1965 AMC 12 Problem 36

Question

Given distinct straight lines

OA

and

OB.

From a point in

OA

a perpendicular is drawn to

OB;

from the foot of this perpendicular a line is drawn perpendicular to

OA.

From the foot of this second perpendicular a line is drawn perpendicular to

OB;

and so on indefinitely. The lengths of the first and second perpendiculars are

a

and

b,

respectively. Then the sum of the lengths of the perpendiculars approaches a limit as the number of perpendiculars grows beyond all bounds. This limit is:

\dfrac b{a-b}

\dfrac a{a-b}

\dfrac{ab}{a-b}

\dfrac{b^2}{a-b}

\dfrac{a^2}{a-b}

Answer: E

Concepts:geometric sequence similarity summation

Difficulty rating: 2190

Small hint:

Successive right triangles formed by the two fixed lines are similar

Big hint:

The perpendicular lengths form a geometric sequence with first term

a

and ratio

\frac{b}{a}

Solution:

Each new right triangle has the same acute angle, so the perpendicular lengths form a geometric sequence. Since the first two lengths are

a,b,

the common ratio is

\frac{b}{a}.

Convergence implies

0\lt \frac{b}{a}\lt1,

and the sum is

\frac{a}{1-\frac{b}{a}}=\frac{a^2}{a-b}.

Therefore, the correct answer is E.