https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Yem567&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T20:04:36ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12B_Problems/Problem_12&diff=1028092019 AMC 12B Problems/Problem 122019-02-15T03:59:30Z<p>Yem567: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
<br />
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>?<br />
<asy><br />
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br />
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</asy><br />
<br />
<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><br />
<br />
<br />
==Solution 1==<br />
<br />
Observe that the "equal perimeter" part implies that <math>BC + BA = 2 = CD + DA</math>. A quick Pythagorean chase gives <math>CD = \frac{1}{2}, DA = \frac{3}{2}</math>.<br />
Use the sine addition formula on angles <math>BAC</math> and <math>CAD</math> (which requires finding their cosines as well), and this gives the sine of <math>BAD</math>. Now, use <math>\sin{2x} = 2\sin{x}\cos{x}</math> on angle <math>BAD</math> to get <math>\boxed{\textbf{(D)} = \frac{7}{9}}</math>.<br />
<br />
Feel free to elaborate if necessary.<br />
<br />
==Solution 1.5 (Little bit of coordinate bash)==<br />
<br />
After using Pythagorean to find <math>AD</math> and <math>CD</math>, we can instead notice that the angle between the y-coordinate and <math>CD</math> is <math>45</math> degrees, and implies that the slope of that line is 1. If we draw a perpendicular from point <math>D</math>, we can then proceed to find the height and base of this new triangle (defined by <math>ADE</math> where <math>E</math> is the intersection of the altitude and <math>AB</math>) by coordinate-bashing, which turns out to be <math>1+\frac{\sqrt{2}}{4}</math> and <math>1-\frac{\sqrt{2}}{4}</math> respectively.<br />
<br />
By double angle formula and difference of squares, it's easy to see that our answer is <math>\boxed{\textbf{(D) }\frac{7}{9}}</math><br />
<br />
~Solution by MagentaCobra<br />
<br />
==Solution 2==<br />
<br />
Let <math>x = \angle BAC</math> and <math>y = \angle CAD</math>, so <math>\angle BAD = x+y</math>.<br />
<br />
<br />
By the double-angle formula, <math>\sin(2\angle BAD)= 2\sin(\angle BAD)\cos(\angle BAD)</math>.<br />
To write this in terms of <math>x</math> and <math>y</math>, we can say that we are looking for <math>2\sin(x+y)\cos(x+y)</math>.<br />
<br />
Using trigonometric addition and subtraction formulas, we know that<br />
<br />
<math>\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)</math><br />
<math>= (\dfrac{1}{\sqrt{2}})(\dfrac{\sqrt{2}}{AD}) + (\dfrac{1}{\sqrt{2}})(\dfrac{CD}{AD})</math><br />
<math>= \dfrac{1}{AD} + \dfrac{CD}{AD\sqrt{2}}</math><br />
<br />
and<br />
<br />
<math>\cos(x+y) = \cos(x)\cos(y) - \sin(x)\sin(y)</math>.<br />
<math>= (\dfrac{1}{\sqrt{2}})(\dfrac{\sqrt{2}}{AD}) - (\dfrac{1}{\sqrt{2}})(\dfrac{CD}{AD})</math><br />
<math>= \dfrac{1}{AD} - \dfrac{CD}{AD\sqrt{2}}</math>.<br />
<br />
So <math>2\sin(x+y)\cos(x+y) = 2[\dfrac{1}{AD} + \dfrac{CD}{AD\sqrt{2}}][\dfrac{1}{AD} - \dfrac{CD}{AD\sqrt{2}}]</math><br />
<math>= \dfrac{2-CD^2}{AD^2}</math>.<br />
<br />
Now we just need to figure out what the numerical answer is.<br />
<br />
From the given information about the triangles' perimeters, we can deduce that <math>CD + AD = 2</math>.<br />
Also, the Pythagorean theorem tell us that <math>CD^2 + 2 = AD^2</math>.<br />
These two equations allow us to write <math>CD and CD^2</math> in terms of <math>AD</math> without redundancy: <math>CD = 2 - AD</math> and <math>CD^2 = AD^2 - 2</math>.<br />
<br />
Plugging these into <math>\dfrac{2-CD^2}{AD^2}</math>, we'll get<br />
<br />
<math>\dfrac{2-CD^2}{AD^2} = \dfrac{2-(2-AD)^2}{AD^2} = \dfrac{-2+4AD-AD^2}{AD^2}</math> <br />
<br />
and<br />
<br />
<math>\dfrac{2-CD^2}{AD^2} = \dfrac{2-(AD^2-2)}{AD^2} = \dfrac{4-AD^2}{AD^2}</math>.<br />
<br />
If we set these equal to each other, now there is an algebraic equation that can be easily solved:<br />
<math>\dfrac{-2+4AD-AD^2}{AD^2} = \dfrac{4-AD^2}{AD^2}</math><br />
<math>-2+4AD-AD^2 = 4-AD^2</math><br />
<math>AD = \dfrac{3}{2}</math><br />
<br />
Now that we know what <math>AD</math> is equal to, we can also figure out <math>CD = \dfrac{1}{2}</math>.<br />
<br />
Thus, <math>\sin(2\angle BAD) = \dfrac{2-CD^2}{AD^2} = \dfrac{2-(\dfrac{1}{2})^2}{(\dfrac{3}{2})^2} = \boxed{(D) \dfrac{7}{9}}</math><br />
<br />
==See Also==<br />
{{AMC12 box|year=2019|ab=B|num-b=11|num-a=13}}</div>Yem567https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12B_Problems/Problem_12&diff=1028082019 AMC 12B Problems/Problem 122019-02-15T03:58:06Z<p>Yem567: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
<br />
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>?<br />
<asy><br />
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br />
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<br />
<br />
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</asy><br />
<br />
<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><br />
<br />
<br />
==Solution 1==<br />
<br />
Observe that the "equal perimeter" part implies that <math>BC + BA = 2 = CD + DA</math>. A quick Pythagorean chase gives <math>CD = \frac{1}{2}, DA = \frac{3}{2}</math>.<br />
Use the sine addition formula on angles <math>BAC</math> and <math>CAD</math> (which requires finding their cosines as well), and this gives the sine of <math>BAD</math>. Now, use <math>\sin{2x} = 2\sin{x}\cos{x}</math> on angle <math>BAD</math> to get <math>\boxed{\textbf{(D)} = \frac{7}{9}}</math>.<br />
<br />
Feel free to elaborate if necessary.<br />
<br />
==Solution 1.5 (Little bit of coordinate bash)==<br />
<br />
After using Pythagorean to find <math>AD</math> and <math>CD</math>, we can instead notice that the angle between the y-coordinate and <math>CD</math> is <math>45</math> degrees, and implies that the slope of that line is 1. If we draw a perpendicular from point <math>D</math>, we can then proceed to find the height and base of this new triangle (defined by <math>ADE</math> where <math>E</math> is the intersection of the altitude and <math>AB</math>) by coordinate-bashing, which turns out to be <math>1+\frac{\sqrt{2}}{4}</math> and <math>1-\frac{\sqrt{2}}{4}</math> respectively.<br />
<br />
By double angle formula and difference of squares, it's easy to see that our answer is <math>\boxed{\textbf{(D) }\frac{7}{9}}</math><br />
<br />
~Solution by MagentaCobra<br />
<br />
==Solution 2==<br />
<br />
Let <math>x = \angle BAC</math> and <math>y = \angle CAD</math>, so <math>\angle BAD = x+y</math>.<br />
<br />
<br />
By the double-angle formula, <math>\sin(2\angle BAD)= 2\sin(\angle BAD)\cos(\angle BAD)</math>.<br />
To write this in terms of <math>x</math> and <math>y</math>, we can say that we are looking for <math>2\sin(x+y)\cos(x+y)</math>.<br />
<br />
Using trigonometric addition and subtraction formulas, we know that<br />
<br />
<math>\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)</math><br />
<math>= (\dfrac{1}{\sqrt{2}})(\dfrac{\sqrt{2}}{AD}) + (\dfrac{1}{\sqrt{2}})(\dfrac{CD}{AD})</math><br />
<math>= \dfrac{1}{AD} + \dfrac{CD}{AD\sqrt{2}}</math><br />
<br />
and<br />
<br />
<math>\cos(x+y) = \cos(x)\cos(y) - \sin(x)\sin(y)</math>.<br />
<math>= (\dfrac{1}{\sqrt{2}})(\dfrac{\sqrt{2}}{AD}) - (\dfrac{1}{\sqrt{2}})(\dfrac{CD}{AD})</math><br />
<math>= \dfrac{1}{AD} - \dfrac{CD}{AD\sqrt{2}}</math>.<br />
<br />
So <math>2\sin(x+y)\cos(x+y) = 2[\dfrac{1}{AD} + \dfrac{CD}{AD\sqrt{2}}][\dfrac{1}{AD} - \dfrac{CD}{AD\sqrt{2}}]</math><br />
<math>= \dfrac{2-CD^2}{AD^2}</math>.<br />
<br />
Now we just need to figure out what the numerical answer is.<br />
<br />
From the given information about the triangles' perimeters, we can deduce that <math>CD + AD = 2</math>.<br />
Also, the Pythagorean theorem tell us that <math>CD^2 + 2 = AD^2</math>.<br />
These two equations allow us to write <math>CD and CD^2</math> in terms of <math>AD</math> without redundancy: <math>CD = 2 - AD</math> and <math>CD^2 = AD^2 - 2</math>.<br />
<br />
Plugging these into <math>\dfrac{2-CD^2}{AD^2}</math>, we'll get<br />
<br />
<math>\dfrac{2-CD^2}{AD^2} = \dfrac{2-(2-AD)^2}{AD^2} = \dfrac{-2+4AD-AD^2}{AD^2}</math> <br />
<br />
and<br />
<br />
<math>\dfrac{2-CD^2}{AD^2} = \dfrac{2-(AD^2-2)}{AD^2} = \dfrac{4-AD^2}{AD^2}</math>.<br />
<br />
If we set these equal to each other, now there is an algebraic equation that can be easily solved:<br />
<math>\dfrac{-2+4AD-AD^2}{AD^2} = \dfrac{4-AD^2}{AD^2}</math><br />
<math>-2+4AD-AD^2 = 4-AD^2</math><br />
<math>AD = \dfrac{3}{2}</math><br />
<br />
Now that we know what <math>AD</math> is equal to, we can also figure out <math>CD = \dfrac{1}{2}</math>.<br />
<br />
Thus, <math>2\sin(x+y)\cos(x+y) = \dfrac{2-CD^2}{AD^2} = \dfrac{2-(\dfrac{1}{2})^2}{(\dfrac{3}{2})^2} = \boxed{(D) \dfrac{7}{9}}</math><br />
<br />
==See Also==<br />
{{AMC12 box|year=2019|ab=B|num-b=11|num-a=13}}</div>Yem567https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12B_Problems/Problem_12&diff=1028072019 AMC 12B Problems/Problem 122019-02-15T03:57:12Z<p>Yem567: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
<br />
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>?<br />
<asy><br />
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br />
import graph; size(8.016233639805293cm); <br />
real labelscalefactor = 0.5; /* changes label-to-point distance */<br />
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br />
pen dotstyle = black; /* point style */ <br />
real xmin = -4.001920114613276, xmax = 4.014313525192017, ymin = -2.552570341575814, ymax = 5.6249093771911145; /* image dimensions */<br />
<br />
<br />
draw((-1.6742337260757447,-1.)--(-1.6742337260757445,-0.6742337260757447)--(-2.,-0.6742337260757447)--(-2.,-1.)--cycle, linewidth(2.)); <br />
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</asy><br />
<br />
<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><br />
<br />
<br />
==Solution 1==<br />
<br />
Observe that the "equal perimeter" part implies that <math>BC + BA = 2 = CD + DA</math>. A quick Pythagorean chase gives <math>CD = \frac{1}{2}, DA = \frac{3}{2}</math>.<br />
Use the sine addition formula on angles <math>BAC</math> and <math>CAD</math> (which requires finding their cosines as well), and this gives the sine of <math>BAD</math>. Now, use <math>\sin{2x} = 2\sin{x}\cos{x}</math> on angle <math>BAD</math> to get <math>\boxed{\textbf{(D)} = \frac{7}{9}}</math>.<br />
<br />
Feel free to elaborate if necessary.<br />
<br />
==Solution 1.5 (Little bit of coordinate bash)==<br />
<br />
After using Pythagorean to find <math>AD</math> and <math>CD</math>, we can instead notice that the angle between the y-coordinate and <math>CD</math> is <math>45</math> degrees, and implies that the slope of that line is 1. If we draw a perpendicular from point <math>D</math>, we can then proceed to find the height and base of this new triangle (defined by <math>ADE</math> where <math>E</math> is the intersection of the altitude and <math>AB</math>) by coordinate-bashing, which turns out to be <math>1+\frac{\sqrt{2}}{4}</math> and <math>1-\frac{\sqrt{2}}{4}</math> respectively.<br />
<br />
By double angle formula and difference of squares, it's easy to see that our answer is <math>\boxed{\textbf{(D) }\frac{7}{9}}</math><br />
<br />
~Solution by MagentaCobra<br />
<br />
==Solution 2==<br />
Let <math>x = \angle BAC</math> and <math>y = \angle CAD</math>, so <math>\angle BAD = x+y</math>.<br />
<br />
By the double-angle formula, <math>\sin(2\angle BAD)= 2\sin(\angle BAD)\cos(\angle BAD)</math>.<br />
To write this in terms of <math>x</math> and <math>y</math>, we can say that we are looking for <math>2\sin(x+y)\cos(x+y)</math>.<br />
<br />
Using trigonometric addition and subtraction formulas, we know that<br />
<br />
<math>\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)</math><br />
<math>= (\dfrac{1}{\sqrt{2}})(\dfrac{\sqrt{2}}{AD}) + (\dfrac{1}{\sqrt{2}})(\dfrac{CD}{AD})</math><br />
<math>= \dfrac{1}{AD} + \dfrac{CD}{AD\sqrt{2}}</math><br />
<br />
and<br />
<br />
<math>\cos(x+y) = \cos(x)\cos(y) - \sin(x)\sin(y)</math>.<br />
<math>= (\dfrac{1}{\sqrt{2}})(\dfrac{\sqrt{2}}{AD}) - (\dfrac{1}{\sqrt{2}})(\dfrac{CD}{AD})</math><br />
<math>= \dfrac{1}{AD} - \dfrac{CD}{AD\sqrt{2}}</math>.<br />
<br />
So <math>2\sin(x+y)\cos(x+y) = 2[\dfrac{1}{AD} + \dfrac{CD}{AD\sqrt{2}}][\dfrac{1}{AD} - \dfrac{CD}{AD\sqrt{2}}]</math><br />
<math>= \dfrac{2-CD^2}{AD^2}</math>.<br />
Now we just need to figure out what the numerical answer is.<br />
<br />
From the given information about the triangles' perimeters, we can deduce that <math>CD + AD = 2</math>.<br />
Also, the Pythagorean theorem tell us that <math>CD^2 + 2 = AD^2</math>.<br />
These two equations allow us to write <math>CD and CD^2</math> in terms of <math>AD</math> without redundancy: <math>CD = 2 - AD</math> and <math>CD^2 = AD^2 - 2</math>.<br />
<br />
Plugging these into <math>\dfrac{2-CD^2}{AD^2}</math>, we'll get<br />
<br />
<math>\dfrac{2-CD^2}{AD^2} = \dfrac{2-(2-AD)^2}{AD^2} = \dfrac{-2+4AD-AD^2}{AD^2}</math> <br />
<br />
and<br />
<br />
<math>\dfrac{2-CD^2}{AD^2} = \dfrac{2-(AD^2-2)}{AD^2} = \dfrac{4-AD^2}{AD^2}</math>.<br />
<br />
If we set these equal to each other, now there is an algebraic equation that can be easily solved:<br />
<math>\dfrac{-2+4AD-AD^2}{AD^2} = \dfrac{4-AD^2}{AD^2}</math><br />
<math>-2+4AD-AD^2 = 4-AD^2</math><br />
<math>AD = \dfrac{3}{2}</math><br />
<br />
Now that we know what <math>AD</math> is equal to, we can also figure out <math>CD = \dfrac{1}{2}</math>.<br />
Thus, <math>2\sin(x+y)\cos(x+y) = \dfrac{2-CD^2}{AD^2} = \dfrac{2-(\dfrac{1}{2})^2}{(\dfrac{3}{2})^2} = \boxed{(D) \dfrac{7}{9}}</math><br />
<br />
==See Also==<br />
{{AMC12 box|year=2019|ab=B|num-b=11|num-a=13}}</div>Yem567https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12B_Problems/Problem_12&diff=1028062019 AMC 12B Problems/Problem 122019-02-15T03:56:45Z<p>Yem567: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
<br />
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>?<br />
<asy><br />
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br />
import graph; size(8.016233639805293cm); <br />
real labelscalefactor = 0.5; /* changes label-to-point distance */<br />
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clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br />
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</asy><br />
<br />
<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><br />
<br />
<br />
==Solution 1==<br />
<br />
Observe that the "equal perimeter" part implies that <math>BC + BA = 2 = CD + DA</math>. A quick Pythagorean chase gives <math>CD = \frac{1}{2}, DA = \frac{3}{2}</math>.<br />
Use the sine addition formula on angles <math>BAC</math> and <math>CAD</math> (which requires finding their cosines as well), and this gives the sine of <math>BAD</math>. Now, use <math>\sin{2x} = 2\sin{x}\cos{x}</math> on angle <math>BAD</math> to get <math>\boxed{\textbf{(D)} = \frac{7}{9}}</math>.<br />
<br />
Feel free to elaborate if necessary.<br />
<br />
==Solution 1.5 (Little bit of coordinate bash)==<br />
<br />
After using Pythagorean to find <math>AD</math> and <math>CD</math>, we can instead notice that the angle between the y-coordinate and <math>CD</math> is <math>45</math> degrees, and implies that the slope of that line is 1. If we draw a perpendicular from point <math>D</math>, we can then proceed to find the height and base of this new triangle (defined by <math>ADE</math> where <math>E</math> is the intersection of the altitude and <math>AB</math>) by coordinate-bashing, which turns out to be <math>1+\frac{\sqrt{2}}{4}</math> and <math>1-\frac{\sqrt{2}}{4}</math> respectively.<br />
<br />
By double angle formula and difference of squares, it's easy to see that our answer is <math>\boxed{\textbf{(D) }\frac{7}{9}}</math><br />
<br />
~Solution by MagentaCobra<br />
<br />
==Solution 2==<br />
Let <math>x = \angle BAC</math> and <math>y = \angle CAD</math>, so <math>\angle BAD = x+y</math>.<br />
<br />
By the double-angle formula, <math>\sin(2\angle BAD)= 2\sin(\angle BAD)\cos(\angle BAD)</math>.<br />
To write this in terms of <math>x</math> and <math>y</math>, we can say that we are looking for <math>2\sin(x+y)\cos(x+y)</math>.<br />
<br />
Using trigonometric addition and subtraction formulas, we know that<br />
<br />
<math>\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)</math><br />
<math>= (\dfrac{1}{\sqrt{2}})(\dfrac{\sqrt{2}}{AD}) + (\dfrac{1}{\sqrt{2}})(\dfrac{CD}{AD})</math><br />
<math>= \dfrac{1}{AD} + \dfrac{CD}{AD\sqrt{2}}</math><br />
<br />
and<br />
<br />
<math>\cos(x+y) = \cos(x)\cos(y) - \sin(x)\sin(y)</math>.<br />
<math>= (\dfrac{1}{\sqrt{2}})(\dfrac{\sqrt{2}}{AD}) - (\dfrac{1}{\sqrt{2}})(\dfrac{CD}{AD})</math><br />
<math>= \dfrac{1}{AD} - \dfrac{CD}{AD\sqrt{2}}</math>.<br />
<br />
So <math>2\sin(x+y)\cos(x+y) = 2[\dfrac{1}{AD} + \dfrac{CD}{AD\sqrt{2}}][\dfrac{1}{AD} - \dfrac{CD}{AD\sqrt{2}}]</math><br />
<math>= \dfrac{2-CD^2}{AD^2}</math>.<br />
Now we just need to figure out what the numerical answer is.<br />
<br />
From the given information about the triangles' perimeters, we can deduce that <math>CD + AD = 2</math>.<br />
Also, the Pythagorean theorem tell us that <math>CD^2 + 2 = AD^2</math>.<br />
These two equations allow us to write <math>CD and CD^2</math> in terms of <math>AD</math> without redundancy: <math>CD = 2 - AD</math> and <math>CD^2 = AD^2 - 2</math>.<br />
<br />
Plugging these into <math>\dfrac{2-CD^2}{AD^2}</math>, we'll get<br />
<br />
<math>\dfrac{2-CD^2}{AD^2} = \dfrac{2-(2-AD)^2}{AD^2} = \dfrac{-2+4AD-AD^2}{AD^2}</math> <br />
<br />
and<br />
<br />
<math>\dfrac{2-CD^2}{AD^2} = \dfrac{2-(AD^2-2)}{AD^2} = \dfrac{4-AD^2}{AD^2}</math>.<br />
<br />
If we set these equal to each other, now there is an algebraic equation that can be easily solved:<br />
<math>\dfrac{-2+4AD-AD^2}{AD^2} = \dfrac{4-AD^2}{AD^2}</math><br />
<math>-2+4AD-AD^2 = 4-AD^2</math><br />
<math>AD = \dfrac{3}{2}</math><br />
<br />
Now that we know what <math>AD</math> is equal to, we can also figure out <math>CD = \dfrac{1}{2}</math>.<br />
Thus, <math>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}}</math><br />
<br />
==See Also==<br />
{{AMC12 box|year=2019|ab=B|num-b=11|num-a=13}}</div>Yem567https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12B_Problems/Problem_12&diff=1028052019 AMC 12B Problems/Problem 122019-02-15T03:55:22Z<p>Yem567: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
<br />
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>?<br />
<asy><br />
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br />
import graph; size(8.016233639805293cm); <br />
real labelscalefactor = 0.5; /* changes label-to-point distance */<br />
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<br />
<br />
draw((-1.6742337260757447,-1.)--(-1.6742337260757445,-0.6742337260757447)--(-2.,-0.6742337260757447)--(-2.,-1.)--cycle, linewidth(2.)); <br />
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label("$C$",(-2.5737405580962416,3.5747806589650395),SE*labelscalefactor,fontsize(14)); <br />
label("$1$",(-2.665881174645728,1.2712652452278765),SE*labelscalefactor,fontsize(14)); <br />
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</asy><br />
<br />
<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><br />
<br />
<br />
==Solution 1==<br />
<br />
Observe that the "equal perimeter" part implies that <math>BC + BA = 2 = CD + DA</math>. A quick Pythagorean chase gives <math>CD = \frac{1}{2}, DA = \frac{3}{2}</math>.<br />
Use the sine addition formula on angles <math>BAC</math> and <math>CAD</math> (which requires finding their cosines as well), and this gives the sine of <math>BAD</math>. Now, use <math>\sin{2x} = 2\sin{x}\cos{x}</math> on angle <math>BAD</math> to get <math>\boxed{\textbf{(D)} = \frac{7}{9}}</math>.<br />
<br />
Feel free to elaborate if necessary.<br />
<br />
==Solution 1.5 (Little bit of coordinate bash)==<br />
<br />
After using Pythagorean to find <math>AD</math> and <math>CD</math>, we can instead notice that the angle between the y-coordinate and <math>CD</math> is <math>45</math> degrees, and implies that the slope of that line is 1. If we draw a perpendicular from point <math>D</math>, we can then proceed to find the height and base of this new triangle (defined by <math>ADE</math> where <math>E</math> is the intersection of the altitude and <math>AB</math>) by coordinate-bashing, which turns out to be <math>1+\frac{\sqrt{2}}{4}</math> and <math>1-\frac{\sqrt{2}}{4}</math> respectively.<br />
<br />
By double angle formula and difference of squares, it's easy to see that our answer is <math>\boxed{\textbf{(D) }\frac{7}{9}}</math><br />
<br />
~Solution by MagentaCobra<br />
<br />
==Solution 2==<br />
Let <math>x = \angle BAC</math> and <math>y = \angle CAD</math>, so <math>\angle BAD = x+y</math>.<br />
<br />
By the double-angle formula, <math>\sin(2\angle BAD)= 2\sin(\angle BAD)\cos(\angle BAD)</math>.<br />
To write this in terms of <math>x</math> and <math>y</math>, we can say that we are looking for <math>2\sin(x+y)\cos(x+y)</math>.<br />
<br />
Using trigonometric addition and subtraction formulas, we know that<br />
<br />
<math>\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)</math><br />
<math>= (\dfrac{1}{\sqrt{2}})(\dfrac{\sqrt{2}}{AD}) + (\dfrac{1}{\sqrt{2}})(\dfrac{CD}{AD})</math><br />
<math>= \dfrac{1}{AD} + \dfrac{CD}{AD\sqrt{2}}</math><br />
<br />
and<br />
<br />
<math>\cos(x+y) = \cos(x)\cos(y) - \sin(x)\sin(y)</math>.<br />
<math>= (\dfrac{1}{\sqrt{2}})(\dfrac{\sqrt{2}}{AD}) - (\dfrac{1}{\sqrt{2}})(\dfrac{CD}{AD})</math><br />
<math>= \dfrac{1}{AD} - \dfrac{CD}{AD\sqrt{2}}</math>.<br />
<br />
So <math>2\sin(x+y)\cos(x+y) = 2[\dfrac{1}{AD} + \dfrac{CD}{AD\sqrt{2}}][\dfrac{1}{AD} - \dfrac{CD}{AD\sqrt{2}}]</math><br />
<math>= \dfrac{2-CD^2}{AD^2}</math>.<br />
Now we just need to figure out what the numerical answer is.<br />
<br />
From the given information about the triangles' perimeters, we can deduce that <math>CD + AD = 2</math>.<br />
Also, the Pythagorean theorem tell us that <math>CD^2 + 2 = AD^2</math>.<br />
These two equations allow us to write <math>CD and CD^2</math> in terms of <math>AD</math> without redundancy: <math>CD = 2 - AD</math> and <math>CD^2 = AD^2 - 2</math>.<br />
<br />
Plugging these into <math>\dfrac{2-CD^2}{AD^2}</math>, we'll get<br />
<br />
<math>\dfrac{2-CD^2}{AD^2} = \dfrac{2-(2-AD)^2}{AD^2} = \dfrac{-2+4AD-AD^2}{AD^2}</math> <br />
<br />
and<br />
<br />
<math>\dfrac{2-CD^2}{AD^2} = \dfrac{2-(AD^2-2)}{AD^2} = \dfrac{4-AD^2}{AD^2}</math>.<br />
<br />
If we set these equal to each other, now there is an algebraic equation that can be easily solved:<br />
<math>\dfrac{-2+4AD-AD^2}{AD^2} = \dfrac{4-AD^2}{AD^2}</math><br />
<math>-2+4AD-AD^2 = 4-AD^2</math><br />
<math>AD = \dfrac{3}{2}</math><br />
<br />
Now that we know what <math>AD</math> is equal to, we can also figure out <math>CD = \dfrac{1}{2}</math>.<br />
Thus, <math>2\sin(x+y)\cos(x+y) = \dfrac{2-CD^2}{AD^2} = \dfrac{2-(\dfrac{1}{2})^2}{(\dfrac{3}{2})^2} = \dfrac{7}{9}</math>.<br />
<br />
==See Also==<br />
{{AMC12 box|year=2019|ab=B|num-b=11|num-a=13}}</div>Yem567https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12B_Problems/Problem_12&diff=1027982019 AMC 12B Problems/Problem 122019-02-15T03:48:01Z<p>Yem567: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
<br />
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>?<br />
<asy><br />
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br />
import graph; size(8.016233639805293cm); <br />
real labelscalefactor = 0.5; /* changes label-to-point distance */<br />
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br />
pen dotstyle = black; /* point style */ <br />
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<br />
<br />
draw((-1.6742337260757447,-1.)--(-1.6742337260757445,-0.6742337260757447)--(-2.,-0.6742337260757447)--(-2.,-1.)--cycle, linewidth(2.)); <br />
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label("$1$",(-2.665881174645728,1.2712652452278765),SE*labelscalefactor,fontsize(14)); <br />
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dot((-0.6404058554606791,4.3595941445393205),linewidth(4.pt) + dotstyle); <br />
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</asy><br />
<br />
<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><br />
<br />
<br />
==Solution 1==<br />
<br />
Observe that the "equal perimeter" part implies that <math>BC + BA = 2 = CD + DA</math>. A quick Pythagorean chase gives <math>CD = \frac{1}{2}, DA = \frac{3}{2}</math>.<br />
Use the sine addition formula on angles <math>BAC</math> and <math>CAD</math> (which requires finding their cosines as well), and this gives the sine of <math>BAD</math>. Now, use <math>\sin{2x} = 2\sin{x}\cos{x}</math> on angle <math>BAD</math> to get <math>\boxed{\textbf{(D)} = \frac{7}{9}}</math>.<br />
<br />
Feel free to elaborate if necessary.<br />
<br />
==Solution 1.5 (Little bit of coordinate bash)==<br />
<br />
After using Pythagorean to find <math>AD</math> and <math>CD</math>, we can instead notice that the angle between the y-coordinate and <math>CD</math> is <math>45</math> degrees, and implies that the slope of that line is 1. If we draw a perpendicular from point <math>D</math>, we can then proceed to find the height and base of this new triangle (defined by <math>ADE</math> where <math>E</math> is the intersection of the altitude and <math>AB</math>) by coordinate-bashing, which turns out to be <math>1+\frac{\sqrt{2}}{4}</math> and <math>1-\frac{\sqrt{2}}{4}</math> respectively.<br />
<br />
By double angle formula and difference of squares, it's easy to see that our answer is <math>\boxed{\textbf{(D) }\frac{7}{9}}</math><br />
<br />
~Solution by MagentaCobra<br />
<br />
==Solution 2==<br />
Let <math>x = \angle BAC</math> and <math>y = \angle CAD</math>, so <math>\angle BAD = x+y</math>.<br />
<br />
By the double-angle formula, <math>\sin(2\angle BAD)= 2\sin(\angle BAD)\cos(\angle BAD)</math>.<br />
To write this in terms of <math>x</math> and <math>y</math>, we can say that we are looking for <math>2\sin(x+y)\cos(x+y)</math>.<br />
<br />
Using trigonometric addition and subtraction formulas, we know that<br />
<br />
<math>\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)</math><br />
<math>= (\dfrac{1}{\sqrt{2}})(\dfrac{\sqrt{2}}{AD}) + (\dfrac{1}{\sqrt{2}})(\dfrac{CD}{AD})</math><br />
<math>= \dfrac{1}{AD} + \dfrac{CD}{AD\sqrt{2}}</math><br />
<br />
and<br />
<br />
<math>\cos(x+y) = \cos(x)\cos(y) - \sin(x)\sin(y)</math>.<br />
<math>= (\dfrac{1}{\sqrt{2}})(\dfrac{\sqrt{2}}{AD}) - (\dfrac{1}{\sqrt{2}})(\dfrac{CD}{AD})</math><br />
<math>= \dfrac{1}{AD} - \dfrac{CD}{AD\sqrt{2}}</math>.<br />
<br />
So <math>2\sin(x+y)\cos(x+y) = 2[\dfrac{1}{AD} + \dfrac{CD}{AD\sqrt{2}}][\dfrac{1}{AD} - \dfrac{CD}{AD\sqrt{2}}]</math><br />
<math>= \dfrac{2-CD^2}{AD^2}</math>.<br />
Now we just need to figure out what the numerical answer is.<br />
<br />
From the given information about the triangles' perimeters, we can deduce that <math>CD + AD = 2</math>.<br />
Also, the Pythagorean theorem tell us that <math>CD^2 + 2 = AD^2</math>.<br />
These two equations allow us to write <math>CD and CD^2</math> in terms of <math>AD</math> without redundancy: <math>CD = 2 - AD</math> and <math>CD^2 = AD^2 - 2</math>.<br />
<br />
Plugging these into <math>\dfrac{2-CD^2}{AD^2}</math>, we'll get<br />
<br />
<math>\dfrac{2-CD^2}{AD^2} = \dfrac{2-(2-AD)^2}{AD^2} = \dfrac{-2+4AD-AD^2}{AD^2}</math> <br />
<br />
and<br />
<br />
<math>\dfrac{2-CD^2}{AD^2} = \dfrac{2-(AD^2-2)}{AD^2} = \dfrac{4-AD^2}{AD^2}</math>.<br />
<br />
==See Also==<br />
{{AMC12 box|year=2019|ab=B|num-b=11|num-a=13}}</div>Yem567https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12B_Problems/Problem_12&diff=1027972019 AMC 12B Problems/Problem 122019-02-15T03:47:37Z<p>Yem567: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
<br />
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>?<br />
<asy><br />
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br />
import graph; size(8.016233639805293cm); <br />
real labelscalefactor = 0.5; /* changes label-to-point distance */<br />
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br />
pen dotstyle = black; /* point style */ <br />
real xmin = -4.001920114613276, xmax = 4.014313525192017, ymin = -2.552570341575814, ymax = 5.6249093771911145; /* image dimensions */<br />
<br />
<br />
draw((-1.6742337260757447,-1.)--(-1.6742337260757445,-0.6742337260757447)--(-2.,-0.6742337260757447)--(-2.,-1.)--cycle, linewidth(2.)); <br />
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draw((2.,-1.)--(-2.,3.), linewidth(2.)); <br />
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label("$B$",(-2.5046350956841272,-0.9861798602345433),SE*labelscalefactor,fontsize(14)); <br />
label("$C$",(-2.5737405580962416,3.5747806589650395),SE*labelscalefactor,fontsize(14)); <br />
label("$1$",(-2.665881174645728,1.2712652452278765),SE*labelscalefactor,fontsize(14)); <br />
label("$1$",(-0.3393306067712029,-1.3547423264324894),SE*labelscalefactor,fontsize(14)); <br />
/* dots and labels */<br />
dot((-2.,3.),linewidth(4.pt) + dotstyle); <br />
dot((-2.,-1.),linewidth(4.pt) + dotstyle); <br />
dot((2.,-1.),linewidth(4.pt) + dotstyle); <br />
dot((-0.6404058554606791,4.3595941445393205),linewidth(4.pt) + dotstyle); <br />
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br />
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</asy><br />
<br />
<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><br />
<br />
<br />
==Solution 1==<br />
<br />
Observe that the "equal perimeter" part implies that <math>BC + BA = 2 = CD + DA</math>. A quick Pythagorean chase gives <math>CD = \frac{1}{2}, DA = \frac{3}{2}</math>.<br />
Use the sine addition formula on angles <math>BAC</math> and <math>CAD</math> (which requires finding their cosines as well), and this gives the sine of <math>BAD</math>. Now, use <math>\sin{2x} = 2\sin{x}\cos{x}</math> on angle <math>BAD</math> to get <math>\boxed{\textbf{(D)} = \frac{7}{9}}</math>.<br />
<br />
Feel free to elaborate if necessary.<br />
<br />
==Solution 1.5 (Little bit of coordinate bash)==<br />
<br />
After using Pythagorean to find <math>AD</math> and <math>CD</math>, we can instead notice that the angle between the y-coordinate and <math>CD</math> is <math>45</math> degrees, and implies that the slope of that line is 1. If we draw a perpendicular from point <math>D</math>, we can then proceed to find the height and base of this new triangle (defined by <math>ADE</math> where <math>E</math> is the intersection of the altitude and <math>AB</math>) by coordinate-bashing, which turns out to be <math>1+\frac{\sqrt{2}}{4}</math> and <math>1-\frac{\sqrt{2}}{4}</math> respectively.<br />
<br />
By double angle formula and difference of squares, it's easy to see that our answer is <math>\boxed{\textbf{(D) }\frac{7}{9}}</math><br />
<br />
~Solution by MagentaCobra<br />
<br />
==Solution 2==<br />
Let <math>x = \angle BAC</math> and <math>y = \angle CAD</math>, so <math>\angle BAD = x+y</math>.<br />
<br />
By the double-angle formula, <math>\sin(2\angle BAD)= 2\sin(\angle BAD)\cos(\angle BAD)</math>.<br />
To write this in terms of <math>x</math> and <math>y</math>, we can say that we are looking for <math>2\sin(x+y)\cos(x+y)</math>.<br />
<br />
Using trigonometric addition and subtraction formulas, we know that<br />
<br />
<math>\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)</math><br />
<math>= (\dfrac{1}{\sqrt{2}})(\dfrac{\sqrt{2}}{AD}) + (\dfrac{1}{\sqrt{2}})(\dfrac{CD}{AD})</math><br />
<math>= \dfrac{1}{AD} + \dfrac{CD}{AD\sqrt{2}}</math><br />
<br />
and<br />
<br />
<math>\cos(x+y) = \cos(x)\cos(y) - \sin(x)\sin(y)</math>.<br />
<math>= (\dfrac{1}{\sqrt{2}})(\dfrac{\sqrt{2}}{AD}) - (\dfrac{1}{\sqrt{2}})(\dfrac{CD}{AD})</math><br />
<math>= \dfrac{1}{AD} - \dfrac{CD}{AD\sqrt{2}}</math>.<br />
<br />
So <math>2\sin(x+y)\cos(x+y) = 2[\dfrac{1}{AD} + \dfrac{CD}{AD\sqrt{2}}][\dfrac{1}{AD} - \dfrac{CD}{AD\sqrt{2}}]</math><br />
<math>= \dfrac{2-CD^2}{AD^2}</math>.<br />
Now we just need to figure out what the numerical answer is.<br />
<br />
From the given information about the triangles' perimeters, we can deduce that <math>CD + AD = 2</math>.<br />
Also, the Pythagorean theorem tell us that <math>CD^2 + 2 = AD^2</math>.<br />
These two equations allow us to write <math>CD and CD^2</math> in terms of <math>AD</math> without redundancy: <math>CD = 2 - AD</math> and <math>CD^2 = AD^2 - 2</math>.<br />
<br />
Plugging these into <math>\dfrac{2-CD^2}{AD^2}</math>, we'll get<br />
<br />
<math>\dfrac{2-CD^2}{AD^2} = \dfrac{2-(2-AD)^2}{AD^2} = \dfrac{-2+4AD-AD^2}{AD^2}</math> and<br />
<br />
<math>\dfrac{2-CD^2}{AD^2} = \dfrac{2-(AD^2-2)}{AD^2} = \dfrac{4-AD^2}{AD^2}</math>.<br />
<br />
==See Also==<br />
{{AMC12 box|year=2019|ab=B|num-b=11|num-a=13}}</div>Yem567https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12B_Problems/Problem_12&diff=1027952019 AMC 12B Problems/Problem 122019-02-15T03:47:15Z<p>Yem567: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
<br />
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>?<br />
<asy><br />
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br />
import graph; size(8.016233639805293cm); <br />
real labelscalefactor = 0.5; /* changes label-to-point distance */<br />
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br />
pen dotstyle = black; /* point style */ <br />
real xmin = -4.001920114613276, xmax = 4.014313525192017, ymin = -2.552570341575814, ymax = 5.6249093771911145; /* image dimensions */<br />
<br />
<br />
draw((-1.6742337260757447,-1.)--(-1.6742337260757445,-0.6742337260757447)--(-2.,-0.6742337260757447)--(-2.,-1.)--cycle, linewidth(2.)); <br />
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label("$A$",(1.9411496528285788,-1.0783204767840298),SE*labelscalefactor,fontsize(14)); <br />
label("$B$",(-2.5046350956841272,-0.9861798602345433),SE*labelscalefactor,fontsize(14)); <br />
label("$C$",(-2.5737405580962416,3.5747806589650395),SE*labelscalefactor,fontsize(14)); <br />
label("$1$",(-2.665881174645728,1.2712652452278765),SE*labelscalefactor,fontsize(14)); <br />
label("$1$",(-0.3393306067712029,-1.3547423264324894),SE*labelscalefactor,fontsize(14)); <br />
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dot((-2.,3.),linewidth(4.pt) + dotstyle); <br />
dot((-2.,-1.),linewidth(4.pt) + dotstyle); <br />
dot((2.,-1.),linewidth(4.pt) + dotstyle); <br />
dot((-0.6404058554606791,4.3595941445393205),linewidth(4.pt) + dotstyle); <br />
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br />
/* end of picture */<br />
</asy><br />
<br />
<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><br />
<br />
<br />
==Solution 1==<br />
<br />
Observe that the "equal perimeter" part implies that <math>BC + BA = 2 = CD + DA</math>. A quick Pythagorean chase gives <math>CD = \frac{1}{2}, DA = \frac{3}{2}</math>.<br />
Use the sine addition formula on angles <math>BAC</math> and <math>CAD</math> (which requires finding their cosines as well), and this gives the sine of <math>BAD</math>. Now, use <math>\sin{2x} = 2\sin{x}\cos{x}</math> on angle <math>BAD</math> to get <math>\boxed{\textbf{(D)} = \frac{7}{9}}</math>.<br />
<br />
Feel free to elaborate if necessary.<br />
<br />
==Solution 1.5 (Little bit of coordinate bash)==<br />
<br />
After using Pythagorean to find <math>AD</math> and <math>CD</math>, we can instead notice that the angle between the y-coordinate and <math>CD</math> is <math>45</math> degrees, and implies that the slope of that line is 1. If we draw a perpendicular from point <math>D</math>, we can then proceed to find the height and base of this new triangle (defined by <math>ADE</math> where <math>E</math> is the intersection of the altitude and <math>AB</math>) by coordinate-bashing, which turns out to be <math>1+\frac{\sqrt{2}}{4}</math> and <math>1-\frac{\sqrt{2}}{4}</math> respectively.<br />
<br />
By double angle formula and difference of squares, it's easy to see that our answer is <math>\boxed{\textbf{(D) }\frac{7}{9}}</math><br />
<br />
~Solution by MagentaCobra<br />
<br />
==Solution 2==<br />
Let <math>x = \angle BAC</math> and <math>y = \angle CAD</math>, so <math>\angle BAD = x+y</math>.<br />
<br />
By the double-angle formula, <math>\sin(2\angle BAD)= 2\sin(\angle BAD)\cos(\angle BAD)</math>.<br />
To write this in terms of <math>x</math> and <math>y</math>, we can say that we are looking for <math>2\sin(x+y)\cos(x+y)</math>.<br />
<br />
Using trigonometric addition and subtraction formulas, we know that<br />
<br />
<math>\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)</math><br />
<math>= (\dfrac{1}{\sqrt{2}})(\dfrac{\sqrt{2}}{AD}) + (\dfrac{1}{\sqrt{2}})(\dfrac{CD}{AD})</math><br />
<math>= \dfrac{1}{AD} + \dfrac{CD}{AD\sqrt{2}}</math><br />
<br />
and<br />
<br />
<math>\cos(x+y) = \cos(x)\cos(y) - \sin(x)\sin(y)</math>.<br />
<math>= (\dfrac{1}{\sqrt{2}})(\dfrac{\sqrt{2}}{AD}) - (\dfrac{1}{\sqrt{2}})(\dfrac{CD}{AD})</math><br />
<math>= \dfrac{1}{AD} - \dfrac{CD}{AD\sqrt{2}}</math>.<br />
<br />
So <math>2\sin(x+y)\cos(x+y) = 2[\dfrac{1}{AD} + \dfrac{CD}{AD\sqrt{2}}][\dfrac{1}{AD} - \dfrac{CD}{AD\sqrt{2}}]</math><br />
<math>= \dfrac{2-CD^2}{AD^2}</math>.<br />
Now we just need to figure out what the numerical answer is.<br />
<br />
From the given information about the triangles' perimeters, we can deduce that <math>CD + AD = 2</math>.<br />
Also, the Pythagorean theorem tell us that <math>CD^2 + 2 = AD^2</math>.<br />
These two equations allow us to write <math>CD and CD^2</math> in terms of <math>AD</math> without redundancy: <math>CD = 2 - AD</math> and <math>CD^2 = AD^2 - 2</math>.<br />
Plugging these into <math>\dfrac{2-CD^2}{AD^2}</math>, we'll get<br />
<math>\dfrac{2-CD^2}{AD^2} = \dfrac{2-(2-AD)^2}{AD^2} = \dfrac{-2+4AD-AD^2}{AD^2}</math> and<br />
<math>\dfrac{2-CD^2}{AD^2} = \dfrac{2-(AD^2-2)}{AD^2} = \dfrac{4-AD^2}{AD^2}</math>.<br />
<br />
==See Also==<br />
{{AMC12 box|year=2019|ab=B|num-b=11|num-a=13}}</div>Yem567https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12B_Problems/Problem_12&diff=1027762019 AMC 12B Problems/Problem 122019-02-15T03:37:12Z<p>Yem567: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
<br />
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>?<br />
<asy><br />
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br />
import graph; size(8.016233639805293cm); <br />
real labelscalefactor = 0.5; /* changes label-to-point distance */<br />
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br />
pen dotstyle = black; /* point style */ <br />
real xmin = -4.001920114613276, xmax = 4.014313525192017, ymin = -2.552570341575814, ymax = 5.6249093771911145; /* image dimensions */<br />
<br />
<br />
draw((-1.6742337260757447,-1.)--(-1.6742337260757445,-0.6742337260757447)--(-2.,-0.6742337260757447)--(-2.,-1.)--cycle, linewidth(2.)); <br />
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label("$A$",(1.9411496528285788,-1.0783204767840298),SE*labelscalefactor,fontsize(14)); <br />
label("$B$",(-2.5046350956841272,-0.9861798602345433),SE*labelscalefactor,fontsize(14)); <br />
label("$C$",(-2.5737405580962416,3.5747806589650395),SE*labelscalefactor,fontsize(14)); <br />
label("$1$",(-2.665881174645728,1.2712652452278765),SE*labelscalefactor,fontsize(14)); <br />
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dot((-2.,3.),linewidth(4.pt) + dotstyle); <br />
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dot((2.,-1.),linewidth(4.pt) + dotstyle); <br />
dot((-0.6404058554606791,4.3595941445393205),linewidth(4.pt) + dotstyle); <br />
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br />
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<br />
<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><br />
<br />
<br />
==Solution 1==<br />
<br />
Observe that the "equal perimeter" part implies that <math>BC + BA = 2 = CD + DA</math>. A quick Pythagorean chase gives <math>CD = \frac{1}{2}, DA = \frac{3}{2}</math>.<br />
Use the sine addition formula on angles <math>BAC</math> and <math>CAD</math> (which requires finding their cosines as well), and this gives the sine of <math>BAD</math>. Now, use <math>\sin{2x} = 2\sin{x}\cos{x}</math> on angle <math>BAD</math> to get <math>\boxed{\textbf{(D)} = \frac{7}{9}}</math>.<br />
<br />
Feel free to elaborate if necessary.<br />
<br />
==Solution 1.5 (Little bit of coordinate bash)==<br />
<br />
After using Pythagorean to find <math>AD</math> and <math>CD</math>, we can instead notice that the angle between the y-coordinate and <math>CD</math> is <math>45</math> degrees, and implies that the slope of that line is 1. If we draw a perpendicular from point <math>D</math>, we can then proceed to find the height and base of this new triangle (defined by <math>ADE</math> where <math>E</math> is the intersection of the altitude and <math>AB</math>) by coordinate-bashing, which turns out to be <math>1+\frac{\sqrt{2}}{4}</math> and <math>1-\frac{\sqrt{2}}{4}</math> respectively.<br />
<br />
By double angle formula and difference of squares, it's easy to see that our answer is <math>\boxed{\textbf{(D) }\frac{7}{9}}</math><br />
<br />
~Solution by MagentaCobra<br />
<br />
==Solution 2==<br />
Let <math>x = \angle BAC</math> and <math>y = \angle CAD</math>, so <math>\angle BAD = x+y</math>.<br />
<br />
By the double-angle formula, <math>\sin(2\angle BAD)= 2\sin(\angle BAD)\cos(\angle BAD)</math>.<br />
To write this in terms of <math>x</math> and <math>y</math>, we can say that we are looking for <math>2\sin(x+y)\cos(x+y)</math>.<br />
<br />
Using trigonometric addition and subtraction formulas, we know that<br />
<br />
<math>\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)</math><br />
<math>= (\dfrac{1}{\sqrt{2}})(\dfrac{\sqrt{2}}{AD}) + (\dfrac{1}{\sqrt{2}})(\dfrac{CD}{AD})</math><br />
<math>= \dfrac{1}{AD} + \dfrac{CD}{AD\sqrt{2}}</math><br />
<br />
and<br />
<br />
<math>\cos(x+y) = \cos(x)\cos(y) - \sin(x)\sin(y)</math>.<br />
<math>= (\dfrac{1}{\sqrt{2}})(\dfrac{\sqrt{2}}{AD}) - (\dfrac{1}{\sqrt{2}})(\dfrac{CD}{AD})</math><br />
<math>= \dfrac{1}{AD} - \dfrac{CD}{AD\sqrt{2}}</math>.<br />
<br />
So <math>2\sin(x+y)\cos(x+y) = 2[\dfrac{1}{AD} + \dfrac{CD}{AD\sqrt{2}}][\dfrac{1}{AD} - \dfrac{CD}{AD\sqrt{2}}]</math><br />
<math>= \dfrac{2-CD^2}{AD^2}</math>.<br />
<br />
==See Also==<br />
{{AMC12 box|year=2019|ab=B|num-b=11|num-a=13}}</div>Yem567https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12B_Problems/Problem_12&diff=1027742019 AMC 12B Problems/Problem 122019-02-15T03:33:45Z<p>Yem567: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
<br />
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>?<br />
<asy><br />
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br />
import graph; size(8.016233639805293cm); <br />
real labelscalefactor = 0.5; /* changes label-to-point distance */<br />
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br />
pen dotstyle = black; /* point style */ <br />
real xmin = -4.001920114613276, xmax = 4.014313525192017, ymin = -2.552570341575814, ymax = 5.6249093771911145; /* image dimensions */<br />
<br />
<br />
draw((-1.6742337260757447,-1.)--(-1.6742337260757445,-0.6742337260757447)--(-2.,-0.6742337260757447)--(-2.,-1.)--cycle, linewidth(2.)); <br />
draw((-1.7696484586262846,2.7696484586262846)--(-1.5392969172525692,3.)--(-1.7696484586262846,3.2303515413737154)--(-2.,3.)--cycle, linewidth(2.)); <br />
/* draw figures */<br />
draw((-2.,3.)--(-2.,-1.), linewidth(2.)); <br />
draw((-2.,-1.)--(2.,-1.), linewidth(2.)); <br />
draw((2.,-1.)--(-2.,3.), linewidth(2.)); <br />
draw((-0.6404058554606791,4.3595941445393205)--(-2.,3.), linewidth(2.)); <br />
draw((-0.6404058554606791,4.3595941445393205)--(2.,-1.), linewidth(2.)); <br />
label("$D$",(-0.9382446143428628,4.887784444795223),SE*labelscalefactor,fontsize(14)); <br />
label("$A$",(1.9411496528285788,-1.0783204767840298),SE*labelscalefactor,fontsize(14)); <br />
label("$B$",(-2.5046350956841272,-0.9861798602345433),SE*labelscalefactor,fontsize(14)); <br />
label("$C$",(-2.5737405580962416,3.5747806589650395),SE*labelscalefactor,fontsize(14)); <br />
label("$1$",(-2.665881174645728,1.2712652452278765),SE*labelscalefactor,fontsize(14)); <br />
label("$1$",(-0.3393306067712029,-1.3547423264324894),SE*labelscalefactor,fontsize(14)); <br />
/* dots and labels */<br />
dot((-2.,3.),linewidth(4.pt) + dotstyle); <br />
dot((-2.,-1.),linewidth(4.pt) + dotstyle); <br />
dot((2.,-1.),linewidth(4.pt) + dotstyle); <br />
dot((-0.6404058554606791,4.3595941445393205),linewidth(4.pt) + dotstyle); <br />
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br />
/* end of picture */<br />
</asy><br />
<br />
<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><br />
<br />
<br />
==Solution 1==<br />
<br />
Observe that the "equal perimeter" part implies that <math>BC + BA = 2 = CD + DA</math>. A quick Pythagorean chase gives <math>CD = \frac{1}{2}, DA = \frac{3}{2}</math>.<br />
Use the sine addition formula on angles <math>BAC</math> and <math>CAD</math> (which requires finding their cosines as well), and this gives the sine of <math>BAD</math>. Now, use <math>\sin{2x} = 2\sin{x}\cos{x}</math> on angle <math>BAD</math> to get <math>\boxed{\textbf{(D)} = \frac{7}{9}}</math>.<br />
<br />
Feel free to elaborate if necessary.<br />
<br />
==Solution 1.5 (Little bit of coordinate bash)==<br />
<br />
After using Pythagorean to find <math>AD</math> and <math>CD</math>, we can instead notice that the angle between the y-coordinate and <math>CD</math> is <math>45</math> degrees, and implies that the slope of that line is 1. If we draw a perpendicular from point <math>D</math>, we can then proceed to find the height and base of this new triangle (defined by <math>ADE</math> where <math>E</math> is the intersection of the altitude and <math>AB</math>) by coordinate-bashing, which turns out to be <math>1+\frac{\sqrt{2}}{4}</math> and <math>1-\frac{\sqrt{2}}{4}</math> respectively.<br />
<br />
By double angle formula and difference of squares, it's easy to see that our answer is <math>\boxed{\textbf{(D) }\frac{7}{9}}</math><br />
<br />
~Solution by MagentaCobra<br />
<br />
==Solution 2==<br />
Let <math>x = \angle BAC</math> and <math>y = \angle CAD</math>, so <math>\angle BAD = x+y</math>.<br />
<br />
By the double-angle formula, <math>\sin(2\angle BAD)= 2\sin(\angle BAD)\cos(\angle BAD)</math>.<br />
To write this in terms of <math>x</math> and <math>y</math>, we can say that we are looking for <math>2\sin(x+y)\cos(x+y)</math>.<br />
<br />
Using trigonometric addition and subtraction formulas, we know that<br />
<br />
<math>\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)</math><br />
<math>= (\dfrac{1}{\sqrt{2}})(\dfrac{\sqrt{2}}{AD}) + (\dfrac{1}{\sqrt{2}})(\dfrac{CD}{AD})</math><br />
<math>= \dfrac{1}{AD} + \dfrac{CD}{AD\sqrt{2}}</math><br />
<br />
and<br />
<br />
<math>\cos(x+y) = \cos(x)\cos(y) - \sin(x)\sin(y)</math>.<br />
<math>= (\dfrac{1}{\sqrt{2}})(\dfrac{\sqrt{2}}{AD}) - (\dfrac{1}{\sqrt{2}})(\dfrac{CD}{AD})</math><br />
<math>= \dfrac{1}{AD} - \dfrac{CD}{AD\sqrt{2}}</math>.<br />
<br />
So <math>2\sin(x+y)\cos(x+y) = [\dfrac{1}{AD} + \dfrac{CD}{AD\sqrt{2}}][\dfrac{1}{AD} - \dfrac{CD}{AD\sqrt{2}}]</math>.<br />
<br />
==See Also==<br />
{{AMC12 box|year=2019|ab=B|num-b=11|num-a=13}}</div>Yem567https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12B_Problems/Problem_12&diff=1027732019 AMC 12B Problems/Problem 122019-02-15T03:32:48Z<p>Yem567: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
<br />
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>?<br />
<asy><br />
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br />
import graph; size(8.016233639805293cm); <br />
real labelscalefactor = 0.5; /* changes label-to-point distance */<br />
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br />
pen dotstyle = black; /* point style */ <br />
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<br />
<br />
draw((-1.6742337260757447,-1.)--(-1.6742337260757445,-0.6742337260757447)--(-2.,-0.6742337260757447)--(-2.,-1.)--cycle, linewidth(2.)); <br />
draw((-1.7696484586262846,2.7696484586262846)--(-1.5392969172525692,3.)--(-1.7696484586262846,3.2303515413737154)--(-2.,3.)--cycle, linewidth(2.)); <br />
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label("$1$",(-2.665881174645728,1.2712652452278765),SE*labelscalefactor,fontsize(14)); <br />
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dot((-0.6404058554606791,4.3595941445393205),linewidth(4.pt) + dotstyle); <br />
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br />
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</asy><br />
<br />
<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><br />
<br />
<br />
==Solution 1==<br />
<br />
Observe that the "equal perimeter" part implies that <math>BC + BA = 2 = CD + DA</math>. A quick Pythagorean chase gives <math>CD = \frac{1}{2}, DA = \frac{3}{2}</math>.<br />
Use the sine addition formula on angles <math>BAC</math> and <math>CAD</math> (which requires finding their cosines as well), and this gives the sine of <math>BAD</math>. Now, use <math>\sin{2x} = 2\sin{x}\cos{x}</math> on angle <math>BAD</math> to get <math>\boxed{\textbf{(D)} = \frac{7}{9}}</math>.<br />
<br />
Feel free to elaborate if necessary.<br />
<br />
==Solution 1.5 (Little bit of coordinate bash)==<br />
<br />
After using Pythagorean to find <math>AD</math> and <math>CD</math>, we can instead notice that the angle between the y-coordinate and <math>CD</math> is <math>45</math> degrees, and implies that the slope of that line is 1. If we draw a perpendicular from point <math>D</math>, we can then proceed to find the height and base of this new triangle (defined by <math>ADE</math> where <math>E</math> is the intersection of the altitude and <math>AB</math>) by coordinate-bashing, which turns out to be <math>1+\frac{\sqrt{2}}{4}</math> and <math>1-\frac{\sqrt{2}}{4}</math> respectively.<br />
<br />
By double angle formula and difference of squares, it's easy to see that our answer is <math>\boxed{\textbf{(D) }\frac{7}{9}}</math><br />
<br />
~Solution by MagentaCobra<br />
<br />
==Solution 2==<br />
Let <math>x = \angle BAC</math> and <math>y = \angle CAD</math>, so <math>\angle BAD = x+y</math>.<br />
<br />
By the double-angle formula, <math>\sin(2\angle BAD)= 2\sin(\angle BAD)\cos(\angle BAD)</math>.<br />
To write this in terms of <math>x</math> and <math>y</math>, we can say that we are looking for <math>2\sin(x+y)\cos(x+y)</math>.<br />
<br />
Using trigonometric addition and subtraction formulas, we know that<br />
<br />
<math>\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)</math><br />
<math>= (\dfrac{1}{\sqrt{2}})(\dfrac{\sqrt{2}}{AD}) + (\dfrac{1}{\sqrt{2}})(\dfrac{CD}{AD})</math><br />
<math>= \dfrac{1}{AD} + \dfrac{CD}{AD\sqrt{2}}</math><br />
<br />
and<br />
<br />
<math>\cos(x+y) = \cos(x)\cos(y) - \sin(x)\sin(y)</math>.<br />
<math>= (\dfrac{1}{\sqrt{2}})(\dfrac{\sqrt{2}}{AD}) - (\dfrac{1}{\sqrt{2}})(\dfrac{CD}{AD})</math><br />
<math>= \dfrac{1}{AD} - \dfrac{CD}{AD\sqrt{2}}</math>.<br />
<br />
==See Also==<br />
{{AMC12 box|year=2019|ab=B|num-b=11|num-a=13}}</div>Yem567https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12B_Problems/Problem_12&diff=1027722019 AMC 12B Problems/Problem 122019-02-15T03:30:35Z<p>Yem567: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
<br />
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>?<br />
<asy><br />
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br />
import graph; size(8.016233639805293cm); <br />
real labelscalefactor = 0.5; /* changes label-to-point distance */<br />
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br />
pen dotstyle = black; /* point style */ <br />
real xmin = -4.001920114613276, xmax = 4.014313525192017, ymin = -2.552570341575814, ymax = 5.6249093771911145; /* image dimensions */<br />
<br />
<br />
draw((-1.6742337260757447,-1.)--(-1.6742337260757445,-0.6742337260757447)--(-2.,-0.6742337260757447)--(-2.,-1.)--cycle, linewidth(2.)); <br />
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draw((2.,-1.)--(-2.,3.), linewidth(2.)); <br />
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label("$D$",(-0.9382446143428628,4.887784444795223),SE*labelscalefactor,fontsize(14)); <br />
label("$A$",(1.9411496528285788,-1.0783204767840298),SE*labelscalefactor,fontsize(14)); <br />
label("$B$",(-2.5046350956841272,-0.9861798602345433),SE*labelscalefactor,fontsize(14)); <br />
label("$C$",(-2.5737405580962416,3.5747806589650395),SE*labelscalefactor,fontsize(14)); <br />
label("$1$",(-2.665881174645728,1.2712652452278765),SE*labelscalefactor,fontsize(14)); <br />
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dot((-2.,-1.),linewidth(4.pt) + dotstyle); <br />
dot((2.,-1.),linewidth(4.pt) + dotstyle); <br />
dot((-0.6404058554606791,4.3595941445393205),linewidth(4.pt) + dotstyle); <br />
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br />
/* end of picture */<br />
</asy><br />
<br />
<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><br />
<br />
<br />
==Solution 1==<br />
<br />
Observe that the "equal perimeter" part implies that <math>BC + BA = 2 = CD + DA</math>. A quick Pythagorean chase gives <math>CD = \frac{1}{2}, DA = \frac{3}{2}</math>.<br />
Use the sine addition formula on angles <math>BAC</math> and <math>CAD</math> (which requires finding their cosines as well), and this gives the sine of <math>BAD</math>. Now, use <math>\sin{2x} = 2\sin{x}\cos{x}</math> on angle <math>BAD</math> to get <math>\boxed{\textbf{(D)} = \frac{7}{9}}</math>.<br />
<br />
Feel free to elaborate if necessary.<br />
<br />
==Solution 1.5 (Little bit of coordinate bash)==<br />
<br />
After using Pythagorean to find <math>AD</math> and <math>CD</math>, we can instead notice that the angle between the y-coordinate and <math>CD</math> is <math>45</math> degrees, and implies that the slope of that line is 1. If we draw a perpendicular from point <math>D</math>, we can then proceed to find the height and base of this new triangle (defined by <math>ADE</math> where <math>E</math> is the intersection of the altitude and <math>AB</math>) by coordinate-bashing, which turns out to be <math>1+\frac{\sqrt{2}}{4}</math> and <math>1-\frac{\sqrt{2}}{4}</math> respectively.<br />
<br />
By double angle formula and difference of squares, it's easy to see that our answer is <math>\boxed{\textbf{(D) }\frac{7}{9}}</math><br />
<br />
~Solution by MagentaCobra<br />
<br />
==Solution 2==<br />
Let <math>x = \angle BAC</math> and <math>y = \angle CAD</math>, so <math>\angle BAD = x+y</math>.<br />
<br />
By the double-angle formula, <math>\sin(2\angle BAD)= 2\sin(\angle BAD)\cos(\angle BAD)</math>.<br />
To write this in terms of <math>x</math> and <math>y</math>, we can say that we are looking for <math>2\sin(x+y)\cos(x+y)</math>.<br />
<br />
Using trigonometric addition and subtraction formulas, we know that<br />
<br />
<math>\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)</math><br />
<math>= (\dfrac{1}{\sqrt{2}})(\dfrac{\sqrt{2}}{AD}) + (\dfrac{1}{\sqrt{2}})(\dfrac{CD}{AD})</math><br />
<math>= \dfrac{1}{AD} + \dfrac{CD}{AD\sqrt{2}}</math><br />
<br />
, and<br />
<math>\cos(x+y) = \cos(x)\cos(y) - \sin(x)\sin(y)</math>.<br />
<br />
==See Also==<br />
{{AMC12 box|year=2019|ab=B|num-b=11|num-a=13}}</div>Yem567https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12B_Problems/Problem_12&diff=1027712019 AMC 12B Problems/Problem 122019-02-15T03:30:19Z<p>Yem567: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
<br />
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>?<br />
<asy><br />
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br />
import graph; size(8.016233639805293cm); <br />
real labelscalefactor = 0.5; /* changes label-to-point distance */<br />
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br />
pen dotstyle = black; /* point style */ <br />
real xmin = -4.001920114613276, xmax = 4.014313525192017, ymin = -2.552570341575814, ymax = 5.6249093771911145; /* image dimensions */<br />
<br />
<br />
draw((-1.6742337260757447,-1.)--(-1.6742337260757445,-0.6742337260757447)--(-2.,-0.6742337260757447)--(-2.,-1.)--cycle, linewidth(2.)); <br />
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draw((2.,-1.)--(-2.,3.), linewidth(2.)); <br />
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label("$B$",(-2.5046350956841272,-0.9861798602345433),SE*labelscalefactor,fontsize(14)); <br />
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label("$1$",(-2.665881174645728,1.2712652452278765),SE*labelscalefactor,fontsize(14)); <br />
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dot((-2.,3.),linewidth(4.pt) + dotstyle); <br />
dot((-2.,-1.),linewidth(4.pt) + dotstyle); <br />
dot((2.,-1.),linewidth(4.pt) + dotstyle); <br />
dot((-0.6404058554606791,4.3595941445393205),linewidth(4.pt) + dotstyle); <br />
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br />
/* end of picture */<br />
</asy><br />
<br />
<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><br />
<br />
<br />
==Solution 1==<br />
<br />
Observe that the "equal perimeter" part implies that <math>BC + BA = 2 = CD + DA</math>. A quick Pythagorean chase gives <math>CD = \frac{1}{2}, DA = \frac{3}{2}</math>.<br />
Use the sine addition formula on angles <math>BAC</math> and <math>CAD</math> (which requires finding their cosines as well), and this gives the sine of <math>BAD</math>. Now, use <math>\sin{2x} = 2\sin{x}\cos{x}</math> on angle <math>BAD</math> to get <math>\boxed{\textbf{(D)} = \frac{7}{9}}</math>.<br />
<br />
Feel free to elaborate if necessary.<br />
<br />
==Solution 1.5 (Little bit of coordinate bash)==<br />
<br />
After using Pythagorean to find <math>AD</math> and <math>CD</math>, we can instead notice that the angle between the y-coordinate and <math>CD</math> is <math>45</math> degrees, and implies that the slope of that line is 1. If we draw a perpendicular from point <math>D</math>, we can then proceed to find the height and base of this new triangle (defined by <math>ADE</math> where <math>E</math> is the intersection of the altitude and <math>AB</math>) by coordinate-bashing, which turns out to be <math>1+\frac{\sqrt{2}}{4}</math> and <math>1-\frac{\sqrt{2}}{4}</math> respectively.<br />
<br />
By double angle formula and difference of squares, it's easy to see that our answer is <math>\boxed{\textbf{(D) }\frac{7}{9}}</math><br />
<br />
~Solution by MagentaCobra<br />
<br />
==Solution 2==<br />
Let <math>x = \angle BAC</math> and <math>y = \angle CAD</math>, so <math>\angle BAD = x+y</math>.<br />
<br />
By the double-angle formula, <math>\sin(2\angle BAD)= 2\sin(\angle BAD)\cos(\angle BAD)</math>.<br />
To write this in terms of <math>x</math> and <math>y</math>, we can say that we are looking for <math>2\sin(x+y)\cos(x+y)</math>.<br />
<br />
Using trigonometric addition and subtraction formulas, we know that<br />
<br />
<math>\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)</math><br />
<math>= (\dfrac{1}{\sqrt{2}})(\dfrac{\sqrt{2}}{AD}) + (\dfrac{1}{\sqrt{2}})(\dfrac{CD}{AD})</math><br />
<math>= \dfrac{1}{AD} + \dfrac{CD}{AD\sqrt{2}}<br />
<br />
, and<br />
</math>\cos(x+y) = \cos(x)\cos(y) - \sin(x)\sin(y)$.<br />
<br />
==See Also==<br />
{{AMC12 box|year=2019|ab=B|num-b=11|num-a=13}}</div>Yem567https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12B_Problems/Problem_12&diff=1027662019 AMC 12B Problems/Problem 122019-02-15T03:25:01Z<p>Yem567: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
<br />
Right triangle <math>ACD</math> with right angle at <math>C</math> is constructed outwards on the hypotenuse <math>\overline{AC}</math> of isosceles right triangle <math>ABC</math> with leg length <math>1</math>, as shown, so that the two triangles have equal perimeters. What is <math>\sin(2\angle BAD)</math>?<br />
<asy><br />
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br />
import graph; size(8.016233639805293cm); <br />
real labelscalefactor = 0.5; /* changes label-to-point distance */<br />
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br />
pen dotstyle = black; /* point style */ <br />
real xmin = -4.001920114613276, xmax = 4.014313525192017, ymin = -2.552570341575814, ymax = 5.6249093771911145; /* image dimensions */<br />
<br />
<br />
draw((-1.6742337260757447,-1.)--(-1.6742337260757445,-0.6742337260757447)--(-2.,-0.6742337260757447)--(-2.,-1.)--cycle, linewidth(2.)); <br />
draw((-1.7696484586262846,2.7696484586262846)--(-1.5392969172525692,3.)--(-1.7696484586262846,3.2303515413737154)--(-2.,3.)--cycle, linewidth(2.)); <br />
/* draw figures */<br />
draw((-2.,3.)--(-2.,-1.), linewidth(2.)); <br />
draw((-2.,-1.)--(2.,-1.), linewidth(2.)); <br />
draw((2.,-1.)--(-2.,3.), linewidth(2.)); <br />
draw((-0.6404058554606791,4.3595941445393205)--(-2.,3.), linewidth(2.)); <br />
draw((-0.6404058554606791,4.3595941445393205)--(2.,-1.), linewidth(2.)); <br />
label("$D$",(-0.9382446143428628,4.887784444795223),SE*labelscalefactor,fontsize(14)); <br />
label("$A$",(1.9411496528285788,-1.0783204767840298),SE*labelscalefactor,fontsize(14)); <br />
label("$B$",(-2.5046350956841272,-0.9861798602345433),SE*labelscalefactor,fontsize(14)); <br />
label("$C$",(-2.5737405580962416,3.5747806589650395),SE*labelscalefactor,fontsize(14)); <br />
label("$1$",(-2.665881174645728,1.2712652452278765),SE*labelscalefactor,fontsize(14)); <br />
label("$1$",(-0.3393306067712029,-1.3547423264324894),SE*labelscalefactor,fontsize(14)); <br />
/* dots and labels */<br />
dot((-2.,3.),linewidth(4.pt) + dotstyle); <br />
dot((-2.,-1.),linewidth(4.pt) + dotstyle); <br />
dot((2.,-1.),linewidth(4.pt) + dotstyle); <br />
dot((-0.6404058554606791,4.3595941445393205),linewidth(4.pt) + dotstyle); <br />
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br />
/* end of picture */<br />
</asy><br />
<br />
<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><br />
<br />
<br />
==Solution 1==<br />
<br />
Observe that the "equal perimeter" part implies that <math>BC + BA = 2 = CD + DA</math>. A quick Pythagorean chase gives <math>CD = \frac{1}{2}, DA = \frac{3}{2}</math>.<br />
Use the sine addition formula on angles <math>BAC</math> and <math>CAD</math> (which requires finding their cosines as well), and this gives the sine of <math>BAD</math>. Now, use <math>\sin{2x} = 2\sin{x}\cos{x}</math> on angle <math>BAD</math> to get <math>\boxed{\textbf{(D)} = \frac{7}{9}}</math>.<br />
<br />
Feel free to elaborate if necessary.<br />
<br />
==Solution 1.5 (Little bit of coordinate bash)==<br />
<br />
After using Pythagorean to find <math>AD</math> and <math>CD</math>, we can instead notice that the angle between the y-coordinate and <math>CD</math> is <math>45</math> degrees, and implies that the slope of that line is 1. If we draw a perpendicular from point <math>D</math>, we can then proceed to find the height and base of this new triangle (defined by <math>ADE</math> where <math>E</math> is the intersection of the altitude and <math>AB</math>) by coordinate-bashing, which turns out to be <math>1+\frac{\sqrt{2}}{4}</math> and <math>1-\frac{\sqrt{2}}{4}</math> respectively.<br />
<br />
By double angle formula and difference of squares, it's easy to see that our answer is <math>\boxed{\textbf{(D) }\frac{7}{9}}</math><br />
<br />
~Solution by MagentaCobra<br />
<br />
==Solution 2==<br />
Let <math>x = \angle BAC</math> and <math>y = \angle CAD</math>, so <math>\angle BAD = x+y</math>.<br />
<br />
By the double-angle formula, <math>\sin(2\angle BAD)= 2\sin(\angle BAD)\cos(\angle BAD)</math>.<br />
To write this in terms of <math>x</math> and <math>y</math>, we can say that we are looking for <math>2\sin(x+y)\cos(x+y)</math>.<br />
<br />
Using trigonometric addition and subtraction formulas, we know that<br />
<math>\sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y)</math>, and<br />
<math>\cos(x+y) = \cos(x)\cos(y) - \sin(x)\sin(y)</math>.<br />
<br />
==See Also==<br />
{{AMC12 box|year=2019|ab=B|num-b=11|num-a=13}}</div>Yem567https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12B_Problems/Problem_12&diff=1027632019 AMC 12B Problems/Problem 122019-02-15T03:18:37Z<p>Yem567: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
<br />
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>?<br />
<asy><br />
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br />
import graph; size(8.016233639805293cm); <br />
real labelscalefactor = 0.5; /* changes label-to-point distance */<br />
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br />
pen dotstyle = black; /* point style */ <br />
real xmin = -4.001920114613276, xmax = 4.014313525192017, ymin = -2.552570341575814, ymax = 5.6249093771911145; /* image dimensions */<br />
<br />
<br />
draw((-1.6742337260757447,-1.)--(-1.6742337260757445,-0.6742337260757447)--(-2.,-0.6742337260757447)--(-2.,-1.)--cycle, linewidth(2.)); <br />
draw((-1.7696484586262846,2.7696484586262846)--(-1.5392969172525692,3.)--(-1.7696484586262846,3.2303515413737154)--(-2.,3.)--cycle, linewidth(2.)); <br />
/* draw figures */<br />
draw((-2.,3.)--(-2.,-1.), linewidth(2.)); <br />
draw((-2.,-1.)--(2.,-1.), linewidth(2.)); <br />
draw((2.,-1.)--(-2.,3.), linewidth(2.)); <br />
draw((-0.6404058554606791,4.3595941445393205)--(-2.,3.), linewidth(2.)); <br />
draw((-0.6404058554606791,4.3595941445393205)--(2.,-1.), linewidth(2.)); <br />
label("$D$",(-0.9382446143428628,4.887784444795223),SE*labelscalefactor,fontsize(14)); <br />
label("$A$",(1.9411496528285788,-1.0783204767840298),SE*labelscalefactor,fontsize(14)); <br />
label("$B$",(-2.5046350956841272,-0.9861798602345433),SE*labelscalefactor,fontsize(14)); <br />
label("$C$",(-2.5737405580962416,3.5747806589650395),SE*labelscalefactor,fontsize(14)); <br />
label("$1$",(-2.665881174645728,1.2712652452278765),SE*labelscalefactor,fontsize(14)); <br />
label("$1$",(-0.3393306067712029,-1.3547423264324894),SE*labelscalefactor,fontsize(14)); <br />
/* dots and labels */<br />
dot((-2.,3.),linewidth(4.pt) + dotstyle); <br />
dot((-2.,-1.),linewidth(4.pt) + dotstyle); <br />
dot((2.,-1.),linewidth(4.pt) + dotstyle); <br />
dot((-0.6404058554606791,4.3595941445393205),linewidth(4.pt) + dotstyle); <br />
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br />
/* end of picture */<br />
</asy><br />
<br />
<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><br />
<br />
<br />
==Solution 1==<br />
<br />
Observe that the "equal perimeter" part implies that <math>BC + BA = 2 = CD + DA</math>. A quick Pythagorean chase gives <math>CD = \frac{1}{2}, DA = \frac{3}{2}</math>.<br />
Use the sine addition formula on angles <math>BAC</math> and <math>CAD</math> (which requires finding their cosines as well), and this gives the sine of <math>BAD</math>. Now, use <math>\sin{2x} = 2\sin{x}\cos{x}</math> on angle <math>BAD</math> to get <math>\boxed{\textbf{(D)} = \frac{7}{9}}</math>.<br />
<br />
Feel free to elaborate if necessary.<br />
<br />
==Solution 1.5 (Little bit of coordinate bash)==<br />
<br />
After using Pythagorean to find <math>AD</math> and <math>CD</math>, we can instead notice that the angle between the y-coordinate and <math>CD</math> is <math>45</math> degrees, and implies that the slope of that line is 1. If we draw a perpendicular from point <math>D</math>, we can then proceed to find the height and base of this new triangle (defined by <math>ADE</math> where <math>E</math> is the intersection of the altitude and <math>AB</math>) by coordinate-bashing, which turns out to be <math>1+\frac{\sqrt{2}}{4}</math> and <math>1-\frac{\sqrt{2}}{4}</math> respectively.<br />
<br />
By double angle formula and difference of squares, it's easy to see that our answer is <math>\boxed{\textbf{(D) }\frac{7}{9}}</math><br />
<br />
~Solution by MagentaCobra<br />
<br />
==Solution 2==<br />
Let <math>x = \angle BAC</math> and <math>y = \angle CAD</math>, so <math>\angle BAD = x+y</math>.<br />
<br />
By the double-angle formula, <math>\sin(2\angle BAD)= 2\sin(\angle BAD)\cos(\angle BAD)</math>.<br />
<br />
==See Also==<br />
{{AMC12 box|year=2019|ab=B|num-b=11|num-a=13}}</div>Yem567https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_12B_Problems/Problem_12&diff=1027622019 AMC 12B Problems/Problem 122019-02-15T03:17:40Z<p>Yem567: /* Solution 2 */</p>
<hr />
<div>==Problem==<br />
<br />
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>?<br />
<asy><br />
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */<br />
import graph; size(8.016233639805293cm); <br />
real labelscalefactor = 0.5; /* changes label-to-point distance */<br />
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br />
pen dotstyle = black; /* point style */ <br />
real xmin = -4.001920114613276, xmax = 4.014313525192017, ymin = -2.552570341575814, ymax = 5.6249093771911145; /* image dimensions */<br />
<br />
<br />
draw((-1.6742337260757447,-1.)--(-1.6742337260757445,-0.6742337260757447)--(-2.,-0.6742337260757447)--(-2.,-1.)--cycle, linewidth(2.)); <br />
draw((-1.7696484586262846,2.7696484586262846)--(-1.5392969172525692,3.)--(-1.7696484586262846,3.2303515413737154)--(-2.,3.)--cycle, linewidth(2.)); <br />
/* draw figures */<br />
draw((-2.,3.)--(-2.,-1.), linewidth(2.)); <br />
draw((-2.,-1.)--(2.,-1.), linewidth(2.)); <br />
draw((2.,-1.)--(-2.,3.), linewidth(2.)); <br />
draw((-0.6404058554606791,4.3595941445393205)--(-2.,3.), linewidth(2.)); <br />
draw((-0.6404058554606791,4.3595941445393205)--(2.,-1.), linewidth(2.)); <br />
label("$D$",(-0.9382446143428628,4.887784444795223),SE*labelscalefactor,fontsize(14)); <br />
label("$A$",(1.9411496528285788,-1.0783204767840298),SE*labelscalefactor,fontsize(14)); <br />
label("$B$",(-2.5046350956841272,-0.9861798602345433),SE*labelscalefactor,fontsize(14)); <br />
label("$C$",(-2.5737405580962416,3.5747806589650395),SE*labelscalefactor,fontsize(14)); <br />
label("$1$",(-2.665881174645728,1.2712652452278765),SE*labelscalefactor,fontsize(14)); <br />
label("$1$",(-0.3393306067712029,-1.3547423264324894),SE*labelscalefactor,fontsize(14)); <br />
/* dots and labels */<br />
dot((-2.,3.),linewidth(4.pt) + dotstyle); <br />
dot((-2.,-1.),linewidth(4.pt) + dotstyle); <br />
dot((2.,-1.),linewidth(4.pt) + dotstyle); <br />
dot((-0.6404058554606791,4.3595941445393205),linewidth(4.pt) + dotstyle); <br />
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br />
/* end of picture */<br />
</asy><br />
<br />
<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><br />
<br />
<br />
==Solution 1==<br />
<br />
Observe that the "equal perimeter" part implies that <math>BC + BA = 2 = CD + DA</math>. A quick Pythagorean chase gives <math>CD = \frac{1}{2}, DA = \frac{3}{2}</math>.<br />
Use the sine addition formula on angles <math>BAC</math> and <math>CAD</math> (which requires finding their cosines as well), and this gives the sine of <math>BAD</math>. Now, use <math>\sin{2x} = 2\sin{x}\cos{x}</math> on angle <math>BAD</math> to get <math>\boxed{\textbf{(D)} = \frac{7}{9}}</math>.<br />
<br />
Feel free to elaborate if necessary.<br />
<br />
==Solution 1.5 (Little bit of coordinate bash)==<br />
<br />
After using Pythagorean to find <math>AD</math> and <math>CD</math>, we can instead notice that the angle between the y-coordinate and <math>CD</math> is <math>45</math> degrees, and implies that the slope of that line is 1. If we draw a perpendicular from point <math>D</math>, we can then proceed to find the height and base of this new triangle (defined by <math>ADE</math> where <math>E</math> is the intersection of the altitude and <math>AB</math>) by coordinate-bashing, which turns out to be <math>1+\frac{\sqrt{2}}{4}</math> and <math>1-\frac{\sqrt{2}}{4}</math> respectively.<br />
<br />
By double angle formula and difference of squares, it's easy to see that our answer is <math>\boxed{\textbf{(D) }\frac{7}{9}}</math><br />
<br />
~Solution by MagentaCobra<br />
<br />
==Solution 2==<br />
Let <math>x = \angle BAC</math> and <math>y = /angle CAD</math>, so <math>/angle BAD = x+y</math>.<br />
<br />
By the double-angle formula, <math>\sin(2\angle BAD)= 2/sin(/angle BAD)/cos(/angle BAD)</math>.<br />
<br />
==See Also==<br />
{{AMC12 box|year=2019|ab=B|num-b=11|num-a=13}}</div>Yem567