Difference between revisions of "2019 AMC 12B Problems/Problem 12"

(Solution 3 (easier finish to Solution 1))
Line 57: Line 57:
==Solution 3 (easier finish to Solution 1)==
==Solution 3 (easier finish to Solution 1)==
Again, use Pythagorean Theorem to find that <math>AD=\frac{3}{2}</math> and <math>CD=\frac{1}{2}</math>. Let <math>\angle DAC=\theta</math>. Note that we want <cmath>\sin{90+2\theta}=\cos{2\theta}</cmath>
Again, use Pythagorean Theorem to find that <math>AD=\frac{3}{2}</math> and <math>CD=\frac{1}{2}</math>. Let <math>\angle DAC=\theta</math>. Note that we want <cmath>\sin{90+2\theta}=\cos{2\theta}</cmath>
which is easy to compute: <math></math>\cos{\theta}=\frac{2\sqrt{2}}{3}\implies \cos{2\theta}=2(\frac{8}{9})-1=\boxed{\textbf{(D) }\frac{7}{9}
which is easy to compute: <cmath>\cos{\theta}=\frac{2\sqrt{2}}{3}\implies \cos{2\theta}=2(\frac{8}{9})-1=\boxed{\textbf{(D) }\frac{7}{9}</cmath>
==See Also==
==See Also==
{{AMC12 box|year=2019|ab=B|num-b=11|num-a=13}}
{{AMC12 box|year=2019|ab=B|num-b=11|num-a=13}}

Revision as of 20:38, 7 March 2019


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 $\sin{\left(2\angle BAD\right)} = \boxed{\textbf{(D) } \frac{7}{9}}$.

Solution 3 (easier finish to Solution 1)

Again, use Pythagorean Theorem to find that $AD=\frac{3}{2}$ and $CD=\frac{1}{2}$. Let $\angle DAC=\theta$. Note that we want \[\sin{90+2\theta}=\cos{2\theta}\]

which is easy to compute:

\[\cos{\theta}=\frac{2\sqrt{2}}{3}\implies \cos{2\theta}=2(\frac{8}{9})-1=\boxed{\textbf{(D) }\frac{7}{9}\] (Error compiling LaTeX. ! File ended while scanning use of \boxed.)

See Also

2019 AMC 12B (ProblemsAnswer KeyResources)
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
Invalid username
Login to AoPS