1984 AMC 12 Problem 27

Question

In

\triangle ABC,

D

is on

AC

and

F

is on

BC.

Also,

AB\perp AC,

AF\perp BC,

and

BD=DC=FC=1.

Find

AC.

\sqrt2

\sqrt3

\sqrt[3]{2}

\sqrt[3]{3}

\sqrt[4]{3}

Let

AC=s

and use the altitude-to-hypotenuse similarity relation

Then compare

\cos\angle ACB

in right triangle

ABC

and isosceles triangle

BDC

Let

AC=s.

The right-triangle projection relation gives

AC^2=FC\cdot BC,

so

BC=s^2.

Since

BD=DC=1,

the numerator and denominator in the Law of Cosines satisfy

\begin{aligned} BC^2+DC^2-BD^2&=s^4,\\ 2(BC)(DC)&=2s^2. \end{aligned}

Hence

\cos\angle BCD=\frac{s^2}{2}.

But

D

lies on

AC,

and in right triangle

ABC,

\cos\angle BCA=\frac{AC}{BC}=\frac{1}{s}.

Thus

\frac{s^2}{2}=\frac{1}{s},

so

s^3=2

and

AC=\sqrt[3]{2}.

Therefore, the correct answer is C.

Problem 27 in other years