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>?
  
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>?
 
 
<asy>
 
<asy>
 
  /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
 
  /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
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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>
 
<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>
  
==Solution 1==
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== Solutions ==
 
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=== 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>.
 
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>.
  
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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>.
 
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)==
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=== 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>.  
 
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>.  
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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>.
 
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)==
+
=== 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: <cmath>\cos{\theta}=\frac{2\sqrt{2}}{3}\implies \cos{2\theta}=2(\frac{8}{9})-1= \boxed{\textbf{(D) } \frac{7}{9}}</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>
  
==See Also==
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== 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 00:14, 19 October 2020

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 $\sin(2\angle BAD)$?

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

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

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
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