2019 AMC 12B Problems/Problem 12

Problem

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 $CD = \frac{1}{2}, DA = \frac{3}{2}$ . 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

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

and

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

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

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

and

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

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, $2\sin(x+y)\cos(x+y) = \dfrac{2-CD^2}{AD^2} = \dfrac{2-(\dfrac{1}{2})^2}{(\dfrac{3}{2})^2} = \boxed{(A) \dfrac{7}{9}}$

2019 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

Page

Toolbox

Search

2019 AMC 12B Problems/Problem 12

Contents

Problem

Solution 1

Solution 1.5 (Little bit of coordinate bash)

Solution 2

See Also