dfrac{1}{\sqrt{2}})(\dfrac{\sqrt{2}}{AD}) + (\dfrac{1}{\sqrt{2}})(\dfrac{CD}{AD})$ $= \dfrac{1}{AD} + \dfrac{CD}{AD\sqrt{2}}$


$\cos(x+y) = \cos(x)\cos(y) - \sin(x)\sin(y)$. $= (\dfrac{1}{\sqrt{2}})(\dfrac{\sqrt{2}}{AD}) - (\dfrac{1}{\sqrt{2}})(\dfrac{CD}{AD})$ $= \dfrac{1}{AD} - \dfrac{CD}{AD\sqrt{2}}$.

So $2\sin(x+y)\cos(x+y) = 2[\dfrac{1}{AD} + \dfrac{CD}{AD\sqrt{2}}][\dfrac{1}{AD} - \dfrac{CD}{AD\sqrt{2}}]$ $= \dfrac{2-CD^2}{AD^2}$.

Now we just need to figure out what the numerical answer is.

From the given information about the triangles' perimeters, we can deduce that $CD + AD = 2$. Also, the Pythagorean theorem tell us that $CD^2 + 2 = AD^2$. These two equations allow us to write $CD and CD^2$ in terms of $AD$ without redundancy: $CD = 2 - AD$ and $CD^2 = AD^2 - 2$.

Plugging these into $\dfrac{2-CD^2}{AD^2}$, we'll get

$\dfrac{2-CD^2}{AD^2} = \dfrac{2-(2-AD)^2}{AD^2} = \dfrac{-2+4AD-AD^2}{AD^2}$


$\dfrac{2-CD^2}{AD^2} = \dfrac{2-(AD^2-2)}{AD^2} = \dfrac{4-AD^2}{AD^2}$.

If we set these equal to each other, now there is an algebraic equation that can be easily solved: $\dfrac{-2+4AD-AD^2}{AD^2} = \dfrac{4-AD^2}{AD^2}$ $-2+4AD-AD^2 = 4-AD^2$ $AD = \dfrac{3}{2}$

Now that we know what $AD$ is equal to, we can also figure out $CD = \dfrac{1}{2}$.

Thus, $\sin(2\angle BAD) = \dfrac{2-CD^2}{AD^2} = \dfrac{2-(\dfrac{1}{2})^2}{(\dfrac{3}{2})^2} = \boxed{(D) \dfrac{7}{9}}$

See Also

2019 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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All AMC 12 Problems and Solutions
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