2018 AMC 12A Problem 17

Question

Farmer Pythagoras has a field in the shape of a right triangle. The right triangle’s legs have lengths of

3

and

4

units. In the corner where those sides meet at a right angle, he leaves a small unplanted square

S

so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortest distance from

S

to the hypotenuse is

2

units. What fraction of the field is planted?

\tfrac{25}{27}

\tfrac{26}{27}

\tfrac{73}{75}

\tfrac{145}{147}

\tfrac{74}{75}

Put the right angle at the origin so the hypotenuse is the line

3x + 4y = 12

The far corner

(s, s)

of the square is distance

2

from that line; solve

\tfrac{|3s + 4s - 12|}{5} = 2

and keep the valid root

Place the right angle at the origin with legs on the axes, so the vertices are

(4, 0),

(0, 3),

(0, 0),

and the square

S

is

[0, s] \times [0, s].

The hypotenuse is

3x + 4y - 12 = 0,

and the distance from its nearest corner

(s, s)

is

\frac{|3s + 4s - 12|}{\sqrt{3^2 + 4^2}} = \frac{|7s - 12|}{5} = 2.

This gives

s = \tfrac{22}{7}

or

s = \tfrac27;

only

s = \tfrac27

keeps the square inside the triangle.

The field has area

\tfrac12 \cdot 3 \cdot 4 = 6

and the unplanted square has area

\left(\tfrac27\right)^2 = \tfrac{4}{49}.

The planted fraction is

1 - \frac{\frac{4}{49}}{6} = 1 - \frac{2}{147} = \frac{145}{147}.

Thus, the correct answer is D.

Problem 17 in other years