Question

Quadrilateral $ABCD$ has right angles at $B$ and $C,$ $\triangle ABC \sim \triangle BCD,$ and $AB \gt BC.$ There is a point $E$ in the interior of $ABCD$ such that $\triangle ABC \sim \triangle CEB$ and the area of $\triangle AED$ is $17$ times the area of $\triangle CEB.$ What is $\dfrac{AB}{BC}?$

$1 + \sqrt{2}$

$2 + \sqrt{2}$

$\sqrt{17}$

$2 + \sqrt{5}$

$1 + 2\sqrt{3}$

Solution:

Set $BC = 1$ and $AB = r \gt 1.$ The similarity $\triangle ABC \sim \triangle BCD$ with the right angles places the figure at $C = (0,0),$ $B = (0,1),$ $A = (r,1),$ $D = \bigl(\tfrac1r, 0\bigr).$ Let $E = (x, y)$ with $x, y \gt 0.$ From $\triangle ABC \sim \triangle CEB$ we get $\dfrac{x}{y} = \tan(\angle ECB) = \tan(\angle BAC) = \dfrac1r$ and $x^2 + y^2 = \dfrac{r}{1 + r^2},$ so $x = \dfrac{r}{1+r^2},$ $y = \dfrac{r^2}{1+r^2}.$ The area of $\triangle CEB$ is $\tfrac12 x,$ and computing $[\triangle AED]$ by the shoelace formula and setting $[\triangle AED] = 17[\triangle CEB]$ simplifies to $r^4 - 18r^2 + 1 = 0.$ Then $r^2 = 9 + 4\sqrt5 = (2 + \sqrt5)^2,$ so $r = 2 + \sqrt5.$

Thus, the correct answer is D.

2017 AMC 12B Problem 24

Solution:

Problem 24 in Other Years