Difference between revisions of "2019 AMC 12B Problems/Problem 12"
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==Problem== | ==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>? | |
+ | <asy> | ||
+ | /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ | ||
+ | import graph; size(8.016233639805293cm); | ||
+ | real labelscalefactor = 0.5; /* changes label-to-point distance */ | ||
+ | pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ | ||
+ | pen dotstyle = black; /* point style */ | ||
+ | real xmin = -4.001920114613276, xmax = 4.014313525192017, ymin = -2.552570341575814, ymax = 5.6249093771911145; /* image dimensions */ | ||
+ | |||
+ | |||
+ | draw((-1.6742337260757447,-1.)--(-1.6742337260757445,-0.6742337260757447)--(-2.,-0.6742337260757447)--(-2.,-1.)--cycle, linewidth(2.)); | ||
+ | draw((-1.7696484586262846,2.7696484586262846)--(-1.5392969172525692,3.)--(-1.7696484586262846,3.2303515413737154)--(-2.,3.)--cycle, linewidth(2.)); | ||
+ | /* draw figures */ | ||
+ | draw((-2.,3.)--(-2.,-1.), linewidth(2.)); | ||
+ | draw((-2.,-1.)--(2.,-1.), linewidth(2.)); | ||
+ | draw((2.,-1.)--(-2.,3.), linewidth(2.)); | ||
+ | draw((-0.6404058554606791,4.3595941445393205)--(-2.,3.), linewidth(2.)); | ||
+ | draw((-0.6404058554606791,4.3595941445393205)--(2.,-1.), linewidth(2.)); | ||
+ | label("$D$",(-0.9382446143428628,4.887784444795223),SE*labelscalefactor,fontsize(14)); | ||
+ | label("$A$",(1.9411496528285788,-1.0783204767840298),SE*labelscalefactor,fontsize(14)); | ||
+ | label("$B$",(-2.5046350956841272,-0.9861798602345433),SE*labelscalefactor,fontsize(14)); | ||
+ | label("$C$",(-2.5737405580962416,3.5747806589650395),SE*labelscalefactor,fontsize(14)); | ||
+ | label("$1$",(-2.665881174645728,1.2712652452278765),SE*labelscalefactor,fontsize(14)); | ||
+ | label("$1$",(-0.3393306067712029,-1.3547423264324894),SE*labelscalefactor,fontsize(14)); | ||
+ | /* dots and labels */ | ||
+ | dot((-2.,3.),linewidth(4.pt) + dotstyle); | ||
+ | dot((-2.,-1.),linewidth(4.pt) + dotstyle); | ||
+ | dot((2.,-1.),linewidth(4.pt) + dotstyle); | ||
+ | dot((-0.6404058554606791,4.3595941445393205),linewidth(4.pt) + dotstyle); | ||
+ | clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); | ||
+ | /* end of picture */ | ||
+ | </asy> | ||
+ | |||
+ | <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> | ||
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==Solution 1== | ==Solution 1== |
Revision as of 18:38, 14 February 2019
Contents
Problem
Right triangle with right angle at
is constructed outwards on the hypotenuse
of isosceles right triangle
with leg length
, as shown, so that the two triangles have equal perimeters. What is
?
Solution 1
Observe that the "equal perimeter" part implies that . A quick Pythagorean chase gives
.
Use the sine addition formula on angles
and
(which requires finding their cosines as well), and this gives the sine of
. Now, use
on angle
to get
.
Feel free to elaborate if necessary.
Solution 2
D 7/9 (SuperWill)
See Also
2019 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 11 |
Followed by Problem 13 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |