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

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==Problem==
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== Problem ==
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Right triangle <math>ACD</math> with right angle at <math>C</math> is constructed outwards on the hypotenuse <math>\overline{AC}</math> of isosceles right triangle <math>ABC</math> with leg length <math>1</math>, as shown, so that the two triangles have equal perimeters. What is <math>\sin(2\angle BAD)</math>?
  
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==Solution==
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<asy>
 +
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
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import graph; size(8.016233639805293cm);
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pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
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pen dotstyle = black; /* point style */
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real xmin = -4.001920114613276, xmax = 4.014313525192017, ymin = -2.552570341575814, ymax = 5.6249093771911145;  /* image dimensions */
  
−
7/9 (SuperWill)
+
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 */
 +
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 +
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draw((-0.6404058554606791,4.3595941445393205)--(-2.,3.), linewidth(2.));
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draw((-0.6404058554606791,4.3595941445393205)--(2.,-1.), linewidth(2.));
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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);
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dot((-2.,-1.),linewidth(4.pt) + dotstyle);
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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>
  
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==See Also==
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<math>\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}</math>
 +
 
 +
== Solutions ==
 +
=== Solution 1 ===
 +
Firstly, note by the Pythagorean Theorem in <math>\triangle ABC</math> that <math>AC = \sqrt{2}</math>. Now, the equal perimeter condition means that <math>BC + BA = 2 = CD + DA</math>, since side <math>AC</math> is common to both triangles and thus can be discounted. This relationship, in combination with the Pythagorean Theorem in <math>\triangle ACD</math>, gives <math>AC^2+CD^2=\left(\sqrt{2}\right)^2+\left(2-DA\right)^2=DA^2</math>. Hence <math>2 + 4 - 4DA + DA^2 = DA^2</math>, so <math>DA = \frac{3}{2}</math>, and thus <math>CD = \frac{1}{2}</math>.
 +
 
 +
Next, since <math>\angle BAC = 45^{\circ}</math>, <math>\sin{\left(\angle BAC\right)} = \cos{\left(\angle BAC\right)} = \frac{1}{\sqrt{2}}</math>. Using the lengths found above, <math>\sin{\left(\angle CAD\right)} = \frac{\left(\frac{1}{2}\right)}{\left(\frac{3}{2}\right)} = \frac{1}{3}</math>, and <math>\cos{\left(\angle CAD\right)} = \frac{\sqrt{2}}{\left(\frac{3}{2}\right)} = \frac{2 \sqrt{2}}{3}</math>.
 +
 
 +
Thus, by the addition formulae for <math>\sin</math> and <math>\cos</math>, we have
 +
<cmath>\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}</cmath>
 +
and
 +
<cmath>\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}</cmath>
 +
 
 +
Hence, by the double angle formula for <math>\sin</math>, <math>\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}}</math>.
 +
 
 +
=== Solution 2 (coordinate geometry) ===
 +
We use the Pythagorean Theorem, as in Solution 1, to find <math>AD=\frac{3}{2}</math> and <math>CD=\frac{1}{2}</math>. Now notice that the angle between <math>CD</math> and the vertical (i.e. the <math>y</math>-axis) is <math>45^{\circ}</math> &ndash; to see this, drop a perpendicular from <math>D</math> to <math>BA</math> which meets <math>BA</math> at <math>E</math>, and use the fact that the angle sum of quadrilateral <math>CBED</math> must be <math>360^{\circ}</math>. Anyway, this implies that the line <math>CD</math> has slope <math>1</math>, so since <math>C</math> is the point <math>(0,1)</math> and the length of <math>CD</math> is <math>\frac{1}{2}</math>, <math>D</math> has coordinates <math>\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)</math>.
 +
 
 +
Thus we have the lengths <math>DE=1+\frac{1}{2\sqrt{2}}</math> (it is just the <math>y</math>-coordinate) and <math>AE = 1-\frac{1}{2\sqrt{2}}</math>. By simple trigonometry in <math>\triangle DAE</math>, we now find <cmath>\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}}</cmath> and <cmath>\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}}</cmath> just as before. We can then use the double angle formula (as in Solution 1) to deduce <math>\sin{\left(2\angle BAD\right)} = \boxed{\textbf{(D) } \frac{7}{9}}</math>.
 +
 
 +
=== 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>
 +
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>
 +
 
 +
== Video Solution1 ==
 +
https://youtu.be/Cx2OmVoFGsw
 +
 
 +
~ Education, the Study of Everything
 +
 
 +
== 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}}
 +
{{MAA Notice}}

Latest revision as of 15:38, 12 September 2022

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]

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

Solutions

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: \[\cos{\theta}=\frac{2\sqrt{2}}{3}\implies \cos{2\theta}=2(\frac{8}{9})-1= \boxed{\textbf{(D) } \frac{7}{9}}\]

Video Solution1

https://youtu.be/Cx2OmVoFGsw

~ Education, the Study of Everything

See Also

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

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