2019 AMC 12B Problem 12

Question

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)?

\dfrac{1}{3}

\dfrac{\sqrt2}{2}

\dfrac{3}{4}

\dfrac{7}{9}

\dfrac{\sqrt3}{2}

\triangle ABC

has legs

1

and hypotenuse

AC=\sqrt2;

set

CD=d

and use equal perimeters

Since

\angle BAC=45^\circ,

use

\sin(2\angle BAD)=\cos(2\angle CAD)

Triangle

ABC

has perimeter

1+1+\sqrt2=2+\sqrt2

and

AC=\sqrt2.

In

\triangle ACD

let

CD=d,

so

AD=\sqrt{2+d^2}

and equal perimeters give

\sqrt2+d+\sqrt{2+d^2}=2+\sqrt2.

Then

\sqrt{2+d^2}=2-d,

so

2+d^2=4-4d+d^2,

giving

d=\dfrac12

and

AD=\dfrac32.

Since

\angle BAC=45^\circ,

writing

\theta=\angle CAD

gives

2\angle BAD=90^\circ+2\theta,

so

\sin(2\angle BAD)=\cos 2\theta.

With

\tan\theta=\dfrac{CD}{AC}=\dfrac{1}{2\sqrt2},

we get

\begin{gathered} \cos2\theta=\dfrac{1-\tan^2\theta}{1+\tan^2\theta} \\ =\dfrac{1-\tfrac18}{1+\tfrac18}=\dfrac{7}{9}. \end{gathered}

Thus, D is the correct answer.

Problem 12 in other years