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2019 AMC 12B Problems/Problem 12

Revision as of 21:52, 18 February 2019 by Sevenoptimus (talk | contribs) (Thoroughly cleaned up the first two solutions for LaTeX, formatting, grammar, and clarity, as well as adding a lot more detail; also removed the third solution which was just the same as the first solution)

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

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 $\sin$, $\sin{\left(2\angle BAD\right)} = 2\sin{\left(\angle BAD\right)}\cos{\left(\angle BAD\right)} = \frac{2(8-1)}{18} = \boxed{\textbf{(D) } \frac{7}{9}}$.

Solution 2 (coordinate geometry)

We use the Pythagorean Theorem, as in Solution 1, to find $AD=\frac{3}{2}$ and $CD=\frac{1}{2}$. Now notice that the angle between $CD$ and the vertical (i.e. the $y$-axis) is $45^{\circ}$ – to see this, drop a perpendicular from $D$ to $BA$ which meets $BA$ at $E$, and use the fact that the angle sum of quadrilateral $CBED$ must be $360^{\circ}$. Anyway, this implies that the line $CD$ has slope $1$, so since $C$ is the point $(0,1)$ and the length of $CD$ is $\frac{1}{2}$, $D$ has coordinates $\left(0+\frac{\left(\frac{1}{2}\right)}{\sqrt{2}}, 1+\frac{\left(\frac{1}{2}\right)}{\sqrt{2}}\right) = \left(\frac{1}{2\sqrt{2}}, 1+\frac{1}{2\sqrt{2}}\right)$.

Thus we have the lengths $DE=1+\frac{1}{2\sqrt{2}}$ (it is just the $y$-coordinate) and $AE = 1-\frac{1}{2\sqrt{2}}$. By simple trigonometry in $\triangle DAE$, we now find \[\sin{\left(\angle BAD\right)} = \frac{\left(1+\frac{1}{2\sqrt{2}}\right)}{\left(\frac{3}{2}\right)} = \frac{\left(2+\frac{1}{\sqrt{2}}\right)}{3} = \frac{2\sqrt{2}+1}{3\sqrt{2}}\] and \[\cos{\left(\angle BAD\right)} = \frac{\left(1-\frac{1}{2\sqrt{2}}\right)}{\left(\frac{3}{2}\right)} = \frac{\left(2-\frac{1}{\sqrt{2}}\right)}{3} = \frac{2\sqrt{2}-1}{3\sqrt{2}}\] just as before. We can then use the double angle formula (as in Solution 1) to deduce $\sin{\left(2\angle BAD\right)} = \boxed{\textbf{(D) } \frac{7}{9}}$.

See Also

2019 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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All AMC 12 Problems and Solutions
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