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 $\sin(2\angle BAD)?$

$\dfrac{1}{3}$

$\dfrac{\sqrt2}{2}$

$\dfrac{3}{4}$

$\dfrac{7}{9}$

$\dfrac{\sqrt3}{2}$

Solution:

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

Thus, D is the correct answer.

2019 AMC 12B Problem 12

Solution:

Problem 12 in Other Years